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1
Reactivity and Control for Organic Synthesis
Reac%vity and Control for Organic Synthesis
O O Me
Which group reacts? Where does it react? How does it react?
O
O O
MgBr
CuBr•SMe2
Me O
chemoselective - enone reacts in preference to lactone (ester) regioselective - 1,4- not 1,2-addition to enone (dia)stereoselective - one major diastereomer formed
︎ Advanced Organic Chemistry: Parts A and B, Francis A. Carey, Richard J. Sundberg Organic Chemistry, Jonathan Clayden, Nick Greeves, Stuart Warren Molecular Orbitals and Organic Chemical Reac
2
Reactivity and Control for Organic Synthesis
︎ there are many different types of selec%vity in organic synthesis: Chemoselec%vity – func%onal group discrimina%on
Regioselec%vity – product structural isomer discrimina%on
Stereoselec%vity – product stereoisomer discrimina%on ︎ we will be primarily concerned with: Chemoselec%vity – i.e. selec%vity between two func%onal groups
O
O
OH
NaBH4
O
OMe
OMe
ketone reduced in preference to ester
Regioselec%vity – i.e. selec%vity between different parts of the same func%onal group
Cl O
Me
Me
Me OH Me
MeMgBr
Me
Me
Cl HNO3, H2SO4
Cl +
Me
NO2
NO2
direct addi%on in preference to conjugate addi%on O
O
O Me O
O
MgBr CuBr•SMe2
Me O
ortho and para products in preference to meta product
chemoselec%ve -‐ enone reacts in preference to lactone (ester) regioselec%ve -‐ 1,4-‐ not 1,2-‐addi%on to enone (dia)stereoselec%ve -‐ one major diastereomer formed
Reactivity and Control for Organic Synthesis
︎ acidity and basicity ︎ nucleophilic alipha%c subs%tu%on nucleophilic aTack on carbonyl groups electrophilic aroma%c subs%tu%on nucleophilic aroma%c subs%tu%on forma%on of rings Hard and SoU direct and conjugate addi%on lithium halogen exchange, and directed lithia%on reduc%ve amina%on
3
4
Reactivity and Control for Organic Synthesis
Brønsted (1879-‐1947) defini%on: an acid is a proton donor a base is a proton acceptor acid HA is the source of a proton H+ HA
H+ + A-‐
remember that the solvent (not usually drawn) acts as the base and deprotonates the acid HA conjugate acid HA + solvent HA + H2O equilibrium constant
K = [________ H3O+][A-‐] [HA][H2O]
solvent•H+ + A-‐
conjugate base
H3O+ + A-‐ Ka =________ [H3O+][A-‐] [HA]
i.e. Ka =K[H2O] water is in such large excess (as solvent) that its concentra%on effec%vely does not change
Ka > 1, equilibrium lies more to the right ∴ stronger acid i.e. HA is a stronger acid than H3O+ and A-‐ is a weaker base than H2O Ka < 1, equilibrium lies more to the leU ∴ weaker acid i.e. A-‐ is a stronger base than H2O and HA is a weaker acid than H3O+ Ka values span a huge range ca. 1012 to 10-‐50 therefore much more convenient to use a logarithmic scale What is the concentra
5
Reactivity and Control for Organic Synthesis
Logarithms – a reminder the logarithm of a number is the exponent to which another fixed value (the base, b) must be raised to produce that number x = by logbx = y we will use log10
log10xa = alog10x log10xy = log10x + log10y
pKa = -‐log10Ka ∴ Ka = 10-‐pKa
higher pKa, smaller Ka, equilibrium lies more to the leU ∴ weaker acid lower pKa, larger Ka, equilibrium lies more to the right ∴ stronger acid
pKa = -‐log10Ka = -‐log10
[A-‐] [H+][A-‐] ________ ____ + = -‐log10[H ] -‐ log10 [HA] [HA] [A-‐]
pH
∴ pKa = pH -‐ log10 ____ [HA] rewri%ng the above gives the Henderson-‐Hasselbalch equa%on:
[A-‐] ____ ∴ pH = pKa + log10 [HA]
if [HA] = [A-‐] then pH = pKa (remember log10 1 = 0)
log10x/y = log10x -‐ log10y
6
Reactivity and Control for Organic Synthesis
pH = -‐log10[H+] at neutral pH, 7 = -‐log10[H+] ∴ [H+] = 10-‐7 M at higher pH, [H+] < 10-‐7 and solu%on is basic (i.e. less acidic) at lower pH, [H+] > 10-‐7 and solu%on is acidic. lower pH – more acidic higher pH – less acidic
for water pH = 7 this refers to the following equilibrium H2O + H2O
H3O+ + OH-‐
Ka = [________ H3O+][HO-‐] [H2O]
[H3O+][HO-‐] = Kw -‐ ionisa%on constant of water and is a constant in aqueous solu%on – its value is easy to find water has pH = 7 ∴ -‐log10[H+] = 7 ∴ [H+] = 10-‐7 M [H+] = [HO-‐] = 10 -‐7 M ∴ [H+][HO-‐] = 10-‐7•10-‐7 = 10-‐14 = Kw ∴ pKw = -‐log10Kw = 14
7
Reactivity and Control for Organic Synthesis
Calculate the pH of a 0.1 M solu
HA
H+ + A-‐
strong acids have Ka > 1 and ∴ pKa < 0 weak acids have Ka < 1 and ∴ pKa > 0 moderately strong acids Ka ≈ 1 and pKa ≈ 0 (generally view acids with pKa -‐2→+2 as moderately strong acids) strong acids include: CF3SO3H HCl H2SO4 + H3O
pKa -‐14 ≈-‐7 ≈-‐3 -‐1.74
weak acids include: ace%c acid NH4+ water HC≡CH NH3 the vast majority of organic compounds are weak acids
Calculate the pKa of H3O+
Note: the strongest base in aqueous solu
any base stronger then HO-‐ deprotonates H2O to give HO-‐
pKa 4.76 9.2 15.74 25 38
8
Reactivity and Control for Organic Synthesis
in a mixture of two acids or two bases: the difference in the pKa’s gives us the log of the equilibrium constant, and the ra%o of the Ka’s gives us the equilibrium constant Worked example: how much acetylene would be deprotonated on treatment with hydroxide in aqueous solu9on? HC≡CH + HO-‐
HC≡C-‐ + H2O
[HC≡C-‐][H2O] Keq = _____________ [HC≡CH][HO-‐] Ka HC≡CH ∴ Keq = ______ Ka H2O
and
[HC≡C-‐][H3O+] Ka HC≡CH = ___________ [HC≡CH]
[H3O+][HO-‐] Ka H2O = ________ [H2O]
= 10-‐25/10-‐15.74 = 1015.74/1025 = 10-‐9.3
i.e. only 1 in 1 billion molecules of acetylene would be deprotonated at equilibrium – to deprotonate acetylene use a solvent which does not have a pKa < 25 and use a stronger base – e.g. NaNH2 in liquid NH3 HC≡CH + H2N-‐
HC≡C-‐ + NH3
pKa NH3 = 38 pKa HC≡CH = 25 ∴ Keq = 10-‐25/10-‐38 = 1013
9
Reactivity and Control for Organic Synthesis
some pKa values in water and DMSO CF3CH2OH
MeOH
H2O
15 (28)
15.74 (31.2)
12.5 (23.5)
Me
Me
NH3 10.6 O
HO
Me
3.6, 10.3
OH
H2 43
9.95 (18.0)
17 (29.4)
Me
Me
~36
F3C
Me Me
N H2 11.05
O
4.76 (12.3)
CH4 48
Me
O OH
tBuOH
Me
H N
10.75
OH
O OH
-0.25
Ph
Ph Ph 23
OH O2N
2.45
Ph
OH 4.2 (11)
HC CH 25
CO2H
O
15
Remember: lower pKa = stronger acid; higher pKa = weaker acid
HO2C
CO2H
3.02, 4.38
HO2C
1.92, 6.23
Reactivity and Control for Organic Synthesis
we are going to generally look at pKa values in water the majority of organic reac%ons are not conducted in water
generally pKa values when measured in organic solvent – typically DMSO – are higher then those measured in water this is a consequence of the organic solvent being less good then water at solva%ng the conjugate base it is generally the case that the trend in pKa values in water and DMSO is very similar pKa H2O in H2O = 15.74 pKa H2O in DMSO = 32; pKa AcOH in H2O = 4.76 pKa AcOH in DMSO = 12.3
when predic%ng or ra%onalising pKa values (i.e. the strengths of organic acids and bases) we need to consider three things: i) strength of the H-‐A bond; ii)
effect of hybridisa%on;
iii)
effect of conjuga%on/delocalisa%on
Most important factor in acid strength is the stability of the conjugate base A-‐
10
11
Reactivity and Control for Organic Synthesis
most important is to draw the equilibrium: then look at the stability of the conjugate base Worked example: explain why phenol is more acidic than methanol Step 1: draw equilibria for both species Step 2: evaluate stability of the conjugate base equilibrium A + H+ CH O CH OH 3
3
OH
O H+
+
equilibrium B
Two factors work to stabilise the phenoxide anion O
i) delocalisa%on one of the oxygen lone pairs is in a ‘p’-‐orbital which can overlap with the π-‐system of the aroma%c ring
O
O
O
O
Remember: these structures are just different ways of drawing the same species – the charge is not actually moving around the ring
Reactivity and Control for Organic Synthesis
ii) induc%ve effect The aroma%c subs%tuent is sp2 hybridized (vs sp3 hybridized in methanol) and hence has more ‘s’ character. The higher propor%on of ‘s’ character means that the electrons see more effec%ve nuclear charge. i.e. sp2 hybridised carbons are more electron-‐withdrawing (electronega%ve) than sp3 hybridised carbons both of the above factors stabilise the phenoxide anion with respect to methoxide ∴ equilibrium B lies further to the right than equilibrium A and hence phenol is the stronger acid most important is to draw the equilibrium: then look at the stability of the conjugate base Predict which of the following two phenols is the stronger acid. OH OH O2N NO2
12
13
Reactivity and Control for Organic Synthesis
Worked example: explain the following order of acid strengths methane (pKa = 48); benzene (pKa = 43); HC≡CH (pKa = 25) Step 1: draw equilibria for the three species Step 2: evaluate stability of the conjugate base CH4 CH3 eq. A anion in sp3 orbital – 25% s-‐character H + H+
H+
+
H+
+
HC C
HC CH
H
H
eq. B anion in sp2 orbital – 33% s-‐character
eq. C anion in sp orbital – 50% s-‐character HC C
acetylide anion more stable than C6H5-‐ which is more stable then CH3-‐ ∴ equilibrium C lies further to the right than equilibrium B which lies further to the right than equilibrium A and hence acidity order is as shown. Explain the acidity of the following compounds CN H3C
pKa (DMSO)
H3C
43
30.8
CH3 44
18
14
Reactivity and Control for Organic Synthesis
how about bases? Brønsted -‐ base is a proton acceptor. There are two ways to deal with bases. Let’s start with a base A-‐ HA + HO-‐
A-‐ + H2O Kb = ________ [HA][HO-‐] [A-‐]
pKb = -‐log10Kb
Kw = [H3O+][HO-‐] ∴ Kb = ________ [HA]Kw [A-‐][H3O+]
Ka =________ [H3O+][A-‐]
∴ Kb =Kw/Ka
[HA]
∴ pKb =14 -‐ pKa
stronger base – lower pKb weaker base – higher pKb
it is inconvenient to have two scales and chemists just use pKa to talk about the strengths of acids and bases i.e. look at the ability of the conjugate base to act as a base.
acid
HA + H2O base
H3O+ + A-‐ conjugate acid
conjugate base
15
Reactivity and Control for Organic Synthesis
How do you find out which is the stronger base – t-‐butoxide, or acetate? look at the pKa’s – here tBuOH holds onto the proton to a much greater extent than ace%c acid, or to put it another way tBuO-‐ much more readily accepts a proton than acetate and hence tBuO-‐ is a stronger base. Higher pKa = weaker acid and tBuOH pKa = 17 + tBuO H+ hence stronger conjugate base. O O + Lower pKa = stronger acid and H+ pKa = 4.76 Me Me OH O hence weaker conjugate base. Example butane is a very weak acid (pKa ~ 43) but butyllithium is a very strong base H2SO4 is a strong acid (pKa -‐3.0) but HSO4-‐ is a very weak base. The problem of amines what do we mean by the ques%on “what is the pKa of ammonia?” strictly speaking this refers to the following equilibrium: NH3
NH2
+
H+
pKa ~ 38 i.e. ammonia is a very weak acid and H2N-‐ is a very strong base
but we might be asking, how good a base is NH3 – which means we need the pKa of ammonium NH4+ NH4
NH3
+
H+
pKa ~ 9.2 i.e. NH3 is a weaker base than HO-‐ (pKa H2O = 15.74) to get around this possible ambiguity we should be specific in asking for the pKa of the conjugate acid of ammonia (some%mes given the symbol pKaH) i.e. of ammonium
16
Reactivity and Control for Organic Synthesis
Important to know some pKa’s compound
pKa (water)
CH4
48
NH3
38
HC CH
25
O Me
OtBu
compound O Me O
NH2 O
MeO
OMe O
O Me
EtO
25
OH
pKa (water) 15
compound HN
13 11 10
NH
6.95
NH
5.21
O Me
pKa (water)
OH
4.76 3.6, 10.3
H2CO3
O Et
Et
tBuOH MeOH
20 17 15
10
CH3NO2
9.24
NH4 O Me
O Me
9
NH3 O F3C
OH
CF3SO3H
4.6 -‐0.25 -‐14
comparing any 2 acids: conjugate base of acid with higher pKa will deprotonate acid with lower pKa e.g. BuLi will deprotonate HC≡CH; NaOMe will deprotonate dimethyl malonate etc. for excellent tabulated pKa values for a large number of organic compounds see: hVp://evans.harvard.edu/pdf/ evans_pka_table.pdf and hVp://www.chem.wisc.edu/areas/reich/pkatable/index.htm
17
Reactivity and Control for Organic Synthesis
1) Explain the following pKa orders: OH
OH
OH
< NO2 most acidic lowest pKa
O Me
<
O
O
Me
OMe
O
<
NO2
O
MeO
most acidic lowest pKa
Me
OMe
least acidic higest pKa
least acidic higest pKa
O Me
<
O Me
Me
<
Me
(d)
Me O least acidic higest pKa
most acidic lowest pKa
Me
Me N H
Me Me
>
Me
O
Me
< Me
(b) O
(c)
>
(a)
N H
Me Me most acidic lowest pKa
Me
N H
Me
least acidic higest pKa
2) Which of the following is more basic? (a)
(b) NH
N H
(c)
Me
3) Explain the pKa’s of maleic and fumaric acid: HO2C
CO2H
3.02, 4.38
O-Na+
CO2H HO2C 1.92, 6.23
Me
S-Na+
N Me
N
Me Me
N
Me
18
Reactivity and Control for Organic Synthesis
4) How might you carryout the following transforma
H
O
Me O
O H
TBDPSO
(b)
O
O
O
Me
H
TBDPSO O
O
Me
OMe
O
O
Me
OMe
OMe Me
O
(c)
O OMe
OH
HO
MeO H N
HO
(e)
OH
HO NH2
(d)
O
O
HO
OH OH
OAc OAc NH2
HO OH OH
Me
NH2
AcO OAc OAc
19
Reactivity and Control for Organic Synthesis
5) Explain the following transforma
(a)
O EtO,
OEt
EtOH
O OEt
OEt O O
(b)
Me O
O EtO,
EtOH
Me
OEt
O OEt
6) In water, the basicity of the amines below is as follows, explain. Me
NH2
<
Me
N
Me
<
Me
7) Predict the product of the following reac
LDA, OMe
Me enantiopure
Br
Me
N H
Me
20
Reactivity and Control for Organic Synthesis
“When two molecules collide, three major forces operate. (i) The occupied orbitals of one repel the occupied orbitals of the other. (ii) Any posi%ve charge on one aTracts any nega%ve charge on the other (and repels any posi%ve). (iii) The occupied orbitals (especially the HOMOs) of each interact with the unoccupied orbitals (especially the LUMOs) of the other, causing an aTrac%on between the molecules.” Molecular Orbitals and Organic Chemical Reac9ons I. Fleming this means that reac%ons generally have both a charge and an orbital component. Reac%ons can be predominantly charge controlled, predominantly orbital controlled, or a mixture of the two. nucleophiles and electrophiles were classified by Pearson as HARD or SOFT – R. G. Pearson, Chemical Hardness, John Wiley & Sons, 1997. Hard nucleophiles (and electrophiles) are small and highly charged and have high electronega%vity (i.e. have a large charge:radius ra%o) – they have a low energy HOMO. Sob nucleophiles (and electrophiles) are larger and have lower electronega%vity and are more polarizable – they have a high energy HOMO. Bases (Nucleophiles)
Acids (Electrophiles)
Hard
Hard
H2O, HO-‐, F-‐, RCO2-‐, Cl-‐, ROH, RO-‐, NH3, RNH2
H+, Li+, Na+, K+, Mg2+, BF3
Intermediate
Intermediate
PhNH2, N3-‐, NC-‐, Br-‐
carboca%ons
SoN
SoN
I-‐, RS-‐, RSe-‐, S2-‐, RSH, RSR, R3P, alkenes, aroma%cs, R-‐
Ag+, Pd2+, I2, Br2, radicals
21
Reactivity and Control for Organic Synthesis
Hard nucleophiles have low energy HOMO’s and a high charge:radius ra%o Hard electrophiles have high energy LUMO’s and a high charge:radius ra%o charge dominates their reac%vity
SoU nucleophiles have high energy HOMO’s and they are polarizable SoU electrophiles have low energy LUMO’s and they are polarizable orbital interac%ons dominates their reac%vity
hard electrophile
soft electrophile
hard:hard charge control
soL:soL orbital control hard nucleophile
Generally the case that: Hard nucleophiles tend to react well with hard electrophiles i.e. the reac
soft nucleophile
22
Reactivity and Control for Organic Synthesis
Generally the case that: Hard nucleophiles tend to react well with hard electrophiles i.e. the reac
Bases (Nucleophiles)
Acids (Electrophiles)
Hard
Hard
H2O, HO-‐, F-‐, RCO2-‐, Cl-‐, ROH, RO-‐, NH3, RNH2
H+, Li+, Na+, K+, Mg2+, BF3
Intermediate
Intermediate
PhNH2, N3-‐, NC-‐, Br-‐
carboca%ons
SoN
SoN
I-‐, RS-‐, RSe-‐, S2-‐, RSH, RSR, R3P, alkenes, aroma%cs, R-‐
Ag+, Pd2+, I2, Br2, radicals
the principle of hard/soU acid bases has been applied to a large number of chemical reac%ons
HO
H H O H
2H2O
faster than
Br Br Br
faster than
HO
Br Br
H H O H
Recently the HSAB theory has been disputed see: H. Mayr, M. Breugst, A. R. Ofial, Angew. Chem. Int. Ed., 2011, 50, 6470
HO Br +
H
Br
23
Reactivity and Control for Organic Synthesis
Ambident nucleophiles alkyla%on of enolates
O H
:Base
O M
RX
O
R +
O-‐alkyla%on
O R
C-‐alkyla%on
with enolates the majority of the charge is, as expected, on the oxygen atom charged electrophiles aTack oxygen e.g. protons, carboca%ons soU electrophiles will generally aTack carbon – largest HOMO coefficient
in general, in enolate reac%ons the oxygen atom is associated with a metal ion and solvent and hence both of these variables affect the ra%o of C:O alkyla%on to maximise C-‐alkyla%on use a lithium base (strong O-‐Li bond) and an alkyl halide in THF (soU-‐soU interac%ons) to maximise O-‐alkyla%on use a highly coordina%ng solvent (e.g. HMPA), a potassium base, and an alkyl sulfonate Ph
O
O +
Ph
Ph
OH
Br
DMF 97% CF3CH2OH 7%
0% 85%
24
Reactivity and Control for Organic Synthesis
Ambident nucleophiles Me K
O
O
Me OEt
O
X
O
HMPA
O
+
O OEt
OEt A
B
Me
X =
A
B
OTs
88%
11%
Cl
60%
32%
Br
39%
38%
I
13%
71%
A. L. Kurts, A. Masias, N. K. Genkina, I. P. Beletskaya, O. A. Reutov, Tetrahedron, 27, 4777
in a similar manner, deprotonated secondary amides alkylate on nitrogen with alkyl halides neutral secondary amines alkylate on oxygen with hard alkyla%ng agents MeO H N
O
CO2tBu O O
R
R
CO2tBu
O
DMF
R'
K2CO3, Et3O BF4 CH2Cl2
N
O
Cl
R
H H N
O
NaH, MeO
CO2tBu O
N
EtO R
H
CO2tBu R'
A. Endo, S. J. Danishefsky, J. Am. Chem. Soc., 2005, 127, 8298.
Explain these two transforma
25
Reactivity and Control for Organic Synthesis
Nucleophilic SubsTtuTon at a saturated carbon SN2 – subs
two limi%ng mechanis%c cases – SN2 and SN1 – mechanis%c con%nuum between these extremes SN1 – subs
example: MeI + NaOH → MeOH + I-‐
Me
example:
Me Me
rate = k[substrate][nucleophile] i.e. rate dependent on both substrate and nucleophile ‡ (-) Nu
Cl
Me
H2O
OH
Me Me
rate = k[substrate] i.e. rate is independent of nucleophile
R
(-) LG R' R''
Me Me
Me
Me R Nu
R'' R'
Nu LG
R Nu
X
Me Me Me
Nu
R'' R'
concerted reac%on, single transi%on state no intermediate is formed
Me Me
favoured by 1° substrates and some 2° substrates requires good nucleophile and leaving group
stepwise reac%on, via an intermediate -‐ the 1st step is rate determining (forma%on of C+), 2nd step is fast
favoured by 3° substrates and some 2° substrates
requires good leaving group and solvent that stabilises carboca%ons
26
Reactivity and Control for Organic Synthesis
SN1 reac%on
enanTomers ‡
X R
R''
X (-)
step 1 R'
R
(+) R
R'
R' Nu
Nu
step 2
R''
R''
R
R''
R R'
R''
R'
Nu
SN1 reac%ons proceeds via a discrete carbenium ion and forma%on of the carbenium ion (step 1) is usually the rate determining step the lowest energy conforma%on of carbenium ions is planar trapping of the carbenium ion by a nucleophile (step 2) is generally fast the nucleophile can trap the carbenium ion from either side, hence enan%oenriched substrates should be expected to give racemic products under SN1 condi%ons – c.f. SN2 reac%ons go with strict inversion of configura%on hence the rate of the reac%on is not affected by the added nucleophile the stability of carbenium ions is in the order ter%ary > secondary > primary due to hyperconjuga%on H CH3
H H
hyperconjuga%on is the overlap of filled C-‐H (or C-‐C) σ-‐bonding orbital with the empty p-‐orbital resul%ng in a lowering in energy of the system i.e. stabilisa%on
CH3
Nomenclature of carboca%ons proposed by Olah J. Am. Chem. Soc., 1972, 94, 808.
27
Reactivity and Control for Organic Synthesis
HyperconjugaTon dona%on of C-‐H σ-‐bond electrons in empty p orbital empty p-orbital H CH3
H H
filled σ C-H orbital
CH3
energy of the bonding electrons reduced system stabilised
greater number of C-‐H (or C-‐C) σ-‐bonds the greater the extent of hyperconjuga%on and the greater stabilisa%on carbenium ion stability therefore goes in the order:
tertiary R
R >
secondary primary R R > R
R
carbenium ions have been observed by NMR and X-‐ray crystal structure analysis
a recent X-‐ray structure of the t-‐butyl ca%on (anion is CHB11Cl11) shows the planar nature of the carbenium ion. E. S. Stoyanov, I. V. Stoyanova, F. S. Tham, C. A. Read; Angew.Chem., Int.Ed. 2012, 51, 9149 conjuga%on with alkenes, arenes and lone pairs, also stabilises carbenium ions
28
Reactivity and Control for Organic Synthesis
conjuga%on with alkenes, arenes and lone pairs, also stabilises carbenium ions
X
benzyl ca%on stabilised by delocalisa%on
ψ3 X
energy of isolated p-‐orbital
allyl ca%on stabilised by delocalisa%on
ψ2
ψ1 RO
X
RO
RO
α-‐heteroatom subs%tuted ca%ons stabilised by delocalisa%on
Which orbitals are overlapping in the stabilisa
allyl ca%on more stable than energy of p-‐orbital – conjuga%on is stabilising
29
Reactivity and Control for Organic Synthesis
the more stable the carbenium ion the faster SN1 reac%on rates of hydrolysis of alkyl chlorides in 50% aqueous ethanol (adapted from Organic Chemistry, Clayden, Greeves and Warren, 2nd Edi%on, OUP 2012) 2° chloride, not that stable C+ not good at SN1
butyl
1° chloride ∴ SN2
iso-propyl
Cl 0.07
91
benzyl Cl
Cl
Cl 1.0
tert-butyl
Cl
1° but benzylic
allyl
0.12
methallyl
allylic ca%on is 2° at one end
1° but allylic
4.0
cinamyl
dimethallyl Cl
Cl 2100
3° chloride very good at SN1
Cl 7700
1° but allylic and benzylic
130000
allylic ca%on is 3° at one end
30
Reactivity and Control for Organic Synthesis
as a carbenium ion is formed during an SN1 reac%on a polar solvent is required for reac%on
the best solvents for promo%ng SN1 reac%ons are polar pro%c solvents such as water and alcohols (they can readily solvate the carbenium ion as well as the leaving group (by hydrogen bonding – see later)) solvent
water
ethanol
ace%c acid
DMSO
DMF
dielectric constant ε
80
25
6.2
46
38
solvent
acetone
EtOAc
THF
ether
hexane
dielectric constant ε
21
6
7.5
4.3
1.9
rela%ve rate of solvolysis (i.e. reac%on with solvent as the nucleophile) of tert-‐butyl bromide is 3 x 104 %mes faster in 50% aqueous ethanol than in neat ethanol Explain the rela
Br Br 1
10-6
Br 10-14
δH In polar solvent the δ+ OH δO carbenium ion is solvated by δH H δ+ polar solvent. It is easier to HO δ+ H LG H cluster water molecules OH H O around the carbenium ion δ- H H δ+ δHO and the leaving group than δethanol molecules
H OδH + H OδH
31
Reactivity and Control for Organic Synthesis
Nucleophilic SubsTtuTon at a saturated carbon SN2 – subs
example: MeI + Na+OH-‐ → MeOH + Na+ I-‐
rate = k[substrate][nucleophile]
‡
H HO
H H
(-) HO
I
H
H
(-) I
HO
H H
H I
H H
H H
I
HOMO of nucleophile (nucleophile lone pair) aVacks the back side of the carbon atom as it is pulng electrons into the C-‐I σ* orbital
at the transi%on state the central carbon atom is bonded to 5 other atoms – hence fundamentally SN2 reac%ons are difficult reac%ons the trajectory of approach is along the path of the bond to the leaving group – evidence from Eschenmoser’s experiments O O S O CH3 NaH
O O S O CH3
S O O H3C
S O O H3C
X H3C
O O S OH
CH3 S O O
reac
32
Reactivity and Control for Organic Synthesis
trajectory results in inversion of configura%on between star%ng material and product ‡ Me Nu
H Et
LG
Me
(-) Nu Et
H
(-) LG
Me Nu
H Et
LG
the rate of an SN2 rec%on is influenced by the nature of the substrate, the nucleophile, the leaving group and, with anionic nucleophiles, the associated counterion, and the solvent What makes a good nucleophile? – i.e. what gives a fast reac%on with an electrophile nucleophilicity is related to basicity but is significantly more complex if the atom we are comparing is the same then nucleophilicity does parallel basicity. O HO PhO > > > H2O > ClO4 Me O basicity is a measure of electron pair dona%on to a proton (generally under equilibra%ng condi%ons) nucleophilicity is electron pair dona%on to another atom, frequently carbon, generally under kine%c condi%ons factors which influence nucleophilicity include: charge, electronega%vity, solvent, size, bond strength
Note: the order of nucleophilici<es is also dependent on the nature of the leaving group
33
Reactivity and Control for Organic Synthesis
Charge charged species are more nucleophilic than their neutral counterparts this is expected as nucleophiles are electron pair donors so the more electron rich the nucleophile the beTer donor it is O
O
Me
O
Me
S
more nucleophilic than
Me
more nucleophilic than
NH OH
Me
SH
NH2 more nucleophilic than
BuLi is obviously more nucleophilic than butane
ElectronegaTvity nucleophilicity is related to basicity, but significantly more complex as it involves dona%on of an electron pair to any atom, whereas basicity is dona%on of an electron pair to H+ in the same row of the periodic table more basic means more nucleophilic ∴ going from leU to right across the periodic table nucleophilicity decreases the more electronega%ve atom is the weaker nucleophile as it holds on to its lone pairs of electrons more %ghtly and is less able to donate an electron pair to form a bond. CH3
>
NH2
most basic most nucleophilic
>
HO
>
F
least basic least nucleophilic
NH3
>
most basic most nucleophilic
H2O
>
HF
least basic least nucleophilic
this does not necessarily mean we will get good yields in SN2 reac%ons with these anions as they are also very basic and hence other reac%on pathways can dominate
Reactivity and Control for Organic Synthesis
34
Solvent in polar pro%c solvents (e.g. water, MeOH, AcOH) nucleophilicity increases going down the group – again the less electronega%ve atom is the more nucleophilic F
<
Cl
<
least nucleophilic
Br
<
I
most nucleophilic
δ-
in polar pro
in polar apro%c solvents (e.g. DMSO and DMF) the order of nucleophilicity can invert when compared with polar pro%c solvents as the solvent has weaker interac%ons with the nucleophile. Frequently reac%ons are much faster in these solvents compared with in water MeI + Cl MeCl + I for the halides under some condi%ons, nucleophilicity now decreases going down the group and again parallels basicity (here the most electronega%ve atom is the best nucleophile). Here charge control appears to be domina%ng the reac%on F > Cl > Br > I most basic most nucleophilic
least basic least nucleophilic
Reactivity and Control for Organic Synthesis
35
with uncharged nucleophiles, nucleophilicity increases going down the group – here orbital control appears to be domina%ng – the nucleophile with the highest energy HOMO reacts the fastest PR3
NR3
>
H2S
increasing nucleophilicity lower energy HOMO H2O least nucleophilic
>
Note: nucleophilicity is complicated and the above should be viewed as guidelines A rule of thumb is that nucleophilicity increases going down a group and increases in moving from right to leU in the periodic table
C
N
O
F
Si
P
S
Cl
Ge As Se Br Sn Sb Te I Pb Bi
Po At
the shape of the nucleophile also influences its nucleophilicity in moving from the star%ng materials to the transi%on state the central carbon goes from 4-‐coordinate to 5-‐ coordinate hence sterically hindered nucleophiles react more slowly MeO fastest
>
O
>
O slowest
conversely, small linear anions such as N3-‐, NC-‐ and RC≡C-‐ are good nucleophiles
increasing nucleophilicity
higher energy HOMO H2Se most nucleophilic
>
36
Reactivity and Control for Organic Synthesis
the following is the order of reac%vity of various nucleophiles with methyl iodide in methanol – all of these anions would be considered good nucleophiles (from Chem. Rev. 1969, 69, 1-‐32) – PR3 are also excellent nucleophiles PhS-‐ > I-‐ > SCN-‐ ≈ CN-‐ > N3-‐ ≈ Br-‐ > Cl-‐ > OAc-‐ in polar pro%c solvents PhS-‐ > CN-‐ > -‐OAc > Cl-‐ ≈ Br-‐ ≈ N3-‐ > I-‐ > SCN-‐ in dipolar apro%c solvents Leaving Group
‡ Me
Nu
H Et
LG
Me
(-) Nu Et
H
(-) LG
Me Nu
H Et
LG
during the SN2 reac%on the bond to the leaving group is broken and the LG departs with a lone pair of electrons i.e. becomes more nega%vely charged in the transi%on state ∴ two factors generally influence the leaving group ability: i) the strength of the bond to carbon ii) the stability of the leaving group
37
Reactivity and Control for Organic Synthesis
leaving group ability relates to pKa i.e. good LG’s are weak bases tosylate TsO-‐
triflate TfO-‐ N2 pKa
>
O F3C
S
-14
O O
>
I
O O S O
>
≈
Br
>
Cl
Me -10
-3
-9
-8
rough order of LG ability – the LG ability depends on the nucleophile and the solvent and the above order can vary; however, very weak bases are good leaving groups iodide is a good leaving group as it forms a weak bond to carbon as well as being a stable anion F-‐ is a very poor leaving group in SN2 reac%ons as it forms a very strong bond to carbon HO-‐ is a very poor leaving group in SN2 reac%ons as it is a strong base (pKa H2O = 15.74) but can be made into a good leaving group by protona%on (pKa H3O+ = -‐1.74) or conversion into a tosylate or triflate
Common leaving groups in SN2 reac%ons tend to have a pKa < 2
38
Reactivity and Control for Organic Synthesis
Nature of the substrate
SN2 reac%ons – back to the transi%on state ‡ R Nu
R''
R
(-) Nu
LG R'
R
(-) LG
Nu
LG
R'' R'
R' R''
the nucleophile has to aTack carbon, hence with larger the R groups the rate of reac%on decreases
in moving from substrate to transi%on state carbon moves from being 4-‐coordinate to being 5-‐coordinate hence as the R groups become larger the rate of the reac%on decreases
SN2 reac%ons ∴ only occur with primary and some secondary substrates – not with ter%ary substrates rela%ve rates of the reac%on of the bromides below with chloride are (Chem. Rev. 1956, 56, 571): methyl Me Br relative rate
ethyl Me
Br
10
60
propyl
iso-propyl Me
Br
Me
neo-pentyl
Me 6.5
Me Me
Br 0.13
Me
tert-butyl Me
Br
0.0003
Me
negligible
neopentyl bromide is par%cularly unreac%ve as the nucleophile is severely hindered from aTacking the necessary carbon atom (Note: neopentyl systems are also unreac
Me Nu
Me Me H H
LG
Me
Me Me
(-) Nu
(-) LG H H
‡ Me Nu
Me Me R'' R'
Me Br
LG
39
Reactivity and Control for Organic Synthesis
Nature of the substrate rate of SN2 reac%ons is increased with substrates which carry an adjacent sp2 (or sp) hybridized atom at the SN2 transi%on state the central carbon is partly bonded to both the nucleophile and the leaving group 3 atoms are sharing 4 electrons i.e. there is a 3-‐centre,4-‐electron bond the central carbon has a partly filled p-‐orbital and the electrons in this orbital can be delocalised into the adjacent π-‐ system which lowers the energy of the transi%on state and the reac%on is faster
H Nu
H
H LG
(-) Nu
delocalisa%on into π-‐system
‡
H (-) LG
H Nu
H LG
40
Reactivity and Control for Organic Synthesis
Nature of the substrate the π-‐system has to be in the correct orienta%on for efficient overlap and consequent transi%on state stabiliza%on
α-‐halo carbonyl compounds are par%cularly reac%ve under SN2 condi%ons as they contain an α sp2 hybridised atom aTached to oxygen and the C=O π* is lower in energy than for an alkene rela%ve rates for reac%on alkyl halides with KI in acetone at 50 °C are given below (from Mechansim in Organic Chemistry, R. W. Alder, R. Baker, J. M. Brown, Wiley, 1971) H
H
H (-) Nu
LG
Nu
‡
H
H
(-) LG
Nu
H LG
delocalisa%on into π-‐system
Me relative rate
Cl 1
Me 200
Cl
Cl
Cl 79
200
MeO
Cl 920
O
N
Cl
Ph 3000
Cl
100,000
41
Reactivity and Control for Organic Synthesis
Summary of structural varia%ons and nucleophilic subs%tu%on taken from Organic Chemistry, Clayden, Greeves and Warren, 2nd Edi%on, OUP 2012. Electrophile
Me X
R
R
X
X
R R
R
X R
R R
X R
methyl
primary
secondary
ter%ary
‘neopentyl’
SN1 mechanism?
✗
✗
✗✓
✓✓
✗
SN2 mechanism?
✓✓
✓
✗✓
✗
✗
X
X
Electrophile
RO
O
O
X R
X
R
X
allyl
benzyl
α-‐alkoxy
α-‐carbonyl
α-‐carbonyl and ter%ary
SN1 mechanism?
✓
✓
✓
✗
✗
SN2 mechanism?
✓
✓
✓
✓✓
possible
✗ = bad ✓ = good, ✓✓ = excellent, ✗✓ = poor
42
Reactivity and Control for Organic Synthesis
1) Explain why the reac
O2N
O KI
O A
O I
O2N
acetone
O2N
KI
OH
NO2
O
B
OH
Ph
NO2
NO2
O
2) For the reac
+
CH3OCH3
NaOMe
+
NaCl
3) Suggest reagents for the following reac
(a)
Me
N
(b)
Me I
HO OH
(c)
N Br
Cl
H
(d)
O
H OH
Br H
Ph
I
acetone
O
NO2
O2N
SMe H
O
43
Reactivity and Control for Organic Synthesis
4) Explain the outcome of the following reac
Cl Et2N
HO
Me NBn2
Cl
(b)
NEt2
NaOH
OH
Me OH
H2O
Bn2N
Et
H
OH
NaOH, H2O
Cl
OH
Et
(c) Br
HBr
OH
(d)
OH Ph
Br
Br
MeO, MeOH
OMe
Br
O
HBr
OH
Ph
Ph
5) Predict the outcome of the following reac
Cl
Br
OMe
MeOH, H
OH
O
MeOH, Et3N
Cl O
H
AlCl3
Br
44
Reactivity and Control for Organic Synthesis
EliminaTon reacTons
mechanis%c con%nuum from E1→E2→E1cB
E2 – elimina
E1 – elimina
example:
example:
R Br
H
+
EtO
+
R
rate = k[substrate][base] i.e. rate dependent on both substrate and base ‡ (-)
EtOH Br
Me Br
Me Me
EtOH Me
Me
+ HBr
rate = k[substrate] i.e. rate is independent of base (which is EtOH in the above case)
X (-) H B
Me Me X H
B
B
X BH
H B
C-H σ to C-X σ* H
HH
X
concerted reac%on, single transi%on state no intermediate is formed, an%periplanar arrangement of proton and leaving group is most favourable for elimina%on requires good base and leaving group 3° substrates give more elimina%on than 2° substrates which give more elimina%on than 1° substrates
Me Me Me
X
Me Me
stepwise reac%on, via an intermediate -‐ the 1st step is rate determining (forma%on of C+), 2nd step is fast
favoured by 3° substrates and some 2° substrates
requires good leaving group and solvent that stabilises carboca%ons
45
Reactivity and Control for Organic Synthesis
EliminaTon reacTons RO
E1cB – elimina%on from the conjugate base
example:
variable kine%cs depending on substrate requires a carbanion stabilising group
acid
O Me
RO
+ HO base
conjugate base
RO
+
O Me
O Me
as the carbanion (an enolate in the above example) helps to expel the leaving group, conjuga%on is developed in the transi%on state leading to the product, HO-‐, and RO-‐ can ∴ be leaving groups requires a base and leaving group SubsTtuTon versus EliminaTon SN1 reac%ons are frequently accompanied by E1 reac%ons if there is an appropriately posi%oned proton – this is unsurprising as both reac%ons proceed through the same intermediate SN2 reac%ons can also be accompanied by E2 reac%ons we need to look at factors affec%ng: i) SN1/E1 product ra%os
ii) SN2/E2 product ra%os iii) change of mechanism i.e. SN1/E1 → SN2/E2
46
Reactivity and Control for Organic Synthesis
primary substrates do SN2 or E2 – SN2 is generally favoured but bulky bases (t-‐BuOK) allow E2 to occur Br
EtO
OEt 91
9
to maximise SN2 – use good nucleophile e.g. RS-‐, X-‐, N3-‐ in dipolar apro%c solvent
ter%ary substrates do SN1 / E1 or E2 with good ionising solvents and no added anionic base then SN1 / E1 will be favoured with added base E2 will be favoured
Br
EtO
OEt <0.1
>97
secondary substrates can do SN1 / E1 or SN2 / E2 with good ionising solvents and no added anionic base or nucleophile then SN1 / E1 will be favoured with added base E2 will be favoured with good nucleophiles, dipolar apro%c solvents SN2 will be favoured
Br
EtO
OEt 20
80
Reactivity and Control for Organic Synthesis
for all substrates moving from primary to secondary to ter%ary favours elimina%on
polar pro%c solvents and base favour E2 over SN2 [base/nucleophile is solvated so easier to aTack outside of the molecule (i,e. remove a proton) than aTack carbon in an SN2 reac%on] heat favours elimina%on over subs%tu%on. EtO Br OEt increased branching at the β-‐posi%on favours elimina%on 40 60 A good overview can be found at: hTp://www.masterorganicchemistry.com/2013/01/18/wrapup-‐the-‐quick-‐n-‐dirty-‐guide-‐to-‐sn1sn2e1e2/
47
48
Reactivity and Control for Organic Synthesis
1) Predict the major product (if any) from the following reac
Br
EtOH
(b)
Br
EtONa EtOH
(c)
Br
EtSNa EtSH
(d)
EtONa Br
EtOH
2) The rate of hydrolysis of tBuCl in water is greatly accelerated by the addi
49
Reactivity and Control for Organic Synthesis
as we have seen, allylic systems are reac%ve under SN2 (stabilisa%on of the transi%on state for subs%tu%on) and under SN1 condi%ons the allylic system has two posi%ons which can be aTacked leading to isomeric products – i.e. there are issues of regioselec%vity Nu
Nu X
Nu
Nu
X
SN2
Nu
Nu
Nu Nu
X
SN2’
SN1
SN1’
sterics and electronics play a role in determining SN/SN’ reac%ons O
O
O
EtO
EtO
OEt
EtO
O OEt
O
+
O
Br
Cl
EtO Cl
PhS
EtO Cl
PhS
50
Reactivity and Control for Organic Synthesis
SN2’ reac%ons are not very common they can also be solvent dependent, I. Fleming, E. J. Thomas, Tetrahedron, 1971, 28, 4989
MeO
SPh Cl
PhS
Cl
DME
Cl MeO
SN2’ Cl MeO
Cl
Cl MeO
Cl
MeO
SPh
SN2 Cl
DME
MeOH
SPh
PhS
Cl
PhS
Cl
MeO
MeO
Cl
Cl
PhS MeOH
MeO
SPh
excellent control of SN2/SN2’ can achieved with organometallic reagents – most notably with copper, palladium and iridium 10 mol% CuCN n-BuMgBr
Bu
OAc
Bu THF 0 °C
94
6
Et2O, 20°C
3
97
J. E. Bäckvall, M. Sellén, B. Grant, J. Am. Chem. Soc., 1990, 112, 6615. For a review on copper-‐catalysed enan%oselec%ve conjugate addi%on and allylic subs%tu%on see: A. Alexakis, J. E. Bäckvall, N. Krause, O. Pàmies, M. Diéguez Chem. Rev. 2008, 108,2 796.
51
Reactivity and Control for Organic Synthesis
O P N O
O Nu +
R
O
Me Me
and related phosphoramidites
Nu
cat. [Ir(COD)Cl]2 OMe
R *
additive
Nu = RNH2, ArO-, malonates, enamines, silylenol ethers, indoles, PhMgBr, NH3, alkenes, vinyltrifluoroborates etc.
high yields high regioselectivity high enantiomeric excess
main names in the field: Helmchen, Alexakis, Hartwig, Carreira, You
1) Explain the outcome of the following reac
O
Br
heat
K2CO3, DMF
2) Suggest reagents and reac
OH
OH
Reactivity and Control for Organic Synthesis
AromaTcity – Electrophilic AromaTc SubsTtuTon and Nucleophilic AromaTc SubsTtuTon
August Kekulé
“I was silng wri
52
53
Reactivity and Control for Organic Synthesis
typical reac%ons of alkenes + Br2
Me
Br
fast
Br
Me
Me
not substitution
Br addition
typical reac%ons of benzene +
Br2
Br
FeBr3 catalyst
Br +
not addition
HBr Br
substitution
retains aroma%c sextet of electrons in subs%tu%on reac%ons does not behave like a “normal” polyene or alkene benzene is both kine
H2/Pt catalyst
H2/Pt catalyst
H2/Pt catalyst
ΔHohydrog= 3 x -‐120 = -‐360 kJmol-‐1 (hypothe%cal, 1,3,5-‐cyclohexatriene)
benzene ≈150 kJmol-‐1 more stable than expected – (represents stability over hypothe%cal 1,3,5-‐ cyclohextriene) – termed the empirical resonance energy (values vary enormously)
54
Reactivity and Control for Organic Synthesis
AromaTcity Hückel’s rule holds for anions, ca%ons and neutrals (4n +2) π-‐electrons for aroma%c compounds; 4n π-‐electrons for an%-‐aroma%c cyclopropenium ca%on -‐ (4n +2), n = 0, 2π electrons
Cl
SbCl5 (Lewis acid)
H H
SbCl6
H
insoluble in non-‐polar solvents; 1 signal in 1H NMR δH = 11.1 ppm -‐ aroma%c and a ca%on compare with cyclopropyl ca%on which is subject to rearrangement to the allyl ca%on Nu
Cl
Ph Ph
Ph
Ph
Ph
Ph
reduced barrier to rotation
Ph Ph
Ph
Ph
Nu
Ph
Ph
Ph
Ph
Ph
Ph
6.3 D
6π-‐aroma%c
2π-‐aroma%c
Ph
Ph
55
Reactivity and Control for Organic Synthesis
Benzene (4n +2), n = 1, 6π electrons δH = 7.26 ppm, planar molecule; bond length = 1.39 Å 1.40 Å H δ = 7.46 H δ = 7.01
H
isoelectronic with pyridine
1.39 Å N
H δ = 8.50
1.34 Å
Cyclopentadienyl Anion (4n +2), n = 1, 6π electrons
H
base
B:
H
H
F3C F3C
pKa = 16
pKa = 43
CF3 H CF3 CF3
pKa < -‐2
2.2 D
C-‐C
sp3-‐sp3
1.54 Å
C-‐C
sp3-‐sp2
1.50 Å
C-‐C
sp3-‐sp
1.47 Å
C-‐C
sp2-‐sp2
1.46 Å
C-‐C
benzene
1.39 Å
C=C
1.34 Å
C≡C
1.21 Å
56
Reactivity and Control for Organic Synthesis
cyclopentadienide anion is isoelectronic with furan pyrrole and thiophene
in each case the (one of the) lone pair(s) is parallel to the p-‐orbitals and part of the π-‐system
S thiophene
furan
pyrrole
0.55 D
0.66 D
1.74 D
X
X
NH
O
X
NH pyrrolidine
X
Electrophilic AromaTc SubsTtuTon
E H
E
E
step 1 is usually rate determining because aroma%city is lost step 2 is fast as aroma%city is regained
σ-‐complex Wheland intermediate arenium ion E
E H
H
57
Reactivity and Control for Organic Synthesis
Electrophilic AromaTc SubsTtuTon
(+) ‡ (+) H E
‡ (+)
step 1
step 2 E H
E
H (+) E
E
H
σ-‐complex Wheland intermediate arenium ion
E
activation energy E
E H
H + E E
Hammond’s postulate: The transi%on state resembles the structure (intermediate or substrate or product) to which it is closest in energy (i.e. transi%on state resembles intermediate arenium ion, therefore what stabilises the arenium ion stabilises the transi%on state.)
58
Reactivity and Control for Organic Synthesis
Mechanis%c Evidence isola%on of intermediates Me EtF, BF3 - 80 °C
Me
Me
Me
Me
Me heat
H BF 4 Me Me stable solid mp -15 °C
Me
SbF5 / FSO3H
SbF6
-120 °C in SO2FCl
Me
δH = 5.6 H H
H H
Me
H
δH = 9.7
H
δH = 8.6
H H
H H
H δH = 9.3
Subs%tuent Effects subs%tuent Y affects both the rate and regiochemistry of the reac%on Y
Y E
Y
Y
E E E
ortho (1,2-‐disubs%tuted)
meta (1,3-‐disubs%tuted)
para (1,4-‐disubs%tuted)
59
Reactivity and Control for Organic Synthesis
Y
Y
Y
Y
E
E
E E
electron dona%ng groups ac%vate the aroma%c ring (i.e. substrate reacts faster than benzene) and are ortho and para direc%ng ACTIVATING group means that the reac%on of the subs%tuted benzene is faster than that of benzene itself O O Typical ac%va%ng groups include: OH, O-‐, O R , HN R ,OR, NH2, NR2, alkyl, Ph OMe
OMe
OMe Br
Br2, AcOH Br 98
kanisole / kbenzene = 109
2
electron withdrawing groups deac%vate the aroma%c ring (i.e. substrate reacts slowed than benzene) and are meta direc%ng DEACTIVATING group means that the reac%on of the subs%tuted benzene is slower than that of benzene itself O O O Typical deac%va%ng groups include: R3N+, CF3, NO2, SO3H, CN, O-‐, R , OR , NR2 halogens are mildly deac%va%ng and direct ortho and para
60
Reactivity and Control for Organic Synthesis
orienta%on of aTack when ring carries an electron dona%ng group, X:, which carries a lone pair (e.g. OMe) ortho and para aTack ortho aTack X:
X:
X:
X E
E
X: E
E
H
X: E
E
ortho
✓✓
para aTack X:
X:
X:
X
X:
X:
E
✓✓
para
meta aTack X:
E
E
X:
X:
HE
E X:
E
X:
E meta
E
E
H
E
E
in the intermediates from ortho and para aTack the carboca%ons are stabliised by overlap with the lone pair of X in the intermediate from meta aTack in the carboca%on is not stabilised by overlap with the lone pair from X
61
Reactivity and Control for Organic Synthesis
reac%on coordinate diagram for aTack on X-‐subs%tuted benzene (X = EDG) TS1
TS2
meta Energy
intermdiate
ortho and para similar energies
more stable intermediate(s) formed faster ∴ ortho and para products predominate benzene reacts slower than these substrates as substrates are more electron rich
X: + E
products
reac%on coordinate Therefore the intermediates (and hence the transi
62
Reactivity and Control for Organic Synthesis
orienta%on of aTack when ring carries an electron withdrawing group, Z, (e.g. NO2) meta aTack ortho aTack Z
Z
Z
Z E
E
E
H
Z E
E
ortho
✗✗
para aTack Z
Z
Z
Z
Z
E
✗✗
para E
HE
E
E
meta aTack Z
Z
Z
Z
Z
E meta
E
E
H
E
E
in the intermediates from ortho and para aTack the carboca%ons are destabilised as next to EWG Z in the intermediate from meta aTack in the carboca%on is never adjacent to EWG
63
Reactivity and Control for Organic Synthesis
reac%on coordinate diagram for aTack on Z-‐subs%tuted benzene (Z = EWG)
ortho and para similar energies
TS1 TS2
Energy
intermediate
less stable intermediate(s) formed slower ∴ meta products predominate benzene reacts faster than these substrates as it is more electron rich
meta
Z + E
products reac%on coordinate
The intermediates (and hence the transi
64
Reactivity and Control for Organic Synthesis
Halogens mildly deac%va%ng as they are electronega%ve and withdraw electron density from the ring through the σ-‐framework (falls off with distance) halogens direct ortho and para as they have lone pairs in high energy orbitals which stabilise the intermediates for ortho/para aTack :OMe
:Cl
OMe
1.2 D
Cl
1.6 D
O
2p –lone pair
good 2p -‐2p overlap MeO – overall electron dona%ng on benzene ring
3p –lone pair
Cl
Me
poor 2p -‐3p overlap Cl – overall electron withdrawing on benzene ring
with anisole the σ-‐electon withdrawing of the oxygen is less than the π-‐ dona%on of the oxygen 2p lone pair and anisole is ac%vated with respect to benzene
N
O
F
P
S
Cl
with chlorobenzene the σ-‐electon withdrawing of the chlorine is greater than the π-‐ dona%on of the chlorine 3p lone pair and chlorobenzene is deac%vated with respect to benzene
Se Br Te I Po At
increasing size of p-‐orbitals
both oxygen and chlorine are electronega%ve
increasing electronega
increasing electronega
65
Reactivity and Control for Organic Synthesis
Ques
why does fluorine react faster than than the other halobenzenes? why does fluorine give the largest amount of the para isomer? product distribu%on %
X
conc. HNO3, conc. H2SO4
X NO2
ortho
meta
para
kbenzene
PhF
12
-‐
87
0.18
PhCl
30
0.9
69
0.064
PhBr
37
1.2
62
0.060
PhI
38
1.8
60
0.12
Examples
+
H
CH3
CH3
CH3
CH3
NO2 59%
<4%
H
H 37%
+
Wheland intermediate for ortho / para aTack is stabilised by hyperconjuga%on – σCH → π
H H
NO2
Me is an electron dona%ng group and hence an ac%va%ng group
H H
O2N
NO2
HNO3 / H2SO4
kArX/
NO2
H
66
Reactivity and Control for Organic Synthesis
what happens if there are two subs%tuents on the benzene ring? subs%tuents can be broadly categorised into three classes
i)
STRONGLY ac%va%ng and ortho and para direc%ng (OH, OR, NH2, NR2)
ii)
mildly ac%va%ng groups such as alkyl groups (ortho and para direc%ng) and halogens (mildly deac%va%ng)
iii) deac%va%ng meta-‐direc%ng groups subs%tuents in group i) will dominate classes ii) and iii) subs%tuents in group ii) will dominate class iii) Examine the electronic effects of subs%tuents then consider sterics
H N
OMe F
MeO: o, p F: o, p MeO dominates ∴ para
Me
O
F3C
Me N Me
Me
AcNH: o, p Me: o, p AcNH dominates ∴ ortho
H MeO
O
MeO
Me2N: o, p CF3: m Me2N dominates ∴ para (sterics)
MeO: o, p CHO: m MeO dominates ∴ para (sterics)
67
Reactivity and Control for Organic Synthesis
opposing -‐ if similar reac%vity will get mixtures of compounds all other things being equal a 3rd group is least likely to enter between two groups meta to one another Cl
HNO3, CO2H H SO 2 4
Cl
OMe
OMe
CO2H
Br2 / AcOH OMe
NO2
Cl: o, p CO2H: m Cl dominates ∴ para OH
Br
MeO: o, p MeO: o, p ∴ ortho / para
Me
OH Me
OMe
Me
Br2, AcOH
Me Cl
Me Cl
HNO3, Ac2O
Cl O2N
25 NO2
Br
HO: o, p Me: o, p HO dominates ∴ para
75
Me: o, p Cl: o, p ∴ mixture
Note: these are guidelines and exact ra
68
Reactivity and Control for Organic Synthesis
subs%tuent effects are important for selec%vity and efficiency when designing a synthe%c route CO2H NO2
CO2H
Me
NO2 or
or
NO2
target material
synthe%cally we want to prepare the target material in a clean, selec%ve and efficient fashion CO2H
Me
NO2
Look at the star%ng materials CO2H, deac%va%ng meta direc%ng
NO2, deac%va%ng meta direc%ng
Me, ac%va%ng ortho / para direc%ng
if possible best to introduce the most deac%va%ng group(s) last in the synthe%c sequence rela%onship of NO2 groups is ortho/para with respect to CO2H ∴ best to use toluene as star%ng material CH3
CH3 HNO3 / H2SO4
CH3 NO2
Me NO2 KMnO 4
HNO3, H2SO4
NO2
nitro group is deac%va%ng ∴ can isolate and separate isomers if required
NO2
CO2H NO2
NO2
69
Reactivity and Control for Organic Synthesis
1) Explain the outcome of the following reac
O
Br N
O
Cl
(b)
Br OH
OMe
AlCl3, heat
OMe
OMe
OMe
(c)
(d) 2 equivalents ClSO2OH
Me
SO2Cl
H
(e)
SO3H SO3H
H2SO4, 160 °C
H2SO4, 80 °C
2) AVempted Friedel-‐Crabs acyla
O +
Cl
AlCl3
+
+ A
B
C
O
70
Reactivity and Control for Organic Synthesis
Ipso a\ack and reversible reac%ons electrophilic aroma%c subs%tu%on is generally an irreversible process all of the above arguments with regard to ortho, meta and para ra%os have been based on the irreversibility of the process i.e. the reac%ons are under kine%c control -‐ but there are some excep%ons not all electrophilic aroma%c subs%tu%on reac%ons are under kine%c control
sulfonyl group is electron withdrawing so we only have mono-‐subs%tu%on
sulfona%on -‐ usual reac%on condi%ons: conc. H2SO4 with SO3
O
S O
O
H
O
S O
OH HO3S
at high temperatures with dilute H2SO4 – sulfona%on is reversible aTack by an electrophile at a posi%on which already carries a non-‐hydrogen subs%tuent is termed ipso-‐subs
Br
OH SO3H
we can use an SO3H group as a blocking group
OH Br
ipso aTack
Br
H2SO4
SO3H
HO3S
H
OH SO3H H
SO3H
Br
OH SO3H
H S O O OH
Br
OH SO3H
H
Br
H
H
71
Reactivity and Control for Organic Synthesis
Friedel CraUs alkyla%on can be under kine%c or thermodynamic control Me
Me tBuCl / AlCl3
thermodynamic Me control Me
Me Me Me
Me
Me tBuCl / AlCl3
high temperature
Me Me
Me
Me Cl: Me
Me AlCl3
Me
Me Cl AlCl3
Me
Me
Me
Me Me
Me
Me
+
Me
Me
H Me
Me Me
Me
H Me
H
kine%c control
low temperature
Me Me
Me
Me Me Me
Me H Me Me
Me Me Me
Me
thermodynamic product all groups Me as far apart as they can be Me
Me
Me Me Me
Me
Me Me
H
72
Reactivity and Control for Organic Synthesis
IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Alkyla%on polyalkyla%on and rearrangement predominate excess MeCl, cat. AlCl3
H
H
Me Me
Me
Me
Me
H H
Cl AlCl3
Me Me
Me
Me
Me
Me
Me Me
Me
Me Me Me Me
with one equivalent of alkyla%ng agent mixtures of products result as the ini%ally formed monoalkyl arene is more reac%ve than the unalkylated arene – alkyl groups are electron dona%ng Me cat. AlCl3, Me Me Me MeCl +
Me
more reac%ve than benzene
more reac%ve than toluene
more reac%ve than toluene
73
Reactivity and Control for Organic Synthesis
IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Alkyla%on polyalkyla%on and rearrangement predominate with primary alkyl halides rearrangement occurs Me Me Me Br Me cat. AlBr3
+
major
primary carboca%ons are very unstable rearrangement to the secondary carboca%on occurs minor
of monoalkylated products
Br: Me
AlBr3
Br AlBr3
H Me
Me
Me
Me H Me
1,2-‐hydride shiU
Me Me
74
Reactivity and Control for Organic Synthesis
IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Acyla%on requires a full equivalent of the Lewis acid mono-‐subs%tu%on predominates as introduced group is electron-‐withdrawing and deac%vates aroma%c ring O
Me
O
Me
Cl
monosubs%tu%on
1 equivalent AlCl3
O
O Me
Cl AlCl3
Me
O
Me
O
O Me
AlCl3
AlCl3
H
carbonyl group can then be removed if required (Clemensen reduc%on, Zn/HCl; Wolf-‐Kishner reduc%on, NH2NH2 then KOH, heat; dithiane than Raney Ni) giving products of a selec%ve Friedel-‐CraUs alkyla%on para-‐isomer generally favoured by steric hindrance O O + MeO
Me
O O
Me
AlCl3, toluene Me
MeO
93% yield
Me
75
Reactivity and Control for Organic Synthesis
IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Acyla%on Fries rearrangement can give access to either the ortho-‐ or para-‐isomer O + HO
Me
O O
Me
O
pyridine Me
O
AlCl3 Me
O
O AlCl3
polar solvent O Me
workup
HO
Me O AlCl3 Me
workup
HO
O Cl3Al
major product in non polar solvents
AlkylaTon cataly
Me solvent-separated ion pair
O Me
O
O
AlCl3
AlCl3 inside solvent cage - tight ion pair
Friedel CraUs summary
AlCl3 Rearrangement subs%tu%on order
Me H O
O
non-polar solvent
O
O
major product in polar solvents e.g. PhNO2 Me O
O
H
AcylaTon stoichiometric no, but loss of CO from R-‐C≡O+ if R+ stable, e.g. Ph3C+ mono
76
Reactivity and Control for Organic Synthesis
IntroducTon of funcTonal groups Blanc chloromethyla%on – related to Fiedel-‐CraUs reac%ons OMe Me O
OMe
(CH2O)n, conc. HCl
Cl Me
Cl Me
O
Me
halogena%on – with ac%vated aroma%cs Lewis acid ac%va%on of the electrophile is not require, with benzene and with deac%vated aroma%cs Lewis acid ac%va%on of the electrophile is required NO2
NO2 Br2, FeBr3 Br
Me
Me
halogena%on can frequently be best achieved using Sandmeyer reac%ons (par%cularly good for introducing I and F as well as Cl, Br and CN) conc. HNO3 conc. H2SO4
NO2
Sn, HCl or H2Pd/C
NH2
HX, NaNO2, 0 °C
X N N
77
Reactivity and Control for Organic Synthesis
IntroducTon of FuncTonal Groups –Synthesis diazonium salts -‐ reac%ons Ar N N X
heat
H2O, 100 °C
Ar N N BF4
Ar OH
SN1 reac%on via carbenium ion
NH2
NaNO2, HX 0 °C
N
Ar F
SN1 reac%on via carbernium ion – Balz Schiemann reac%on
N -N 2
v. high energy intermediate, offset by the forma%on of N2 carbenium not stabilised by π-‐system as is orthogonal to π-‐system
slow
empty sp2 orbital
Ar N N X
cat. CuX, KX X = Br, Cl, CN
radical reac%on
Ar N N X
Ar N N X
radical reac%on
Ar N N X
Ar H
radical reac%on
Me
Ar X
N Ar
HO
H3PO2
O
KI
Ar N N X
Ar X
N
electrophilic aroma%c subs%tu%on
O
O OR
HO
O
Me
OR N
NHAr
via enol
78
Reactivity and Control for Organic Synthesis
Nucleophilic AromaTc SubsTtuTon SN2: alipha%c vs aroma%c AliphaTc nucleophile aTacks C-‐I σ* resul%ng in inversion of configura%on ‡ Me Nu + H Et
I
(-) Nu
Me (-) I H Et
remember SN2 reac%ons at sp2 hybridised centres (i.e. alkenes and arenes are incredibly rare)
Me Nu
H Et
+
I
AromaTc no possibility of nucleophile aTacking backside of C-‐LG σ* (transi%on geometry impossible) Lowest Unoccupied Molecular Orbital (LUMO) is π* not σ* aTacking electron rich arene with electron rich nucleophile SN1: alipha%c vs aroma%c Me Me Nu AliphaTc Me Me X Nu Me Me Me carbenium stabilised by hyperconjuga%on Me
LG
Me
AromaTc possible but very high energy intermediate (see Sandmeyer reaca%ons)
X
Nu
Nu
empty sp2 orbital
v. high energy intermediate, offset by the forma%on of N2 carbenium not stabilised by π-‐system as is orthogonal to π-‐system
79
Reactivity and Control for Organic Synthesis
Nucleophilic AromaTc SubsTtuTon SNAr – Addi%on – Elimina%on Mechanism LG LG Nu LG Nu Nu rate determining step
Nu
LG Nu -LG
Meisenheimer intermediate
nucleophile aTacks LUMO, electron withdrawing groups lower energy of LUMO and stabilise the nega%ve charge in the intermediate best to have electron withdrawing group(s), ortho and / or para to the leaving group Evidence isola%on of intermediates Cl O2N
OMe NO2
MeO
K
O2N
NO2
the nega%ve charge is delocalised ortho and para to leaving group
NO2
O
N
MeO
K
O
MeO OMe O N O2N
O
MeO OMe O N O2N
O
N
N
O
O
H. Ueda, M. Sakabe, J. Tanaka, Bull. Chem. Soc. Jpn., 1968, 41, 2866-‐2871.
O
O
80
Reactivity and Control for Organic Synthesis
Nucleophilic AromaTc SubsTtuTon SNAr – Addi%on – Elimina%on Mechanism for halogens as leaving groups, rate of reac%on usually follows kF > kCl > kBr (c.f. rate of SN2 reac%ons kI > kBr > kCl > kF) NO2
NO2 X
MeO
50 °C
OMe
X =
F
Cl
Br
I
krel
2810
3.1
2.1
1
rate determining step is generally aTack of nucleophile on aroma%c ring therefore bond strength to leaving group is not so important in influencing the rate fluorine is the most electronega%ve element and enhances the electrophilicity of the carbon being aTacked increasing the rate of aTack by the nucleophile O
N
O
O F
Nu
N
O
O F Nu
N
O Nu
rate = k[substrate][nucleophile]
1st step usually rate determining leaving group ability does depend on the nucleophile, nevertheless leaving groups can broadly be divided into three classes: taken from Physical and Mechanis
81
Reactivity and Control for Organic Synthesis
reac%vity is complementary to Pd-‐catalysed cross-‐coupling reac%ons of halobenzenes where regioselec%vity is generally governed by the rate of oxida%ve addi%on into the Ar-‐X bond which depends on bond strength General Trends in Oxida OTf > Br >> Cl. with bidentate phosphines rate of oxida%ve addi%on increases with decreasing bite angle. low oxida%on state metals are electron rich (nucleophilic) therefore good donor ligands i.e. H-‐, R-‐, R3P. promote oxida%ve addi%on oxida%ve addi%on to alkyl halides is slow as precomplexa%on is less favourable. bulky ligands can be good as they lead to dissocia%on and more reac%ve metal complex. metal is oxidised and hence substrate is reduced therefore electron deficient substrates react faster than electron rich substrates. reac%on proceeds with reten%on of olefin geometry for sp2 electrophiles. ‡ I
I
+
PdL2
PdL2
PdI
MeO
MeO2C Cl
Cl
82
Reactivity and Control for Organic Synthesis
Nucleophilic AromaTc SubsTtuTon the ac%va%ng group
OMe
Cl
NO2
Me3N+
SO2Ph
O=CPh
CF3
H
krel
114000
2130
18400
2700
800
1
O2N
O2N
MeO
Y
Y
Y
Explain the outcome of the following reac
NO2 Me
Me
Me
NH3, MeOH
NO2
Me
NO2
the nucleophile – typically good nucleophiles in SNAr reac%ons include: RS-‐, HO-‐, RO-‐, PO-‐, RNH2 Synthesis of fluoxe
NHMe
HO Ph
NaH, O Me
N Me
Me
83
Reactivity and Control for Organic Synthesis
total synthesis of vancomycin – glycopep%de an%bio%c currently the ‘last line of defence’ to treat methicillin-‐ resistant staphylococcus aureus [MRSA]. Me HO H2N Me O
OH HO O
O
O
O HO O
H N H H
HO
OH OH
HO
Cl
O O
NH HO2C
O
Cl O H N
NO2
OH
N H O
O
OH O
H N O
N H
H N H H
FOH
MeHNOC
Na2CO3
HO O
NHBoc
OH
O
Oallyl
O
Cl O H H N N H H
OMs
OH
NH MeHNOC
OMe OMe
MeO
NH2
NO2
OMs
Cl O H N
NH
NHMe
Oallyl
NHBoc
O
OMe OMe
MeO
Na2CO3 OR O Cl F O
N O
D. A. Evans et al. Angew. Chem. Int. Ed. Engl. 1998, 37, 2700-‐2704
OR O N O F O Cl
OR OR
84
Reactivity and Control for Organic Synthesis
total synthesis of vancomycin – glycopep%de an%bio%c currently the ‘last line of defence’ to treat methicillin-‐ resistant staphylococcus aureus [MRSA]. Me HO H2N Me O
OH HO O
O
O HO O
Cl O H H N N H H
NH
HO
HO2C
OH OH
HO
O
Cl
O O
O
OH
O
F
allylO
N H O
O
OH O
H N O
N H
OH
Cl O H H N N H H
O
O
NH
NHMe
MeHNOC BnO
OH O
H N
N H O
O
N H
Boc NMe
NHDdm
OBn OBn
CsF, DMSO
NH2
allylO
O HO O
Cl O H H N N H H
NH MeHNOC BnO
Ddm = MeO
NO2
OMe
D. A. Evans et al. Angew. Chem. Int. Ed. Engl. 1998, 37, 2700-‐2704
OBn OBn
O O O
N H O
Cl OH O
H N O
NHDdm
N H
Boc
NMe
85
Reactivity and Control for Organic Synthesis
HeteroaromaTcs and Nucleophilic subsTtuTon 4
pyridine is electron deficient at C-‐2 and C-‐4 and it is prone to aTack by nucleophiles
3
N 1
2
N
N
HOMO of pyridine is nitrogen lone pair pyridine undergoes electrophilic aroma%c subs%tu%on only very slowly as reac%on with electrophiles occurs on nitrogen lone pair E N
E
E N E
v. slow
N E
NO2
c. HNO3, c. H2SO4, 300 °C, 24 h N
high energy intermediate -‐ electrophile reac%ng with posi%vely charged nucleophile
N
6%
N H
86
Reactivity and Control for Organic Synthesis
HeteroaromaTcs and Nucleophilic subsTtuTon
N
N
pyridine is electron deficient at C-‐2 and C-‐4 and is prone to aTack by nucleophiles
N
HOMO of pyridine is nitrogen lone pair MeO Na
POCl3 N H
N
O
OMe
Cl
N
N
Cl
OMe
nucleophilic aroma%c subs%tu%on
O R
compare with MeO R
Cl
O OMe
the leaving group needs to be posi%oned ortho or para to the pyridine nitrogen atom below are the rela%ve rates of reac%on with MeO-‐ in MeOH at 50 °C Cl Cl N
10-‐5
CF3
Cl
1
NO2 N
Cl
5
3,000
Cl
N
82,000
Cl
700,000
reac%on of the corresponding N-‐oxides and N-‐methyl pyridinium salts is significantly faster than for the parent chloropyridines
87
Reactivity and Control for Organic Synthesis
below are the rela%ve rates of reac%on with MeO-‐ in MeOH at 50 °C
N
N O
Cl
3,000
N Me
Cl
6 x 107
Cl
1.3 x 1012
Cl
Cl
Cl
N
N O
N Me
82,000
9 x 107
M. Liveris, J. Miller, J. Chem. Soc., 1963, 3486-‐3492
as with benzenoid aroma%cs fluoride is a beTer ac%vator (leaving group) than chloride rela%ve rate of reac%on with EtO-‐ in EtOH F3C N
1
Cl
Cl
N
65
Cl
N
320
F
N
Cl
2800
M. Schlosser, T. Rausis, Helv. Chimica Acta, 2005, 88, 1240-‐1249
4.2 x 1010
88
Reactivity and Control for Organic Synthesis
pyrimidines and related heterocycles are more reac%ve than 2-‐halopyridines towards nucleophilic aroma%c subs%tu%on increasing reac%vity toward nucleophilic aroma%c subs%tu%on
X
X N
N
>
N
N
> X
N
X X
>
N
N
>
N
N N
>
>
X
N
N
X
taken from “Heterocyclic Chemistry” 5th Edi%on, J. A. Joule and K. Mills, Wiley 2010. O N N
Cl
Me
Ph
N
LiHMDS, toluene
Cl
N
Cl
N
BuNH2
O Ph
Bu N Ph
pTSA
N
Et
Et
D. S. Chekmarev, S. V. Shorshnev, A. E. Stepanov, A. N. Kasatkin, Tetrahedron 2006, 62, 9919-‐9930 there are not always ‘back of the envelope’ explana%ons of selec%vity Cl
OMe MeO
N R
N
Cl
Cl
N R
N
N
+ Cl
R
N
OMe
A B Y. Goto and co-‐workers, Bull. Chem. Soc. Jpn., 1989, 37, 2892
R
CO2Me
Cl
A:B
96:4
92:8
H
Ph
Me
OMe
85:15 84:16 76:24 8:92
89
Reactivity and Control for Organic Synthesis
HeteroaromaTcs pyridine N-‐oxides – much more suscep%ble to electrophilic aTack at 2 and 4 posi%ons (and to nucleophilic addi%on at 2 and 4 posi%ons) 4 3
N
H2O2, CH3CO2H
2
N
N
N
N
O
O
O
O
promotes electrophilic subs%tu%on at 2 and 4 posi%ons
nitra%on of pyridine N-‐oxide NO2
NO2 Me
Me
HNO3, H2SO4,100 °C
N
N
O
O
H
NO2 Me
NO2
NO2
Me
PCl3 N
POCl3
66% yield, c.f. nitra%on of pyridine in acid (6% yield)
NO2 Me
+
NO2 Me
PCl3
N
N
N
O
O
O
Me N
Cl P Cl Cl
90
Reactivity and Control for Organic Synthesis
HeteroaromaTcs pyridine N-‐oxides – N-‐deoxygena%on with rearrangement O Me
Me
O O
Me
Me, 100 °C
H
N
O
O
Me H
N O
Me
O
N
Me
O O
2
1
3
N
Me
1O
O
O3 2
Me
Me
pyridine N-‐oxides – conversion to chloro compounds R
O P Cl Cl Cl
R
R
R H
N
N
O
O
O P Cl Cl
Cl
N O
Cl O
P Cl Cl
N
Cl
91
Reactivity and Control for Organic Synthesis
HeteroaromaTcs pyridones
N H
N
O
OH
O
OH
N H
N
nitra%on NO2
O
O NO2
HNO3, H2SO4 N H
O
N H
Cl
N H
H
O NO2 POCl3
NO2 N H
NO2
POCl3 N
O Cl P Cl O Cl NO2
NO2
NaBH4 N
O P Cl O Cl Cl NO2
N
N
H
H
NaBH4 Cl NO2 N
92
Reactivity and Control for Organic Synthesis
HeteroaromaTcs pyrrole, thiophene and furan all three have aroma%c proper%es in each case the (one of the) lone pair(s) is parallel to the p-‐orbitals and part of the π-‐system the aroma%c heterocycles are electron rich β 3
α 2
X S1
O
NH
thiophene
furan
pyrrole
0.55 D
0.66 D
1.74 D
X
X
X
NH pyrrolidine
pyridine is electron poor Nu
N
order of aroma%city is: thiophene > pyrrole > furan (enol ether like)
sulfur is the largest atom and hence is beTer matched for bonding to sp2-‐hybridised carbon atoms in a 5-‐membered ring leading to thiophene being the most aroma%c
93
Reactivity and Control for Organic Synthesis
HeteroaromaTcs ReacTons electrophilic subs%tu%on – kine%c reac%on at the 2-‐posi%on is favoured over reac%on at the 3-‐posi%on more reac%ve than benzene – e.g. pyrrole similar reac%vity to aniline E
H
X
H
X
E X
H
X
E X
E
H
E
E
X
E
X
E
X H
E
X
more delocalised intermediate ∴ more stable ∴ 2-‐subs%tu%on favoured
less delocalised intermediate ∴ less stable ∴ 3-‐subs%u%on disfavoured
subs%tuents already present on the aroma%c heterocycle exert less direc%ng effect than the corresponding subs%tuents in benzene E
EWG
X
H
EDG
X
X E
H
EDG
E
EWG E
X
E
EWG H
X
X
EDG
E
X
EDG
with EWG at α-‐posi%on β’-‐subs%tu%on favoured E
with EDG at α-‐posi%on α’-‐subs%tu%on favoured
94
Reactivity and Control for Organic Synthesis
HeteroaromaTcs ReacTons nitra%on O S
Me
O
O N
S
O
O
Me
O
O
O N
Me
O
O
NO2
AcOH, -10 °C
H
AcO
NO2
O
H
H N + 13% NO2
51%
O O
H N
O
trace NO2
60% O N
Me
+
AcOH, 0 °C
O
H N
S
NO2
O
N O
NO2
NO2
AcOH,-25 °C
addi%on product acyla%on and formyla%on O H N
Cl3C
Cl
H N
O CCl3 90% HNO3, -50 °C
α' O2N
O Ph S
N Me
H N
O CCl3
β’ > α’ or β for α-‐EWG
β' O
H, POCl3, 35 °C
S
O
H N H
then hydrolysis 78%
Me
N H, POCl3, RT Me then hydrolysis
H N
O H
83%
95
Reactivity and Control for Organic Synthesis
HeteroaromaTcs indole 3β 2α
N1 H
N H
O benzofuran
S
benzothiophene
more enamine-‐like than pyrrole aTack of electrophiles at the β posi%on is the lowest energy pathway
E N H
E N H H H E
E N H E
E N H
N H
minor
N H
major
aTack at β posi%on retains aroma%c sextet of benzenoid ring
96
Reactivity and Control for Organic Synthesis
HeteroaromaTcs indole 3β 2α
N1 H
N H
O benzofuran
S
benzothiophene
if the β-‐posi%on is blocked α-‐aTack occurs α-‐aTack can occur via β-‐aTack followed by rearrangement (● = CT2 i.e. a tri%ated methylene group)
BF3•OEt2 N H
OH
direct aTack at α-‐posi%on would give solely
N H
O H F3B
N H
N H
N H H
N H
N H
N H H
N H
1:1 mixture
97
Reactivity and Control for Organic Synthesis
Explain the outcome of the following reac
Cl
Me
H N
N
OH
Me N
O
OH
NaH,
F
N
Me N
O
H
120 °C, 15 hours
OH
H
(b)
MsCl, Et3N
NMe CO2tBu N H single enantiomer
Me
Me
(d) N iPr3Si
N H racemate O2N
(c)
i) F
O
Br N
Me
NO2
ii) aq. NaOH
Me
N
Me
Br
O N iPr3Si
O
Me N
OH
CO2tBu
H
98
Reactivity and Control for Organic Synthesis
Vicarious Nucleophilic Subs%tu%on -‐ (nucleophilic subs%tu%on of hydrogen) Reviews: M. Mąkosza, J. Winiarski, Acc. Chem. Res. 1987, 20, 282; M. Mąkosza Pure & Appl. Chem. 1997, 69, 559 H O2N
Cl
H
SO2Ph
O2N
O2N SO2Ph
KOH, DMSO
H
PhO2S
O
O N
mechanism
O
O N
O
SO2Ph
O N Cl
H SO2Ph Cl
O
H HO
O N
O
SO2Ph
O N
SO2Ph
rate determining step
SO2Ph
LG
rate determining step is elimina%on of H-‐X (HCl) from σ-‐adduct
EWG
For VNS require a nucleophile which carries a leaving group
LG = Cl, Br, PhO, PhS, RO-‐ etc; EWG = SO2Ph, SO2NR2, SO2OPh, POPh2, CN, CO2Et How can we explain the following results?
NO2 MeO
NO2
MeO DMSO
Cl
Cl
NO2
SO2Ph
KOH, DMSO
Cl
69% SO2Ph
99
Reactivity and Control for Organic Synthesis
Vicarious Nucleophilic Subs%tu%on orienta%on of addi%on depends on: structure of the carbanion; structure of the arene; reac%on condi%ons NO2
Cl
NO2
SO2Ph
KOH, DMSO
F
SO2Ph
F
18%
for aTack on nitrobenzene, as the bulk of the nucleophile increases the amount of para isomer increases
X NO2
SO2Ph R
NO2
NO2
SO2Ph
KOH, DMSO
PhO2S
X
R
yield / %
ortho
para
F
H
63
74
26
Cl
H
75
53
47
Cl
Et
68
100
Cl
Ph
93
100
M. Makosza, J. Goliński, J. Baran, J. Org. Chem., 1984, 49 1488
100
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
ortho-‐lithia%on by lithium halogen exchange – faster then deprotona%on generally requires an organolithium base an aryl / alkenyl bromide or iodide. OMe
OMe
OMe Br
Bu Li
anion in an sp3 orbital
Li
Br + BuBr
Bu
via “ate” complex
anion in an sp2 orbital
mechanism involves aTack of alkyl lithium at the halogen via an intermediate “ate” complex reac%on is an equilibrium process which favours the more stable anion (remember, anion order is sp3>sp2>sp – the stability of the anion is in the order of the pKa of the corresponding hydrocarbon) in the above example an sp3 anion (butyl lithium) gives an sp2 anion D. E. Applequist, D. F. O’Brien, J. Am. Chem. Soc., 1963, 85, 743.
101
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
LDA and other amide bases are good for deprotona%on but NOT for halogen lithium exchange
OMe Br
lithium halogen exchange with LDA would lead to the forma%on of a very weak halogen-‐ nitrogen bond – reac%on is thermodynamically in the wrong direc%on
OMe
N Li
X
Li
I
N Br
+
I I
LDA, I2
Me tBuLi, MeI
N SO2Ph
N SO2Ph
N SO2Ph
Mark G. Saulnier and Gordon W. Gribble J. Org. Chem. 1982, 47, 757
O
Br
Bu Li
O
Li
R X
O
R
102
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
in general rate of exchange I > Br >> Cl
to make aryllithiums by lithium halogen exchange – generally use n-‐butyllithium; tert-‐butyllithium may also be used to make vinyllithiums and alkyllithiums one frequently uses tert-‐butyllithium with primary alkyl halides it is necessary to use two equivalents of tert-‐butyllithium Me
Me
Me I
O
O
Me
2 equiv. tert-BuLi
OTBS
H
Me Li
O
PMP
Me
O
Me + I
Me Me Li Me
Me
Me
H Me
OTBS
PMP
Me
Me
Me H
with one equivalent of tert-‐butyllithium protodeiodina%on is likely to occur
O
O
OTBS
PMP
lithium halogen exchange is a very fast reac%on which can outcompete deprotona%on of OH groups and addi%on to C=O groups Br
O O
O
NR O
O O
nBuLi, THF, -78 °C RHN
O
L. Ollero, L. Castedo, D. Dominguez, Tetrahedron, 1999, 55, 4445-‐4456
Me Me Me
103
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
synthesis of morphine – J. E. Toh, P. L. Fuchs, J. Org. Chem. 1987, 52, 473–475 Provide a mechanism for the reac
OMe Br
OMe O
2.2 equiv. BuLi, -78 °C
OH
O OH
O
OH
Me N H
SO2Ph
PhO2S
morphine OH
H
H
annula%on forming benzocyclobutanes – I. A. Aidhen, J. R. Ahuja, Tetrahedron LeV. 1992, 33, 5431-‐5432. I
MeO
O
t-BuLi OMe N Me
MeO
O
MeO MeO
selec%ve halogen metal exchange is possible Ar Br
Br N
OMe
n-BuLi, Et2O, -100 °C
Br
Li N
OMe
H O
OH Br N
OMe
Cl
73%
104
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on it is also possible to make Grignard reagents by lithium halogen exchange generally using iPrMgBr or iPrMgCl
For reviews see: P. Knochel et al. Angew. Chem. Int. Ed. 2003, 42, 4302; Chem. Commun. 2006, 583; Heterocycles 2014, 88, 827. CO2Me
CO2Me iPrMgCl, THF, -10 °C
PhCHO
I
Br
MgCl
Br iPrMgCl•LiCl, THF -50 °C
Br
Ph
Br
OH Br
O MgCl
tBu
N
Br
OH tBu
H
Br
Br Br
CO2Me
Br O O MgCl S Ph CN
iPrMgCl•LiCl, THF -50 °C Br
N
Br
CN Br
N
Br
iPrMgCl•LiCl, THF -50 °C
CN Br
N
MgCl
105
Reactivity and Control for Organic Synthesis
AromaTc organometallics for aroma%c heterocycles metal halogen exchange shows the following selec%vity: with 5-‐ring heterocycles the 2-‐posi%ons undergoes exchange faster than the 3 posi%on with 6-‐ring heterocycles, the 3 posi%on undergoes exchange faster than the 2-‐posi%on iodine metal exchange is faster than bromine metal exchange summary 5-‐ring 2>3; 6-‐ring 3>2; I>Br
X
Br Br
S
Br
R M
X
Br
M N
Br
EtMgCl. THF, RT
S
Br
MgCl
M
R M
Br
N
tBuN C O
Br O
S
NHtBu
Br
76%
Br
49%
O S
Br I
EtMgCl. THF, RT
S
Br
Me2N
H
S
Br O
MgCl H
l. Christophersen, M. Begtrup, S. Ebdrup, H. Petersen, P. Vedsø J. Org. Chem., 2003, 68, 9513
106
Reactivity and Control for Organic Synthesis
AromaTc organometallics directed ortho-‐metalla%on reviews: V. Snieckus, Chem. Rev. 1990, 90, 879-‐933; J. Clayden in “Organolithiums: Selec
OMe H
O
Li
BuLi
Me
Bu H
O
O Li
H
NMe2
MeO H
OLi NMe2
DMF
H, H2O
use amides as electrophiles: reac
OMe
BuLi,
Me2N
BuLi
NMe2
MeO H
NMe2 H
MeO
OMe Li
NMe2
NMe2
various direc%ng groups can be used Me Me COCl HO
Me NH2
then dehydrate
O
Me
Me
N
O BuLi
Me
Me
N
O Li
Br
R
O H
Li NMe2
O H
Me
N R
107
Reactivity and Control for Organic Synthesis
increasing ability to direct ortho-‐lithia%on
condi%on dependent order Me NR2 O
O
O-‐carbamates
R2N
O
3° amides
R
O S O
sulfones
R2N
O S O
sulfonamides
O
Me N
oxazloines
O
O
MOM ethers
OR
ethers
X
halogens
O
benzylic alkoxides
O RN
O
2° amides
O
S
tBu
RN
sulfoxides
imines
most powerful directors
N
OtBu
N-‐carbamates O
NR2
anilines
NR2
-‐78 °C
-‐78 °C
-‐78 °C
-‐50 °C
-‐20 °C
trifluoromethyl NR ( )n 2
aminomethyl
-‐78 °C
CF3
remote amines
0 °C
>0 °C
temperature (°C) of ortho-‐lithia%on with RLi in THF or ether
Adapted from: J. Clayden in “Organolithiums: Selec
>20 °C
108
Reactivity and Control for Organic Synthesis
with ortho-‐directors which are also electrophiles, the precise reac%on condi%ons including: the nature of the base, addi%ve and order of addi%on can influence the outcome of the reac%on see: P. Beak, R. A. Brown, J. Org Chem., 1982, 47, 34-‐46 O
O
Me
NEt2
n-BuLi
O
NEt2
s-BuLi, TMEDA
O
O NEt2
s-BuLi, TMEDA
NEt2 Li
F
F Me3SiCl
Me
O
O NEt2 s-BuLi, TMEDA
SiMe3 F
NEt2
then MeI
SiMe3 F
R. J. Mills, N. J. Taylor, V. Snieckus, J. Org. Chem., 1989, 54, 4372
Li
O
NEt2 E
E
Electrophile
Yield (%)
D2O
88
MeI
77
EtI
70
B(OMe)3; H2O2 (adds an OH)
56
acetone
54
PhCHO
79
CH2=CHCH2Br
60
109
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on synthesis of Fredericamycin – T. R. Kelly, S. H. Bell, N. Ohashi, R. J. Armstrong-‐Chong, J. Am. Chem. Soc. 1988, 110, 6471-‐6480 MeO
O
TBDMSO s-BuLi MeO Li
O
O Cl
NEt2 O NEt2
TBDMSO
NEt2 O
OMe t-BuLi
O
MeI
O Li
TBDMSO
NEt2 O
OMe
Me
OMe
O
O
Me
fredericamycin
O
OMe
HN
O O HO HN
O
O
TBDMSO Me Me
N Li
EtO N EtO
OMe
TBDMSO Me Me
NEt2 O
OMe
O
EtO O
OEt
TBDMSO
TBDMSO
110
Reactivity and Control for Organic Synthesis
lithia%on of 5-‐membered heterocycles
lithia%on occurs preferen%ally α to the heteroatom due to induc%ve effect of heteroatom with, in some instances a DOM effect furan and thiophene can be readily metalled α to the metal S
S
n-BuLi
O
Li
O
n-BuLi
Li
ether, reflux
-10 °C, ether
with pyrrole itself, the N-‐deprotona%on occurs first – the more ionic the N-‐metal bond the greater the percentage aTack at nitrogen with a more covalent N-‐M bond C-‐aTack occurs. H N
H N
NaNH2
N
Na
Me
EtMgBr
OEt
N
BuLi
Me N
Et
Me N
I
O
MgBr N
I
Me N
H
H O
workup H
H N
O H
Et
111
Reactivity and Control for Organic Synthesis
with pyrroles bearing an N-‐EWG on nitrogen α-‐metalla%on occurs
O
OtBu N
O
Me N Me Me Li Me
OtBu N
SnMe3
then Me3SnCl
SO2Ph N
LDA, then B(OMe)3
SO2Ph N B(OH)
2
then HCl, water
selec%vity can be achieved using LDA or butyllithium
no lithium halogen exchange as would make weak N-‐Br bond most acidic proton removed by directed metalla%on
iPr S
Li Br
N Li
iPr S
H Bu Li Br
S
E Li
S E
112
Reactivity and Control for Organic Synthesis
pyridines
pyridines are electron deficient aroma%cs and pyridines which carry a direc%ng group (halogen, CN, CO2H etc.) undergo ready metalla%on 2, and 4-‐subsistuted pyridines metallate in the 3 posi%on 3-‐subsituted pyridines generally metallate in the 4 posi%on 1 eq. BuLi, 3 eq.
Me N Me Me Li Me
N
CO2H
then CO2
CO2H
CO2H N
CO2H
1 eq. BuLi, 3 eq.
CO2H
then PhCHO, then H2SO4
Me N Me Me Li Me
then CO2
N
O
O
Me N Me Me Li Me
N
1 eq. BuLi, 3 eq.
Ph N
F. Mongin, F. Trécourt, G. Quéguiner, Tetrahedron LeV., 1999, 40, 5438
CO2H CO2H N
113
Reactivity and Control for Organic Synthesis
“Halogen dance” term introduced by BunneT for isomerisa%on reac%ons which can accompany deprotona%on of halogenated aroma%cs J. F. BunneT, Acc. Chem. Res. 1972, 5, 139. Br
Br PhNHK, NH3 Br
40-‐60% Br
Br
Br
for halogen dance to be synthe%cally useful the isomerisa%on must be thermodynamically favourable S
NaNH2, NH3
Br
S
Br Br
S S
H
Br
S
S
+
Br
+ S
S Br
Br
Br
l. Brandsma, R. L. P. de Jong, Synth. Commun. 1990, 20, 1697-‐1700
Li I N
Cl
I
LDA, THF - 70 °C N
O
I
Cl
Li N
Cl
H
I
O
OEt
H N
H
70%
Cl
F. Guillier, F. Nivoliers, A. Conchennee, A. Godard, F. Marsais, G. Queguiner Synth. Commun. 1996, 23, 4412-‐4436
66-‐72%
114
Reactivity and Control for Organic Synthesis
reac%ons of alkenes -‐ depending on subs%tuents alkenes can be: electron rich and hence nucleophilic increasing nucleophilicity
NR2
OR
O E E
E
E
alkene
enol ether
E
E
enamine
HOMO = π bond of alkene LUMO = σ* or π* on electrophile
enolate
electron poor and hence electrophilic increasing electrophilicity
Nu
O N
Nu O
O
Nu
Nu
O OR
HOMO = lone pair on nucleophile LUMO = π* on alkene
Nu
O
N NR2
115
Reactivity and Control for Organic Synthesis
Overview of reacTons of (electron rich) alkenes Br Br
Br2
:Br Br
Br
Br
Br
build up of par%al posi%ve charge H
Me
Br2, H2O
bromina%on stereospecific an%
Br
SN2
Me OH
Me Br
2O: (+)
SN2 / SN1 borderline
Me Br
Br
halohydrin forma%on stereospecific an%
Me OH Br
long, weak bond
Me
Me OH
OsO4 O
N
O Me
Me
O
Me O
O Os
OH
O
O
O
Os(VIII) O
O Os
O
O
Os
Me OH
O H O 2 O
O
OH N
O Me
+ HO HO
Os
O O
Os(VI)
dihydroxyla%on stereospecific syn
116
Reactivity and Control for Organic Synthesis
Overview of reacTons of alkenes
Me
m-CPBA
Me O , then PPh 3 3
ozonolysis
Me O
Me
O O
Ar
H O
O O
Me O
+
Me O
O
O O Me
O O PPh3
Me
Me O
Me O
O
epoxida%on stereospecific syn
Me O
O
O O
O Me
PPh3
O
O O
O
PPh3
117
Reactivity and Control for Organic Synthesis
Overview of reacTons of alkenes
Me
BH3 then H2O2, NaOH
Me
Me H
OH
B
H
Me H
Me H H
B R
R O O
Me
H2O2, NaOH
OH
Me HBr, water
Br
Me
Me
Me Me
H H2, Pd/C H
Me Me
Me H
stereospecific migra%on with reten%on of configura%on
O BR2 Br
Me H
H
R B R O OH
hydrobora%on / oxdia%on
Br
Me
ionic reac%on with HBr
hydrogena%on stereospecific syn
118
Reactivity and Control for Organic Synthesis
1) How would you carry out the following transforma
2) Explain the following transforma
O HS
(b)
O
O +
OEt
OEt
O
EtO S
O
O H2O2, NaOH, MeOH
O
3) Explain the following: Treatment of the enolate A with B at -‐78 °C followed by quenching the reac
OLi MeO
OPh
Me CO Me 2 HO OPh
O
Me A
B
C
D
PhO
Me CO2Me
119
Reactivity and Control for Organic Synthesis
Conjugate AddiTon vs Direct AddiTon H O 1 2
4
1 O 3 2
O Nu
Nu
Nu
Nu
HO Nu
Nu
O
O
direct addi%on 1,2-‐addi%on
conjugate addi%on Michael addi%on 1,4-‐addi%on
H
conjugate addi%on requires the presence of an electron-‐withdrawing group which results in the lowering of the energy of all of the π-‐orbitals of the system and hence the alkene is less nucleophilic and more electrophilic O
O
i.e. alkene is electron poor
Evidence of delocalisa%on O
1678 cm-‐1 1628 cm-‐1
143 ppm
O
1712 cm-‐1
118 ppm O
1653 cm-‐1
133 ppm
infra red – remember ν ∝ √k/μ i.e. higher stretching frequency = stronger bond
13C NMR
133 ppm
120
Reactivity and Control for Organic Synthesis
examples of conjugate addi%on O
NC
O
KCN (cat)., HCN OMe
NC
regenerates cyanide anion – cataly%c HCN too weak an acid to protonate carbonyl group ∴ need a good nucleophile, cyanide anion, to aTack neutral substrate
OMe H
O NC
OMe
H O:
H CN
O
HCl Cl
HCl protonates carbonyl oxygen making the whole system more electrophilic
O
Cl
H
O Cl
H
O
H
Cl H
O
OH
O
NaOH cat.
O
O
H
H
NaOH
H
H OH
O
O H H
O
enolate generated by conjugate addi%on reacts with the alcohol to regenerate alkoxide for conjugate addi%on
121
Reactivity and Control for Organic Synthesis
conjugate addi%on O
Nu
Nu
faster conjugate addi%on
Nu
EWG
pKa
H
O N
Nu
O
O
the beTer the ability to stabilise the nega%ve charge the beTer the conjugate acceptor is and hence the faster the reac%on. This can be related to pKa slower conjugate addi%on
increasing electrophilicity
Nu
O
Nu
O
10
20
O NC
R2N
24
N NR2
O RO
R
Nu
O
OR
O O2N
Nu
O
25
25
for some nucleophiles conjugate addi%on is the major pathway, for other nucleophiles direct addi%on is the major pathway whereas for others slight varia%on in condi%ons can alter the course of the reac%on
122
Reactivity and Control for Organic Synthesis
examples of conjugate addi%on O Ph
heat
+ OEt
O
O Ph
N H
Ph
N
OEt
Bu OH
BuMgBr
O
O
BuMgBr
irreversible reac%on 1,2-‐addi%on
NaCN, HCN
Bu
irreversible reac%on 1,4-‐addi%on
NC OH
O
NaCN, HCN 80 °C
5-10 °C
formed faster kine%c product lower ac%va%on energy
O
Ph
N H
1 % CuCl
O
N
but if given the choice amines do conjugate addi%on
irreversible reac%on
O
heat
+
O NC
more stable thermodynamic product
OEt
123
Reactivity and Control for Organic Synthesis
conjugate addi%on product is generally the thermodynamically more stable product with respect to the direct addi%on product. a rough comparison of bond energies supports the above conjecture gained O-‐H
lost C=O O
NaCN, HCN
NC O
H
5-10 °C
lost
kJmol-‐1
gained
kJmol-‐1
C=Oπ
370
C-‐C
350
overall gain kJmol-‐1
90
gained C-‐C lost C=C O
NaCN, HCN 80 °C
O
C=Cπ
270
O-‐H
460
C-‐C
350 130
NC H
gained C-‐C
gained C-‐H
C-‐H
400
conjugate addi%on product is generally the thermodynamically more stable product with respect to the direct addi%on product because it retains the strong carbonyl double bond – this is general for most α,β-‐unsaturated systems in the above example: the direct addi%on product is the kine%c product i.e. the fastest formed product and hence the product formed by the pathway with the lowest ac%va%on energy
124
Reactivity and Control for Organic Synthesis
why is the direct addi%on product formed fastest? O
O
delocalisa%on indicates that the carbonyl carbon (and one of the alkenyl carbons) bear par%al posi%ve charge
O
β α
more electron deficient carbon charged nucleophiles will aTack the carbonyl carbon faster than the β-‐carbon (Hard-‐Hard interac%on) carbonyl carbon carries the larger posi%ve charge as it is closer to the electronega%ve oxygen atom charged nucleophiles will aTack the β-‐carbon but more slowly
energy
thermodynamic control
ac%va%on energy NC O
O NC
able to reverse at 80 °C
difficult to reverse even at 80 °C
O NaCN + HCN
O NC
kine%c control lower ac%va%on energy
thermodynamic product lower in energy more stable extent of reac%on
at 80 °C cyanohydrin is reversible at 0 °C cyanohydrin is irreversible at 80 °C kine%c product is s%ll formed first but reverts to star%ng materials and slower conjugate addi%on occurs
NC OH kine%c product formed faster
if direct addi%on is reversible conjugate addi%on will result
125
Reactivity and Control for Organic Synthesis
what is actually happening in the addi%on of HCl to methyl vinyl ketone H O:
thermodynamically controlled reac%on most stable product is formed charged nucleophiles usually do direct addi%on
O
HCl Cl
O O
Cl
H
O
H
O
Cl
H
HO Cl
H
Cl
Cl
H H
not all products arising from conjugate addi%on are the result of ini%al reversible direct addi%on for certain nucleophiles conjugate addi%on is the kine%cally most favoured pathway in these instances the kine%c product also happens to be the thermodynamic product
O Ph
OEt
O
O
+ N H
irreversible reac%on
Ph
N
Ph
N
+ OEt
N H
Ph
irreversible reac%on
O OEt
126
Reactivity and Control for Organic Synthesis
Orbital Controlled Reac%ons (SoU-‐SoU interac%ons) if the nucleophile has a high energy HOMO this will be close in energy to the LUMO of the α,β-‐unsaturated system therefore conjugate aTack will occur at the β-‐carbon – the reac%on is under orbital control for fast reac%on we require a small HOMO / LUMO gap HOMO = Nu lone pair Nu Nu:
O
O
LUMO = π*
O Ph
N
+ OEt
N H
Ph
O OEt
for amines: uncharged, direct addi%on not favoured (lone pair of intermediate energy) conjugate addi%on is the major pathway in the above example the reac%on is under orbital control
1. Generally 2nd row elements (e.g. P, S) favour conjugate addi%on as they have high -‐ energy 3s/3p lone pairs that are a good energy match for the LUMO of the substrate. 2. If the nucleophile is uncharged then conjugate addi%on oUen results.
127
Reactivity and Control for Organic Synthesis
Predict the outcome of the following reac
PhSH, base
O
BuMgBr
O
LiAlH4
O
BuMgBr 1% CuCl
the conjugate acceptor the more electrophilic the carbonyl group the more likely to undergo direct addi%on – charge control dominates O
O
R2NH
very reac%ve carbonyl group, charge control therefore en%rely 1,2-‐addi%on
NR2
Cl
mainly direct addi%on
O
nearly always conjugate addi%on O
Cl
O H
O
O OR
always conjugate addi%on
N NR2
O N
O
128
Reactivity and Control for Organic Synthesis
sterics can influence the outcome of the reac%ons O O
MeMgBr
Me
O O
conjugate addi%on even though hard Grignard reagent
Summary: more reac%ve nucleophiles (RLi, RMgBr, LiAlH4) prefer direct addi%on more reac%ve electrophiles prefer direct addi%on less reac%ve nucleophiles prefer conjugate addi%on less reac%ve electrophiles prefer conjugate addi%on watch out for: reversible direct addi%on which leads to conjugate addi%on i.e. kine%c versus thermodynamic control
129
Reactivity and Control for Organic Synthesis
Carbonyl Groups – difference in reacTvity towards nucleophiles O R
O H
R
O Cl
R
O O
O R
R
O Me
R
O R
OMe
O OH
R
NR2
in general there are two types of mechanism with aldehydes and ketones addi%on occurs – frequently followed by subsequent reac%on with esters, acids, amides etc. addi%on and subsequent elimina%on occurs O Nu:
O
Nu O
Structure of the carbonyl group C-‐O σ-‐bond and C-‐O π-‐bond
Nu:
O
O
Nu O X
O
π-‐bonding orbital
X
Nu
O
π*-‐an%bonding orbital
the π-‐bonding orbital is polarised towards oxygen the more electronega%ve atom ∴ the π* an%bonding orbital is polarised towards carbon i.e. the large lobe is on carbon in the mechanism of aTack the HOMO of the nucleophile overlaps with the LUMO (π*) of the carbonyl group
HOMO
LUMO
Nu O
130
Reactivity and Control for Organic Synthesis
typical addi%ons reac%ons of aldehydes and ketones are as follows: hydra%on – aldehydes are prone to hydra%on – ketones far less so O R
H
+
H2O
HO
OH
R
H
Keq hexafluoro-‐ acetone formaldehyde chloral
F3C
O
F3C H O H Cl3C
O
H
Keq
1.2 x 106
acetaldehyde
2280
acetone
2000
How hydrated would you expect the following compounds to be? O O
O O
Me O
1.06
O
0.001
H Me Me
131
Reactivity and Control for Organic Synthesis
in a similar manner, aldehydes and ketones react with alcohols under acid catalysis to give acetals the mechanism of this reac%on is very frequently drawn incorrectly! loss of water
electrophilic O
MeOH
OH
H MeO
OH
±H
MeO
electrophilic OMe
OH2
H MeOH
H
+
MeO
H MeO
OMe
OMe
acid catalysis is required so that the intermediate in the red box can expel water – if no acid were present HO-‐ would be the leaving group – MeO is not a good enough ‘pusher’ to kick out hydroxide the intermediates in blue boxes are closely related – they are much more electrophilic versions of the original carbonyl group and so are readily aTacked by MeOH which is a very weak nucleophile the whole process is in equilibrium and the most stable product is therefore formed the reac%on can be readily reversed using acidic water
aldehydes readily form acetals with simple alcohols
with ketones the equilibrium greatly favours the ketone – using diols allows efficient acetal forma%on why is this?
O
OH
HO H
O
O
132
Reactivity and Control for Organic Synthesis
acetal forma%on is mechanis%cally closely related to numerous other reac%ons of carbonyl compounds aldehydes and ketones readily react with primary amines and related nitrogen nucleophiles to give imines and related compounds – these reac%ons are generally catalysed by acid
O
PhNH2
O
H H N O Ph
±H
Ph
H N
H
OH
Ph
H N
OH2
acid catalysed
H
N
+
Ph
H
Explain shape of the pH rate profile for oxime forma
O
+
NH2OH
NOH
rate
2
4
6
pH
8
N
Ph
133
Reactivity and Control for Organic Synthesis
secondary amines also react with aldehydes and ketones to give iminium ions and subsequently enamines O
Me Me
N H
Me
N
Me
Me
N
Me
H
iminium ion enamine aldehydes and ketones are more electrophilic than the corresponding imines – oxygen is more electronega%ve than carbon iminium ions are more electrophilic than than the corresponding aldehyde or ketone N R
R' H
R''
O R
H
N
R
R' H
more electrophilic
least electrophilic N R
R' R
R''
O R
R
R
N
R' R
Provide a mechanism for the following reac
NH2
+
O
NaBH4, AcOH, NaOAc
HN
134
Reactivity and Control for Organic Synthesis
Carboxylic acids and related groups mul%ple types of delocalisa%on are possible O R
R
Me
R
O X
O
X
X
C
OEt
Me
O
X p-‐type lone pair → C-‐O π*
O
1743 cm-‐1
Me N Me
1646 cm-‐1
most important for esters (X= OR) and amides (X = NR2)
O and N have 2p orbitals which are a good size and energy match for C=O π* orbital (itself made up of overlap of two 2 p orbitals) nitrogen is less electronega%ve than oxygen and hence with amides there is greater p-‐type lone pair → C-‐O π* dona%on than with esters ∴ amides are less reac%ve towards nucleophiles than esters rough order of importance NR2 > OR > Cl evidence: esters and amides are planar i.e. there is substan%al double bond character between X and the carbonyl carbon
O Me
Me N Me
135
Reactivity and Control for Organic Synthesis
Carboxylic acids and related groups mul%ple types of delocalisa%on are possible O R
R
O
X
R
C
O Cl
O
O X
O
νmax = 1715 cm-‐1
X
Cl Cl
1766 cm-‐1
F
F F
1771 cm-‐1
O sp2 lone pair → C-‐X σ* implica%on is carbonyl group has par%al triple bond character and is also very electrophilic oxygen is more electronega%ve than nitrogen and hence the C-‐O σ* is lower in energy than the C-‐N σ* and a beTer energy match for O sp2 lone pair rough order of importance X = F > OR > Cl > NR2 evidence from IR stretching frequencies and X-‐ray crystal structure analysis
126.4.4 ° 117.4 °
O O Me
O OEt
1743 cm-‐1
Me
Me
CF3
N Me
1646 cm-‐1 E. J. Corey, J. O. Link, S. Sarshar, Y. Shao, Tetrahedron LeV. 1992, 33, 7103 the balance of these effects determines the reac%vity of the system
136
Reactivity and Control for Organic Synthesis
Carboxylic acids and related groups esters predominantly exist in the (Z)-‐conforma%on third type of delocalisa%on affects the conforma%on of esters O R
O
O
O
R R
O sp2 lone pair → C-‐O σ*
C
R
O
Ph
O O
Me
Ph
Z-‐ester
O Me
E-‐ester
A .A. Yakovenko, J. H. Gallegos, M. Yu. An%pin, A. Masunov, T. V. Timofeeva, Cryst.Growth Des. 2011, 11, 3964 in terms of reac%vity towards nucleophiles -‐ rough order of reac%vity is: increasing electrophilicity O R
O R
Cl
O O
O R
R
O Me
R
O OMe
R
O OH
R
NR2
conversely in terms of reac%vity towards electrophiles – amides are the most reac%ve E O R
NR2
E R
O NR2
137
Reactivity and Control for Organic Synthesis
chemoselec%vity in the reduc%on of carbonyl compounds. OH
NaBH4
H2, Pd/C
O
for chemo and regioselec%ve reduc%on it is important to choose the correct reagent OH
HO MnO2 O HO
O
LiAlH4
O EtO O
NaBH4 CeCl3
OH EtO O
138
Reactivity and Control for Organic Synthesis
summary of reducing agents for carbonyl groups adapted from Organic Chemistry, Clayden, Greeves and Warren, 2nd Edi%on, OUP 2012. O
R
reduced
O
aldehyde
reduced slowly not usually reduced
NR
NR
R
H
H iminium ion R
O
O R
H
aldehyde
R
OH
O R
R
ketone
R
1° alcohol
BH3•NH3, LDA
DIBAL H
R
H
aldehyde
via acid chloride O
O OR
ester
H
R
NR2
R
NR2
R
amide
OH
acid
NaCNBH3 NaBH4 LiBH4 LiAlH4 BH3 OH R
NHR
amine
R
OH
1° alcohol
R
R
2° alcohol
R
OH
1° alcohol
amine
R
OH
1° alcohol
139
Reactivity and Control for Organic Synthesis
selec%vity in reduc%on why are esters reduced with LiBH4 but only slowly with NaBH4? O R
Li
O
BH4 R
OR
Li OR
O R
H
H B H H
H
Li
O R
OR
R
H
OH
H
H B H H
Li+ has a higher charge/radius ra%o compared with Na+ as it is smaller ∴ Li+ is a more potent Lewis acid than Na+ Li+ serves to ac%vate the ester carbonyl for reduc%on how can we reduce an ester to an aldehyde? DIBAL-‐H – diisobutylaluminium hydride
Al H
Al
H H
Al
exists as an H-‐bridged dimer but reacts as a monomer Al has an empty p-‐orbital, the monomer is electrophilic
DIBAL-‐H only becomes a good reducing agent once it has been ac%vated by complexa%on by a Lewis base
140
Reactivity and Control for Organic Synthesis
DIBAL-‐H – diisobutylaluminium hydride – commercially available as solu%ons in various organic solvents work-‐up of DIBAL-‐H reac%ons can be complicated by gela%on due to the amphoteric nature of AlIII salts
par%%oning the reac%on mixture between an organic solvent and aqueous Rochelle salt (Na+K+ tartrate) coupled with vigorous s%rring is a useful method of solubilising these gels
R Al H R O R
R Al
O
DIBAL-H R
OR
R H
O R
OR
R Al H OR
R
addi%on of acid destroys excess hydride and protonates tetrahedral intermediate
MeOH then H
O R
tetrahedral intermediate stable at low temperature (-‐78 °C)
H H OR H
O R
H
ester reduced to aldehyde with DIBAL-‐H at low temperature
at higher temperature, DIBAL-‐H reduces esters to alcohols R Al H R O R
OR
R O
DIBAL-H R
R Al
OR
H
O R
R Al H OR
aldehyde much more reac%ve than ester R
tetrahedral intermediate not stable at RT
O R
DIBAL-H H
R
OH
rapid reduc%on to alcohol
141
Reactivity and Control for Organic Synthesis
DIBAL-‐H – diisobutylaluminium hydride – commercially available as solu%ons in various organic solvents O O
O N
DIBAL-H, toluene
O
N
O
O
OH
it can be very difficult to reduce an α,β-‐unsaturated ester to an aldehyde with DIBAL-‐H
lactols are very readily prepared by reduc%on of lactones with DIBAL-‐H H
H
O
DIBAL-H, -78 °C
O
OH O
O OH
H
H
H
DIBAL is also very useful for reducing nitriles to aldehydes what is the mechanism of this reac
H
C
N
DIBAL-H then H2O, H
H
O H
H
142
Reactivity and Control for Organic Synthesis
if we add a Grignard reagent or organolithium to an aldehyde or ketone, monoaddi%on occurs but with esters double addi%on is the general outcome O R
O R''
R'' MgBr OR'
or R''
R
Li
O
R'' MgBr
OR'
R
O R"
R
R"
R
R"
ketone more electrophilic than ester in order to obtain mono-‐addi%on use amides as the electrophile Me
OMe H
BuLi
O
Li
Me
Bu H
O
O H
Li
NMe2 DMF
MeO H
OH
H2O, H
R"
R"
ter%ary alcohol
OLi NMe2
H, H2O
the most versa%le solu%on is to use Weinreb amides
MeO H
O H NMe2 H
MeO
O H
Me
O R
OMe
O N Me H • HCl
iPrMgCl, or AlMe3
O R
OMe N Me
note: aluminium and magnesium amides are par<cularly nucleophilic towards esters
Weinreb amide for use of iPrMgCl in the synthesis of Weinreb amides see: J. M. Williams, R. B. Jobson, N. Yasuda, G. Marchesini, U.-‐H. Dolling,E. J. J. Grabowski, Tetrahedron LeV., 1995, 36, 5461-‐5464
143
Reactivity and Control for Organic Synthesis
Weinreb amides – a very reliable ketone synthesis. For a review see: M. Mentzel and H. M. R. Hoffmann, J. Prakt. Chem. 1997, 339, 517-‐524.
O R
R' MgBr
OMe N Me
or R'
Li
O Mg OMe R N R' Me
OH
H , H2 O R
O
OMe
Me ClMg N Me THF, -20 °C to RT, O then HCl (aq) O
OMe N R' H Me
R
tetrahedral intermediate protonated on work up and collapses to generate ketone
stable chelated tetrahedral intermediate
Boc N
O
Boc N OMe
O O
93%
VITAE PHARMACEUTICALS, INC. Patent: WO2007/117560 A2, 2007. O MeO
N Me
OH
O OTBDPS
Me
H
OH
MgBr THF, 77%
OTBDPS H
D. A. Evans, J. T. Starr, Angew. Chem. Int. Ed., 2002, 41, 1787-‐1790
Me
R'
144
Reactivity and Control for Organic Synthesis
DIBAL-‐H reduc%on of esters to aldehydes can, at %mes, be difficult to control – using a Weinreb amide overcomes this problem O MeO
OTBDMS
O Me
N Me
OTBDMS Me
H
Me
Me
DIBAL-H, THF Br
Br
TBDMSO
TBDMSO
TBDPSO
TBDPSO
OTBDMS Me
OTBDMS Me
D. A. Evans, J. T. Starr, Angew. Chem. Int. Ed., 2002, 41, 1787-‐1790 enolates will also add to Weinreb amides Me
OLi
O Me N OMe
OtBu
O Li MeN O
O CO2tBu
H, H2O
O OtBu
83%
145
Reactivity and Control for Organic Synthesis
lithium aluminium hydride – LiAlH4 -‐ all four hydrides are ac%ve
powerful and frequently non-‐selec%ve reducing agent: will reduce aldehydes, ketones, esters to primary alcohols and amides to amines LiAlH4 both in solu%on and as a solid is highly flammable – requires anhydrous solvents work-‐up of LiAlH4 reduc%ons can be tricky due to the amphoteric nature of AlIII salts a useful ‘anhydrous’ work up introduced by Feiser involves, for n grams of LiAlH4 adding dropwise n mL of water, n mL of 15% NaOH solu%on, and then 3n mL of water. In favourable cases a granular precipitate is produced which can be filtered. L. F. Fieser, M. Fieser, M. Reagents for Organic Synthesis 1967, 581-‐595. another safe method for neutralising excess LiAlH4 involves quenching the reac%on with EtOAc reduc%on of esters tetrahedral intermediate collapses to give an aldehyde O R
Li
O
AlH4
OR
R H
H Al H H
Li OR
O R
H
Li
O R
OR H
H Al H H
H
R
OH
146
Reactivity and Control for Organic Synthesis
reduc%on of esters O R
Li
tetrahedral intermediate collapses to give an aldehyde O
AlH4
OR
R
Li OR
O R
H
H Al H H
H
Li
O R
OR
H
R
OH
H
H Al H H
reduc%on of amides AlH3 O R
Li
O
AlH4
NR2
R H
Li NR2
R
H
Al H H
O H
Li NR2
O R
H
AlH3
NR2 R
NR2
tetrahedral intermediate collapses to give an iminium ion
H
H
H Al H H
why this difference in reac%on outcome? RO-‐ is a beTer leaving group then R2N-‐ lone pair of amine is higher in energy than O and hence R2N is a beTer ‘pusher’ than RO
R
NR2
147
Reactivity and Control for Organic Synthesis
O
H
H
N Me
Me O
N Me LiAlH4, THF
H
O H
H
O
O
H
O
O
Me HO
LiAlH4, THF
Me
HO
O
Me
OH
Me
amide reduced to amine 1,2-‐reduc%on of α,β-‐unsaturated ketone (hard nucleophile)
Me
lactone (ester) reduc%on leads to diol
lithium borohydride – will reduce esters to primary alcohols – see above (can be prepared from cheap NaBH4 and LiCl, LiBr or LiI)
sodium borohydride – frequently used in alcoholic solvents such as MeOH or EtOH generally does not reduce esters, epoxides, lactones, nitriles. All four hydrides are ac%ve NaBH4 reacts with pro%c solvents to generate alkoxy borohydrides
Explain the stereoselec
O O
H
O
NaBH4, MeOH Cl
H
HO O
O Cl
H 87%
148
Reactivity and Control for Organic Synthesis
Luche reduc%on NaBH4 is not selec%ve for 1,2-‐ versus 1,4-‐reduc%on – addi%on of CeCl3•7H2O increases the amount of 1,2-‐reduc%on O
NaBH4 or NaBH4, CeCl3•7H2O
OH
1,2-‐reduc%on 1,4-‐reduc%on
OH +
MeOH
NaBH4, MeOH
51%
49%
NaBH4, CeCl3•7H2O, MeOH
99%
trace
it appears that CeCl3 accelerates the reac%on of pro%c solvents with NaBH4 to generate alkoxy borohydrides NaBH(4-‐n)OMen which are harder reducing agents CeCl3 acidifies the MeOH allowing it to ac%vate the carbonyl oxygen making the carbonyl carbon more posi%ve Reagent is harder, substrate is harder, therefore 1,2-‐reduc%on -‐ A L. Gemal, J. –L. Luche, J. Am. Chem. Soc., 1981, 103, 5454-‐5459 MeO H B MeO OMe
O
H
Me O CeIII
OH
O MeMe
HO Me
Me
NaBH4, CeCl3•7H2O, MeOH
MeMe
Me
Me
Me
Me 58%
Givaudan Roure (Interna%onal) SA Patent: US5929291 A1, 1999
149
Reactivity and Control for Organic Synthesis
in a related manner the use of anhydrous cerium(III) chloride in the presence of Grignard reagents and organolithium reagents allows the addi%on of organometallics to highly enolisable aldehydes and ketones T. Imamoto, N. Takiyama, K. Nakamura, T. Hatajima, Y. Kamiya, J . Am. Chem. Soc. 1989, 111 , 4392-‐4398; N. Takeda, T. Imamoto Org. Synth. 1999, 76, 228 methods to dry CeCl3•7H2O -‐ W. H Bunnelle, B. A. Narayanan, Org. Synth., Coll. Vol. VIII, 1993, 602. O
HO
Bu
BuM, THF
product
recovered star%ng material
BuLi
26%
55%
BuLi, CeCl3
92-‐97%
BuMgBr
28%
BuMgBr, CeI3
96%
23%
150
Reactivity and Control for Organic Synthesis
other modified borohydrides: NaBH3CN, NaBH(OAc)3 reagents of choice for reduc%ve amina%on
O N H
Ph
O
CH2O, NaBH3CN
N Me
pH 5
Me
TBDMSO
AcO O
N H H
+
Me O
R
Ph
in each case, reduc%on of the intermediate iminium ion is more rapid than the reduc%on of the corresponding aldehyde
Me
NaBH(OAc)3, SnCl2
TBDMSO O N H
H
R AcO
Me
this is one method to solve the problem of polyalkyla%on when aTemp%ng to alkylate amines
R
NH2
MeI
R
NHMe
MeI
at least as nucleophilic as star%ng amine
R
NMe2
MeI
at least as nucleophilic 1° and 2° amine
R
NMe3 I
polyalkyla%on occurs
151
Reactivity and Control for Organic Synthesis
borane complexed to a Lewis base, THF•BH3, Me2S•BH3 is a good reducing agent for carboxylic acids and amides H O R
O
B H
H
H O
H
R
H B O
O
H H
R
H B
H LB B O H
:LB H
O
R
R
OH
R
O
OBR2 R
Ar'
HO2C
BH3•THF, THF, 0 °C
Ar
EtO2C
R
O
R work-up
O H
O
H R
H HO
Ar'
EtO2C
Ar
LB
O
R
R B
H B
B H
R
H
98%
P. C. Lobben, S. S. –W. Leung, S. Tummala, Org. Process Res. Dev. 2004, 8, 1072–1075. H O R
B H
NR2
H
H O R
H B
O
H
NHR2
R
H
H B
H
NR2
R
B H
R
NR2 R
H
R
NR2 BR2 work-up
R
NR2
152
Reactivity and Control for Organic Synthesis
ReducTon examples MeO
MeO BH3•THF OH
O
OH O
O
O. Hoshino,Y. Mizuno,M. Murakata, H. Yamaguchi Chem. Pharm. Bull., 1999, 47, 1380-‐1383
O Me H
HO2C H O
O Me
Me CH2N2
H
H
OH
HO
H
MeO2C
O Me
Me
H
H
MeO2C
pTSA
H
O
O
O Me
H
O LiAlH4 O Me HO
Me HO
H H
O Me H
water, pTSA H
O
D. N. Kirk, M. S. Rajagopalan, M. J. Varley, J. Chem. Soc., Perkin 1, 1983, 2225-‐2228
H
O O
H
O Me
153
Reactivity and Control for Organic Synthesis
ReducTon examples OMe
Me
O
(COCl)2, DMF
O
O
O
OMe
Me NaBH4, THF
O
CO2H
O
OMe
Me
Cl
O O
O
O
HO
T. P. O'Sullivan, H. Zhang, L. N. Mander, Org. Biomol. Chem., 2007, 5, 2627-‐2635 Recently ChareVe reported the chemoselec
O
N
O
Tf2O, then EtO2C Me
H
H
N H
H
Ph
N
CO2Et Me
90%
N
then Et3SiH then basic workup MeO2C
MeO2C
O
F
Tf2O,
N H
N
H
then Et3SiH then acid workup MeO2C
MeO2C
Ph
O
Tf2O,
N H
O
F
O 86%
95%
154
Reactivity and Control for Organic Synthesis
Rings and ring strain classifica%on of rings classifica%on
small
normal
medium
large
number atoms in ring
3, 4
5, 6
7-‐12
>12
types of ring strain
108°
H
angle strain (Baeyer strain) – distor%on of angles from the idealised values
Me
H
Me
111°
H
H
120° H
Me
larger to relief of strain between geminal methyl groups angle strain – most important in small rings OH Na CO , MeOH 2 3
O
O
H O
~60°
88°
MeO H
155
Reactivity and Control for Organic Synthesis
torsional strain arises from devia%on from an ideal staggered arrangement HH HH
H
H
H
H
H
H
H H
major feature of 3 and 4-‐membered rings H
H
H
H
H H
H
H
H H
H
H
HH
H H
H
H
H H
H H
H H
H
in its chair form cyclohexane has no torsional strain
H H
H
H
H
HH
H H
H H
HH
the lowest energy conforma%on of cyclopentane is an envelope which has some torsional strain
as a result of torsional strain 6-‐membered rings are generally more stable than 5-‐membered rings bond length strain – arises from devia%on of bond lengths from their ideal values
H Me
transannular strain– arises from proximity on non-‐bonded atoms frequently important in medium rings H H
H
C-‐H = 1.09 Å
Me
C-‐C = 1.54 Å
156
Reactivity and Control for Organic Synthesis
general features of ring closure
K or k X:
Y
X
if ring closure is thermodynamically controlled (K) then energy of product will be important ΔG = ΔH -‐ TΔS ring strain considera%ons mean K is generally only favourable for 5-‐ and 6-‐ membered ring if ring closure is kine%cally controlled (k) then the energy of the transi%on state will be important ΔG‡ = ΔH‡ -‐ TΔS‡
k =
kBT -‐ΔG‡/RT e h
ΔH‡ -‐ enthalpy of ac%va%on includes bond breaking/making enthalpic considera%ons and the change in strain energy in reaching the transi%on state ΔS‡ -‐ reflects the difference in the levels of organisa%on between star%ng material(s) and transi%on state
157
Reactivity and Control for Organic Synthesis
rates of cyclisa%on of ω-‐bromo malonates: M. A. Casadei, C. Galli, L. Mandolini, J. Am. Chem. Soc., 1984, 106, 1051-‐1056 6
es%mated value
k =
increasing ΔS‡
4
CO2Et
2
log k
kBT -‐ΔG‡/RT e h
EtO2C
( )n
Br
NaH, DMSO k
0 -‐2
EtO2C
small
-‐4
CO2Et ( )n
-‐6 2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22
ring size
Small rings: 3-‐membered rings are formed fast; 4-‐membered rings more slowly ΔS‡ favourable as liTle preorganisa%on is required – the ends are already close to one another ΔH‡ unfavourable due to developing strain
158
Reactivity and Control for Organic Synthesis
◾ rates of cyclisa%on of ω-‐bromo malonates: M. A. Casadei, C. Galli, L. Mandolini, J. Am. Chem. Soc., 1984, 106, 1051-‐1056 6
es%mated value
k =
increasing ΔS‡
4
CO2Et
2
log k
kBT -‐ΔG‡/RT e h
EtO2C
( )n
Br
NaH, DMSO k
0 -‐2
normal
-‐4
medium
large
EtO2C
CO2Et ( )n
-‐6 2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22
ring size
Normal rings: 5-‐membered rings generally formed fastest ΔS‡ is becoming less favourable as more preorganisa%on is required – the ends are less close to one another ΔH‡ is fairly consistent – 5-‐7 membered rings are rela%vely unstrained Medium rings: ΔS‡ is s%ll increasing but propor%onally less as ring size increases – the ends are less close to one another ΔH‡ becomes dominant – transannular strain reflected in TS and hence rate of cyclisa%on Large rings: ΔS‡ is unfavourable as the ends are unlikely to meet – similar to an intermolecular reac%on. Solu%on do reac%ons under high dilu%on ΔH‡ no ring strain so not important -‐ large rings are similar to acyclic compounds
159
Reactivity and Control for Organic Synthesis
correct alignment of orbitals is also key for efficient ring forma%on LUMO C-‐Br σ*
O
Br Br
O
O
HOMO enolate
X
O
O
good overlap O-‐alkyla%on occurs
poor overlap so no C-‐alkyla%on similarly O O S O CH3 NaH
O O S O CH3
S O O H3C
S O O H3C
Br
X
O O S OH CH3 S O O
H3C
Sir Jack Baldwin proposed a set of guidelines (Baldwin’s rules) to asses the likelihood that a given, kine%cally controlled cyclisa%on would be feasible
160
Reactivity and Control for Organic Synthesis
“Ring-‐forming reac
Sir Jack Baldwin
Waynflete Professor of Organic Chemistry, Oxford 1978-‐2005 Baldwin’s Rules Biosynthesis of penicillins
modes of cyclisa%on X Y Nu:
Y
Y X Nu:
X Nu
exo – bond being broken is outside the ring being formed
X Nu
endo – bond being broken is inside the ring being formed
X
X Nu:
X
X
Nu:
Nu: X
exo-‐tet
sp3 hybridised
exo-‐trig
sp2 hydridised
X
exo-‐dig sp hybridised
Nu:
endo-‐tet
sp3 hybridised
Nu:
endo-‐trig
sp2 hydridised
Nu:
endo-‐dig sp hybridised
161
Reactivity and Control for Organic Synthesis
Baldwin’s rules
TET
TRIG
DIG
ring size
3
4
5
6
exo
✓
✓
✓
✓
endo
-‐
-‐
✗
✗
exo
✓
✓
✓
✓
endo
✗
✗
✗
✓
exo
✗
✗
✓
✓
endo
✓
✓
✓
✓
in general all exo-‐tet and exo-‐trig cyclisa%ons are favoured
5-‐exo-‐trig is faster than 6-‐endo trig How would you assign this reac
O O S O CH3
S O O H3C
S O O H3C
X H3C
Is this likely to be an efficient transforma
O O S OH
NaH I
CH3 S O O
O O S
S O O H3C
S O O H3C
162
Reactivity and Control for Organic Synthesis
1) The following amine undergoes cyclisa
OMe
2) Explain the contras
O
HO H
HO
O
3) Explain the following reac
O
O NaOH, H2O
Me Me
Me
HO
HO H
Me
Me
H
O
O
163
Reactivity and Control for Organic Synthesis
Thorpe-‐Ingold effect; M. E. Jung, G. Piizzi, Chem. Rev. 2005, 105, 1735−1766
the increased rate of cyclisa%on when puœng a geminal dialkyl group in the cyclising chain is known as the Thorpe Ingold effect. a good explana%on for the Thorpe Ingold effect concerns reac%ve rotamers
O Me
krel
R R O
R = Me Br
R=H
H
H
O
H
H
Me
1
O
O Me
Me H
O
O
39
O
O
O
O
H
O
O
major conforma%on, ends held far apart, cyclisa%on cannot occur
O Me
Me H
H
Br
Br
H
H Me
Me Br
Br
reac%ve rotamers
for gem-‐dimethyl-‐subs%tuted substrate all of the staggered conforma%ons are of similar energy and in two of the conformers cyclising groups are in close proximity
164
Reactivity and Control for Organic Synthesis
Examples O
cat. neat 25 °C
oligomers
iPr F3C F3C
O O
cat. neat 25 °C
iPr N Ph
O Mo O F3C
CF3
Explain the following rates of cyclisa
Br
krel
1
NH2
Br
2.2
Br
NH2
158
Br
NH2
0.16
Br
iPr NH2
9190
165
Reactivity and Control for Organic Synthesis
From Corey’s synthesis of longifolene, J. Am. Chem. Soc., 1964, 86, 478. Explain the various aspects of selec
O O HO
OH TsOH
O
O O
PPh3
O OsO4 O O
O O LiClO4
O
TsCl. pyridine
TsO HCl (aq) O O
O O
OH
HO
OH
166
Reactivity and Control for Organic Synthesis
Appendix Hybridisa%on and bonding -‐ a brief recap Hybridisa%on is a useful concept used by organic chemists to describe the bonding in organic molecules A quick method to work our the hybridisa%on of an atom is to count the number of subs%tuents on that atom (including lone pairs of electrons), remembering that in the bonded environment first row elements generally have 8 electrons around them 4 subs%tuents = sp3 hybridised, 3 subs%tuents = sp2 hybridised, 2 subs%tuents = sp hybridised z
z
z y
y
y
x
x
x +
+
py
px
pz
+ s
sp3 hybrid orbitals are made up from one s orbital and three p orbitals giving four hybrid orbitals which point to the corners of a regular tetrahedron. This is the bonding arrangement found in methane (bond angle = 109°) where the sp3 hybrid orbitals overlap with the hydrogen 1s orbitals (not shown) sp3 hybrid orbitals H H
C H
H H
H
H H
167
Reactivity and Control for Organic Synthesis
Similarly, the nitrogen atom in ammonia can be viewed as sp3 hybridised as can the oxygen atom in water although the H-‐X-‐H bond angle is slightly less than 109° due to lone pair–bond pair repulsion lone pair in sp3 orbital
N H
H H
N H
H
O
H
H
H
O
H
H
N-‐atom is sp3 hybridised
O-‐atom is sp3 hybridised
For sp2 hybridisa%on we mix two p-‐orbitals and one s-‐orbital to give three sp2 hybrid orbitals (and leave one p-‐orbital) z
z
z
y
y
x pz
y
x py
x +
+ px
s
The three sp2 hybrid orbitals are arranged 120° apart This is the bonding arrangement found in ethene with the sp2 hybrids overlapping with the hydrogen 1s orbitals (not shown) the remaining pz orbital(s) overlap to form the π-‐bond H H
H H
H
H
H
H
pz orbital
C-‐atom is sp2 hybridised
sp2 hybrid orbitals
168
Reactivity and Control for Organic Synthesis
Similarly, this is the hybridisa%on in carbonyl compounds and imines H
H
lone pair in sp2 orbital
O
N
H
H
H
N and C-‐atoms are sp2 hybridised
O and C-‐atoms are sp2 hybridised
For sp hybridisa%on we mix one p-‐orbitals and one s-‐orbital to give two sp hybrid orbitals (and leave two p-‐orbital) z
z
z
y
y
y
x
x
pz
x +
py
px
s
The two sp hybrid orbitals are arranged 180° apart This is the bonding arrangement found in ethyne with the sp hybrids overlapping with the hydrogen 1s orbitals (not shown) the remaining p orbitals overlapping to form the two π-‐bonds lone pair in sp orbital
p orbitals H
H
C-‐atom is sp2 hybridised
H
H
H
N
in nitriles the N and C-‐atoms are sp hybridised
sp hybrid orbitals
169
Reactivity and Control for Organic Synthesis
So to recap, in general, a quick method to work our the hybridisa%on of an atom is to count the number of subs%tuents on that atom (including lone pairs of electrons), remembering that in the bonded environment first row elements generally have 8 electrons around them -‐ sp3 = 4 sp2 = 3 sp = 2 What is the hybridisa%on of the red atoms in the following examples?
NH 4+
H 3O+
CO N
H H C C C H H
O CO 2
O
Let’s look at the bonding in amides.
All other things being equal, amides are planar molecules and we are happy to draw the delocalisa%on of the nitrogen lone pair as shown to indicate the par%al double bond character of the C-‐N bond O R A
O N R
R
R
NHR 2 B
Following the discussion above the hybridisa%on of the C and O atoms is sp2 but what is the hybridisa%on of the nitrogen atom? Again, following the above discussion, and looking at the form of the amide on the leU hand side (A), the nitrogen atom has 4 subs%tuents, 2 x R, C=O and a lone pair and ∴ is sp3 hybridised However, most organic chemists would say the N atom is sp2 hybridised. Why is this?
170
Reactivity and Control for Organic Synthesis
O R A
N R
R
R
NHR 2 B
R
R
O R
N
C
O
N
R
R
N-‐sp2 hybridised N-‐lone pair in p-‐orbital
C
O
R
N-‐sp3 hybridised N-‐lone pair in sp3-‐orbital
The curly arrows above represent the overlap of the nitrogen lone pair with the C-‐O π-‐orbitals (the an%bonding π* orbital). The best overlap therefore is if the N-‐atom is sp2 hybridised resul%ng in the N-‐lone pair being in a p-‐ orbital with excellent overlap with the p-‐orbitals of the C–O π-‐system If the N-‐atom were sp3 hybridised then the N-‐lone pair would be in an sp3 orbital which would result in poorer overlap with the adjacent C-‐O π-‐system – Generally beTer overlap = greater stabilisa%on In general if a π-‐system has an adjacent atom which carries a lone pair then most organic chemists would view the hybridisa%on of the adjacent atom as sp2 with the lone pair in a p-‐orbital to maximise overlap with the adjoining π-‐ system. What is the hybridisa%on of each of the heteroatoms in the following molecules? O OMe R
O
R
O
NMe 2
OMe
N O
Perhaps we should not be too concerned about this as some molecules, for example anilines, are frequently not perfectly planar and the hybridisa%on of nitrogen is somewhere between perfectly sp2 and perfectly sp3 Addi%onally we should be aware that other effects (e.g. sterics) can override electronic effects and hence the hydridisa%on may not be as expected
Reactivity and Control for Organic Synthesis
Reac%vity and Control for Organic Synthesis
O O Me
Which group reacts? Where does it react? How does it react?
O
O O
MgBr
CuBr•SMe2
Me O
chemoselective - enone reacts in preference to lactone (ester) regioselective - 1,4- not 1,2-addition to enone (dia)stereoselective - one major diastereomer formed
︎ Advanced Organic Chemistry: Parts A and B, Francis A. Carey, Richard J. Sundberg Organic Chemistry, Jonathan Clayden, Nick Greeves, Stuart Warren Molecular Orbitals and Organic Chemical Reac
2
Reactivity and Control for Organic Synthesis
︎ there are many different types of selec%vity in organic synthesis: Chemoselec%vity – func%onal group discrimina%on
Regioselec%vity – product structural isomer discrimina%on
Stereoselec%vity – product stereoisomer discrimina%on ︎ we will be primarily concerned with: Chemoselec%vity – i.e. selec%vity between two func%onal groups
O
O
OH
NaBH4
O
OMe
OMe
ketone reduced in preference to ester
Regioselec%vity – i.e. selec%vity between different parts of the same func%onal group
Cl O
Me
Me
Me OH Me
MeMgBr
Me
Me
Cl HNO3, H2SO4
Cl +
Me
NO2
NO2
direct addi%on in preference to conjugate addi%on O
O
O Me O
O
MgBr CuBr•SMe2
Me O
ortho and para products in preference to meta product
chemoselec%ve -‐ enone reacts in preference to lactone (ester) regioselec%ve -‐ 1,4-‐ not 1,2-‐addi%on to enone (dia)stereoselec%ve -‐ one major diastereomer formed
Reactivity and Control for Organic Synthesis
︎ acidity and basicity ︎ nucleophilic alipha%c subs%tu%on nucleophilic aTack on carbonyl groups electrophilic aroma%c subs%tu%on nucleophilic aroma%c subs%tu%on forma%on of rings Hard and SoU direct and conjugate addi%on lithium halogen exchange, and directed lithia%on reduc%ve amina%on
3
4
Reactivity and Control for Organic Synthesis
Brønsted (1879-‐1947) defini%on: an acid is a proton donor a base is a proton acceptor acid HA is the source of a proton H+ HA
H+ + A-‐
remember that the solvent (not usually drawn) acts as the base and deprotonates the acid HA conjugate acid HA + solvent HA + H2O equilibrium constant
K = [________ H3O+][A-‐] [HA][H2O]
solvent•H+ + A-‐
conjugate base
H3O+ + A-‐ Ka =________ [H3O+][A-‐] [HA]
i.e. Ka =K[H2O] water is in such large excess (as solvent) that its concentra%on effec%vely does not change
Ka > 1, equilibrium lies more to the right ∴ stronger acid i.e. HA is a stronger acid than H3O+ and A-‐ is a weaker base than H2O Ka < 1, equilibrium lies more to the leU ∴ weaker acid i.e. A-‐ is a stronger base than H2O and HA is a weaker acid than H3O+ Ka values span a huge range ca. 1012 to 10-‐50 therefore much more convenient to use a logarithmic scale What is the concentra
5
Reactivity and Control for Organic Synthesis
Logarithms – a reminder the logarithm of a number is the exponent to which another fixed value (the base, b) must be raised to produce that number x = by logbx = y we will use log10
log10xa = alog10x log10xy = log10x + log10y
pKa = -‐log10Ka ∴ Ka = 10-‐pKa
higher pKa, smaller Ka, equilibrium lies more to the leU ∴ weaker acid lower pKa, larger Ka, equilibrium lies more to the right ∴ stronger acid
pKa = -‐log10Ka = -‐log10
[A-‐] [H+][A-‐] ________ ____ + = -‐log10[H ] -‐ log10 [HA] [HA] [A-‐]
pH
∴ pKa = pH -‐ log10 ____ [HA] rewri%ng the above gives the Henderson-‐Hasselbalch equa%on:
[A-‐] ____ ∴ pH = pKa + log10 [HA]
if [HA] = [A-‐] then pH = pKa (remember log10 1 = 0)
log10x/y = log10x -‐ log10y
6
Reactivity and Control for Organic Synthesis
pH = -‐log10[H+] at neutral pH, 7 = -‐log10[H+] ∴ [H+] = 10-‐7 M at higher pH, [H+] < 10-‐7 and solu%on is basic (i.e. less acidic) at lower pH, [H+] > 10-‐7 and solu%on is acidic. lower pH – more acidic higher pH – less acidic
for water pH = 7 this refers to the following equilibrium H2O + H2O
H3O+ + OH-‐
Ka = [________ H3O+][HO-‐] [H2O]
[H3O+][HO-‐] = Kw -‐ ionisa%on constant of water and is a constant in aqueous solu%on – its value is easy to find water has pH = 7 ∴ -‐log10[H+] = 7 ∴ [H+] = 10-‐7 M [H+] = [HO-‐] = 10 -‐7 M ∴ [H+][HO-‐] = 10-‐7•10-‐7 = 10-‐14 = Kw ∴ pKw = -‐log10Kw = 14
7
Reactivity and Control for Organic Synthesis
Calculate the pH of a 0.1 M solu
HA
H+ + A-‐
strong acids have Ka > 1 and ∴ pKa < 0 weak acids have Ka < 1 and ∴ pKa > 0 moderately strong acids Ka ≈ 1 and pKa ≈ 0 (generally view acids with pKa -‐2→+2 as moderately strong acids) strong acids include: CF3SO3H HCl H2SO4 + H3O
pKa -‐14 ≈-‐7 ≈-‐3 -‐1.74
weak acids include: ace%c acid NH4+ water HC≡CH NH3 the vast majority of organic compounds are weak acids
Calculate the pKa of H3O+
Note: the strongest base in aqueous solu
any base stronger then HO-‐ deprotonates H2O to give HO-‐
pKa 4.76 9.2 15.74 25 38
8
Reactivity and Control for Organic Synthesis
in a mixture of two acids or two bases: the difference in the pKa’s gives us the log of the equilibrium constant, and the ra%o of the Ka’s gives us the equilibrium constant Worked example: how much acetylene would be deprotonated on treatment with hydroxide in aqueous solu9on? HC≡CH + HO-‐
HC≡C-‐ + H2O
[HC≡C-‐][H2O] Keq = _____________ [HC≡CH][HO-‐] Ka HC≡CH ∴ Keq = ______ Ka H2O
and
[HC≡C-‐][H3O+] Ka HC≡CH = ___________ [HC≡CH]
[H3O+][HO-‐] Ka H2O = ________ [H2O]
= 10-‐25/10-‐15.74 = 1015.74/1025 = 10-‐9.3
i.e. only 1 in 1 billion molecules of acetylene would be deprotonated at equilibrium – to deprotonate acetylene use a solvent which does not have a pKa < 25 and use a stronger base – e.g. NaNH2 in liquid NH3 HC≡CH + H2N-‐
HC≡C-‐ + NH3
pKa NH3 = 38 pKa HC≡CH = 25 ∴ Keq = 10-‐25/10-‐38 = 1013
9
Reactivity and Control for Organic Synthesis
some pKa values in water and DMSO CF3CH2OH
MeOH
H2O
15 (28)
15.74 (31.2)
12.5 (23.5)
Me
Me
NH3 10.6 O
HO
Me
3.6, 10.3
OH
H2 43
9.95 (18.0)
17 (29.4)
Me
Me
~36
F3C
Me Me
N H2 11.05
O
4.76 (12.3)
CH4 48
Me
O OH
tBuOH
Me
H N
10.75
OH
O OH
-0.25
Ph
Ph Ph 23
OH O2N
2.45
Ph
OH 4.2 (11)
HC CH 25
CO2H
O
15
Remember: lower pKa = stronger acid; higher pKa = weaker acid
HO2C
CO2H
3.02, 4.38
HO2C
1.92, 6.23
Reactivity and Control for Organic Synthesis
we are going to generally look at pKa values in water the majority of organic reac%ons are not conducted in water
generally pKa values when measured in organic solvent – typically DMSO – are higher then those measured in water this is a consequence of the organic solvent being less good then water at solva%ng the conjugate base it is generally the case that the trend in pKa values in water and DMSO is very similar pKa H2O in H2O = 15.74 pKa H2O in DMSO = 32; pKa AcOH in H2O = 4.76 pKa AcOH in DMSO = 12.3
when predic%ng or ra%onalising pKa values (i.e. the strengths of organic acids and bases) we need to consider three things: i) strength of the H-‐A bond; ii)
effect of hybridisa%on;
iii)
effect of conjuga%on/delocalisa%on
Most important factor in acid strength is the stability of the conjugate base A-‐
10
11
Reactivity and Control for Organic Synthesis
most important is to draw the equilibrium: then look at the stability of the conjugate base Worked example: explain why phenol is more acidic than methanol Step 1: draw equilibria for both species Step 2: evaluate stability of the conjugate base equilibrium A + H+ CH O CH OH 3
3
OH
O H+
+
equilibrium B
Two factors work to stabilise the phenoxide anion O
i) delocalisa%on one of the oxygen lone pairs is in a ‘p’-‐orbital which can overlap with the π-‐system of the aroma%c ring
O
O
O
O
Remember: these structures are just different ways of drawing the same species – the charge is not actually moving around the ring
Reactivity and Control for Organic Synthesis
ii) induc%ve effect The aroma%c subs%tuent is sp2 hybridized (vs sp3 hybridized in methanol) and hence has more ‘s’ character. The higher propor%on of ‘s’ character means that the electrons see more effec%ve nuclear charge. i.e. sp2 hybridised carbons are more electron-‐withdrawing (electronega%ve) than sp3 hybridised carbons both of the above factors stabilise the phenoxide anion with respect to methoxide ∴ equilibrium B lies further to the right than equilibrium A and hence phenol is the stronger acid most important is to draw the equilibrium: then look at the stability of the conjugate base Predict which of the following two phenols is the stronger acid. OH OH O2N NO2
12
13
Reactivity and Control for Organic Synthesis
Worked example: explain the following order of acid strengths methane (pKa = 48); benzene (pKa = 43); HC≡CH (pKa = 25) Step 1: draw equilibria for the three species Step 2: evaluate stability of the conjugate base CH4 CH3 eq. A anion in sp3 orbital – 25% s-‐character H + H+
H+
+
H+
+
HC C
HC CH
H
H
eq. B anion in sp2 orbital – 33% s-‐character
eq. C anion in sp orbital – 50% s-‐character HC C
acetylide anion more stable than C6H5-‐ which is more stable then CH3-‐ ∴ equilibrium C lies further to the right than equilibrium B which lies further to the right than equilibrium A and hence acidity order is as shown. Explain the acidity of the following compounds CN H3C
pKa (DMSO)
H3C
43
30.8
CH3 44
18
14
Reactivity and Control for Organic Synthesis
how about bases? Brønsted -‐ base is a proton acceptor. There are two ways to deal with bases. Let’s start with a base A-‐ HA + HO-‐
A-‐ + H2O Kb = ________ [HA][HO-‐] [A-‐]
pKb = -‐log10Kb
Kw = [H3O+][HO-‐] ∴ Kb = ________ [HA]Kw [A-‐][H3O+]
Ka =________ [H3O+][A-‐]
∴ Kb =Kw/Ka
[HA]
∴ pKb =14 -‐ pKa
stronger base – lower pKb weaker base – higher pKb
it is inconvenient to have two scales and chemists just use pKa to talk about the strengths of acids and bases i.e. look at the ability of the conjugate base to act as a base.
acid
HA + H2O base
H3O+ + A-‐ conjugate acid
conjugate base
15
Reactivity and Control for Organic Synthesis
How do you find out which is the stronger base – t-‐butoxide, or acetate? look at the pKa’s – here tBuOH holds onto the proton to a much greater extent than ace%c acid, or to put it another way tBuO-‐ much more readily accepts a proton than acetate and hence tBuO-‐ is a stronger base. Higher pKa = weaker acid and tBuOH pKa = 17 + tBuO H+ hence stronger conjugate base. O O + Lower pKa = stronger acid and H+ pKa = 4.76 Me Me OH O hence weaker conjugate base. Example butane is a very weak acid (pKa ~ 43) but butyllithium is a very strong base H2SO4 is a strong acid (pKa -‐3.0) but HSO4-‐ is a very weak base. The problem of amines what do we mean by the ques%on “what is the pKa of ammonia?” strictly speaking this refers to the following equilibrium: NH3
NH2
+
H+
pKa ~ 38 i.e. ammonia is a very weak acid and H2N-‐ is a very strong base
but we might be asking, how good a base is NH3 – which means we need the pKa of ammonium NH4+ NH4
NH3
+
H+
pKa ~ 9.2 i.e. NH3 is a weaker base than HO-‐ (pKa H2O = 15.74) to get around this possible ambiguity we should be specific in asking for the pKa of the conjugate acid of ammonia (some%mes given the symbol pKaH) i.e. of ammonium
16
Reactivity and Control for Organic Synthesis
Important to know some pKa’s compound
pKa (water)
CH4
48
NH3
38
HC CH
25
O Me
OtBu
compound O Me O
NH2 O
MeO
OMe O
O Me
EtO
25
OH
pKa (water) 15
compound HN
13 11 10
NH
6.95
NH
5.21
O Me
pKa (water)
OH
4.76 3.6, 10.3
H2CO3
O Et
Et
tBuOH MeOH
20 17 15
10
CH3NO2
9.24
NH4 O Me
O Me
9
NH3 O F3C
OH
CF3SO3H
4.6 -‐0.25 -‐14
comparing any 2 acids: conjugate base of acid with higher pKa will deprotonate acid with lower pKa e.g. BuLi will deprotonate HC≡CH; NaOMe will deprotonate dimethyl malonate etc. for excellent tabulated pKa values for a large number of organic compounds see: hVp://evans.harvard.edu/pdf/ evans_pka_table.pdf and hVp://www.chem.wisc.edu/areas/reich/pkatable/index.htm
17
Reactivity and Control for Organic Synthesis
1) Explain the following pKa orders: OH
OH
OH
< NO2 most acidic lowest pKa
O Me
<
O
O
Me
OMe
O
<
NO2
O
MeO
most acidic lowest pKa
Me
OMe
least acidic higest pKa
least acidic higest pKa
O Me
<
O Me
Me
<
Me
(d)
Me O least acidic higest pKa
most acidic lowest pKa
Me
Me N H
Me Me
>
Me
O
Me
< Me
(b) O
(c)
>
(a)
N H
Me Me most acidic lowest pKa
Me
N H
Me
least acidic higest pKa
2) Which of the following is more basic? (a)
(b) NH
N H
(c)
Me
3) Explain the pKa’s of maleic and fumaric acid: HO2C
CO2H
3.02, 4.38
O-Na+
CO2H HO2C 1.92, 6.23
Me
S-Na+
N Me
N
Me Me
N
Me
18
Reactivity and Control for Organic Synthesis
4) How might you carryout the following transforma
H
O
Me O
O H
TBDPSO
(b)
O
O
O
Me
H
TBDPSO O
O
Me
OMe
O
O
Me
OMe
OMe Me
O
(c)
O OMe
OH
HO
MeO H N
HO
(e)
OH
HO NH2
(d)
O
O
HO
OH OH
OAc OAc NH2
HO OH OH
Me
NH2
AcO OAc OAc
19
Reactivity and Control for Organic Synthesis
5) Explain the following transforma
(a)
O EtO,
OEt
EtOH
O OEt
OEt O O
(b)
Me O
O EtO,
EtOH
Me
OEt
O OEt
6) In water, the basicity of the amines below is as follows, explain. Me
NH2
<
Me
N
Me
<
Me
7) Predict the product of the following reac
LDA, OMe
Me enantiopure
Br
Me
N H
Me
20
Reactivity and Control for Organic Synthesis
“When two molecules collide, three major forces operate. (i) The occupied orbitals of one repel the occupied orbitals of the other. (ii) Any posi%ve charge on one aTracts any nega%ve charge on the other (and repels any posi%ve). (iii) The occupied orbitals (especially the HOMOs) of each interact with the unoccupied orbitals (especially the LUMOs) of the other, causing an aTrac%on between the molecules.” Molecular Orbitals and Organic Chemical Reac9ons I. Fleming this means that reac%ons generally have both a charge and an orbital component. Reac%ons can be predominantly charge controlled, predominantly orbital controlled, or a mixture of the two. nucleophiles and electrophiles were classified by Pearson as HARD or SOFT – R. G. Pearson, Chemical Hardness, John Wiley & Sons, 1997. Hard nucleophiles (and electrophiles) are small and highly charged and have high electronega%vity (i.e. have a large charge:radius ra%o) – they have a low energy HOMO. Sob nucleophiles (and electrophiles) are larger and have lower electronega%vity and are more polarizable – they have a high energy HOMO. Bases (Nucleophiles)
Acids (Electrophiles)
Hard
Hard
H2O, HO-‐, F-‐, RCO2-‐, Cl-‐, ROH, RO-‐, NH3, RNH2
H+, Li+, Na+, K+, Mg2+, BF3
Intermediate
Intermediate
PhNH2, N3-‐, NC-‐, Br-‐
carboca%ons
SoN
SoN
I-‐, RS-‐, RSe-‐, S2-‐, RSH, RSR, R3P, alkenes, aroma%cs, R-‐
Ag+, Pd2+, I2, Br2, radicals
21
Reactivity and Control for Organic Synthesis
Hard nucleophiles have low energy HOMO’s and a high charge:radius ra%o Hard electrophiles have high energy LUMO’s and a high charge:radius ra%o charge dominates their reac%vity
SoU nucleophiles have high energy HOMO’s and they are polarizable SoU electrophiles have low energy LUMO’s and they are polarizable orbital interac%ons dominates their reac%vity
hard electrophile
soft electrophile
hard:hard charge control
soL:soL orbital control hard nucleophile
Generally the case that: Hard nucleophiles tend to react well with hard electrophiles i.e. the reac
soft nucleophile
22
Reactivity and Control for Organic Synthesis
Generally the case that: Hard nucleophiles tend to react well with hard electrophiles i.e. the reac
Bases (Nucleophiles)
Acids (Electrophiles)
Hard
Hard
H2O, HO-‐, F-‐, RCO2-‐, Cl-‐, ROH, RO-‐, NH3, RNH2
H+, Li+, Na+, K+, Mg2+, BF3
Intermediate
Intermediate
PhNH2, N3-‐, NC-‐, Br-‐
carboca%ons
SoN
SoN
I-‐, RS-‐, RSe-‐, S2-‐, RSH, RSR, R3P, alkenes, aroma%cs, R-‐
Ag+, Pd2+, I2, Br2, radicals
the principle of hard/soU acid bases has been applied to a large number of chemical reac%ons
HO
H H O H
2H2O
faster than
Br Br Br
faster than
HO
Br Br
H H O H
Recently the HSAB theory has been disputed see: H. Mayr, M. Breugst, A. R. Ofial, Angew. Chem. Int. Ed., 2011, 50, 6470
HO Br +
H
Br
23
Reactivity and Control for Organic Synthesis
Ambident nucleophiles alkyla%on of enolates
O H
:Base
O M
RX
O
R +
O-‐alkyla%on
O R
C-‐alkyla%on
with enolates the majority of the charge is, as expected, on the oxygen atom charged electrophiles aTack oxygen e.g. protons, carboca%ons soU electrophiles will generally aTack carbon – largest HOMO coefficient
in general, in enolate reac%ons the oxygen atom is associated with a metal ion and solvent and hence both of these variables affect the ra%o of C:O alkyla%on to maximise C-‐alkyla%on use a lithium base (strong O-‐Li bond) and an alkyl halide in THF (soU-‐soU interac%ons) to maximise O-‐alkyla%on use a highly coordina%ng solvent (e.g. HMPA), a potassium base, and an alkyl sulfonate Ph
O
O +
Ph
Ph
OH
Br
DMF 97% CF3CH2OH 7%
0% 85%
24
Reactivity and Control for Organic Synthesis
Ambident nucleophiles Me K
O
O
Me OEt
O
X
O
HMPA
O
+
O OEt
OEt A
B
Me
X =
A
B
OTs
88%
11%
Cl
60%
32%
Br
39%
38%
I
13%
71%
A. L. Kurts, A. Masias, N. K. Genkina, I. P. Beletskaya, O. A. Reutov, Tetrahedron, 27, 4777
in a similar manner, deprotonated secondary amides alkylate on nitrogen with alkyl halides neutral secondary amines alkylate on oxygen with hard alkyla%ng agents MeO H N
O
CO2tBu O O
R
R
CO2tBu
O
DMF
R'
K2CO3, Et3O BF4 CH2Cl2
N
O
Cl
R
H H N
O
NaH, MeO
CO2tBu O
N
EtO R
H
CO2tBu R'
A. Endo, S. J. Danishefsky, J. Am. Chem. Soc., 2005, 127, 8298.
Explain these two transforma
25
Reactivity and Control for Organic Synthesis
Nucleophilic SubsTtuTon at a saturated carbon SN2 – subs
two limi%ng mechanis%c cases – SN2 and SN1 – mechanis%c con%nuum between these extremes SN1 – subs
example: MeI + NaOH → MeOH + I-‐
Me
example:
Me Me
rate = k[substrate][nucleophile] i.e. rate dependent on both substrate and nucleophile ‡ (-) Nu
Cl
Me
H2O
OH
Me Me
rate = k[substrate] i.e. rate is independent of nucleophile
R
(-) LG R' R''
Me Me
Me
Me R Nu
R'' R'
Nu LG
R Nu
X
Me Me Me
Nu
R'' R'
concerted reac%on, single transi%on state no intermediate is formed
Me Me
favoured by 1° substrates and some 2° substrates requires good nucleophile and leaving group
stepwise reac%on, via an intermediate -‐ the 1st step is rate determining (forma%on of C+), 2nd step is fast
favoured by 3° substrates and some 2° substrates
requires good leaving group and solvent that stabilises carboca%ons
26
Reactivity and Control for Organic Synthesis
SN1 reac%on
enanTomers ‡
X R
R''
X (-)
step 1 R'
R
(+) R
R'
R' Nu
Nu
step 2
R''
R''
R
R''
R R'
R''
R'
Nu
SN1 reac%ons proceeds via a discrete carbenium ion and forma%on of the carbenium ion (step 1) is usually the rate determining step the lowest energy conforma%on of carbenium ions is planar trapping of the carbenium ion by a nucleophile (step 2) is generally fast the nucleophile can trap the carbenium ion from either side, hence enan%oenriched substrates should be expected to give racemic products under SN1 condi%ons – c.f. SN2 reac%ons go with strict inversion of configura%on hence the rate of the reac%on is not affected by the added nucleophile the stability of carbenium ions is in the order ter%ary > secondary > primary due to hyperconjuga%on H CH3
H H
hyperconjuga%on is the overlap of filled C-‐H (or C-‐C) σ-‐bonding orbital with the empty p-‐orbital resul%ng in a lowering in energy of the system i.e. stabilisa%on
CH3
Nomenclature of carboca%ons proposed by Olah J. Am. Chem. Soc., 1972, 94, 808.
27
Reactivity and Control for Organic Synthesis
HyperconjugaTon dona%on of C-‐H σ-‐bond electrons in empty p orbital empty p-orbital H CH3
H H
filled σ C-H orbital
CH3
energy of the bonding electrons reduced system stabilised
greater number of C-‐H (or C-‐C) σ-‐bonds the greater the extent of hyperconjuga%on and the greater stabilisa%on carbenium ion stability therefore goes in the order:
tertiary R
R >
secondary primary R R > R
R
carbenium ions have been observed by NMR and X-‐ray crystal structure analysis
a recent X-‐ray structure of the t-‐butyl ca%on (anion is CHB11Cl11) shows the planar nature of the carbenium ion. E. S. Stoyanov, I. V. Stoyanova, F. S. Tham, C. A. Read; Angew.Chem., Int.Ed. 2012, 51, 9149 conjuga%on with alkenes, arenes and lone pairs, also stabilises carbenium ions
28
Reactivity and Control for Organic Synthesis
conjuga%on with alkenes, arenes and lone pairs, also stabilises carbenium ions
X
benzyl ca%on stabilised by delocalisa%on
ψ3 X
energy of isolated p-‐orbital
allyl ca%on stabilised by delocalisa%on
ψ2
ψ1 RO
X
RO
RO
α-‐heteroatom subs%tuted ca%ons stabilised by delocalisa%on
Which orbitals are overlapping in the stabilisa
allyl ca%on more stable than energy of p-‐orbital – conjuga%on is stabilising
29
Reactivity and Control for Organic Synthesis
the more stable the carbenium ion the faster SN1 reac%on rates of hydrolysis of alkyl chlorides in 50% aqueous ethanol (adapted from Organic Chemistry, Clayden, Greeves and Warren, 2nd Edi%on, OUP 2012) 2° chloride, not that stable C+ not good at SN1
butyl
1° chloride ∴ SN2
iso-propyl
Cl 0.07
91
benzyl Cl
Cl
Cl 1.0
tert-butyl
Cl
1° but benzylic
allyl
0.12
methallyl
allylic ca%on is 2° at one end
1° but allylic
4.0
cinamyl
dimethallyl Cl
Cl 2100
3° chloride very good at SN1
Cl 7700
1° but allylic and benzylic
130000
allylic ca%on is 3° at one end
30
Reactivity and Control for Organic Synthesis
as a carbenium ion is formed during an SN1 reac%on a polar solvent is required for reac%on
the best solvents for promo%ng SN1 reac%ons are polar pro%c solvents such as water and alcohols (they can readily solvate the carbenium ion as well as the leaving group (by hydrogen bonding – see later)) solvent
water
ethanol
ace%c acid
DMSO
DMF
dielectric constant ε
80
25
6.2
46
38
solvent
acetone
EtOAc
THF
ether
hexane
dielectric constant ε
21
6
7.5
4.3
1.9
rela%ve rate of solvolysis (i.e. reac%on with solvent as the nucleophile) of tert-‐butyl bromide is 3 x 104 %mes faster in 50% aqueous ethanol than in neat ethanol Explain the rela
Br Br 1
10-6
Br 10-14
δH In polar solvent the δ+ OH δO carbenium ion is solvated by δH H δ+ polar solvent. It is easier to HO δ+ H LG H cluster water molecules OH H O around the carbenium ion δ- H H δ+ δHO and the leaving group than δethanol molecules
H OδH + H OδH
31
Reactivity and Control for Organic Synthesis
Nucleophilic SubsTtuTon at a saturated carbon SN2 – subs
example: MeI + Na+OH-‐ → MeOH + Na+ I-‐
rate = k[substrate][nucleophile]
‡
H HO
H H
(-) HO
I
H
H
(-) I
HO
H H
H I
H H
H H
I
HOMO of nucleophile (nucleophile lone pair) aVacks the back side of the carbon atom as it is pulng electrons into the C-‐I σ* orbital
at the transi%on state the central carbon atom is bonded to 5 other atoms – hence fundamentally SN2 reac%ons are difficult reac%ons the trajectory of approach is along the path of the bond to the leaving group – evidence from Eschenmoser’s experiments O O S O CH3 NaH
O O S O CH3
S O O H3C
S O O H3C
X H3C
O O S OH
CH3 S O O
reac
32
Reactivity and Control for Organic Synthesis
trajectory results in inversion of configura%on between star%ng material and product ‡ Me Nu
H Et
LG
Me
(-) Nu Et
H
(-) LG
Me Nu
H Et
LG
the rate of an SN2 rec%on is influenced by the nature of the substrate, the nucleophile, the leaving group and, with anionic nucleophiles, the associated counterion, and the solvent What makes a good nucleophile? – i.e. what gives a fast reac%on with an electrophile nucleophilicity is related to basicity but is significantly more complex if the atom we are comparing is the same then nucleophilicity does parallel basicity. O HO PhO > > > H2O > ClO4 Me O basicity is a measure of electron pair dona%on to a proton (generally under equilibra%ng condi%ons) nucleophilicity is electron pair dona%on to another atom, frequently carbon, generally under kine%c condi%ons factors which influence nucleophilicity include: charge, electronega%vity, solvent, size, bond strength
Note: the order of nucleophilici<es is also dependent on the nature of the leaving group
33
Reactivity and Control for Organic Synthesis
Charge charged species are more nucleophilic than their neutral counterparts this is expected as nucleophiles are electron pair donors so the more electron rich the nucleophile the beTer donor it is O
O
Me
O
Me
S
more nucleophilic than
Me
more nucleophilic than
NH OH
Me
SH
NH2 more nucleophilic than
BuLi is obviously more nucleophilic than butane
ElectronegaTvity nucleophilicity is related to basicity, but significantly more complex as it involves dona%on of an electron pair to any atom, whereas basicity is dona%on of an electron pair to H+ in the same row of the periodic table more basic means more nucleophilic ∴ going from leU to right across the periodic table nucleophilicity decreases the more electronega%ve atom is the weaker nucleophile as it holds on to its lone pairs of electrons more %ghtly and is less able to donate an electron pair to form a bond. CH3
>
NH2
most basic most nucleophilic
>
HO
>
F
least basic least nucleophilic
NH3
>
most basic most nucleophilic
H2O
>
HF
least basic least nucleophilic
this does not necessarily mean we will get good yields in SN2 reac%ons with these anions as they are also very basic and hence other reac%on pathways can dominate
Reactivity and Control for Organic Synthesis
34
Solvent in polar pro%c solvents (e.g. water, MeOH, AcOH) nucleophilicity increases going down the group – again the less electronega%ve atom is the more nucleophilic F
<
Cl
<
least nucleophilic
Br
<
I
most nucleophilic
δ-
in polar pro
in polar apro%c solvents (e.g. DMSO and DMF) the order of nucleophilicity can invert when compared with polar pro%c solvents as the solvent has weaker interac%ons with the nucleophile. Frequently reac%ons are much faster in these solvents compared with in water MeI + Cl MeCl + I for the halides under some condi%ons, nucleophilicity now decreases going down the group and again parallels basicity (here the most electronega%ve atom is the best nucleophile). Here charge control appears to be domina%ng the reac%on F > Cl > Br > I most basic most nucleophilic
least basic least nucleophilic
Reactivity and Control for Organic Synthesis
35
with uncharged nucleophiles, nucleophilicity increases going down the group – here orbital control appears to be domina%ng – the nucleophile with the highest energy HOMO reacts the fastest PR3
NR3
>
H2S
increasing nucleophilicity lower energy HOMO H2O least nucleophilic
>
Note: nucleophilicity is complicated and the above should be viewed as guidelines A rule of thumb is that nucleophilicity increases going down a group and increases in moving from right to leU in the periodic table
C
N
O
F
Si
P
S
Cl
Ge As Se Br Sn Sb Te I Pb Bi
Po At
the shape of the nucleophile also influences its nucleophilicity in moving from the star%ng materials to the transi%on state the central carbon goes from 4-‐coordinate to 5-‐ coordinate hence sterically hindered nucleophiles react more slowly MeO fastest
>
O
>
O slowest
conversely, small linear anions such as N3-‐, NC-‐ and RC≡C-‐ are good nucleophiles
increasing nucleophilicity
higher energy HOMO H2Se most nucleophilic
>
36
Reactivity and Control for Organic Synthesis
the following is the order of reac%vity of various nucleophiles with methyl iodide in methanol – all of these anions would be considered good nucleophiles (from Chem. Rev. 1969, 69, 1-‐32) – PR3 are also excellent nucleophiles PhS-‐ > I-‐ > SCN-‐ ≈ CN-‐ > N3-‐ ≈ Br-‐ > Cl-‐ > OAc-‐ in polar pro%c solvents PhS-‐ > CN-‐ > -‐OAc > Cl-‐ ≈ Br-‐ ≈ N3-‐ > I-‐ > SCN-‐ in dipolar apro%c solvents Leaving Group
‡ Me
Nu
H Et
LG
Me
(-) Nu Et
H
(-) LG
Me Nu
H Et
LG
during the SN2 reac%on the bond to the leaving group is broken and the LG departs with a lone pair of electrons i.e. becomes more nega%vely charged in the transi%on state ∴ two factors generally influence the leaving group ability: i) the strength of the bond to carbon ii) the stability of the leaving group
37
Reactivity and Control for Organic Synthesis
leaving group ability relates to pKa i.e. good LG’s are weak bases tosylate TsO-‐
triflate TfO-‐ N2 pKa
>
O F3C
S
-14
O O
>
I
O O S O
>
≈
Br
>
Cl
Me -10
-3
-9
-8
rough order of LG ability – the LG ability depends on the nucleophile and the solvent and the above order can vary; however, very weak bases are good leaving groups iodide is a good leaving group as it forms a weak bond to carbon as well as being a stable anion F-‐ is a very poor leaving group in SN2 reac%ons as it forms a very strong bond to carbon HO-‐ is a very poor leaving group in SN2 reac%ons as it is a strong base (pKa H2O = 15.74) but can be made into a good leaving group by protona%on (pKa H3O+ = -‐1.74) or conversion into a tosylate or triflate
Common leaving groups in SN2 reac%ons tend to have a pKa < 2
38
Reactivity and Control for Organic Synthesis
Nature of the substrate
SN2 reac%ons – back to the transi%on state ‡ R Nu
R''
R
(-) Nu
LG R'
R
(-) LG
Nu
LG
R'' R'
R' R''
the nucleophile has to aTack carbon, hence with larger the R groups the rate of reac%on decreases
in moving from substrate to transi%on state carbon moves from being 4-‐coordinate to being 5-‐coordinate hence as the R groups become larger the rate of the reac%on decreases
SN2 reac%ons ∴ only occur with primary and some secondary substrates – not with ter%ary substrates rela%ve rates of the reac%on of the bromides below with chloride are (Chem. Rev. 1956, 56, 571): methyl Me Br relative rate
ethyl Me
Br
10
60
propyl
iso-propyl Me
Br
Me
neo-pentyl
Me 6.5
Me Me
Br 0.13
Me
tert-butyl Me
Br
0.0003
Me
negligible
neopentyl bromide is par%cularly unreac%ve as the nucleophile is severely hindered from aTacking the necessary carbon atom (Note: neopentyl systems are also unreac
Me Nu
Me Me H H
LG
Me
Me Me
(-) Nu
(-) LG H H
‡ Me Nu
Me Me R'' R'
Me Br
LG
39
Reactivity and Control for Organic Synthesis
Nature of the substrate rate of SN2 reac%ons is increased with substrates which carry an adjacent sp2 (or sp) hybridized atom at the SN2 transi%on state the central carbon is partly bonded to both the nucleophile and the leaving group 3 atoms are sharing 4 electrons i.e. there is a 3-‐centre,4-‐electron bond the central carbon has a partly filled p-‐orbital and the electrons in this orbital can be delocalised into the adjacent π-‐ system which lowers the energy of the transi%on state and the reac%on is faster
H Nu
H
H LG
(-) Nu
delocalisa%on into π-‐system
‡
H (-) LG
H Nu
H LG
40
Reactivity and Control for Organic Synthesis
Nature of the substrate the π-‐system has to be in the correct orienta%on for efficient overlap and consequent transi%on state stabiliza%on
α-‐halo carbonyl compounds are par%cularly reac%ve under SN2 condi%ons as they contain an α sp2 hybridised atom aTached to oxygen and the C=O π* is lower in energy than for an alkene rela%ve rates for reac%on alkyl halides with KI in acetone at 50 °C are given below (from Mechansim in Organic Chemistry, R. W. Alder, R. Baker, J. M. Brown, Wiley, 1971) H
H
H (-) Nu
LG
Nu
‡
H
H
(-) LG
Nu
H LG
delocalisa%on into π-‐system
Me relative rate
Cl 1
Me 200
Cl
Cl
Cl 79
200
MeO
Cl 920
O
N
Cl
Ph 3000
Cl
100,000
41
Reactivity and Control for Organic Synthesis
Summary of structural varia%ons and nucleophilic subs%tu%on taken from Organic Chemistry, Clayden, Greeves and Warren, 2nd Edi%on, OUP 2012. Electrophile
Me X
R
R
X
X
R R
R
X R
R R
X R
methyl
primary
secondary
ter%ary
‘neopentyl’
SN1 mechanism?
✗
✗
✗✓
✓✓
✗
SN2 mechanism?
✓✓
✓
✗✓
✗
✗
X
X
Electrophile
RO
O
O
X R
X
R
X
allyl
benzyl
α-‐alkoxy
α-‐carbonyl
α-‐carbonyl and ter%ary
SN1 mechanism?
✓
✓
✓
✗
✗
SN2 mechanism?
✓
✓
✓
✓✓
possible
✗ = bad ✓ = good, ✓✓ = excellent, ✗✓ = poor
42
Reactivity and Control for Organic Synthesis
1) Explain why the reac
O2N
O KI
O A
O I
O2N
acetone
O2N
KI
OH
NO2
O
B
OH
Ph
NO2
NO2
O
2) For the reac
+
CH3OCH3
NaOMe
+
NaCl
3) Suggest reagents for the following reac
(a)
Me
N
(b)
Me I
HO OH
(c)
N Br
Cl
H
(d)
O
H OH
Br H
Ph
I
acetone
O
NO2
O2N
SMe H
O
43
Reactivity and Control for Organic Synthesis
4) Explain the outcome of the following reac
Cl Et2N
HO
Me NBn2
Cl
(b)
NEt2
NaOH
OH
Me OH
H2O
Bn2N
Et
H
OH
NaOH, H2O
Cl
OH
Et
(c) Br
HBr
OH
(d)
OH Ph
Br
Br
MeO, MeOH
OMe
Br
O
HBr
OH
Ph
Ph
5) Predict the outcome of the following reac
Cl
Br
OMe
MeOH, H
OH
O
MeOH, Et3N
Cl O
H
AlCl3
Br
44
Reactivity and Control for Organic Synthesis
EliminaTon reacTons
mechanis%c con%nuum from E1→E2→E1cB
E2 – elimina
E1 – elimina
example:
example:
R Br
H
+
EtO
+
R
rate = k[substrate][base] i.e. rate dependent on both substrate and base ‡ (-)
EtOH Br
Me Br
Me Me
EtOH Me
Me
+ HBr
rate = k[substrate] i.e. rate is independent of base (which is EtOH in the above case)
X (-) H B
Me Me X H
B
B
X BH
H B
C-H σ to C-X σ* H
HH
X
concerted reac%on, single transi%on state no intermediate is formed, an%periplanar arrangement of proton and leaving group is most favourable for elimina%on requires good base and leaving group 3° substrates give more elimina%on than 2° substrates which give more elimina%on than 1° substrates
Me Me Me
X
Me Me
stepwise reac%on, via an intermediate -‐ the 1st step is rate determining (forma%on of C+), 2nd step is fast
favoured by 3° substrates and some 2° substrates
requires good leaving group and solvent that stabilises carboca%ons
45
Reactivity and Control for Organic Synthesis
EliminaTon reacTons RO
E1cB – elimina%on from the conjugate base
example:
variable kine%cs depending on substrate requires a carbanion stabilising group
acid
O Me
RO
+ HO base
conjugate base
RO
+
O Me
O Me
as the carbanion (an enolate in the above example) helps to expel the leaving group, conjuga%on is developed in the transi%on state leading to the product, HO-‐, and RO-‐ can ∴ be leaving groups requires a base and leaving group SubsTtuTon versus EliminaTon SN1 reac%ons are frequently accompanied by E1 reac%ons if there is an appropriately posi%oned proton – this is unsurprising as both reac%ons proceed through the same intermediate SN2 reac%ons can also be accompanied by E2 reac%ons we need to look at factors affec%ng: i) SN1/E1 product ra%os
ii) SN2/E2 product ra%os iii) change of mechanism i.e. SN1/E1 → SN2/E2
46
Reactivity and Control for Organic Synthesis
primary substrates do SN2 or E2 – SN2 is generally favoured but bulky bases (t-‐BuOK) allow E2 to occur Br
EtO
OEt 91
9
to maximise SN2 – use good nucleophile e.g. RS-‐, X-‐, N3-‐ in dipolar apro%c solvent
ter%ary substrates do SN1 / E1 or E2 with good ionising solvents and no added anionic base then SN1 / E1 will be favoured with added base E2 will be favoured
Br
EtO
OEt <0.1
>97
secondary substrates can do SN1 / E1 or SN2 / E2 with good ionising solvents and no added anionic base or nucleophile then SN1 / E1 will be favoured with added base E2 will be favoured with good nucleophiles, dipolar apro%c solvents SN2 will be favoured
Br
EtO
OEt 20
80
Reactivity and Control for Organic Synthesis
for all substrates moving from primary to secondary to ter%ary favours elimina%on
polar pro%c solvents and base favour E2 over SN2 [base/nucleophile is solvated so easier to aTack outside of the molecule (i,e. remove a proton) than aTack carbon in an SN2 reac%on] heat favours elimina%on over subs%tu%on. EtO Br OEt increased branching at the β-‐posi%on favours elimina%on 40 60 A good overview can be found at: hTp://www.masterorganicchemistry.com/2013/01/18/wrapup-‐the-‐quick-‐n-‐dirty-‐guide-‐to-‐sn1sn2e1e2/
47
48
Reactivity and Control for Organic Synthesis
1) Predict the major product (if any) from the following reac
Br
EtOH
(b)
Br
EtONa EtOH
(c)
Br
EtSNa EtSH
(d)
EtONa Br
EtOH
2) The rate of hydrolysis of tBuCl in water is greatly accelerated by the addi
49
Reactivity and Control for Organic Synthesis
as we have seen, allylic systems are reac%ve under SN2 (stabilisa%on of the transi%on state for subs%tu%on) and under SN1 condi%ons the allylic system has two posi%ons which can be aTacked leading to isomeric products – i.e. there are issues of regioselec%vity Nu
Nu X
Nu
Nu
X
SN2
Nu
Nu
Nu Nu
X
SN2’
SN1
SN1’
sterics and electronics play a role in determining SN/SN’ reac%ons O
O
O
EtO
EtO
OEt
EtO
O OEt
O
+
O
Br
Cl
EtO Cl
PhS
EtO Cl
PhS
50
Reactivity and Control for Organic Synthesis
SN2’ reac%ons are not very common they can also be solvent dependent, I. Fleming, E. J. Thomas, Tetrahedron, 1971, 28, 4989
MeO
SPh Cl
PhS
Cl
DME
Cl MeO
SN2’ Cl MeO
Cl
Cl MeO
Cl
MeO
SPh
SN2 Cl
DME
MeOH
SPh
PhS
Cl
PhS
Cl
MeO
MeO
Cl
Cl
PhS MeOH
MeO
SPh
excellent control of SN2/SN2’ can achieved with organometallic reagents – most notably with copper, palladium and iridium 10 mol% CuCN n-BuMgBr
Bu
OAc
Bu THF 0 °C
94
6
Et2O, 20°C
3
97
J. E. Bäckvall, M. Sellén, B. Grant, J. Am. Chem. Soc., 1990, 112, 6615. For a review on copper-‐catalysed enan%oselec%ve conjugate addi%on and allylic subs%tu%on see: A. Alexakis, J. E. Bäckvall, N. Krause, O. Pàmies, M. Diéguez Chem. Rev. 2008, 108,2 796.
51
Reactivity and Control for Organic Synthesis
O P N O
O Nu +
R
O
Me Me
and related phosphoramidites
Nu
cat. [Ir(COD)Cl]2 OMe
R *
additive
Nu = RNH2, ArO-, malonates, enamines, silylenol ethers, indoles, PhMgBr, NH3, alkenes, vinyltrifluoroborates etc.
high yields high regioselectivity high enantiomeric excess
main names in the field: Helmchen, Alexakis, Hartwig, Carreira, You
1) Explain the outcome of the following reac
O
Br
heat
K2CO3, DMF
2) Suggest reagents and reac
OH
OH
Reactivity and Control for Organic Synthesis
AromaTcity – Electrophilic AromaTc SubsTtuTon and Nucleophilic AromaTc SubsTtuTon
August Kekulé
“I was silng wri
52
53
Reactivity and Control for Organic Synthesis
typical reac%ons of alkenes + Br2
Me
Br
fast
Br
Me
Me
not substitution
Br addition
typical reac%ons of benzene +
Br2
Br
FeBr3 catalyst
Br +
not addition
HBr Br
substitution
retains aroma%c sextet of electrons in subs%tu%on reac%ons does not behave like a “normal” polyene or alkene benzene is both kine
H2/Pt catalyst
H2/Pt catalyst
H2/Pt catalyst
ΔHohydrog= 3 x -‐120 = -‐360 kJmol-‐1 (hypothe%cal, 1,3,5-‐cyclohexatriene)
benzene ≈150 kJmol-‐1 more stable than expected – (represents stability over hypothe%cal 1,3,5-‐ cyclohextriene) – termed the empirical resonance energy (values vary enormously)
54
Reactivity and Control for Organic Synthesis
AromaTcity Hückel’s rule holds for anions, ca%ons and neutrals (4n +2) π-‐electrons for aroma%c compounds; 4n π-‐electrons for an%-‐aroma%c cyclopropenium ca%on -‐ (4n +2), n = 0, 2π electrons
Cl
SbCl5 (Lewis acid)
H H
SbCl6
H
insoluble in non-‐polar solvents; 1 signal in 1H NMR δH = 11.1 ppm -‐ aroma%c and a ca%on compare with cyclopropyl ca%on which is subject to rearrangement to the allyl ca%on Nu
Cl
Ph Ph
Ph
Ph
Ph
Ph
reduced barrier to rotation
Ph Ph
Ph
Ph
Nu
Ph
Ph
Ph
Ph
Ph
Ph
6.3 D
6π-‐aroma%c
2π-‐aroma%c
Ph
Ph
55
Reactivity and Control for Organic Synthesis
Benzene (4n +2), n = 1, 6π electrons δH = 7.26 ppm, planar molecule; bond length = 1.39 Å 1.40 Å H δ = 7.46 H δ = 7.01
H
isoelectronic with pyridine
1.39 Å N
H δ = 8.50
1.34 Å
Cyclopentadienyl Anion (4n +2), n = 1, 6π electrons
H
base
B:
H
H
F3C F3C
pKa = 16
pKa = 43
CF3 H CF3 CF3
pKa < -‐2
2.2 D
C-‐C
sp3-‐sp3
1.54 Å
C-‐C
sp3-‐sp2
1.50 Å
C-‐C
sp3-‐sp
1.47 Å
C-‐C
sp2-‐sp2
1.46 Å
C-‐C
benzene
1.39 Å
C=C
1.34 Å
C≡C
1.21 Å
56
Reactivity and Control for Organic Synthesis
cyclopentadienide anion is isoelectronic with furan pyrrole and thiophene
in each case the (one of the) lone pair(s) is parallel to the p-‐orbitals and part of the π-‐system
S thiophene
furan
pyrrole
0.55 D
0.66 D
1.74 D
X
X
NH
O
X
NH pyrrolidine
X
Electrophilic AromaTc SubsTtuTon
E H
E
E
step 1 is usually rate determining because aroma%city is lost step 2 is fast as aroma%city is regained
σ-‐complex Wheland intermediate arenium ion E
E H
H
57
Reactivity and Control for Organic Synthesis
Electrophilic AromaTc SubsTtuTon
(+) ‡ (+) H E
‡ (+)
step 1
step 2 E H
E
H (+) E
E
H
σ-‐complex Wheland intermediate arenium ion
E
activation energy E
E H
H + E E
Hammond’s postulate: The transi%on state resembles the structure (intermediate or substrate or product) to which it is closest in energy (i.e. transi%on state resembles intermediate arenium ion, therefore what stabilises the arenium ion stabilises the transi%on state.)
58
Reactivity and Control for Organic Synthesis
Mechanis%c Evidence isola%on of intermediates Me EtF, BF3 - 80 °C
Me
Me
Me
Me
Me heat
H BF 4 Me Me stable solid mp -15 °C
Me
SbF5 / FSO3H
SbF6
-120 °C in SO2FCl
Me
δH = 5.6 H H
H H
Me
H
δH = 9.7
H
δH = 8.6
H H
H H
H δH = 9.3
Subs%tuent Effects subs%tuent Y affects both the rate and regiochemistry of the reac%on Y
Y E
Y
Y
E E E
ortho (1,2-‐disubs%tuted)
meta (1,3-‐disubs%tuted)
para (1,4-‐disubs%tuted)
59
Reactivity and Control for Organic Synthesis
Y
Y
Y
Y
E
E
E E
electron dona%ng groups ac%vate the aroma%c ring (i.e. substrate reacts faster than benzene) and are ortho and para direc%ng ACTIVATING group means that the reac%on of the subs%tuted benzene is faster than that of benzene itself O O Typical ac%va%ng groups include: OH, O-‐, O R , HN R ,OR, NH2, NR2, alkyl, Ph OMe
OMe
OMe Br
Br2, AcOH Br 98
kanisole / kbenzene = 109
2
electron withdrawing groups deac%vate the aroma%c ring (i.e. substrate reacts slowed than benzene) and are meta direc%ng DEACTIVATING group means that the reac%on of the subs%tuted benzene is slower than that of benzene itself O O O Typical deac%va%ng groups include: R3N+, CF3, NO2, SO3H, CN, O-‐, R , OR , NR2 halogens are mildly deac%va%ng and direct ortho and para
60
Reactivity and Control for Organic Synthesis
orienta%on of aTack when ring carries an electron dona%ng group, X:, which carries a lone pair (e.g. OMe) ortho and para aTack ortho aTack X:
X:
X:
X E
E
X: E
E
H
X: E
E
ortho
✓✓
para aTack X:
X:
X:
X
X:
X:
E
✓✓
para
meta aTack X:
E
E
X:
X:
HE
E X:
E
X:
E meta
E
E
H
E
E
in the intermediates from ortho and para aTack the carboca%ons are stabliised by overlap with the lone pair of X in the intermediate from meta aTack in the carboca%on is not stabilised by overlap with the lone pair from X
61
Reactivity and Control for Organic Synthesis
reac%on coordinate diagram for aTack on X-‐subs%tuted benzene (X = EDG) TS1
TS2
meta Energy
intermdiate
ortho and para similar energies
more stable intermediate(s) formed faster ∴ ortho and para products predominate benzene reacts slower than these substrates as substrates are more electron rich
X: + E
products
reac%on coordinate Therefore the intermediates (and hence the transi
62
Reactivity and Control for Organic Synthesis
orienta%on of aTack when ring carries an electron withdrawing group, Z, (e.g. NO2) meta aTack ortho aTack Z
Z
Z
Z E
E
E
H
Z E
E
ortho
✗✗
para aTack Z
Z
Z
Z
Z
E
✗✗
para E
HE
E
E
meta aTack Z
Z
Z
Z
Z
E meta
E
E
H
E
E
in the intermediates from ortho and para aTack the carboca%ons are destabilised as next to EWG Z in the intermediate from meta aTack in the carboca%on is never adjacent to EWG
63
Reactivity and Control for Organic Synthesis
reac%on coordinate diagram for aTack on Z-‐subs%tuted benzene (Z = EWG)
ortho and para similar energies
TS1 TS2
Energy
intermediate
less stable intermediate(s) formed slower ∴ meta products predominate benzene reacts faster than these substrates as it is more electron rich
meta
Z + E
products reac%on coordinate
The intermediates (and hence the transi
64
Reactivity and Control for Organic Synthesis
Halogens mildly deac%va%ng as they are electronega%ve and withdraw electron density from the ring through the σ-‐framework (falls off with distance) halogens direct ortho and para as they have lone pairs in high energy orbitals which stabilise the intermediates for ortho/para aTack :OMe
:Cl
OMe
1.2 D
Cl
1.6 D
O
2p –lone pair
good 2p -‐2p overlap MeO – overall electron dona%ng on benzene ring
3p –lone pair
Cl
Me
poor 2p -‐3p overlap Cl – overall electron withdrawing on benzene ring
with anisole the σ-‐electon withdrawing of the oxygen is less than the π-‐ dona%on of the oxygen 2p lone pair and anisole is ac%vated with respect to benzene
N
O
F
P
S
Cl
with chlorobenzene the σ-‐electon withdrawing of the chlorine is greater than the π-‐ dona%on of the chlorine 3p lone pair and chlorobenzene is deac%vated with respect to benzene
Se Br Te I Po At
increasing size of p-‐orbitals
both oxygen and chlorine are electronega%ve
increasing electronega
increasing electronega
65
Reactivity and Control for Organic Synthesis
Ques
why does fluorine react faster than than the other halobenzenes? why does fluorine give the largest amount of the para isomer? product distribu%on %
X
conc. HNO3, conc. H2SO4
X NO2
ortho
meta
para
kbenzene
PhF
12
-‐
87
0.18
PhCl
30
0.9
69
0.064
PhBr
37
1.2
62
0.060
PhI
38
1.8
60
0.12
Examples
+
H
CH3
CH3
CH3
CH3
NO2 59%
<4%
H
H 37%
+
Wheland intermediate for ortho / para aTack is stabilised by hyperconjuga%on – σCH → π
H H
NO2
Me is an electron dona%ng group and hence an ac%va%ng group
H H
O2N
NO2
HNO3 / H2SO4
kArX/
NO2
H
66
Reactivity and Control for Organic Synthesis
what happens if there are two subs%tuents on the benzene ring? subs%tuents can be broadly categorised into three classes
i)
STRONGLY ac%va%ng and ortho and para direc%ng (OH, OR, NH2, NR2)
ii)
mildly ac%va%ng groups such as alkyl groups (ortho and para direc%ng) and halogens (mildly deac%va%ng)
iii) deac%va%ng meta-‐direc%ng groups subs%tuents in group i) will dominate classes ii) and iii) subs%tuents in group ii) will dominate class iii) Examine the electronic effects of subs%tuents then consider sterics
H N
OMe F
MeO: o, p F: o, p MeO dominates ∴ para
Me
O
F3C
Me N Me
Me
AcNH: o, p Me: o, p AcNH dominates ∴ ortho
H MeO
O
MeO
Me2N: o, p CF3: m Me2N dominates ∴ para (sterics)
MeO: o, p CHO: m MeO dominates ∴ para (sterics)
67
Reactivity and Control for Organic Synthesis
opposing -‐ if similar reac%vity will get mixtures of compounds all other things being equal a 3rd group is least likely to enter between two groups meta to one another Cl
HNO3, CO2H H SO 2 4
Cl
OMe
OMe
CO2H
Br2 / AcOH OMe
NO2
Cl: o, p CO2H: m Cl dominates ∴ para OH
Br
MeO: o, p MeO: o, p ∴ ortho / para
Me
OH Me
OMe
Me
Br2, AcOH
Me Cl
Me Cl
HNO3, Ac2O
Cl O2N
25 NO2
Br
HO: o, p Me: o, p HO dominates ∴ para
75
Me: o, p Cl: o, p ∴ mixture
Note: these are guidelines and exact ra
68
Reactivity and Control for Organic Synthesis
subs%tuent effects are important for selec%vity and efficiency when designing a synthe%c route CO2H NO2
CO2H
Me
NO2 or
or
NO2
target material
synthe%cally we want to prepare the target material in a clean, selec%ve and efficient fashion CO2H
Me
NO2
Look at the star%ng materials CO2H, deac%va%ng meta direc%ng
NO2, deac%va%ng meta direc%ng
Me, ac%va%ng ortho / para direc%ng
if possible best to introduce the most deac%va%ng group(s) last in the synthe%c sequence rela%onship of NO2 groups is ortho/para with respect to CO2H ∴ best to use toluene as star%ng material CH3
CH3 HNO3 / H2SO4
CH3 NO2
Me NO2 KMnO 4
HNO3, H2SO4
NO2
nitro group is deac%va%ng ∴ can isolate and separate isomers if required
NO2
CO2H NO2
NO2
69
Reactivity and Control for Organic Synthesis
1) Explain the outcome of the following reac
O
Br N
O
Cl
(b)
Br OH
OMe
AlCl3, heat
OMe
OMe
OMe
(c)
(d) 2 equivalents ClSO2OH
Me
SO2Cl
H
(e)
SO3H SO3H
H2SO4, 160 °C
H2SO4, 80 °C
2) AVempted Friedel-‐Crabs acyla
O +
Cl
AlCl3
+
+ A
B
C
O
70
Reactivity and Control for Organic Synthesis
Ipso a\ack and reversible reac%ons electrophilic aroma%c subs%tu%on is generally an irreversible process all of the above arguments with regard to ortho, meta and para ra%os have been based on the irreversibility of the process i.e. the reac%ons are under kine%c control -‐ but there are some excep%ons not all electrophilic aroma%c subs%tu%on reac%ons are under kine%c control
sulfonyl group is electron withdrawing so we only have mono-‐subs%tu%on
sulfona%on -‐ usual reac%on condi%ons: conc. H2SO4 with SO3
O
S O
O
H
O
S O
OH HO3S
at high temperatures with dilute H2SO4 – sulfona%on is reversible aTack by an electrophile at a posi%on which already carries a non-‐hydrogen subs%tuent is termed ipso-‐subs
Br
OH SO3H
we can use an SO3H group as a blocking group
OH Br
ipso aTack
Br
H2SO4
SO3H
HO3S
H
OH SO3H H
SO3H
Br
OH SO3H
H S O O OH
Br
OH SO3H
H
Br
H
H
71
Reactivity and Control for Organic Synthesis
Friedel CraUs alkyla%on can be under kine%c or thermodynamic control Me
Me tBuCl / AlCl3
thermodynamic Me control Me
Me Me Me
Me
Me tBuCl / AlCl3
high temperature
Me Me
Me
Me Cl: Me
Me AlCl3
Me
Me Cl AlCl3
Me
Me
Me
Me Me
Me
Me
+
Me
Me
H Me
Me Me
Me
H Me
H
kine%c control
low temperature
Me Me
Me
Me Me Me
Me H Me Me
Me Me Me
Me
thermodynamic product all groups Me as far apart as they can be Me
Me
Me Me Me
Me
Me Me
H
72
Reactivity and Control for Organic Synthesis
IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Alkyla%on polyalkyla%on and rearrangement predominate excess MeCl, cat. AlCl3
H
H
Me Me
Me
Me
Me
H H
Cl AlCl3
Me Me
Me
Me
Me
Me
Me Me
Me
Me Me Me Me
with one equivalent of alkyla%ng agent mixtures of products result as the ini%ally formed monoalkyl arene is more reac%ve than the unalkylated arene – alkyl groups are electron dona%ng Me cat. AlCl3, Me Me Me MeCl +
Me
more reac%ve than benzene
more reac%ve than toluene
more reac%ve than toluene
73
Reactivity and Control for Organic Synthesis
IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Alkyla%on polyalkyla%on and rearrangement predominate with primary alkyl halides rearrangement occurs Me Me Me Br Me cat. AlBr3
+
major
primary carboca%ons are very unstable rearrangement to the secondary carboca%on occurs minor
of monoalkylated products
Br: Me
AlBr3
Br AlBr3
H Me
Me
Me
Me H Me
1,2-‐hydride shiU
Me Me
74
Reactivity and Control for Organic Synthesis
IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Acyla%on requires a full equivalent of the Lewis acid mono-‐subs%tu%on predominates as introduced group is electron-‐withdrawing and deac%vates aroma%c ring O
Me
O
Me
Cl
monosubs%tu%on
1 equivalent AlCl3
O
O Me
Cl AlCl3
Me
O
Me
O
O Me
AlCl3
AlCl3
H
carbonyl group can then be removed if required (Clemensen reduc%on, Zn/HCl; Wolf-‐Kishner reduc%on, NH2NH2 then KOH, heat; dithiane than Raney Ni) giving products of a selec%ve Friedel-‐CraUs alkyla%on para-‐isomer generally favoured by steric hindrance O O + MeO
Me
O O
Me
AlCl3, toluene Me
MeO
93% yield
Me
75
Reactivity and Control for Organic Synthesis
IntroducTon of FuncTonal Groups –Synthesis Friedel CraUs Acyla%on Fries rearrangement can give access to either the ortho-‐ or para-‐isomer O + HO
Me
O O
Me
O
pyridine Me
O
AlCl3 Me
O
O AlCl3
polar solvent O Me
workup
HO
Me O AlCl3 Me
workup
HO
O Cl3Al
major product in non polar solvents
AlkylaTon cataly
Me solvent-separated ion pair
O Me
O
O
AlCl3
AlCl3 inside solvent cage - tight ion pair
Friedel CraUs summary
AlCl3 Rearrangement subs%tu%on order
Me H O
O
non-polar solvent
O
O
major product in polar solvents e.g. PhNO2 Me O
O
H
AcylaTon stoichiometric no, but loss of CO from R-‐C≡O+ if R+ stable, e.g. Ph3C+ mono
76
Reactivity and Control for Organic Synthesis
IntroducTon of funcTonal groups Blanc chloromethyla%on – related to Fiedel-‐CraUs reac%ons OMe Me O
OMe
(CH2O)n, conc. HCl
Cl Me
Cl Me
O
Me
halogena%on – with ac%vated aroma%cs Lewis acid ac%va%on of the electrophile is not require, with benzene and with deac%vated aroma%cs Lewis acid ac%va%on of the electrophile is required NO2
NO2 Br2, FeBr3 Br
Me
Me
halogena%on can frequently be best achieved using Sandmeyer reac%ons (par%cularly good for introducing I and F as well as Cl, Br and CN) conc. HNO3 conc. H2SO4
NO2
Sn, HCl or H2Pd/C
NH2
HX, NaNO2, 0 °C
X N N
77
Reactivity and Control for Organic Synthesis
IntroducTon of FuncTonal Groups –Synthesis diazonium salts -‐ reac%ons Ar N N X
heat
H2O, 100 °C
Ar N N BF4
Ar OH
SN1 reac%on via carbenium ion
NH2
NaNO2, HX 0 °C
N
Ar F
SN1 reac%on via carbernium ion – Balz Schiemann reac%on
N -N 2
v. high energy intermediate, offset by the forma%on of N2 carbenium not stabilised by π-‐system as is orthogonal to π-‐system
slow
empty sp2 orbital
Ar N N X
cat. CuX, KX X = Br, Cl, CN
radical reac%on
Ar N N X
Ar N N X
radical reac%on
Ar N N X
Ar H
radical reac%on
Me
Ar X
N Ar
HO
H3PO2
O
KI
Ar N N X
Ar X
N
electrophilic aroma%c subs%tu%on
O
O OR
HO
O
Me
OR N
NHAr
via enol
78
Reactivity and Control for Organic Synthesis
Nucleophilic AromaTc SubsTtuTon SN2: alipha%c vs aroma%c AliphaTc nucleophile aTacks C-‐I σ* resul%ng in inversion of configura%on ‡ Me Nu + H Et
I
(-) Nu
Me (-) I H Et
remember SN2 reac%ons at sp2 hybridised centres (i.e. alkenes and arenes are incredibly rare)
Me Nu
H Et
+
I
AromaTc no possibility of nucleophile aTacking backside of C-‐LG σ* (transi%on geometry impossible) Lowest Unoccupied Molecular Orbital (LUMO) is π* not σ* aTacking electron rich arene with electron rich nucleophile SN1: alipha%c vs aroma%c Me Me Nu AliphaTc Me Me X Nu Me Me Me carbenium stabilised by hyperconjuga%on Me
LG
Me
AromaTc possible but very high energy intermediate (see Sandmeyer reaca%ons)
X
Nu
Nu
empty sp2 orbital
v. high energy intermediate, offset by the forma%on of N2 carbenium not stabilised by π-‐system as is orthogonal to π-‐system
79
Reactivity and Control for Organic Synthesis
Nucleophilic AromaTc SubsTtuTon SNAr – Addi%on – Elimina%on Mechanism LG LG Nu LG Nu Nu rate determining step
Nu
LG Nu -LG
Meisenheimer intermediate
nucleophile aTacks LUMO, electron withdrawing groups lower energy of LUMO and stabilise the nega%ve charge in the intermediate best to have electron withdrawing group(s), ortho and / or para to the leaving group Evidence isola%on of intermediates Cl O2N
OMe NO2
MeO
K
O2N
NO2
the nega%ve charge is delocalised ortho and para to leaving group
NO2
O
N
MeO
K
O
MeO OMe O N O2N
O
MeO OMe O N O2N
O
N
N
O
O
H. Ueda, M. Sakabe, J. Tanaka, Bull. Chem. Soc. Jpn., 1968, 41, 2866-‐2871.
O
O
80
Reactivity and Control for Organic Synthesis
Nucleophilic AromaTc SubsTtuTon SNAr – Addi%on – Elimina%on Mechanism for halogens as leaving groups, rate of reac%on usually follows kF > kCl > kBr (c.f. rate of SN2 reac%ons kI > kBr > kCl > kF) NO2
NO2 X
MeO
50 °C
OMe
X =
F
Cl
Br
I
krel
2810
3.1
2.1
1
rate determining step is generally aTack of nucleophile on aroma%c ring therefore bond strength to leaving group is not so important in influencing the rate fluorine is the most electronega%ve element and enhances the electrophilicity of the carbon being aTacked increasing the rate of aTack by the nucleophile O
N
O
O F
Nu
N
O
O F Nu
N
O Nu
rate = k[substrate][nucleophile]
1st step usually rate determining leaving group ability does depend on the nucleophile, nevertheless leaving groups can broadly be divided into three classes: taken from Physical and Mechanis
81
Reactivity and Control for Organic Synthesis
reac%vity is complementary to Pd-‐catalysed cross-‐coupling reac%ons of halobenzenes where regioselec%vity is generally governed by the rate of oxida%ve addi%on into the Ar-‐X bond which depends on bond strength General Trends in Oxida
I
+
PdL2
PdL2
PdI
MeO
MeO2C Cl
Cl
82
Reactivity and Control for Organic Synthesis
Nucleophilic AromaTc SubsTtuTon the ac%va%ng group
OMe
Cl
NO2
Me3N+
SO2Ph
O=CPh
CF3
H
krel
114000
2130
18400
2700
800
1
O2N
O2N
MeO
Y
Y
Y
Explain the outcome of the following reac
NO2 Me
Me
Me
NH3, MeOH
NO2
Me
NO2
the nucleophile – typically good nucleophiles in SNAr reac%ons include: RS-‐, HO-‐, RO-‐, PO-‐, RNH2 Synthesis of fluoxe
NHMe
HO Ph
NaH, O Me
N Me
Me
83
Reactivity and Control for Organic Synthesis
total synthesis of vancomycin – glycopep%de an%bio%c currently the ‘last line of defence’ to treat methicillin-‐ resistant staphylococcus aureus [MRSA]. Me HO H2N Me O
OH HO O
O
O
O HO O
H N H H
HO
OH OH
HO
Cl
O O
NH HO2C
O
Cl O H N
NO2
OH
N H O
O
OH O
H N O
N H
H N H H
FOH
MeHNOC
Na2CO3
HO O
NHBoc
OH
O
Oallyl
O
Cl O H H N N H H
OMs
OH
NH MeHNOC
OMe OMe
MeO
NH2
NO2
OMs
Cl O H N
NH
NHMe
Oallyl
NHBoc
O
OMe OMe
MeO
Na2CO3 OR O Cl F O
N O
D. A. Evans et al. Angew. Chem. Int. Ed. Engl. 1998, 37, 2700-‐2704
OR O N O F O Cl
OR OR
84
Reactivity and Control for Organic Synthesis
total synthesis of vancomycin – glycopep%de an%bio%c currently the ‘last line of defence’ to treat methicillin-‐ resistant staphylococcus aureus [MRSA]. Me HO H2N Me O
OH HO O
O
O HO O
Cl O H H N N H H
NH
HO
HO2C
OH OH
HO
O
Cl
O O
O
OH
O
F
allylO
N H O
O
OH O
H N O
N H
OH
Cl O H H N N H H
O
O
NH
NHMe
MeHNOC BnO
OH O
H N
N H O
O
N H
Boc NMe
NHDdm
OBn OBn
CsF, DMSO
NH2
allylO
O HO O
Cl O H H N N H H
NH MeHNOC BnO
Ddm = MeO
NO2
OMe
D. A. Evans et al. Angew. Chem. Int. Ed. Engl. 1998, 37, 2700-‐2704
OBn OBn
O O O
N H O
Cl OH O
H N O
NHDdm
N H
Boc
NMe
85
Reactivity and Control for Organic Synthesis
HeteroaromaTcs and Nucleophilic subsTtuTon 4
pyridine is electron deficient at C-‐2 and C-‐4 and it is prone to aTack by nucleophiles
3
N 1
2
N
N
HOMO of pyridine is nitrogen lone pair pyridine undergoes electrophilic aroma%c subs%tu%on only very slowly as reac%on with electrophiles occurs on nitrogen lone pair E N
E
E N E
v. slow
N E
NO2
c. HNO3, c. H2SO4, 300 °C, 24 h N
high energy intermediate -‐ electrophile reac%ng with posi%vely charged nucleophile
N
6%
N H
86
Reactivity and Control for Organic Synthesis
HeteroaromaTcs and Nucleophilic subsTtuTon
N
N
pyridine is electron deficient at C-‐2 and C-‐4 and is prone to aTack by nucleophiles
N
HOMO of pyridine is nitrogen lone pair MeO Na
POCl3 N H
N
O
OMe
Cl
N
N
Cl
OMe
nucleophilic aroma%c subs%tu%on
O R
compare with MeO R
Cl
O OMe
the leaving group needs to be posi%oned ortho or para to the pyridine nitrogen atom below are the rela%ve rates of reac%on with MeO-‐ in MeOH at 50 °C Cl Cl N
10-‐5
CF3
Cl
1
NO2 N
Cl
5
3,000
Cl
N
82,000
Cl
700,000
reac%on of the corresponding N-‐oxides and N-‐methyl pyridinium salts is significantly faster than for the parent chloropyridines
87
Reactivity and Control for Organic Synthesis
below are the rela%ve rates of reac%on with MeO-‐ in MeOH at 50 °C
N
N O
Cl
3,000
N Me
Cl
6 x 107
Cl
1.3 x 1012
Cl
Cl
Cl
N
N O
N Me
82,000
9 x 107
M. Liveris, J. Miller, J. Chem. Soc., 1963, 3486-‐3492
as with benzenoid aroma%cs fluoride is a beTer ac%vator (leaving group) than chloride rela%ve rate of reac%on with EtO-‐ in EtOH F3C N
1
Cl
Cl
N
65
Cl
N
320
F
N
Cl
2800
M. Schlosser, T. Rausis, Helv. Chimica Acta, 2005, 88, 1240-‐1249
4.2 x 1010
88
Reactivity and Control for Organic Synthesis
pyrimidines and related heterocycles are more reac%ve than 2-‐halopyridines towards nucleophilic aroma%c subs%tu%on increasing reac%vity toward nucleophilic aroma%c subs%tu%on
X
X N
N
>
N
N
> X
N
X X
>
N
N
>
N
N N
>
>
X
N
N
X
taken from “Heterocyclic Chemistry” 5th Edi%on, J. A. Joule and K. Mills, Wiley 2010. O N N
Cl
Me
Ph
N
LiHMDS, toluene
Cl
N
Cl
N
BuNH2
O Ph
Bu N Ph
pTSA
N
Et
Et
D. S. Chekmarev, S. V. Shorshnev, A. E. Stepanov, A. N. Kasatkin, Tetrahedron 2006, 62, 9919-‐9930 there are not always ‘back of the envelope’ explana%ons of selec%vity Cl
OMe MeO
N R
N
Cl
Cl
N R
N
N
+ Cl
R
N
OMe
A B Y. Goto and co-‐workers, Bull. Chem. Soc. Jpn., 1989, 37, 2892
R
CO2Me
Cl
A:B
96:4
92:8
H
Ph
Me
OMe
85:15 84:16 76:24 8:92
89
Reactivity and Control for Organic Synthesis
HeteroaromaTcs pyridine N-‐oxides – much more suscep%ble to electrophilic aTack at 2 and 4 posi%ons (and to nucleophilic addi%on at 2 and 4 posi%ons) 4 3
N
H2O2, CH3CO2H
2
N
N
N
N
O
O
O
O
promotes electrophilic subs%tu%on at 2 and 4 posi%ons
nitra%on of pyridine N-‐oxide NO2
NO2 Me
Me
HNO3, H2SO4,100 °C
N
N
O
O
H
NO2 Me
NO2
NO2
Me
PCl3 N
POCl3
66% yield, c.f. nitra%on of pyridine in acid (6% yield)
NO2 Me
+
NO2 Me
PCl3
N
N
N
O
O
O
Me N
Cl P Cl Cl
90
Reactivity and Control for Organic Synthesis
HeteroaromaTcs pyridine N-‐oxides – N-‐deoxygena%on with rearrangement O Me
Me
O O
Me
Me, 100 °C
H
N
O
O
Me H
N O
Me
O
N
Me
O O
2
1
3
N
Me
1O
O
O3 2
Me
Me
pyridine N-‐oxides – conversion to chloro compounds R
O P Cl Cl Cl
R
R
R H
N
N
O
O
O P Cl Cl
Cl
N O
Cl O
P Cl Cl
N
Cl
91
Reactivity and Control for Organic Synthesis
HeteroaromaTcs pyridones
N H
N
O
OH
O
OH
N H
N
nitra%on NO2
O
O NO2
HNO3, H2SO4 N H
O
N H
Cl
N H
H
O NO2 POCl3
NO2 N H
NO2
POCl3 N
O Cl P Cl O Cl NO2
NO2
NaBH4 N
O P Cl O Cl Cl NO2
N
N
H
H
NaBH4 Cl NO2 N
92
Reactivity and Control for Organic Synthesis
HeteroaromaTcs pyrrole, thiophene and furan all three have aroma%c proper%es in each case the (one of the) lone pair(s) is parallel to the p-‐orbitals and part of the π-‐system the aroma%c heterocycles are electron rich β 3
α 2
X S1
O
NH
thiophene
furan
pyrrole
0.55 D
0.66 D
1.74 D
X
X
X
NH pyrrolidine
pyridine is electron poor Nu
N
order of aroma%city is: thiophene > pyrrole > furan (enol ether like)
sulfur is the largest atom and hence is beTer matched for bonding to sp2-‐hybridised carbon atoms in a 5-‐membered ring leading to thiophene being the most aroma%c
93
Reactivity and Control for Organic Synthesis
HeteroaromaTcs ReacTons electrophilic subs%tu%on – kine%c reac%on at the 2-‐posi%on is favoured over reac%on at the 3-‐posi%on more reac%ve than benzene – e.g. pyrrole similar reac%vity to aniline E
H
X
H
X
E X
H
X
E X
E
H
E
E
X
E
X
E
X H
E
X
more delocalised intermediate ∴ more stable ∴ 2-‐subs%tu%on favoured
less delocalised intermediate ∴ less stable ∴ 3-‐subs%u%on disfavoured
subs%tuents already present on the aroma%c heterocycle exert less direc%ng effect than the corresponding subs%tuents in benzene E
EWG
X
H
EDG
X
X E
H
EDG
E
EWG E
X
E
EWG H
X
X
EDG
E
X
EDG
with EWG at α-‐posi%on β’-‐subs%tu%on favoured E
with EDG at α-‐posi%on α’-‐subs%tu%on favoured
94
Reactivity and Control for Organic Synthesis
HeteroaromaTcs ReacTons nitra%on O S
Me
O
O N
S
O
O
Me
O
O
O N
Me
O
O
NO2
AcOH, -10 °C
H
AcO
NO2
O
H
H N + 13% NO2
51%
O O
H N
O
trace NO2
60% O N
Me
+
AcOH, 0 °C
O
H N
S
NO2
O
N O
NO2
NO2
AcOH,-25 °C
addi%on product acyla%on and formyla%on O H N
Cl3C
Cl
H N
O CCl3 90% HNO3, -50 °C
α' O2N
O Ph S
N Me
H N
O CCl3
β’ > α’ or β for α-‐EWG
β' O
H, POCl3, 35 °C
S
O
H N H
then hydrolysis 78%
Me
N H, POCl3, RT Me then hydrolysis
H N
O H
83%
95
Reactivity and Control for Organic Synthesis
HeteroaromaTcs indole 3β 2α
N1 H
N H
O benzofuran
S
benzothiophene
more enamine-‐like than pyrrole aTack of electrophiles at the β posi%on is the lowest energy pathway
E N H
E N H H H E
E N H E
E N H
N H
minor
N H
major
aTack at β posi%on retains aroma%c sextet of benzenoid ring
96
Reactivity and Control for Organic Synthesis
HeteroaromaTcs indole 3β 2α
N1 H
N H
O benzofuran
S
benzothiophene
if the β-‐posi%on is blocked α-‐aTack occurs α-‐aTack can occur via β-‐aTack followed by rearrangement (● = CT2 i.e. a tri%ated methylene group)
BF3•OEt2 N H
OH
direct aTack at α-‐posi%on would give solely
N H
O H F3B
N H
N H
N H H
N H
N H
N H H
N H
1:1 mixture
97
Reactivity and Control for Organic Synthesis
Explain the outcome of the following reac
Cl
Me
H N
N
OH
Me N
O
OH
NaH,
F
N
Me N
O
H
120 °C, 15 hours
OH
H
(b)
MsCl, Et3N
NMe CO2tBu N H single enantiomer
Me
Me
(d) N iPr3Si
N H racemate O2N
(c)
i) F
O
Br N
Me
NO2
ii) aq. NaOH
Me
N
Me
Br
O N iPr3Si
O
Me N
OH
CO2tBu
H
98
Reactivity and Control for Organic Synthesis
Vicarious Nucleophilic Subs%tu%on -‐ (nucleophilic subs%tu%on of hydrogen) Reviews: M. Mąkosza, J. Winiarski, Acc. Chem. Res. 1987, 20, 282; M. Mąkosza Pure & Appl. Chem. 1997, 69, 559 H O2N
Cl
H
SO2Ph
O2N
O2N SO2Ph
KOH, DMSO
H
PhO2S
O
O N
mechanism
O
O N
O
SO2Ph
O N Cl
H SO2Ph Cl
O
H HO
O N
O
SO2Ph
O N
SO2Ph
rate determining step
SO2Ph
LG
rate determining step is elimina%on of H-‐X (HCl) from σ-‐adduct
EWG
For VNS require a nucleophile which carries a leaving group
LG = Cl, Br, PhO, PhS, RO-‐ etc; EWG = SO2Ph, SO2NR2, SO2OPh, POPh2, CN, CO2Et How can we explain the following results?
NO2 MeO
NO2
MeO DMSO
Cl
Cl
NO2
SO2Ph
KOH, DMSO
Cl
69% SO2Ph
99
Reactivity and Control for Organic Synthesis
Vicarious Nucleophilic Subs%tu%on orienta%on of addi%on depends on: structure of the carbanion; structure of the arene; reac%on condi%ons NO2
Cl
NO2
SO2Ph
KOH, DMSO
F
SO2Ph
F
18%
for aTack on nitrobenzene, as the bulk of the nucleophile increases the amount of para isomer increases
X NO2
SO2Ph R
NO2
NO2
SO2Ph
KOH, DMSO
PhO2S
X
R
yield / %
ortho
para
F
H
63
74
26
Cl
H
75
53
47
Cl
Et
68
100
Cl
Ph
93
100
M. Makosza, J. Goliński, J. Baran, J. Org. Chem., 1984, 49 1488
100
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
ortho-‐lithia%on by lithium halogen exchange – faster then deprotona%on generally requires an organolithium base an aryl / alkenyl bromide or iodide. OMe
OMe
OMe Br
Bu Li
anion in an sp3 orbital
Li
Br + BuBr
Bu
via “ate” complex
anion in an sp2 orbital
mechanism involves aTack of alkyl lithium at the halogen via an intermediate “ate” complex reac%on is an equilibrium process which favours the more stable anion (remember, anion order is sp3>sp2>sp – the stability of the anion is in the order of the pKa of the corresponding hydrocarbon) in the above example an sp3 anion (butyl lithium) gives an sp2 anion D. E. Applequist, D. F. O’Brien, J. Am. Chem. Soc., 1963, 85, 743.
101
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
LDA and other amide bases are good for deprotona%on but NOT for halogen lithium exchange
OMe Br
lithium halogen exchange with LDA would lead to the forma%on of a very weak halogen-‐ nitrogen bond – reac%on is thermodynamically in the wrong direc%on
OMe
N Li
X
Li
I
N Br
+
I I
LDA, I2
Me tBuLi, MeI
N SO2Ph
N SO2Ph
N SO2Ph
Mark G. Saulnier and Gordon W. Gribble J. Org. Chem. 1982, 47, 757
O
Br
Bu Li
O
Li
R X
O
R
102
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
in general rate of exchange I > Br >> Cl
to make aryllithiums by lithium halogen exchange – generally use n-‐butyllithium; tert-‐butyllithium may also be used to make vinyllithiums and alkyllithiums one frequently uses tert-‐butyllithium with primary alkyl halides it is necessary to use two equivalents of tert-‐butyllithium Me
Me
Me I
O
O
Me
2 equiv. tert-BuLi
OTBS
H
Me Li
O
PMP
Me
O
Me + I
Me Me Li Me
Me
Me
H Me
OTBS
PMP
Me
Me
Me H
with one equivalent of tert-‐butyllithium protodeiodina%on is likely to occur
O
O
OTBS
PMP
lithium halogen exchange is a very fast reac%on which can outcompete deprotona%on of OH groups and addi%on to C=O groups Br
O O
O
NR O
O O
nBuLi, THF, -78 °C RHN
O
L. Ollero, L. Castedo, D. Dominguez, Tetrahedron, 1999, 55, 4445-‐4456
Me Me Me
103
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on
synthesis of morphine – J. E. Toh, P. L. Fuchs, J. Org. Chem. 1987, 52, 473–475 Provide a mechanism for the reac
OMe Br
OMe O
2.2 equiv. BuLi, -78 °C
OH
O OH
O
OH
Me N H
SO2Ph
PhO2S
morphine OH
H
H
annula%on forming benzocyclobutanes – I. A. Aidhen, J. R. Ahuja, Tetrahedron LeV. 1992, 33, 5431-‐5432. I
MeO
O
t-BuLi OMe N Me
MeO
O
MeO MeO
selec%ve halogen metal exchange is possible Ar Br
Br N
OMe
n-BuLi, Et2O, -100 °C
Br
Li N
OMe
H O
OH Br N
OMe
Cl
73%
104
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on it is also possible to make Grignard reagents by lithium halogen exchange generally using iPrMgBr or iPrMgCl
For reviews see: P. Knochel et al. Angew. Chem. Int. Ed. 2003, 42, 4302; Chem. Commun. 2006, 583; Heterocycles 2014, 88, 827. CO2Me
CO2Me iPrMgCl, THF, -10 °C
PhCHO
I
Br
MgCl
Br iPrMgCl•LiCl, THF -50 °C
Br
Ph
Br
OH Br
O MgCl
tBu
N
Br
OH tBu
H
Br
Br Br
CO2Me
Br O O MgCl S Ph CN
iPrMgCl•LiCl, THF -50 °C Br
N
Br
CN Br
N
Br
iPrMgCl•LiCl, THF -50 °C
CN Br
N
MgCl
105
Reactivity and Control for Organic Synthesis
AromaTc organometallics for aroma%c heterocycles metal halogen exchange shows the following selec%vity: with 5-‐ring heterocycles the 2-‐posi%ons undergoes exchange faster than the 3 posi%on with 6-‐ring heterocycles, the 3 posi%on undergoes exchange faster than the 2-‐posi%on iodine metal exchange is faster than bromine metal exchange summary 5-‐ring 2>3; 6-‐ring 3>2; I>Br
X
Br Br
S
Br
R M
X
Br
M N
Br
EtMgCl. THF, RT
S
Br
MgCl
M
R M
Br
N
tBuN C O
Br O
S
NHtBu
Br
76%
Br
49%
O S
Br I
EtMgCl. THF, RT
S
Br
Me2N
H
S
Br O
MgCl H
l. Christophersen, M. Begtrup, S. Ebdrup, H. Petersen, P. Vedsø J. Org. Chem., 2003, 68, 9513
106
Reactivity and Control for Organic Synthesis
AromaTc organometallics directed ortho-‐metalla%on reviews: V. Snieckus, Chem. Rev. 1990, 90, 879-‐933; J. Clayden in “Organolithiums: Selec
OMe H
O
Li
BuLi
Me
Bu H
O
O Li
H
NMe2
MeO H
OLi NMe2
DMF
H, H2O
use amides as electrophiles: reac
OMe
BuLi,
Me2N
BuLi
NMe2
MeO H
NMe2 H
MeO
OMe Li
NMe2
NMe2
various direc%ng groups can be used Me Me COCl HO
Me NH2
then dehydrate
O
Me
Me
N
O BuLi
Me
Me
N
O Li
Br
R
O H
Li NMe2
O H
Me
N R
107
Reactivity and Control for Organic Synthesis
increasing ability to direct ortho-‐lithia%on
condi%on dependent order Me NR2 O
O
O-‐carbamates
R2N
O
3° amides
R
O S O
sulfones
R2N
O S O
sulfonamides
O
Me N
oxazloines
O
O
MOM ethers
OR
ethers
X
halogens
O
benzylic alkoxides
O RN
O
2° amides
O
S
tBu
RN
sulfoxides
imines
most powerful directors
N
OtBu
N-‐carbamates O
NR2
anilines
NR2
-‐78 °C
-‐78 °C
-‐78 °C
-‐50 °C
-‐20 °C
trifluoromethyl NR ( )n 2
aminomethyl
-‐78 °C
CF3
remote amines
0 °C
>0 °C
temperature (°C) of ortho-‐lithia%on with RLi in THF or ether
Adapted from: J. Clayden in “Organolithiums: Selec
>20 °C
108
Reactivity and Control for Organic Synthesis
with ortho-‐directors which are also electrophiles, the precise reac%on condi%ons including: the nature of the base, addi%ve and order of addi%on can influence the outcome of the reac%on see: P. Beak, R. A. Brown, J. Org Chem., 1982, 47, 34-‐46 O
O
Me
NEt2
n-BuLi
O
NEt2
s-BuLi, TMEDA
O
O NEt2
s-BuLi, TMEDA
NEt2 Li
F
F Me3SiCl
Me
O
O NEt2 s-BuLi, TMEDA
SiMe3 F
NEt2
then MeI
SiMe3 F
R. J. Mills, N. J. Taylor, V. Snieckus, J. Org. Chem., 1989, 54, 4372
Li
O
NEt2 E
E
Electrophile
Yield (%)
D2O
88
MeI
77
EtI
70
B(OMe)3; H2O2 (adds an OH)
56
acetone
54
PhCHO
79
CH2=CHCH2Br
60
109
Reactivity and Control for Organic Synthesis
AromaTc organometallics Ortho-‐directed electrophilic aroma%c subs%tu%on synthesis of Fredericamycin – T. R. Kelly, S. H. Bell, N. Ohashi, R. J. Armstrong-‐Chong, J. Am. Chem. Soc. 1988, 110, 6471-‐6480 MeO
O
TBDMSO s-BuLi MeO Li
O
O Cl
NEt2 O NEt2
TBDMSO
NEt2 O
OMe t-BuLi
O
MeI
O Li
TBDMSO
NEt2 O
OMe
Me
OMe
O
O
Me
fredericamycin
O
OMe
HN
O O HO HN
O
O
TBDMSO Me Me
N Li
EtO N EtO
OMe
TBDMSO Me Me
NEt2 O
OMe
O
EtO O
OEt
TBDMSO
TBDMSO
110
Reactivity and Control for Organic Synthesis
lithia%on of 5-‐membered heterocycles
lithia%on occurs preferen%ally α to the heteroatom due to induc%ve effect of heteroatom with, in some instances a DOM effect furan and thiophene can be readily metalled α to the metal S
S
n-BuLi
O
Li
O
n-BuLi
Li
ether, reflux
-10 °C, ether
with pyrrole itself, the N-‐deprotona%on occurs first – the more ionic the N-‐metal bond the greater the percentage aTack at nitrogen with a more covalent N-‐M bond C-‐aTack occurs. H N
H N
NaNH2
N
Na
Me
EtMgBr
OEt
N
BuLi
Me N
Et
Me N
I
O
MgBr N
I
Me N
H
H O
workup H
H N
O H
Et
111
Reactivity and Control for Organic Synthesis
with pyrroles bearing an N-‐EWG on nitrogen α-‐metalla%on occurs
O
OtBu N
O
Me N Me Me Li Me
OtBu N
SnMe3
then Me3SnCl
SO2Ph N
LDA, then B(OMe)3
SO2Ph N B(OH)
2
then HCl, water
selec%vity can be achieved using LDA or butyllithium
no lithium halogen exchange as would make weak N-‐Br bond most acidic proton removed by directed metalla%on
iPr S
Li Br
N Li
iPr S
H Bu Li Br
S
E Li
S E
112
Reactivity and Control for Organic Synthesis
pyridines
pyridines are electron deficient aroma%cs and pyridines which carry a direc%ng group (halogen, CN, CO2H etc.) undergo ready metalla%on 2, and 4-‐subsistuted pyridines metallate in the 3 posi%on 3-‐subsituted pyridines generally metallate in the 4 posi%on 1 eq. BuLi, 3 eq.
Me N Me Me Li Me
N
CO2H
then CO2
CO2H
CO2H N
CO2H
1 eq. BuLi, 3 eq.
CO2H
then PhCHO, then H2SO4
Me N Me Me Li Me
then CO2
N
O
O
Me N Me Me Li Me
N
1 eq. BuLi, 3 eq.
Ph N
F. Mongin, F. Trécourt, G. Quéguiner, Tetrahedron LeV., 1999, 40, 5438
CO2H CO2H N
113
Reactivity and Control for Organic Synthesis
“Halogen dance” term introduced by BunneT for isomerisa%on reac%ons which can accompany deprotona%on of halogenated aroma%cs J. F. BunneT, Acc. Chem. Res. 1972, 5, 139. Br
Br PhNHK, NH3 Br
40-‐60% Br
Br
Br
for halogen dance to be synthe%cally useful the isomerisa%on must be thermodynamically favourable S
NaNH2, NH3
Br
S
Br Br
S S
H
Br
S
S
+
Br
+ S
S Br
Br
Br
l. Brandsma, R. L. P. de Jong, Synth. Commun. 1990, 20, 1697-‐1700
Li I N
Cl
I
LDA, THF - 70 °C N
O
I
Cl
Li N
Cl
H
I
O
OEt
H N
H
70%
Cl
F. Guillier, F. Nivoliers, A. Conchennee, A. Godard, F. Marsais, G. Queguiner Synth. Commun. 1996, 23, 4412-‐4436
66-‐72%
114
Reactivity and Control for Organic Synthesis
reac%ons of alkenes -‐ depending on subs%tuents alkenes can be: electron rich and hence nucleophilic increasing nucleophilicity
NR2
OR
O E E
E
E
alkene
enol ether
E
E
enamine
HOMO = π bond of alkene LUMO = σ* or π* on electrophile
enolate
electron poor and hence electrophilic increasing electrophilicity
Nu
O N
Nu O
O
Nu
Nu
O OR
HOMO = lone pair on nucleophile LUMO = π* on alkene
Nu
O
N NR2
115
Reactivity and Control for Organic Synthesis
Overview of reacTons of (electron rich) alkenes Br Br
Br2
:Br Br
Br
Br
Br
build up of par%al posi%ve charge H
Me
Br2, H2O
bromina%on stereospecific an%
Br
SN2
Me OH
Me Br
2O: (+)
SN2 / SN1 borderline
Me Br
Br
halohydrin forma%on stereospecific an%
Me OH Br
long, weak bond
Me
Me OH
OsO4 O
N
O Me
Me
O
Me O
O Os
OH
O
O
O
Os(VIII) O
O Os
O
O
Os
Me OH
O H O 2 O
O
OH N
O Me
+ HO HO
Os
O O
Os(VI)
dihydroxyla%on stereospecific syn
116
Reactivity and Control for Organic Synthesis
Overview of reacTons of alkenes
Me
m-CPBA
Me O , then PPh 3 3
ozonolysis
Me O
Me
O O
Ar
H O
O O
Me O
+
Me O
O
O O Me
O O PPh3
Me
Me O
Me O
O
epoxida%on stereospecific syn
Me O
O
O O
O Me
PPh3
O
O O
O
PPh3
117
Reactivity and Control for Organic Synthesis
Overview of reacTons of alkenes
Me
BH3 then H2O2, NaOH
Me
Me H
OH
B
H
Me H
Me H H
B R
R O O
Me
H2O2, NaOH
OH
Me HBr, water
Br
Me
Me
Me Me
H H2, Pd/C H
Me Me
Me H
stereospecific migra%on with reten%on of configura%on
O BR2 Br
Me H
H
R B R O OH
hydrobora%on / oxdia%on
Br
Me
ionic reac%on with HBr
hydrogena%on stereospecific syn
118
Reactivity and Control for Organic Synthesis
1) How would you carry out the following transforma
2) Explain the following transforma
O HS
(b)
O
O +
OEt
OEt
O
EtO S
O
O H2O2, NaOH, MeOH
O
3) Explain the following: Treatment of the enolate A with B at -‐78 °C followed by quenching the reac
OLi MeO
OPh
Me CO Me 2 HO OPh
O
Me A
B
C
D
PhO
Me CO2Me
119
Reactivity and Control for Organic Synthesis
Conjugate AddiTon vs Direct AddiTon H O 1 2
4
1 O 3 2
O Nu
Nu
Nu
Nu
HO Nu
Nu
O
O
direct addi%on 1,2-‐addi%on
conjugate addi%on Michael addi%on 1,4-‐addi%on
H
conjugate addi%on requires the presence of an electron-‐withdrawing group which results in the lowering of the energy of all of the π-‐orbitals of the system and hence the alkene is less nucleophilic and more electrophilic O
O
i.e. alkene is electron poor
Evidence of delocalisa%on O
1678 cm-‐1 1628 cm-‐1
143 ppm
O
1712 cm-‐1
118 ppm O
1653 cm-‐1
133 ppm
infra red – remember ν ∝ √k/μ i.e. higher stretching frequency = stronger bond
13C NMR
133 ppm
120
Reactivity and Control for Organic Synthesis
examples of conjugate addi%on O
NC
O
KCN (cat)., HCN OMe
NC
regenerates cyanide anion – cataly%c HCN too weak an acid to protonate carbonyl group ∴ need a good nucleophile, cyanide anion, to aTack neutral substrate
OMe H
O NC
OMe
H O:
H CN
O
HCl Cl
HCl protonates carbonyl oxygen making the whole system more electrophilic
O
Cl
H
O Cl
H
O
H
Cl H
O
OH
O
NaOH cat.
O
O
H
H
NaOH
H
H OH
O
O H H
O
enolate generated by conjugate addi%on reacts with the alcohol to regenerate alkoxide for conjugate addi%on
121
Reactivity and Control for Organic Synthesis
conjugate addi%on O
Nu
Nu
faster conjugate addi%on
Nu
EWG
pKa
H
O N
Nu
O
O
the beTer the ability to stabilise the nega%ve charge the beTer the conjugate acceptor is and hence the faster the reac%on. This can be related to pKa slower conjugate addi%on
increasing electrophilicity
Nu
O
Nu
O
10
20
O NC
R2N
24
N NR2
O RO
R
Nu
O
OR
O O2N
Nu
O
25
25
for some nucleophiles conjugate addi%on is the major pathway, for other nucleophiles direct addi%on is the major pathway whereas for others slight varia%on in condi%ons can alter the course of the reac%on
122
Reactivity and Control for Organic Synthesis
examples of conjugate addi%on O Ph
heat
+ OEt
O
O Ph
N H
Ph
N
OEt
Bu OH
BuMgBr
O
O
BuMgBr
irreversible reac%on 1,2-‐addi%on
NaCN, HCN
Bu
irreversible reac%on 1,4-‐addi%on
NC OH
O
NaCN, HCN 80 °C
5-10 °C
formed faster kine%c product lower ac%va%on energy
O
Ph
N H
1 % CuCl
O
N
but if given the choice amines do conjugate addi%on
irreversible reac%on
O
heat
+
O NC
more stable thermodynamic product
OEt
123
Reactivity and Control for Organic Synthesis
conjugate addi%on product is generally the thermodynamically more stable product with respect to the direct addi%on product. a rough comparison of bond energies supports the above conjecture gained O-‐H
lost C=O O
NaCN, HCN
NC O
H
5-10 °C
lost
kJmol-‐1
gained
kJmol-‐1
C=Oπ
370
C-‐C
350
overall gain kJmol-‐1
90
gained C-‐C lost C=C O
NaCN, HCN 80 °C
O
C=Cπ
270
O-‐H
460
C-‐C
350 130
NC H
gained C-‐C
gained C-‐H
C-‐H
400
conjugate addi%on product is generally the thermodynamically more stable product with respect to the direct addi%on product because it retains the strong carbonyl double bond – this is general for most α,β-‐unsaturated systems in the above example: the direct addi%on product is the kine%c product i.e. the fastest formed product and hence the product formed by the pathway with the lowest ac%va%on energy
124
Reactivity and Control for Organic Synthesis
why is the direct addi%on product formed fastest? O
O
delocalisa%on indicates that the carbonyl carbon (and one of the alkenyl carbons) bear par%al posi%ve charge
O
β α
more electron deficient carbon charged nucleophiles will aTack the carbonyl carbon faster than the β-‐carbon (Hard-‐Hard interac%on) carbonyl carbon carries the larger posi%ve charge as it is closer to the electronega%ve oxygen atom charged nucleophiles will aTack the β-‐carbon but more slowly
energy
thermodynamic control
ac%va%on energy NC O
O NC
able to reverse at 80 °C
difficult to reverse even at 80 °C
O NaCN + HCN
O NC
kine%c control lower ac%va%on energy
thermodynamic product lower in energy more stable extent of reac%on
at 80 °C cyanohydrin is reversible at 0 °C cyanohydrin is irreversible at 80 °C kine%c product is s%ll formed first but reverts to star%ng materials and slower conjugate addi%on occurs
NC OH kine%c product formed faster
if direct addi%on is reversible conjugate addi%on will result
125
Reactivity and Control for Organic Synthesis
what is actually happening in the addi%on of HCl to methyl vinyl ketone H O:
thermodynamically controlled reac%on most stable product is formed charged nucleophiles usually do direct addi%on
O
HCl Cl
O O
Cl
H
O
H
O
Cl
H
HO Cl
H
Cl
Cl
H H
not all products arising from conjugate addi%on are the result of ini%al reversible direct addi%on for certain nucleophiles conjugate addi%on is the kine%cally most favoured pathway in these instances the kine%c product also happens to be the thermodynamic product
O Ph
OEt
O
O
+ N H
irreversible reac%on
Ph
N
Ph
N
+ OEt
N H
Ph
irreversible reac%on
O OEt
126
Reactivity and Control for Organic Synthesis
Orbital Controlled Reac%ons (SoU-‐SoU interac%ons) if the nucleophile has a high energy HOMO this will be close in energy to the LUMO of the α,β-‐unsaturated system therefore conjugate aTack will occur at the β-‐carbon – the reac%on is under orbital control for fast reac%on we require a small HOMO / LUMO gap HOMO = Nu lone pair Nu Nu:
O
O
LUMO = π*
O Ph
N
+ OEt
N H
Ph
O OEt
for amines: uncharged, direct addi%on not favoured (lone pair of intermediate energy) conjugate addi%on is the major pathway in the above example the reac%on is under orbital control
1. Generally 2nd row elements (e.g. P, S) favour conjugate addi%on as they have high -‐ energy 3s/3p lone pairs that are a good energy match for the LUMO of the substrate. 2. If the nucleophile is uncharged then conjugate addi%on oUen results.
127
Reactivity and Control for Organic Synthesis
Predict the outcome of the following reac
PhSH, base
O
BuMgBr
O
LiAlH4
O
BuMgBr 1% CuCl
the conjugate acceptor the more electrophilic the carbonyl group the more likely to undergo direct addi%on – charge control dominates O
O
R2NH
very reac%ve carbonyl group, charge control therefore en%rely 1,2-‐addi%on
NR2
Cl
mainly direct addi%on
O
nearly always conjugate addi%on O
Cl
O H
O
O OR
always conjugate addi%on
N NR2
O N
O
128
Reactivity and Control for Organic Synthesis
sterics can influence the outcome of the reac%ons O O
MeMgBr
Me
O O
conjugate addi%on even though hard Grignard reagent
Summary: more reac%ve nucleophiles (RLi, RMgBr, LiAlH4) prefer direct addi%on more reac%ve electrophiles prefer direct addi%on less reac%ve nucleophiles prefer conjugate addi%on less reac%ve electrophiles prefer conjugate addi%on watch out for: reversible direct addi%on which leads to conjugate addi%on i.e. kine%c versus thermodynamic control
129
Reactivity and Control for Organic Synthesis
Carbonyl Groups – difference in reacTvity towards nucleophiles O R
O H
R
O Cl
R
O O
O R
R
O Me
R
O R
OMe
O OH
R
NR2
in general there are two types of mechanism with aldehydes and ketones addi%on occurs – frequently followed by subsequent reac%on with esters, acids, amides etc. addi%on and subsequent elimina%on occurs O Nu:
O
Nu O
Structure of the carbonyl group C-‐O σ-‐bond and C-‐O π-‐bond
Nu:
O
O
Nu O X
O
π-‐bonding orbital
X
Nu
O
π*-‐an%bonding orbital
the π-‐bonding orbital is polarised towards oxygen the more electronega%ve atom ∴ the π* an%bonding orbital is polarised towards carbon i.e. the large lobe is on carbon in the mechanism of aTack the HOMO of the nucleophile overlaps with the LUMO (π*) of the carbonyl group
HOMO
LUMO
Nu O
130
Reactivity and Control for Organic Synthesis
typical addi%ons reac%ons of aldehydes and ketones are as follows: hydra%on – aldehydes are prone to hydra%on – ketones far less so O R
H
+
H2O
HO
OH
R
H
Keq hexafluoro-‐ acetone formaldehyde chloral
F3C
O
F3C H O H Cl3C
O
H
Keq
1.2 x 106
acetaldehyde
2280
acetone
2000
How hydrated would you expect the following compounds to be? O O
O O
Me O
1.06
O
0.001
H Me Me
131
Reactivity and Control for Organic Synthesis
in a similar manner, aldehydes and ketones react with alcohols under acid catalysis to give acetals the mechanism of this reac%on is very frequently drawn incorrectly! loss of water
electrophilic O
MeOH
OH
H MeO
OH
±H
MeO
electrophilic OMe
OH2
H MeOH
H
+
MeO
H MeO
OMe
OMe
acid catalysis is required so that the intermediate in the red box can expel water – if no acid were present HO-‐ would be the leaving group – MeO is not a good enough ‘pusher’ to kick out hydroxide the intermediates in blue boxes are closely related – they are much more electrophilic versions of the original carbonyl group and so are readily aTacked by MeOH which is a very weak nucleophile the whole process is in equilibrium and the most stable product is therefore formed the reac%on can be readily reversed using acidic water
aldehydes readily form acetals with simple alcohols
with ketones the equilibrium greatly favours the ketone – using diols allows efficient acetal forma%on why is this?
O
OH
HO H
O
O
132
Reactivity and Control for Organic Synthesis
acetal forma%on is mechanis%cally closely related to numerous other reac%ons of carbonyl compounds aldehydes and ketones readily react with primary amines and related nitrogen nucleophiles to give imines and related compounds – these reac%ons are generally catalysed by acid
O
PhNH2
O
H H N O Ph
±H
Ph
H N
H
OH
Ph
H N
OH2
acid catalysed
H
N
+
Ph
H
Explain shape of the pH rate profile for oxime forma
O
+
NH2OH
NOH
rate
2
4
6
pH
8
N
Ph
133
Reactivity and Control for Organic Synthesis
secondary amines also react with aldehydes and ketones to give iminium ions and subsequently enamines O
Me Me
N H
Me
N
Me
Me
N
Me
H
iminium ion enamine aldehydes and ketones are more electrophilic than the corresponding imines – oxygen is more electronega%ve than carbon iminium ions are more electrophilic than than the corresponding aldehyde or ketone N R
R' H
R''
O R
H
N
R
R' H
more electrophilic
least electrophilic N R
R' R
R''
O R
R
R
N
R' R
Provide a mechanism for the following reac
NH2
+
O
NaBH4, AcOH, NaOAc
HN
134
Reactivity and Control for Organic Synthesis
Carboxylic acids and related groups mul%ple types of delocalisa%on are possible O R
R
Me
R
O X
O
X
X
C
OEt
Me
O
X p-‐type lone pair → C-‐O π*
O
1743 cm-‐1
Me N Me
1646 cm-‐1
most important for esters (X= OR) and amides (X = NR2)
O and N have 2p orbitals which are a good size and energy match for C=O π* orbital (itself made up of overlap of two 2 p orbitals) nitrogen is less electronega%ve than oxygen and hence with amides there is greater p-‐type lone pair → C-‐O π* dona%on than with esters ∴ amides are less reac%ve towards nucleophiles than esters rough order of importance NR2 > OR > Cl evidence: esters and amides are planar i.e. there is substan%al double bond character between X and the carbonyl carbon
O Me
Me N Me
135
Reactivity and Control for Organic Synthesis
Carboxylic acids and related groups mul%ple types of delocalisa%on are possible O R
R
O
X
R
C
O Cl
O
O X
O
νmax = 1715 cm-‐1
X
Cl Cl
1766 cm-‐1
F
F F
1771 cm-‐1
O sp2 lone pair → C-‐X σ* implica%on is carbonyl group has par%al triple bond character and is also very electrophilic oxygen is more electronega%ve than nitrogen and hence the C-‐O σ* is lower in energy than the C-‐N σ* and a beTer energy match for O sp2 lone pair rough order of importance X = F > OR > Cl > NR2 evidence from IR stretching frequencies and X-‐ray crystal structure analysis
126.4.4 ° 117.4 °
O O Me
O OEt
1743 cm-‐1
Me
Me
CF3
N Me
1646 cm-‐1 E. J. Corey, J. O. Link, S. Sarshar, Y. Shao, Tetrahedron LeV. 1992, 33, 7103 the balance of these effects determines the reac%vity of the system
136
Reactivity and Control for Organic Synthesis
Carboxylic acids and related groups esters predominantly exist in the (Z)-‐conforma%on third type of delocalisa%on affects the conforma%on of esters O R
O
O
O
R R
O sp2 lone pair → C-‐O σ*
C
R
O
Ph
O O
Me
Ph
Z-‐ester
O Me
E-‐ester
A .A. Yakovenko, J. H. Gallegos, M. Yu. An%pin, A. Masunov, T. V. Timofeeva, Cryst.Growth Des. 2011, 11, 3964 in terms of reac%vity towards nucleophiles -‐ rough order of reac%vity is: increasing electrophilicity O R
O R
Cl
O O
O R
R
O Me
R
O OMe
R
O OH
R
NR2
conversely in terms of reac%vity towards electrophiles – amides are the most reac%ve E O R
NR2
E R
O NR2
137
Reactivity and Control for Organic Synthesis
chemoselec%vity in the reduc%on of carbonyl compounds. OH
NaBH4
H2, Pd/C
O
for chemo and regioselec%ve reduc%on it is important to choose the correct reagent OH
HO MnO2 O HO
O
LiAlH4
O EtO O
NaBH4 CeCl3
OH EtO O
138
Reactivity and Control for Organic Synthesis
summary of reducing agents for carbonyl groups adapted from Organic Chemistry, Clayden, Greeves and Warren, 2nd Edi%on, OUP 2012. O
R
reduced
O
aldehyde
reduced slowly not usually reduced
NR
NR
R
H
H iminium ion R
O
O R
H
aldehyde
R
OH
O R
R
ketone
R
1° alcohol
BH3•NH3, LDA
DIBAL H
R
H
aldehyde
via acid chloride O
O OR
ester
H
R
NR2
R
NR2
R
amide
OH
acid
NaCNBH3 NaBH4 LiBH4 LiAlH4 BH3 OH R
NHR
amine
R
OH
1° alcohol
R
R
2° alcohol
R
OH
1° alcohol
amine
R
OH
1° alcohol
139
Reactivity and Control for Organic Synthesis
selec%vity in reduc%on why are esters reduced with LiBH4 but only slowly with NaBH4? O R
Li
O
BH4 R
OR
Li OR
O R
H
H B H H
H
Li
O R
OR
R
H
OH
H
H B H H
Li+ has a higher charge/radius ra%o compared with Na+ as it is smaller ∴ Li+ is a more potent Lewis acid than Na+ Li+ serves to ac%vate the ester carbonyl for reduc%on how can we reduce an ester to an aldehyde? DIBAL-‐H – diisobutylaluminium hydride
Al H
Al
H H
Al
exists as an H-‐bridged dimer but reacts as a monomer Al has an empty p-‐orbital, the monomer is electrophilic
DIBAL-‐H only becomes a good reducing agent once it has been ac%vated by complexa%on by a Lewis base
140
Reactivity and Control for Organic Synthesis
DIBAL-‐H – diisobutylaluminium hydride – commercially available as solu%ons in various organic solvents work-‐up of DIBAL-‐H reac%ons can be complicated by gela%on due to the amphoteric nature of AlIII salts
par%%oning the reac%on mixture between an organic solvent and aqueous Rochelle salt (Na+K+ tartrate) coupled with vigorous s%rring is a useful method of solubilising these gels
R Al H R O R
R Al
O
DIBAL-H R
OR
R H
O R
OR
R Al H OR
R
addi%on of acid destroys excess hydride and protonates tetrahedral intermediate
MeOH then H
O R
tetrahedral intermediate stable at low temperature (-‐78 °C)
H H OR H
O R
H
ester reduced to aldehyde with DIBAL-‐H at low temperature
at higher temperature, DIBAL-‐H reduces esters to alcohols R Al H R O R
OR
R O
DIBAL-H R
R Al
OR
H
O R
R Al H OR
aldehyde much more reac%ve than ester R
tetrahedral intermediate not stable at RT
O R
DIBAL-H H
R
OH
rapid reduc%on to alcohol
141
Reactivity and Control for Organic Synthesis
DIBAL-‐H – diisobutylaluminium hydride – commercially available as solu%ons in various organic solvents O O
O N
DIBAL-H, toluene
O
N
O
O
OH
it can be very difficult to reduce an α,β-‐unsaturated ester to an aldehyde with DIBAL-‐H
lactols are very readily prepared by reduc%on of lactones with DIBAL-‐H H
H
O
DIBAL-H, -78 °C
O
OH O
O OH
H
H
H
DIBAL is also very useful for reducing nitriles to aldehydes what is the mechanism of this reac
H
C
N
DIBAL-H then H2O, H
H
O H
H
142
Reactivity and Control for Organic Synthesis
if we add a Grignard reagent or organolithium to an aldehyde or ketone, monoaddi%on occurs but with esters double addi%on is the general outcome O R
O R''
R'' MgBr OR'
or R''
R
Li
O
R'' MgBr
OR'
R
O R"
R
R"
R
R"
ketone more electrophilic than ester in order to obtain mono-‐addi%on use amides as the electrophile Me
OMe H
BuLi
O
Li
Me
Bu H
O
O H
Li
NMe2 DMF
MeO H
OH
H2O, H
R"
R"
ter%ary alcohol
OLi NMe2
H, H2O
the most versa%le solu%on is to use Weinreb amides
MeO H
O H NMe2 H
MeO
O H
Me
O R
OMe
O N Me H • HCl
iPrMgCl, or AlMe3
O R
OMe N Me
note: aluminium and magnesium amides are par<cularly nucleophilic towards esters
Weinreb amide for use of iPrMgCl in the synthesis of Weinreb amides see: J. M. Williams, R. B. Jobson, N. Yasuda, G. Marchesini, U.-‐H. Dolling,E. J. J. Grabowski, Tetrahedron LeV., 1995, 36, 5461-‐5464
143
Reactivity and Control for Organic Synthesis
Weinreb amides – a very reliable ketone synthesis. For a review see: M. Mentzel and H. M. R. Hoffmann, J. Prakt. Chem. 1997, 339, 517-‐524.
O R
R' MgBr
OMe N Me
or R'
Li
O Mg OMe R N R' Me
OH
H , H2 O R
O
OMe
Me ClMg N Me THF, -20 °C to RT, O then HCl (aq) O
OMe N R' H Me
R
tetrahedral intermediate protonated on work up and collapses to generate ketone
stable chelated tetrahedral intermediate
Boc N
O
Boc N OMe
O O
93%
VITAE PHARMACEUTICALS, INC. Patent: WO2007/117560 A2, 2007. O MeO
N Me
OH
O OTBDPS
Me
H
OH
MgBr THF, 77%
OTBDPS H
D. A. Evans, J. T. Starr, Angew. Chem. Int. Ed., 2002, 41, 1787-‐1790
Me
R'
144
Reactivity and Control for Organic Synthesis
DIBAL-‐H reduc%on of esters to aldehydes can, at %mes, be difficult to control – using a Weinreb amide overcomes this problem O MeO
OTBDMS
O Me
N Me
OTBDMS Me
H
Me
Me
DIBAL-H, THF Br
Br
TBDMSO
TBDMSO
TBDPSO
TBDPSO
OTBDMS Me
OTBDMS Me
D. A. Evans, J. T. Starr, Angew. Chem. Int. Ed., 2002, 41, 1787-‐1790 enolates will also add to Weinreb amides Me
OLi
O Me N OMe
OtBu
O Li MeN O
O CO2tBu
H, H2O
O OtBu
83%
145
Reactivity and Control for Organic Synthesis
lithium aluminium hydride – LiAlH4 -‐ all four hydrides are ac%ve
powerful and frequently non-‐selec%ve reducing agent: will reduce aldehydes, ketones, esters to primary alcohols and amides to amines LiAlH4 both in solu%on and as a solid is highly flammable – requires anhydrous solvents work-‐up of LiAlH4 reduc%ons can be tricky due to the amphoteric nature of AlIII salts a useful ‘anhydrous’ work up introduced by Feiser involves, for n grams of LiAlH4 adding dropwise n mL of water, n mL of 15% NaOH solu%on, and then 3n mL of water. In favourable cases a granular precipitate is produced which can be filtered. L. F. Fieser, M. Fieser, M. Reagents for Organic Synthesis 1967, 581-‐595. another safe method for neutralising excess LiAlH4 involves quenching the reac%on with EtOAc reduc%on of esters tetrahedral intermediate collapses to give an aldehyde O R
Li
O
AlH4
OR
R H
H Al H H
Li OR
O R
H
Li
O R
OR H
H Al H H
H
R
OH
146
Reactivity and Control for Organic Synthesis
reduc%on of esters O R
Li
tetrahedral intermediate collapses to give an aldehyde O
AlH4
OR
R
Li OR
O R
H
H Al H H
H
Li
O R
OR
H
R
OH
H
H Al H H
reduc%on of amides AlH3 O R
Li
O
AlH4
NR2
R H
Li NR2
R
H
Al H H
O H
Li NR2
O R
H
AlH3
NR2 R
NR2
tetrahedral intermediate collapses to give an iminium ion
H
H
H Al H H
why this difference in reac%on outcome? RO-‐ is a beTer leaving group then R2N-‐ lone pair of amine is higher in energy than O and hence R2N is a beTer ‘pusher’ than RO
R
NR2
147
Reactivity and Control for Organic Synthesis
O
H
H
N Me
Me O
N Me LiAlH4, THF
H
O H
H
O
O
H
O
O
Me HO
LiAlH4, THF
Me
HO
O
Me
OH
Me
amide reduced to amine 1,2-‐reduc%on of α,β-‐unsaturated ketone (hard nucleophile)
Me
lactone (ester) reduc%on leads to diol
lithium borohydride – will reduce esters to primary alcohols – see above (can be prepared from cheap NaBH4 and LiCl, LiBr or LiI)
sodium borohydride – frequently used in alcoholic solvents such as MeOH or EtOH generally does not reduce esters, epoxides, lactones, nitriles. All four hydrides are ac%ve NaBH4 reacts with pro%c solvents to generate alkoxy borohydrides
Explain the stereoselec
O O
H
O
NaBH4, MeOH Cl
H
HO O
O Cl
H 87%
148
Reactivity and Control for Organic Synthesis
Luche reduc%on NaBH4 is not selec%ve for 1,2-‐ versus 1,4-‐reduc%on – addi%on of CeCl3•7H2O increases the amount of 1,2-‐reduc%on O
NaBH4 or NaBH4, CeCl3•7H2O
OH
1,2-‐reduc%on 1,4-‐reduc%on
OH +
MeOH
NaBH4, MeOH
51%
49%
NaBH4, CeCl3•7H2O, MeOH
99%
trace
it appears that CeCl3 accelerates the reac%on of pro%c solvents with NaBH4 to generate alkoxy borohydrides NaBH(4-‐n)OMen which are harder reducing agents CeCl3 acidifies the MeOH allowing it to ac%vate the carbonyl oxygen making the carbonyl carbon more posi%ve Reagent is harder, substrate is harder, therefore 1,2-‐reduc%on -‐ A L. Gemal, J. –L. Luche, J. Am. Chem. Soc., 1981, 103, 5454-‐5459 MeO H B MeO OMe
O
H
Me O CeIII
OH
O MeMe
HO Me
Me
NaBH4, CeCl3•7H2O, MeOH
MeMe
Me
Me
Me
Me 58%
Givaudan Roure (Interna%onal) SA Patent: US5929291 A1, 1999
149
Reactivity and Control for Organic Synthesis
in a related manner the use of anhydrous cerium(III) chloride in the presence of Grignard reagents and organolithium reagents allows the addi%on of organometallics to highly enolisable aldehydes and ketones T. Imamoto, N. Takiyama, K. Nakamura, T. Hatajima, Y. Kamiya, J . Am. Chem. Soc. 1989, 111 , 4392-‐4398; N. Takeda, T. Imamoto Org. Synth. 1999, 76, 228 methods to dry CeCl3•7H2O -‐ W. H Bunnelle, B. A. Narayanan, Org. Synth., Coll. Vol. VIII, 1993, 602. O
HO
Bu
BuM, THF
product
recovered star%ng material
BuLi
26%
55%
BuLi, CeCl3
92-‐97%
BuMgBr
28%
BuMgBr, CeI3
96%
23%
150
Reactivity and Control for Organic Synthesis
other modified borohydrides: NaBH3CN, NaBH(OAc)3 reagents of choice for reduc%ve amina%on
O N H
Ph
O
CH2O, NaBH3CN
N Me
pH 5
Me
TBDMSO
AcO O
N H H
+
Me O
R
Ph
in each case, reduc%on of the intermediate iminium ion is more rapid than the reduc%on of the corresponding aldehyde
Me
NaBH(OAc)3, SnCl2
TBDMSO O N H
H
R AcO
Me
this is one method to solve the problem of polyalkyla%on when aTemp%ng to alkylate amines
R
NH2
MeI
R
NHMe
MeI
at least as nucleophilic as star%ng amine
R
NMe2
MeI
at least as nucleophilic 1° and 2° amine
R
NMe3 I
polyalkyla%on occurs
151
Reactivity and Control for Organic Synthesis
borane complexed to a Lewis base, THF•BH3, Me2S•BH3 is a good reducing agent for carboxylic acids and amides H O R
O
B H
H
H O
H
R
H B O
O
H H
R
H B
H LB B O H
:LB H
O
R
R
OH
R
O
OBR2 R
Ar'
HO2C
BH3•THF, THF, 0 °C
Ar
EtO2C
R
O
R work-up
O H
O
H R
H HO
Ar'
EtO2C
Ar
LB
O
R
R B
H B
B H
R
H
98%
P. C. Lobben, S. S. –W. Leung, S. Tummala, Org. Process Res. Dev. 2004, 8, 1072–1075. H O R
B H
NR2
H
H O R
H B
O
H
NHR2
R
H
H B
H
NR2
R
B H
R
NR2 R
H
R
NR2 BR2 work-up
R
NR2
152
Reactivity and Control for Organic Synthesis
ReducTon examples MeO
MeO BH3•THF OH
O
OH O
O
O. Hoshino,Y. Mizuno,M. Murakata, H. Yamaguchi Chem. Pharm. Bull., 1999, 47, 1380-‐1383
O Me H
HO2C H O
O Me
Me CH2N2
H
H
OH
HO
H
MeO2C
O Me
Me
H
H
MeO2C
pTSA
H
O
O
O Me
H
O LiAlH4 O Me HO
Me HO
H H
O Me H
water, pTSA H
O
D. N. Kirk, M. S. Rajagopalan, M. J. Varley, J. Chem. Soc., Perkin 1, 1983, 2225-‐2228
H
O O
H
O Me
153
Reactivity and Control for Organic Synthesis
ReducTon examples OMe
Me
O
(COCl)2, DMF
O
O
O
OMe
Me NaBH4, THF
O
CO2H
O
OMe
Me
Cl
O O
O
O
HO
T. P. O'Sullivan, H. Zhang, L. N. Mander, Org. Biomol. Chem., 2007, 5, 2627-‐2635 Recently ChareVe reported the chemoselec
O
N
O
Tf2O, then EtO2C Me
H
H
N H
H
Ph
N
CO2Et Me
90%
N
then Et3SiH then basic workup MeO2C
MeO2C
O
F
Tf2O,
N H
N
H
then Et3SiH then acid workup MeO2C
MeO2C
Ph
O
Tf2O,
N H
O
F
O 86%
95%
154
Reactivity and Control for Organic Synthesis
Rings and ring strain classifica%on of rings classifica%on
small
normal
medium
large
number atoms in ring
3, 4
5, 6
7-‐12
>12
types of ring strain
108°
H
angle strain (Baeyer strain) – distor%on of angles from the idealised values
Me
H
Me
111°
H
H
120° H
Me
larger to relief of strain between geminal methyl groups angle strain – most important in small rings OH Na CO , MeOH 2 3
O
O
H O
~60°
88°
MeO H
155
Reactivity and Control for Organic Synthesis
torsional strain arises from devia%on from an ideal staggered arrangement HH HH
H
H
H
H
H
H
H H
major feature of 3 and 4-‐membered rings H
H
H
H
H H
H
H
H H
H
H
HH
H H
H
H
H H
H H
H H
H
in its chair form cyclohexane has no torsional strain
H H
H
H
H
HH
H H
H H
HH
the lowest energy conforma%on of cyclopentane is an envelope which has some torsional strain
as a result of torsional strain 6-‐membered rings are generally more stable than 5-‐membered rings bond length strain – arises from devia%on of bond lengths from their ideal values
H Me
transannular strain– arises from proximity on non-‐bonded atoms frequently important in medium rings H H
H
C-‐H = 1.09 Å
Me
C-‐C = 1.54 Å
156
Reactivity and Control for Organic Synthesis
general features of ring closure
K or k X:
Y
X
if ring closure is thermodynamically controlled (K) then energy of product will be important ΔG = ΔH -‐ TΔS ring strain considera%ons mean K is generally only favourable for 5-‐ and 6-‐ membered ring if ring closure is kine%cally controlled (k) then the energy of the transi%on state will be important ΔG‡ = ΔH‡ -‐ TΔS‡
k =
kBT -‐ΔG‡/RT e h
ΔH‡ -‐ enthalpy of ac%va%on includes bond breaking/making enthalpic considera%ons and the change in strain energy in reaching the transi%on state ΔS‡ -‐ reflects the difference in the levels of organisa%on between star%ng material(s) and transi%on state
157
Reactivity and Control for Organic Synthesis
rates of cyclisa%on of ω-‐bromo malonates: M. A. Casadei, C. Galli, L. Mandolini, J. Am. Chem. Soc., 1984, 106, 1051-‐1056 6
es%mated value
k =
increasing ΔS‡
4
CO2Et
2
log k
kBT -‐ΔG‡/RT e h
EtO2C
( )n
Br
NaH, DMSO k
0 -‐2
EtO2C
small
-‐4
CO2Et ( )n
-‐6 2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22
ring size
Small rings: 3-‐membered rings are formed fast; 4-‐membered rings more slowly ΔS‡ favourable as liTle preorganisa%on is required – the ends are already close to one another ΔH‡ unfavourable due to developing strain
158
Reactivity and Control for Organic Synthesis
◾ rates of cyclisa%on of ω-‐bromo malonates: M. A. Casadei, C. Galli, L. Mandolini, J. Am. Chem. Soc., 1984, 106, 1051-‐1056 6
es%mated value
k =
increasing ΔS‡
4
CO2Et
2
log k
kBT -‐ΔG‡/RT e h
EtO2C
( )n
Br
NaH, DMSO k
0 -‐2
normal
-‐4
medium
large
EtO2C
CO2Et ( )n
-‐6 2
3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22
ring size
Normal rings: 5-‐membered rings generally formed fastest ΔS‡ is becoming less favourable as more preorganisa%on is required – the ends are less close to one another ΔH‡ is fairly consistent – 5-‐7 membered rings are rela%vely unstrained Medium rings: ΔS‡ is s%ll increasing but propor%onally less as ring size increases – the ends are less close to one another ΔH‡ becomes dominant – transannular strain reflected in TS and hence rate of cyclisa%on Large rings: ΔS‡ is unfavourable as the ends are unlikely to meet – similar to an intermolecular reac%on. Solu%on do reac%ons under high dilu%on ΔH‡ no ring strain so not important -‐ large rings are similar to acyclic compounds
159
Reactivity and Control for Organic Synthesis
correct alignment of orbitals is also key for efficient ring forma%on LUMO C-‐Br σ*
O
Br Br
O
O
HOMO enolate
X
O
O
good overlap O-‐alkyla%on occurs
poor overlap so no C-‐alkyla%on similarly O O S O CH3 NaH
O O S O CH3
S O O H3C
S O O H3C
Br
X
O O S OH CH3 S O O
H3C
Sir Jack Baldwin proposed a set of guidelines (Baldwin’s rules) to asses the likelihood that a given, kine%cally controlled cyclisa%on would be feasible
160
Reactivity and Control for Organic Synthesis
“Ring-‐forming reac
Sir Jack Baldwin
Waynflete Professor of Organic Chemistry, Oxford 1978-‐2005 Baldwin’s Rules Biosynthesis of penicillins
modes of cyclisa%on X Y Nu:
Y
Y X Nu:
X Nu
exo – bond being broken is outside the ring being formed
X Nu
endo – bond being broken is inside the ring being formed
X
X Nu:
X
X
Nu:
Nu: X
exo-‐tet
sp3 hybridised
exo-‐trig
sp2 hydridised
X
exo-‐dig sp hybridised
Nu:
endo-‐tet
sp3 hybridised
Nu:
endo-‐trig
sp2 hydridised
Nu:
endo-‐dig sp hybridised
161
Reactivity and Control for Organic Synthesis
Baldwin’s rules
TET
TRIG
DIG
ring size
3
4
5
6
exo
✓
✓
✓
✓
endo
-‐
-‐
✗
✗
exo
✓
✓
✓
✓
endo
✗
✗
✗
✓
exo
✗
✗
✓
✓
endo
✓
✓
✓
✓
in general all exo-‐tet and exo-‐trig cyclisa%ons are favoured
5-‐exo-‐trig is faster than 6-‐endo trig How would you assign this reac
O O S O CH3
S O O H3C
S O O H3C
X H3C
Is this likely to be an efficient transforma
O O S OH
NaH I
CH3 S O O
O O S
S O O H3C
S O O H3C
162
Reactivity and Control for Organic Synthesis
1) The following amine undergoes cyclisa
OMe
2) Explain the contras
O
HO H
HO
O
3) Explain the following reac
O
O NaOH, H2O
Me Me
Me
HO
HO H
Me
Me
H
O
O
163
Reactivity and Control for Organic Synthesis
Thorpe-‐Ingold effect; M. E. Jung, G. Piizzi, Chem. Rev. 2005, 105, 1735−1766
the increased rate of cyclisa%on when puœng a geminal dialkyl group in the cyclising chain is known as the Thorpe Ingold effect. a good explana%on for the Thorpe Ingold effect concerns reac%ve rotamers
O Me
krel
R R O
R = Me Br
R=H
H
H
O
H
H
Me
1
O
O Me
Me H
O
O
39
O
O
O
O
H
O
O
major conforma%on, ends held far apart, cyclisa%on cannot occur
O Me
Me H
H
Br
Br
H
H Me
Me Br
Br
reac%ve rotamers
for gem-‐dimethyl-‐subs%tuted substrate all of the staggered conforma%ons are of similar energy and in two of the conformers cyclising groups are in close proximity
164
Reactivity and Control for Organic Synthesis
Examples O
cat. neat 25 °C
oligomers
iPr F3C F3C
O O
cat. neat 25 °C
iPr N Ph
O Mo O F3C
CF3
Explain the following rates of cyclisa
Br
krel
1
NH2
Br
2.2
Br
NH2
158
Br
NH2
0.16
Br
iPr NH2
9190
165
Reactivity and Control for Organic Synthesis
From Corey’s synthesis of longifolene, J. Am. Chem. Soc., 1964, 86, 478. Explain the various aspects of selec
O O HO
OH TsOH
O
O O
PPh3
O OsO4 O O
O O LiClO4
O
TsCl. pyridine
TsO HCl (aq) O O
O O
OH
HO
OH
166
Reactivity and Control for Organic Synthesis
Appendix Hybridisa%on and bonding -‐ a brief recap Hybridisa%on is a useful concept used by organic chemists to describe the bonding in organic molecules A quick method to work our the hybridisa%on of an atom is to count the number of subs%tuents on that atom (including lone pairs of electrons), remembering that in the bonded environment first row elements generally have 8 electrons around them 4 subs%tuents = sp3 hybridised, 3 subs%tuents = sp2 hybridised, 2 subs%tuents = sp hybridised z
z
z y
y
y
x
x
x +
+
py
px
pz
+ s
sp3 hybrid orbitals are made up from one s orbital and three p orbitals giving four hybrid orbitals which point to the corners of a regular tetrahedron. This is the bonding arrangement found in methane (bond angle = 109°) where the sp3 hybrid orbitals overlap with the hydrogen 1s orbitals (not shown) sp3 hybrid orbitals H H
C H
H H
H
H H
167
Reactivity and Control for Organic Synthesis
Similarly, the nitrogen atom in ammonia can be viewed as sp3 hybridised as can the oxygen atom in water although the H-‐X-‐H bond angle is slightly less than 109° due to lone pair–bond pair repulsion lone pair in sp3 orbital
N H
H H
N H
H
O
H
H
H
O
H
H
N-‐atom is sp3 hybridised
O-‐atom is sp3 hybridised
For sp2 hybridisa%on we mix two p-‐orbitals and one s-‐orbital to give three sp2 hybrid orbitals (and leave one p-‐orbital) z
z
z
y
y
x pz
y
x py
x +
+ px
s
The three sp2 hybrid orbitals are arranged 120° apart This is the bonding arrangement found in ethene with the sp2 hybrids overlapping with the hydrogen 1s orbitals (not shown) the remaining pz orbital(s) overlap to form the π-‐bond H H
H H
H
H
H
H
pz orbital
C-‐atom is sp2 hybridised
sp2 hybrid orbitals
168
Reactivity and Control for Organic Synthesis
Similarly, this is the hybridisa%on in carbonyl compounds and imines H
H
lone pair in sp2 orbital
O
N
H
H
H
N and C-‐atoms are sp2 hybridised
O and C-‐atoms are sp2 hybridised
For sp hybridisa%on we mix one p-‐orbitals and one s-‐orbital to give two sp hybrid orbitals (and leave two p-‐orbital) z
z
z
y
y
y
x
x
pz
x +
py
px
s
The two sp hybrid orbitals are arranged 180° apart This is the bonding arrangement found in ethyne with the sp hybrids overlapping with the hydrogen 1s orbitals (not shown) the remaining p orbitals overlapping to form the two π-‐bonds lone pair in sp orbital
p orbitals H
H
C-‐atom is sp2 hybridised
H
H
H
N
in nitriles the N and C-‐atoms are sp hybridised
sp hybrid orbitals
169
Reactivity and Control for Organic Synthesis
So to recap, in general, a quick method to work our the hybridisa%on of an atom is to count the number of subs%tuents on that atom (including lone pairs of electrons), remembering that in the bonded environment first row elements generally have 8 electrons around them -‐ sp3 = 4 sp2 = 3 sp = 2 What is the hybridisa%on of the red atoms in the following examples?
NH 4+
H 3O+
CO N
H H C C C H H
O CO 2
O
Let’s look at the bonding in amides.
All other things being equal, amides are planar molecules and we are happy to draw the delocalisa%on of the nitrogen lone pair as shown to indicate the par%al double bond character of the C-‐N bond O R A
O N R
R
R
NHR 2 B
Following the discussion above the hybridisa%on of the C and O atoms is sp2 but what is the hybridisa%on of the nitrogen atom? Again, following the above discussion, and looking at the form of the amide on the leU hand side (A), the nitrogen atom has 4 subs%tuents, 2 x R, C=O and a lone pair and ∴ is sp3 hybridised However, most organic chemists would say the N atom is sp2 hybridised. Why is this?
170
Reactivity and Control for Organic Synthesis
O R A
N R
R
R
NHR 2 B
R
R
O R
N
C
O
N
R
R
N-‐sp2 hybridised N-‐lone pair in p-‐orbital
C
O
R
N-‐sp3 hybridised N-‐lone pair in sp3-‐orbital
The curly arrows above represent the overlap of the nitrogen lone pair with the C-‐O π-‐orbitals (the an%bonding π* orbital). The best overlap therefore is if the N-‐atom is sp2 hybridised resul%ng in the N-‐lone pair being in a p-‐ orbital with excellent overlap with the p-‐orbitals of the C–O π-‐system If the N-‐atom were sp3 hybridised then the N-‐lone pair would be in an sp3 orbital which would result in poorer overlap with the adjacent C-‐O π-‐system – Generally beTer overlap = greater stabilisa%on In general if a π-‐system has an adjacent atom which carries a lone pair then most organic chemists would view the hybridisa%on of the adjacent atom as sp2 with the lone pair in a p-‐orbital to maximise overlap with the adjoining π-‐ system. What is the hybridisa%on of each of the heteroatoms in the following molecules? O OMe R
O
R
O
NMe 2
OMe
N O
Perhaps we should not be too concerned about this as some molecules, for example anilines, are frequently not perfectly planar and the hybridisa%on of nitrogen is somewhere between perfectly sp2 and perfectly sp3 Addi%onally we should be aware that other effects (e.g. sterics) can override electronic effects and hence the hydridisa%on may not be as expected
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