Chap-1-alkane-theory.pdf

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CHEMISTRY ALLEN Study Package For – JEE (Advanced)

JEE-Chemistry

HYDROCARBON ALKANE 

Intro duct ion (i)

:

Branched and unbranched aliphatic saturated hydro carbons are called member of alkane. The structural formula of alkane have only single bonds or all bonds in alkane is only  bonds.

(ii)

Alkanes does not reacts with chemical reagants such as dil. and conc. HCl, dil. & conc. H 2SO 4 , dil. & conc. HNO 3 , Caustic soda, acidic & basic K 2 Cr 2 O 7 , KMnO 4 etc. That is why alkanes are called paraffins. (Parum=little, affins = reactivity).

Prop er t y

Characteristics of

Prop er t y

Characteristics

alkane

of

alkane

General formula

C nH 2n+2

C—C Bond length

1.54 A°

C—C Bond energy

82.67 kcal/mole

C—H Bond length

1.112 A°

C—H Bond energy

98.67 kcal/mole

Hybridisation on C

sp 3

Bond angle

109°.28 '

shape

Tetrahedral



General Met ho ds of Preparat ions :

1.

From alkene s a nd alkyne s (Sabat ier a nd Sa ndrens react ion) or (By hy drogenat ion of alkene s

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and alkynes) : Alkenes and alkynes on catalytic hydrogenation gives alkanes. R—CH

CH—R + H 2

Catalyst   R — C H 2— C H 2— R

Alkene R—C

Alkane Catalyst   R — C H 2— C H 2— R

C—R + 2H 2 Alkyne

Catalyst

2.

:

(a)

Pd/Pt at ordinary temp. and pressure

(b)

Ni, 200–300° C (sabatier)

(c)

Raney Nicker at room temp.

(d)

Raney nickel is obtained by boiling Ni/Al with NaOH. Al dissolved & Ni obtained in finally divided state.

(e)

Methane can not be prepared by this method (From unsaturated hydrocarbon).

From alkyl Halide s (By reduct ion) : 2H  

R—X Catalyst



(Nascent Hydrogen )

R—H + HX

:

(i) Zn + HCl

(ii) Zn + CH 3 COOH

(iv) Red P + HI

(v) Al + Hg + ethanol

Mechanism

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

: Zn  Zn +2 + 2e  –

+

(iii) Zn—Cu couple in C 2 H 5 OH





2e R — X R+ X

–

–

R + H Cl  R—H + Cl 

Zn +2 + 2Cl  ZnCl 2 .

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JEE-Chemistry (a)

Alkyl halides can also be reduced to alkane by H 2 /Pd or LiAlH 4 or H 2 /Ni.

(b)

Reduction is due to the electron transfer from the metal to the substrate (R- X)

(c)

If any alkyl halide is asked, the H-atom of any carbon atom of given alkane is removed by halogen atom.

3.

From alkyl halide (By Wur tz react ion): A solution of alkyl halide in ether on heating with sodium gives alkane. R — X  2Na

(a)

Dry   R— R

X— R



 2NaX

ether

Two moles of alkyl halide treated with Na in presence of dry ether. If ether is wet then we obtain alcohol. 2Na +

H 2O



2NaOH + H 2

C H 3I +

NaOH



CH 3 OH + NaI



(b)

Methane can not be prepared by this method. The alkane produced is higher and symmetrical i.e. it contains double the number of carbon atoms present in the alkyl halide taken.

(c)

Two different alkyl halides, on wur tz raction give all possible alkanes.

(d)

The seperation of mixture in to individual members is not easy because their B.P. are near to each other and thus wurtz reaction is not suitable for the synthesis of alkanes containing odd number of carbon atom.

(e)

This reaction generally fails with tertiary alkyl halide.

Mecha nism : Two mechanism have been proposed for this react ion. (a)

Ionic

Mechanism: 

2Na  2Na   + 2e 











2e R — X  R+ X



R — X + R R — R + X–

2Na  + 2X   2NaX

Product Example

: 2 C 2H 5— I

+

2 N a 

C 2 H 5 —C 2 H 5 + 2NaI n–butane

(b)

Free radical mecha nism : 

Na Na  + e 





.

e R — X R+ X



Na   X   NaX

R   R   R — R Product

Free radicals also undergo Disproportionation i.e. one radical gains hydrogen at the expense of the other which loss hydrogen. H .

.

.

.

CH2 — CH2 + C2H5  C2H6

+

Ethane

C2H4

Ethylene

This explains the presence of ethylene and ethane in the butane obtained by Wur tz reaction.

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Methanol

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JEE-Chemistry Ex.

If two moles of Isopropyl chloride reacts with Na in presence of dr y ether. Which alkane is obtained.

S o l . 2, 3-Dimethyl butane. Ex.

If isopropyl chloride and ethyl chloride both react with Na in presence of dry ether which alkanes are obtained.

S o l . n-Butane, 2-Methyl butane and 2, 3-Dimethyl butane. Ex.

Which of the following compound can not obtained from wurtz reaction. (A) ethane

(B) butane

(C) isobutane

(D) hexane

Sol.

(C) [Hint : In wurtz reaction unsymmetrical alkane can not be obtained.

Ex.

When ethyl chloride and n-propyl chloride undergoes wur tz reaction which is not obtained. (A) n–butane

(B) n–pentane

(C) n–hexane

(D) isobutane

S o l . (D) C 2 H 5 —Cl + C 3 H 7 Cl

Na   C H —C H + C H —C H + C H —C H 2 5 2 5 3 7 3 7 2 5 3 7 dry ether

4.

:

Core y-House

Synt he sis

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This method is suitable for the preparation of unsymmetrical alkanes i.e. those of type R—R' (i)

RX + Li  RLi + LiX

(ii)

2RLi + CuX  R 2 CuLi + LiX

(iii)

R 2 CuLi + R ' X  R—R' + RCu + LiX (1 or2  )

Note : In Corey-house reaction symmetrical and unsymmetrical alkane both can be formed. 5.

From Frankland Reagent:

If Zn is used in place of Na, the reaction is named as Frankland reaction.

R—X + 2Zn +RX  R 2 Zn + ZnX 2 Frankland reagent R 2 Zn + R—X  R—R + RZnX 6.

From Carboxylic Acid (By decarboxylation) : Saturated monocarboxylic acid salt of sodium or potassium on dry distillation with soda lime give alkane.  RCOONa + NaOH   R—H + Na 2 CO 3 Cao

Soda Lime (a)

The process of elimination of Carbon-di-oxide from Carboxylic acid called decarboxylation.

(b)

Replacement of -COOH by hydrogen is known as decarboxylation. The alkane formed always contains one carbon atom less than the original acid.

(c)

This reaction is employed for stepping down a homologous series.

(d)

Soda lime is prepared by soaking quick lime CaO with NaOH solution and then drying the products.

(e)

Decarboxylat ion of sodium formate gives H 2   HCOONa + NaOH (CaO)   H 2 + Na 2 CO 3    CH 3 COONa + NaOH + CaO  CH 4 + Na 2 CO 3

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Mecha nisim : Decarboxylation proceeds via. the formation of carbanion i ntermediate as follows.

OH

CH3 C O + OH(NaOH)



CH3

C

O

ONa

ONa OH CH3

O

C

CH3 + HO C O (NaHCO3)



ONa(Acidic)

ONa

CH O C 3 + H

ONa



CH3 H +O C ONa Product O



Na 2 C O 3

O 

O Na + Na

O C O

(a)

If in a compound two carboxylic groups are present and they are attached to same carbon atom then also decarboxylation of one of the carboxylic groups takes place simply on heating.

Ex.

COOH COOH

 CH 3 COOH + CO 2 

(b)

CH 4 can be prepared by CH 3 COOH.

(c)

C 2 H 6 can be prepared by CH 3 CH 2 COOH.

(d)

CH 3 —CH 2 —CH 3 can be prepared by Butanoic acid and 2–Methyl propanoic acid.

How many acids can be taken to obtain isobutane from decarboxylation ? (A) 4

S o l . (C)

(B) 3

(C) 2

(D) 5

To obtain isobutane the acids are (i)

CH3 CH CH2 COOH  CH3 CH CH2 H CH3

CH3

COOH (ii)

CH3

C

H

CH3  CH3

CH3

C

CH3

CH3

So two acids can be taken.

Reactivity of acid  stability of carbanion Presence of electron attracting group (–I) in the hydrocarbon part of the fatty acid increases the decarboxylation. If – I is more effective group then weak base may be taken. Example

:

(i)

3  R CH CH2COOH  

CH CH2 H

OH

OH

NaHCO

R

NaHCO

(ii)

R

3 CH CH2COOH  R

CH CH2 H

NO2

NO2



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JEE-Chemistry (iii)

-Keto acids are decarboxylated readily simply on heating (soda lime is not required)

R

Ex.

 C CH2COOH  R

C CH3

O

O

Give reactivity order for decarboxylation ? C H 3— C H 2— C O O H

CH2

CH—COOH

I

II

(A) I > II > III S o l . (B)

CH

C —COOH III

(B) III > II > I

(C) III > I > II

(D) None is correct

In decarboxylation intermediate s are,

CH 3 — CH  2

CH 

CH 2

I

CH

II

C

III

The stability order of carbanion – III > II > I So reactivity order for acid is – III > II > I 7.

From carboxylic acid (By Kolbe's proce ss) : Alkanes are formed on electrolysis of concentrated aqueous solution of sodium or potassium salt of saturated monocarboxylic acids. Electrolysis — R  2CO  2NaOH  H 2 2RCOONa + 2H 2 O   R 2   

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At Anode

At Cathode

Electrolysis of Sodium propionate solution give n-butane, ethylene, ethane and ethyl propionate as followselectro. 2C 2 H 5 — COONa   C 2 H 5 — C 2 H 5  2CO 2  2NaOH  H 2



Mechanism

:



—COO  + N a

electro. C 2 H 5 —C OON a   C 2H 5 H2O

(Ionization) – e

 C 2 H 5 — COO  C 2 H 5 —COO   

At Anode :



C2H5

Fragmentation .  C 2 H 5 + CO 2 C O 

O 



C 2 H 5 + C 2 H 5  C 2 H 5 —C 2 H 5 Product H

..

.

.

CH2 — CH2 + C2H5  CH 3 —CH 2 —H + CH 2

CH 2

(minor products) An ester is also formed. 



C 2 H 5 — CO O + C H 5  C 2 H 5 —COOC 2 H 5 2 (minor products) At cathode :

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Na  + e   Na Na + H 2 O 

NaOH +

1

2

H2

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JEE-Chemistry (a) (b)

Methane can not be prepared by this method. Electrolysis of an acid salt gives symmetrical alkane, However in case of a mixture of Carboxylic acid salts, all probable alkanes are formed. Electrolysis

R'COOK + R"COOK  (R'—R" + R'—R' + R"—R") +2CO 2 + H 2 +2NaOH

8.

 - position decrease the yield of alkanes.

(c)

Presence of alkyl groups in

(d) (e) (f) (g)

Tr ue aromatic acids do not undergo Kolbe's electroly tic reaction. Free radical mechanism has been suggested for Kolbe reaction. At anode alkane (major) and CO 2 gas is formed while at cathode NaOH and H 2 gas is formed. The concentration of NaOH in solution is increased with time so pH of solution is also increased.

From alka nol, alka nals, Alka none a nd alka noic acid (By reduct ion) : The reduction of either of the above in presence of red P and HI gives corresponding alkane. R—OH

+2HI

Re d P   R—H+H 2 O+ I 2 150  C

R—CHO + 4HI

Re d P   RCH 3 +H 2 O + 2I 2 150  C

R—CO—R+4HI

Re d P   R—CH 2 —R+H 2 O + 2I 2 150  C

RCOOH + 6HI

Re d P   R—CH 2 —R+H 2 O + 3I 2 150  C

R—CH 3 + I 2 2P + 3I 2 9.

From

alka none s

R—CH 2 —I + HI 2PI 3



(By

Clemmensen's

met ho d)

:

Carbonyl compound (Preferably ketones) may also be reduced with Zinc amalgam and concentrated HCl (Zn—Hg/HCl), this reaction is called Clemmensen reduction. R—CO—R'+4H

Zn / Hg   R—CH —R'+H O 2 2 con. HCl

CH 4 , CH 3 —CH 3 , isobutane and neopentane are not obtained from Ketones because these alkane do not contain 10.

CH 2 group.

From alka nals a nd alka none s (By Wolf Kishner react ion) : C

O + NH 2 N H 2

C



Hydrazine 

N.NH 2

Glycol / KOH    

CH 2 + N 2

Hydrazone

From G.R. : (a) For mat ion of alka ne s w it h same number of C atoms : With same number of C-atoms as G.R. react with compound containing active hydrogen alkanes is obtained. R

Mg

X+ H

O

H  R H + Mg (OH) X

+R

O

H  R H + Mg (OR) X

+R

(b)

NH

H  R H + Mg (NHR) X

This reaction is used to determine the number of active H-atoms in the compound this is known as Zerewitnoff's method. G.R. react w it h alkyl halide to give h igher alka ne s : RMgX + R'—X  R—R' + MgX 2

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In the above reaction I 2 is formed which acts as reducing agent and may reduce alkane and form alkyl halide. So red P is added i n the react ion to remove I 2 formed in the react ion.

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JEE-Chemistry Ex.

Which of the following does not give alkane with R—Mg—X. (A) Ph—OH (B) Cl —NH 2 (C) CH 3 COOH (D) HCl S o l . (B) [Hint : Except Cl—NH 2 all have active hydrogen, but Cl—NH 2 when reacts with R—Mg—X the product is R—NH 2 .] 

11.

From metal carbide (By hy drolysis) : Only CH 4 can be obtained by the hydrolysis of Be or Al carbides Al 4 C 3  12H 2 O

  

4 Al(OH ) 3  3CH 4

Be 2 C  4H 2 O

  

2Be(OH ) 2  CH 4

Physical

Prop er t ie s

(i)

C 1 to C 4 gases, Neopentane also gas but n-pentane and isopentane are low B.P. liquids.

(ii)

Next member s C 5 to C 17 are Colourless liquids and above C 17 are Wa xy solids.

(iii)

Density : The density of alkanes increases with increase in molecular weight and becomes constant at 0.8 g/mL. Thus all alkanes are lighter than water.

(iv)

Solubility : Alkanes being non polar and thus insoluble in water but soluble in non-polar solvents Example

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:

(v)

 i.e.

: C 6 H 6 , CCl 4 ,ether etc.



The solubility of alkanes decreases with increase in molecular weight



Liquid alkanes are themselves good non-polar solvents.

Boiling point -

 molecular weight (for n-alkanes)

Vanderwaals force of attraction  molecular weight  surface area of molecule. boiling point

Pentane < hexane < heptane

Also boiling point 

1 number of side chain

because the shape approaches to spherical which results in decrease in Vanderwaals forces (as sur face area decreases) Thus boiling point n–Pentane > Isopentane > neopentane (vi)

Melt i ng Poi nt : M.P. of alkanes do not show regular trend. Alkanes with even number of carbon atoms have higher M.P. than their alkanes of odd number of carbon atoms. The abnormal trend in M.P. is due to the fact that alkanes with odd carbon atoms have their carbon atom on the same side of the molecule and in even carbon atom alkane the end Carbon atom on opposite side. Thus alkanes with even carbon atoms are packed closely in crystal lattice to permit greater intermolecular at tractions.

C C

C C

C C

C

C

<

C

Odd number of carbon

C C

C C

Even number of carbon

E x . Alkanes are inert is nature, why ? S o l . Alkanes are quite inert substances with highly stable nature. Their inactiveness has been explained as: (i) Alkanes have all the C—C and C—H bonds being stronger  bonds and are not influenced by acid, oxidants under ordinary conditions. (ii) The C—C bond is completely non polar and C—H is weak polar. Thus polar species i.e. electrophiles or nucleophiles are unable to attack these bonds under ordinary conditions.

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JEE-Chemistry 12.

Chemical

Prop er t ie s

:



Oxidation : Complete oxidat ion or combust ion : Burn readily w ith non-luminous flame in pre sence of air  or oxygen to give CO 2 and water with evolution of heat. Therefore, alkanes are used as fuels.

 3n  1   O 2  nCO 2 + (n+1) H 2 O + Q; (H =–ve)  2 

C n H 2 n 2   

Incomplete oxidation : In limited supply of air gives carbon black and CO. 2CH 4 + 3O 2



2CO + 4H 2 O

CH 4 + O 2



C + 2H 2 O

C–black (used in printing) 

(i)

Catalytic oxidation : Alkanes are easily converted to alcohols and aldehydes under controlled catalytic oxidation. Red hot Cu or Fe tube   2CH OH 2CH 4 + O 2  High P and T 3

Mo 2 O 3   HCHO + H O 2

CH 4 + O 2 (ii)

230 0 C,100 atm

Alkanes on oxidation in presence of manganese acetate give fatty acids. (CH COO ) Mn

(iii)

Ter tiar y alkanes are oxidized to give ter tiar y alcohols by KMnO 4 .

CH3 CH3 C

CH3 [O]

 CH3 H 

CH3 Ex.

OH

CH3

How many litre of Oxygen required for complete conbustion of 6.0 g ethane at NTP ?

2C 2 H 6  7O 2  4CO 2  6H 2O

Sol.

60 g ethane required O 2 (at NTP) = 7 ×



C

KMnO 4

22.4 litre

1 g ethane required O 2 (at NTP) =

7  22.4 litre 2  30

6 g ethane required O 2 (at NTP) =

7  22.4  6  15.68 litre 2  30

Substitution Reactions : Substitution reaction in alkanes shows free radical mechanism. They give following substitution reaction. (a) Halogenation : Replacement of H-atom by halogen atom R—H + X 2  R—X + HX Halogenation is made on exposure to (halogen + alkane) mixture to UV or at elevated temp. The reactivity order for halogens shows the order. F2 > Cl 2 > Br2 > I 2

Tertiary C – H > Sec. C – H > primary C – H

Reactivity order of hydrogen atom in alkane is (i)

Fluorination : Reacts explosively even in dark. Fluorination can be achieved without violence when alkane is treated with F 2 diluted with an inert gas (like N 2 ) By the action of HgF 2 on bromo or iodo derivatives. C 2 H 5 I + HgF 2  C 2 H 5 F + HgI 2

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3 2 CH 3 (CH 2 ) n CH 3   CH 3 (CH 2 ) n COOH high Temp

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JEE-Chemistry (ii)

Chlorination

:

Cl

Cl

Cl

Cl

2 2 2 2 CH 4   CH 3 Cl   CH 2 Cl 2   CHCl 3   CCl 4

The monochloro derivative of alkane is obtained by taking alkane in large excess. When chlorine is in excess, a mixture of mono, di, tri, tetra and perchloro derivatives is obtained. Explosively CH 4 + Cl 2  C + HCl 

Mechanism

UV for CH 4 + Cl 2  

Step I

Chain

initiation

step

Step II

Chai n

propagat ion

CH 3 Cl + HCl UV

 Cl   Cl  Cl : Cl  or 

:

step

. . : Cl + H : CH3  H: Cl + CH3 Methane

Methyl radical

. . CH3 + Cl : Cl  CH3Cl + Cl Step III

Chai n

ter mi nat ion

step

: Cl   Cl   Cl 2 , CH 3   Cl  CH 3 Cl ,

CH 3   CH 3  CH 3 CH 3 A





Cl

can also at tack CH 3 Cl to form chloromethyl ( CH 2 Cl ) free radical. T his free radical

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participates further in the chain reaction to yield methylene chloride (dichloromethane).

H . Cl + H: C

Cl

Cl

and

H

H . C + Cl : Cl H

Simi larly, chloroform and CCl 4 are obtained by fur ther chain reaction. (iii) Bromi nat ion : Br 2 reacts with alkane s in a simi lar manner but le ss vigorously. (iv) Iodination : Iodine reacts with alkanes reversibly. HI formed as the by product is a power ful reducing agent and is capable of reducing the CH 3 I to CH 4 . Iodination may be carried out in the presence of an oxidising agent such as HIO 3 , HIO 4 , HNO 3 , HgO etc. Which destroy HI,

CH 4  I 2

CH 3 I  HI

5HI + HIO 3  3I 2 + 3H 2 O Iodination is ver y slow because energy of activation of the reaction is ver y large 

CH 4  I   HI + C H 3 Note

: Halogenation is inhibited in presence of oxygen because oxygen reacts with alkyl free radicals to form less react ive peroxy alkyl radical R–O–O • which can not propagate the chain.



Reactivity (i)

selectivity

Principle

:

Probability factor : The factor is based on the number of each kind of H atom in the molecule. For example is CH 3—CH 2 —CH 2 —CH 3 there are six equivalent 1° H's and four equivalent 2° H's. The probability of abstracting 1° H's to 2° H's is 6 to 4. i.e., 3 to 2.

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JEE-Chemistry (ii)

(iii)

Reactivity of halogen free radical : the more reactive chlorine free radical is less selective and more influenced by the probability factor. On the other hand, the less reactive Br radical is more selective and less influenced by the probability factor (Reactivity selectivity principle). Reactivity of alkanes (ease of abstration of 'H' atoms) : Since the rate determining step in halogenations is abstraction of hydrogen by a halogen atom be the formation of alkyl radical, halogenation of alkanes follows order of stability of free radical is 3° > 2° > 1° > CH 3 . Reactivity ratio of H atom for Chlorination (1° : 2° : 3° H) 1 : 3.8 : 5 Reactivity ratio of H atom for bromination (1 : 82 : 1600) T he above order of stabi lity of radicals is due to the ease of their format ion from the cor responding alkane which in turn is due to difference in the value of H. CH 3 —H  •CH 3  H •

H 1

C H 3 — C H 2 — H  •CH 3 — CH 2  H •

H 2

• • CH3—CH—CH3  CH —CH—CH +H

H 3

3

3

H CH3

CH3 •

CH3—C—CH3  CH3—C—CH 3+H •

H 4

Ex.

Reactivity of any H-atom  number of H atoms of that kind × reactivity of that H. Thus the amount of energy required to form the various classes of radicals decreases in the order CH 3 > 1° > 2° > 3° (H 1 >H 2 >H 3 >H 4 ). Therefore, it easie st to form 3° radical and it is most difficult to form CH 3. We can also interpret this in an alternative way the case of abstraction of H atoms from hydrocarbon fallows the sequence 3° > 2° > 1° CH 4 which should also be the case of formation of free radicals. The above order of stability is in accordance with the stability of free radicals on the basis of delocalization of odd electron. Order of stability of free radical is : Allyl, benzyl > 3° > 2° > 1° > methyl, vinyl. What is the percentage of products obtained from monobromination of isobutane ?

Br S o l . CH3 CH CH3 + Br2  CH3 C

CH3 (I)

CH3

Pr oduct (I) = Pr oduct (II)

CH3 + CH3 CH CH2 Br

No.of primary H × No.of tertiary H

reactivity of primary H reactivity of tertiary H

% of product (I) =

9 100  0.56% 1600  9

% of product (II) =

1600 100  99.44% 1600  9

(b)

Nitrat ion

CH3 (II)

=

9 1 9   1 1600 1600

: (Vapour phase nitration) This involve s the substitut ion of a hydrogen atom of

alkane with -NO 2 group.

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JEE-Chemistry At ordinary temperature, alkanes do not react with HNO 3 . But reacts with vapours of Conc. HNO 3 at 450 ° C. 0

400  500 C R — H  HO — NO 2   R — NO 2  H 2 O

Since the reaction is carried at high temp. the C—C bonds of alkanes break during the reaction and a mixture of nitroalkane s is formed. Example

450 0 C : CH 3 —CH 3 + HNO 3   CH 3 CH 2 NO 2 + CH 3 NO 2 + H 2 O

450°C

CH3CH2CH3 + HNO3



1–Nitro propane 2–Nitro propane Nitro ethane Nitromethane

25% 40% 10% 25%

Mechanism : (Free Radical substitution) 



 OH NO HO  NO 2  2

Step – I

– R





H + OH  R + H2O





R + HO – NO 2  R – NO 2 + OH

Step – II Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Che\Unit-10\ALKENE, ALKANE, ALKYNE-AROMATIC HYDROCARBON\Eng\1.Alkene, alkene, alkyne\1. theory.p65

(Pr oduct ) 

Step – III



R + NO2  R – NO 2 



R + O   R – OH (minor) 



OH + N   HONO 2 (c)

Sulphonation : Replacement of H atom of alkane by –SO 3 H is known sulphonation. Alkane react with fuming H 2 SO 4 or oleum (H 2 S 2 O 7 ). The branched lower alkanes and higher alkanes react to give alkane sulphonic acid. Example

:

CH3 CH3

CH3 H + HO SO3H CH3 C

C CH3

CH3

2–Methyl propane

2–methyl propane–2–sulphonic acid

The reactivity order for sulphonation is 

SO3H + H2O

tert. H > Sec. H > prim. H

Mechanism : (Free Radical substitution) 



400 C HOSO 3 H   HO   SO 3 H 



C 6 H 13 H  OH  C 6 H 13  H 2 O 



C 6 H 13  S O 3 H  C 6 H 13 SO 3 H

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JEE-Chemistry Lower members such as propane, butane, pentane etc. react with SO 3 in vapour phase to form sulphonic acids. C 3 H 8 + SO 3  C 3 H 7 – SO 3 H (d)

Chlorosulphonat ion (Reed react ion) : Reaction with a mixture of SO 2 and Cl 2 at ordinar y temp. in the presence of UV light is called chlorosulphonation. UV  

C 3 H 8  SO 2  Cl 2

C 3 H 7 SO 2 Cl  HCl Propane sulphonyl Chloride

Further hydrolysis of alkane sulphonyl chloride gives alkane sulphonic acid. H O

2  

C 3 H 7 SO 2 Cl

C 3 H 7 SO 3 H  HCl

propane sulphonic acid C 3 H 7 SO 3 H  NaCl

C 3 H 7 SO 3 Na  HCl



Sodium salt of sulphonic acid (used as detergent) 

Isomerization

:

Unbranched chain alkanes on heating with AlCl 3 + HCl / 200 0 C are converted in to branched chain alkanes

CH 3 — CH 2 — CH 2 — C H 3

3  

CH3 CH CH3

n-butane

Isobutane

Branched chain alkanes converted to more branched alkane.

CH3 CH3 CH CH2

CH3 CH3 AlCl  HCl

3 CH2 CH3   CH3 CH CH CH3

Isomerisation of alkanes is of great importance in petroleum industry to increase the octane number of petrol (gasoline). 

Pyrolysis

or

Cracki ng

or

t her mal

decomposit ion

:

When alkanes are heated to 500-700 0 C they are decomposed in to lower hydrocarbon. This decomposition is called pyrolysis. In petroleum industry it is also termed as cracking.Cracking is used for the manufacture of petrol, petrol gas/oil gas etc. Example

1000  C : CH 4   C  H2

500 C CH 3  CH 3   CH 2  CH 2  H 2 absence of air

CH3CH2CH3

CH2 CH3

CH2 + CH4 CH CH2 + H2

Cracking n-Butane   1-Butene + 2–Butene + Ethane + Ethene + Propene + CH 4 + H 2

The mechanism of pyrolysis occurs via free radicals.

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CH3 AlCl  HCl

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JEE-Chemistry 

Hydroforming or dehydrogenat ion or cyclisat ion or cataly t ic reforming or aromat izat ion : Unbranched higher alkanes (from 6 to 10 carbon atoms) when heated in presence of oxides of Cr, Mo, V on Al 2 O 3 suppor t at 500 0 C aromat ic hydrocarbons are formed. Cr O / Al O

2 3 2 3  n  hexane  500  C

+

4H2

CH3 Cr O / Al O

2 3 2 3 CH 3 (CH 2 ) 5  CH 3  

500 C

n-heptane

+ 4H2 Toluene

Cr O / Al O

2 3 2 3 CH 3 (CH 2 ) 6 – CH 3    500  C

CH3 CH3

n-octane o-xylene It provides an excellent method of passing from aliphatic to aromatic series. Chlorinolysis

:

0

300  400 C CH 3  CH 2  CH 3  Cl 2   C 2 Cl 6  CCl 4  HCl Pr essure (g) (s) ( )

Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Che\Unit-10\ALKENE, ALKANE, ALKYNE-AROMATIC HYDROCARBON\Eng\1.Alkene, alkene, alkyne\1. theory.p65



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JEE-Chemistry SOLVED

Which of the following reactions can be employed for getting unsymmetrical alkanes in good yield ? (A) Wurtz reaction

(B) Corey–House reaction

(C) Both

(D) None of these

Sol.

Wurtz reaction is suitable for symmetrical alkanes

Ex.2

Sodium propionate on decarboxylation with sodalime gives (A) Propane

(B) Ethane

Ans.(B)

(C) Butane

(D) Pentane

Ans.(B)

Sol.

Decarboxylation with soda lime results in the formation of alkane with one carbon less than the starting compounds

Ex.3

Which of the following alkanes cannot be produced by Kolbe electrolysis of sodium or potassium salts of carboxylic acids ? (A) Methane

(B) Ethane

(C) Butane

(D) Hexane

Ans.(A)

Sol.

In Kolbe electrolysis, the alkane is formed by union of two alkyl groups. The alkane formed has, thus, two or more carbon atoms.

Ex.4

The homolytic fission of hydrocarbon results in the formation of (A) Free radicals

(B) Carbocations

(C) Carbanions

(D) Carbenes.

Ans.(A)

Sol.

Homolytic fission results in the formation of free radicals.

Ex.5

n-Heptane when heated to a temperature of about 800 K under high pressure in the presence of Cr 2O3/Al2O3 catalyst gives (A) 1-heptene

Sol.

(B) 2-Methylhexane

 4H 2 CH 3 —(CH 2 ) 5 —CH 3  

(C) Toluene

(D) Xylene.

Ans.(C)

CH3 Toluene

Ex.6

The reaction conditions leading to the best yield of C2H5Cl are Dark

UV light (A) C2H6 (excess) + Cl2  

  (B) C2H6 + Cl2 room temperature

UV light (C) C2H6 + Cl2 (excess)  

UV light (D) C2H6 + Cl2  

Ans.(D)

Sol.

C2H6 should be used in excess, otherwise polychlorination will take place

Ex.7

Number of isomer which can be theoretically obtained on monochlorination of 2-methylbutane is (A) 1 1

Sol.

(B) 2 2

3

(C) 3

(D) 4

(C) Ethanal

(D) CO2 and H2O

Ans.(D)

4

CH3—CH—CH2—CH3 1

CH3 Ex.8

Complete oxidation of ethane yields (A) Ethanol

(B) Ethanoic acid

Ans.(D) Sol.

2C 2H6 + 7O 2  4CO 2 + 6H 2O

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Ex.1

EXA MPLES

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JEE-Chemistry Ex. 9

In iso-pentane, the H atom that can be most easily substituted is on 1

2

3

4

CH3—CH—CH2—CH3 CH3 (A) C—1

(B) C—2

(C) C—3

(D) C—4

Ans.(B)

Sol.

Ease of substitution of various types of H atoms is 30 > 20 > 10.

Ex.10

8 c.c. of gaseous hydrocarbon requires 40 c.c. of O 2 for complete combustion. Identify hydrocarbon.

Sol.

Volume of hydrocarbon = 8 c.c. ;

8 2 = 40 3n+ 1

Formula No. 1,

1 2 = 5 3n+ 1

or

Volume of O 2

3n + 1 = 10 or

= 40 c.c.

(For alkane)

3n = 10 - 1 = 9

,

n= 3

The value of n comes in whole number from 1st formula it means hydrocarbon is Alkane and it is of 3C atom.

Node-6\E:\Data\2014\Kota\JEE-Advanced\SMP\Che\Unit-10\ALKENE, ALKANE, ALKYNE-AROMATIC HYDROCARBON\Eng\1.Alkene, alkene, alkyne\1. theory.p65



Hydrocarbon is C 3 H 8 (Propane)

Ex.11

10 mL of a mixture of CH 4 and C 3 H 8 requires 41 mL of oxygen for complete combustion. What is the volume of CH 4 and C 3 H 8 in the mixture.

Sol.

Suppose the volume of CH 4 in (CH 4 + C 3 H 8 ) mix = x c.c. = Volume of C 3 H 8 w ill be = 10 – x c.c. For CH 4 CH 4  2O 2  CO 2  2H 2 O

  For C 3 H 8

 

1 Vol. of CH 4 requires 2 vol. of O 2 for complete combustion x c.c. of CH 4 , 2x c.c. of O 2 C 3 H 8 +5O 2

 3CO 2 + 4H 2 O

1 volume of C 3 H 8 requires 5 ml of O 2 for complete combustion (10 – x) c.c. of C 3 H 8 requires 5(10 – x) c.c. of O 2

Total Volume of O 2 = 2x + 5 (10 – x) it is equivalent to 41 (according to question)

 

2x + 5 (10 – x) = 41 x = 3 c.c.

Volume of CH 4 is 3 c.c. a nd volume of C 3 H 8 is 7 c.c. Ex.12

If 5 g C 2H 5I reacts with Na (Metallic) in presence of ether, and the yield is 60% then how many grams of n-butane will you get.

Sol.

2C 2 H 5 I  2Na  C 4 H 10  2NaI Molecular weight of 2C 2 H 5 I = 24 + 5 + 127 = 156 Molecular weight of C 4 H 10 = 48 + 10 = 58 Two molecule of C 2 H 5 I are taking par t in above reaction.

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JEE-Chemistry 

We get 58 g of C 4 H 10 from 2 x 156 g of C 2 H 5 I



We get

58 2x156

g C 4 H 10 from 1 g of C 2 H 5 I



We get

58 x5 2x156

g C 4 H 10 from 5 g of C 2 H 5 I

yield is 60% So the quantity of C 4 H 10 will be

58 x5 60 x 2x156 100

g = 0.55 g

Ex.13

The density of one hydrocarbon at N.T.P. is 1.964 g/litre. Identify the hydrocarbon.

Sol.

Molecular weight of Hydrocarbon = density of 1 lit. x 22.4 = 1.964 x 22.4 = 44 So Molecular weight of hydrocarbon = 44

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So the hydrocarbon is C 3 H 8 (Propane).

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