Jp Xii Organic Chemistry (06).pdf

ORGANIC CHEMISTRY

TARGET : JEE (Main + Advanced) 2016 Course : VIJETA (JP)

Date : 04-05-2015

NO. 07 & 08

TEST IN FORMATI ON DATE : 17.05.2015

PART TEST (PT) - 1 (ADVANCED)

Syllabus : Stereoisomerism and Organic reaction Mechanisms-I,Cheminfo (Till date)

This DPP is to be discussed in the week (04-05-2015 to 09-05-2015)

DPP No. # 07 (JEE-MAIN) Total Marks : 60

Max. Time : 40 min.

Single choice Objective ('–1' negative marking) Q.1 to Q.20

(3 marks, 2 min.)

[60, 40]

ANSWER KEY DPP NO. # 07 (JEE-MAIN) 1.

(A)

2.

(D)

3.

(C)

4.

(D)

5.

(B)

6.

(B)

7.

(D)

8.

(A)

9.

(A)

10.

(C)

11.

(D)

12.

(C)

13.

(D)

14.

(C)

15.

(C)

16.

(D)

17.

(A)

18.

(C)

19.

(B)

20.

(D)

DPP No. # 08 (JEE-ADVANCED) 1.

(D)

2.

(B)

3.

(D)

4.

(A)

5.

(C)

6.

(D)

7.

(i).(C)

(ii).

(B)

8.

(C)

9.

3

10.

32

11.

(B)

12.

(C)

13.

(D)

14.

(B)

1.

Correct IUPAC name of the compound

(A*)2-Methylbutenedioic anhydride (C) 2-Methyl-1,4-diketobutene epoxy

;kSfxd

(B) 3-Methylbutenedioic anhydride (D) 2-Methylcyclopentanoxy-1,4-dione

dk lgh IUPAC uke gksxk

(A*) 2-esfFkyC;wVhuMkbZvksbd ,ugkbMªkbM (C) 2-esfFky-1,4-MkbZdhVksC;wVhu bikWDlh 2.

is

(B) 3-esfFkyC;wVhuMkbZvksbd ,ugkbMªkbM (D) 2-esfFkylkbDyksiUs VuksDlh-1,4-MkbZvkWu

Molecular formula C3H9N represents : (A) Only primary amine (B) Only secondary amine (C) Three primary amine, two secondary amine and one tertiary amine (D*) Two primary amine, one secondary amine and one tertiary amine Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 1

v.kqlw=k C3H9N iznf'kZr djrk gS % (A) dsoy izkFkfed ,ehu (B) dsoy f}rh;d ,ehu (C) rhu izkFkfed ,ehu, nks f}rh;d ,ehu rFkk ,d r`rh;d ,ehu (D*) nks izkFkfed ,ehu, ,d f}rh;d ,ehu rFkk ,d r`rh;d ,ehu 3.

In the given sequence of reactions which of the following is correct structure of compound A. fn;s x;s vfHkfØ;k Øe esa fuEu esa ls dkSulh ;kSfxd A dh lgh lajpuk gSA

(A)

(B)

(C*)

(D)

O

3 

Sol.

4.

Zn / H2O

Which of the following compound does not give iodoform reaction.

fuEu esa ls dkSu vk;ksMksQkeZ vfHkfØ;k ugh nsrk gS \ (A) CH3COCH3

(B) C2H5OH

(C) CH3CHO

(D*) PhCOPh

5.

What is the % s character in hybridisation of carbon when it exerts strongest –I effect ? tc dkcZu ijek.kq lcls izcy –I izHkko n'kkZrk gS rks blds ladj.k esa % s y{k.k D;k gksxk \ (A) 25% (B*) 50% (C) 75% (D) 100%

6.

Which of the following statement is true about resonance. (A) In resonating structure hybridisation of atom will be change. (B*) Cannonical structures are imaginary (C) Cannonical structure explains all features of a molecule (D) In resonating structures position of nuclei change.

vuqukn ds fo"k; esa fuEu esa ls dkSulk dFku lgh gS \ (A) vuquknh lajpuk esa] ijek.kq dk ladj.k ifjofrZr gks tkrk gSA (B*) dSuksfudy lajpuk,sa dkYifud gksrh gSA (C) dSuksfudy lajpuk fdlh v.kq ds leLr y{k.kksa dh O;k[;k djrh gSA (D) vuquknh lajpukvksa esa] ukfHkd dh fLFkfr ifjofrZr gksrh gSA 7.

Hyperconjugation is possible in :

fuEu esa ls fdl v.kq esa vfrla;qXeu lEHko gS \

(A)

Sol. Sol.

(B) CH2 = CH2

(C) C6H5 – CH = CH2 (D*) CH3 – CH2 – CH = CH2 CH3 – CH2 – CH = CH2 has two -hydrogen for hyperconjugation. CH3 – CH2 – CH = CH2 esa vfrla;qXeu ds fy, nks -gkbMªkstu ijek.kq mifLFkr gSA

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 2

8.

Select the correct stability order of the following carbanions.

fuEu dkcZ_.kk;uksa ds LFkkf;Ro dk lgh Øe D;k gS\ (i)

(ii)

(A*) (i) > (ii) > (iii) > (iv) (B) (ii) > (iii) > (i)> (iv) 9.

(iii)

(iv)

(C) (iii) > (iv) > (ii) > (i)

(D) (iv)> (ii) > (i) > (iii)

Which of the following is most stable carbocation intermediate.

fuEu esa ls lcls vf/kd LFkk;h e/;orhZ dkcZ/kuk;u dkSulk gSA (A*) Sol.

10.

11.

(B)

Carbocation stablized by + M effect. + M izHkko ds dkj.k dkcZ/kuk;u LFkk;h gksrk Ph–CH2–Br

(C)

(D)

gSA

Product , Product is ¼mRikn]

mRikn gS %½

(A) Ph–CH3

(B) Ph–CH2–Ph

(C*) Ph–CH2–CH2–Ph

(D) Ph–H

Which of the following is an electrophile ?

fuEu esa ls dkSulk ,d bysDVªkWuLusgh gS \ (A) H2O 12.

(B) NH3

 (C) NH 4

(D*) AlCl3

 (C*) H3 O

(D) Cl

(C) NaNH2

(D*) All of these ¼mijksDr

 H (C*) C 2

(D)

Which of the following is not an electrophile ?

fuEu esa ls dkSulk ,d bysDVªkWuLusgh ugha gS \ (A) BH3 13.

 (B) NO 2

Which of the following is nucleophile ?

fuEu esa ls dkSulk ukfHkdLusgh gS \ (A) I– 14.

(B) CH3O–

lHkh½

Which of the following is not a nucleophile ?

fuEu esa ls dkSulk ukfHkdLusgh ugha gS \ (A) 15.

(B)

Which among the following species is an ambident nucleophile ?

fuEu esa ls dkSulh iztkfr mHk;/kehZ ukfHkdLusgh gSA (A) C6H5O–

(B)

(C*)

 H (D) C 2

Sol.

CN– is an ambident nucleophile because it has two lone pair donar atom (C and N). CN– mHk;/kehZ ukfHkdLusgh gS D;ksafd blesa nks ,dkdh bysDVªkWu ;qXe nkrk ijek.kq gSA (C rFkk N)

16.

Which of the following is best leaving group ?

fuEu esa ls dkSulk lcls vPNk fu"dklu lewg gS \

(A)

(B) CH3–O–

(C) OH–

(D*) CF3SO3–

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 3

17.

Identify correct order of leaving group ability for given leaving groups ?

fn;s x, fu"dklu lewgksa dh fu"dklu ;ksX;rk dk ?kVrk gqvk lgh Øe gS \

18.

(I) OH–

(II)

(III) CH3COO–

(IV) CH3SO3–

(A*) IV > III > II > I

(B) III > IV > II > I

(C) II > IV > III > I

(D) IV > II > III > I

Leaving group ability order amongst the following

fuEu dk fu"dklu ;ksX;rk dk lgh Øe gS

Hint :

(I) C6H5O–, (II) p-(CH3) C6H4O– (III) p-(OCH3)C6H4O– (IV) p-(NO2)C6H4O– (A) I > II > III > IV (B) III > II > I > IV (C*) IV > I > II > III (D) IV > III > II > I Conjugated base of strong acid is weak base and behave as better leaving group.

izcy vEy dk la;qXeh {kkj nqcZy {kkj gksrk gS rFkk nqcZy {kkj vPNs fu"dklu lewg gksrs gSaA 19.

Identify the correct set of aprotic solvent .

vizksfVd foyk;dksa dk lgh lewg igpkfu, (A) Water, DMSO ty] DMSO (C) Ethanol, Acetone ,sFksukWy] ,sflVksu esfFky,ehu 20.

(B*) DMSO, Acetone DMSO ,,sflVksu (D) Diethylether, Methyl amine MkbZ,sfFky

bZFkj ]

Identify the non polar solvents.

fuEu esa ls v/kzqoh; foyk;d gSA (A)

(B) CCl4

(C)

(D*) All of these ¼mijksDr

lHkh½

DPP No. # 08 (JEE-ADVANCED) Total Marks : 45

Max. Time : 30 min.

Single choice objective ('–1' negative marking) Q.1 to Q.6 (3 marks, 2 min.) Comprehension ('–1' negative marking) Q.7 (i to ii) (3 marks, 2 min.) Matching List Type (Only One options correct) ('–1' negative marking) Q.8 (3 marks, 2 min.) Integer Type Questions Q.9 to Q.10 (3 marks, 2 min.) ChemINFO : 4 Questions ('–1' negative marking) Q.11 to Q.14 (3 marks, 2 min.)

1.

[18, 12] [06, 04] [03, 02] [06, 04] [12, 08]

Phenyl magnesium bromide reacts with phenyl acetylene, which of the following is not expected as one of the product. (A) Benzene (B) Hydrocarbon (C) Grignard reagent (D*) Alkyne.

Qsfuy esXuhf'k;e czksekbM] Qsfuy ,lhfVyhu ds lkFk vfHkfØ;k djrk gS rks fuEu esa ls dkSulk mRikn ugha curk gS \ (A) csUthu (B) gkbkMªksdkcZu (C) fxzU;kj vfHkdeZd (D*) ,Ydkbu 2.

Which of the following compound gives methane on treatment with CH3MgI.fuEu

esa ls dkSulk ;kSfxd CH3MgI

ds lkFk fØ;k djds esFksu xSl nsrk gSA (A)

Sol.

(B*)

CH Mg

3   

(C)

(D)

+ CH4

Aromatic anion ¼,sjksesfVd

_.kk;u½

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 4

3.

The relative rates of following compounds towards nucleophilic addition reaction follow the order.

fuEufyf[kr ;kSfxdks esa ukfHkdLusgh ;ksxkRed vfHkfØ;k ds fy, vkisf{kd vfHkfØ;k nj dk Øe gksxk

Sol.

(A) I > II > III > IV (B) II > IV > I > III (C) III > IV > II > I (D*) IV > II > III > I Relative rate towards nucleophilic addition depends on following factors. A. Electronic factor i.e. more amount of positive charge on carbonyl carbon more reactive the carbonyl compound. B. Steric factor i.e. more the crowdness, less reactive the molecule.

vkisf{kd vfHkfØ;k dh nj ukfHkdLusgh ;ksxkRed vfHkfØ;k ij fuHkZj gS A. bys D Vª k W f ud iz H kko i.e. mu dkcks Z fuy ;kS fxdks a ]ftu ij /kukRed vkos ' k vf/kd gks mudh fØ;k'khyrk vf/kd gksrh gS A B. f=kfoe izHkko i.e. vf/kd f=kfoe ck/kk gks rks ;kSfxd dh fØ;k'khyrk de gks tkrh gSA 4.

If pentan-2-one is reduced with NaBH4 followed by hydrolysis with D2O, the product will be : ;fn isUVsu-2-vksu dk NaBH4 dh mifLFkfr esa vip;u ds ckn D2O }kjk ty vi?kVu ls izkIr mRikn gksxk : (A*) CH3–CH(OD)CH2CH2CH3 (B) CH3CD(OH)CH2CH2CH3 (C) CH3CH(OH)CH2CH2CH3 (D) CH3CD(OD)CH2CH2CH3

5.

W hich of the following give

when heated with propanone ?

(A) CH 3 MgI then D 2O

(B) Li Al H 4 then D 3O+

(C*) Li Al D 4 then H3O+

(D) Na BH 4 then D 3O+

izksisukWu dks fuEu esa ls fdlds lkFk xeZ djus ij

6.

(A) CH3 MgI rFkk D 2O

(B) Li Al H 4 rFkk D 3O +

(C*) Li Al D 4 rFkk H 3O+

(D) Na BH 4 rFkk D 3O+

Compound

;kSfxd

Sol.

curk gS \

can be prepared by

cuk;k tk ldrk gSA

(A)

(B)

(C) Ph – COCH3

(D*) PhCHO

(A)

(B)

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 5

(C)

(D)

7.

Comprehension The relative comparision of equilibrium constants of addition of RMgBr on four carbonyl compounds has been shown by the graph below. pkj dkcksZfuy ;kSfxdksa ij RMgBr ds ;ksx ls lkE; fu;rkdksa ds lkis{k rqyuk fuEu xzkQ }kjk iznf'kZr fd;k x;k gSA

(i).

Which compound corresponds to H if the four given compounds are : fn;s x;s pkj ;ksfxdksa esa ls dkSulk ;ksfxd H dks iznf'kZr djrk gSA (A) p – NO2C6H4CHO (B) C6H5CHO (C*) p – MeOC6H4CHO

(ii).

Sol.

8.

Which compound corresponds to G if the four given compounds are : fn;s x;s pkj ;ksfxdksa esa ls dkSulk ;kSfxd G dks iznf'kZr djrk gS (A) CH3CHO (B*) C6H5COCH3 (C) C6H5CHO Compounds Keq CH3CHO very large p-NO2C6H4CHO 1420 C6H5CHO 210 p-MeOC6H4CHO 32 CH3COCH2CH3 38 C6H5COCH3 0.8 C6H5COC6H5 very small

H2O

+ CH3MgCl 

(Q)

H2O  

(R) HO

+ NaBH4 2

(S) Codes (A) (C*)

Ans.

(D) C6H5COC6H5

Match List I (Reaction) with List II (Product) and select the correct answer using the code given below the lists : lwph I (vfHkfØ;k) dks lwph II ¼mRikn½ ls lqesfyr dhft, rFkk lwfp;ksa ds uhps fn;s x;s dksM dk iz;ksx djds lgh mÙkj pqfu;s: List I lwph I List II lwph II HO (P) CH3–CH2–CHO + CH3MgCl 2

9.

(D) CH3CH2CHO

(1) CH3–CH2–CH2–CH2–OH (2)

(3)

(4)

¼dksM½ : P 4 2

Q 2 4

R 1 1

S 3 3

(B) (D)

P 3 1

Q 2 4

R 4 2

S 1 3

How many carbonyl compounds will give secondary alcohol with molecular formula C 5H12O after reaction with LiAlH4. v.kwlw=k C5H12O okys fdrus dkcksZfuy ;kSfxd LiAlH4 ds lkFk vfHkfØ;k dj f}rh;d ,YdksgkWy nsaxsA 3

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 6

Sol.

(i)

(ii)

10.

 No. of fractions (Y) CH3  C  C  CH3     2 No. of product (X)     ( excess ) || || O O

PhMgBr

(iii)

H O

Fractional distillation

Report your answer as XY PhMgBr

H O

CH3  C  C  CH3    2 ( excess ) || || O O

mRiknksa dh la[;k (X) 

izHkktksa dh la[;k (Y)

viuk mÙkj XY ds :i esa nhft,sA Ans.

Sol.

32

,

,

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 7

ChemINFO-1.4

Oxidation Oxidative cleavage of C=C/CC

Daily Self-Study Dosage for mastering Chemistry

KMnO4 and K2Cr2O7 are amongst the strongest oxidising agent. They completely rupture the C=C or CC and form oxidised products as Acids and ketones depending on structure of C=C and CC. Observe the following oxidative cleavage of C=C and CC by oxidising agents (O.A) as (a) KMnO4 (b) K2Cr2O7 or CrO3/H (c) O3/H2O or O3/H2O2 Examples :

General Reactions : O.A .  

O.A .  

O.A .



O.A .  

O. A .



+ CO2+ H2O

O.A .  

O. A .

 2RCOOH

O.A .



+ H2 O

+ CO2+ H2O

O. A .   2CH COOH 3

+ CO2 + H2O

O. A .



+ CO2+H2O

Note : 1. Alkene & alkyne having = CH2 &  C–H will give CO2 as one of the oxidation product by oxidising agent KMnO 4 and K2Cr2O7 or CrO3/H+ . O.A.

2. HCOOH(Formic Acid) or H2C2O4(Oxalic Acid)  CO2 + H2O. Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. K 2 Cr2 O 7   P ; Products is :

11.

(A) CH3 – CH2 – CHO 12.

(B*)

(C) CH3–CH2–COOH

(D) CH3–COOH

Which Product obtained by the following reaction. 

KMnO4 / H  

(A)

13.

(B)

(D)

KMnO / H

A 4 CH3COOH + CH3CH2COOH. The Alkyne 'A' can be : (A) CH3–CH=CH–CH3 (C) CH3–CH=CH–CH2–CH3

14.

(C*)

(B) CH3–C  C–CH3 (D*) CH3–C  C–CH2–CH3

From which of the following alkene, CO 2 is one of the oxidation product by KMnO 4 / K2Cr2O7 /H+. (A) CH3–CH=CH–CH3 (B*) CH3–CH=CH2 (C)

(D)

Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 8

ChemINFO-1.4

Oxidation C=C/CC

Daily Self-Study Dosage for mastering Chemistry

dk vkWDlhdkjh fo[k.Mu

KMnO4 rFkk K2Cr2O7 izcyre

vkWDlhdkjd vfHkdeZd gksrs gSaA ;s C=C ;k CC dks iw.kZ:i ls rksM+rs gS rFkk C=C ;k CC dh lajpuk ds vkèkkj ij vkWDlhd`r mRikn tSls vEy rFkk dhVksu cukrs gSA vkWDlhdkjh vfHkdeZd (O.A) }kjk C=C ;k CC dk vkWDlhdkjh fo[k.Mu fuEu vkWDlhdkjh inkFkZ }kjk gksrk gSA (b) K2Cr2O7 or CrO3/H

(a) KMnO4

(c) O3/H2O or O3/H2O2 Examples :

General Reactions : O.A .  

O.A .  

O.A .



O.A .  

O. A .



+ CO2+ H2O

O.A .  

O. A .

 2RCOOH

O.A .



+ H2 O

+ CO2+ H2O

O. A .   2CH COOH 3

+ CO2 + H2O

O. A .



+ CO2+H2O

uksV : 1. ,Ydhu o ,YdkbZu tks = CH2 ;k  C–H j[krs gS tks vkWDlhdkjh vfHkdeZd KMnO4 rFkk K2Cr2O7 ;k CrO3/H+ ds }kjk CO2 ds :i esa ,d vkWDlhdkjh mRikn nsrs gSA O.A. 2. HCOOH (QkfeZd vEy) ;k H2C2O4(vkDtsfyd vEy)  CO2H2O Memorize this theory as soon as you get the DPP. Revise it regularly and master this concept by practice. K 2 Cr2 O 7   P ;

11.

(A) CH3 – CH2 – CHO 12.

mRikn gS :

(B*)

(C) CH3–CH2–COOH

(D) CH3–COOH

fuEu vfHkfØ;k esa dkSu lk mRikn izkIr gksrk gS \ KMnO / H

 4

(A) 13.

14.

(B) KMnO / H

A 4 CH3COOH + CH3CH2COOH.

(C*)

(D)

,YdkbZu 'A' gks ldrh gS %

(A) CH3–CH=CH–CH3

(B) CH3–C  C–CH3

(C) CH3–CH=CH–CH2–CH3

(D*) CH3–C  C–CH2–CH3

fuEu esa ls dkSulh ,Ydhu KMnO4 / K2Cr2O7 /H+ ds lkFk vfHkfØ;k djds vkWDlhdkjh mRikn ds :i esa CO2 nsrh gSA (A) CH3–CH=CH–CH3

(B*) CH3–CH=CH2

(C)

(D) Corporate Office : CG Tower, A-46 & 52, IPIA, Near City Mall, Jhalawar Road, Kota (Raj.)-324005 Website : www.resonance.ac.in | E-mail : [email protected] Toll Free : 1800 200 2244 | 1800 258 5555| CIN: U80302RJ2007PTC024029

PAGE NO.- 9