Isomerism Alkyl Halide - Dr. Sangeeta · PDF fileIsomerism Alkyl Halide ......
Transcript of Isomerism Alkyl Halide - Dr. Sangeeta · PDF fileIsomerism Alkyl Halide ......
Dr. Sangeeta Khanna Ph.D 1 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
Isomerism
Alkyl Halide
Method of Preparation of Organic Halide
Nucleophilic Substitution
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 2 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
Test Date – 30.7.2016 Topic: Isomerism & Alkyl Halide
READ INSTRUCTIONS CAREFULLY 1. The test is of 2 hour duration. 2. The maximum marks are 311. 3. This test consists of 75 questions. 4. Keep your mobiles switched off during Test in the Halls.
(Single Correct Choice Type) Negative Marking [-1]
This Section contains 45 multiple choice questions. Each question has four choices A), B), C) and D) out
of which ONLY ONE is correct. Marks: 45 × 4 = 180
1. The configuration of the chiral centre and the geometry of the double bond in the following molecule can be described by
a. R and E b. S and E c. R and Z d. S and Z
C
2. XCHCHNaSCH 23
Θ
3
The reaction is fastest when X is:
a. – OH b. – F c. – I d. 3
| |
CH
O
CO
C
3. Given the number of N and M? a. 6, 6 b. 6, 4 c. 4, 4 d. 7, 3 B Sol.
F
C
C
OH HOOC
C
Me Me
H
CH3
CH3
H3C
Cl2, h
CH2Cl
CH3
H3C
(i) and (ii) (+) and (-)-isomer
CH3
CH3
H3C +
Cl
+
CH3
H3C Cl
CH3
+
(iii)
CH3
CH2Cl
H3C
(iv) and (v) (+) and (-) isomer
(vi)
CH3
CH3
H3C
Cl2, h C5H11Cl (N isomeric products)
Fractional
distillation M (isomeric products)
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 3 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
Thus, a total of six isomers will be formed. Since enantiomers have same b.p., they cannot be separated by fractional distillation, i.e., (i) and (ii) will distil together and similarly (iv) and (v) will distil together. Thus, out of six isomers formed, four can be separated by fractional distillation
4. Rate of reaction with aqueous ethanol follows the order: a. P > Q > S > R b. Q > P > R > S c. P > R > Q > S d. R > P > S > Q A 5. The reaction proceed by the mechanism:
a. SN1 b. SN2 c. SE2 d. SNi B
6. Consider the following nucleophiles
IVIIIIII
Θ| |
3
ΘΘ| |
| |
Θ| |
| |3 O
O
CCH ,O-Ph ,O
O
O
SPh ,O
O
O
SCF
when attached to sp3 hybridized carbon, their leaving group ability in nucleophilic substitution reactions decreases in the order: a. I > II > III > IV b. I > II > IV > III c. IV > I > II > III d. IV > III > II > I
B 7. Find the product of the following reaction: a. b. c. d. B 8. The rate of SN1 reaction is fastest with :
a. b. c. d. A
O – CH2 – Br
(P)
CH3 Br
(Q) (R)
Br
(S)
Br
OH
+ SOCl2
N H
+ SO2 + HCl
H Cl
Br CH3OH
OCH3
OCH3
CH
Br
CH
Br
CH
Br
NO2 CH2 Br
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 4 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
9. The order of decreasing nucleophilicities of the following species is :
a. Θ
3
Θ
3
Θ
3 OCOCHOCHSCH b. Θ
3
Θ
3
Θ
3 OCHSCHOCOCH
c. Θ
3
Θ
3
Θ
3 OCHOCOCHSCH d. Θ
3
Θ
3
Θ
3 SCHOCOCHOCH
A
10. Arrange
Increasing order of reactivity in SN1 reactions a. I < II < III < IV b. I < II < IV < III c. IV < II < III < I d. I < III < II < IV
A Sol. First compound is a secondary halide and the other three are tertiary halides. Also the reactivity follows the
order: chloride < bromide < iodide. Thus the reactivity order is I < II < III < IV. 11. A dextrorotatory optically active alkyl halide undergoes hydrolysis by SN2 mechanism. The resulting
alcohol is
a. dextrorotatory b. laevorotary c. optically inactive due to racemization d. may be dextro-or laevorotatory D
Sol. SN2 reactions occur with inversion of configuration. Therefore, an optically active reactant gives an optically active product whose sign of rotation cannot be predicted.
12. CH3 – Br + Nu– CH3 – Nu + Br – The decreasing order of the rate of the above reaction with nucleophiles (Nu–) A to D is
[Nu– = (A) PhO–, (B) CH3COO–, (C) HO–, (D) CH3O–]
a. D > C > A > B b. D > C > B > A c. A > B > C > D d. B > D > C > A A
Sol. PhO– is less resonance stabilized than AcO– ion and hence is a stronger nucleophile than AcO–. Therefore, the overall nucleophilicity decreases in the order: CH3O
– (D) > HO– (C) > PhO– (A) > AcO– (B)
13. The SN1 reactivity of the following halides will be in the order (i) (CH3)3CBr (ii) (C6H5)2CHBr (iii) (C6H5)2C(CH3)Br (iv) (CH3)2CHBr (v) C2H5Br a. (v) > (iv) > (i) > (ii) > (iii) b. (ii) > (i) > (iii) > (v) > (iv) c. (i) > (iii) > (v) > (ii) > (iv) d. (iii) > (ii) > (i) > (iv) > (v) D
Sol. More stable the carbocation, more reactive is the alkyl halide in SN1 reactions. Since the stability of the carbocations decreasing in the order:
5223332563256 HCHC)CH(C)CH(HC)HC()CH(C)HC(
, therefore, reactivity of the alkyl
halides decreases in the order: (iii) > (ii) > (i) > (iv) > (v) 14. Of the following alkyl halide, one with the lowest boiling point is
a. Ethyl bromide b. Isopropyl bromide c. n-Butyl bromide d. Methyl bromide D
Sol. For the same halogen, smaller the size of the alkyl group, weaker the van der Waals forces of attraction and hence lower is the b.p. Therefore, option
15. SN1 reactions are favoured by
a. non-polar solvents b. bulky groups on the carbon atom attached to the halogen atom
Cl
CH3
(II),
CH3
(III), Br
Cl
(I),
CH3
(IV) in I
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 5 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
c. small groups on carbon atom attached to the halogen atom d. Electron withdrawing group on substrate B
Sol. Bulky groups o the atom attached to the halogen atom will hinder attack by SN2 mechanism due to steric hindrance and hence would favour SN1 mechanism.
16. Neopentyl alcohol on treatment with HBr gives
a. Neopentyl bromide b. 2-Bromo-2-methybutane c. 2-Methyl-2-butene d. 2-Methyl-1-butene B
Sol.
etanmethy lbu2Bromo2
32
3|
|3
Br32
3|
3shif t Methy l
2,12
3
2
|
|3
O2H
H
alcohol Neopenty l
2
3
3
|
|3 CHCH
CH
Br
CCHCHCH
CH
CCHHC
CH
CH
CCHOHCH
CH
CH
CCH
17. Find the major product of the following reaction: a. b. c. d. D 18. Find the product of the following reaction:
a. b. c. d.
A
Sol. In presence of neutral nucleophile, mechanism is SN1
19. Which of the following is least reactive towards nucleophilic displacement reaction when treated with
aqueous KOH?
a. 2, 4, 6-Trinitrochlorobenzene b. 2, 4-Dinitrochlorobenzene
c. 4-Nitrochlorobenzene d. 3-Nitrochlorobenzene
D
Sol. Cl is activated towards nucleophilic substitution reaction by presence of NO2 groups at o- and p-
positions and not at m-position.
20. A compound has vapour density 29. On warming with a solution PCl5, it gives non terminal gem
dihalides the compound is:
a. CH3CH2CHO b. CH3COCH3 c. CH3CHOHCH3 d. CH2 = CHCH2Br.
B
Sol. If V.D. is 29, the Mol. wt, is 2 × 29 = 58. Thus, the compound in CH3COCH3.
21. 1, 4-Dibromobutane (0.1 mole) is treated with Na2S (0.1 mole) is aqueous ethanol, the product formed
is
a. BrCH2CH2CH2CH2SH b. HSCH2CH2CH2CH2SH
CH2 – Br KSH
Br
CH2 – Br
SH
S
CH2 – SH
SH
CH2 – SH
Br
Cl
H2O
OH
OH
OH
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 6 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
c. BrCH2CH2CH2S
–Na+ d.
D
Sol.
22. In the reaction,
252Ni
2HKCN NHHCBA
a. A is CH3I b. B is CH3NC c. A is C2H5I d. B is C2H5NC A
Sol. 252]H[
)B(3
KCN
)A(3 NHHCCNCHICH
23. In dinitrochlorobenzene, chlorine is readily replaced. This is because
a. NO2 makes the ring electron rich at ortho and para positions b. NO2 withdraws e– from meta-position c. NO2 donates e– at m-position
d. NO2 withdraws e– from ortho/para-positions D
Sol. NO2 group withdraws electrons from o- and p-positions and hence activates the Cl towards nucleophilic substitution reactions
24. The correct increasing order of reactivity of halides for SN1 reaction is
a. CH3CH2X < (CH3)2CHX < CH2 = CHCH2X b. (CH3)2CHX < CH3CH2X < CH2 = CHCH2X c. (CH3)2CHX < CH2 = CHCH2X < CH3CH2X d. CH2 = CHCH2X < (CH3)2CHX < CH3CH2X A
Sol. The correct increasing order of reactivity follows the sequence: 1° < 2° < allyl < benzyl
25. Rate of SN1 attack will be
a. b. c. d.
B 26. Which statement is not correct for nucleophilic substitution reaction?
a. Nucleophilic substitution reaction is mainly of two types: unimolecular or bimolecular b. Nucleophilicity of incoming group should be more than the nucleophilicity of leaving group c. Substitution reaction is carried out in the presence of polar solvent d. Aryl SN2 reaction takes place via formation of carbocation D
S
Br – CH2
CH2
CH2 – Br
CH2
Na2S
-NaBr Br – CH2
CH2 CH2
CH2 -NaBr
Na+ - S
CH2 CH2
CH2 CH2
S
Cl
<
Cl
<
Cl
Cl
<
Cl
<
Cl
Cl
<
Cl
<
Cl
Cl
<
Cl
<
Cl
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 7 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
27. ACH
CH
CCHKCN )2(
HBr )1(2
3|
3 . Product ‘A’ is
a. 2, 2-Dimethyl propane nitrile b. 3-Methylbutane nitrile c. 2-Methyl propane isocyanide d. 2- Methyl propane nitrile A
Sol. 3
3|
|33
3|
|3 CH
CH
CN
CCH ,CH
CH
Br
CCH
28. In the given reaction:
[X] will be: a. b.
c. d. C
Sol. 29. The structure of the major product formed in the following reaction is a. b. c. d. D Sol. Aralkyl halides are more reactive than aryl halides, therefore, only the halogen in the side chain is
displaced.
30. CH3CH2CH2 – I NaCN
CH3CH2CH2CN
Which of the following solvent can be used in this reaction
a. Hexane b. H2O c. alcohol d. 3
3
CH
CHN
O
CH /\||
l CH3 HOH
[X]
OH CH3 CH3
OH
CH3
Cl
I
NaCN
DMF
CN
CN
Cl
I
NC
CN
Cl CN
I
I
CH2Cl NaCN
DMF
I
CH2CN
H3C H3C
OH and CH3 CH3
OH
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 8 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
D Sol. It is SN2 & aprotic polar solvent will favour. 31. Arrange the following in decreasing order of reactivity towards SN1 reaction. a. P > Q > R > S b. S > R > P > Q c. Q > S > P > R d. Q > S > R > P C 32. Which of the SN2 reaction is fastest?
a. CH3Br + HC C– CH3C CH + Br–
b. CH3Br + HC CH CH3C CH + HBr
c. CH3CH2Br + HC C– CH3CH2C CH + Br–
d. CH3CH2Br + HC CH CH3CH2C CH + Br–
A 33. What is the correct increasing order of reactivity of the following in the SN2 reaction?
I. (CH3)2CHCH2Cl II. CH3CH2CH(Cl)CH3
III. C6H5Cl IV. p – O2N – C6H4 – CH2Cl
a. I < II < III < IV b. III < II < I < IV c. II < III < IV < I d. III < I < II < IV D
34. Which of the following are the examples of strong nucleophiles but weak base in protic solvents? I. CH3S
– II. CH3O– III. I– IV. H2O V. F–
a. Both I and IV b. Both IV and V c. Both I and III d. Both III and IV C
35. The specific rotation of optically pure (R)-2-butanol is – 13.52° at 25°C. An optically pure sample of (R)-2—bromobutane was treated with aqueous NaOH in order to form 2-butanol via SN2 reaction. What would be the specific rotation of the product assuming 100% yield?
a. 0° b. – 13.52° c. + 13.52° c. None of these C
36. Which of the following reaction cannot be used to prepare arylhalide;
a. b. c. d.
B 37. Cyanide anion has two atoms that have lone pair of electrons. Either could act as nucleophile in the
reaction. Yet in the vast majority of the cases, carbon acts as nucleophile and forms a bond to the
substrate, why?
a. Nitrogen is strongly solvated in hydroxylic solvents, it will be less reactive
b. Carbon is more reactive in polar aprotic solvents, normally used for these SN2 reactions
c. Carbon is less electronegative than N
d. The lone pair of electrons on C is in sp hybrid orbital
C
Br
CH2 – Br
Br
Br
(P) (Q) (R) (S)
N2Cl KI + I2
+ Cl2 FeCl3
+ ICl
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 9 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
38. When CH3CH2Br is treated with aqueous NaCN, CH3CH2CN is formed while CH3CH2NC is formed
when AgCN is used in the similar reaction. it is due to
a. negative charge on carbon atom in NaCN
b. covalent nature of Ag C bond in AgCN
c. covalent nature of C – N bond in NaCN
d. ionic nature of Ag – C bond in AgCN
A,B
39. Which one of the two iodine will be more reactive in the SN1 and SN2 reactions?
a. A will be faster in SN1 reaction but slower in SN2
b. A will be a faster both in SN1 and SN2 reactions
c. A and B will be equally reactive
d. B will be faster in both SN1 and SN2 reactions
B
40. In SN1 reaction, racemisation occurs if the reaction occurs at a stereogenic centre, however, 50:50
mixture of enantiomers are rarely obtained, why?
a. Usually one enantiomer is more stable than other
b. Retention of configuration is always favoured in reaction
c. Hydroxylic solvent favour retention of configuration
d. There is steric hindrance to the approach of nucleophile from the front side as some of the leaving
group are not departed completely which favour inversion of configuration more than retention
D
41. Pick out the most reactive alkyl halide for an SN1 reaction.
a. b.
c. d.
A
42. Pick out the compound which reacts fastest in the presence of AgNO3.
a. (CH3)3CCl b. (CH3)2CHCH2Cl c. (CH3)2CHCl d. CH3CH2Cl A
43. Which of the following statement(s) is/are true for SN1 reaction?
I. The rate of SN1 reaction depends on concentration of alkyl halide
II. The rate of SN1 reaction depends on concentration of nucleophile
III. SN1 reaction of alkyl halides are favoured by non-polar solvents
a. Only I b. Only II c. Only III d. Both I and III A
I I
A
B
Ph
Br
Ph
Br
Br
Ph
Ph
Br
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 10 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
44. Pick out the strongest substrate(s) for a SN1 reaction. a. Only I b. Only II c. Only III d. Only IV
D 45. Pick out the following factor(s) which promote a SN1 reaction:
I. Temperature II. Concentration of nucleophile III. Concentration of alkyl halide IV. Aprotic solvents a. Both I and II b. Both II and III c. Both III and IV d. Both I and III D
SECTION – B (ASSERTION & REASON)
This Section contains 5 multiple choice questions. Each question has four choices A), B), C) and D)
out of which ONLY ONE is correct. (5 × 4 = 20 Marks)
(a) Both Statement I and Statement II are correct and Statement II is the correct explanation of
Statement I (b) Both Statement I and Statement II are correct but Statement II is not the correct explanation of
Statement I (c) Statement I is correct but Statement II is incorrect (d) Statement II is correct but Statement I is incorrect
1. Consider the following two bromide I and II, undergoing solvolysis reaction in boiling ethanol:
Statement I A is less reactive than B in the given solvolysis reaction. Statement II Carbocation of (B) is more stable due to methyl shift. a. (a) b. (b) c. (c) d. (d) A
2. Statement I When 3-bromo propene, which contain a labelled 13C at C-1 position is refluxed with methanol, following products were obtained.
32
13
232
13
2Boil
OHCH2
13
2 OCHHCCHCHOCHCHCHCHBrCHCHCH 3
Statement II Methanol has an acidic proton bonded to oxygen. a. (a) b. (b) c. (c) d. (d) B
3. Statement – I: Optically active 2-iodobutane on treatment with NaI in acetone undergoes racemization Statement – II: Repeated Walden inversions on the reactant and its products eventually gives a racemic mixture.
a. (a) b. (c) c. (c) d. (d)
Br
C2H5
I.
C2H5 Br
II.
Br
C6H5
III.
IV. C6H5
Br
CH3 H3C
H3C
CH3
Br Ethanol
Product A.
B.
Product Ethanol
CH3 H3C
CH3
Br
CH3
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 11 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
A Sol. Statement – II is the correct explanation of Statement – I.
)(
32
|
|3
)(32
|
3 CHCH
H
CCH -
CHCHHCCH
I
II
4. Statement – I: SN2 reaction of an optically active alkyl halide with an aqueous solution of KOH always gives an alcohol with opposite sign of rotation. Statement – II: SN2 reactions always proceed with inversion of configuration.
a. (a) b. (c) c. (c) d. (d) D
Sol. Correct Assertion: SN2 reaction of an optically active halide with aqueous solution of KOH always gives an alcohol which is also optically active but its rotation can be +ve or –ve.
5. Statement – I : A Bulky anion is a good nucleophile Statement – II: Steric hindrance oppose nucleophilic attack. a. (a) b. (c) c. (c) d. (d) D
SECTION – C (Paragraph Type)
This Section contains 3 paragraph. Each of these questions has four choices A), B), C) and D) out of which ONLY ONE is correct. 10 × 4 = 40 Marks
Paragraph – 1 Optical isomerism arises due to chirality of molecules. Chiral molecules may or may not contain chiral carbons. If a molecule contains only one chiral carbon atom, it exist in two stereoisomers which are always optically active. If, however, the molecule contains two chiral carbon atoms, the number of stereoisomers increases to either 3 or 4 depending up whether the two chiral carbon atoms are similar or dissimilar. The number of stereoisomers which are non-superimposable mirror images of each other are called enantiomers while those which are not mirror images of each other are called diatereomers. Enantiomers are always chiral molecules whereas diastereomers may or may not be.
1. Which of the following compounds can have non-superimposable mirror image?
a. b. c. d.
D
Sol. a, b & c contain plane of symmetry and are optically inactive. So, they form superimposable mirror image.
2. Which of the following are optically inactive?
a. b. c. d.
D Sol.
OH
OH
OH
H
H
H
CH3
CH3
OH
OH
H
H
H
CH3
CH3
HO
OH
OH
H
H
H
CH3
CH3
Br
OH H
H
CH3
CH3
H
HO
HO
Cl
H
C = C = C = C = C
C2H5
H
H3C
H
H
CH3
CH3 – CH – COOH
D
C = C = C = C
H3C
H
CH3
H
Cl
H
C = C = C = C = C
C2H5
H
Optically active
;
H
CH3
H3C
H
; H3C – CH – COOH
D Optically active
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 12 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
‘d’ will show geometrical Isomerism
3. How many optically active stereoisomers are possible for butane-2, 3-diol?
a. 1 b. 2 c. 3 d. 4
B
Sol. Butane-2, 3-diol (CH3 - 3CHHOHCHOHC
) has two similar chiral carbon atoms. Therefore, like
tartaric acid it has three stereisomeric forms (d-, - and meso). Out of these only two (d- and ) are
optically active
4. The number of enantiomers of the compound CH3CHBrCHBrCOOH is
a. 0 b. 1 c. 3 d. 4
D
Sol. The compound HBrCOOHCHBrCCH3
contains two dissimilar chiral carbon atoms and hence 22 = 4
enantiomers are possible. Passage – 2
In a SN2 reaction, back side attack takes place which leads to inversion of configuration at -carbon which is known as Walden’s inversion:
5. What is true regarding the following reaction? a. The product would be a dextro isomer
b. The product would be a laevo isomer c. The product would be a racemic mixture d. The product would be either laevo or dextro isomer D
6. What can be correctly predicted regarding product of the following reaction?
a. The product would be achiral
b. The product would be optically active c. The product would be racemic mixture d. The product would contain unequal amount of enantiomers A
CN–
+ C X NC - - - - C - - - - X
–
–
NC C + X–
Transition state
CH3
H Br
C2H5
(laevo)
+ H C C–Na
+ H C C H + NaBr
CH3
C2H5
H
Cl
CH3
(1.0 mol)
+ NaCN(aq) (1.0 mol)
H
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 13 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
7. What can be predicted correctly about product of the following SN2 reaction? a. Product would be enantiomer of starting compound
b. Product would be a pure enantiomer c. Product would be a racemic mixture d. Product would be achiral
C Passage – 3
Nucleophilic aliphatic substitution reaction proceeds either by SN1 mechanism or by SN2 mechanism. The SN1 is a two step unimolecular reaction in which a carbocation is first formed as an intermediate in a slow rate determining step followed by an attack by a nucleophile in a fast step. Since carbocation is planar, the attack by a nucleophile can take place from either side of the plane. An optical active substrate will give rise to the formation of both (+) and (-) forms of the product. In most of the cases, the product usually coexists of 5-20% inverted product and 80-95% racemic mixture. The more stable the carbocation, the greater is the proportion of racemisation.
8. Which of the following compound will give SN1 reaction?
(I) C6H5 – CH2 – Br (II) CH2 = CH – CH2 – Br (III) CH3 – CH2 – Br (IV) BrCH
CH
CH
CCH 2
3
3
|
|3
Select the correct answer from the codes given below: a. (I), (II) and (III) b. (I), (II) and (IV) c. (II), (III) and (IV) d. (I), (III) and (IV) B
9. Which one of the following substrates will give maximum racemisation in the following reaction?
OH
R
R
CRX
R
R
CR|
|HX
O2H|
|
a. 32
3
3
|
|56 CHCH
CH
CH
CHC b. Br
CH
HC
CCHCH
3
52
|
|2
c. d. C Sol. It will form most stable carbocation (a) X = CH3 No leaving group
CH3
CH3
Br H
H OH + NaOH (aq)
C6H5 C
Br
CH3
OCH3 C6H5 C
Br
NH3
NO2
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 14 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
10. Consider the following reactions:
(I) OHH
CH
CHCClH
CH
CHC)(
3
|56
water%5
Acetone%95
3
|56
(II) OHH
CH
CHCClH
CH
CHC)(
3
|56
water%10
Acetone%90
3
|56
(III) OHH
CH
CHCClH
CH
CHC)(
3
|56
water%20
Acetone%80
3
|56
(IV) OHH
CH
CHCClH
CH
CHC)(
3
|56
water%100
3
|56
The correct decreasing order of proportion of inverted product would be a. (I) > (II) > (III) > (IV) b. (II) > (I) > (III) > (IV) c. (III) > (II) > (I) > (IV) d. (IV) > (III) > (II) > (I) A
Sol. It will form most stable carbocation
SECTION – D (More than One Answer Type) No Negative Marking
This Section contains 8 multiple choice questions. Each question has four choices A), B), C) and D) out of which One or More than one answer may be correct. 8 × 5 = 40
1. Which of the following compounds will not give SN2 reaction?
a. b. H2C = CH - Cl c. d.
A, B, C, D 2. Which of the following compounds will give SN1 reaction?
a. H3C – CH2 – Br b. c. ClCHOCH 23
d.
C, D 3. Which of the following cannot be prepared by direct halogenation of benzene with X2 in presence of
Lewis Acid
a. Iodobenzene b. Chlorobenzene c. Bromobenzene d. Fluorobenzene A,D
4. Which of the following statements regarding the SN1 reaction shown by alkyl halide is correct?
a. The added nucleophile plays no kinetic role in SN1 reaction. b. The SN1 reaction involves only inversion of configuration of the optically active substrate c. The SN1 reaction on the chiral starting material ends up with racemization of the product. d. The more stable the carbocation intermediate faster is the SN1 reaction A,C,D
Sol.SN1 reaction involve racemisation of configuration. 5. Which reaction results in the formation of a pair of enantiomers?
a. b. c. d. B,D
Cl
Br Br
Br
Ph
I
CH2CH3
H3C
Br H
I
Acetone
Br H
H2O
Br H
CH3OH OH H
HBr
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 15 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
6. The correct statement(s) concerning the structure E, F and G is (are)
a. E, F and G are resonance structures b. E, F and E, G are tautomers c. F and G are geometrical isomers d. F and G are diastereomers
B,C, D 7. In which of the following reaction(s), the major product is a primary alkyl bromide?
a. b.
c. d.
B,C,D Sol. (A)
(B)
(C)
(D)
8. In which of the following pair(s), the first one is stronger nucleophile than the second one?
a. CH3O–, (CH3)3CO– b. (CH3)2N
–, 2NH
c. C6H5O–, (CH3)2CHO– d. HS–, HO–
A,D
CH3
O
H3C
H3C
(E)
CH3
OH
H3C
H3C
(F)
CH3
OH H3C
H3C
(G)
+ Br2 h COOAg Br2/CCl4
CH2
+ HBr (C6H5CO)2O2 OH
PBr3
+ Br2 h
Br
Secondary (major)
COOAg + Br2
Br
Primary + AgBr + CO2
CH2
+ HBr (C6H5CO)2O2
Br
Anti-Markownikoff’s addition (primary)
OH PBr3 Br
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 16 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
SECTION – E (Matrix Type) No Negative Marking
This Section contains 2 questions. Each question has four choices (A, B, C and D) given in Column I and five statements (p, q, r, and s) in Column II. Any given statement in Column I can have correct matching with one or more statement(s) given in Column II. 8 × 2 = 16 Marks
1. Match the Column – I with Column – II.
Column – I Column – II
A.
(P) Plane of symmetry
B.
(Q)
Optically inactive
C.
(R) Optically active
D.
(S)
Geometrical isomerism
Sol. A r; B p, q; C p, q, s; D p, q 2. Match the Column – I with Column – II.
Column – I Column – II
A.
dark
HBr2
|
|
3 CH
CH
CH
CH
CCH
3
3
(P)
Carbocation Intermediate
B.
(Q)
Free Radical Reaction
C.
(R)
Rearrangement
D.
H
HOH2
|
|56 CHCH
CH
HC
CHC
3
52
(S)
Markovnikoff’s addition
Sol. A P,S; B q; C p; D p, r, s
C = C = C
HO
CH3
Br
Cl
CH3
CH3
Br
OH
H3C
Cl C = C = C = C
CH3
Cl
H
H
Cl
CH3
Cl
H3C
CH = CH – CH3 HBr
Peroxide
CH3 Cl2
FeCl3
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 17 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
SECTION – F (Integer Type) No Negative Marking
This Section contains 5 questions. The answer to each question is a single digit integer ranging from 0 to 10. 5 × 3 = 15
1. Consider the following solvolysis reaction.
Heat
5232
|
|
|
|56 OHHCCHCH
H
Br
C
CH
CH
CHC
3
3
How many principal products would be formed by SN1 reaction? Sol. 6 Intermediate are
52
|
|HCHC
Me
Me
CPh
and
Isomer 4
52|
|
|
Nu52
|
|
HCH
Me
C
Me
Nu
CPhHCH
Me
C
Me
CPh
2. How many of the following undergo solvolysis reaction faster than benzyl chloride? Sol. 5 [iii, iv, v, vi, vii] 3. If 2, 4-dimethyl pentane is subjected to free radical chlorination reaction, how many different
monochlorinated products would be formed? Sol. 4
CHCl2 CHCl2
Cl
CH2Cl
H3C
i.
ii.
iii.
CH2Cl
MeO
CH2Cl ClCH2
OCOCH3
iv.
v.
vi.
CH3 Cl
CH3 O CH2Cl CH2Cl
O2N
vii.
viii.
ix.
CH2Cl
ClH2C
x.
+ Cl2 hv
Cl
(±) +
Cl
+
Cl
Dr. Sangeeta Khanna Ph.D
Dr. Sangeeta Khanna Ph.D 18 CHEMISTRY COACHING CIRCLE D:\Important Data\2016\+2\Org\Test\Grand Test\+2 Grand test-IV.doc
4. If 1,3-butadiene is treated with excess of bromine in CCl4, how many different tetrabromides would be formed?
Sol. 3 5. Consider the following reaction, How many different monobromo derivatives would be produced? Sol. 4
(I) has two optically active enantiomers and (II) has two geometrical isomers.
H3C CH2
H3C CH2
C = CH2 + N Br
O
O
hv
CCl4
CH3 CH2
CH3 CH2
C = CH2 + NBS
CH3 CH
CH3 CH2
C = CH2 +
Br
CH3 CH
CH3 CH2 C CH2
Br
(I) (II)
+ Br2 CCl4 H
H
Br
Br
CH2Br
CH2Br
+ H
H
Br
(±)
Br
CH2Br
CH2Br