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North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: 011-65462244, 65662255
South Delhi : 28-A/11, Jia Sarai, Near-IIT Hauz Khas, New Delhi-16, Ph : 011-26851008, 26861009
1
2. Section-A 30ONLY ONE Q.1 to Q.10 Q.11 to Q.3
Section-B 10ONE or MORE than ONE
2 marks.
-C 20 Q.41 to Q.50 1 MarkQ.51 to Q.60 2 Marks
1 mark 1/3 marks2 marks 2/3 marks
contains Multiple Choice Questions (MCQ). Each question has 4 choices (a), (b), (c) and (d), for its answer, out of which is correct. From carries 1 Marks and 0 carries 2 Marks each.
3. contains Multiple Select Questions(MSQ). Each question has 4 choices (a), (b), (c) and (d) for its answer, out of which is/are correct. For each correct answer you will be awarded
4. Section contains Numerical Answer Type (NAT) questions. From carries each and carries each. For each NAT type question, the value of answer in between 0 to 9.
5. In all sections, questions not attempted will result in zero mark. In Section–A (MCQ), wrong answer will result in negative marks. For all questions, will be deducted for each wrong answer. For all questions, will be deducted for each wrong answer. In Section–B (MSQ),there is no negative and no partial marking provision. There is no negative marking in Section –C (NAT) as well.
Regn. No.
E: [email protected], W : www . careerendeavour.inJIA SARAI
[SOLUTIONS]
North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: 011-65462244, 65662255
South Delhi : 28-A/11, Jia Sarai, Near-IIT Hauz Khas, New Delhi-16, Ph : 011-26851008, 26861009
2
Section-A : Multiple Choice Questions (MCQ)
Q.1 to Q.10: Carry 1 Mark each.1. The major product (P) formed will be
CH3
O O
OROH
(P)
(a) O
CH3
O
(b)
O
CH3
O (c)
CH3
O
O(d) COOH
COOH
CH3
Soln. It is an example of Baeyer villiger oxidation and 2º alkyl group has more migrating aptitude as compare to 1ºalkyl group.Correct option is (b)
2. Which of the following is most reactive as a nucleophile(a) PhO (b) PhS (c) 2Ph CH O (d) 2 2Ph CH NH
Soln. Basicity and Nucleophilicity move anti-parallel in a group from up to down.Correct option is (b).
3. Which compound is the major product of the following reaction
ClCl+ NaI MeOH Product
(a) ICl (b) ClI (c) II (d) OMeCl
Soln. This reaction with a powerfull nucleophile I–, will be by the SN2 mechanism. In substrate, an allylic chloride anda vinylic chloride, allylic system are very reactive by the SN2 mechanism and vinylic systems are unreactive.Correct option is (a).
4. Which ketone has the largest equilibrium constant for hydration
(a)
O
(b)
O
(c)
O
(d) O
Soln. The carbonyl carbon charges from sp2 to sp3 hybridisation in the conversion of a ketone of its hydrate, so thereis a relief of ring strain upon the hydration of cyclobutanone.Correct option is (a).
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5. Decreasing order of heat of hydrogenation in the following Alkenes.
(1) (2) (3) (4)
(a) 1 > 2 > 3 > 4 (b) 4 > 3 > 2 > 1 (c) 4 > 2 > 1 > 3 (d) 2 > 3 > 4 > 1
Soln. Heat of hydrogenation 1
-hydrogen
Correct option is (c)
6. The incorrect statement(s) about the furan is/are(a) Furan gives electrophilic substitution reaction faster than benzene(b) Furan is less aromatic then benzene(c) Furan gives poly bromination with Br2 in non-alcoholic solvents.(d) Furan does not gives Diels-Alder reaction
Soln. Furan gives Diels-Alder reaction, act as Diene.Correct option is (d)
7. The major product (P) is
O
OHH2N
MeS(i) (Boc)2/Base
(ii) NaHCO3/Bn-Cl(iii) H+
(P)
(a)
O
OBn(Boc)HN
MeS
(b)
O
OBnH2N
MeS
(c)
O
OHH2N
MeS
(d)
O
OH(Boc)HN
MeS
Soln.
O
OHBocHN
MeSBn–Cl
O
OBnBocHN
MeS
H+
O
OBnH2N
MeS
(Boc)2OBase NaHCO3
O
OHH2N
MeS
Correct option is (b)
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8. How many signals would you expect in the 13C NMR of the following compounds , respectively
(A) (B) O
O
(C)
(a) 3, 2, 3 (b) 2, 2, 3 (c) 3, 4, 2 (d) 2, 4, 3Soln. 13C NMR in
A = 2 signalB = 2 signalC = 3 signal
Correct option is (b)
9. The major product formed in the following reaction
O
NMe2PhLiAlH4, –AlCl3 (3 : 1)
Et2O(P)
(a)
O
HPh (b) OHPh
(c) NMe2Ph (d) NMe2Ph
Soln. Reduction with lithium aluminium hydride-aluminium chloride (3 : 1) provide a good route from , unsatur-ated carbonyl compounds to unsaturated amines.Correct option is (c).
10. The reaction given below is an example ofMeO NO2
NaOEt
NO2
(a) E2-elimination (b) E1-elimination(c) syn-elimination (d) E1CB-elimination
Soln. Chemical reaction involved in the above transformation can be illustrated as
MeO NO2
H HNaOEtE1CB
MeO NO2 NO2
Carbanion (Stable)
Correct answer: (d)
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Q.11 to Q.30: Carry 2 Marks each.11. Order of basicity in following compound will be
(1) N
H
(2)
NH2
(3)
HN
(4)
HN
(a) 1 > 2 > 3 > 4 (b) 4 > 3 > 2 > 1 (c) 4 > 3 > 1 > 2 (d) 2 > 1 > 4 > 3
Soln. If the lonepair of N involves in resonase than basicity of N is decrease. Hence, in the compound (4) lone pairof nitrogen atom does not involve in the resonance, so most basic in nature.Correct option is (c)
12.NH
O
(i) NaNH2(ii) MeOTs(iii) LDA(iv) EtBr
Product will be ?
(a)
O
(b)
O
NH2
(c) N
O
Me
(d) N
O
Soln.NH
O
NaNH2N
O
Me OTsN
OH Me
LDA
N
O
CH2CH3Br
MeN
O
Me
Correct option is (c)
13. Which is the major product of the following reaction of an acyl azide?
Ph
O
N3(i) Toluene, (ii) H2O
(P)
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(a)
O
NH2
(b)
Ph NH2
(c)
Ph
O
OH(d)
H
O
Ph
Soln.Ph
O
N3
PhO
NPh N C O
Wolffrearrangement
Ph NH2
H2O–CO2
Correct option is (b).
14. Which combination of compounds in a-d identifies A and B in the following reactionOMe
Na, Liq. NH3
EtOH, Et2O(A) (B)
HCl, H2O
(a)
OMe
A =
O
B =, (b)
OMe
A =
O
B =,
(c)
OMe
A =
O
B =, (d)
OMe
A =
O
B =,
Soln.OMe
Birchreduction
OMe OH O O
conjugatedenone
non-conjugatedenone
HCl H+ isomerise
Correct option is (c).
15. Which is obtained as the main product upon reaction of m–t-butylanisole (1-t-butyl-3-methoxybenzene) withconc. (HNO3 + H2SO4)
(a)
OCH3
O2N(b)
OCH3O2N
(c) OCH3
NO2
(d)
OCH3
NO2
Soln. In between t-butyl group and methoxy group, OCH3 is strong activating group and –NO2 group occupy to itsortho position.Correct option is (c)
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16.
OH
CH2I2, Zn-Cu
Et2OProduct
The major product formed will be
(a) OH
(b) O
(c) OH
(d) O
Soln. Since –OH is polar group. So, Simmons smith reagent will attack from the same side of –OH group duechelation between O and Zn metal.Correct option is (c)
17.
I
O
Bu3SnH, AIBNPh-H, 80ºC
(P)
The major product formed in the above reaction is
(a)
O
(b) C11H23
O
(c)
CH3
O
(d) none of these
Soln.
I
OBu3Sn
AIBN
O
•
O
•
Bu3SnH
O
Product
•
Correct option is (a)
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18. The major product (P) isOMe
OMe
+
O
MeMe(P)
(a)
OMe
C
OMe
Me
O
Me (b)
OMe
C
OMe
Me
O
Me (c)
OMe
C
OMe
Me
O
Me (d)
OMe
C
OMe
Me
O
Me
Soln.
OMe
OMe
+Me
Me
O
(4+2)
(endo-approach)
OMe
C
OMe
Me
O
Me
Correct option is (a)
19. The product (P) is
O
R1R2
*
(P)
(a)
OH
R1R2
*
(b)
OH
R1R2
*
(c)
OH
R1R2
* (d)
OH
R1R2
*
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Soln.
O
R1R2
*
[3, 3] S.T.R.
2
31
32
1
O
R2 R1
*
[3, 3] S.T.R.
O
R2 R1
H*
enolization
OH
R1R2
*
Correct option is (a)20. The product (P) is
NH
(i) Me2NH/H+/HCHO(ii) Me–I(iii) Hydrolysis
(P)
(a) NH
NMe2 (b) NH
NMe2
(c) NH
OH (d) NH
OH
Soln.NH
Me2NH/H+/HCHO
NH
NMe2NH
NMe3
hydrolysis
NH
OH
Me–I
Correct option is (c)
21. The product (P) is
NH
(i) BuLi(ii) (Boc)2O(iii) EtO2C C C CO2Et /
(P)
(a) NH
BocC
O
CCEtO2C(b)
N C
Boc
O
CC CO
2 Et
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(c)
N
CO2Et
CO2Et
Boc
(d)
N
CO2Et
CO2Et
Boc
EtO2C
EtO2C
Soln.NH
BuLi
N
Li
(Boc)2O
N
Boc
N
+
CO2Me
CO2Me
Boc
N
CO2Me
CO2Me
Boc
(4+2)cyclo-addition
Correct option is (c)22. The product (P) is
O
OH
HOHO
OH OH
(i) MeOH/HCl(ii) MeCHO/p-TSA
(P)
(a) OOMe O
HO
OHOMe
(b)
O
O
OHOMe
OMe
OH
(c) O
OMe O
HO
OH
OMe
(d)
OO
Me
O
HO
OH
OMe
Soln.
O
OH
HOHO
OH OH
H+/MeOH O
OH
HOHO
OHOMe
MeCHOPTSA
OOMe O
HO
OHOMe
Correct option is (a)
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23. Suggests structure for the products of these reactions, interpreting the spectroscopic data
O
Br MeOH(P)
MeO
Molecular formula : C6H12O2
1max cm 1745 ; C ppm :179, 52, 39, 27 ; H ppm ,1.20 9H, s , 3.67 3H,s
(a) O
O Me(b) O
Me
O
(c) OMe
O(d)
O
OH
Soln.
O
Br
H MeO
Br
O
O
MeOHOMe
OH3C
CH3
H3COMe
1.20 ppm 3.67 ppm
Correct option is (b)
24. The configuration at the two stereocentres in the compound given below are
OMe
H
14
(a) 1R, 4R (b) 1R, 4S (c) 1S, 4R (d) 1S, 4S
Soln.O
Me
H
(3)(4)
(2)
1 42
(1)
1R-Configuration
(3)
(1)
(4)
(2)
31
42
4R-Configuration
3
Correct answer: (a)25. Among the carbocations given below
HH
(A) (B) (C)(a) A is homoaromatic, B is antiaromatic and C is aromatic.(b) A is aromatic, B is antiaromatic and C is homoaromatic.(c) A is antiaromatic, B is aromatic and C is harmoaromatic.(d) A is homoaromatic, B is aromatic and C is antiaromatic.
Soln. HH
+
H H
+
HH
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(A) Homoaromatic character
(Aromatised by bypassing the sp3-carbon atom)
(B)
(+)
+ (C) +
• Delocalisation • Delocalisation• sp2 - carbons (planar)• 4n = 4 ( - electron) • (4n+2) = 2 ( -electron)
n = 1 n = 0• Antiaromatic character. • Aromatic character.
Correct answer: (a)
26. The order of carbonyl stretching frequency in the IR spectra of ketone, amide and anhydride is:(a) Anhydride > amide > ketone (b) Ketone > amide > anhydride(c) Amide > anhydride > ketone (d) Anhydride > ketone > amide
Soln. Because C–O of anhydride = 1800 –1900 cm–1.
C–O of ketone = 1720 cm–1. C–O of amide = 1600–1700 cm–1.Correct answer: (d)
27. In the 1H NMR spectrum recorded at 293 K, an organic compound (C3H7NO), exhibited signals at 7.8(1H, s), 2.8 (3H, s) and 2.6 (3H, s). The compound is
(a)
O
H NMe2(b)
H
N NMe2Me (c)
O
NH
MeMe (d)
NH
Me OMeSoln. DMF appears in two forms as shown below
C
O
NHCH3
CH3
C N
O
H
CH3
CH3
Form-1 at higher temperature Form-2 at 293 K
At lower temperature DMF remains in form-2 preferentiallyIn form-2 methyl groups are chemically non-equivalent and they appears as singlets at 2.8 and 2.6 ppm.Singlet at 7.8 ppm appears due to olefinic proton of form-2.Correct answer is (a)
28. Consider the following statements for [18]-annulene(A) It is aromatic(B) The inner protons resonate at 9.28 in its 1H NMR spectrum(C) There are six protons in the shielded zone.(a) A, B, C (b) A and B only (c) B and C only (d) A and C only
Soln. [18] annulene is aromatic compound and shows ring current• The ring current produces strong induced magnetic field.• Outer 12 protons appears at 8.9 ppm because they remains in deshielding zone.Inner six protons appears at – 1.8 ppm because they remains in shielding zone (up-field)Hence, statement A and C are correct.
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HH H
HH
H
H H
H
H
H
H
HH
H
H
H
H
Outer hydrogen H~9.828 ppm
Inner hydrogen H~ -1.8 ppm
Correct answer is (d).
29. Among the structures given below, the most stable conformation for the following compound is
(a) (b) (c) (d)
Soln. In conformation, all methyl group in same direction. A and C having all group in same side. But in case (A)1-3, diaxial interaction occur. So, (C) is more stable as compare to (A).Correct option is (c).
30. The gauche conformation 60º of n-butane posseses
(a) plane of symmmetry; and is achiral (b) C2-axis of symmetry; and is chiral(c) centre of symmetry; and is achiral (d) plane of symmetry; and is chiral
Soln. The Gauche conformations of n-Butane are non-superimposable mirror image of each other, thus Gaucheconformation is chiral.
Me
H
Me
Mirror
Me
HMe
If contains neither plane nor centre of symmetry.A C2 axis passes through the mid-point of C2–C3 bond and bisecting the dihedral angle between the
two methyls.Correct answer is (b).
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Section-B : Multiple Select Questions (MSQ)
Q.31 to Q.40: Carry 2 Marks each.31. Select the compounds undergoing inter or intra molecular Cannizaro reaction
(a)
O
PhCl
ClBr
(b)
H O
OH
(c)
O
CHOPh(d) OHC CH2 OH
Soln. Compound in which -hydrogen atoms are absent, give Cannizaro reaction.Correct options are (a), (b) and (c)
32. In which reactions benzyne will be formed as intermediate
(a)
F
Br
+ Mg (b)
NH2
COOH
(i) HNO2(ii)
(c) S
N
N
OO
(d) F
NaNH2
Soln. In structure, (a), (b) and (c) formed Benzyne intermediate.Correct options are (a), (b) and (c).
33. The incorrect statement(s) about the heterocyclic compounds is/are(a) Pyrrole gives electrophilic substitution reaction while as pyridine give nucleophilic substitution reactions(b) Pyrrole does not act as base while as pyridine does.(c) Furan and thiophine gives electrophilic substitution reactions with equal rate(d) Pyrrole gives nucleophilic addition reaction with Grignard reagent.
Soln. 3rd and 4th statement are incorrect.Correct option (c) and (d).
34. In which the product is formed through dis-rotation
(a)
(b)
H
H
(c)
(d)
Soln. (a)
(conc.)
(4n)
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(b)
H
H
Dis-rotation.
(4n+2)
(c)
Dis
(4n+2)Correct option is (b) and (c)
35. The correct structure of the D-glucose is/are
(a)
O
OH
OH
OH
OH
OH(b)
O
OH
OH
OHHO
OH
(c) O
OH
OH OH
HOHO (d)
CHO
H OH
HO H
H OH
H OH
CH2OH
Soln. All forms are the correct structures of D-glucose.Correct option is (a), (b), (c) and (d)
36. Which of the following equation shown as correct product
(a)
N
NH2
O
Br2, NaOH
H2ON
NH2
(b) COClCH2N2
H2O CH2 COOH
(c)
O
C(CH3)3CH2
H3CCF3COOOH
O
O(H3C)3C CH2 CH3
(d) COClNaN3
H2ONH2 + CO2
Soln. The reaction include (a) Hoffmann rearrangement, (b) a wolff rearrangement, (d) curtius rearrangement.Reaction (c) is possible Baeyer-villiger oxidation, but t-butyl has a higher migratory aptitude than ethyl togive the alternative ester as the major product.Correct options are (a), (b) and (d).
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37. The products formed in the following reaction is/are
D-ribose(i) NaCN/HCN(ii) H2/Pd/BaSO4
(iii) H3O+
(P)
(a) D-Allose (b) D-Altrose (c) D-Glucose (d) D-Mannose
Soln.
CHO
H OH
H OH
CH2OH
H OH
C
H OH
H OH
H OH
N
NaCN/HCN
H3O+
H OHCH2OH
C
HO H
H OH
H OH
N
H OHCH2OH
HC
H OH
H OH
H OH
NH
H OHCH2OH
HC
HO H
H OH
H OH
NH
H OHCH2OH
CHO
H OH
H OH
H OH
H OHCH2OH
D- Allose
CHO
HO H
H OH
H OH
H OHCH2OH
D- Altrose
H2, Pd/BaSO4
+
Correct options are (a) and (b)
38. The correct statement(s) about the Diels-Alder reaction is/are(a) electron donating groups on diene increases, the rate of Diels-Alder reaction(b) electron donating groups on Dienophile decreases, the rate of Diels-Alder reaction(c) Anthracene gives Diels-Alder reaction while as benzene does not give(d) Aromatic character is directly proportional to the reactivity of Diels-Alder reactions
Soln. In presence of electron donating group on the Diene, the energy of HOMO increases while electronwithdrawing group on Dienophile decreases the energy of LUMO. Hence, HOMO-LUMO gap de-creases and rate of reaction increases.Antracene gives the Diels-Alder reaction because it is less aromatic.Statements (1), (2) and (3) are correct.
Correct options are (a), (b) and (c).
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39. The number of compounds which under goes [3, 3] sigmatropic shift
(a)
O
(b) (c) (d)
Soln. Compound first, third and fourth undergoes [3, 3] sigmatropic rearrangement and for [3, 3] sigmatropicrearrangement compound should be in 1, 5-hexadiene form.Correct options are (a), (c) and (d).
40. The correct statement(s) about the given compounds
(A)
O
CO2MeAr (B)
O
CO2MeAr (C)
O
CO2MeAr (D)
O
CO2MeAr
(a) A and B are enantiomers (b) B and D are diastereomers(c) B and C are homomers (d) C and D are enantiomers
Soln. A and B Enantiomers
B and D DiastereoisomersB and C DiastereoisomersC and D EnantiomersCorrect options are (a), (b) and (d).
Section-C : Numerical Answer Type (NAT)Q.41 to Q.50: Carry 1 Mark each.41. Among the following, how many incorrect statement for the following reaction is/are _________
O
1. MeMgBr, Et2O
2. H3PO4+
(A) (B)
(1) A is the major product and it will have five signals in the proton decoupled 13C NMR spectrum(2) A is the minor product and it will have eight signals in the proton decoupled 13C NMR spectrum(3) B is the major product and it will have five signals in the proton decoupled 13C NMR spectrum(4) B is the minor product and it will have five signals in the proton decoupled 13C NMR spectrum
Soln. Chemical reaction involved in the above transformation can be illustrated as
MeMgBrCH3
HO H H2PO4H3PO4
CH3
OH2
O
H2C
HB
HA
By loss of HBSytzeff's product
more stableMajor
By loss of HAHoffmann product
less stable
Minor
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e
d c b
dc
ePlane of symmetry
Five distinct carbon atoms, thus five signals in protondecoupled 13C NMR spectrum.
a
e
d c b
gh
f
aEight distinct carbon atoms, thus eight signals in proton decoupled NMR spectrum.
Correct answer is (3)
42. Nucleophilic substitution reactions of pyridine occurs at which position ____________Soln. Pyridine is -sink and gives nucleophilic substitution reaction at second position.
Correct answer is (2)
43. The number of compounds which undergoes Diels-Alder reaction
(1) (2) (3) (4)
(5) (6)
Soln. For Diels-Alder’s reaction Diene should be in cisoid form. The nephthelene is highly aromatic, hencedoes not give the Diels-Alder reaction.Correct answer is (2)
44. How many in the following molecules are chiral
HOOC
HOOC
(A) (B) C C
HOOC
H
C
C6H5
H
(C) C C
H
H
C
C6H5
H
(D) CH3
CH3H
Cl(E)
Br
Cl
Cl
BrSoln. (a) Achiral, due to the presence of plane of symmetry
(b) Chiral(c) Achiral, plane of symmetry present(d) Achiral, Plane of symmetry present(e) Achiral, due to the presence of centre of symmetry.Correct answer is (1)
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45. In the IR spectrum of p-nitrophenyl acetate, the carbonyl absorption band appears at ____________cm–1
Soln. C = O structure occurs at cm–1.
1750 – 1735cm–1 for aliphatic esters.1740 – 1715cm–1 if C = O conjugated with aromatic.1765 – 1762cm–1 if oxygen atom is conjugated with alkene or aromatic.Example: Phenyl acetate 1765 cm–1
p-nitrophenyl acetate 1761 cm–1.
O C
O
CH3
O2N
Correct answer is (1755 to 1765)46. Among the following, how many are polar protic solvents?
(i) H2O (ii) CH3OH (iii) CH3CH2OH (iv) NH3(v) Acetone (vi) DMF (vii) Benzene (viii) DMSO (ix) Cyclohexane
Soln. Molecule having polarity with acidic proton known as polarprotic solvent. (Proton attached with more elec-tronegative element containing polar molecule).
2 3 3 2 3H O, CH OH, CH CH OH, NH are polar protic solvent.Correct answer is (4)
47. How many of following rearrangements form isocyanate ________________(1) Wolf rearrangement (2) Hoffmann rearrangement(3) Beckmann rearrangement (4) Schmidt rearrangement(5) Lossen rearrangement (6) Curtus rearrangement (7) Claisen rearrangement
Soln. In the above rearrangements, Curtius, Hoffman, Lossen and Schmidt form isocyanate intermediate.Correct answer is (4)
48. HOOCC
Cl
C N
NO2
O
In the above structure, how many functional groups not reduce by borane.
Soln. HOOC Cl
CN
NO2
O
Reduction by borane CCl
NO2
O
HO
NH2
In the above structure, RCOCl and R-NO2 are not reduce by Borane.
Correct answer is (2)
49. Calculate the pI of the following salt of amino acid having pKa value x = 2.2, y = 3.9 and z = 7.9 respectively.
O
H3N
HO
OH
O
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Soln. O
H3N
HO
OH
O
2.23.9
7.9PI of an amino acid having ionizible side chain is the average pKa values of similarly ionizable groups.
2.2 3.9 6.1 3.052 2
Correct answer is (3.05)
50. How many products will be formed upon the bromination of furan in alcoholic and non-alcoholic solvent.Soln. Furan upon bromination in alcoholic solvent gives addition product. While as with non-alcoholic solvent gives
polybromination products.Correct answer is (2)
Q.51 to Q.60: Carry 2 Marks each.
51.
CH3
OCH3
+H+
(P)
How many product will be formed in above reaction.
Soln.+ H
CH3
OCH3
+
CH3
OCH3
Correct answer is (1)
52. How many compounds gives cannizaro reaction
(1) Cl3C CHO (2) Cl3C C CCl3
O
(3) OHC CHO (4) Cl3C C CH2
O
CCl3
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(5) OO
C(CH3)3 (6) N
HO
(7) O
Soln. Compounds in which -hydrogen atoms are absent give cannizaro reaction.Correct answer is (5)
53. Reaction,OH
H+
How many intermediate (carbocations) will be form in above conversion.
Soln.OH
H+Ring
expension
(1)
(3)
OH2
(2)
Ringexpension
Total carbocation is (3)Correct answer is (3)
54. How many products will be formed on ozonolysis of following compound
CH3
Ozonolysis (P)
H
CH3
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Soln. (1)
(2)
(3)
(4)
(5)
Correct answer is (5)
55. CEt
O
CCl
O
CH
O
NH2
O
Number of moles of CH3MgBr required to neutrals above compound is/are________________
Soln. Acid derivatives reacts two mole of CH3MgBr and carbonyl compound (aldehyde or ketone) react with onemole of CH3MgBr).Correct option is (7).
56. Among the following neutral amino acids is/are(1) Lyscine (2) Arginine (3) Valine (4) Aspartic acid(5) isoleucine (6) Traphtophane (7) Leucine
Soln. Lyscine, Arginine are the basic amino acids.Aspartic acid is the acidic amino acid.Valine, isoleucine, Traphtophane and Leucine are the neutral amino acids. Correct answer is (4)
57. The []D of a 90% optically pure 2-arylpropanoic acid solution is +135º. On treatment with a base at RT forone hour, []D changed to +120º. The optical purity is reduced to 40% after 3 hours. If so, its specific rotationafter 3 hours would be ________________degree.
Soln. • Optically purity also known as enantiomeric excess a compound of 90% B.E. shows 135º specific rotation.
So, its 100% optically pure isomer will show 135 100 150º specific rot.90
• At one hour the specific rotation reduced to 120º. So, E.E, or optical purity = 120 100 80%150
• At three hours optical purity is 40%. So, specific rotation = 150 40 60º100
Correct answer is (60)
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58. Among the following, how many are aromatic in nature?
(A) O
(B) O
N (C) S
N
(D)
O
O
(E) N
(F) B
H
(G) (H)
(I) HH (J)
O
Soln. B, D, H aromatic in natureCorrect answer is (3)
59. For the below compound
The max value is ________ mµSoln. Basic value = 217 mµ
2-Ring residues = 10 mµ (2 × 5)2-Exo cyclic double bonds = 10 mµ1-bicyclic system = 15 mµ
= 252 mµ
Correct answer is (252)
60. C
HNEtO
O O
NO2
OOH
In the above structure, how many functional groups are reduced by LiAlH4.Soln. Total number of five functional groups reduced by LiAlH4.
HNHO
OH
NO2
OHOH
H
Correct answer is (5)
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IIT-JAM CHEMISTRY-CYDate : 08-01-2017
TEST SERIES - 2(Organic Chemistry)
ANSWER KEY
Section-A : Multiple Choice Questions (MCQ)
1. (b) 2. (b) 3. (a) 4. (a) 5. (c)6. (d) 7. (b) 8. (b) 9. (c) 10. (d)11. (c) 12. (c) 13. (b) 14. (c) 15. (c)16. (c) 17. (a) 18. (a) 19. (a) 20. (c)21. (c) 22. (a) 23. (b) 24. (a) 25. (a)26. (d) 27. (a) 28. (d) 29. (c) 30. (b)
Section-B : Multiple Select Questions (MSQ)
31. (a),(b), (c) 32. (a),(b), (c) 33. (c),(d) 34. (b), (c)35. (a),(b),(c), (d) 36. (a), (b), (d) 37. (a),(b) 38. (a),(b), (c)39. (a), (c), (d) 40. (a), (b), (d)
Section-C : Numerical Answer Type (NAT)
41. (3) 42. (2) 43. (2) 44. (1)45. (1755 to 1765) 46. (4) 47. (4) 48. (2)49. (3.05) 50. (2) 51. (1) 52. (5)53. (3) 54. (5) 55. (7) 56. (4)57. (60) 58. (3) 59. (252) 60. (5)
Booklet : B