CENTERS : MUMBAI / DELHI /AKOLA / LUCKNOW /NASHIK / PUNE / NAGPUR / BOKARO / DUBAI # 96 FINAL LAP - 2019 GENERAL ORGANIC CHEMISTRY RESONANCE & AROMATICITY Q.1 Which of the following statements is (are) true about resonance. (a) Resonance is an intramolecular process. (b) Resonance involves delocalization of both s and p electrons. (c) Resonance involves delocalization of p electrons only. (d) Resonance decreases potential energy of a molecule. (e) Resonance has no effect on the potential energy of a molecule. (f) Resonance is the only way to increase molecular stability. (g) Resonance is not the only way to increase molecular stability. (h) Any resonating molecule is always more stable than any nonresonating molecule. (i) The canonical structure explains all features of a molecule. (j) The resonance hybrid explains all features of a molecule. (k) Resonating structures are real and resonance hybrid is imaginary. (l) Resonance hybrid is real and resonating structures are imaginary. (m) Resonance hybrid is always more stable than all canonical structures. Q.2 Consider structural formulas A, B and C: (A) (B) (C) (a) Are A, B and C constitutional isomers, or are they resonance forms? (b) Which structures have a negatively charged carbon? (c) Which structures have a positively charged carbon? (d) Which structures have a positively charged nitrogen? (e) Which structures have a negatively charged nitrogen? (f) What is the net charge on each structure? (g) Which is a more stable structure, A or B? Why? (h) Which is a more stable structure, B or C? Why? Q.3 In each of the following pairs, determine whether the two represent resonance forms of a single species or depict different substances. If two structures are not resonance froms, explain why. (a) and (b) and (c) and Q.4 Which of the following statement is incorrect ? (A) Contributing structures contributes to the resonance hybrid in proportion of their energies. (B) Equivalent contributing structure make the resonance very important. (C) Contributing structures represent hypothetical molecules having no real existance. (D) Contributing structures are less stable than the resonance hybrid.
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RESONANCE & AROMATICITY...GENERAL ORGANIC CHEMISTRY FINAL LAP - 2019 RESONANCE & AROMATICITY Q.1 Which of the following statements is (are) true about resonance. (a) Resonance is an
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Q.1 Which of the following statements is (are) true about resonance.(a) Resonance is an intramolecular process.(b) Resonance involves delocalization of both s and p electrons.(c) Resonance involves delocalization of p electrons only.(d) Resonance decreases potential energy of a molecule.(e) Resonance has no effect on the potential energy of a molecule.(f) Resonance is the only way to increase molecular stability.(g) Resonance is not the only way to increase molecular stability.(h) Any resonating molecule is always more stable than any nonresonating molecule.(i) The canonical structure explains all features of a molecule.(j) The resonance hybrid explains all features of a molecule.(k) Resonating structures are real and resonance hybrid is imaginary.(l) Resonance hybrid is real and resonating structures are imaginary.(m) Resonance hybrid is always more stable than all canonical structures.
Q.2 Consider structural formulas A, B and C:
(A) (B) (C)(a) Are A, B and C constitutional isomers, or are they resonance forms?(b) Which structures have a negatively charged carbon?(c) Which structures have a positively charged carbon?(d) Which structures have a positively charged nitrogen?(e) Which structures have a negatively charged nitrogen?(f) What is the net charge on each structure?(g) Which is a more stable structure, A or B? Why?(h) Which is a more stable structure, B or C? Why?
Q.3 In each of the following pairs, determine whether the two represent resonance forms of a single speciesor depict different substances. If two structures are not resonance froms, explain why.
(a) and
(b) and
(c) and
Q.4 Which of the following statement is incorrect ?(A) Contributing structures contributes to the resonance hybrid in proportion of their energies.(B) Equivalent contributing structure make the resonance very important.(C) Contributing structures represent hypothetical molecules having no real existance.(D) Contributing structures are less stable than the resonance hybrid.
Q.13 Write stability order of following intermediates:
(i) 23 CHCH
(b) 33 CHCHCH
(c)
3
3
3
CH|CCH|
CH
(ii) (a) (b) (c)
(iii) (a) (b) (c)
(iv) (a)
CHC (b)
CHCH2 (c)
23 CHCH
(v) (a) (b) (c)
Q.14 Among the following molecules, the correct order of C – C bond lenght is(A) C2H6 > C2H4 > C6H6 > C2H2 (B) C2H6 > C6H6 > C2H4 > C2H2(C6H6 is benzene)(C) C2H4 > C2H6 > C2H2 > C6H6 (D) C2H6 > C2H4 > C2H2 > C6H6
Q.15 Which of the following is (are) the correct order of bond lengths :(A) C – C > C = C > C C > C N (B) C = N > C = O > C = C(C) C = C > C = N > C = O (D) C – C > C = C > C º C > C – H
Q.16 C1 – C2 bond is shortest in
(A) (B) (C) (D)
Q.17 In which of the following resonance is possible(A) CH2 = CH – CH2 – CHO (B) CH2 = CH – CH = O(C) CH3COCH3 (D) CH2 = CH – CH2 – CH = CH2
Q.18 Which of the following compounds have delocalised electrons ?
Q.22 In which of the following pairs, indicated bond having less bond dissociation energy :(a)
BrCHCH 23 &
ClCHCH 23
(b) BrCHCHCH3 &
Br|
CHCHCH 33
(c) &
(d) &
(e) &
(f) &
Q.23 Discuss the following observations:(a) C–Cl bond in vinyl chloride is stronger than in chloroethane.(b) Carbon-carbon bond length in ethene is shorter than in CH2 = CHOCH3(c) CH3SH is stronger acid than CH3OH(d) CH3CH2NH2 is stronger base than CH2 = CHNH2.
Q.24 A canonical structure will be more stable if(a) it has more number of p bonds than if it has less number of p bonds.(b) the octets of all atoms are complete than if octets of all atoms are not complete.(c) it involves cyclic delocalization of (4n + 2) p – electrons than if it involves acyclic delocalization of (4n +
2) p – electrons.(d) it involves cyclic delocalization (4n) p – electrons than if it involves acyclic delocalizationof
(4n) p – electrons.(e) +ve charge is on more electronegative atom than if +ve charge is on less electronegative atoms.(f) –ve charge is on more electronegative atom than if –ve charge is on less electronegative atom.
Q.25 Resonance energy will be more if(a) canonical structures are equivalent than if canonical structures are non-equivalent.(b) molecule is aromatic than if molecule is not aromatic.
Q.26 In each set of species select the aromatic species.
(i) (a) (b) S
(c) +
(d) +
(ii) (a) +
(b)
(c) (d)
(iii) (a) +
(b)
(c)
(d) N
H
Q.27 Consider the given reaction:
+ 3H2 C/Pd
In the above reaction which one of the given ring will undergo reduction?
Q.28
Br
3AgNO AA
Select the correct statement about product A.(A) Product is aromatic (B) Product has high dipole moment.(C) Product has less resonance energy (D) Product is soluble in polar solvent.
Q.29 Which of the following is incorrectly orderd for resonance stability
Q.41 Which one of the following molecules has all the effect, namely inductive, mesomeric and hyperconjugative?(A) CH3Cl (B) CH3–CH = CH2
(C)
O||
CHCCHCHCH 33 (D) CH2 = CH – CH = CH2
Q.42 Hyperconjugation is best described as:(A) Delocalisation of p electrons into a nearby empty orbital.(B) Delocalisation of s electrons into a nearby empty orbital.(C) The effect of alkyl groups donating a small amount of electron density inductively into a carbocation.(D) The migration of a carbon or hydrogen from one carbocation to another.
Q.43 Select the correct statement.(i) Delocalisation of s-electron is hyperconjugation.(ii) Delocalisation of p-electron is resonance.(iii) Partial displacement of s-electron is inductive effect.(A) i & iii (B) ii & iii (C) i & ii (D) i, ii, iii
Q.44 Arrange following compounds in decreasing order of electrophilic substitution.
(i)
CH3
(ii)
CH3
H—C—CH3
(iii)
CH3
CH3—C—CH3
(iv)
H
H—C—CH3
(A) i > ii > iii > iv (B) iii > iv > ii > i (C) i > iv > ii > iii (D) i > ii > iv > iii
FINAL LAP - 2019GENERAL ORGANIC CHEMISTRYQ.45 Select correct statement:
(A) –NO2 and –COOH group deactivates benzene nucleus for attack of E+ at o– and p– sites.(B) –NH2 and –OMe group activates benzene nucleous for attack of E+ at o– and p– sites.(C) –NH2 and –COOH group activates benzene nucleous for attack of E+ at o– and p– sites.(D) –NO2 and –OMe group activates benzene nucleous for attack of E+ at o– and p– sites.
Q.46 In which of the following pairs, indicated bond is of greater strength :
(a) and 22 CHCH (b)
CHCCH3 and
CHHC
(c) and ClCHCH 23
(d)
22 CHCHCHCH and
3222 CHCHCHCH
(e)
22 CHCHCHCH and 22 NOCHCH
(f) and
Q.47 Choose the more stable alkene in each of the following pairs. Explain your reasoning.(a) 1-Methylcyclohexene or 3-methylcyclohexene(b) Isopropenylcyclopentane or allylcyclopentane
(c) or
Q.48 Match each alkene with the appropriate heat of combustion:Heats of combustion (kJ/mol) : 5293 ; 4658; 4650; 4638; 4632(a) 1-Heptene (b) 2,4-Dimethyl-1-pentene(c) 2,4-Dimethyl-2-pentene (d) 4,4-Dimethyl-2-pentene(e) 2,4,4-Trimethyl-2-pentene
Q.49 Compare heat of hydrogenation (Decreasing order)
Q.2 a = Resonance form, b= A, c = C, d= A & B, e=B&C, f = 0, g = B, h = B
Q.3 (a) is resonance form; (b) is not resonance form due to different number of l.p. and b.p.;(c) is not resonance form due to different number of l.p. and b.p
Q.4 A Q.5 D Q.6 (b), (d), (e) Q.7 (b), (d), (e) Q.8 (i) a (ii) c
Q.9 (a) 23 CHCCH|
¯O
(b) 2CHCHCHCH|
¯O
(c)
CH2 CH2CH2 CH2CH2+ +
+ +
+
(d) + (e) O¯ O O O
(f) NH+
(g) O+(h)
O¯
(i) 33 CHCHCHCHCHCHCH
(j) 23 CHCHCHCHCHCH
]
Q.10 (i) c > b > a (ii) b > c > a (iii) c > b > a (iv) b > c > a(v) d>c > b>a (vi) a < b < c
Q.11 (i) a > b > c (ii) a > b > c (iii) a > c > b (iv) a > b > c
Q.12 (i) b < d < a < c (ii) b > c > a (iii) c > a > b (iv) c > a > b(v) a > b
Q.13 (i) c > b > a (ii) c > b > a (iii) b > c > a (iv) c > b > a(v) a > c > b
Q.14 B Q.15 A, C, D Q.16 D Q.17 B Q.18 B Q.19 iii > ii > i
Q.6 Arrange the following compound in decreasing order of their basicity.(i) (a) H2C = CHNa (b) CH3CH2Na (c) CH3CH2ONa (d) HC CNa
(ii) (a) NH2 (b) CH – NH22 (c) NH2
NO2
(d) C – NH2
O(iii) (a) HO¯ (b) NH3 (c) H2O
Q.7 Basicity order in following compound is :
N
N
NH
O CH3
CH3
CH2 – NH – C – CH 3H2N– C – CH 2b
d
a
c
CH3
CH3
(A) b > d > a > c (B) a > b > d > c (C) a > b > c > d (D) a > c > b > d
Q.8 Consider the following bases:(I) o-nitroaniline (II) m-nitroaniline (III) p-nitroanilineThe decreasing order of basicity is:(A) II > III > I (B) II > I > III (C) I > II >III (D) I > III > II
Q.9 Consider the basicity of the following aromatic amines:(I) aniline (II) p-nitroaniline (III) p-methoxyaniline (IV) p-methylanilineThe correct order of decreasing basicity is:(A) III > IV > I > II (B) III > IV > II > I(C) I > II > III > IV (D) IV > III > II > I
Q.19 Rank the amines in each set in order of increasing basicity.
(a)NH2 NH2 N
H
(b)
HN NH N
(c) N NN N H HH
Q.20
Pyrimidine Imidazole PurineAmong the following which statement(s) is/are ture:(A) Both N of pyrimidine are of same basic strength(B) In imidazole protonation takes places on N–3.(C) Purine has 3 basic N.(D) Pyrimidine imidazole and purine all are aromatic
Q.4 Give the correct order of initials T or F for following statements. Use T if statement is true and F if it isfalse.I. Me–CH=C=C=CH–Br is optically active.II. All optically active compound are chiral.III. All chiral pyramidal molecules are optically inactive.
IV CH3–CH2–CH2–COOH and
COOH|
CHCHCH 33 are positional isomers.
(A) TTTF (B) FTFT (C) FTFF (D) TFTT
Q.5 From the following four structures select:(a) The optically active isomers (b) Optically inactive isomers(c) Enantiomers pairs (d) Distereomer pairs
Q.10 For each of the following molecules, indicate, whether they are chiral, achiral or meso compound:
(I) (II) (III) (IV)
(V) (VI) (VII)
Q.11 For each of the following pair of structures, indicate, if the compounds are identical, constitutional isomers,enantiomers, distereomers, diffrerent.
(a) and (b) and
Q.12 How many stereoisomer may have this natural occuring compound.
(A) 8 (B) 16 (C) 64 (D) 128
Q.13 How many optically active stereoisomers are possible for butane-2, 3-diol(A) 1 (B) 2 (C) 3 (D) 4
Q.14 Identify whether each of following pair of compounds are identical or enantiomers, diastereoisomer orconstitutional isomers.
FINAL LAP - 2019ISOMERSQ.23 Truxillic acid can exist theretically in five stereosiomeric form all of which are known an optically inactive
explain,where a = – CO2H, b = – C6H5
Q.24 The most stable enol-form of compound is
(A) (B) (C) (D)
Q.25(a) A cyclobutandicarboxylic acid exist in two stereo-isomeric forms in which one is polar but non-resolvablewhile other is non-polar but resolvable into enantiomers. Deduce structures of all these compounds.
(b)Select resolvable compounds.
(i) (ii) (iii)
3CH|
OSPh
(iv) (v) MeCHBrCH2Me (vi)
(vii) (viii) (ix) BrDHNMe 2
(x)
OH|
MeCHCHMeCH 22 (xi) (xii) C=C=CH2
Q.26 Which of the Newman projections shown below represents the most stable conformation about theC1–C2 bond of 1-iodo-2-methyl propane?
(A) I, II only (B) III, IV only (C) I, II, III (D) All
2. No. of functional groups present in the following compound is :
(A) 5 (B) 7 (C) 6 (D) 8
3. Quinine is the most important alkaloid obtained from cinchona bark. It’s molecular formula is C20H24N2O2. Itmay contain(A) 5 double bond & 6 ring (B) 6 double bond & 4 ring(C) 6 double bond & 3 ring (D) 7 double bond & 5 ring
4. In the structure of 4-Isopropyl-2,4,5-trimethylheptane, number of 10, 20 & 30 H’s are respectively(A) 18, 5, 4 (B) 21, 4, 3 (C) 18, 4, 3 (D) 21, 5, 4
5. How many 1° amines are possiple for the molecular formula C7H9N which contain benzene ring also(A) 1 (B) 4 (C) 3 (D) 2
6. Find out the total number of geometrical isomers of the following compound(CH3)CH = C = C(CH3)(C2H5)(A) 0 (B) 2 (C) 3 (D) 4
7. Which of the following will not show geometrical isomerism
(A) (B) (C) (D)
8. How many products are formed by monochlorination of
FINAL LAP - 2019ISOMERS15. Which statement is correct about the following structures
BrClH
OCH3
CH3
C H2 5
I
BrCl
HH O3C
CH3
C H2 5
II
Br
Cl
HH O3C
CH3
C H2 5
III(A) I & III are structural isomers(B) I & II & I & III are enantiomers(C) I, III are enantiomers and I, II are structural isomers(D) I, II & III are stereoisomers
16. Indicate the number of chiral carbon atoms in the following molecule.
HO(A) 6 (B) 7 (C) 8 (D) 9
17. 3, 5-Dimethylcyclohexanol has total stereoisomers(A) 3 (B) 4 (C) 8 (D) 12
18. 2, 4-Dimethylcyclohexanol has total stereoisomers(A) 4 (B) 3 (C) 8 (D) 12
19. In the number of stereoisomers are
(A) 26 (B) 25 (C) 24 (D) 23
20. Ph – CH2 – CH2 – COOH PdRe
Br2 PhCH2CHBrCOOH
The product is(A) racemic mixture (B) pure (+) (C) pure (–) (D) none of these
21. and are
(A) enantiomers (B) diastereomers (C) mesomers (D) all of these
22. can have
(A) geometrical isomer (B) optical isomers (C) epimer (D) none of these
Select the correct statement/s(A) All are identical compounds (B) III and IV are identical and meso compounds(C) I and II are optically active compounds(D) I and II are configurational isomers while II and IV are conformational isomers
39. Draw the most stable conformation of 1-Chloropropane. Which statement is correct about thisconformation.(A) It is the most polar form (B) It has maximum torsional strain(C) It has minimum steric strain (D) A and C both
FINAL LAP - 2019ISOMERS53. Establish the relationship between the following pairs, as enantiomers, identicals, diastereomers or struc-
tural isomers.
(i) (ii)
(iii) (iv)
54. Observe the structures of compounds A, B, C and D. Write the relationship between the given pairs ofcompounds.
(I) A and B are .......... (II) A and C are ............(III) A and D are ......... (IV) C and D are ..........
55. Compounds with molecular formula C2H7N can be(A) Functional isomers and metamers (B) Only functional isomers(C) Functional and positional isomers (D) Chain, positional, functional and metamers
56. Total number of isomers having molecular formula C3H4O and only functional group can be,
(A) 1 (B) 2 (C) 3 (D) 4
57. How many products are formed by monochlorination of (Ignore stereoisomers)
(A) 4 (B) 2 (C) 5 (D) 3
58. The total number of 4° (quaternary) carbon atoms present in cholestanol is
(A) 3 (B) 8 (C) 4 (D) 2
59. Total number of phenol isomers of the compound C8H10O is(A) 10 (B) 9 (C) 8 (D) 6
60. The following two compounds I and II are ...................... isomers.
and
(A) Geometrical (B) Geometrical and positional(C) Positional (D) Geometrical, optical and positional
(A) I, II & III have CnH2n-2 general formula (B) I , II & III have same empirical formula(C) I, II & III have same molecular formula (D) I , II are identical and homologue of compound III.
67. Which of the following will show geometical isomer.
(A) (B) (C) (D)
68. Total number of geometrical isomers in the following compound,
Ph–CH=CH– –CH2–CH= –OH
(A) 2 (B) 4 (C) 8 (D) 16
69. Which of the following will not show geometrical isomerism.
74. Consider the following four statementsS1 : Geometrical isomers are not mirror image isomer.S2 : A compound having double bond (restricted bond) always show geometrical isomerismS3 : Acyclic compounds having single bond does not show geometrical isomerismS4 : Cyclodecene can exist in cis & trans form
(A) T F F T (B) T T F T (C) F F F T (D) T F T T
75. Which will not show geometrical isomerism
(A) (B) (C) ClCH = CHCl (D) Ph—N=N—Ph
76. Which of the following compound posses plane of symmetry -
(A) (B) (C) (D)
77. Which of the following compounds are optically active ?
(A) (B) (C) (D)
78. Well known pain killer Nurofen is an ibuprofen how many stereoisomers it would have ?
If we have a racemic mixture of ibuprofen which one of the following can be used to resolve the isomers.
(II) Cyclic chiral pyramidal molecules are optically active.
(IV) CH3–CH2–CH2–COOH and
COOH|
CHCHCH 33 are chain isomers.
Q.5 (a) II and IV are chiral, hence optically active, (b) I and III are achiral, posses plane of symmetry, henceoptically inactive, (c) There is no enantiomer pair, both II and IV are identical structure, (d) I and II, IIand III, I and III are pair of distereomers.
Q.6 Compound I is optically inactive since it contain a plane of symmetry. Compound II is enantiomeric sinceit does not contain plane of symmetry, hence chiral. Also compound I is polar while II is non polar.]
Q.7 (a) Both I and II are optically, but they are not mirror image of one another, hence, they are distereomers(b) Distereomers ,(c) Enantiomers,(d) I and V are enantiomers,(e) IV and VI are distereomers]
Q.15 This compound has two chiral carbon, and a double bond capable of showing geometrical isomerismgiving rise to total eight different configurations possible for the molecule as shown below:
Q.16 (a) Both are similar structures, (b) They are positional isomers, (c) They are enantiomers, (d) chain ,(e) Metamers
Q.1 2-chloropentane on halogenation with chlorine gives 2,3, dichloropentane. What will be the structure offree radical species formed in the reaction?(A) Planar (B) Trigonal planar (C) Square planar (D) Pyramidal
Q.2 The correct order of rate of Wurtz recation.
(I) CH2–F ether
Na CH2–CH2
(II) CH2–Cl ether
Na CH2–CH2
(III) CH2–Br ether
Na CH2–CH2
(IV) CH2–I ether
Na CH2–CH2
(A) I > II > III > IV (B) II > I > III > IV(C) IV > III > II > I (D) In all rate of Wurtz reaction is same
Q.6 Find out the correct order of rate of reaction towards allylic substitution.
(I) CH3–CH = CH2 (II) CH3–CH2–CH=CH2 (III) 23
3
CHCHCHCH|
CH
(A) I > II > III (B) II > I > III(C) III > II > I (D) III > I > II
Q.7 What will be the major product, when 2-methyl butane undergoes bromination in presence of light?(A) 1-bromo-2-methyl butane(B) 2-bromo-2-methyl butane(C) 2-bromo-3-methyl butane(D) 1-bromo-3-methyl butane
Q.8 Which can not be the possible product of the given reaction
FINAL LAP - 2019REACTION INTERMEDIATESQ.9 Pick the correct statement for monochlorination of R-secbutyl Bromide.
Me
Et
HHBr C300
Cl2
(A) There are five possible product ; four are optically active one is optically inactive(B) There are five possible product ; three are optically inactive & two are optically active(C) There are five possible product ; two are optically inactive & three are optically active(D) None of these
Q.10 Correct order of rate of photochlorination for following compounds is:
CH3–CH3 CD3–CD3
3
33
3
CH|
CHCCH|HC
(I) (II) (III)(A) II < I < III (B) I < II < III (C) III < I < II (D) II < III < I
Q.11
,CCl
NBS
4Allylic brominated products
Find out the incorrect statement.(A) It gives total 9 allylic brominated products(B) All allylic brominated products are optically active(C) Substrate has 7 allylic hydrogens(D) NBS gives Br2 constantly to reaction mixture.
Q.12 Which of the following carbocation is most stable?
(A) (B)
(C) (D)
Q.13 Which carbocation is least likely to form as an intermediate?
FINAL LAP - 2019REACTION INTERMEDIATESQ.14 For the reactions
(I) + Clr , o1H (II) + Clr , o
2H
(III) + Cl , o
3H (IV) + Cl , o4H
The correct decreasing order of enthalpies of reaction for producing carbocation is
(A) o1H > o
2H > o3H > o
4H (B) o4H > o
1H > o2H > o
3H
(C) o3H > o
2H > o1H > o
4H (D) o2H > o
1H > o4H > o
3H
Q.15Br
(I), which is not the correct statement
(A) I is more soluble than bromocyclopropane(B) I gives pale yellow ppt. on addition with AgNO3(C) I is having lower dipole moment than bromocyclopropane
(D) I is more ionic than
Q.16 Which one of the following carbocation would you expect to rearrange.
(A) (B) (C)
(D)
Q.17 How many 1,2-Shifts are involved during the course of following reaction:
(A) 1 (B) 2 (C) 3 (D) 4
Q.18 How many 1,2-Shifts are involved during the course of following reaction:
FINAL LAP - 2019REACTION INTERMEDIATESEXERCISE-IIQ.1 Which of the following can be produced by Wurtz reaction in good yield.
(A) (B) (C) (D)
Q.2 Select true statement(s):(A) Instead of radical substitution, cyclopropane undergoes electrophilic addition reactions in sun light.(B) In general, bromination is more selective than chlorination.(C) The 2,4,6-tri-tert, butylphenoxy radical is resistant to dimerization.(D) The radical-catalysed chlorination, ArCH3 ArCH2Cl, occurs faster when Ar = phenyl than whenAr = p-nitrophenyl.
Q.3 Choose all alkane that give only one monochloro derivative upon reaction with chlorine in sun light.
(A) (B) (C) (D)
Q.4
h/CCl
NBS
4 HBr
(X) + (Y) enantiomeric pair
(A) (B) (C)
Br
Br(D)
Br
Br
Q.5 Select correct statement about the product (P) of the reaction:
42 CCl/Br P
(A) P is optically inactive due to internal compensation(B) P is optically inactive due to the presence of plane of symmetry in the molecule(C) The structure of P can have three optical isomers possible.(D) P can have four possible optical isomers.
Q.6 Products formed when HCl adds to 2,4- hexadiene is:(A) 4-chloro-2-hexene (B) 2-chloro-3-hexene(C) 2-chloro-4-hexene (D) 1-chloro-2-hexene
Q.3 Identify missing products in the given reaction sequence.CH3 – CH2 – CH3 hv/Br2 (A) KOHaq (B)
42SOH (C)
4
2
CCl
Br (D) = ?
Q.4 Find out the total no. of products (including stereo) in the given reaction :CH3
NBS, CCl4
Peroxide, Products.
Q.5 With the help of following data show HBr exhibits the peroxide effect.H1
0/kJ mol–1 H20/kJ mol–1
H–X X
+ CH2 = CH2 X CH2 – C
H2 XCH2– C
H2 + H–X XCH2CH3 + X
HCl –67 + 12.6HBr –25.1 – 50.2HI +46 –117.1
Q.6 Addition of small amount of (C2H5)4Pb to a mixture of methane and chlorine, starts the reaction at140°C instead of the usual minimum 250°C. Why?
Q.7 On chlorination, an equimolar mixture of ethane and neopentane yields neopentyl chloride andethyl chloride in the ratio 2.3 : 1. How does the reactivity of 1° hydrogen in neopentane compare withthat of a 1° hydrogen in ethane?
Q.8 It required 0.7 g of a hydrocarbon (A) to react completely with Br2 (2.0 g) and form a non resolvableproduct. On treatment of (A) with HBr it yielded monobromo alkane (B). The same compound (B) wasobtained when (A) was treated with HBr in presence of peroxide. Write down the structure formula of(A) and (B) and explain the reactions involved.
Q.9 Give product(s) in each of the following reactions .
(a) CH3 –
3CH|CH – CH2 – CH2– CH3 hv/Br2 (A) (major)
(b) + NBS 256 )COOHC( (B)
(c) CH3 – CH2 – CH = CH2 + Me3COCl hv (C) + (D)
(d) C6H5 – CH2 – CH2 – CH3
3
3
3
CH|
/ClOCCH|
CH
(E) (major)
Q.10 We saw that acid-catalyzed dehydration of 2,2-dimethyl-cyclohexanol afforded 1,2-dimethylcyclohexene.To explain this product we must write a mechanism for the reaction in which a methyl shift transforms asecondary carbocation to a tertiary one. Another product of the dehydration of 2,2-dimethylcyclohexanolis isopropylidencyclopentane. Write a mechanism to rationalize its formation.
FINAL LAP - 2019REACTION INTERMEDIATESQ.12 Assuming that cation stability governs the barrier for protonation in H – X additions, predict which
compound in each of the pairs in parts (a) & (b) will be more rapidly hydrochlorinated in a polar solvent.(I) (II)
(a) CH2 = CH2 or
(b) or
Q.13 Choose the member of the following pairs of unsaturated hydrocarbons that is more reactive towardsacid-catalysed hydration and predict the regiochemistry of the alcohols formed from thi s compound.
(a)
(I)
or
(b) or
(c) or
Q.14 Give product in the following reaction.
(i) NH2
HCl
NaNO2 A (ii) NH2 HCl
NaNO2 B
(iii)
CH2NH2
HCl
NaNO2 C (iv)
NH2
HCl
NaNO2 D
(v) NH2OH
HCl
NaNO2 E
Q.15 What are the products of the following reactions ?
Q.3 On conversion into Grignard followed by treatment with ethanol, how many alkyl halides (excludingstereoisomers) would yield 2-methyl butane.(A) 2 (B) 3 (C) 4 (D) 5
Q.4 Which of the following reacts with Grignard reagent to give alkane?(A) nitro ethane (B) acetyl acetone (C) acetaldehyde (D) acetone
Q.5 How many litres of methane would be produced when 0.595 g of CH3MgBr is treated with excess ofC4H9NH2(A) 0.8 litre (B) 0.08 litre (C) 0.112 litre (D) 1.12 litre
Q.6 How many litres of ethene would be produced when 2.62 g of vinyl magnesium bromide is treated with224 ml of ethyne at STP.(A) 0.224 litre (B) 0.08 litre (C) 0.448 litre (D) 1.12 litre
Q.7
MgBr OH
+ AA
(A) (B) (C) (D)
O – Ph
Q.8 In which of the following reactions 3°alcohol will be obtained as a product.
Q.10 Compounds are shown with the no. of RMgX required for complete reaction, select the incorrect option(A) CH3COOC2H5 1(B) CH3COCl 2(C) HOCH2COOC2H5 3
(D) 4
Q.11 What will be the order of reactivity of the following carbonyl compounds with Grignard's reagent?
(I) C = OH
H(II) C = O
H
CH3
(III) C = OCH3
CH3(IV) C = O
Me C3
Me C3(A) I > II > III > IV (B) IV > III > II > I(C) II > I > IV > III (D) III > II > I > IV
Q.12 Carbonyl compound (X) + Grignard reagent (Y) Me PhOH
Q.15 The number of moles of grignard reagent consumed per mole of the compound
is:
(A) 4 (B) 2 (C) 3 (D) 1
Q.16 Select the correct statement:(A) 1,4-dibromobutane react with excess of magnesium in ether to generate di-Grignard reagent.(B) 1,2-dichlorocyclohexane treated with excess of Mg in ether produces cyclohexene.(C) Vicinal dihalides undergo dehalogenation to give alkene when heated with Zn dust or Mg.(D) 1,3-dichloropropane by treatment with Zn dust or Mg forms cyclopropane.
(A) is ethyl acetate(B) further react with CH3MgBr/H2O+ to give acetone(C) further react with CH3MgBr/H2O+ to give t-butyl alcohol(D) Can give pinacol when treated with Mg followed by H2O
Q.21 Order of rate of reaction of following compound with phenyl magnesium bromide is:
IClCMe
||O
II
HCMe||O
III
EtOCMe||O
(A) I > II > III (B) II > III > I (C) III > I > II (D) II > I > III
Q.22 Select the correct order of decreasing reactivity of the following compounds towards the attack ofGrignard reagent(I) Methyl benzoate (II) Benzaldehyde (III) Benzoylchloride (IV) Acetophenone(A) II > III > I > IV (B) III > IV > II > I(C) III > II > IV > I (D) II > IV > I > III
CH3(A) The product is optically active(B) The product contains plane of symmetry(C) The product shows geometrical isomerism.(D) The product shows optical isomerism.
Q.43 The reaction of 1 mole each of p-hydroxy acetophenone and methyl magnesium iodide will give
(A) CH4 + IMgO C—CH3
O(B) CH –O3 C—CH3
O
(C) CH –C3 OH
OMgI
CH3
(D) CH O3 C—CH3
OMgI
Q.44 (i) O
+ Ph Mg Br 1r Ph CH2 CH2 OH
(ii) O
+ Ph Mg Br 2r Ph CH2 CH2 CH2 OH
(A) r2 > r1 (B) r1 > r2 (C) r1 = r2 (D) r1 = 2r2
Q.45 How many moles of Grignard reagent will be required by one mole of given compound?
C – OEt
C – Cl
O
OCH –CH2 2
Cl
SHHO
(A) 7 (B) 6 (C) 8 (D) 5
Q.46 Consider the given organometallic compound.(I) (CH3)2Hg (II) (CH3)2Zn (III) (CH3)2Mg (IV) CH3LiThe correct decreasing order of ionic character is(A) I > II > III > IV (B) II > I > III > IV (C) I > III > II > IV (D) IV > III > II > I
For Q. No.48 to Q. No. 50Consider the given reaction and answer the following questions
COOCH3
OCH3
SH||OO
O
)excess(
MeMgBr Products
Q.48 No. of RMgX consumed in the reaction is(A) 4 (B) 5 (C) 6 (D) 7
Q.49 How many product will be fromed in given reaction (excluding stereo)(A) 2 (B) 3 (C) 4 (D) 5
Q.50 Which of the following reaction will give the same Hydrocarbon formed as one of the product in theabove reaction.(A) EtMgBr + Me – OH (B) PhMgBr + Me – OH (C) MeMgBr + Ph – OH (D) MeMgBr + CH3 – CHO
Q.51 Compare the two methods shown for the preparation of carboxylic acids:
Which of the following statements correctly describes this conversion?
(A) Both method 1 and method 2 are appropriate for carrying out this conversion.(B) Neither method 1 nor method 2 is appropriate for carrying out this conversion.(C) Method 1 will work well, but method 2 is not appropriate.(D) Method 2 will work well, but method 1 is not appropriate.
Q.52 Which of the given compound can not show acid-base reaction with Grignard reagent.
(A) CH3–NO2 (B) (C) Cl (D)
Q.53 (a)
O
O
O
OClCl
Cl
SHOMe
C
)excess(
RMgCl Number of moles of Grignard reagent consumed.
(b) Number of mole of Grignard consumed in given molecule.(When Grignard reagent is in excess)
OXIDATION OF ALKENES, ALCOHOLS & CARBONYL COMPOUNDS
(I) OXIDATION OF ALKENES
R–CH=CR2 R–CH—CR2
R–CH—CR2
OsO4
Cold dil.alkalineKMnO4
H2OOH
OH
OH
OH* Cold dil. alkaline KMnO4 is called as Bayer’s reagent.* Overall syn addition* Given by alkenes & alkynes* Benzene & Cyclopropane can not give this reaction.If we use acidic KMnO4 or warm KMnO4 or too concentrated KMnO4 the oxidative cleavage ofGlycol occurs resulting in mixture of Carboxylic acids & Ketones.
R–CH = CR2
4KMnO,H RCOOH + R2C = O
Hot acidic KMnO4, Hot acidic K2Cr2O7 & hot acidic NaIO4 gives same result with alkene. The effectis similar to that of oxidative ozonolysis on alkenes.
Preilschaive reaction :Epoxidation of alkenes is reaction of alkenes with peroxyacids.
CH2=CH2 + HOOCCH||
O
3 CH –CH2 2
O + OHCCH
||O
3
With the decrease in nucleophilicity of double bond, rate of reaction decreases. With the decrease in ewithdrawing substituents in leaving group, rate decreases.
(1) Cu / 300°C (or Red hot Cu tube) (2) H /KMnO , 4 (Strong oxidising agent)(3) H /K Cr O , 2 2 7 (Strong oxidising agent) (4) PCC (Pyridinium chloro chromate)
HN CrO3 Cl or N CrO3 + HCl
(5) Collin’s reagent (6) Sarett reagent (i.e. PCC in CH2Cl2)
(N
(2 mol) + CrO3 + CH2Cl2)N
+ CrO3 + HCl + CH2Cl2
(7))dichromatePyridinium(
PDC
2
Cr O2 7
N
(8) Jones reagent (H2CrO4 in Anhydrous acetone)
or CrO3 + H2SO4 in acetone.Sufficiently mild so that it oxidises alcoholswithout oxidising or rearranging double bonds(8 or 9)
(9) TsCl + DMSO + NaHCO3 (10) MnO2-Oxidises only allylic or benzylic–OH.i.e.
RCH2OH ClTs RCH2OTss3NaHCO
DMSO RCHO 1° Allylic or benzylic OH 2MnO Aldehyde
R2CHOH ClTs R2CH–OTss3NaHCO
DMSO R2CO 2° Allylic or benzylic OH 2MnO Ketone
R3COH ClTs R3C–OTss3NaHCO
DMSO × No effect on 3° ROH and on Carbon-carbon
multiple bond.(11) Periodic cleavage (12) NBS
A similar oxidation is obtained incase of HIO4known as periodic cleavage. (13) Openaur oxidation
R–CH–OH
R– CH — O – I = O
R–CH
R C–OH2
R C–O–H2
+ +
+
HO–I=O
O
O
O
O
O
HIO3
R C=O2
R – CH – R
(R CH–O) 2 2
R CH–OH + 2
R – C – R
CH –C – CH3 3OH
O
O
Al(OCMe )3 3
Al(OCMe )3 3
Me COH + Al(OCHR )3 2 3
Al–O
3R C=O + (Me C–O)2 2 3
3Me C2
O CR2H
AlBut reaction is only observed for Vic-diols. Oxidation of alcohol with aluminium tertiary
Ketons are not easy to oxidize so they do not give these 5 tests. These five tests can be used to distinguishaldehyde and ketones. Both gives 2,4 DNP test
KETONES ARE DIFFICULT TO OXIDIZEKetones can be oxidized from their enolic form at high temperature with very strong oxidizing agent.Oxidation of ketones is sometimes governed by Popoff’s rule. According to this rule carbonyl groupremains with the smaller alkyl group. More electron rich alkene will be easy to oxidized.
Me – C – Me
O
MeCOOH + CO + H2 2O[O]
Allylic oxidationSeO2 is a selective oxidizing agent with converts –CH2– group adjacent to carbonyl group into carbonylgroup.The reagent, in general, oxidises active methylene and methyl groups to ketonic and aldehydicgroups respectively.
CCH||
O
2 2SeO
CC||||
OO
; 3CHC||
O
2SeO CHOC||
O
Double bonds, triple bonds and aromatic rings may also activate the methylene group.The methylene ormethyl group to the most highly substitued end of the double bond is hydroxylated according to theorder of preference of oxidation CH2 > CH3 > CH groups.
CH3= CH–CH3 2SeO 22 CHCHCH|
OH
2° C – H > 1° C – H > 3° C – HRate of reactivity orderCH3 – CH = CH – CH2–CH3 2SeO
(1) ALDOL CONDENSATIONThe -hydrogen of carbonyl compounds are acidic due to the fact that the anion (enolate ion) is stabilizedby resonance.
CH – C – H2||O
|H OH (base)
CH = C – H2 |O
CH – C – H2 ||O
(enolate anion) (enolate anion)(3 Hydrogens)
Base catalysed AldolIn aqueous base, two acetaldehyde molecules react to form a -hydroxy aldehyde called aldol. Thereaction is called Aldol condensation. The enolate ion is the intermediate in the aldol condensation ofaldehyde and ketone. Acetaldehyde for instance, forms a dimeric product aldol in presence of a dilutebase (10% NaOH).
2CH3CHO C5
HOH/HO
OH|
CHOCHCHCH 23 OH2
CH3 – CH = CH – CHO
-hydroxy butyraldehyde (Aldol)Mechanism :
H–C–CH2 H–C–CH2–CH–CH3
H–C–CH–CH–CH3
H–C–CH3
H–C=CH2
O O
O O
O
O
O
O
OHOH
HH–O–H
HOHH–C=CH–CH–CH3
–OH
H–C–CH=CH–CH3
rds
OH
Aldols are stable and may be isolated. They, however can be dehydrated easily by heating the basicreaction mixture or by a separate acid catalyzed reaction. Thus if the above reaction is heated theproduct is dehydrated to 2-butenal (crotonaldehyde).
FINAL LAP - 2019NAME REACTIONSAcid catalysed AldolIn acid catalysed aldol condensation enol form of carbonyl is the nucelophile in place of enolate.Mechanism :
CH –C–CH3 3
CH –C–CH3 3 CH –C–CH3 3
CH –C–CH3 3
CH –C=CH3 2
CH –C=CH3 2
CH3
CH3
CH3
CH3CH3H
CH –C=CH–C–CH3 3
CH –C–CH=C–CH3 3
CH –C—CH –C–CH3 2 3
CH –C—CH –C–CH3 2 3CH –C—CH–C–CH3 3
O
O O–H
O–H
OH
O–H
O
O
O–H
OO
OH
OHOH2
H / H O,+2
H+
( unsaturated carbonyl compound)
+
–H O2
–H
H
(Aldol)Q.1 Write the product and mechanism for given reactions.
Esters undergo SNAE Reaction, when attacked by a Nu� generated by the interaction of a base(usually base related to the Alkoxy anion of ester) with one of the molecule of ester and this Nu� attackson another molecule. The reaction over all is considered as condensation of esters known as claisenester condensation.
ORCMe2||
O
ionAcidificat)ii(RONa)i(
ORCCHCMe||||
OO
2
(-keto ester)Mechanism :
ORCCH||
O
3 ROH
RO
RO
Na
Na
Na
(rds)
Me–C=CH–COOR Me–C–CH=C–OR
O OO
Acidification
ORCCHCMe||||
OO
2
Some times, when two ester groups are present within the molecule then the condensation occursintramolecule then cyclization caused thus is known as Dieckmann cyclization or Dieckmann'scondensation.
FINAL LAP - 2019NAME REACTIONS(3) PERKIN CONDENSATION
In Perkin reaction, condensation has been effected between aromatic aldehydes and aliphatic acidanhydrides in the presence of sodium or potassium salt of corresponding acid of that anhydride, to yield, -unsaturated aromatic acids.
FINAL LAP - 2019NAME REACTIONS(5) REFORMATSKY REACTION
-halo esters when treated with Zn in gives organometallic halo ester which provides the attacking Nu�for the another reactant, which is a carbonyl compound. When Nu� attacks on carbonyl compound itgives an intermediate which upon acidic hydrolysis followed by heating, results in formation of -unsaturated acid. The overall reaction is known as Reformatsky reaction.
(6) CANNIZARO REACTIONThis reaction is given by aldehyde having no - hydrogens in the presence of conc. NaOH/ or KOH/.
HCH||
O
KOH OCH
||O
K+ + CH3OH
Mechanism :
OH
O O
H HH–C–H C r.d.s O–CH3 + H–O–C–H
||O
OHCH3 + O–C–H
||O
In the presence of a very strong concentration of alkali, aldehyde first forms a doubly charged anion (I)from which a hydride anion is transferred to the second molecule of the aldehyde to form acid and analkoxide ion. Subsequently, the alkoxide ion acquires a proton from the solvent.
H–C
H–C
H–C–H H–C–H
H–C–H
O
O
O O
O H
H
O
OH
H
O
O
OH OH
(I)
(I)+ C = O + H–C–O
H
H
HCH O2 HCH OH2H O2
From solvent + OH
Q.1 Which of following will not undergo Cannizaro reaction
FINAL LAP - 2019NAME REACTIONS(7) BENZIL-BENZILIC REARRANGEMENT OR BENZILIC ACID REARRANGEMENT
The base catalysed reaction of 1, 2-diketones to a salt of -2-hydroxy carboxylic acid is known asBenzilic acid rearrangement, this reaction is mainly applicable when aryl group is present on both carbonylcarbons.
FINAL LAP - 2019NAME REACTIONS(8) MICHAEL ADDITION
Michael Addition : unsaturated carbonyl compound undergo michael reaction with compoundshaving active methylene. Many different nucleophile can add to unsaturated carbonyl compound.When the nucleophile is an enolate the addition reaction has a special name MICHAEL REACTION.Mechanism :
R – C – CH – C – R2
||O
OH (base)
– H O2
||O
R – C – CH – C – R||O
||O R – CH= CH – C – R
||O
unsaturatedcarbonyl compound
HOH– OH
CH – CH – CH = C – RCH – CH – CH – C – R2
O
RR
R–CR–C
R–CR–C
OOO
OO
Q.1 CH – C – CH – C – CH + CH = CH – CH3 2 3 2 || || ||O O O
FINAL LAP - 2019NAME REACTIONS(9) TISCHENKO REACTION / TISCHENKO CONDENSATION
(1) This reaction takes place between two molecules of aldehydes.(2) Reaction is catalysed by aluminium ethoxide.(3) This is a two-step reaction, i.e., redox reaction followed by ester formation. Thus this reaction is
FINAL LAP - 2019NAME REACTIONS(10) COREY HOUSE SYNTHESIS
Reaction of Gilman’s reagent with alkyl halide gives alkane as one of the product which is known asCorey House synthesis,
[R Cu] Li2 R – X' R – R + RCu + LiX'
Gilman reagent(Lithium dialkyl cuprate)
To obtained good yield of alkane, the alkyl halide R’–X must be either a methyl halide, a primary alkylhalide, or a secondary cycloalkyl halide.(Best CH3X, if possible)
Preparation of Gilman reagent:
R – X XLi
Li2
um)iLith(alkylmole)(2
LiR XLi
,XCu
reagentGilman
2 LiCu][R
Complete the following reaction sequence :
Q.1 CH3– I OEt
Li
2
(A) CuI (B)
ICHCHCHCHCH 22223 (C)
CH3CH2CH2CH2BrOEt
Li
2
(D)CuI
(E)
BrCHCHCHCHCH 22223 (F)
Q.2 (A) hCl2 (B) Li (C)
CuCl (D)
ClCH3
3
323
CH|
CHCHCHCH
Q.3 CH3 – CH2 – Br
,CuBr)ii(
Li)i( (A) ClCH3 (B)
Q.4 I
CuI)ii(Li)i(
(A)
BrCHCH 23 (B)
Q.5 CH3 – CH2 – Br Li (A) CuBr
Cl|
CHCHCH 33
(ii)
(i)
H
O
(C)
(D)
(B)
Q.6 How can you prepare following compounds using corey house synthesis.
The reaction of alkene or Alkyne with ozone (O3) followed by hydrolysis is known as ozonolysis.It is two types : (I) Reductive ozonolysis In presence of reducing agent
(II) Oxidative ozonolysis In presence of oxidising agentReducing agents : Zn, H2O or Zn, CH3COOH or (CH3)2S or (Ph)3P etc.Oxidising agents : H2O2 or
O||
HOOCR or Ag2O etc.
Example 1 :
C=C C=OC CR RRR’ R’O
O OR RRH H
O3 Zn H2O/
H2O2
step I–70°C
Reductiveozonolysis
Oxidativeozonolysis
+ R’–C–H
R–C–R R’–C–OH
O
O O
+
[SCT- Cut the double bondand paste two oxgen atomsand vice versa]
Mechanism :
C=OR’
H
R–C–R R’–C–H
O O
+
C
C
R
R’
R
H
+ +O
O
O
C
C
R R
H
H H
O
OO
OO
R
R’R C
CO
OR
R
R’O=C
R’
R—C—O
R
OC
O
Note : In case of oxidative ozonolysis aldehyde (not ketone) further undergoes oxidation which givesacid as product.
FINAL LAP - 2019NAME REACTIONS(12) OXYMERCURATION-DEMERCURATION
OMDM is a hydration process of alkene according to Markovnikov’s rule with no rearrangement of cyclicmercuinium ion. When reaction with that reagent is complete, sodium borohydride and hydroxide ion are addedto the reaction mixture.
Sodium borohydride (NaBH4) converts the carbon-mercury bond into a carbon-hydrogen bond. Becausethe reaction results in the loss of mercury, it is called demercuration.
OH|CHCHCH 23 —Hg—OAc
¯HO
NaBH4
OH|CHCHCH 33 + Hg + AcO¯
Q.1 OH,NaBH)ii(
OHCH)OCOCH(Hg)i(
4
33 Q.2 OH,NaBH)ii(
OH,)OAc(Hg)i(
4
22
Q.3
CH3
¯HO,NaBH)ii(
OHCH,)COOCF(Hg)i(
4
323 Q.4
OH,NaBH)ii(
HOCH,)OAc(Hg)i(
4
32
Q.5OH
OH,NaBH)ii(
OH,)OAc(Hg)i(
4
22 Q.6
C CH
OH,NaBH)ii(
OH,)OAc(Hg)i(
4
22
Q.7 How could each of the following compounds be synthesized from an alkene by OMDM?
FINAL LAP - 2019NAME REACTIONS(13) HYDROBORATION-OXIDATION
Hydroboration has been developed by brown as a reaction of tremendous synthetic utility because alkylboranes are able to undergo a variety of transformation. Hydroboration is a one step, four centre, cisaddition process in accordance with Markovnikov’s rule but after oxidation it seems to be appear toviolate Markovnikov’s rule.
FINAL LAP - 2019NAME REACTIONS (14) RIEMMER TIEMENN REACTION
It is believed that chloroform and hydroxide ion react to produce an electron deficient intermediatedichloro carbene : CCl2 (DCC)
HHO— H2O + : :CCl2 + Cl
Treatment of phenol with DCC in basic medium introduces an aldehyde group, onto the aromatic ring.This reaction is known as Reimmer Tiemenn reaction.
H)iii(,CCl:)ii(
HO)i(
2 +
The Reimmer–Tiemenn reaction involves electrophilic substitution on highly reactive phenoxide ring. Theelectrophilic reagent is dichloro carbene : CCl2
+ :CCl2 HO—
H
o-product is major product because :(i) there are two o-positions available as compared to one para.(ii) o-product is more stable due to the formation of six membered chelate formation.
If 1 ° amines (aliphatic and aromatic) react with DCC in basic medium it yield isocyanide or carbylamine.This reaction is known as carbylamine reaction:
The product is known as isocyanide & it is a foul smelling substance.
Q.1 Step marked by is :(A) Aromatization reaction (B) intramolecular acid base reaction(C) both of the above (D) none of the above
Q.2 Addition of NaCl in aqueous solution will make Riemmer Tiemann reaction:(A) slower (B) faster (C) no effect on rate (D) can't be predicted
Q.3 If is used instead of during Riemmer Tiemenn reaction the product formed is:
(A) (B) (C) (D)
Q.4 If CCl4 is used in place of CHCl3 during Riemmer Tiemenn reaction, the product formed is:
(A) (B) (C) (D)
Q.5 o/p ratio in Riemmer Tiemenn reaction follows following order, if NaOH & KOH are used one by oneas base(A) NaOH = KOH (B) NaOH > KOH(C) NaOH < KOH (D) can't be decided on the basic of information given here
Q.6 Which of the following will not give carbylamine reaction(A) t-butyl amine (B) aniline (C) sec. butylamine (D) N-methyl methanamine
Q.7 Correct order of rate of carbylamine reaction for following compounds is:
(I) (II) (III)(A) I > II > III (B) II > I > III (C) II > III > I (D) III > I > II
Q.8 Rate of carbylamine reaction for Me–NH2 will be :(A) CHF3 > CHCl3 (B) CHCl3 = CHF3 (C) CHCl3 CHF3 (D) CHF3 < CHCl3
Q.2 On complete hydrolysis of starch, we finally get(A) Glucose (B) Fructose(C) Glucose and fructose (D) Sucrose
Q.3 The term anomers of glucose refers to(A) Isomers of glucose that differ in configurations at carbons one and four (C–1 and C-4)(B) A mixture of (D)-glucose and (L)-glucose(C) Enantiomers of glucose(D) Isomers of glucose that differ in configuration at carbon one (C–1)
Q.4 Which of the following is an example of aldopentose?(A) Erythrose (B) Ribose (C) Fructose (D) Dihydroxyacetone
Q.5 Glucose and fructose form(A) Same osazone (B) Same acid on oxidation(C) Same alcohol when reduced (D) Different osazone
Q.6 The change of optical rotation of glucose solution with time is refered to as:(A) Mutarotation (B) Inversion (C) Specific rotation (D) Autorotation
Q.7 To become a carbohydrate a compound must contain at least:(A) 2 carbon atoms (B) 3 carbon atoms (C) 4 carbon atoms (D) 6 carbon atoms
Q.9 The correct name of 'sucrose' is(A) -D-glucopyranosyl--D-fructofuranoside(B) -D-glucopyranosyl--D-fructofuranoside(C) -D-glucopyranosyl--D-fructofuranoside(D) -D-glucopyranosyl--D-fructofuranoside
Q.10 Which one of the following is laevorotatory(A) Glucose (B) Sucrose (C) Fructose (D) None of these
Q.11 Which of the following is a disaccharide(A) Lactose (B) Starch (C) Cellulose (D) Glucose
Q.15 Which pair is different for reaction with Fehling solution:(A) Glucose, Fructose (B) HCHO, CH3CHO(C) CH3COCH3, C6H5CHO (D) Glucose, Sucrose
Q.16 Glucose contains in addition to aldehyde group(A) One secondary OH and four primary OH groups(B) One primary OH and four secondary OH groups(C) Two primary OH and three secondary OH groups(D) Three primary OH and two secondary OH groups
Q.17 Glucose reacts with excess of phenyl hydrazine and forms(A) Glucosazone (B) Glucose phenyl hydrazine(C) Glucose oxime (D) Sorbitol
Q.18 Which set of terms correctly identifies the carbohydrate shown
HOH C2
H
H
OH H
OH
OH
H
O
1. Pentose 2. Hexose3. Aldose 4. Ketose5. Pyranose 6. Furanose(A) 1, 3 and 6 (B) 1, 3 and 5 (C) 1, 4 and 6 (D) 2, 3 and 6
Q.19 Hydrolysis of sucrose is called:(A) Inhibition (B) Saponification (C) Inversion (D) Hydration
Q.20 Which of the following pentoses will be optically active
Q.21 The secondary structure of a protein refers to(A) -helical backbone (B) Hydrophobic interactions(C) Sequence of -amino acids (D) Fixed configuration of the polypeptide backbone
Q.22 A tripeptide is written as Glycine-Alanine-Glycine. The correct structure of the tripeptide is
(A) NH2
O
NH
CH3
O
NH
CH3
COOH
(B) NH2
O
NHO
NH
CH3
COOH
CH3
(C) NH2
O
NHO
NH COOHCH3
(D) NH2
O
NHO
NH COOH
CH3
CH3
Q.23 Which compound can exist in a dipolar (zwitter ion) state(A) C6H5CH2CH(N = CH2) COOH (B) (CH3)2CH·CH(NH2)COOH(C) C6H5CONHCH2COOH (D) HOOC·CH2CH2COCOOH
Q.24 Which of the following statement is incorrect for maltose.(A) It is a disaccharide (B) It undergoes mutarotation(C) It is a reducing sugar (D) It does not have hemiacetal group.
Q.25 Identify the correct statement about lactose.(A) It consists of one galactose and one glucose unit(B) Mutarotation is not possible(C) Anomeric carbon of galactose is attached to 4' carbon of glucose which is –1, 4'-glycoside bond.(D) Lactase is not used to cleave the –1, 4'-glycoside bond.
Q.26 Which of the following carbohydrates would be most abundant in the diet of strict vegetarian?(A) Amylose (B) Glycogen (C) Cellulose (D) Maltose
Q.28 D-Ribose when treated with dilute HNO3 forms.
(A)
HO
HO
O
O
HHH
OHOHOH , Achiral (B)
OH
HO
O
O
HH
H
OHHO
OH , Chiral
(C)
HO
O
O
HH
H
OHHO
HO
OH , Achiral (D)
OH
HO
O
O
HHH
OHOHOH
, Chiral
Q.29 Consider the given process.
H
HH
HOOHOH
CH OH2
H–C–OH
CHO
(I)
OH¯
H O2 H
HH
HOOHOH
CH OH2
C–OH
CHOH
(II)
OH¯
H O2 H
HH
HOOHOH
CH OH2
HO–C–H
CHO
(III)and identify the incorrect statement.(A) Configuration at C–2 is lost on enolisation(B) I and III are epimers(C) Proton transfer from water to C–1 converts ene diol to an aldose.(D) D-glucose can isomerise to D-fructose through enol intermediate.
Q.30 Glucose HI/P)ii(
OH/HCN)i( 3
P
P is :(A) n- heptanoic acid (B) 2-methyl hexanoic acid(C) n-heptane (D) 2-methyl hexane
Q.31 When methyl D-glucopyranoside is treated with HIO4 how many moles of HIO4 are consumed with permole of the sugar ?(A) 2 (B) 3 (C) 4 (D) 5
Q.32 The configuration of the C-2 epimer of D-glucose is-(A) 2R, 3S, 4R, 5S (B)2S, 3S, 4R, 5R(C) 2S, 3R, 4S, 5R (D)2R, 3S, 4R, 5R
Q.34 Same osazone derivative is obtained in case of D-glucose, D-Mannose and D-Fructose due to(A) the same configuration at C-5(B) the same constitution.(C) the same constitution at C-1 and C-2(D) The same constitution and same configuration at C-3, C-4, C-5 and C-6 but different constitutionand configuration at C-1 and C-2 which becomes identical by osazone formation.
Q.35NaBH4
NaBH4
(P)
(R)
D(–) –Erythrose
D(–) –Threose Which of the following statement is correct about P and R ?(A) Both are optically active(B) Both are optically inactive(C) P is optically inactive and R is optically active(D) Neither P nor R has asymmetric carbon.
Q.36 Which of the following compounds will not show mutarotation:(A) -D (+) glucopyranose (B) -D(+) glucopyranose(C) Methyl--D-glucopyranoside (D) -D(+) galactopyranose
Q.37 Amylose and cellulose both are linear polymers of glucose. The difference between them is:(A) Amylose has (1 4') linkage and cellulose has (1 4') linkage(B) Amylose has (1 4') linkage and cellulose has (1 4') linkage(C) Amylose has (1 4') linkage and cellulose has (1 6') linkage(D) Amylose has (1 4') linkage and cellulose has (1 6') linkage
Q.38 Glycogen on hydrolysis gives:(A) Lactose and Glucose (B) Only Glucose(C) Glucose and Fructose (D) Glucose and Maltose
Q.39 An example of disaccharide made up of two unit of the same monosaccharide.(A) Maltose (B) Sucrose (C) lactose (D) None
Q.40 The colour of the precipitate formed when a reducing sugar is heated with Fehling solution is:(A) Brown (B) Red (C) Blue (D) Green
Q.41 Osazone formation involves only 2 carbon atoms of glucose because of(A) Oxidation (B) Chelation (C) Reduction (D) Hydrolysis
Q.43 Hydrolysis of lactose with dilute acid yield(A) Equimolar mixture of D-glucose and D-glucose.(B) Equimolar mixture of D-glucose and D-galactose.(C) Equimolar mixture of D-glucose and D-fructose.(D) Equimolar mixture of D-galactose and D-galactose.
Q.44 Celluose is a straight chain polysaccharide composed of(A) D-glucose units joined by -glycosidic linkage(B) D-glucose units joined by -glycosidic linkage(C) D-galactose units joined by -glycosidic linkage(D) D-galactose units joined by -glycosidic linkage
Q.45 - amino acid when heated with BaO forms -(A) - unsaturated acid (B) - unsaturated amine(C) Carboxylic acid (D)Amine
Q.46 The pH of the solution containing following zwitter ion species is NH3
R
COO¯
H
(A) 4 (B)6 (C) 8 (D)9
Q.47 Peptide linkage is -
(A) –C – O –
O(B)–C – NH2
O
(C) –C – NH–
O
(D)–C – NH–NH2
O
Q.48 The monomer of nucleic acids are held together by(A) Phosphoester linkage (B) Amide linkage(C) Glycosidic linkage (D) Ester linkage
Q.49 Test used to identify peptide linkage in protein is:(A) Biuret (B) Ninhydrin test (C) Molisch test (D) 2,4-DNP test
Q.50 Which one of the following structures represents the peptide chain:
Q.52 The monomer that undergo radical polymerisation most easily is
(A) CH2=CH2 (B) C6H5CH=CH2 (C) CH2=CMeMe
(D) CH3–CH=CH2
Q.53 Select the incorrect statement about Nylon 2-nylon-6.(A) It is a copolymer.(B) It is biodegradable.(C) It is an alternating polyamide.(D) It is made up of
2
3
NH|
COOHCHCH and H2N(CH2)5COOH.
Q.54 The polymer formed as a result of following sequence of reaction is
Q.59 Which of the following compound(s) will give blue colour when it is converted into Lassaigne's extractand FeSO4 is added followed by FeCl3.
(I) (II) (III) NH2–OH (IV) 22 NHCNHNH||
O
(A) I and IV (B) IV only (C) I, III & IV (D) I, II, III & IV
Q.60 Select reagent which is used in laboratory to differentiate 1°, 2° and 3° amines from each other:(A) NaOH, I2 (B) PhSO2Cl (C) CHCl3 , KOH (D) CS2, HgCl2
Q.1 Carbohydrates may be :(A) Sugars (B) Starch(C) Polyhydroxy aldehyde/ ketones (D) Compounds that can be hydrolysed to sugar
Q.2 Select the correct statement:(A) S-glyceraldehyde is also known as L-glyceraldehyde(B) The configuration of the stereocenter most distant from the carbonyl group determines whether amonosaccharide is D or L.(C) Glucose and all naturally occurring sugars are D-sugars(D) D-erythrose and D-threose are diastereomers.
Q.3 Select the incorrect statement.(A) Monosaccharide are insoluble in organic solvents like diethyl ether.(B) Anomers of a cyclic monosaccharides differ in the position of the OH group at the hemiacetalcarbon.(C) D-ribose the OH group used to form the five membered furanose ring is located on C4.(D) Aldopentoses and ketohexoses form pyranose rings in solution.
Q.4 The peptide bond is a key feature in :(A) Vitamins (B) Proteins (C) Nucleotides (D) Polypeptides
Q.5 Which is not correct about monosaccharides.(A) Optically active polyhydroxy carbonyl compounds.(B) Fructose & glucose can not be distinguish by Br2/H2O(C) Glucose and mannose are anomers(D) Fructose & glucose can be distinguish by Fehling solution
Q.6 Which of the following is disaccharides ?(A) Lactose (B) Sucrose (C) Cellulose (D) Maltose
Q.7 Select the correct statement.(A) Proteins upon hydrolysis gives -amino acid only.(B) Except glycine, all other naturally occuring -amino acids are optically active.(C) In fibrous proteins polypeptide chains are held together by hydrogen and disulphide bonds.(D) Fats upon hydrolysis gives -amino acids
Q.8 Select the correct statement.(A) Coiling of polypeptide chain form fibrous protein.(B) Quarternary structure of protein also exist.(C) Lysine is an amino acid with basic side chain.
(D) The absolute configuration of NH3 – CH(CH2OH)COO¯ (L-serine) is S.
Q.9 Select the correct statement.(A) All proteins are polyamides formed by joining amino acids together.(B) All L-amino acids except cysteine have the S-configuration.(C) Proline is 1° amine consisting 6 membered ring.(D) Proline is a 2° amine consisting of five membered ring.
Q.10 Select the correct option.(A) Isoelectric point is the pH at which an amino acid exists primarily in its neutral form.(B) Isoelectric point is the average of pK a values of -COOH amino -
3NH groups [valid only forneutral amino acid](C) Glycine is characterised by two pKa values.(D) For neutral amino acid the concentration of zwitter ion is maximum at its isoelectric point.
Q.11 Amino acids are synthesised from.(A) -Halo acids by reaction with NH3.(B) Aldehydes by reaction with NH3 and cyanide ion followed by hydrolysis.(C) Alkyl halides by reaction with the enolate anion derived from diethyl acetamidomalonate & hydrolysis.(D) Alcohols by reaction with NH3 and CN¯ ion followed by hydroysis.
Q.12 Which of the following carbohydrates developes blue colour on treatment with iodine solution ?(A) Glucose (B) Amylose (C) Starch (D) Fructose
Q.13 Which of the following are correct.(A) Br2/H2O can be used to differentiate between aldose & ketose.(B) All monosaccharides are reducing sugar(C) Osazone formation destroys the configuration about C2 of an aldose but not effect the rest ofmolecule(D) Mono saccharides undergoes mutarotation.
Q.14 Match the column :Column I Column II
(A) Sucrose (P) Two acetals(B) Maltose (Q) No hemiacetal(C) Lactose (R) 1,4'-glycosidic bond(D) Cellulose (S) Hydrolysis product is glucose
POLYMERSQ.15 Select the correct statement.
(A) High density polythene is a linear polymer.(B) Low density polythene is a branched chain polymer.(C) Chain growth polymers are also known as addition polymer.(D) Step growth polymer is also known condensation polymer.
Q.16 Select the correct statement.(A) Elastomers have the weakest intermolecular forces.(B) Buna-N is an elastomer with crosslinks.(C) Some fibres have crystalline nature.(D) Thermoplastic polymers have stronger intermolecular forces than fibres
Q.1 The pKa values for the three acidic group P,Q,R are 4.3, 9.7 and 2.2 respectively
HOOC–CH –CH –COOH2
NH3
(R) (P)
(Q)
+
Calculate the isoelectric point of the amino acid ?
Q.2 Statement 1 : Bromine water changes glucose to gluconic acid.Statement 2 : Bromine water acts as oxidising agent.(A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.(B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.(C) Statement-1 is true, statement-2 is false.(D) Statement-1 is false, statement-2 is true.
Comprehension (3 to 5)Consider the following reversible process for a reaction of D-glucose.
Q.8 Which of the following statement is not true considering the process given above.(A) The general class of polymer formed is known as homopolymer(B) The polymer obtained is stereoregular(C) Buna–N can be prepared using above process(D) Synthetic rubber can be formed by above process using 1,3- butadiene.
Q.9 Match the column :Column I Column II(Carbohydrate) (Properties)
(A) Starch (P) Mutarotation(B) Sucrose (Q) Non reducing sugar(C) Lactose (R) -glycosidic bond(D) Maltose (S) -glycosidic bond
Q.12 Match the columnColumn I Column II(Functional group) (Test used or complex formed during confirmatory test)
(A) Aldehydic (P) [(C6H5O)6Fe]–3 (violet)
(B) Phenolic (Q)
CH — C–O–
CH — C–O–
CuOO
OO(blue)
(C) Alcohol (R)
RCH(OH)OSONH
RCH(OH)OSONHC NH
violet red
(S) (ROH)2Ce(NO3)4 (Red)
(T) Molisch's Test
Q.13 Read following statements.(a) Protein on complete hydrolysis give.(b) Hormons belong to the class of(c) Carbohydrates are stored in the body as(d) The excess of glucose is stored in the liver asWhich set of terms correctly identifies the statements given above.(i) Lipids (ii) -amino acid (iii) Peptides (iv) Fats(v) Proteins (vi) Cellulose (vii) Glycogen (viii) Peptones