Questions Chapter 1-10

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SECTION - I : STRAIGHT OBJECTIVE TYPE

1.1 IUPAC name of compound

CI Br is (A) 3-Bromo7-chloro-7ethyl-5-(1,1-dimethyethyl)-5-(2-methylpropyl)-3-methylnonane (B) 3-Bromo7-chloro-5-(1,1-dimethyethyl)-7-ethyl-3methyl-5-(2-methylpropyl)nonane (C) 3-Bromo7-chloro-7ethyl-3-,methyl-5-(1,1-dimethyethyl)-5-(2-methylpropyl)nonane (D) 3-Bromo-5-(1,1-dimethyethyl)-5-(2-methylpropyl)-7-chloro-7ethyl-5--3-methylnonane

1.2 Which of the following structure\s is the correct structure of 3-ethyl-5, 5-diisopropyl-7-methlnonane

(A) (B)

(C) (D)

1.3 The correct IUPAC name of the folllowing compound is

(A) 5,6-Diethyl-8-methyl dec-6-ene (B) 5,6-Diethyl-3-methyl dec-4-ene (C) 5,6-Diethyl-3-methyl dec-4-ene (D) 2,4,5-Triethylnon-3-ene

1.4 Correct IUPAC name of the following compound is :

HO

Br

(A) 3-(Hepta-2,4,6-trienyl)-4 bromo cyclopenta-2, 4, -dien-1-ol (B) 7-(2-Bromo-4-hydroxy cyclopenta-1,4-dienyl)hepta-1,3,5-triene

General Organic Chemistry-I

(GOC-I)

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(C) 7-(5-Bromo-3-hydroxycyclopenta-1,4-dienyl)hepta-1,3,5-triene (D) 3-Bromo-4-(hepta-2,4,6-trienyl)cyclopenta-2,4-dien-1-oll

1.5 The IUPAC name of the compound

Br

will be :

(A) Tropyluim bromide (B) 1-Bromocyclohepta-2,4,6-triene (C) 3-Bromocyclohepta-1,4,6-triene (D) 7-Bromocyclohepta-1,3,5-triene

1.6 Correct IUPAC name of the co,pound H C – CH – O – C₃ ₂ O – C – CH – CH₂ ₃

O O

(A) 4-(Ethyl methanolyonxy)phenylpropanoate (B) Ethyl 4-propanoyloxybenzenecarboxylate (C) 4-(1-Oxo-2-oxabutyl)phenylpropanoate

(D) 1-(1-Oxo-2-oxbutyl)-4-(1-oxopropoxy)benzene

1.7 Correct IUPAC name of the compound

Et

Me O

O is O

(A) 2-Ethyl-3-methylbut-2-ene-1,4-dioic anhydride (B) 3-Ethyl_2-methylbut-2-enedioic anhydride (C) 2-Ethyl-3-Methyl-1,4-diketobut-2-enoic anhydride (D) 2-Ethyl-3-methylcyclopenatanoxy-1,4-dione

1.8 The IUPAC name of the following compound is

COCl COOC H₂ ₅

(A) 2-(Ethoxycarbonyl) benzalychloride (B) Ethyl 2-(Chloroformyl)benzoate

(C) Ethyl 2-(chloromethanoyl)benzoate (D) Ethyl2-(Chorocarbonyl)benzene carboxylate.

1.9 IUAC name of Cl Br CH₃ C H₂ ₅ (A) 4-Bromo-6-chloro-2-ethyl-1-methylcyclohex-1-ene (B) 5-Bromo-1-chloro-3-ethyl-2-methylcyclohex-2-ene (C) 5-Bromo-3-chloro-1-ethyl-2-methylcyclohex-1-ene (D) 1-Bromo-5-chloro-3-ethyl-4-methylcyclohex-3-ene

1.10 A hydrocarbon (R) has six membered ring in which there is no unsaturation. Two alkyl groups are atttached to the ring adjacent to each other. One group has 3 carbon atoms with branching at 1st carbon atom of chain and another has 4 carbon atoms. The larger alkyl group has main chain of three carbon atoms of which second carbon is substituted. Correct IUPAC name of compound (R) is

(A) 1-(1-Methylethyl)-2-(1-methylpropyl)cyclohexane (B) 1-(2-Methylethyl)-2-(1-methylpropyl)cyclohexane

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(C) 1-(1-Methylethyl)-2-(2-methylpropyl)cyclohexane (D) 1-(1-Methylethyl)-2-butylcyclohexane

1.11 Identify the structure of x,

CH – C – CH ₃ – C – H + H – C – H + CH – C – C – CH – C – H ₂ ₂ O O O O O O CH₃ CH₃ CH₃ H /Ni2 Zn/H O2 O3 X (A) (B) (C) (D) H C₃

1.12 In the given sequence reaction which of the following is the correct structure of compounds A.

O O O O H H + HCHO H / Ni₂ (i) O₃ (ii) Zn / H₂O A(C H )₁₀ ₁₄ (A) (B) (C) (D) H

1.13 For the following reactions sequenceHOOC COOH O₃

H₂O₂ C7H10 H₂/ Ni excess Y C₇H₁₂ X

The structure consistent with X and Y are:

(A)

(B)

(C)

(D)

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1.14 An organic hydrocarbon on oxidative ozonolysis produces oxalic acid and butanedioic acid. Its structure is (A) (B) (C) (D) 1.15 C–O O H–C–O O and are

(A) Position isomers (B) Chain isomers

(C) Functional isomers (D) Metamers

1.16 In which reaction a chiral reactant is giving a chiral product.

C = O + H₂ H – C – OH Hint : (D) H Cl C = O CH₃ H Cl CH₃ H / Ni2 H CH – CH₂ ₃ CH = CH₂ CH₂OH O / Zn / H O 3 2 H C = C HOC CH₃ (B) H CHO CH₃ H H KMnO / OH /₄ – C – C H₂ ₅ CH₃ H Reductive ozonolysis (i) O (ii) Zn / H O 3 2

1.17 Which of the following statements is true abnout the follownig conformer (X)?

I CH₃ COOH I COOH (X) CH₃

(A) (X) is the most stable conformer of meso-2,3-Diiodo-2,3-dimethylbutanedioic acid

(B) The most stable conformation will be

CH₃ CH3 COOH COOH I I (Y)

(C) The dipole moment of (X)is not zero but that fo Y is zero. (D) None

1.18 An unsaturated hydrocarbon on jcomplete hydrogenation gives 1-isopropyl-3 methylcyclohexane, after ozonolysis it gives one mole of formaldehyde, one mole of acetone and one moleof2,4-Dioxohexanedial. The possible structure\s of the hydrocarbon maybe

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(A) (B) (C) (D)

1.19 How manyh assymmetric carbon atoms are present in (i) 2-Dimethyl cyclohexane

(ii) 3-Methyl cyclopentene (iii) 3-Methylcyclohexene

(A) 2,1,1 (B) 1,1,1 (C) 2,0.2 (D) 2,0,1

1.20 Which of the following statements is not correct?

(A) A compound whose molecule has D configuration will always be dextrorotatory

(B) A compound whose molecule has D configuration may be dextrorotatory or levorotatory (C) A compound whose molecule has R configuration may be dexrotatory or levorotatory (D) A compound whose molecule has L configuration may be dextrorotatory or levorotatory

1.21 Identify the pair of enantiomers amongst the given pairs:

(A) D OH, COOH CH₃ HOOC OH CH₃ D (B) HO CH ,₃ Ph H HO Ph CH₃ H (C) H OH ,₃ COOH CH HOOC CH₃ OH H (D) H OH, CH₃ HO Ph CH₃ CH₃ CH₃

1.22 The stereochemical formula of deiastereomer 'Y' of optically active compound 'X' is: X=2,3-Dihydroxbutanedioic acid. (A)_HOOC _H OH OH H COOH (B)HOOC H OH H COOH OH (C) H OH COOH HO COOH H (D)HO COOH H H OH COOH

SECTION-II : MULTIPLE CORRECT ANSWER TYPE

1.23 Which of the followning statements are not correct?

(A) A meso compound has chiral centres but exhibits no optical activity

(B) A racemic mixture is optically inctive becaure of two equal and opposite rotation of same molecules in mixture.

(C) A meso compound has molecules which are superimpossable on their mirror images even though they contain chiral centres

(D) A meso compound is optically inctive because the rotation caused by any molecule is cancelled by and equal and opposite rotation caused by another molecules that is the mirror image of the first

1.24 Consider following compounds

OH C H₂ ₅ (I) OCH₃ CH₃ (III) OH CH – CH₃ (II) CH₂OCH₃ (IV)

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Choose the correct statement(s)from the following

(A) I,II and III are functional isomers (B) I and II re position isomers (C) III and IV are chain isomers (D) III and IV are metamers

1.25 Which of the following statement\s is \are true about the following compounds

O

(I) (II) (III)

O O

(A) (I)and (III)are structural isomers. (B) (I)and (II)are geometrical diastereomers (C) (II)and (III)are strukctureal isomers (D) (I)and (II)are jidentical

1.26 Which of the following is\are a meso compoumd.

(A) OH OH O CH OH₂ CH OH₂ (B) COOH H COOH NH₂ NH₂ H (C) CI F Br CI Br F COOH (D) COOH H OH HO H

1.27 Which of the following pair represents the correct relationship

I II (A) NH₂ CI OH _CI _OH NH2 Relationship (B) Positional Isomers Chain Isomer (C) (D) NH – C H₃ ₇ _{H C – N – C H} 3 2 5 Functional Isomers Metamer Isomers CH –CH –CH –C–OCH₃ ₂ ₂ ₃ O CH –CH₃ ₂–C–OCH –CH₂ ₃ O

SECTION-III:REASONING TYPE

1.28 Statement-1:Restricted rotation about a bond is the necessary condition for geometrical isomerism.

Statement-2:Two different orientations are possible due to restricted rotation about a bond if theend groups are different.

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for

Statement-1.

(C) Statement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement -2 is True.

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1.29 Statement-1: Aracemic mixture of 2-Chloropropanoic acid is treated with excess of (+)-2-Butanol. The reaction can be represented as follows:

CH3–CH–COOH + CH3–CH–CH2–CH3CH3–CH–COO–CH–CH2–CH3+CH3–CH–CH2–CH3 CI OH

(+)excess

CI OH_{(+)(Left unreacted)} Statement-2: The solution kjof reaction mixture at time(t = 0), will be dexterorotatory because of (+)2-butanol

(A) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for

Statement-1.

(C) Statement-1 is True, Statement-2 is False. (D) State,emt-1is False, Statement-2 is True.

SECTION-IV : TRUE & FALSE STATEMENT TYPE

1.30 The products(ester)of following reactiongive diastereomers

Me H D Me Et H COOH + HO H SO2 4 Comprehension #1

Compound X(C7H14O4)on ozonolysis gives (Y) and (Z),(Z) is the aldhyde which gives only one

oxime with NH2-OH. On treatment with I2\NaOH,(Y)gives yellow solid CHI3 alongwith compound

given below. HO H H OH OH H COONa CH –OH₂

When (X)is treated with D₂\Ni, it gives two optically active compound.(V)and (W). 1.31 compound(V)and (W)are

(A) Enantiomers (B) Diastereomers (C) Identical (D) Nuclear Isomers

1.32 Compound (Y)can give... type of oximes on treatment with NH₂-OH

(A) 2 (B) 6 (C) 8 (D) 7

1.33 Which of the following statement is true

(A)'X' gives positive test with 2,4-DNP and Br2 solution

(B)Y gives positive with both 2,4-DNP and tollens reagent (C)Compound(V)give positive test with NaHCO3

(D)

CH₃ NH₂ C O

is the isomeric amide of oxime of next higher homolog of Z.

Comprehension#2

Structural isomers have different covalent linkage of atoms. Stereoisomers are compounds that have same sequence of covalent bonds but differ in the relative dispositions of their atoms in space. Geometrical and optical isomers are the two important types of configurational isomers. The compound with double bonds or ring structure have restricated rotation, so exist in two

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geometrical forms. The double bonds in larger rings (ring size 10 carbon or large) can also cause geometrical isomerism. The optical isomers rotate the plane of plane-polarised light. Asp3

hybridised carbon atom bearting four different types of substituents is called and asymmeric centre or chiral centre. A chiral object or molecule cannot be superimposed on its mirror image. Stereoisomers that are mirror images of each other are called diastereomers. Deastereomersw have different physical properties.

A racemic mixture is optically inctive and contains equal amounts of both the enantiomers . Resolution refers to method of separating a racemic mexture into two pure enantiomers. A mesocompound is and opticall incactive stereoisoker, which is achiral due to the presence of and internalk plane of syummetry or centre of symmetry within the molecule.

1.34 The pair showing identical species is

H H COOH OH OH HOOC (C) and COOH OH H OH H HOOC C CH₃ Cl Br H C H₃C Cl Br H and (D) (B)H D OH Br Me Et D H Et OH Br Me and (A) H Cl H Cl and

1.35 Observe the following reaction

H CH₃ NH₂ * + HOOC CH₃ Cl H C*

(R')-1-Phenylethylamine (R+S)-2-Chloropropanoic acid

(R' R) + (R' S) (1) (R')+ (R) (R')+ (S) (R'R) (R'S) hydrolysis hydrolysis (3) (3) separation by recrystallisation (2)

Which statement is not correct about the above observation. (A)The product mixture os step-1 is optically active

(B)The products R'R and R' S have identical structural formul. (C)R'Ris nonsuperimposable on R' S

(D)R'Rand R' S have same solubillity in water.

1.36 The number of chiral centres present in the following compound is

H H OH H H H HO O OH CH OH₂ O O HOH C₂ H H OH H OH CH OH₂ D (+) - Sucrose

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(A) 7 (B) 8 (C) 9 (D) 10

SECTION-VI:MATRIX-MATCH TYPE

1.37 Column I may match with more than one conditions of column II.

Column I Column II

Ph – CH = CH – Ph

(A) (p) Ozonolysis followed by reaction with

NH2 OH leads to more than one oxime product

(CH ) C = CH – CH = Cl₃₂ CH₃ (B)

(q) Can exhibit geometrical isomers

CH = CH – CH = CH₂ ₂ (r) Compounds with this structure formula

can be separated into different fractions upon fractional distillation

CHO – CH – CH – CHO OH OH

(s) Is capable of showing stereoisomerism

1.38 Column I Column II (A) HO D D D OH CH₃ OH D D D OH H₃C

(p) can be separated by fractional

crystallisation (B) CHO C C CH₃ CH₃ CHO CHO C C CH₃ CHO CH₃

(q) Can not be separated by

fractional crystallisation (C) COOH OH OH COOH H H H H OH OH COOH

COOH (r) Optically resolvable

Cl OH₃ OH₂I D Br H CH₃ Br CH₂Cl I D H

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SECTION-VII: SUBJECTIVE ANSWIR TYPE

SHORT SUBJECTIVE:

1.39 Total number of stereoisomes possible for molecule A will be

H C₃ CH = CH CH = CH

1.40 Cl / hv₂ _A Fractional _B

distillation

Number of possible fractions of B are:

1.41 The number of isomers for the compound with molecular formula C₂HDFCI is : 1.42 Number of sp2_-sp2_{sigma bonds in viven compound A is:}

1.43 Write lowest molecular weight of saturated cyclic hydrocarbon which has four substtituents. Molecular weight=98.

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SECTION-I: STARINGT OBJECTIVE TYPE

2.1 Choose the correct statement

(A) I effect transfers e from one carbon atom to another (B) I effect operates in both \p bond

(C) I effect creates not charge in moecule

(D) I effect cteates prtial cjarges and it is distance dependent

2.2 The correct stability order of following species is :

(x) _(y) (z)

C

(w)

(A)x>Y>w>z (B)y>x>w>z (C)x>w>z>y (D)z>x>y>w

2.3 Which of the following does not represent the resonating structure of

(A) (B) – + + – + – (C) (D)

2.4 The most stable resonating structure of CH – O = CH = CH₃ ₂:

(A) (B) H C – O = CH – CH3 2 (C) H C – O = CH – CH3 2(D) H C – O = CH – CH3 2

+ + + +

2.5 Ordinarily the barrier to rotation about a carbon-carbon double bond is quite high but in compound P double bond between two rings was observed by NMR to have a rotational energy barrier of only about 20 cal.\mol., showing that it has lot of single bond charcter.

nC H₃ ₇

The reason for this is

(A) Double bond having partial triple bond charcter because of resonance (B) Doule bond undergo flipping

(C) Double bond having very high single bond charcter because of aromaticity gained in both three and five membered rings.

(D) +I effect of nC3H7 groups makes double bond having partial single bond charcter.

General Organic Chemistry-II

(GOC-II)

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2.6 Most contributing structure in nitroethene is (A)CH = CH – N₂ + O O– CH = CH – N₂ + O O– + – CH – CH – N₂ + O O– + – CH – CH = N₂ + O– O– + (B) (C) (D)

2.7 In which of the followig molecules all the effects namely inductive, mesomeric and hyperconjjgation operate:? (A) Cl (B) CH₃ CH₃ (C) COCH3 CH₃ (D) 2.8 Between N CH₃ CH₃ H₃C ₊ N H H H + and (l) (ll)

(A) N(I)has more 's' charcter in N–CH3 bonds

(B) N(II)has more 's' charcter in N–H bonds (C) N(I)has less 's' charcter in N–CH3 bonds

(D) None of these

2.9 The acid strength order is:

O O OH OH I II OH C O OH HO III C O CH₃ CH₃ IV

(A)I> IV> II> III (B)III> I> II> IV (C)II >III> I>IV (D)I >III >II> IV

2.10 Acid strenght of the conjugate acids of the following

are-N NH N NH N

H

(l) (ll) (lll) (lV)

N

(A)I> II> III >IV (B)III> II> I> IV (C) IV> III> II> I (D)None of these

2.11 The acid dissociation constants of the following acids are given as under:

Compound Ka CICH2COOH 136 X 10–5 CH3CH2– CI CHCOOH 139 X 10–5 CH₃ CI CHCH₂COOH 8.9 x10–5 CI CH₃CHCH₂COOH 2.96 x10–5 CH₃CHCH₂COOH 1.52 X 10–5

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From this data, the following obnservations can be made. Mark the correct statements for above mentioned compoumds.

(i) The above variation in acidities of the above acids are due to inductive effect only. (ii) The above variation are both due to inductibve and resonance effects.

(iii) Inductive effect varies shrply with distance. (iv) -I effect of chlorine is not much.

(A) i and ii (B) i and ii (C) ii and iii (D) iii and iv

2.12 The correct order of acidic strength is :

(l) COOH OH (ll) COOH OH OH (lll) COOH CH₃ COOH CH₃ OH (lV)

(A) III> IV> I> II (B) II> I> III> IV (C)II> III> I> IV (D)II> I> IV> III

2.13 Which one of the following reaction is not possible? (A) CH3COONa + HCI  CH3COOH + NaCI

(B) CH3 –SO3 H + H – C = C – Na  CH3SONa + H – C = C – H (C) R – C = C – H + PhONa  PhOH + R – C = C – Na (D) H – C = C – H + NaH₂ H – C = C – Na + NH3 2.14 If S O P O Q R H CH₃

is mixed with NaOH solution. Acid base reaction occurs and HO snatches H from

organic molecule. Which carbon will loose H easily?

(A) P (B) Q (C) R (D) S 2.15 OH COOH (x) OH COOH (y) (z) HOOC HO

The correct acidic strength order of acidic hydrogen x,y and z is respectively. (A) x > z > y (B) x > y > z (C) z > y > x (D) y > z > x 2.16 H N (p) NH(r) N H O O (q)

The correct basicity order of atoms p,q and r is :

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2.17 N H I N H II O N H III N IV

The order of basicity is

(A) I > IV > II > III (B) I > II > IV > III (C) III > II > IV > I (D) II > III > IV > I

2.18 The correct basic strengtyh order is:

C O CH₃ H₂N C O CH₂ NH₂ CH₃ NH O NH C O CH₃ I II III IV

(A) I > II > IV > III (B) IV > III > II > I (C) III > II > IV > I (D) III > IV > II > I

2.19 The correct order of acid and basic strength for the following pair of compounds should be?

COOH COOH CH₃ and COOH COOH CH₃ and Acid strength :

(I) (II) (III) (IV)

; Basic strength : NH₂ NH₂ CH₃ and NH₂ NH2 CH₃ and

(V) (VI) (VII) (VIII)

;

(A) I > II ; III > IV ; V > VI ; VII < VIII (B) I < II ; III > IV ; V < VI ; VII > VIII (C) I > II ; III > IV ; V > VI ; VII > VIII (D) I < II ; III > IV ; V < VI ; VII < VIII

2.20 Observe the foolowing reaction :

COOH + COONa OH NO₂ + O Na NO₂ – ₊ OH NO₂ ONa NO₂ + NaHCO₃ + H CO₂ ₃ ONa OH + H CO₂ ₃ + NaHCO₃

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Which of the following is the correct order of acid strength : COOH (A) OH NO₂ OH > > H CO >₂ ₃ COOH (B) OH NO₂ OH > > > H CO₂ ₃ COOH (C) OH NO₂ OH >H CO₂ ₃> > COOH (D) OH NO₂ OH > > > H CO₂ ₃

2.21. Which of the following statement is CORRECT regarding the inductive effect?

(A) donating inductive effect (+ l effect) is generally more powerfull than electron-withdrawing inductive effect (–l effet t)

(B) it implies the shifting of s electrons from more eletronegative atom to the lesser eletronegative atom in a molecule.

(C) it implies the shifting of r electrons from less electronegativbe atom to the more electronegative atom in a molecule

(D) it increases with increase in distance.

2.22 Which of the following statement regarding resonance is NOT correct?

(A) the different resonating structures of a molecule have fixed arrangement of atomic nuclei. (B) the different resonbating strcutues differ in the arrangement of electrons.

(C) the hybrid structure has equal contribution from all athe resonationg structures always. (D) None of the individual resonating structure explains all charcteristics of the molecule.

SECTION-II: MULTIPLE CORRECT ANSWER TYPE

2.23 Which of the following statements would be incorrect about this compoud?

NO2 NO2 NO2 Br 1 5 3

(A) All three C – N bond are of same length

(B) C₁ – N and C₃ – N bonds are of same length but longer than C₃ – N bond (C) C1 – N and C5 – N bonds are of same length but longer than C3 – N bond

(D) C1 – N and C3 – N bonds are of different length but both are longer than C5 – N bond

2.24 Choose the incorrect statement:

(A) Salicylic acid (o–Hydroxybenzoic acid) is much stronger than its m-,p-isomers and benzoic acid itself.

(B) Acidity of salicylic acid is due to steric inbibition of re3sonance, as – OH group forces – COOH out of the plane of ring

(C) The orbitals which are in the same plane take part in resonance (D) All the resonating structures have real existence

2.25 In which of the following pairs the first one is the stronger base than second.

(A) CH3CHOO, HCOO (B) HO,NH

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2.26 Which statement among the followijng are corect?

(A) Hydration effect stabilises dimethyl ammonium ion more than trimethyl ammonium ion (B) In chlorobenzene as there is no hydration effect so, trimethyl ammonium ion gets less

stabillised than dimethyl ammonium ion

(C) RCONH2 is feebly acidic with respect to RCOOH

(D) CH3 > NH2 > OH is the basicityh order

2.27 Resonance structures of a molecule should have.

(A) Identical arrangement of atoms (B) Nearly the same energy content (C) The same number of paired electrons (D) Identical bonding

SECTION-III : ASSERTION AND REASON TYPE

2.28 Statement-1: Ortho iodobenzoic acid is strongest acid among all ortho halobenzoic acids. Statement-2: Iodine exerts maximum ortho effect (steric effect) so the acid weakening resonace effect of aromatic romg os decreased/

(A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for

Statement-1

(C) Satement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True

2.29 Statement-1: Salicylic acid is much stronger than its m-, p-isomers adnsf benzoic acid itself. Statement-2: It is due to steric inhibition of resonance, as – OH group forces – COOH out of the plane of ringt.

Statement-1

SECTION - IV : TURE AND FALSE TYPE

2.30 Give the correct order of initials T or F following statements. Use T if statement is true and F if it is false .

S1 : Both the C – O bond lengths in HCOOK are equals

NH₂ both have same dipole moment value.

S₂ : HO OH and NH₂ NH₂

S3: Propyl chloride shows maximum dipolemoment in anti conformer.

S4 :

N

N NH₂ is stronger base then

(A) T T T T (B) T F T F (C) T T F F (D) F F T T

SECTION - V : COMPREHENSION TYPE

comprehension # 1

The concept of resonance explains various properties of compounds. The molecules with conjugated system of p bons, are stabilized by resonace and have low heat of hydrogenation.

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Hyperconjugative stabilization also decreases heat of hydrogenation. In aromatic rings a functional group with a lone pair of electron exerts +m effect. Some functional groups like –NO, –NC,–CH = CH2 can function both as electron releasing (+m, +R) or electron withdrawing (–m, –R) groups. More extended conjugation provides more stabilization.

2.31 The correct heat of hydrogenation order is

(p) 1,3-Pentadiene (q) 1,3-Butadiene

(r) 2,3-Dimethyl-1, 3-butadiene (s) Propadiene (A) p > q > r > s (B) s > q > p > r (C) q > s > p > r (D) s > q > p > r

2.32 The most stable carbocation is

(A)

(C)

(B)

(D)

2.33 The most stable resonating structure of following compound is O = N N = O

N = O O – N (A) N = O O = N (C) N – O O – N (B) N = O O – N (D) Comprehension # 2

The key concepts of resonance are :

Resonance occurs because of the overlaping of orbitals. Double onds are made up of pi bonds, formed from the overlap of 2p orbitals. The electrons in these pi orbitals will be spread over mor4 than two atoms, and hence are delocalized . Both paired and unshared electrons may be delocalized , but all the electrons must be conjugated in a pi system. If the orbitals do not overlap (such as in orthogonal orbitals) the sturctures are not true resonance structures and do not mix. Molecles or species with resonance structures are generally considered to be more stable than those without them. The delocalization of the electrons lower the orbital energies, imparting this this stability. The resonhance in benzene gives rise to the property of aromaticity . The gain in stability is called the resonance energy. All resonance structures for the same molecute must have the same sigma framework (sigma bond form from the "head on' overlap of hybridized orbitals). furthermore, they must be correct Lewis structures with the same number of electrons(and consquent charge) as well as the same number of unpaired electrons. Resonancestructures with arbitrary separation of chargfe of charge are unimportant, as are those with fewer covalent bonds . theese unimportant resonance structures only contribute minimum (or not at all) to the overall. From the above theory of resonance answer the followings. 2.34 The correct resonation structre of 1, 3butakiene is

-(A) CH — CH — CH = CH₂ ₂ (C) CH — CH = CH — CH₂ ₂ + – + – (B) CH — CH — CH = CH₂ ₂ (D) None of these + –

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2.35 The correct stable resonating structure of benzene is -(B) – – (A) ₊ + (C) + – – + – + (D) None of these

2.36 Which resonating sturcture is not correct

CH =CH—CH₂ ₂ (A) + CH — CH = CH+ ₂ ₂ (C) CH —C—NH₃ CH —C=NH₃ O O NH —C—NH₂ ₂ + NH₂ NH — C = NH₂ ₂ NH₂ (B) (D) N H N H +

SECTION - VI : MATRIX - MATCH TYPE

2.37 Match the following

(A) CH₃ CH₃ CH₃ CH3 N (B) Cl Cl O– N=O + O N₂ (C) CH3 CH3 NH₂ O N2 (C) CH3 CH₃ N O N₂ CH₃ CH₃ (p)  is more (q)  is less (r) Larger C– N compared to C – NO2

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2.38 Match the pKa values with the given compounds Acid pK_a NO₂ OH NO₂ NO₂ (A) (B) CH COOH₃ (C) O O NH Phthalimide COOH (D) (p) 4 (q) 9 (r) 5 (s) 0.74

SECTION - VII : SUBJECTIVE ANSWER TYPE

SHORT SUBJECTIVE :

2.39 For the compound :

OH

Give the number 1 for presence of resonance only, 2 for presence of resonance and hyperconjugation only, 3 for presence of resonance, hyperconjugation and inductive effect and 4 for presence of resonance hyperconjugation, inductive effect and electromaric efffect.

2.40 OH CH – C – H 2 Br Br OHC HOOC x mole NH₂

value of X used for complition of reaction will be :

2.41 Incorrect statement among these four :

(i) I effect is prermanent polarisation of sigma bond pair of electrons in the molecule. (ii) In resonation structures the hybridisation of atoms do not change.

(iii) In Hyperconjugative structures the hybridisation of carbon atom change. (iv) Presence of methyl group on an anion always destabilises ith anion.

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2.43 In the following set of resonationg structures which sets have the second resonation structure more contributing than first :

NH₂ OCH₃ + NH₂ OCH₃ + N O + N O O O O O + Cl Cl N O N O I : II : III : IV :

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SECTION - I : STRAIGHT OBJECTIVE TYPE

3.1 Read the following road map carefully

CH3 Ph–CH₂CHOK + gas EtOTs II CH3 Ph–CH2–CHOTs IV C H₂ ₅OK CH₃ Ph–CH₂CHOH K I 1-Phenyl-2-propanol III TsCl S.R = +33.02° CH₃ Ph–CH₂–CH–OC H₂ ₅ ethyl–1–phenyl–2–propylether CH3 Ph–CH₂–CH–OC H₂ ₅ Ethyl–1–phenyl–2–propylether

(A) Both the ethers obtained by the two routes have opposite but equal optical rotation. (B) One of the ether is obtained as a recemic mixture.

(C) Step II & III both are SN2 reaction and both have inversion. (D) Step II has inversion but step III has retention.

3.2 A compond A has the molecular formula C5H9CI. It does not react with bromine in crabon tetrachloride. On treatement with strong base it produces a single compound B. B shas a molecular formula C5H8 and reacts with bromine in carbon tetrachloride. ozonolysis of B produces a compound cC which has a molecular formula C5H8O2. Which of the following structures is that of A? (A) Cl (B) Cl CH₃ (C) Cl CH₃ (D) Cl 3.3 Me SO2 4 NaOH PhOH P, P is

(A) Ph–O–SO2OMe (B) PhOMe (C) PhOSO2OPh (D) PhMe

3.4 P = _Br COCH COCH₂ ₃ H D NaOH P

Reaction Mechanism

3

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Which of the following statement is correct P = C –CH₃ O O H D (B) P = OH COCH COCH₂ ₃ D (A) H P = O D (D) H P = (C) CH3 O O D H 3.5 OH Br NaOH (B) O (C) OH OH (D) O (A) OH OH 3.6 CH – CH SH₃ ₂ (i) CH O3

(ii) ethylene oxide (iii) H O₂ Product, Product is : (A) CH – CH – S – CH – ₃ ₂ ₂ CH – OH₂ ( ) CH – CH – C ₃ ₂ O – CH – ₂ CH – OH₂ ( ) CH – CH – D ₃ ₂ S ( ) CH – CHB ₃ ₂O CH₂ CH₂ 3.7 Br Ph + Ph – CH – NH₂ ₂ O EtO Product, Product is (A) OEt Ph O (B) Ph O NH Ph (D) Ph NH Ph (C) Br Ph OH NH Ph 3.8 In the reaction A. OH OH CH₃

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-O CH₃ CH₃ (A) CH₃ COCH₃ (C) CH₃ CH₃ (B) (D) CH₃ CH₃ O 3.9 C H₂ ₅ CH (C₃ H )₂₆ C OH H TsCl _A Kl, DMSO _{B (Major);}

(A) R (B) S (C) R,S both (D) None of these

3.10 C C OH OH CH₃ NO₂ NO2 CH₃ H+ A ; C C CH₃ NO₂ NO₂ O CH₃ (A) NO2 C C NO₂ NO₂ NO O (B) C CH₃ NO₂ NO₂ CH(OH) CH₃ (A) CH₃

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3.11 The products of hydrolysis of

O – C H₂ ₅ O

, is

H O3

(A) HOCH₂CH₂CH₂CH₂ CHO + CH₃CHO (B) HOCH₂CH₂CH₂CH₂OH + CH₃CHO (C) HOCH2CH2CH2CH2CHO + C2H5OH (D) HOCH2CH2CH2CH2CH2OH + C2H5OH

3.12 CH – OH2 A. H SO2 4/H The product A is : CH – HSO2 4 CH2 CH2 CH2 (A) (B) (C) (D) OH 3.13 (CH3)3CCI + (CH3)3CO–K+  Product

(A) SN Product will be more (B) E2 Product will be more

(C) both will be same (D) None of these

3.14 Neopentyl iodide is treated with aq. AgNO₃ solution, a yellow precipitate is formed along with other compound which is

(A) (CH ) CCH ONO₃₃ ₂ ₂ (B) (CH ) C – CH CH_{3 2} ₂ ₃ OH

(C) (CH ) CCH OH_{3 3} ₂ (D) (CH ) C–CH CH_{3 2} ₂ ₃ ONO₂

3.15 The major end product of the following reaction is

OH CH Cl2 H C3 aq. AgNO₃ OH H C₃ CH2– OH (A) CH₃ H₃C O (B) (C)H₃C O (D) CH₃ O

3.16 The major product P of the following reaction is

NH₂ _CH 3 CH – CH = C CH3 NH₂ (D) NH₂ _CH 3 CH – CH – CH CH₃ NH₂ (C) OCH₃ NH₂ _CH 3 CH – CH – C₂ CH₃ OCH₃ NH₂ (B) NH₂ _CH 3 C – CH – CH2 CH₃ NH₂ (A) OCH₃ NH₂ _CH 3 CH – CH – CH CH3 NH₂ Cl CH OH3 _(P)

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3.17 The product of following reacting is 14 Ar O OH (A) 14 Ar O O H O₃ 14 OH (C) + ArOH O 14 _{+ ArOH} OH O (B) 14 O (D) + ArOH O 3.18 O CH₂ H₂C 14 H /H O + ₁₈ 2

Which can not be the product.

H C₂ 14CH₂ 18 OH OH (A) CH₂ 14 CH₂ 18 OH OH (B) CH₂ CH₂ 18 OH (C) 18 OH (D) A and B both

3.19 The correct order of SN2 \ E2 ration for the % yield of product of the following halide is

-CH – -CH – C – -CH₃ ₃ Ph I CH – CH – CH – CH₃ ₃ I Ph CH – CH – ₃ ₂ I CH – CH – CH – CH₃ ₃ (P) (Q) (R) (S) I (A) R > S > Q > P (B) R > Q > S > P (C) P > R > S > Q (D) Q > P > R > S

3.20 The poduct in the given reaction is

H ONa + Me Ph H Me Ph Br I II P H O Me Ph Me Ph H (A) H O Ph Me Me H (C) Ph H O Me Ph Me Ph H (B) H O Me Ph H Me Ph (B)

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3.21

Ph

reaction _{Rearranged most stable Carbocation is}

Ph Ph Ph _Ph

(A) (B) (C) (D)

3.22 H Br

Ag+

Rearranged Carbocation + AgBr

Rearranged carbocation is : + (D) (C) (B) (A) CH₃

SECTION - II : MULTIPLE CORRECT ANSWER TYPE

3.23 The correct statement(s) about solvent effect is\are :

(A) Decreasing solvent polarty causes a large increase in therate of the SN2 attack by ammonia

on an alkyl halide

RX + NH₃ RNH₃ + X +

(B) Increasing solvent polarity causes a large decrease in the rate of the SN2 attack by hydroxide

ion on trimethyl sulphonmium ion

HO (C + H_{3 3}) S CH OH + (CH ) S₃ _{3 3}

DMS Trimethyl

sulphonium ion

(C) Increasing solvent polarty causes a small decrease in the rat3e of the SN2 attack by

trimethylamins on trimethylsulfonium ion.

(CH ) N_{3 3} + (CH ) S₃₃+ CH N(CH ) + (CH ) S₃+ _{3 3} _{3 2}

(D) all ar3e incorrect

3.24 Which of the following reaction(s) is\are posible (A) _{CH – C ClCH NEt}

3 H 2 2 CH C (NEt )CH – OH3 H 2 2

OH

(B) Either CH – C₃ H – C H₂– SEt or CH CH(sEt)CH OH₃ ₂ HCl CH C ClCH₃ H ₂SEt OH

(C) Treatment of either epoxide I or epoxide II with aqueous _OH gives the same product III

CH – CH – CH – CH Br₃ ₂ or CH – CH – CH – CH₃ ₂ O Br I II _III O CH – CH – CH – CH₃ ₂– OH O OH

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(D) OH OEt OH OEt + OH OEt EtOH EtO – 3.25 OH OH NaOH 1eq P Q R 'S' O CH –C–Cl2 Cl NaOH LiAlH₄ O O O R = (A) O Q = OH C = O CH Cl₂ (C) O Q = OH CH – C – Cl₂ O (B) S = OH OH OH (D) 3.26 OH NH₂ CH₃C H₆ ₄ Ph H Ph ? HNO₂ H Ph O Ph pCH C₃ ₆H₅ O Ph H Ph pCH C₃ ₆H₅ (A) (B) H Ph O Ph pCH C₃ ₆H₅(C) Ph O H Ph pCH C₃ ₆H₅ (D)

3.27 Which of the following reaction will go faster if the concentration of the nucleophile is increased?

H _Br + CH O₃ (CH ) CCl_{3 3} CH COOH₃ H O₂ + CH S₃ Br Br Acetone KI (A) (B)

SECTION - III : ASSERTION AND REASON TYPE

3.28 Statement-1 : (CH ) C CH_{3 3} OH CH₃ (CH ) CCH=CH +_{3 3} ₂ CH₃ CH₃ CH₃ CH₃ C=C _{+ CH} 2=C CH₃ CH(CH₃)₂ H SO₂ ₄ 40% 20% 40%

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Statement-1

3.29 Statement-1 : Ethers are prepared from alcohols in acid medium through S_N1_{or S}

N2 mechanisms

depending upon nature of alcohol. Statement-2 : _{ROH + H}+ _{R – OH} 2 + R – O H R – OH+ ₂ _{R – O – R + H O} 2 ROR + H O3 + slow ₊ H or, R – OH + H+ R OH+₂ R + H O+ ₂ R + O – R + R – O – R+ slow H H R OR + H O₃ + H O2 fast

Statement-1

SECTION - IV : TRUE AND FALSE TYPE

3.30

Ph OH

CH₃ H SO2 4 CH₃ Ph (Elimination follows E mechanism.)₁

Heat

SECTION - V : COMPREHENSION TYPE

Comprehension # 1

Read the following reaction

O H+ P Cl_h2 Q NaOEt / R NBS S NaOH T 3.31 Compound 'T' is OH OH OH OH OH HO OH OH (A) (B) (C) (D)

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3.32 Compound 'Q' is Ol OH Cl OH Cl OH Cl (A) (B) (C) (D)

3.33 When treation 'S' with strong base, product obtained is

OH OH OH

OH

Br Comprehension # 2

A hydrocarbon (X) of the formula C₆H₁₂ does not react with bromine water but reacts with bromine in presence of light, forming compound (Y) . Compound (Y) on treatment with A|c. KOH gives compound [Z] which on ozonolysos gives (T) of the formula C6H10O2. compound (T

reduces Tollens reagent and gives compound (W).(W) gives iodoform t4st and produce compound (U) which when heated with P2O5 forms a cyclic anhydride (V).

3.34 Compound V is O O O CH₃ O O O CH – C – CH – ₃ ₂ CH – CH = O₂ O (A) (B) (C) (D) CHO – CH = CH – CHO 3.35 Compound W is

(A) COOH – (CH ) – COOH_{2 2} (B) COOH COOH (C) COOH COCH₃ (D) CH – CH – CH – COOH₃ ₂ CH = O 3.36 Compound 'X' is (A) (C) (B) (D) CH₃ CH₃ CH₃ CH₃ CH₃

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SECTION - VI : MATRIX - MATCH TYPE

3.37 Match the column I with column II

CH₃ H Cl Ph C = C KNH₂ CH₃ OH Ph H SOCl₂ CH₃ N – CH₃ CH₃ C O Ph + NaOH CH –CH –CH₂ ₂ ₃ CH –N–CH –CH₃ ₂ ₃–CH₃ CH₂–CH₃ + _NaOH (P) -elimination (Q) S_N2 (R) -elimination (S) SN₁ (Y) (Z) (W) (X) Column - ( ) Reactions I Column - ( ) Reactions II 3.38 CH CH CH₃ ₂ ₂ CH CH₂ ₃ C—Br + H O₂ H C₃ CH CH₂ ₃ C—Br + H O₂ H C₃ CH =CH—C — Cl + SH / (CH COCH₂ ₃ ₃) CH₃ CH₃ (A) (B) CH CHDCl + OH₃ – (C) (D) (p) Retention (q) Racemisation (r) Inversion

(s) Product with -bond shift

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SECTION - VII : SUBJECTIVE ANSWER TYPE

SHORT SUBJECTIVE :

3.39 Correct statement about nucleophiles and leaving groups is/are

(i) Nucleoghiles have an unshared electron pair and can make use of theis to react with and electron rich species.

(ii) The nucleophillicity of an element generally inhcreases on goiing left to right in periodic table.

(iii) A nucleophile is an electron deficient species

(iv) Leaving group ability kincreased in moving from L__R and T  B in periodic table. 3.40 CH –CH3 –C–CH –Br2

CH₃ CH₃

CH₃

NaOH/H O₂

Which of the following stat4ements is incorrect about the given reaction (i) The reation is S_N1 reaction

(ii) The reaction intermediate is planar (sp2₎

(iii) The major product will be 2-ethyl-3-methylbutan-2-ol (iv) The major product has two stereogenic centre

3.41 which of the following will not give SN1_{reaction with aq. AgNO} 3. (i) (ii) CH Br₂ Br Ph Ph Br_Ph Ph Br (iii) (iv) Br

3.42 In the given reaction following products are expected.

CH – Br₂ CH3 + CH₂–OCH₃ CH₃ OCH₃ + + + OCH₃

(I) (II) (III) (IV) (V)

CH OH3

Which observetions seen to be incorrect

(i) Ist_{is the major product obtained by E1 reaction}

(ii) IV is the major product obtained by SN1 reaction

(iii) formation of II or V involves a strained carbocation intermediate (iv) In the solvolysis reactions a carbocation intermediate is formed.

3.43 Observe the following flow chart and answer the number of x,y, z, w.

Positional Isomers = (X) Monochlorides Isopentane Total Isomers = (Y)

Total number of 1° alcohols = (Z) including stereoisomers

KOH/DMSO (SN )2

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SECTION - I : STRAIGHT OBJECTIVE TYPE

4.1 2-Phenylcycloprop-2-en-1-one is allowed to react with phenylmagnesium bromide and kthe reaction mixture is hydrolysed with prechloric acid. The product formed is

(A) Ph – C – C – CHO Ph OH O (B) Ph Ph O (C) OH Ph Ph Ph Ph (D)

4.2 Which of the following reactions will give 2° chiral alcohol as one or more of major organic products? (A) H D CH₃ MgI O2 H O3 (B) CH MgI₃ + H – C– OCH O CH₃ – C H₂ ₅ H O3 (excess) (C) H CH₃ MgBr + CH₂ O CH₂ – H O3 (D) None 4.3 CH₃–C C CH– – ₃ O O PhMgBr (excess) H O₂

No of product (X) Fractional distillation no. of fractios (Y)

(A) 3,2 (B)3,3 (C) 4,2 (D) 4,3

4.4 The end product of following reaction is

O O O CH MgBr₃ PhMgBr (1) (2) (3) H

(A) ,-diketone (B) -Hydroxy acid (C)1,2-Diol (D) -Hydroxy acid

4.5 Observe the following reaction sequence

X Br / hv2 _Y Mg / Ether _Z O W dil.H SO2 4 U O /Zn/H O3 2 _O X can be (A) (B) (C) (D) CH3

Grignard Reagent

Reduction & Alkane

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4.6 O (x) CH OH2 pyridine, SOCl₂ Mg (C H ) O₂ _{5 2} _H+ O CH – C(CH )₂ _{3 2} Product

Product of the reaction is

CH – C – CH – CH₂ ₂ ₃ OH CH₃ O _{CH – C –} _CH 2 H C( 3)2 O O CH – C₂ H₂– C O– H O CH₃ CH₃ CH – C – ₂ CH₂ – OH O CH₃ CH₃ (A) (B) (D) (C)

4.7 Observe the following sequence of reactions.

CH –CH –CH₃ ₂ ₃ (X) (Y) (1) Br / hv₂ (2) Mg/ether (3) CH –Br₃ (4) Cl / hv₂ (5) CH OH₃ (Z) (W) (R) The product R is :

(A) CH – CH – ₃ CH – OCH (B) CH –CH –CH –CH –OH (C) CH –C–CH (D) CH –C –CH₂ ₃ ₃ ₂ ₂ ₂ ₃ ₃ ₃ ₃

CH₃ CH₃ CH₃ OH OCH₃ 4.8 Ph – N = C = O (i) CH MgBr3 (i ) Hi ₂O LiAlH₄ P Q (A) P is Ph–NH–C –CH and Q is Ph–NH–CH –CH₃ ₂ ₃ ( ) P is Ph– NH – CH and Q is HB ₃ N ₃ O ( ) P is C PhN=C — OH and Q is Ph – NH – CH₃ CH₃ ( ) P is D Ph – NH – CH – CH₃ and Q is Ph – NH – CH – CH₃ CH₃ OH₃

4.9 Which of the following reduction methods is not suitable for preparing and alcohol?

(A) CH COOC H + ₃ ₂ ₅ NaBH₄ ( ) C CH C₃ H C Cl+ ₂ O LiAlB₄ ( ) B CH COOC H + ₃ ₂ ₅ Na/EtoH ( ) CH COOD ₃ H +H₂ Ni 4.10 _C _{C – C C} Ph – CH Br H C = C H CH – Ph Br Li / NH₃ Br₂(1equivalent) X, 'X' is :

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4.11 Compound O O O N C Cl O

on reduction with LiAIH4 produces

HO N CH OH₂ CH OH₂ CH OH₂ (A) HO O OH N CH OH₂ (B) O OH N CH OH₂ (C) N CH OH₂ CH OH₂ CH OH₂ (D)

4.12 Which one is the correct energy profile for CI reaction ?

(A) R.coord E n e rg y (B) R.coord E n e rg y (C) R.coord E n e rg y (D) All of these

4.13 Which of the following is correct statement regarding relative acidic charcter of cyuclopropane and propane ?

(A) Cyclopropane is more acidic than propane (B) Propane is more acidic than cyclopropane (C) Both are equclly acidic

(D) Bothe are neutral

4.14

No . of products and No. of fractions are respectively

(A) 6,5 (B) 6,4 (C) 5,4 (D) 6,3

4.15 Which of the following is correct potential energy diagram for the given chin propagatiog step.

CH – H+F₃ B.E.=435kJ/mol

CH + H– F₃

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4.16 Consider the following reaction CH –CH₃ –CH–CH + Br₃ D CH₃ 'X' + HBr, X can be CH –CH₃ –CH–CH₂ CH₃ D CH –CH₃ –C–CH₃ CH₃ D CH –C₃ –CH–CH₃ CH₃ D CH –CH₃ –CH–CH₃ CH₃ (A) (B) (C) (D)

SECTION - II : MULTIPLE CORRECT ANSWER TYPE

4.17 Br Mg NBS B A C D E + F Zn AgNO / H O₃ ₂ O / H O₃ ₂ 14 (i) CH COCH₃ ₃ (ii) H (iii) (A) Product C is Br 14 (A) Product D is OH 14 _O _OH OH O 14 & only (D) F is (B) E is CH = O & CH = O₂ 14 ₂ 4.18 Ph – CH – CH – CH₂ ₃ CH₃ (monochlorination) Cl ₂/ h

Which statements is /are correct about photochemical chlorination of theabove compound [More than one correct]

(A) The major product wil be chiral carbon atom having optically inactive compound (B) The intermediate free radical of the major product is resonance stabilised (C) The intermediate free radical is tertiary

(D) THe intermediate free redical is planer , and stabilised by only hyperconjugation

4.19 Which reaction is/are correct.

(A) C N (i) Mg / Ether tertiary alcohol (ii) H O₃ (B) Cl _COOH Mg ether CO₂ H O₃ + F Cl MgF MgCl Mg ether (C) (D) + CH – C₃ O CH CH MgBr3 HO C C – CH₃ H O₃ + +

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4.20 Point out the following incorrect Grignard synthesis. N H Br N H CH – Ph OH (i) Mg, ether (ii) Ph – CH = O (iii) H O₃ Br O O HOCH₂ O O MgBr (C) _OH (B) (A) (i) Mg, ether (ii) PhCH = O (iii) H O3 OH (D) O HO CH CH₂ ₃ OH (i) CH – C – CH₃ ₃ (ii) H O3 O (i) CH CH MgBr (1eq.)₃ ₂ (ii) H O₃

SECTION - III : ASSERTION AND REASON TYPE

4.21 Stetement-1 : Cyclopropane has the highest heat of combustion per methylene group. Statement-2 : Its potential energy is raised by angle strain.

Statement-1.

(C) Satement-1 is True, Statement-2 is False. (D) Statement-1 is False, Statement-2 is True.

4.22 Statement-1 : Branched alkanes have lower boiling point than their unbranched isomers. Statement-2 : Branched alkanes has relatively small surface area, so less London's dispersion force act among molecules.

Statement-1

4.23 Statement-1: Alkanes float on the surface of water

Statement-2 : Density of alkanes is in the ranbe of 0.6 – 0.9 g/ml, which is lower than water. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for

Statement-1

4.24 Statement-1 : Grignard reagent can be prepared in all nonpolar solvent. Statement-2 : Diethyl ether solvates the Grignard reagent.

Statement-1

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4.25 Statement-1 : The preparation of G.R. occurs in solution phase.

Statement-2 : Teh reaction will be explosive in solid phase. GR. is stable only in solution phase. (A) Statement-1 is True, Statement-2 is True; Statement-2 is acorrect explanation for Statement-1. (B) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for

Statement-1

SECTION - IV : TRUE AND FALSE TYPE

4.26 Observe the following natural product and choose the correct statement

(x)HOOC COOCH₃ C H₆ ₅ N (w) H (y) O N₂ O (z) O (r)

(1) LiAIH₄ will reduced x, y, z, w, r (2) NaBH4 will reduced r

(3) Na/C2H5OH will reduced r, w

(4) (CH3)2CHOH + [(CH3)2CHO]3AI, (MPV) will reduced r

(A) TFTF (B) TTTT (C) FTFT (D) FFFT

4.27 Diethyl ether is the solvent for Griganrd reagent but not for RLi, because RLi reacts with C2H5OC2H5 and givesw CH2 = CH2.

SECTION - V : COMPREHENSION TYPE

Read teh following reaction

CH CH P Mg Q Ether NaOCl R S Cl 2CH – CHO₂ KOH NaOH H O₂ T (ii) Dil.KMnO₄ (i) H / Pd. BaSO₂ ₄

4.28 'S' in the above reaction

(A) C C CH₂ O H CH₂ O H (B) C C CH₂ O H MgBr (C) C CMgBr CH₂–OH OH H (D) C C OH H CH₂– OH H HO CH₂– OH 4.29 'U' is H OH H H H OH OH OH CH – OH₂ CH – OH₂ (A) H OH H H H OH HO HO CH – OH₂ CH – OH₂ (B) H OH HO OH OH H H H CH – OH₂ CH – OH₂ (B) H OH COOH CH – OH₂ (D)

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4.30 'Q' is

(A) CIMgC  CMgCI (B) CH  C – MgCI

(C) CI–O–C  C–MGCI (D) CI–O–C  C–OCI

considering the following reations (I,II,III) and give your answer for following questions

HO _C–Cl O (i) CH MgBr₃ (ii) H O₃ + C–OEt O (i) CH MgBr (excess)₃ (ii) H O₃ + CH – Cl₂ EtO – C – OEt O (i) Ph MgBr (excess) (ii) H+ (a) EtO – C – OEt O (i) CH MgBr (excess)₃ (ii) H+ (b) H – C – OEt O (i) Ph MgBr (excess) (ii) H+ (c) H – C – OEt O (i) CH MgBr (excess)₃ (ii) H+ (d) ClCH COOEt₂ (i) CH MgBr (excess)₃ (ii) H+ (e)

4.31 If CH3MgBr is taken in excess in reaction (I), how many moles of CH3MgBr will be consumed

in reaction (I)

(A) 2 (B) 3 (C) 4 (D) 5

4.32 The product in the reaction (II) will be

C CH3 O Cl C CH₃ Cl OEt OH CH – C – CH₃ ₃ OH CH₂–Cl CH – C – CH₃ ₃ OH CH₂– CH₃ (A) (B) (C) (D)

4.33 In which seft of the reactions in (III) the product will be 2° alcohol ?

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4.34 Match teh following consider the reaction of CH3MgBr with the compounds in column A and

product are written in column B.

Column A Column B

(A) CINH2 (p) CH3CN

(B) CICN (q) CH3NH2

(C) CH₃COCI (r) C₂H₅COCH₃ (D) C2H5COOC2H5 (s) CH3COCH3

4.35 Match the following :

O Cl – C O C COOH O – C – CH₃ CH = CH₂ O (i) [Al(OiPr) ] + CH – CH – CH₃ ₃ ₃ (ii) H O₂ OH (1) (2) (P) (Q) NaBH (C H OH + H O)₄ ₂ ₅ ₂

Na/EtOH (low temp.)

(3) _{(4) (ii) H O} 2 (R) (i) LiAlH / Et O4 2 (S) H C OH COOH HOH C₂ O O – C – CH₃ CH = CH2 H C OH COOH Cl – C O O – C – CH₃ CH = CH₂ O H C OH COOH HOH C₂ O O – C – CH₃ CH = CH₂ O H C OH COOH HOH C₂ OH CH = CH₂ O (X) (P) (Y) (Z) (Q) (R) (W) (S) Column - I Column - II

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4.36 Match the following : Column A Column B (A) CH3CHCI   3 2 Li CH CH CI Cul ? (p) (CH3)2CHCH2CH(CH3)2 (B) (CH3)2CHI   3 2 Li (CH )CH Br Cul ? (q) CH3CH2CH2CH3 (C) CH3CH2C(CH3)2Br   3 2 Li CH CH Br Cul (r) (CH3)3CCH2CH3 (D) (CH3)3CI  3 2  Li CH CH –Br Cul (s) CH3CH2C(CH3)2CH2CH3

4.37 Match the following :

Process Graph (A) Fluorination (p) P o te n ti a l e n e rg y Reaction CO-ordination A B C (B) Chlorination (q) A B C (C) Bromination (r) A B C (D) Iodination (s) A B C

Position (A) is for CH₄,X, (B) is for CH₃, HX, (C) is for CH₃–X,X

SECTION - VII : SUBJECTIVE ANSWER TYPE

SHORT SUBJECTIVE :

4.38 An isomer of C5H12 gives total six isomeric products on monkochlorination. Calculate the

percentage yield of the primary monochloride which is chiral. Consider the following relative reactivity of C–H bonds for chlorination.

1 3 5 2° C – H 3° C – H 1° C – H Degree of C – H Relative reactivity for chlorination (RR)

(41)

4.39 How many molar equivalent of R-MgX is required in reaciton with formic anhydride completely? 4.40 HS C – NH₂ C CH O C – OH O excess CH MgBr₃

the numbers of CH gas formed in this reaction₄

4.41 Observe the following experiment Mg powder An alkyl halide (P) ethereal solution (lce) A sweet smelling liquid (Q) H O2 (R) (1) ₍₂₎

(I) (II) (III)

(a) If the reactant 'P' is ethl chloride then the product R has the numbers of optically active compound:

(b) If the loquid Q is

H – C – OC H₂ ₅

O then the product R can be (P can be any other halide)

CH – CH – CH₃ ₃ OH (1) CH – C – CH₃ ₃ OH (2) CH₃ H – C – C H₂ ₅ OH (3) H C H – C – C H₂ ₅ ₂ ₅ OH (4) C H₂ ₅ (c) If R is C — CH3 CH₃ OH

then the numbers of monochlorination product are:

4.42 O

O O

1 mole CH – Mg – X₃

+ H O₂ X

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SECTION - I : STRAIGHT OBJECTIVE TYPE

5.1 The major product of the following reaction would be

NH Br₂ NH Br Br N H Br N H Br N Br (A) (B) (C) (D)

5.2 In the following series of reactions the major products P and S are respectively.

Br ₂/ h

(P) tBuO , (Q)(i) HOBr/H

+ (ii) ThO₂, (iii) alc.KOH, (R)(i) HOBr/H + (ii) ThO₂, (iii) alc.KOH, (S) (A) Br ; (C) Br ; (B) Br ; (D) Br ;

5.3 End product D of the following reaction will be

Cl A B C D H O₂ ZnCl HCl₂ Mg/ether D O₂ D (A) (B) OD _(C) D (D) D OD

5.4 consider the following reactions

H C₅ ₂ C₂H₅ H C₂H₅ H C = C H C₅ ₂ H H C = C C₂H₅ CH₂ C H C₅ ₂ O 2CH₃ CH₂ COOH R₁ _R 2 R₃ R₄ H C —C ₅ ₂ C—C H₂ ₅

The correct set of reagents for these reactions is

Alkenes & Alkynes

5

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R

(A) H /Lindlar catalyst (B) H /Lindlar catalyst (C) (i) O , (ii) H O (D) H O,H SO ,HgSO 1 2 2 3 2 2 2 4 4 R Na / Iiq.NH Na / liq.NH H O,H SO ,HgSO H /Lindlar catalyst 2 3 3 2 2 4 4 2 R (i) O , (ii) H O H O,H SO ,HgSO Na / Iiq.NH (i) O , (ii) H O 3 3 2 2 2 4 4 3 3 2 R H O,H SO ,HgSO (I) O , (ii) H O H /Lindlar catalyst Na / Iiq.NH 4 2 2 4 4 3 2 2 3

5.5 The major product of the following reaction is :

Br Cl Na / ether₍₁₎ Br 2₍₂₎/ h C H O2 5

–

/ C H OH, ₂ ₅

(3)

(A) (B) (C) (D)

5.6 Themost ap0propriatew major product of the followning sequence of reactions would be.

CH CH₃ ₂ H₃C C C CH₃ H CH – C – O – O – H₃ O LiAlD₄ H₃O+ Me Me D OH(±) Et H (A) Me CH₃ DH H (±) Et D (B) Me Me OH D (±) Et H (C) CH₃ CH₃ DH OD CH CH₃ ₂ H (D)

5.7 Compound'P' of the following reaction swequence can be

(P) O O [C H O ]₈ ₆ ₃ O H O₂ + COOH [C H O]₈ ₈ _{CH – COOH} 2 CrO / Acetone / H O₃ ₂ [C H₁₆ ₂₁B] HOO– B–H C C (A) (B) CH₂– C CH (C) C CH (C) _{CH = CH}₂

(44)

5.8 The alkene limonene has the following structure,

Which product resutls from the reaction of limonene and 1 molar equivalent chlorine water?

OH CI (A) OH CI (B) OH CI CI (C) OH (D)HO CI CI

5.9 The end product 'W' in the follownign sequence of reactions is :

CH CH₂ ₃ NBS (X) W alc. KOH aq. KOH Y Z Hg(OAc)2 NaBH₄ (A) Ph – CH – CH – O – CH – Ph₂ ₂ CH₃ ( ) Ph – CH – C ₂ O – CH – ₂ CH₂ – Ph ( ) Ph– CH –B O – CH – Ph CH₃ CH₃ (D) Ph – CH – CH₃ CH = CH₂

5.10 Which of the following corrently represents tthe rate of acid-catalysed hydration of followning alkenes. C = CH – CH₃ CH₃ CH = CH – C H₂ ₅ PhCH = CH – C H₂ ₅ CH – C = CH₃ ₂ CH₃ (l) (ll) (lll) (lV) (A) lll > l > ll > IV (B) lV > lll > l > Il (C) l > lll> lll > IV (D) lV > lll > lll > I 5.11 Ph–C C–Ph HOCl/H + Na/NH₃ Ph Ph A B ; B is H Cl OH H (A) (B) Ph Ph HO Cl H H (C) Ph Ph H H OH Cl (D) B and C both

5.12 The most stable conformation of the product of followning reaction

C CH HBr / R O2 2 (1 eqivalent) HBr / dark H Br Br H H Ph (A) Ph Br Br H H H (B) Ph Br Br Br H H (C) H H H Br Ph Br (D)

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5.13 In the given reaction,

CH₃

(X)

B H , THF₂ ₆ H₂O₂, NaOH TSCl t–BuO K– + _(Y)

The product 'Y' is

(A) A positional isomer of X (B) Identical to X

(C) Chain isomer of X (D) An oxidation product of X.

5.14 Which of the following compouinds would liberate two moles of methane when treated with methyl magnesium bromide?

CH – CH – CH – C CH₃ ₂ (A) OH CH – C – CH – C CH₃ – ₃ (C) O CH₃ O (B) (D) CH – C – CH – ₃ ₂ CH OH₂ O OH COOC H₂ ₅

5.15 The best yiest of product 'X' ca be obtained byt using which one of the foloowing sequence of reagents and reactants

X = CH – C C – CH – CH (CH )₃ _{3 2} OH

(A) CH – C CH₃ NaNH2 (CH ) CHCHO3 2 H3O + (A) CH – C CH₃ NaNH2 (CH ) CHCHO3 2 H3O

+ ( ) CHB ( ₃) CH₂ – C CH – C – H NaNH2 CH3I ( ) CHC ₃– C – H CC C – H –₂ CH₃ NaNH2 CH3I OH OH ( ) CHD ( ₃) CH₂ – C CH – C – H CH3MgI CH3I OH 5.16 H₃C – C – CH – CH₂ ₃ CH₃ H₃C – CH₂ – CH – ₂ N H₃C – C – CH₃ CH₃ CH₃– C – Cl O X, X is AlCl₃ (i) CH₃I (ii) AgOH,

(46)

(A) CH – CH – CH – Cl₃ ₂ COCH₃ ( ) CH – CH – CH – CB ₃ ₂ – CH₃ Cl O ( ) CH – CH –C ₃ C – CH – ₂ CH₃ Cl ( ) CH – CH – CH – CB ₃ ₂ – CH₃ Cl O O CH₃

5.17 Fastest rate of electrophilic addition will take place in

(A) HO CH = CH2

(C) CH₃O CH = CH₂

(B)ON CH = CH₂

(D) _{CH = CH}

2

5.18 Which of the following will be the correct product (P) for the given reaction ?

OH Conc. H PO₃ ₄ (P) OH (A) OH (B) (C) (D) 5.19 CH₂ HBr X (major) R O₂ ₂ aq NaOH Y (major) H2SO4/ _{Z (major)} Identify the correct option

BrCH3 BrCH3 OHCH3 OHCH3 CH3 CH₃ CH₂ OH CH₃ Br CH₃ OH CH₂ CH₃ Br CH₃ X Y Z (A) (B) (C) (D*)

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5.20 X HOOC – C – CH – CH – CH – CH – COOH₂ ₂ ₂ ₂ O (i) KMnO ₄/ H O ₂ / (ii) H O₃ + X may be (A) COOH (B) COOH (C) COOH O (D) O 5.21 CH3 CH₃ _CH 3 C C = C H D D (1) C H CO H₆ ₅ ₃ ( ) 2 CH MgBr₃ _{( )}_{3 H O/H} 2 + Product; Product is : D C C H (CH ) C_{3 3} D OH OH (A) D C C H (CH ) C_{3 3} D OH₃ OH₃ (B) D C C H (CH ) C_{3 3} D OH₃ OH (C) D C C H (CH ) C_{3 3} D OH OH₃ (C)

5.22 The final product of the following reaction is

Me H H o–COCH₃ Ph Ph BrCl/CCl₄ Br Me H Cl Ph Ph (A) Me _Cl H Br Ph Ph (B) Br _Me H Cl Ph Ph (C) Br _Me H Br Ph Ph (D)

SECTION - II : MULTIPLE CORRECT ANSWER TYPE

5.23 Which statement is correct about the end product of the following reaction series,

C CH HOCI (excess)

(1) (2)

conc. NaOH,

(A) It is optically inctive hydroxy detone (B) It is a resolvable hydroxy acid

(C) It is a nonresolvable aldehyde (D) It is an optically inactive hydroxy acid

5.24 The following synthesis can not be carried out be:

CH = CH₂ _{CH = CH} 2 I CI I (A) CI₂ (1) (2)

ICI / CH COOH₃ ICI / CH COOH₃ (3)

Zn dust (4)

(B) HOCI / H₍₁₎ CI / Fe2₍₂₎ ₍₃₎

ICI / ZnCI (excess)₂

(4) NaOH,

(48)

(C) (B) (1) HOBr / H (2) CI / Fe₂ (3) ICI / ZnCl (excess)₂ (4) Zn dust, CH COOH₃ Br / CH COOH₂ ₃ (1) (2) CI / Fe₂ (3)

ICI / CH COOH (excess)₃

(4) NaNH₂ 5.25 Bu – C CH LiNH₂ A (i) PhCHO (ii) H O₂ B MnO2 C

Compound C of the above reaction can not be :

(A) CHO C CBu (B) CHO C CBu (C) CHO C CBu (D) O C – C CBu

5.26 Acetone (CH3COCH3) is the major product in :

I II III CH = C = CH₂ ₂ H O3 CH C CH₃ H SO / HgSO / H O2 4 4 2 CH C CH₃ BH THF 3-H O / O3-H₂ ₂

(A) I (B) II (C) III (D) none

SECTION - III : ASSERTION AND REASON TYPE

5.27 Statement-1 : CH₂ CH₂ HBr 40°C CH₃ Br CH₂ Minor Major CH₂ + CH₂ Br Minor Major – 20°C CH₃ Br CH₂ CH₃ + CH₂ Br Statement-2 : CH₂ CH2 H + CH₃ CH₂ + Allylic cation Br Br CH₃ CH₂ CH₂ CH₃ + Br

major at low temp major at of high temp CH₃

CH₂ Br