Question 1

Why is the chlorine atom in vinyl chloride comparatively less reactive?

Accepted Answer

Because the C–Cl bond acquires some partial double-bond character. In vinyl chloride the chlorine lone pair delocalises into the C=C, giving the C-Cl bond partial double-bond character (shorter, stronger, less reactive). The stored 'sp3 carbon' reason is wrong - the carbon is sp2.

Question 2

What is the product obtained when neopentyl bromide undergoes the Wurtz reaction?

Accepted Answer

2,2,5,5-tetramethylhexane. Wurtz coupling of neopentyl bromide (CH3)3CCH2Br gives (CH3)3C-CH2-CH2-C(CH3)3 = 2,2,5,5-tetramethylhexane, not 2,2,4,4-tetramethylpentane.

Question 3

Which of the following reactions is most likely to produce a hydrocarbon in high yield with ease?

Accepted Answer

RCOOK subjected to electrolytic oxidation. Electrolytic oxidation of RCOOK typically leads to the decarboxylation of the carboxylate, resulting in the formation of a hydrocarbon in high yield, as the process effectively removes the carboxyl group.

Question 4

Which of the following reactions proceeds by an S_N2 mechanism?

Accepted Answer

CH3Br reacts with OH− to form CH3OH and Br−. CH3Br is a primary (methyl) substrate with no steric hindrance, so its reaction with OH- proceeds cleanly by SN2. Tertiary (CH3)3C-Br goes by SN1, and the dehydration of ethanol is an elimination.

Question 5

Identify the incorrect statement among the following:

Accepted Answer

Methyl magnesium bromide reacts with acetone to produce butan-2-ol.. The statement is incorrect because methyl magnesium bromide (a Grignard reagent) reacts with acetone to produce 2-propanol, not butan-2-ol, as the reaction involves the addition of the methyl group to the carbonyl carbon of acetone.

Question 6

When chloroform is exposed to air and sunlight by keeping it open, what product is formed?

Accepted Answer

carbonyl chloride. When chloroform is exposed to air and sunlight, it undergoes photodecomposition, leading to the formation of carbonyl chloride, which is a result of the breakdown of chloroform in the presence of oxygen.

Question 7

In the Wurtz–Fittig reaction, which pair of compounds undergoes condensation?

Accepted Answer

One molecule each of an aryl halide and an alkyl halide. The Wurtz-Fittig reaction couples an aryl halide with an alkyl halide using sodium in dry ether to give an alkylarene (e.g. C6H5Br + CH3Br + 2Na -> C6H5CH3). It is the mixed aryl/alkyl coupling.

Question 8

When 1,1,1-trichloroethane is treated with silver powder, which major organic product is obtained?

Accepted Answer

2-Butyne. Silver removes all chlorines from two molecules of 1,1,1-trichloroethane (CH3CCl3) with coupling of the two carbons bearing the halogens: 2 CH3CCl3 + 6Ag -> CH3-C(triple bond)C-CH3 + 6AgCl, giving 2-butyne.

Question 9

In the nucleophilic substitution reaction R−Br + Cl− (DMF)→ R−Cl + Br−, which substrate will show complete inversion of configuration?

Accepted Answer

C6H5CHCH3Br. The substrate C6H5CHCH3Br has a chiral center that allows for complete inversion of configuration during the nucleophilic substitution reaction. The presence of the bulky groups on either side of the chiral center facilitates this inversion, making it the only option that exhibits this behavior.

Question 10

Chlorobenzene needs very harsh conditions for substitution of Cl by OH, whereas in 2,4-dinitrochlorobenzene the chlorine is displaced much more easily. The reason is that

Accepted Answer

the NO2 groups withdraw electron density from the ortho and para positions. The -NO2 groups at the 2- and 4-positions withdraw electron density (by -I and -M effects) from the ortho and para carbons, stabilising the Meisenheimer intermediate and so making nucleophilic displacement of Cl by OH much easier.

Question 11

What is the principal product obtained when 1,1,1-trichloropropane is heated with aqueous potassium hydroxide?

Accepted Answer

Propionic acid. 1,1,1-Trichloropropane (CH3CH2CCl3) has three halogens on the same carbon. Aqueous KOH hydrolyses this gem-trihalide all the way to the carboxylic acid, giving CH3CH2COOH (propionic acid).

Question 12

Consider the following substitutions:
(i) (CH3)2CH–CH2Br reacts with C2H5OH to give (CH3)2CH–CH2OC2H5 and HBr
(ii) (CH3)2CH–CH2Br reacts with C2H5O− to give (CH3)2CH–CH2OC2H5 and Br−
The reaction pathways for (i) and (ii), respectively, are:

Accepted Answer

SN1 and SN2. With a neutral weak nucleophile (C2H5OH) the substitution proceeds by an SN1-type (solvolysis) pathway, while with the strong nucleophile ethoxide (C2H5O-) it goes by SN2. So (i) is SN1 and (ii) is SN2.

Question 13

In the reaction sequence CH3CH2Cl a0--NaCN--> X a0--Ni/H2--> Y a0--acetic anhydride--> Z, what is the compound Z?

Accepted Answer

CH3CH2CH2NHCOCH3. CH3CH2Cl + NaCN -> CH3CH2CN (X). Ni/H2 reduces the nitrile to the primary amine CH3CH2CH2NH2 (Y). Acetic anhydride acetylates this amine to give CH3CH2CH2NHCOCH3 (Z, N-propylacetamide).

Question 14

For the reaction CH3Br + Nu− → CH3−Nu + Br−, the nucleophiles A to D give the following decreasing order of reaction rate: [Nu− = (A) PhO−, (B) AcO−, (C) HO−, (D) CH3O−]. Which sequence is correct?

Accepted Answer

D > C > B > A. For SN2 on CH3Br nucleophilicity follows conjugate-acid basicity: CH3O- (pKa ~15.5) > HO- (15.7, but stronger nucleophile here) > PhO- (10) > AcO- (4.8). Order D > C > A > B.

Question 15

The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate:

Accepted Answer

CH3 – CH+ – CH2 – Cl. The correct option represents a carbocation intermediate where the chlorine atom has added to the propene, resulting in a positively charged carbon adjacent to the chlorine. This structure is consistent with the electrophilic addition mechanism of HOCl to propene, where the formation of a carbocation is a key step.

Question 16

Which of the following species shows the highest reactivity?

Accepted Answer

ICl. ICl is more reactive than the other species because it is a polar molecule with a significant dipole moment, which allows it to engage in more vigorous reactions compared to the nonpolar diatomic molecules like I2, Cl2, and Br2.

Question 17

Which organic chloroalkane undergoes complete inversion of configuration in an S_N2 reaction?

Accepted Answer

CH3Cl. SN2 rate decreases with steric hindrance: methyl > primary > secondary > tertiary. CH3Cl (methyl chloride) is the least hindered, so it undergoes the fastest clean SN2 with complete inversion (backside attack).

Question 18

Given the following alkyl bromides:
(A) CH3–CH2–CH2–CH2–Br
(B) CH2=CH–CH(Br)–CH3
(C) CH3–CH(Br)–CH3
Which sequence correctly represents their order of S_N1 reactivity?

Accepted Answer

B > C > A. SN1 rate follows carbocation stability. (B) CH2=CH-CHBr-CH3 gives a resonance-stabilised allylic secondary cation (most stable), (C) isopropyl is a simple secondary cation, and (A) n-butyl is primary. Order: B > C > A.

Question 19

An organic compound (A) has the molecular formula C8H9Br and on warming with alcoholic AgNO3 it produces a yellow precipitate. When (A) is oxidized, it yields an acid (B) with formula C8H6O4. Acid (B) readily cyclizes to its anhydride on heating. Which of the following is compound (A)?

Accepted Answer

o-CH3C6H4CH2Br. A gives AgBr (benzylic C-H2Br) with alcoholic AgNO3; oxidation of both the CH3 and CH2Br groups of o-CH3C6H4CH2Br gives o-phthalic acid C8H6O4, which readily forms phthalic anhydride on heating. A para isomer would give terephthalic acid, which does not cyclize.

Question 20

For an S_N2 substitution, which sequence correctly arranges the following chlorides in decreasing reactivity: CH3Cl, CH3CH2Cl, (CH3)2CHCl, and (CH3)3CCl?

Accepted Answer

CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl. SN2 rate is governed by backside-attack steric hindrance, so reactivity falls as alpha-substitution rises: CH3Cl > CH3CH2Cl > (CH3)2CHCl > (CH3)3CCl.

Question 21

When 1,1,1-trichloroethane is treated with silver powder, the principal organic product obtained is

Accepted Answer

2-Butyne. 1,1,1-trichloroethane (CH3CCl3) with Ag powder undergoes Wurtz-type coupling at the CCl3 carbons, joining two units and eliminating halogen to form CH3-C(=)C-CH3, i.e. 2-butyne. Answer: option 2.

Question 22

Which method is most suitable for preparing alkyl fluorides?

Accepted Answer

Swarts reaction. The Swarts reaction treats an alkyl chloride/bromide with a metallic fluoride (AgF, Hg2F2, CoF2, SbF3) to give the alkyl fluoride, e.g. CH3Br + AgF -> CH3F + AgBr. Free-radical fluorination is too violent/uncontrolled to be useful.

Question 23

Which of the following on heating with aqueous KOH, produces acetaldehyde?

Accepted Answer

CH3CHCl2. Heating CH3CHCl2 with aqueous KOH leads to dehydrohalogenation, resulting in the formation of acetaldehyde through an elimination reaction where a hydrogen and a chlorine atom are removed.

Question 24

A solution of (−)-1-chloro-1-phenylethane in toluene racemises slowly in the presence of a small amount of SbCl5, due to the formation of -

Accepted Answer

carbocation. The presence of SbCl5 facilitates the formation of a carbocation by stabilizing the positive charge, allowing for the racemization process to occur as the molecule can interconvert between enantiomers.

Question 25

The reaction of propene with HOCl (Cl2 + H2O) proceeds through the intermediate:
(1) CH3–CH2–CH2–OH
(2) CH3–CH+–CH2–Cl
(3) CH3–CH(OH)–CH2+
(4) CH3–CHCl–CH2+

Accepted Answer

CH3–CH+–CH2–Cl. The correct option, CH3–CH+–CH2–Cl, represents a carbocation intermediate formed during the electrophilic addition of HOCl to propene, where the double bond reacts with the electrophile, leading to the formation of a positively charged carbon atom.

Question 26

The increasing order of the reactivity of the following halides for the S_N1 reaction is:
(I) CH3CH(Cl)CH2CH3
(II) CH3CH2CH2Cl
(III) p-CH3O–C6H4–CH2Cl

Accepted Answer

(II) < (I) < (III). The reactivity of halides in S_N1 reactions is influenced by the stability of the carbocation formed during the reaction. In this case, (II) has a primary carbocation which is the least stable, while (I) has a secondary carbocation, and (III) has a carbocation that is stabilized by the electron-donating methoxy group, making it the most reactive.

Question 27

An 'A assertion' and a 'R reason' are given below. Choose the correct answer from the following options:
A assertion (A): Vinyl halides do not undergo nucleophilic substitution easily.
Reason (R): Even though the intermediate carbocation is stabilized by loosely held p-electrons, the cleav

Accepted Answer

Both (A) and (R) are correct statements and (R) is the correct explanation of (A). Vinyl halides resist nucleophilic substitution because the lone pair on halogen conjugates with the C=C, giving the C-X bond partial double-bond character (shorter, stronger). Both (A) and (R) are correct and (R) is the correct explanation of (A).

Question 28

Increasing order of reactivity of the following compounds for S_N1 substitution is:
(A) (CH3)2CH–CH2–Cl
(B) CH3–CH2–CH2–Cl
(C) p-OCH3–C6H4–CH2–Cl
(D) C6H5–CH2–CH2–Cl

Accepted Answer

(B) < (A) < (D) < (C). SN1 rate follows carbocation stability. (C) p-methoxybenzyl gives a resonance- and OMe-stabilized benzylic cation (most reactive). (A), (B), (D) all ionize to primary cations; the accepted order is (B) < (A) < (D) < (C).

Question 29

The major product of the following reaction is - CH3CH2CHBrCH2Br
(i) KOH alc.
(ii) NaNH2 in liq. NH3

Accepted Answer

CH3CH2C≡CH. 1,2-dibromobutane on double dehydrohalogenation with alc. KOH gives a bromoalkene/alkyne mixture; NaNH2 in liq. NH3 then drives it fully to the terminal alkyne but-1-yne, CH3CH2C(triple)CH. So the product is CH3CH2C≡CH.

Question 30

The major product obtained from E2-elimination of 3-bromo-2-fluoropentane is:

Accepted Answer

CH3CH2CH=C(F)–CH3. Br at C3 leaves and a beta-H at C2 is removed to give the more stable internal alkene 2-fluoro-2-pentene, CH3CH2CH=C(F)CH3. The stored option still retains Br and is terminal, so it is not an elimination product.

Question 31

Arrange the following bonds according to their average bond energies in descending order: C–Cl, C–Br, C–F, C–I

Accepted Answer

C–F > C–Cl > C–Br > C–I. Carbon-halogen bond energy decreases as the halogen gets larger: C-F > C-Cl > C-Br > C-I.

Question 32

In Carius method, halogen containing organic compound is heated with fuming nitric acid in the presence of:

Accepted Answer

AgNO3. In the Carius method, silver nitrate (AgNO3) is used to facilitate the conversion of halogenated organic compounds into their corresponding silver halides, allowing for the determination of halogen content through subsequent analysis.

Question 33

Which of the following compound is added to the sodium extract before addition of silver nitrate for testing of halogens?

Accepted Answer

Nitric acid. Before adding silver nitrate in the halogen test, the sodium (Lassaigne) extract is boiled with dilute nitric acid to decompose any NaCN/Na2S that would otherwise give false precipitates. The added compound is nitric acid.

Question 34

Among the following compounds I-IV, which one forms a yellow precipitate on reacting sequentially with (i) NaOH (ii) dil. HNO3 (iii) AgNO3?

I: p-chlorotoluene
II: m-chlorotoluene
III: o-bromotoluene
IV: o-chlorobenzyl iodide

Accepted Answer

IV. The correct option is IV because o-chlorobenzyl iodide can undergo a nucleophilic substitution reaction with NaOH to form a phenolic compound, which upon treatment with dilute HNO3 can be oxidized to a corresponding phenol. This phenol then reacts with AgNO3 to form a yellow precipitate of silver halide, indicating the presence of the halogen.

Question 35

The gas 'A' is having very low reactivity reaches to stratosphere. It is non-toxic and non-flammable but dissociated by UV-radiations in stratosphere. The intermediates formed initially from the gas 'A' are:

Accepted Answer

Cl· + CF2Cl. Gas A is a chlorofluorocarbon (Freon) such as CF2Cl2: unreactive, non-toxic, non-flammable, photodissociated in the stratosphere. UV cleaves the weakest C-Cl bond homolytically, giving Cl(.) and .CF2Cl as the initial intermediates.

Question 36

The IUPAC name of ethylidene chloride is -

Accepted Answer

1,1-Dichloroethane. The correct option, 1,1-Dichloroethane, accurately reflects the structure of ethylidene chloride, which has two chlorine atoms attached to the same carbon atom in a two-carbon alkane chain.

Question 37

In the given conversion the compound A is: [Starting compound: o-bromophenol] (CH3)3CLi → [A] (i) CO2 (ii) H3O+ → product: o-hydroxybenzoic acid

Accepted Answer

A dilithiated intermediate with Li at the former Br position and OLi at the former OH position. The correct option describes a dilithiated intermediate formed when the lithium reagent replaces the bromine atom and the hydroxyl group is converted to an alkoxide, allowing for further reactions with CO2 and H3O+ to yield the final product.

Question 38

Two isomers (A) and (B) with Molar mass 184 g/mol and elemental composition C 52.2%; H 4.9% and Br 42.9% gave benzoic acid and p-bromobenzoic acid, respectively on oxidation with KMnO4. Isomer ‘A’ is optically active and gives a pale yellow precipitate when warmed with alcoholic AgNO3. Iso

Accepted Answer

H3C — CHBr — C6H5 and 1-bromo-4-ethylbenzene. Isomer A, H3C — CHBr — C6H5, is optically active due to the presence of a chiral center, while isomer B, 1-bromo-4-ethylbenzene, does not have a chiral center and thus is not optically active. The reaction with alcoholic AgNO3 indicates the presence of a benzylic bromide in isomer A, which forms a pale yellow precipitate, confirming its structure.

Question 39

The graph which represents the following reaction is:
(C6H5)3C–Cl OH− / Pyridine (C6H5)3C–OH

Accepted Answer

rate vs [(C6H5)3C–Cl] is a straight line through origin. The reaction is first-order with respect to the substrate (C6H5)3C–Cl, meaning that the rate of the reaction is directly proportional to the concentration of (C6H5)3C–Cl. Therefore, plotting the rate against the concentration of (C6H5)3C–Cl yields a straight line through the origin.

Question 40

Which of the following compounds is an example of freon?

Accepted Answer

C2Cl2F2. C2Cl2F2 is classified as a freon because it contains both chlorine and fluorine atoms, which are characteristic of chlorofluorocarbons (CFCs) used as refrigerants.

Question 41

Assertion A: Hydrolysis of an alkyl chloride is a slow reaction but in the presence of NaI, the rate of the hydrolysis increases.
Reason R: I⁻ is a good nucleophile as well as a good leaving group.
In the light of the above statements, choose the correct answer from the options given below

Accepted Answer

Both A and R are true and R is the correct explanation of A. The hydrolysis of alkyl chlorides is indeed slow, but the presence of NaI enhances the reaction rate because I⁻ acts as a strong nucleophile, facilitating the substitution reaction. Additionally, I⁻ is a good leaving group, which further supports the reaction mechanism, making R a valid explanation for A.

Question 42

The one giving maximum number of isomeric alkenes on dehydrohalogenation reaction is (excluding rearrangement)
(1) 1-Bromo-2-methylbutane
(2) 2-Bromopentane
(3) 2-Bromo-3, 3-dimethylpentane
(4) 2-Bromopropane

Accepted Answer

2-Bromopentane. 2-Bromopentane can undergo dehydrohalogenation to form multiple alkenes due to the presence of two different β-hydrogens that can be eliminated, leading to various structural isomers. This results in a higher number of possible alkenes compared to the other options.

Question 43

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): Haloalkanes react with KCN to form alkyl cyanides as a main product while with AgCN form isocyanide as the main product.
Reason (R): KCN and AgCN both are highly ionic c

Accepted Answer

(A) is correct but (R) is not correct.. The assertion is accurate because haloalkanes do indeed react with KCN to yield alkyl cyanides and with AgCN to produce isocyanides. However, the reason provided is incorrect; while KCN and AgCN are ionic, their ionic nature does not explain the differing products formed in the reactions.

Question 44

The correct statement regarding nucleophilic substitution reaction in a chiral alkyl halide is:
(1) Retention occurs in S_N1 reaction and inversion occurs in S_N2 reaction
(2) Racemisation occurs in both S_N1 and S_N2 reactions
(3) Racemisation occurs in S_N1 reaction and inversion occurs

Accepted Answer

(3) Racemisation occurs in S_N1 reaction and inversion occurs in S_N2 reaction. In an S_N1 reaction, the formation of a planar carbocation intermediate allows for the possibility of attack from either side, leading to racemisation. In contrast, S_N2 reactions involve a direct backside attack by the nucleophile, resulting in inversion of configuration at the chiral center.

Question 45

Which among the following halide/s will not show S_N1 reaction ? (A) H2C = CH – CH2Cl (B) CH3 – CH = CH – Cl (C) C6H5 – CH2 – Cl (D) (CH3)2CH – Cl Choose the most appropriate answer form the options given below: (1) (B) and (C) only (2) (B) only (3) (A) and (B) only (4) (A), (B) and (D) on

Accepted Answer

(B) only. Option (B) is correct because the presence of a double bond adjacent to the leaving group in CH3 – CH = CH – Cl destabilizes the carbocation that would form in an S_N1 reaction, making it less likely to occur.

Question 46

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: Aryl halides cannot be prepared by replacement of hydroxyl group of phenol by halogen atom.
Reason R: Phenols react with halogen acids violently.
In the light of the above sta

Accepted Answer

A is true but R is false. Assertion A is true because aryl halides cannot be formed by simply replacing the hydroxyl group in phenol due to the stability of the phenolic structure. Reason R is false as phenols do not react violently with halogen acids; instead, they typically undergo electrophilic substitution reactions.

Question 47

Given below are two statements:
Statement I: High concentration of strong nucleophilic reagent with secondary alkyl halides which do not have bulky substituents will follow S_N2 mechanism.
Statement II: A secondary alkyl halide when treated with a larger excess of ethanol follows S_N1 mech

Accepted Answer

Both Statement I and Statement II are true. Both statements are accurate because a strong nucleophile can facilitate an S_N2 reaction with secondary alkyl halides lacking bulky groups, while a secondary alkyl halide in a solvent like ethanol can undergo an S_N1 reaction due to the solvent's ability to stabilize the carbocation intermediate.

Question 48

The product (C) in the below mentioned reaction is:

CH3 – CH2 – CH2 – Br
KOH(alc), Δ → B
A HBr → B
KOH(aq) → C

Accepted Answer

Propan-2-ol. The reaction sequence involves the elimination of bromine to form an alkene, followed by hydration, which results in the formation of an alcohol. In this case, the product is propan-2-ol, as the hydration of the alkene leads to the formation of the secondary alcohol.

Question 49

Identify the correct set of reagents or conditions ‘X’ and ‘Y’ in the following set of transformation.
CH3 – CH2 – CH2 – Br —X→ Product —Y→ CH3 – CH(Br) – CH3

Accepted Answer

X = conc.alc. NaOH, 80°C, Y = HBr/acetic acid. The correct option involves using concentrated alcoholic NaOH at high temperature to facilitate elimination, forming an alkene intermediate, which is then treated with HBr in acetic acid to achieve the desired bromination at the more substituted carbon, resulting in the final product.

Question 50

During the detection of acidic radical present in a salt, a student gets a pale yellow precipitate soluble with difficulty in NH4OH solution when sodium carbonate extract was first acidified with dil. HNO3 and then AgNO3 solution was added. This indicates presence of:

Accepted Answer

Br−. The formation of a pale yellow precipitate that is only slightly soluble in ammonium hydroxide indicates the presence of bromide ions (Br−), as this behavior is characteristic of silver bromide (AgBr), which is less soluble compared to other silver halides.

JEE Main Chemistry: Haloalkanes and Haloarenes questions with solutions

Questions