Question 1

When hydrogen halides react with alcohols, alkyl group rearrangement does not typically happen with most primary alcohols. What is the most accurate reason for this?

Accepted Answer

Primary carbocations are not stable and therefore do not form.. Primary carbocations are not stable and therefore do not form, which is the reason that alkyl group rearrangement does not typically happen with most primary alcohols during reactions with hydrogen halides.

Question 2

Which of the following statements accurately describes the behavior of the given compound?

Accepted Answer

The electron-withdrawing group (–C≡N) is delocalized with the carbon-carbon double bond, allowing the carbocation intermediate formed during nucleophilic attack to be stabilized by resonance. Therefore, such alkenes participate in nucleophilic addition reactions.. The electron-withdrawing group (–C≡N) delocalizes with the carbon-carbon double bond, allowing the carbocation intermediate to be stabilized by resonance, thus facilitating nucleophilic addition reactions.

Question 3

Tertiary and secondary carbocations readily participate in elimination reactions when treated with strong alkoxide bases, while aryl and vinyl halides resist nucleophilic substitution. Which of the following reactions is correct?

Accepted Answer

Tert-butyl bromide reacts with sodium methoxide to form isobutylene.. Tert-butyl bromide reacts with sodium methoxide to form isobutylene through an elimination reaction, which is a characteristic of tertiary carbocations when treated with strong alkoxide bases.

Question 4

Although neopentyl alcohol is a primary alcohol, it follows the SN1 mechanism due to steric hindrance caused by a bulky alkyl group. Identify its structure.

Accepted Answer

CH3C(CH3)2CH2OH. Neopentyl alcohol follows the SN1 mechanism due to steric hindrance caused by its bulky alkyl group, and its structure is CH3C(CH3)2CH2OH.

Question 5

In the following sequence of reactions, identify the structures of intermediates Y, Z, W, and M: Y --(NaBH4)--> Z --(conc. HBr)--> W Y --(N2H4 / KOH, heat)--> M M --(NBS, heat/light)--> W W --(1) Mg/ether, (2) CO2, (3) H3O+)--> Ph-CH(COOH)-CH2-CH3 Which of the following statements about th

Accepted Answer

Z is (plus/minus) Ph-CH(OH)-CH2-CH3. Working backwards: W = Ph-CH(Br)-CH2-CH3 (secondary benzylic bromide). Z is obtained by NaBH4 reduction of ketone Y, giving the racemic secondary alcohol Ph-CH(OH)-CH2-CH3. Therefore Z = (+-) Ph-CH(OH)-CH2-CH3, making statement 1 correct. Y is Ph-CO-CH2-CH3 (propiophenone), not Ph-CH(OH)-CH3, so statement 2 is false.

Question 6

Consider the compound CH3-CH2-CH(X)-CH3, where X is a halogen. When treated with alcoholic KOH and heated, it undergoes elimination to give alkene products. For which choice of X will the proportion of the Hofmann (less substituted) alkene product be the greatest?

Accepted Answer

X = F. Fluorine is the poorest leaving group among halogens; the E2 transition state with C-F bond breaking is early and reactant-like, reducing the preference for the more substituted (Saytzeff) product and maximising the proportion of the Hofmann (less substituted) alkene.

Question 7

How many of the following reactions give a major product predominantly via an elimination mechanism? (1) Ethyl chloride (CH3CH2Cl) treated with NaOCH3 / CH3OH (2) 2-chlorobutane (CH3CH2CHClCH3) treated with potassium tert-butoxide (t-BuOK)

Accepted Answer

1. Reaction (1): Ethyl chloride is a primary alkyl halide. NaOCH3/CH3OH is a moderately basic, non-bulky nucleophile/base. Primary substrates strongly prefer SN2 over E2. Major product: CH3CH2OCH3 (SN2). Not elimination. Reaction (2): 2-Chlorobutane is a secondary alkyl halide. t-BuOK is a bulky strong base. Bulky bases cannot attack the carbon backside effectively (steric), so they abstract a beta-H via E2. Major product: but-2-ene (E2). This is elimination. Count = 1.

Question 8

Starting from p-toluidine (a benzene ring bearing -NH2 and -CH3 groups at para positions), the following reaction sequence is carried out: Step 1: NaNO2 + HCl at 0-5 deg C, then Cu2Cl2 + HCl gives compound [A]. Step 2: [A] treated with Cl2 under UV light (hv) gives compound [B]. Step 3: [B

Accepted Answer

ClCH2-C6H4-C6H4-CH2Cl (with Cl on both benzylic carbons). p-Toluidine is converted to p-chlorotoluene [A] by Sandmeyer reaction, then free-radical chlorination with Cl2/hv substitutes the benzylic CH3 to give p-chlorobenzyl chloride [B] (Cl-C6H4-CH2Cl). Wurtz coupling of two molecules of [B] with sodium gives [C] = ClCH2-C6H4-C6H4-CH2Cl.

Question 9

How many moles of HI are required to completely react with one mole of the following compound: a benzene ring bearing four substituents — an allyl group (CH2=CH-CH2-), a 2-hydroxyethyl group (CH2CH2OH), a hydroxyl group (OH), and a methoxy group (OCH3)?

Accepted Answer

4. The compound has four HI-reactive sites: (1) the allyl C=C double bond, (2) the primary alcohol CH2CH2OH, (3) the phenolic OH, and (4) the aryl methyl ether OCH3. Each consumes exactly one mole of HI, giving a total of 4 moles.

Question 10

How many of the following molecules/compounds are chiral? (A) Unsubstituted cyclohexane (B) trans-1,2-dibromocyclopropane (C) CH3-CH=C=CH-CH3 (penta-2,3-diene) (D) trans-1,4-dimethylcyclohexane (E) 2-bromobutane (CH3-CHBr-CH2-CH3) (F) CH3-CH=C=C=CH-CH3 (hexa-2,3,4-triene)

Accepted Answer

3. Analyzing each: (A) cyclohexane - achiral. (B) trans-1,2-dibromocyclopropane - chiral (non-superimposable on mirror image, no plane of symmetry in trans form). (C) penta-2,3-diene CH3CH=C=CHCH3 - has identical groups at each end, achiral. (D) trans-1,4-dimethylcyclohexane - has a center of inversion, achiral. (E) 2-bromobutane - has one chiral carbon (C2 bonded to H, Br, CH3, CH2CH3), chiral. (F) hexa-2,3,4-triene has even number of double bonds (3 cumulated), so same geometry as cumulene with even count - ends are in same plane, achiral. Thus chiral molecules: B and E = 2. But this interpr

Question 11

Which of the following statements about nucleophilic substitution reactions are correct?

Accepted Answer

CH3CH2Cl reacts more slowly than PhCH2Cl in bimolecular nucleophilic substitution (SN2). Ag2O in water generates Ag⁺ which removes the leaving group (Br⁻) as AgBr, promoting SN1 ionization — it acts as a Lewis acid, not by producing nucleophilic OH⁻. PhCH2Cl (benzyl chloride) has enhanced SN2 reactivity relative to ethyl chloride due to partial carbocation stabilization and empty d-like orbital overlap in the transition state. The SN2 order is CH3Cl > CH3CH2Cl > secondary > tertiary, so option D is wrong. Options A and C are correct.

Question 12

How many of the following compounds will give a white precipitate of silver salt when treated with ammoniacal AgNO3 solution (Tollens-type test for terminal alkynes and active halides)? (a) CH3-CH2-C≡CH (1-butyne) (b) CH3-C≡C-CH3 (2-butyne) (c) CH2=CH-CH2-Cl (allyl chloride) (d) CH2=C(Cl)-

Accepted Answer

4. Ammoniacal AgNO3 reacts with: (1) Terminal alkynes: the terminal C-H is acidic (pKa ~25), forming Ag-C≡C- precipitate. (2) Halides that can undergo SN1 (active halides): allylic, benzylic, propargylic halides form carbocations readily and react with Ag+. Vinyl halides (sp2 C-X) are unreactive toward SN1. Internal alkynes have no acidic C-H. Analysis: (a) 1-butyne (terminal alkyne) -> YES (white/buff AgC≡C salt). (b) 2-butyne (internal alkyne) -> NO. (c) Allyl chloride (allylic) -> YES. (d) 2-chloropropene (vinyl halide) -> NO. (e) 3-chlorocyclohexene (allylic chloride) -> YES. (f) Benzyl ch

Question 13

Which of the following molecules can exhibit optical isomerism?

Accepted Answer

CH3-CH(OH)-COOH (lactic acid). Optical isomerism arises when a molecule lacks a plane of symmetry. CH3-CH(OH)-COOH has a chiral carbon (C-2 bears H, OH, CH3, COOH — four different substituents), so it is optically active. 1-propanol has no chiral centre. Acetone has a plane of symmetry. Buta-1,2-diene (CH3-CH=C=CH2) is an allene; for an allene to be chiral, the two terminal groups on each sp-carbon must be different. Here the terminal carbons bear H,H and H,CH3 — the CH3 end has two different groups (H and CH3) but the other end has two identical H atoms, so the molecule is NOT chiral. Therefo

Question 14

Consider two chair conformations of 1,2-dimethylcyclohexane in which both methyl groups are equatorial in one conformation and both are axial in the other. These two structures are best described as:

Accepted Answer

Conformers. Two chair conformations of the same compound (e.g., both-equatorial vs both-axial arrangement of substituents) are interconverted by ring flipping without breaking any bonds. They represent the same constitutional structure with different conformations — these are called conformers (or conformational isomers). Enantiomers and diastereomers are configurational stereoisomers requiring bond-breaking to interconvert. They are not identical either (they differ in 3D arrangement at a given instant). Hence the correct relationship is conformers.

Question 15

The compound 2-methylpent-3-ene (CH3-CH=CH-CH(CH3)-CH3) is treated with NBS in CCl4 under UV light. Which of the following products CANNOT be formed in this reaction?

Accepted Answer

CH3-CH=CH-CH(CH2Br)-CH3. In CH3-CH=CH-CH(CH3)-CH3 the double bond is between C2 and C3. Allylic positions are C1 (the CH3 attached to C2) and C4 (the CH bearing the methyl at C4). Abstracting H from C1 gives radical at C1 delocalised to C3, producing BrCH2-CH=CH-CH(CH3)-CH3 (option B) or CH3-CH=CH-CBr(CH3)... wait, the allylic radical from C1 spans C1-C2=C3 giving product at C1 or C3 but not C4. Abstracting H from C4 gives radical spanning C4-C3=C2, yielding bromination at C4 (option D) or C2 (option C). The CH2Br group in option A would require abstraction of a hydrogen from the methyl substi

Question 16

Which of the following statements about nucleophilic substitution reactions is/are correct? (i) The rate of hydrolysis of tert-butyl bromide increases upon addition of Ag2O. (ii) Aqueous Ag2O acts as a source of nucleophilic OH- ions. (iii) CH3CH2Cl is less reactive than PhCH2Cl in a bimol

Accepted Answer

Only (i) and (ii) are correct. tert-Butyl bromide undergoes SN1; Ag+ removes Br- and aqueous Ag2O provides nucleophilic OH-, so both (i) and (ii) are correct. PhCH2Cl is actually more reactive than CH3CH2Cl in SN2 (benzylic activation), so (iii) is wrong. SN2 rate order is CH3Cl > CH3CH2Cl > CH3CH(Cl)CH3 (steric hindrance increases), so (iv) is also wrong.

Question 17

Consider the following eight reactions. How many of them produce an alkyl halide as one of the final products? (a) C2H5OH + SOCl2 -> (b) R-I + AgF -> (c) EtOH + NaI + H2SO4 -> (d) EtOH + NaI + H3PO4 -> (e) CH3-CH=CH2 + HBr -> (f) CH3-CH2-Cl + NaI (in acetone) -> (g) cyclopentyl-CH(Cl)-CH3

Accepted Answer

5. (a) EtOH + SOCl2 -> C2H5Cl (ethyl chloride) + SO2 + HCl. YES, alkyl halide. (b) R-I + AgF -> R-F + AgI. YES, alkyl halide (fluoride). (c) EtOH + NaI + H2SO4 -> C2H5I + NaHSO4 + H2O. YES, alkyl halide. (d) EtOH + NaI + H3PO4: H3PO4 is a weaker acid than H2SO4 and does not protonate ethanol well enough for NaI substitution. This reaction does NOT give alkyl iodide efficiently. NO. (e) CH3-CH=CH2 + HBr -> CH3-CHBr-CH3 (Markovnikov). YES, alkyl halide. (f) C2H5Cl + NaI in acetone -> C2H5I + NaCl (Finkelstein reaction). YES, alkyl halide. (g) Cyclopentyl-CH(Cl)-CH3 + moist Ag2O -> cyclopentyl-CH

Question 18

Consider the compound CH3CH2-C(Br)(CH3)-CH3. When this compound is treated with C2H5OH in the presence of heat, what is the major product formed?

Accepted Answer

CH3-CH=C(CH3)2. C2H5OH at high temperature promotes E2 elimination. Zaitsev's rule predicts the major product is the more substituted alkene, which is 2-methylbut-2-ene (CH3-CH=C(CH3)2) formed by losing HBr toward the more substituted carbon.

Question 19

The correct order of dipole moments of ortho-, meta-, and para-dichlorobenzene is

Accepted Answer

ortho > meta > para. For para-dichlorobenzene the two C-Cl dipoles point in exactly opposite directions and cancel (mu = 0). For ortho (60 deg apart) the resultant is sqrt(3)*mu₀, and for meta (120 deg apart) it is mu₀. Hence the order is ortho > meta > para.

Question 20

Consider the following sequence of reactions starting from compound X (molecular formula C6H10O): X + (i) CH3MgBr, (ii) H2O -> Q Q + 20% H3PO4, 360 K -> R (major product) Q + conc. HCl -> S (major product) R + HBr / benzoyl peroxide, heat -> U (major product) R + (i) H2/Ni, (ii) Br2/hv ->

Accepted Answer

Q is an alcohol. X is cyclohexanone; Grignard addition gives 1-methylcyclohexanol (Q, a tertiary alcohol). H3PO4/360 K dehydrates Q to 1-methylcyclohexene (R, an alkene). Conc. HCl converts Q to 1-chloro-1-methylcyclohexane (S) via SN1 — not dehydration — so option B is wrong. HBr/peroxide on R gives anti-Markovnikov product U (correct, D). Q being an alcohol (A) and R being an alkene (C) and U being anti-Markovnikov (D) are all correct; B is incorrect because S is a substitution product, not a dehydration product.

Question 21

Consider the following two statements about the dipole moments of dichlorobenzene isomers: Statement 1: The dipole moment of para-dichlorobenzene is zero because of its symmetrical structure. Statement 2: ortho-dichlorobenzene has a larger dipole moment than meta-dichlorobenzene because of

Accepted Answer

(A) Both Statement 1 and Statement 2 are correct. Statement 1: In para-DCB the two C-Cl bonds point in exactly opposite directions (180 deg apart), so the vector sum of their dipole moments is zero. Statement 1 is correct. Statement 2: The resultant of two equal vectors of magnitude mu at angle theta is 2*mu*cos(theta/2). For ortho (theta ≈ 60 deg): 2*mu*cos 30 ≈ 1.73*mu. For meta (theta ≈ 120 deg): 2*mu*cos 60 = mu. So ortho > meta, and the reason (smaller angle between bond dipoles) is correct. Both statements are correct.

Question 22

Examine the following claim and reasoning about an organic reaction: Claim: When 1-butene reacts with HBr in the presence of a peroxide, the product is 1-bromobutane. Reasoning: The reaction proceeds via formation of a primary carbon radical. Choose the correct option:

Accepted Answer

Both the claim and the reasoning are correct, and the reasoning correctly explains the claim. In the presence of peroxides, HBr adds via a free-radical mechanism (anti-Markovnikov). The bromine radical adds to the less substituted (terminal) carbon of 1-butene, generating the more stable secondary radical at C2, and then H adds to C2, giving 1-bromobutane. The claim is correct; however the reasoning states a primary radical is formed, which is incorrect — a secondary radical intermediate is formed.

Question 23

2-Methylpropyl bromide (isobutyl bromide) undergoes two separate reactions: (i) with sodium ethoxide (C2H5O-) and (ii) with ethanol (C2H5OH). What are the mechanisms and the products A and B formed in reactions (i) and (ii) respectively?

Accepted Answer

SN2 in (i), A = isobutyl ethyl ether; SN1 in (ii) with rearrangement, B = tert-butyl ethyl ether. With the strong nucleophile ethoxide, SN2 occurs directly at the primary carbon giving isobutyl ethyl ether. With the weak nucleophile ethanol under protic conditions, SN1 proceeds via a hydride-shifted tertiary carbocation, giving tert-butyl ethyl ether.

Question 24

Which of the following statements about SN2 reactions is/are correct?

Accepted Answer

CH3CH2CH2I reacts more readily than (CH3)2CHI in SN2 reactions. Option A: n-propyl iodide (primary) reacts faster than isopropyl iodide (secondary) in SN2 — correct. Option B: Cl is a worse leaving group than Br, so propyl chloride reacts more slowly than propyl bromide — incorrect. Option C: n-butyl bromide is primary and less hindered than neopentyl-type (CH3)3CCH2Br (which has severe 1,2-steric strain) — n-butyl bromide reacts faster — correct. Option D: methoxy group (OCH3) is electron-donating, which destabilises the transition state for SN2 at the benzylic position relative to the electr

Question 25

Which of the following reactions produces a chiral centre in the product?

Accepted Answer

CH3CH=CH2 + HOBr ->. Addition of HOBr to propene (CH3CH=CH2) follows Markovnikov rule: OH adds to the more substituted carbon (C2) and Br to C1, giving CH3CH(OH)CH2Br. C2 has CH3, OH, H, CH2Br as four different groups, creating a chiral centre. In other options, addition is either to symmetric alkenes or gives symmetric products without chiral centres.

Question 26

Identify the INCORRECT statement(s) among the following:

Accepted Answer

Higher temperatures generally favor substitution reactions over elimination reactions.. Statement C is incorrect because higher temperatures actually favor elimination (E2) over substitution. Statement D is also incorrect because triphenylchloromethane (trityl chloride) hydrolyzes very easily due to the exceptional stability of the triphenylmethyl carbocation.

Question 27

Consider the following two-step reaction sequence starting from propanoic acid (CH3CH2COOH): Step 1: Treatment with Br2 in the presence of red phosphorus, followed by hydrolysis with H2O, gives compound P. Step 2: Compound P is treated first with NaOH, then acidified with H3O+, giving comp

Accepted Answer

When Q is treated with a solution of NaNO2 in aqueous HCl, nitrogen gas is liberated.. Step 1 (HVZ) converts CH3CH2COOH to 2-bromopropanoic acid (P). Step 2 with NaOH causes SN2 inversion giving 2-hydroxypropanoic acid, i.e., lactic acid (Q). NaBH4 does not reduce -COOH, so statement 1 is wrong; NH4OH on an acid gives amide not lactic acid, so statement 2 is wrong; NaNO2/HCl liberates N2 from primary amines (not -OH), so statement 3 is wrong; alpha-bromo acids are more acidic than the parent acid due to electron withdrawal, so statement 4 is correct.

Question 28

Which of the following reagents or reaction conditions would give 2-bromo-1-phenylpropane (C6H5CH2CHBrCH3) as the product?

Accepted Answer

C6H5CH2CH(OH)CH3 treated with PBr3. PBr3 converts secondary alcohols to alkyl bromides via an SN2-like mechanism. C6H5CH2CH(OH)CH3 has the hydroxyl at the correct position; replacement with Br gives 2-bromo-1-phenylpropane directly. Option B (anti-Markovnikov HBr addition) would put Br at C-1 of the alkene, giving C6H5CHBrCH2CH3 (1-bromo-1-phenylpropane). Option C gives radical bromination at the benzylic position predominantly.

Question 29

Chlorobenzene labelled with C-14 at position 1 (the carbon bearing Cl) is treated with NaNH2 in liquid ammonia. Which of the following correctly describes the product(s) formed?

Accepted Answer

Aniline with C-14 only at position 1 (50%) and aniline with C-14 at position 2 (50%). The reaction proceeds via benzyne mechanism: NaNH2 eliminates HCl from chlorobenzene-1-C14 to give 1,2-didehydrobenzene (benzyne) with C-14 at position 1. The benzyne triple bond is between C1 (C-14) and C2 (C-12). Nucleophilic addition of NH2⁻ can occur at either C1 or C2 with equal probability, giving two products: aniline with NH2 at C-14 position (50%) and aniline with NH2 at the adjacent unlabelled carbon with C-14 still in ring (50%). This is the classic isotopic scrambling evidence for the benzyne mech

Question 30

Which of the following statements about organic reaction mechanisms is/are correct? (A) In acidic medium, an alcohol R-OH gets protonated to form R-OH2+. (B) Monochlorination of 2-methylbutane with Cl2 under UV light gives exactly two monochloro products in the ratio 33.3%: 66.6%. (C) Elim

Accepted Answer

A, C and D only. Statements A, C, and D are correct. Statement B is wrong because 2-methylbutane has four distinct sets of hydrogen atoms, so monochlorination can give four different monochloro products, not just two.

Question 31

Which of the following reactions proceeds with inversion of configuration? (A) (S)-2-butanol + Na gives the sodium alkoxide, then reacts with CH3Br via SN2. (B) (S)-2-butanol + TsCl (tosyl chloride) gives the tosylate, then tosylate reacts with CH3ONa via SN2. (C) (S)-2-butanol + PCl5 via

Accepted Answer

Reaction B only. Reaction A: (S)-2-butanol reacts with Na to form sodium (S)-2-butoxide. This breaks only the O-H bond, so configuration at chiral center is retained. The sodium butoxide then reacts with CH3Br: the oxygen (nucleophile) attacks the methyl carbon of CH3Br in SN2. The chiral center (C2 of butoxide) is NOT attacked. Configuration at C2 is retained. No inversion at chiral center. Reaction B: (S)-2-butanol + TsCl: the O-H bond is replaced by O-Ts, C-O bond is retained. Configuration at chiral center is retained (still S). The tosylate is a good leaving group. CH3ONa then attacks...

Question 32

Styrene (Ph-CH=CH2) reacts with BrCCl3 in the presence of a peroxide initiator. What is the major product?

Accepted Answer

Ph-CH(Br)-CH2-CCl3. Peroxide-initiated radical chain: CCl3 radical (from BrCCl3) adds to the terminal CH2 of styrene, forming a stable benzylic radical at the Ph-CH position, which then abstracts Br from another BrCCl3, giving Ph-CH(Br)-CH2-CCl3.

Question 33

Which of the following statements about the reaction of an alkyl halide with aqueous silver oxide (Ag2O) is/are correct? (A) The reaction is faster because Ag+ assists in removing Br- through formation of insoluble AgBr, making the C-Br bond easier to break. (B) Aqueous Ag2O provides AgOH

Accepted Answer

Both A and B. Statement A: Ag+ precipitates Br- as AgBr (insoluble), pulling the equilibrium forward and assisting in breaking the C-Br bond. This is a well-established reason why Ag2O is more effective than NaOH for this conversion. Statement A is correct. Statement B: Ag2O + H2O -> 2 AgOH. AgOH then acts as the nucleophile in an SN2 reaction. The equation is correct and this is the accepted mechanism. Statement B is correct. Both A and B are correct.

Question 34

In which of the following pairs is the first compound more reactive than the second in an SN2 reaction?

Accepted Answer

Isobutyl chloride; tert-Butyl chloride. Chlorobenzene is much less reactive than benzyl chloride in SN2. Isobutyl chloride is a primary halide while tert-butyl chloride is tertiary, so isobutyl is indeed more reactive — this pair is correct. 2-Chlorobutane and 3-Chlorobutane are both secondary, roughly equal. Option D lists the same compound twice.

Question 35

Chlorobenzene labelled with C-14 at carbon-1 reacts with NaNH2 in liquid ammonia to give aniline-type products where the amino group is found at both C-1 and C-2 positions (scrambling). Which mechanistic statement best explains this observation?

Accepted Answer

The reaction proceeds via an elimination-addition pathway involving a benzyne intermediate. The C-14 label appears equally at C-1 and C-2 in the product, which is only possible if the reaction goes through a symmetrical benzyne intermediate formed by elimination of HCl. Addition of NH2- to either end of the triple bond in benzyne gives the observed scrambling.

Question 36

Propanoic acid undergoes the following sequence of reactions: Step 1: Treatment with red phosphorus and Br2, followed by hydrolysis (Hell-Volhard-Zelinsky reaction) gives a compound more acidic than propanoic acid. Step 2: The product from Step 1 is treated with NH2OH, then acidified with

Accepted Answer

CH3CH(NH2)COOH. The HVZ reaction on propanoic acid gives 2-bromopropanoic acid; substitution of Br by NH2 (via NH3 or related nucleophile) yields 2-aminopropanoic acid (alanine, Q). The NaNO2/HCl treatment (diazotization) confirms Q has a primary amine group.

Question 37

Consider the following two reactions: Reaction 1: C6H5-CH2-CH2-CH3 is treated with Br2 under UV light (hv). Reaction 2: C6H5-CH=CH-CH3 is treated with HBr in the presence of benzoyl peroxide. What are the major products of Reaction 1 and Reaction 2 respectively?

Accepted Answer

C6H5-CH(Br)-CH2-CH3 and C6H5-CH2-CH(Br)-CH3. Free radical bromination at the benzylic position is highly selective, giving C6H5-CH(Br)-CH2-CH3. Anti-Markovnikov HBr addition (peroxide) places Br at the less hindered end: Br goes to C3 (the CH3-bearing carbon of the double bond? — re-examine). In Ph-CH=CH-CH3, Br radical adds to the less substituted carbon (the CH end, i.e., the one bearing Ph is more stable radical site on the other carbon), giving Ph-CH2-CH(Br)-CH3.

Question 38

Styrene (Ph-CH=CH2) is treated with BrCCl3 in the presence of a peroxide initiator. Which of the following correctly describes the mechanism and/or the product of this reaction?

Accepted Answer

The •CCl3 radical adds to styrene to give the intermediate Ph-CH•-CH2-CCl3. The peroxide generates RO• which abstracts Br from BrCCl3 giving •CCl3. This adds to the less hindered terminal carbon of styrene forming the stable benzylic radical Ph-CH•-CH2-CCl3. The benzylic radical then abstracts Br from another BrCCl3 molecule to give the anti-Markovnikov-like product Ph-CH(Br)-CH2-CCl3 and regenerates •CCl3.

Question 39

In which of the following reactions is the product formed in accordance with Hofmann's rule (elimination preferentially gives the less substituted alkene)? (A) CH3CH2CH(Br)CH3 treated with alcoholic KOH under heat (B) CH3CH2CH(Br)CH3 treated with CH3ONa in CH3OH under heat (C) CH3CH2CH(CH3

Accepted Answer

Options C and D only. Reactions A and B involve a secondary alkyl bromide; with alcoholic KOH (A) or sodium methoxide (B), Zaitsev elimination is favoured giving the more substituted but-2-ene. Reactions C and D involve quaternary ammonium and sulfonium salts respectively, where the bulky leaving group forces attack at the less hindered beta carbon, yielding the Hofmann product (less substituted alkene).

Question 40

How many alkyl halides (including all stereoisomers) on reaction with Mg in dry ether, followed by treatment with H+ / H2O, yield 2-methylbutane as the product?

Accepted Answer

4. Protonolysis of a Grignard reagent simply replaces MgX with H, restoring the alkyl group. So the alkyl halide must have the same carbon skeleton as 2-methylbutane. Enumerate all unique positions for the halide in the 2-methylbutane framework and count stereoisomers at chiral centers.

Question 41

How many distinct monochlorinated products (including stereoisomers) can be obtained by free-radical chlorination of methylcyclobutane when heated with Cl2?

Accepted Answer

6. Methylcyclobutane has distinct H sets: CH3 (primary, 1 product), C1-H (tertiary, 1 product), C2/C4 axial/equatorial positions (give cis and trans 1-methyl-2-chlorocyclobutane = 2 stereoisomers), and C3 (gives 1-methyl-3-chlorocyclobutane with cis/trans = 2 stereoisomers). Total = 1 + 1 + 2 + 2 = 6 distinct products.

Question 42

But-2-yne (CH3-C≡C-CH3) is treated along two parallel pathways. Path 1: catalytic hydrogenation over Pd-BaSO4 gives compound X; X is then reacted with Br2 in CCl4 to give product Z. Path 2: reduction with Na in liquid NH3 gives compound Y; Y is then reacted with Br2 in CCl4 to give product

Accepted Answer

3. Lindlar hydrogenation yields cis-but-2-ene; anti addition of Br2 gives the meso product (Z = 1 compound). Birch reduction gives trans-but-2-ene; anti addition of Br2 gives the racemic pair (A = two enantiomers). Total distinct compounds Z + A = 1 + 2 = 3.

Question 43

How many of the following compounds CANNOT undergo Friedel-Crafts reaction: toluene, nitrobenzene, xylene, cumene, aniline, chlorobenzene, m-nitroaniline, m-dinitrobenzene?

Accepted Answer

3. Friedel-Crafts reactions require a sufficiently nucleophilic ring and a Lewis acid catalyst (like AlCl3). Compounds that CANNOT react: (1) Nitrobenzene — ring heavily deactivated by -NO2 (meta director, strong EWG). (2) Aniline — -NH2 group complexes with AlCl3 to give PhNH3⁺ AlCl4⁻, completely deactivating the ring. (3) m-Nitroaniline — has both -NO2 and -NH2 (amine complexes with catalyst, and ring is further deactivated). (4) m-Dinitrobenzene — two -NO2 groups make ring too deactivated. That gives 4 compounds. However the standard answer is 3. Reconsidering: aniline and m-nitroaniline bo

Question 44

Among the following species, which one has the highest nucleophilicity?

Accepted Answer

CH3(-). Nucleophilicity increases as the electronegativity of the atom bearing the lone pair decreases. Carbon is the least electronegative among C, N, O, so CH3(-) (carbanion) holds its electron pair most loosely and attacks electrophiles most readily.

Question 45

Phenyl benzoate (Ph-O-CO-Ph) is subjected to Friedel-Crafts chlorination using Cl2/AlCl3. What is the major product?

Accepted Answer

Phenyl benzoate with Cl at the meta position on the benzoyl ring. Phenyl benzoate: the benzoate ring (C6H5-C=O-O-) has the -COO- group as meta director; the phenoxy ring (-O-CO-C6H5) has the -O- as ortho/para director. Under Cl2/AlCl3, electrophilic aromatic substitution occurs preferentially on the more activated ring (phenoxy ring, bearing -O-). Cl substitutes para to -O-, giving the para-chloro phenoxy product. However, if the question intends the structure as Ph-COO-Ph where the left Ph is directly attached to C=O (benzoyl), then the COO- group deactivates and directs meta. Standard JEE tr

Question 46

Which of the following compounds undergoes the fastest SN2 reaction with NaI in acetone? (A) CH3Cl, (B) CH3CH2Cl, (C) (CH3)2CHCl, (D) (CH3)3CCl

Accepted Answer

CH3Cl (methyl chloride). In SN2 reactions, the nucleophile attacks the carbon from the backside simultaneously as the leaving group departs. Steric hindrance around the electrophilic carbon is the dominant factor. Methyl chloride (CH3Cl) has no alkyl groups on the carbon bearing Cl, so it has the least steric hindrance and reacts fastest. Ethyl chloride (primary) is slower, isopropyl chloride (secondary) is much slower, and tert-butyl chloride (tertiary) essentially does not react by SN2 at all (it prefers E2 or SN1). Order: CH3Cl > CH3CH2Cl > (CH3)2CHCl >> (CH3)3CCl.

Question 47

Two cyclic ethers are treated with excess HI: (I) Tetrahydropyran (a six-membered ring with one oxygen atom) (II) Chroman (a benzene ring fused with a dihydropyran ring containing one oxygen) Which compound(s) give a di-iodo product on reaction with excess HI?

Accepted Answer

I and II both. In tetrahydropyran (THP), HI protonates the oxygen, then I- acts as a nucleophile to open the ring at the C-O bond. With excess HI the resulting iodo-alcohol reacts again to replace the OH with I, giving 1,5-diiodopentane. Chroman similarly undergoes C-O bond cleavage at both positions of the heterocyclic ring with excess HI, producing an aryl iodo-ethanol first and then a diiodo product. The aromatic ring is not cleaved. Both compounds give di-iodide.

Question 48

Cyclopropane undergoes the following two reactions: (i) Cyclopropane + HBr gives product [A] (ii) Cyclopropane + Br2 gives product [B] Identify [A] and [B] correctly.

Accepted Answer

[A] is 1-bromopropane and [B] is 1,3-dibromopropane. Cyclopropane undergoes ring-opening addition reactions. With HBr, the C-C bond breaks and HBr adds across it to give 1-bromopropane (all carbons of cyclopropane are equivalent, so only one mono-bromo product is possible). With Br2, ring opening gives 1,3-dibromopropane via a carbocation intermediate or concerted mechanism.

Question 49

In the reaction: CH3-CH2-S-CH2-CH2-Br + HOH (water) -> [X], the product [X] is:

Accepted Answer

An episulfonium ion intermediate that opens to give CH3-CH2-S-CH2-CH2-OH. The thioether sulfur (nucleophilic) attacks the carbon bearing Br intramolecularly to form a cyclic episulfonium ion (three-membered S-containing ring). Water then acts as a nucleophile, opening the ring to give CH3-CH2-S-CH2-CH2-OH (2-(ethylthio)ethanol) as product X. The original options A and B both show CH3-CH2-S-CH2-CH2-OH, which is the correct connectivity of the final product.

Question 50

2-Nitropropane reacts with formaldehyde in the presence of NaOH (Henry reaction). The major organic product obtained is:

Accepted Answer

2-Nitro-2-methylpropan-1-ol (CH3)2C(NO2)CH2OH. NaOH abstracts the acidic alpha-H from (CH3)2CHNO2 at the carbon bearing NO2, generating the resonance-stabilised nitronate anion (CH3)2C^(-)(NO2). This nucleophile attacks formaldehyde (HCHO) to give the beta-nitroalcohol (CH3)2C(NO2)CH2OH after protonation.

JEE Advanced Chemistry: Haloalkanes and Haloarenes questions with solutions

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