JEE Chemistry Haloalkanes and Haloarenes MCQs Set 21

Answer: (a) the cyclic ether formed by intramolecular displacement of fluorine (iodine retained)

Explanation:
1. The nitro group activates the positions ortho and para to it: iodine is ortho and fluorine is para to the nitro group.
2. In SNAr the rate-determining step is the attack on the ring carbon, which is fastest for fluorine (most electronegative), so F is displaced much faster than I.
3. The alkoxide, formed from the side-chain OH, is held close to the C-F carbon and closes a six-membered ring (a chromane); iodine remains.

Teacher's Note:
1. Leaving-group order in SNAr: F >> Cl > Br > I.

Answer: (d) (1S,3S)-trans-3-methylcyclohexanol

Explanation:
1. SN2 inverts the configuration at the carbon bearing Br, so the cis bromide gives the trans alcohol.
2. The priorities at C-1 stay in the same order (OH replaces Br as the top priority), so inversion changes the descriptor R to S at C-1; C-3 is unchanged (S). The product is (1S,3S).

Teacher's Note:
1. SN2 inverts the stereocentre: cis becomes trans in a ring.

Answer: (b) R, S

Explanation:
1. The benzylic carbon is not attacked (the S- nucleophile attacks the other carbon), so it stays R (S still has the highest priority).
2. The butyl carbon is attacked by SN2 with inversion; with sulphur replacing bromine (both highest priority), the descriptor changes from R to S.
3. Result: (1) R, (2) S.

Teacher's Note:
1. Only the carbon that bears the leaving group is inverted.

Answer: (c) r1 > r3 > r2

Explanation:
1. Tertiary halides ionise much faster than secondary ones, so (1) and (3) beat (2).
2. Between the two tertiary halides, bromide is a better leaving group than chloride, so r1 > r3. Order: r1 > r3 > r2.

Teacher's Note:
1. SN1 rate: tertiary > secondary; Br > Cl as leaving group.

Answer: (c) 1-methylcyclohexan-1-ol

Explanation:
1. \( \ce{Ag+} \) removes chloride and the primary cation formed is unstable, so a hydride from the ring CH shifts to the CH2 carbon as the chloride leaves.
2. This gives the tertiary 1-methylcyclohexyl cation, which is captured by water to give 1-methylcyclohexan-1-ol. Ring expansion would give only a secondary cycloheptyl cation, which is less stable.

Teacher's Note:
1. Primary cation rearranges by 1,2-hydride shift to the tertiary cation.

Answer: (d) 2-chlorobutane, SN1

Explanation:
1. OH is replaced by Cl at the same carbon, so the product is 2-chlorobutane.
2. Under the strongly acidic ionising conditions of Lucas reagent a secondary alcohol reacts through the SN1 mechanism (carbocation).

Teacher's Note:
1. Lucas reagent: SN1 for secondary and tertiary alcohols.

Answer: (b) 1-(2,4-dinitrophenyl)piperidine

Explanation:
1. The ring carbon carrying Cl is activated by ortho and para nitro groups; piperidine (a good N nucleophile) adds there (SNAr) and chloride is lost.
2. The product is the tertiary aromatic amine, with the chlorine replaced, in the same way as in Sanger's reagent chemistry.

Teacher's Note:
1. 2,4-Dinitrohalobenzenes are readily attacked by amines.

Answer: (d) sodium cyclohexoxide + \( \ce{CH3I} \)

Explanation:
1. Williamson synthesis works best with a methyl (or primary) halide and the more hindered alkoxide: the alkoxide attacks \( \ce{CH3I} \) in SN2.
2. Iodocyclohexane is secondary and would give mostly cyclohexene with the strongly basic methoxide; with methanol alone it is too slow, and cyclohexanol itself is too weak a nucleophile.

Teacher's Note:
1. Use the less hindered halide in the Williamson synthesis.

Answer: (a) 1-methoxy-1-phenylpropane

Explanation:
1. Triflate is an exceptionally good leaving group and the substrate is a secondary benzylic compound: ionisation gives the resonance-stabilised benzylic cation (SN1, solvolysis).
2. Methanol captures the cation at the benzylic carbon, giving the methyl ether (racemic).

Teacher's Note:
1. Solvolysis of benzylic triflates gives substitution through the benzylic cation.

Answer: (d) 1-ethoxypropan-2-ol

Explanation:
1. A strong nucleophile opens the epoxide by SN2 at the less hindered carbon (CH2), so the ethoxide attacks the CH2 carbon.
2. The alkoxide formed on the other carbon is protonated by ethanol: \( \ce{CH3CH(OH)CH2OC2H5} \).

Teacher's Note:
1. Basic opening of epoxides: attack at the less substituted carbon.

Answer: (b) The strength of a nucleophile generally decreases on going down a group of the periodic table (in protic solvents)

Explanation:
1. Nucleophiles have lone pairs (or pi electrons) that form the new bond; both cyanide (C or N) and nitrite (O or N) are ambident.
2. In protic solvents nucleophilicity increases down a group (I- > Br- > Cl- > F-; RS- > RO-) because the larger, more polarisable ions are less strongly solvated, so the statement is false.

Teacher's Note:
1. Down a group: nucleophilicity increases in protic solvents.

Answer: (c) p-nitroiodobenzene

Explanation:
1. In SNAr the rate-determining step is the addition of the nucleophile to the ring carbon bearing the halogen, which is promoted by the electronegativity of the halogen.
2. The reactivity order is F > Cl is about Br > I, so the iodo compound is the least reactive.

Teacher's Note:
1. SNAr leaving group order: F > Cl is about Br > I.

Answer: (c) trans-1,2-dichloro-1-hexene

Explanation:
1. Acetylide displaces bromide from 1-bromobutane (SN2): \( \ce{HC#C-CH2CH2CH2CH3} \), 1-hexyne (A).
2. Chlorine adds to the triple bond by anti addition (via a chloronium ion), giving the trans-dihaloalkene 1,2-dichloro-1-hexene (B).

Teacher's Note:
1. Halogen addition to an alkyne is anti: trans-dihaloalkene.

Answer: (d) 2-butanol

Explanation:
1. The Grignard reagent is a strong nucleophile and a base: it opens the epoxide by SN2 attack on the less hindered carbon (the CH2).
2. This gives \( \ce{CH3CH(O-)CH2CH3} \), and protonation gives butan-2-ol.

Teacher's Note:
1. Grignard reagents attack the less substituted epoxide carbon.

Answer: (b) sodium cyanide

Explanation:
1. Cyanide is a good nucleophile but only a weak base, so it substitutes the secondary bromide.
2. Ethoxide is a strong base and gives mainly elimination (propene). Sulphate is a very poor nucleophile and chloride exchange is not a preparatively useful substitution.

Teacher's Note:
1. Strong nucleophile and weak base favours substitution on secondary halides.

Answer: (d) \( \ce{CH3CH2CH2NHCOCH3} \)

Explanation:
1. Cyanide displaces chloride: \( \ce{CH3CH2CN} \) (propanenitrile).
2. Catalytic hydrogenation reduces the nitrile to the primary amine \( \ce{CH3CH2CH2NH2} \).
3. Acetic anhydride acetylates the amine on nitrogen: N-propylacetamide, \( \ce{CH3CH2CH2NHCOCH3} \).

Teacher's Note:
1. R-CN reduces to R-CH2-NH2; amines are acetylated by anhydrides.

Answer: (a) an alkyl nitrate

Explanation:
1. Silver nitrate supplies the nitrate ion, \( \ce{NO3-} \), whose oxygen attacks the carbocation-like carbon while AgX precipitates, giving \( \ce{R-ONO2} \), an alkyl nitrate.
2. (Nitroalkanes arise from AgNO2 and alkyl nitrites from NaNO2.)

Teacher's Note:
1. AgNO3 gives R-ONO2; AgNO2 gives R-NO2.

Answer: (d) II > III > I

Explanation:
1. SN1 reactivity depends on the ease of forming the carbocation. The bridgehead chloride cannot give a planar carbocation in the rigid bicyclic skeleton, so it is the least reactive.
2. Between the two tertiary chlorides, the tert-amyl chloride (II) gives the more substituted, more stable cation (more hyperconjugation and inductive stabilisation) than tert-butyl chloride (III). Order: II > III > I.

Teacher's Note:
1. More alkyl groups stabilise the cation; bridgehead cations cannot become planar.

Answer: (b) iii > ii > i > iv

Explanation:
1. The rate is controlled by cation stability. (iii) gives a tertiary diphenylmethyl cation (two phenyl rings plus a methyl), the most stable.
2. (ii) is a secondary diphenylmethyl cation stabilised by two rings, more stable than the tert-butyl cation (i); (iv) gives a secondary alkyl cation and is the slowest.

Teacher's Note:
1. Each phenyl group stabilises a carbocation more than an alkyl group does.

Answer: (c) biphenyl

Explanation:
1. Sodium in dry ether couples two aryl halide molecules (Wurtz-Fittig/Fittig reaction): \( \ce{2 ArBr + 2 Na -> Ar-Ar + 2 NaBr} \).
2. Two phenyl groups join to give biphenyl.

Teacher's Note:
1. Fittig reaction: aryl halide + Na gives biaryl.

Answer: (c) 1-chloro-2-butene

Explanation:
1. 1-Chloro-2-butene, \( \ce{ClCH2CH=CHCH3} \), is an allylic primary chloride: the adjacent pi bond stabilises the SN2 transition state.
2. 1-Chloro-1-butene and 2-chloro-1-butene are vinylic (inert), and 4-chloro-1-butene is only an ordinary primary chloride.

Teacher's Note:
1. Allylic halides are the most reactive primary halides in SN2.

Answer: (d) 1-bromobutane

Explanation:
1. Unhindered primary halides react fastest in SN2.
2. The secondary bromide is slower because of steric hindrance, the highly branched halide is much slower, and the vinylic chloride (sp2 carbon) does not react by SN2 at all.

Teacher's Note:
1. SN2: primary >> secondary >> tertiary; vinylic inert.