JEE Chemistry Haloalkanes and Haloarenes MCQs Set 16

Answer: (c) In acid the OH group is protonated to \( \ce{-OH2+} \), which leaves as water, a good leaving group

Explanation:
1. The hydroxide ion is a very poor leaving group, so it will not depart from the neutral alcohol in basic or neutral medium.
2. An acid protonates the oxygen to give \( \ce{R-OH2+} \); water is then lost easily to give a carbocation (E1).
3. So acid catalysis is needed.

Teacher's Note:
1. Leaving-group ability: \( \ce{H2O} \) is much better than \( \ce{OH-} \).
2. Always ask whether the group that must leave is a stable, weakly basic species.

Answer: (d) Pent-2-ene

Explanation:
1. The secondary cation at C-2 is formed.
2. Loss of H from C-3 gives pent-2-ene (disubstituted, more stable), while loss from C-1 gives pent-1-ene (monosubstituted).
3. Pent-2-ene (cis and trans, trans mainly) is the major product.

Teacher's Note:
1. Saytzeff product is the major alkene.
2. Internal alkenes are more stable than terminal ones.

Answer: (b) 2-Methylbut-2-ene

Explanation:
1. The unstable primary cation \( \ce{(CH3)3C-CH2+} \) rearranges at once by a methyl shift to the tertiary cation \( \ce{(CH3)2C+CH2CH3} \).
2. Loss of a proton from the CH2 group gives \( \ce{(CH3)2C=CHCH3} \), 2-methylbut-2-ene.

Teacher's Note:
1. Primary carbocations rearrange to more stable ones wherever possible.
2. 1,2-methyl shift is common in neopentyl systems.

Answer: (d) 3-Iodo-2-methylpentane

Explanation:
1. 3-Iodo-2-methylpentane, \( \ce{(CH3)2CH-CHI-CH2CH3} \), can lose H from C-2 giving \( \ce{(CH3)2C=CHCH2CH3} \) (2-methylpent-2-ene), or from C-4 giving \( \ce{(CH3)2CH-CH=CHCH3} \) (4-methylpent-2-ene).
2. 2-Iodo-2-methylpentane would give 2-methylpent-1-ene as well, and 2-iodo-4-methylpentane would give 4-methylpent-1-ene, so they do not fit.

Teacher's Note:
1. Write the double bond positions of each alkene and look for the common carbon bearing the halogen.
2. Both alkenes contain the C2-C3 / C3-C4 bonds around C-3.

Answer: (a) 2-Methylbut-2-ene and 2-methylbut-1-ene

Explanation:
1. Ethoxide is a small base and removes a hydrogen from the CH2 group of the ethyl chain, leading to the more substituted alkene (Saytzeff): 2-methylbut-2-ene, about 70 percent.
2. The bulky tert-butoxide removes the more accessible hydrogen of the methyl group, giving the less substituted alkene (Hofmann): 2-methylbut-1-ene, about 70 percent.

Teacher's Note:
1. Small base gives Saytzeff, bulky base gives Hofmann product.
2. The size of the base decides the regiochemistry.

Answer: (d) 1-Chloro-2-methylbutane

Explanation:
1. 1-Chloro-2-methylbutane, \( \ce{ClCH2-CH(CH3)-CH2CH3} \), has only one beta-carbon (C-2) which carries one H.
2. Elimination gives \( \ce{CH2=C(CH3)CH2CH3} \), 2-methylbut-1-ene, and nothing else.
3. The other chlorides can give more than one alkene, and the tertiary one gives mostly 2-methylbut-2-ene.

Teacher's Note:
1. Only one beta-carbon with H means only one alkene.
2. Draw the alkene and add HCl back across the double bond at the terminal carbon.

Answer: (b) An acidic beta-hydrogen and a relatively poor leaving group

Explanation:
1. In E1cB a base first removes an acidic beta-hydrogen, forming a carbanion (conjugate base) as intermediate.
2. The leaving group then departs in a slower second step; this is usual when it is a poor leaving group such as OH or F.
3. So an acidic beta-H and a poor leaving group are required.

Teacher's Note:
1. E1cB: carbanion first, leaving group later.
2. Acidity is usually produced by a C=O, NO2 or similar electron-withdrawing group.

Answer: (c) \( \ce{CH3CH(OH)CH3 ->[conc. H2SO4][heat] CH3CH=CH2} \)

Explanation:
1. Elimination removes two atoms or groups from adjacent carbons and forms a multiple bond.
2. Propan-2-ol loses water to give propene, an elimination.
3. The others are addition (Grignard), substitution (PBr3) and acid-base reactions.

Teacher's Note:
1. Look for the formation of a C=C from a saturated compound.
2. Dehydration and dehydrohalogenation are the standard examples.

Answer: (c) It is a two-step process, rearrangement of the intermediate is possible, and a good leaving group favours it

Explanation:
1. E1 involves slow ionisation to a carbocation and then loss of a proton, so it has two steps.
2. The carbocation can rearrange, and a good leaving group speeds up the ionisation.
3. The single-step, carbanion and stereospecific descriptions belong to E2 or E1cB.

Teacher's Note:
1. E1: carbocation, rearrangement possible, no strict stereospecificity.
2. E2: concerted and stereospecific (anti).

Answer: (d) 1,2-Dimethylcyclohexene

Explanation:
1. Loss of water would give a primary cation, which is unfavourable, so a hydride shift from ring carbon C-1 occurs at the same time, giving the tertiary ring cation carrying a CH3 at C-1 (next to the C-2 methyl).
2. Loss of the proton from C-2 gives the tetrasubstituted alkene, 1,2-dimethylcyclohexene, the most stable one.

Teacher's Note:
1. Primary cation to tertiary cation by 1,2-hydride shift.
2. Tetrasubstituted alkenes are the most stable.

Answer: (b) 1-Bromo-2,2-dimethylpropane (neopentyl bromide)

Explanation:
1. E2 needs a hydrogen on the carbon adjacent to the one bearing the halogen.
2. In neopentyl bromide \( \ce{(CH3)3C-CH2Br} \) the beta-carbon is quaternary and has no hydrogen.
3. The other three each have a beta-hydrogen.

Teacher's Note:
1. Check every beta-carbon for hydrogen.
2. No beta-H, no beta-elimination.

Answer: (d) 3-Methylbut-1-ene

Explanation:
1. The bulky base removes the more accessible hydrogens from the CH3 group (C-1) rather than the hindered tertiary H on C-3.
2. This gives the Hofmann product, \( \ce{(CH3)2CH-CH=CH2} \), 3-methylbut-1-ene.

Teacher's Note:
1. Bulky base: less hindered beta-hydrogen is removed.
2. Strong base and heat: E2, not SN2.

Answer: (a) X = 2,3-dimethylbut-2-ene, Y = 2,3-dimethylbut-1-ene

Explanation:
1. Methoxide is a small base, so the Saytzeff product, the tetrasubstituted 2,3-dimethylbut-2-ene, is major.
2. The bulky tert-pentoxide removes a hydrogen from a methyl group, giving the Hofmann product \( \ce{CH2=C(CH3)CH(CH3)2} \), 2,3-dimethylbut-1-ene.

Teacher's Note:
1. Small base: Saytzeff. Bulky base: Hofmann.
2. Tetrasubstituted alkenes are the most stable.

Answer: (d) Pent-3-en-2-one

Explanation:
1. The hydrogens on the carbon between C=O and the C-F carbon are acidic.
2. KOH removes one, giving an enolate (carbanion), and then fluoride, a poor leaving group, leaves: E1cB.
3. The conjugated enone \( \ce{CH3COCH=CHCH3} \) is formed.

Teacher's Note:
1. Beta-halo ketones eliminate through E1cB.
2. The conjugated product is the stable one.

Answer: (b) \( \ce{O2N-CH=CH-CH3} \)

Explanation:
1. The hydrogen on the carbon bearing the nitro group is the most acidic; the carbanion is stabilised by the nitro group.
2. Hydroxide removes it (E1cB) and chloride leaves, giving the conjugated nitroalkene \( \ce{O2N-CH=CH-CH3} \).

Teacher's Note:
1. Electron-withdrawing groups such as NO2 make the alpha-hydrogen acidic.
2. The carbanion intermediate decides the regiochemistry.

Answer: (c) IV > III > I > II

Explanation:
1. The rate depends on the stability of the carbocation formed on losing water.
2. IV gives a tertiary doubly benzylic cation (most stable); III gives a tertiary benzylic cation; I gives a secondary cation that can shift H to a benzylic tertiary one; II would give a primary cation (least stable).
3. Order: IV > III > I > II.

Teacher's Note:
1. Rate of E1 follows carbocation stability.
2. Benzylic, tertiary cations are very stable.

Answer: (c) 2-Iodo-2-methylbutane

Explanation:
1. E1 rate depends on carbocation stability and leaving-group ability.
2. Tertiary substrates ionise most easily, and iodide is a better leaving group than bromide.
3. So 2-iodo-2-methylbutane is the most reactive.

Teacher's Note:
1. Tertiary substrate and best leaving group win in E1.
2. Leaving ability: I > Br > Cl > F.

Answer: (d) SN2

Explanation:
1. SN1, E1 and E2 are all fastest for tertiary halides (stable cation, or a more substituted alkene).
2. In SN2 the rate falls as steric hindrance rises, so the order is 1 degree > 2 degree > 3 degree, the reverse.
3. Therefore SN2 is the odd one out.

Teacher's Note:
1. SN2 is the only one that prefers unhindered substrates.
2. Crowding in the transition state slows SN2.

Answer: (b) 23

Explanation:
1. Dehydration of butan-2-ol gives but-1-ene, cis-but-2-ene and trans-but-2-ene, so Y = 3.
2. Butan-2-ol with Lucas reagent (HCl/\( \ce{ZnCl2} \)) gives 2-chlorobutane, which has a stereocentre and is formed as two enantiomers, so Z = 2.
3. ZY = 23.

Teacher's Note:
1. Butan-2-ol is the smallest optically active alcohol.
2. Count enantiomers and cis/trans isomers separately.

Answer: (d) 5

Explanation:
1. The secondary cation at C-2 can lose H from C-1 to give 3-phenylbut-1-ene (one stereocentre, 2 enantiomers) and from C-3 to give 2-phenylbut-2-ene (cis and trans, 2).
2. A hydride shift gives the tertiary benzylic cation at C-3, which gives 2-phenylbut-2-ene and 2-phenylbut-1-ene (1).
3. Total = 2 + 2 + 1 = 5.

Teacher's Note:
1. Check for rearrangement to a benzylic cation.
2. Count enantiomers of the chiral alkene.

Answer: (a) Relative to a small base, a bulky base gives more of the less substituted alkene

Explanation:
1. With a small base, the more substituted alkene (Saytzeff product) predominates because it is more stable.
2. With a bulky base, the more accessible hydrogens on the less substituted carbon are removed, increasing the Hofmann product.
3. E2 reactivity is 3 degree > 2 degree > 1 degree.

Teacher's Note:
1. Bulky base, Hofmann product.
2. E2 rate parallels alkene stability.

Answer: (c) The product is the more stable, more substituted alkene (Saytzeff rule)

Explanation:
1. Alcoholic KOH gives E2 elimination.
2. The transition state resembles the alkene, so the more stable disubstituted pent-2-ene forms faster than the monosubstituted pent-1-ene.
3. This is the Saytzeff rule.

Teacher's Note:
1. Saytzeff: more substituted alkene predominates with small bases.
2. Pent-2-ene is disubstituted, pent-1-ene monosubstituted.