JEE Chemistry Haloalkanes and Haloarenes MCQs Set 15

Answer: (c) 2-Phenylpropene

Explanation:
1. Grignard addition to the C=O of propanone gives the alkoxide, and hydrolysis gives 2-phenylpropan-2-ol (a tertiary alcohol).
2. Heating with \( \ce{H3PO4} \) dehydrates the tertiary alcohol by the E1 route; the only available beta-hydrogens are on the methyl groups.
3. The product is \( \ce{PhC(CH3)=CH2} \), 2-phenylpropene.

Teacher's Note:
1. Grignard reagents convert ketones into tertiary alcohols.
2. Tertiary alcohols dehydrate easily with acid and heat.

Answer: (d) 1-Methylcyclohexene

Explanation:
1. Acid protonates the OH group and water leaves, giving a secondary carbocation at C-1 (E1).
2. Loss of a proton from C-2 (the carbon carrying the methyl group) gives the more substituted, trisubstituted alkene (Saytzeff rule), whereas loss from C-6 gives the disubstituted 3-methylcyclohexene.
3. The major product is 1-methylcyclohexene.

Teacher's Note:
1. Alcohol dehydration follows Saytzeff's rule: the more substituted alkene is the major product.
2. Compare the number of alkyl groups on the double bond.

Answer: (b) Alcohols dehydrate under strongly basic conditions by the E1 mechanism

Explanation:
1. E2 needs an anti-periplanar H and leaving group, so it is stereospecific; the SN2 reaction proceeds with inversion and gives one isomer; 3-hydroxypropanal has an acidic alpha-H, so E1cB operates in base.
2. E1 dehydration of an alcohol needs the OH to be protonated to a good leaving group (water), which requires acidic conditions, not basic ones.
3. Hence the statement about E1 dehydration in strongly basic conditions is incorrect.

Teacher's Note:
1. Alcohol dehydration is acid catalysed (E1 for 2 degree and 3 degree alcohols).
2. OH is a poor leaving group unless protonated.

Answer: (a) trans-Pent-2-ene

Explanation:
1. A strong base in ethanol and heat favours E2 elimination over substitution.
2. The Saytzeff product, the more substituted pent-2-ene, is the major product.
3. The trans isomer is more stable than cis, so trans-pent-2-ene is major.

Teacher's Note:
1. Small strong bases follow Saytzeff's rule.
2. trans-Alkenes are more stable than cis-alkenes.

Answer: (d) N,N-Dimethylpent-4-en-1-amine

Explanation:
1. Exhaustive methylation gives 1,1,2-trimethylpyrrolidinium iodide, which becomes the hydroxide with \( \ce{Ag2O} \).
2. Hofmann elimination removes the most accessible (least hindered, most acidic) beta-hydrogen: those of the methyl group on C-2.
3. The C2-N bond breaks, opening the ring and forming \( \ce{CH2=CH-CH2CH2CH2-N(CH3)2} \), N,N-dimethylpent-4-en-1-amine.

Teacher's Note:
1. Hofmann elimination gives the least substituted alkene.
2. Ring-containing quaternary ammonium hydroxides open the ring on elimination.

Answer: (a) Methylenecyclohexane (exocyclic alkene, Hofmann product)

Explanation:
1. A bulky base removes the most accessible hydrogens, those of the methyl group, rather than the more hindered ring hydrogens.
2. This gives the less substituted exocyclic alkene (Hofmann product), methylenecyclohexane.

Teacher's Note:
1. Bulky bases such as t-BuOK favour the Hofmann (less substituted) alkene.
2. Small bases such as ethoxide favour the Saytzeff alkene.

Answer: (c) \( \ce{PhSO2CH=CH2} \)

Explanation:
1. Deuterium appears in the recovered reactant at the carbon next to \( \ce{SO2Ph} \), showing that this hydrogen is removed reversibly to give a carbanion (E1cB).
2. The carbanion expels the poor leaving group \( \ce{-OCH3} \), forming \( \ce{PhSO2CH=CH2} \), phenyl vinyl sulfone.

Teacher's Note:
1. The sulfonyl group acidifies the adjacent C-H.
2. H/D exchange in recovered starting material is evidence for an E1cB carbanion.

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

Explanation:
1. Only a chloride with a 2-methylbutane skeleton can be the answer: 1-chloropentane and 1-chloro-2,2-dimethylpropane do not have this skeleton (1-chloropentane would give pentane; the neopentyl chloride has no beta-hydrogen).
2. 2-Chloro-2-methylbutane gives two alkenes (2-methylbut-2-ene and 2-methylbut-1-ene).
3. 1-Chloro-2-methylbutane has one beta-carbon (C-2) with one H, and gives only 2-methylbut-1-ene, which hydrogenates to 2-methylbutane.

Teacher's Note:
1. Write the alkene after elimination and check that only one is possible.
2. A halide on a carbon next to a quaternary carbon cannot undergo beta-elimination.

Answer: (b) Cyclopentanecarbaldehyde

Explanation:
1. The tertiary OH is protonated and lost, giving a tertiary ring carbocation (the more stable of the two possible).
2. A hydride shifts from the neighbouring \( \ce{CH2OH} \) carbon to the cationic ring carbon, giving a protonated aldehyde (pinacol-type shift).
3. Loss of the proton gives cyclopentanecarbaldehyde.

Teacher's Note:
1. In a 1,2-diol the cation forms at the carbon giving the more stable cation, and a group migrates to it.
2. Hydride shift from a carbinol carbon gives a carbonyl compound.

Answer: (a) P > Q > S > R

Explanation:
1. SN2 is hindered by branching at the carbon bearing the halogen, while E2 is favoured by more beta-hydrogens and more substituted alkenes.
2. The tertiary halide P gives only E2 with a base, then secondary Q, then S (primary), and R has no beta-hydrogen so no E2 at all.
3. Order: P > Q > S > R.

Teacher's Note:
1. Branching raises E2/SN2, since SN2 is blocked.
2. Methyl halides have no beta-H and can only undergo substitution.

Answer: (d) Alcoholic KOH; then HBr with peroxide; then KI in acetone

Explanation:
1. Alcoholic KOH converts 2-bromopropane into propene.
2. HBr with peroxide adds anti-Markovnikov to give 1-bromopropane.
3. The Finkelstein reaction (KI in acetone) converts 1-bromopropane into 1-iodopropane by SN2.

Teacher's Note:
1. Move a halogen from C-2 to C-1: eliminate, add anti-Markovnikov, then exchange.
2. HI or HCl adds Markovnikov and would give the 2-halide.

Answer: (a) 2-Bromo-2-methylbutane

Explanation:
1. E2 is fastest for tertiary halides because of the more stable, more substituted alkene being formed and the relief of strain in the transition state.
2. 1-Bromo-2,2-dimethylpropane has no beta-hydrogen, 2-bromopentane and 2-bromo-3-methylbutane are secondary.
3. So 2-bromo-2-methylbutane, a tertiary halide, reacts most readily.

Teacher's Note:
1. Tertiary > secondary > primary for E2.
2. The neopentyl type cannot eliminate.

Answer: (c) Propene

Explanation:
1. Alcoholic KOH gives the strongly basic alkoxide ion, which promotes dehydrohalogenation (E2).
2. Loss of HBr from \( \ce{CH3CH2CH2Br} \) gives propene.
3. Aqueous KOH, not alcoholic, would give mainly propan-1-ol.

Teacher's Note:
1. Alcoholic KOH: elimination; aqueous KOH: substitution.
2. One molecule of HBr lost gives an alkene.

Answer: (d) trans-Pent-2-ene

Explanation:
1. Dehydration proceeds through a secondary carbocation (E1).
2. Loss of a proton from C-3 gives pent-2-ene (Saytzeff product), preferred to pent-1-ene.
3. trans-Pent-2-ene is more stable than the cis isomer and so predominates.

Teacher's Note:
1. Saytzeff: the more substituted alkene is major.
2. Among geometrical isomers trans is usually major.

Answer: (b) 3-Methylpent-1-ene

Explanation:
1. Oxidative ozonolysis giving acetic acid and butanone means Z = \( \ce{CH3CH=C(CH3)CH2CH3} \), 3-methylpent-2-ene.
2. 3-Methylpent-1-ene with HI gives (Markovnikov) 2-iodo-3-methylpentane, which with ethanolic KOH eliminates towards the tertiary C-3 hydrogen (Saytzeff) to give 3-methylpent-2-ene.
3. The other options cannot lead to Z: 2-methylpent-2-ene would give acetone and propanoic acid by ozonolysis.

Teacher's Note:
1. Work backwards from the ozonolysis products to the alkene.
2. HI adds to a terminal alkene to give the 2-iodo compound.

Answer: (a) E1cB

Explanation:
1. The carbanion is already formed before the leaving group leaves.
2. Loss of LG from the carbanion, with the lone pair forming the new pi bond, is the second step of the E1cB mechanism.

Teacher's Note:
1. E1cB: conjugate base is formed first, then LG leaves.
2. A nitro group stabilises the carbanion.

Answer: (d) (Z)-Tetradec-9-en-1-ol

Explanation:
1. Excess \( \ce{NaNH2} \) forms the alkoxide and the acetylide; the acetylide (pKa about 25) is the more reactive nucleophile in the alkylation step because the alkoxide (pKa about 17) is the weaker base.
2. The acetylide displaces bromide from n-butyl bromide, giving \( \ce{HOCH2(CH2)7C#C-C4H9} \).
3. Lindlar catalyst reduces the alkyne to the cis alkene: (Z)-tetradec-9-en-1-ol.

Teacher's Note:
1. Lindlar catalyst gives cis-alkenes; sodium in liquid ammonia gives trans.
2. The conjugate base of the weaker acid reacts as the nucleophile.

Answer: (a) \( \ce{CHD2CH2OH} \) and \( \ce{CD2=CH2} \)

Explanation:
1. \( \ce{CHD2MgI} \) adds to \( \ce{CH2=O} \), giving after hydrolysis the primary alcohol \( \ce{CHD2-CH2-OH} \) (X).
2. Dehydration removes OH from C-1 and a beta-hydrogen from the \( \ce{CHD2} \) carbon. Because C-H breaks faster than C-D (isotope effect), H is lost in preference to D, giving \( \ce{CD2=CH2} \) (Y).

Teacher's Note:
1. Grignard + HCHO gives a primary alcohol with one carbon more.
2. The kinetic isotope effect: C-H cleaves faster than C-D.

Answer: (c) A and B only

Explanation:
1. Bridgehead halides cannot form planar carbocations (SN1) and the back side is shielded (SN2); (A) is true.
2. Both SN1 and E1 begin with ionisation of the C-X bond to a carbocation; (B) is true.
3. SN2 occurs with inversion, not retention (C false); E2 is favoured by strong bases (D false).

Teacher's Note:
1. SN1 and E1 share the carbocation step.
2. SN2 gives inversion (Walden inversion).

Answer: (d) 33

Explanation:
1. The secondary carbocation is formed first and can be trapped by ethanol, giving the unrearranged ether as a pair of enantiomers (2 products).
2. A methyl shift gives the tertiary cation, trapped by ethanol to give one achiral ether. So X = 3.
3. Alkenes: 3,3-dimethylbut-1-ene (from the secondary cation), 2,3-dimethylbut-2-ene and 2,3-dimethylbut-1-ene (from the tertiary cation): Y = 3.
4. XY = 33.

Teacher's Note:
1. Count enantiomers separately when a stereocentre is formed.
2. Carbocations rearrange before the final step.

Answer: (b) X = nitro

Explanation:
1. E1cB needs the formation of a carbanion at the benzylic carbon, which is stabilised by electron-withdrawing groups.
2. The nitro group has the strongest -I and -M effects and so makes the benzylic hydrogen most acidic.
3. Hence the nitro-substituted sulfonium ion reacts fastest.

Teacher's Note:
1. Electron-withdrawing substituents speed up E1cB.
2. Electron donors (methyl, methoxy) slow it.