JEE Chemistry Hydrocarbons MCQs Set 06

Answer: (b) \(\ce{Na/NH3}\) (liquid)

Explanation:
1. Sodium in liquid ammonia reduces an alkyne by anti addition of hydrogen through a radical anion and vinyl radical.
2. This places the two alkyl groups on opposite sides, giving the trans-alkene.
3. Pd-C, Pd and Ni with hydrogen would add hydrogen on the same face and usually go on to the alkane.

Teacher's Note:
1. Dissolving metal reduction gives trans alkenes.
2. Lindlar catalyst gives cis alkenes.

Answer: (b) hydroboration-oxidation, oxymercuration-demercuration, acid-catalysed hydration (with rearrangement)

Explanation:
1. 2-Cyclohexylethanol has the OH on the terminal carbon, which is the anti-Markovnikov product of hydroboration-oxidation (A).
2. 1-Cyclohexylethanol is the Markovnikov alcohol without rearrangement, given by oxymercuration-demercuration (B).
3. In acid-catalysed hydration the secondary carbocation shifts a hydride from the ring carbon to give a tertiary carbocation, which gives 1-ethylcyclohexanol (C).

Teacher's Note:
1. Oxymercuration and hydroboration never rearrange.
2. Acid-catalysed hydration goes through a carbocation, so rearrangement is possible.

Answer: (c) \(\ce{Hg(OCOCH3)2.H2O/NaBH4}\)

Explanation:
1. The product is the Markovnikov alcohol with no rearrangement.
2. Oxymercuration-demercuration gives Markovnikov addition through a mercurinium ion, without any carbocation shift.
3. Acid-catalysed hydration would shift a hydride to a tertiary benzylic cation, giving a different alcohol, and hydroboration gives the primary alcohol.

Teacher's Note:
1. To get a Markovnikov alcohol from a branched alkene without rearrangement, use oxymercuration-demercuration.
2. The benzylic position makes rearrangement very likely with acids.

Answer: (a) 1-bromo-1-methyl-1,2,3,4-tetrahydronaphthalene

Explanation:
1. NBS brominates by a free radical mechanism at the position that gives the most stable radical.
2. The tertiary benzylic hydrogen on the carbon carrying the methyl group gives the most stable (tertiary and benzylic) radical.
3. So bromine enters that carbon.

Teacher's Note:
1. NBS prefers allylic and benzylic C - H bonds.
2. Tertiary benzylic is the most reactive site.

Answer: (b) benzene and \(\ce{Mg(OMe)Br}\)

Explanation:
1. A Grignard reagent behaves as a strong base (carbanion source) and takes a proton from any acidic hydrogen.
2. Methanol supplies the hydrogen of its O - H group, so \(\ce{C6H5MgBr + CH3OH -> C6H6 + Mg(OCH3)Br}\).

Teacher's Note:
1. Grignard reagents are destroyed by water, alcohols and acids.
2. The R group of RMgX becomes R - H.

Answer: (b) propanedial

Explanation:
1. Birch reduction of benzene gives cyclohexa-1,4-diene.
2. Reductive ozonolysis cleaves both double bonds.
3. Each C = C cleavage gives two CHO groups, and each CH2 is flanked by two former alkene carbons, so the only product (two molecules) is \(\ce{OHC-CH2-CHO}\), propanedial.

Teacher's Note:
1. Birch reduction of benzene gives the 1,4-diene.
2. Trace each carbon after cleavage to name the product.

Answer: (c) benzoic acid

Explanation:
1. Alkaline or acidic \(\ce{KMnO4}\) oxidises any alkyl side chain that has a benzylic hydrogen completely to a carboxylic group.
2. The side chain \(\ce{-CH2CH3}\) is cleaved and the ring carbon ends up as \(\ce{-COOH}\).
3. So the product is benzoic acid.

Teacher's Note:
1. The length of the side chain does not matter, only a benzylic H is needed.
2. The product is always \(\ce{C6H5COOH}\).

Answer: (b) trans-1-phenyl-3-methylbut-1-ene, \(\ce{C6H5-CH=CH-CH(CH3)2}\)

Explanation:
1. Dehydration follows Saytzeff's rule and gives the most stable alkene.
2. Loss of the benzylic hydrogen gives a double bond conjugated with the benzene ring.
3. The trans isomer is more stable than the cis isomer because of less steric repulsion, so the trans conjugated alkene is the major product.

Teacher's Note:
1. Conjugation with the benzene ring makes the styrene-type alkene most stable.
2. Trans alkenes are usually more stable than cis.

Answer: (c) cis-but-2-ene

Explanation:
1. Double dehydrohalogenation of the vicinal dichloride gives but-2-yne (A) as the major product.
2. Partial hydrogenation over Lindlar catalyst (Pd-\(\ce{CaCO3}\)) adds hydrogen on the same face.
3. So B is cis-but-2-ene.

Teacher's Note:
1. Lindlar catalyst is poisoned palladium and gives cis alkenes.
2. Vicinal dihalides give alkynes with a strong base.

Answer: (d) all of these

Explanation:
1. Converting \(\ce{C=O}\) to \(\ce{CH2}\) is a reduction of the carbonyl group.
2. HI with red phosphorus, Clemmensen reduction (Zn-Hg/HCl) and Wolff-Kishner reduction (\(\ce{NH2NH2}\)/KOH) all achieve this for aryl ketones.
3. So all three work.

Teacher's Note:
1. Friedel-Crafts acylation followed by reduction is the way to make straight-chain alkylbenzenes.
2. This avoids the rearrangement that happens in direct alkylation.

Answer: (a) benzene

Explanation:
1. Three molecules of ethyne polymerise (cyclic trimerisation) when passed through a red-hot iron tube at about 600 to 700 C.
2. The product is benzene, \(\ce{3C2H2 -> C6H6}\).

Teacher's Note:
1. Tetramerisation to cyclooctatetraene needs a nickel cyanide catalyst.
2. Remember red hot tube means benzene.

Answer: (c) 2,3-dimethylbut-2-ene

Explanation:
1. Stability of alkenes increases with the number of alkyl groups on the double bond carbons (hyperconjugation).
2. 2,3-Dimethylbut-2-ene is tetrasubstituted and has 12 hyperconjugative C - H bonds.
3. Ethene is unsubstituted, 2-methylpropene is disubstituted, and but-2-ene is also disubstituted.

Teacher's Note:
1. More substituted alkene means more stable.
2. Tetrasubstituted is the most stable.

Answer: (c) \(\ce{CH2=CH-CH2-Br}\)

Explanation:
1. The Wurtz reaction needs an alkyl halide where the C - X bond is reactive.
2. Vinyl, aryl and alkynyl halides have the halogen on an sp2 or sp carbon and are unreactive toward sodium in dry ether (Wurtz).
3. Allyl bromide has the halogen on an sp3 carbon and reacts to give hexa-1,5-diene.

Teacher's Note:
1. Vinyl and aryl halides do not take part in simple Wurtz coupling.
2. The Wurtz-Fittig reaction couples an aryl halide with an alkyl halide.

Answer: (d) but-2-ene

Explanation:
1. The peroxide effect shows up only when the two double bond carbons are different, so that two different products are possible.
2. But-2-ene is symmetrical, so Markovnikov and anti-Markovnikov addition give the same 2-bromobutane.
3. The other three are unsymmetrical.

Teacher's Note:
1. Anti-Markovnikov addition is only meaningful for unsymmetrical alkenes.
2. Peroxide effect works only with HBr.

Answer: (d) ammoniacal \(\ce{AgNO3}\)

Explanation:
1. Propyne is a terminal alkyne and gives a white precipitate of silver propynide with ammoniacal silver nitrate.
2. Propene gives no precipitate.
3. Both decolourise bromine and Baeyer's reagent, so only (d) distinguishes them.

Teacher's Note:
1. Terminal alkynes give acetylide precipitates.
2. Bromine and \(\ce{KMnO4}\) tests cannot separate an alkene from an alkyne.

Answer: (c) \(\ce{Cl2CHCHO}\)

Explanation:
1. HOCl adds to the triple bond to give \(\ce{Cl-CH=CH-OH}\), which tautomerises to chloroethanal.
2. A second molecule of HOCl adds and gives dichloroacetaldehyde after tautomerism, with loss of water.
3. The net product is \(\ce{Cl2CH-CHO}\).

Teacher's Note:
1. Think addition of HOCl then tautomerism.
2. Excess HOCl gives the dichloro compound.

Answer: (c) 1-chloro-4-(1-methylcyclohexyl)benzene

Explanation:
1. Protonation of 1-methylcyclohexene gives the tertiary 1-methylcyclohexyl carbocation.
2. This carbocation is the electrophile for Friedel-Crafts alkylation of chlorobenzene.
3. Chlorine is ortho/para directing, and the bulky tertiary electrophile attacks mainly at the para position.

Teacher's Note:
1. Alkenes with acid give carbocations that alkylate aromatic rings.
2. Chlorine deactivates the ring but directs ortho and para.

Answer: (c) 2-chloro-2,3-dimethylpentane

Explanation:
1. The proton adds to the carbon farther from the tert-butyl group to give a secondary carbocation next to the quaternary carbon.
2. A methyl group shifts to this cationic carbon, giving the more stable tertiary carbocation \(\ce{(CH3)2C+-CH(CH3)-CH2CH3}\).
3. Chloride adds to it to give 2-chloro-2,3-dimethylpentane.

Teacher's Note:
1. Carbocations rearrange by hydride or methyl shifts to get more stable.
2. A carbocation next to a quaternary carbon gives a methyl shift.

Answer: (b) cis-but-2-ene

Explanation:
1. Debromination by zinc is an anti-elimination, with the two bromine atoms anti-periplanar.
2. For the d (or l) racemic form, placing the two bromines anti puts the two methyl groups on the same side.
3. So the product is cis-but-2-ene (the meso form would give the trans isomer).

Teacher's Note:
1. Anti elimination links stereochemistry of the reactant and the alkene.
2. Meso gives trans, racemic gives cis in this reaction.

Answer: (b) 1-methylcyclohexene

Explanation:
1. Soda lime decarboxylation generates a carbanion at the carbon that carried \(\ce{COO-}\).
2. The chlorine on the adjacent carbon is a leaving group, so the carbanion expels chloride in an elimination.
3. This forms the C = C bond between those two carbons, giving 1-methylcyclohexene.

Teacher's Note:
1. A carbanion with a beta leaving group eliminates to give an alkene.
2. Ordinary soda lime decarboxylation replaces COOH with H.

Answer: (a) 1-phenylpropan-2-ol

Explanation:
1. Oxymercuration-demercuration adds water across the double bond by Markovnikov's rule.
2. The OH group goes to the more substituted carbon, the middle carbon of the propene chain.
3. So the product is \(\ce{C6H5CH2-CH(OH)-CH3}\), 1-phenylpropan-2-ol.

Teacher's Note:
1. Hydroboration gives the anti-Markovnikov alcohol, 3-phenylpropan-1-ol.
2. Oxymercuration gives the Markovnikov alcohol.

Answer: (d) 4

Explanation:
1. Monochlorination gives four structural isomers: 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 3-chloro-2-methylbutane and 1-chloro-3-methylbutane.
2. 1-Chloro-2-methylbutane and 2-chloro-3-methylbutane each have one chiral carbon and give two enantiomers each.
3. The other two are achiral, so the total number of chiral compounds is 2 + 2 = 4.

Teacher's Note:
1. Count stereocentres in each isomer separately.
2. Each chiral carbon gives a pair of enantiomers.

Answer: (d) 2-methylpent-2-ene

Explanation:
1. Join the carbonyl carbons of the products by a double bond: \(\ce{(CH3)2C=O}\) and \(\ce{O=CH-CH2CH3}\) give \(\ce{(CH3)2C=CH-CH2CH3}\).
2. This is 2-methylpent-2-ene.

Teacher's Note:
1. Reverse ozonolysis by joining the two carbonyl carbons.
2. Ketones come from the more substituted end.

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

Explanation:
1. The strongest activating group is \(\ce{-NH2}\) (aniline), followed by \(\ce{-OH}\) (phenol).
2. In acetanilide the nitrogen lone pair is partly delocalised into the carbonyl, so it activates less.
3. In anilinium ion \(\ce{-NH3+}\) is strongly deactivating, so it is least reactive. Hence I > III > II > IV.

Teacher's Note:
1. Compare the electron-donating strength of the substituent.
2. Positive charge on the ring substituent deactivates it.