JEE Chemistry Hydrocarbons MCQs Set 05

Answer: (a) \(\ce{AgNO3 + NH4OH}\) (Tollens' reagent)

Explanation:
1. But-1-yne is a terminal alkyne with an acidic hydrogen, so it gives a white precipitate of the silver salt with ammoniacal silver nitrate.
2. But-2-yne has no acidic hydrogen and gives no precipitate.
3. Both decolourise bromine water and Baeyer's reagent, so only (a) distinguishes them.

Teacher's Note:
1. Only terminal alkynes react with ammoniacal silver nitrate or ammoniacal cuprous chloride.
2. The test is for the acidic C - H bond.

Answer: (c) 2-methylpropan-2-ol, \(\ce{(CH3)3C-OH}\)

Explanation:
1. Butane isomerises to isobutane (2-methylpropane) with \(\ce{AlCl3/HCl}\).
2. \(\ce{KMnO4}\) oxidises the tertiary C - H bond of isobutane to give the tertiary alcohol.
3. So Y is 2-methylpropan-2-ol.

Teacher's Note:
1. Alkanes with a tertiary hydrogen are oxidised by \(\ce{KMnO4}\) to tertiary alcohols.
2. Isomerisation of n-alkanes needs anhydrous \(\ce{AlCl3}\) and \(\ce{HCl}\).

Answer: (a) benzyl bromide, \(\ce{C6H5CH2Br}\)

Explanation:
1. n-Heptane undergoes aromatisation over \(\ce{Cr2O3 + Al2O3}\) at high temperature and pressure to give toluene (X).
2. N-bromosuccinimide with light brominates the benzylic position by a free radical mechanism.
3. So Y is benzyl bromide.

Teacher's Note:
1. NBS brominates allylic and benzylic positions.
2. Ring bromination needs \(\ce{Br2}\) with a Lewis acid, not NBS.

Answer: (b) Wurtz reaction

Explanation:
1. The Wurtz reaction with two different alkyl halides gives a mixture of three alkanes, so it is not suitable for unsymmetrical alkanes.
2. Decarboxylation, Clemmensen reduction and the Corey-House synthesis give a single alkane.

Teacher's Note:
1. Wurtz reaction works well only for symmetrical alkanes.
2. The Corey-House synthesis uses lithium dialkylcuprate and can join two different alkyl groups.

Answer: (b) \(\ce{CH3CHO}\)

Explanation:
1. Calcium carbide with water gives acetylene, \(\ce{HC#CH}\) (X).
2. Hydration of acetylene with dilute \(\ce{H2SO4}\) and \(\ce{HgSO4}\) gives vinyl alcohol, which tautomerises to acetaldehyde.

Teacher's Note:
1. Hydration of acetylene gives an aldehyde and that of other alkynes gives ketones.
2. Kucherov's reaction uses \(\ce{Hg^{2+}}\) and dilute \(\ce{H2SO4}\).

Answer: (c) \(\ce{CH3CH=O}\)

Explanation:
1. Elimination follows Saytzeff's rule, so the major product is the more substituted alkene but-2-ene, \(\ce{CH3CH=CHCH3}\).
2. Reductive ozonolysis breaks the double bond to give two molecules of ethanal, \(\ce{CH3CHO}\).

Teacher's Note:
1. Alcoholic KOH gives dehydrohalogenation, and the more substituted alkene is major.
2. Ozonolysis of a symmetrical alkene gives one carbonyl compound.

Answer: (b) 2-bromobutane, \(\ce{CH3CH2CHBrCH3}\)

Explanation:
1. Kolbe electrolysis of potassium propanoate gives butane (P) at the anode.
2. Free radical bromination is highly selective and brominates the secondary carbon preferentially.
3. So the major product is 2-bromobutane.

Teacher's Note:
1. Kolbe's electrolysis joins two alkyl groups from two carboxylate ions.
2. Bromine is much more selective than chlorine.

Answer: (d) a racemic mixture of the (2R,3R) and (2S,3S) isomers

Explanation:
1. Lindlar's catalyst reduces the alkyne to the cis alkene, cis-but-2-ene.
2. Bromine adds anti across the double bond, and the anti addition to a cis alkene gives the pair of enantiomers (2R,3R) and (2S,3S) in equal amounts.
3. The product is therefore a racemic mixture.

Teacher's Note:
1. Anti addition to a cis alkene gives a racemic mixture, and to a trans alkene gives a meso compound.
2. Sodium in liquid ammonia would give the trans alkene.

Answer: (d) \(\ce{CH#CH}\)

Explanation:
1. Acidity depends on the s-character of the carbon bearing the hydrogen.
2. The s-character is 50% in the sp carbon of ethyne, 33% in ethene and benzene, and 25% in ethane.
3. The sp carbon is the most electronegative, so ethyne is the most acidic.

Teacher's Note:
1. The order of acidity is alkyne > alkene > alkane.
2. More s-character means a more stable carbanion.

Answer: (b) toluene

Explanation:
1. Acetylene polymerises in a red hot iron tube to give benzene (X).
2. Benzene undergoes Friedel-Crafts alkylation with \(\ce{CH3Cl}\) and \(\ce{AlCl3}\) to give toluene.

Teacher's Note:
1. Three molecules of ethyne give one molecule of benzene.
2. Friedel-Crafts alkylation puts an alkyl group on the ring.

Answer: (b) position isomers

Explanation:
1. Propan-2-ol dehydrates to propene (X).
2. Hydroboration-oxidation gives the anti-Markovnikov alcohol, propan-1-ol (Y).
3. Propan-2-ol and propan-1-ol differ only in the position of the OH group, so they are position isomers.

Teacher's Note:
1. Hydroboration-oxidation gives the anti-Markovnikov product.
2. Position isomers have the same carbon chain and the same functional group at different positions.

Answer: (d) both (a) and (b)

Explanation:
1. Cold dilute alkaline \(\ce{KMnO4}\) hydroxylates an alkene to a glycol, and its purple colour is discharged.
2. Alkanes do not react, so it can differentiate an alkene from an alkane.
3. Alkynes also decolourise the reagent, so it cannot differentiate an alkyne from an alkene.

Teacher's Note:
1. Baeyer's reagent is cold, dilute and alkaline \(\ce{KMnO4}\).
2. It is a test for unsaturation.

Answer: (d) benzaldehyde

Explanation:
1. Chromium trioxide in acetic anhydride oxidises the methyl group of toluene to the diacetate of the aldehyde, \(\ce{C6H5CH(OCOCH3)2}\), and prevents further oxidation to the acid.
2. Hydrolysis with water in the second step gives benzaldehyde.

Teacher's Note:
1. This is the Etard-type partial oxidation of toluene to benzaldehyde.
2. The acetic anhydride protects the aldehyde from over oxidation.

Answer: (d) benzene

Explanation:
1. A Grignard reagent is a very strong base and phenol is an acid.
2. The Grignard reagent takes the acidic hydrogen of the OH group and gives benzene, together with \(\ce{C6H5OMgBr}\).

Teacher's Note:
1. Grignard reagents react with any compound that has an acidic hydrogen (water, alcohol, phenol, acids).
2. They give the corresponding hydrocarbon.