JEE Chemistry Hydrocarbons MCQs Set 37

Answer: (a) Pentanedioic acid (glutaric acid)

Explanation:
1. The ring opens at the double bond giving a dicarbonyl compound with five carbons.
2. With water (oxidative work-up) the aldehyde groups are oxidised to carboxylic acids.

Teacher's Note:
1. Ring opening keeps all carbon atoms in one chain.

Answer: (b) 1,3,5-Trimethylbenzene (mesitylene)

Explanation:
1. Cyclic trimerisation of three propyne molecules gives a benzene ring bearing three methyl groups.
2. Symmetrical joining gives 1,3,5-trimethylbenzene.

Teacher's Note:
1. Trimerisation of an alkyne gives a trisubstituted benzene.

Answer: (b) \( \ce{C6H5CH(Br)CH=CHCH3} \)

Explanation:
1. NBS brominates through a radical chain at the most reactive allylic or benzylic C-H.
2. Abstracting the C-H that is both benzylic and allylic gives a radical stabilised by the ring and the double bond, so Br enters at that carbon.

Teacher's Note:
1. The most stabilised radical forms fastest.

Answer: (d) 2,3-Dimethylbut-2-ene

Explanation:
1. \( \ce{(CH3)2C=C(CH3)2} \) is symmetrical, and each half becomes \( \ce{(CH3)2C=O} \).

Teacher's Note:
1. A symmetrical alkene gives one product.

Answer: (c) \( \ce{CH3CH2CH(Br)-C6H4-OH} \) (para)

Explanation:
1. H+ adds to the carbon farther from the ring so that the benzylic carbocation forms; it is further stabilised by the para OH group.
2. Br- then attacks the benzylic carbon, giving \( \ce{CH3CH2CH(Br)-C6H4-OH} \); the phenolic OH is unaffected.

Teacher's Note:
1. Benzylic carbocations are stabilised by resonance.

Answer: (c) Ethylcyclohexane

Explanation:
1. A beta-keto acid loses \( \ce{CO2} \) on heating: X = cyclohexyl methyl ketone.
2. Clemmensen reduction converts C=O to \( \ce{CH2} \): Y = ethylcyclohexane.

Teacher's Note:
1. Beta-keto acids decarboxylate on heating.

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

Explanation:
1. Protonation gives the most stable carbocation: a tertiary diphenyl carbocation from III, a tertiary benzylic one from II, a secondary benzylic one from I.
2. Nitroethene has a strongly electron-withdrawing group and is the least reactive.

Teacher's Note:
1. The more stable the carbocation intermediate, the faster the addition.

Answer: (d) Benzyl chloride

Explanation:
1. Light promotes free radical substitution on the side chain (benzylic C-H).
2. The first product is benzyl chloride, \( \ce{C6H5CH2Cl} \).

Teacher's Note:
1. Light or heat means side chain; Fe or AlCl3 means ring substitution.

Answer: (a) Hex-1-ene

Explanation:
1. Baeyer's reagent is a test for unsaturation.
2. Alkenes are oxidised to cis-diols, so the purple colour disappears, while benzene, alkanes and \( \ce{CCl4} \) do not react.

Teacher's Note:
1. The test is positive for C=C and C#C.

Answer: (b) Acetone and carbon dioxide

Explanation:
1. The double bond is cleaved. The \( \ce{(CH3)2C=} \) end gives acetone.
2. The \( \ce{=CH2} \) end is oxidised to formic acid and then to \( \ce{CO2} + \ce{H2O} \).

Teacher's Note:
1. Terminal =CH2 becomes CO2 under strong oxidation.

Answer: (a) II > I > III

Explanation:
1. Electrophilic addition is fastest when the alkene is electron rich and gives a stable carbocation.
2. Isobutene gives a stable tertiary carbocation (II). Vinyl chloride has some +M donation from Cl (I). The CHO group deactivates the double bond strongly (III).

Teacher's Note:
1. Electron-donating groups increase and electron-withdrawing groups decrease the rate of electrophilic addition.

Answer: (a) \( \ce{BrCH2-CH2-CCl3} \)

Explanation:
1. The CCl3 group has a strong -I effect, so the carbocation \( \ce{CH3-CH+-CCl3} \) is destabilised.
2. Instead H+ adds to the carbon next to CCl3 giving the primary cation \( \ce{+CH2-CH2-CCl3} \), and Br- gives \( \ce{BrCH2CH2CCl3} \) (anti-Markovnikov product).

Teacher's Note:
1. An EWG next to the double bond reverses the orientation of addition.

Answer: (a) Propadiene

Explanation:
1. \( \ce{CH2=C=CH2} \): cleaving both double bonds gives \( \ce{HCHO} \) from each end and \( \ce{CO2} \) from the central carbon.

Teacher's Note:
1. The central carbon of the 1,2-diene becomes CO2.

Answer: (c) Deuteriobenzene, \( \ce{C6H5D} \)

Explanation:
1. The Grignard reagent is strongly basic; the carbon-magnesium bond is protonated (here deuterated) by D2O.
2. The product is \( \ce{C6H5D} \) and Mg(OD)Br.

Teacher's Note:
1. Grignard reagents react with water or D2O to give the hydrocarbon.

Answer: (b) \( \ce{NH2NH2 / C2H5ONa} \) (Wolff-Kishner)

Explanation:
1. Only the C=O must be reduced to \( \ce{CH2} \) while the OH group stays.
2. Zn-Hg/HCl or HI would also convert the alcohol group (to chloride or iodide), while the basic Wolff-Kishner conditions leave the OH intact.

Teacher's Note:
1. Choose basic conditions when an acid-sensitive OH is present.

Answer: (b) (a-s), (b-r), (c-q), (d-p)

Explanation:
1. HBr gives 2-bromopropane by Markovnikov addition.
2. NBS gives allylic bromination: 3-bromopropene.
3. \( \ce{Br2/CCl4} \) gives the vicinal dibromide.
4. HBr followed by aqueous hydrolysis gives propan-2-ol.

Teacher's Note:
1. Match each reagent with its characteristic reaction.

Answer: (d) Hexadeuteriobenzene, \( \ce{C6D6} \)

Explanation:
1. \( \ce{CaC2 + 2D2O -> C2D2 + Ca(OD)2} \), so A = \( \ce{C2D2} \).
2. Three molecules of ethyne trimerise in a red hot tube to give benzene: B = \( \ce{C6D6} \).

Teacher's Note:
1. Cyclic polymerisation of acetylene gives benzene.

Answer: (d) a racemic mixture of (2S,3R)- and (2R,3S)-3-bromobutan-2-ol

Explanation:
1. HOBr adds through a bromonium ion followed by anti attack of water (OH).
2. Anti addition to the trans alkene places OH and Br on the same side in the Fischer projection (erythro), which corresponds to the (2S,3R)/(2R,3S) pair.

Teacher's Note:
1. Anti addition to a trans alkene gives the erythro pair.

Answer: (b) Neoprene

Explanation:
1. Vinylacetylene adds HCl to give 2-chlorobuta-1,3-diene (chloroprene), P.
2. Polymerisation of chloroprene gives neoprene.

Teacher's Note:
1. Chloroprene is the monomer of neoprene.

Answer: (c) (c) and (e)

Explanation:
1. Step (a) is initiation.
2. The real propagation steps are (c), in which Cl radical abstracts H, and (e), in which methyl radical abstracts Cl; the chain carriers Cl. and CH3. are regenerated.
3. Steps (b) and (d) involve an H atom, which is energetically unfavourable and does not occur.

Teacher's Note:
1. Propagation consumes one radical and produces another.

Answer: (d) All of \( \ce{AgNO3 + NH4OH} \), \( \ce{Cu2Cl2 + NH4OH} \) and sodium

Explanation:
1. But-1-yne is a terminal alkyne with an acidic hydrogen.
2. It gives a white silver acetylide with ammoniacal AgNO3, a red copper acetylide with ammoniacal Cu2Cl2, and liberates \( \ce{H2} \) with sodium.

Teacher's Note:
1. Terminal alkynes have acidic hydrogen.

Answer: (c) Two molecules of ethanoic acid

Explanation:
1. Ozonolysis of an alkyne cleaves the triple bond and gives carboxylic acids.
2. Each \( \ce{CH3-C#} \) end becomes \( \ce{CH3COOH} \).

Teacher's Note:
1. Alkynes give acids on ozonolysis.