JEE Chemistry Hydrocarbons MCQs Set 24

Answer: (b) Maleic anhydride

Explanation:
1. Vapour phase catalytic oxidation breaks the ring
2. Maleic anhydride is the product

Teacher's Note:
1. Cold dilute alkaline KMnO4 gives a diol from an alkene, while hot acidic KMnO4 cleaves the double bond.
2. Terminal alkynes are separated from other hydrocarbons by their acidic hydrogen.

Answer: (a) Electrophilic substitution reactions

Explanation:
1. The delocalised pi cloud attracts electrophiles
2. Substitution keeps the aromatic stabilisation, so electrophilic substitution is typical

Teacher's Note:
1. Use Huckel's rule (4n + 2 pi electrons in a planar cyclic conjugated ring) to judge aromaticity.
2. Side chain reactions in light go through radicals, while ring reactions with a Lewis acid go through cations.

Answer: (b) Phenol

Explanation:
1. The OH group donates electrons into the ring and activates it
2. Nitrobenzene and benzoic acid are deactivated

Teacher's Note:
1. Aromatic rings keep their stability by electrophilic substitution, not addition.
2. Activating groups direct ortho and para, while strongly withdrawing groups direct meta.

Answer: (a) Activating groups

Explanation:
1. Groups that donate electrons by delocalisation or hyperconjugation activate the ring and direct ortho and para
2. Only the halogens are o,p-directing and deactivating

Teacher's Note:
1. Work out the electrophile first (NO2+, SO3, R+, RCO+, Cl+) and then where it attacks.
2. Alkyl groups on the ring raise reactivity, while nitro and carbonyl groups lower it.

Answer: (c) Toluene

Explanation:
1. The methyl group activates the ring
2. Benzene is less reactive, and benzoic acid and nitrobenzene are deactivated

Teacher's Note:
1. Aromatic rings keep their stability by electrophilic substitution, not addition.
2. Activating groups direct ortho and para, while strongly withdrawing groups direct meta.

Answer: (b) Toluene

Explanation:
1. The methyl group activates the ring towards electrophiles
2. Nitro and chloro groups deactivate it

Teacher's Note:
1. Work out the electrophile first (NO2+, SO3, R+, RCO+, Cl+) and then where it attacks.
2. Alkyl groups on the ring raise reactivity, while nitro and carbonyl groups lower it.

Answer: (d) Nitrobenzene

Explanation:
1. The nitro group strongly withdraws electrons from the ring
2. This makes nitrobenzene the least reactive to electrophilic substitution

Teacher's Note:
1. Draw the possible intermediates and pick the most stable one (tertiary, then secondary, then primary).
2. Only HBr shows the peroxide effect, not HCl or HI.

Answer: (a) Nitrobenzene

Explanation:
1. The nitro group is a strong deactivator
2. Nitration of nitrobenzene needs more drastic conditions

Teacher's Note:
1. Use Huckel's rule (4n + 2 pi electrons in a planar cyclic conjugated ring) to judge aromaticity.
2. Side chain reactions in light go through radicals, while ring reactions with a Lewis acid go through cations.

Answer: (c) There is cyclic delocalisation of pi electrons in benzene

Explanation:
1. The six pi electrons are delocalised over the whole ring, which makes benzene resist addition
2. All C-C bonds are equal and monosubstitution gives one product

Teacher's Note:
1. Ozonolysis splits C=C into two carbonyl groups, so join the carbonyl carbons to find the alkene.
2. Hydroboration-oxidation gives the anti-Markovnikov alcohol, while oxymercuration gives the Markovnikov alcohol.

Answer: (c) (i) anhydrous \( \ce{AlCl3} \) and (ii) direct sunlight

Explanation:
1. A Lewis acid gives ring substitution to chlorobenzene
2. Sunlight gives addition to benzene hexachloride

Teacher's Note:
1. Electrophilic addition goes through the more stable carbocation, so H adds to the carbon with more hydrogens.
2. With HBr and a peroxide the radical path reverses the orientation.

Answer: (d) \( \ce{C6H6 + CH3Cl} \)

Explanation:
1. Alkylation needs an alkyl halide and benzene
2. \( \ce{CH3COCl} \) would give acylation

Teacher's Note:
1. Work out the electrophile first (NO2+, SO3, R+, RCO+, Cl+) and then where it attacks.
2. Alkyl groups on the ring raise reactivity, while nitro and carbonyl groups lower it.

Answer: (d) Electrophilic displacement

Explanation:
1. The attacking \( \ce{NO2+} \) replaces a hydrogen of the ring
2. So it is electrophilic substitution (displacement)

Teacher's Note:
1. Use Huckel's rule (4n + 2 pi electrons in a planar cyclic conjugated ring) to judge aromaticity.
2. Side chain reactions in light go through radicals, while ring reactions with a Lewis acid go through cations.

Answer: (d) All of these

Explanation:
1. Benzene undergoes substitution (nitration, halogenation)
2. It adds hydrogen or chlorine under forcing conditions and burns in air

Teacher's Note:
1. Draw the possible intermediates and pick the most stable one (tertiary, then secondary, then primary).
2. Only HBr shows the peroxide effect, not HCl or HI.

Answer: (b) Trimerisation of ethyne

Explanation:
1. Three ethyne molecules cyclise in a red hot tube to benzene
2. Ethene with butadiene gives cyclohexene and reduction of benzaldehyde gives toluene

Teacher's Note:
1. Link each named reaction to its reagent and the change in carbon number.
2. Elimination gives alkenes, reduction gives alkanes, and coupling doubles the alkyl chain.

Answer: (a) Benzene

Explanation:
1. A hydrocarbon contains only carbon and hydrogen
2. Benzene is \( \ce{C6H6} \)
3. The others contain N or O

Teacher's Note:
1. Work out the electrophile first (NO2+, SO3, R+, RCO+, Cl+) and then where it attacks.
2. Alkyl groups on the ring raise reactivity, while nitro and carbonyl groups lower it.

Answer: (a) They have a relatively high percentage of carbon

Explanation:
1. The carbon content is high (benzene is 92.3% C)
2. There is not enough oxygen for complete combustion, so carbon particles glow as soot

Teacher's Note:
1. Use Huckel's rule (4n + 2 pi electrons in a planar cyclic conjugated ring) to judge aromaticity.
2. Side chain reactions in light go through radicals, while ring reactions with a Lewis acid go through cations.

Answer: (c) Benzene

Explanation:
1. Benzene is a planar regular hexagon with sp2 carbons
2. Methane and isobutane are tetrahedral, and cyclohexane is puckered

Teacher's Note:
1. Aromatic rings keep their stability by electrophilic substitution, not addition.
2. Activating groups direct ortho and para, while strongly withdrawing groups direct meta.

Answer: (c) The rate of nitration of benzene is greater than that of hexadeuterobenzene

Explanation:
1. Breaking the C-H or C-D bond is not the slow step, so there is no isotope effect
2. Benzene and hexadeuterobenzene nitrate at nearly the same rate, so (c) is false

Teacher's Note:
1. Work out the electrophile first (NO2+, SO3, R+, RCO+, Cl+) and then where it attacks.
2. Alkyl groups on the ring raise reactivity, while nitro and carbonyl groups lower it.

Answer: (b) \( \ce{KMnO4} \)

Explanation:
1. Hot alkaline or acidic \( \ce{KMnO4} \) oxidises a benzylic alkyl group completely
2. The product is \( \ce{-COOH} \)

Teacher's Note:
1. Decide which functional groups can react with the reagent and which cannot.
2. A test that works for both compounds cannot distinguish them.

Answer: (c) Benzene

Explanation:
1. Nitration of benzene gives nitrobenzene (B)
2. Sn/HCl reduces it to aniline

Teacher's Note:
1. Identify the functional group that is removed or formed and count the carbon atoms of the product.
2. Wurtz and Kolbe couple two alkyl groups, soda lime removes one carbon, and Grignard reagents pick up an acidic proton.

Answer: (c) \( \ce{C6H6Cl6} \)

Explanation:
1. In sunlight chlorine adds to benzene by a radical mechanism
2. Benzene hexachloride is formed

Teacher's Note:
1. Ozonolysis splits C=C into two carbonyl groups, so join the carbonyl carbons to find the alkene.
2. Hydroboration-oxidation gives the anti-Markovnikov alcohol, while oxymercuration gives the Markovnikov alcohol.

Answer: (a) Dissolves easily

Explanation:
1. Naphthalene is non-polar and kerosene is a non-polar hydrocarbon solvent
2. Like dissolves like

Teacher's Note:
1. Fractions of crude oil come off in order of boiling point: gases, petroleum ether, gasoline, kerosene, diesel, lubricating oil.
2. Cracking breaks large molecules into smaller, more useful ones.

Answer: (b) Anthracene

Explanation:
1. Anthracene is a linear fusion of three benzene rings
2. Naphthalene has two, and triphenylmethane has three separate rings

Teacher's Note:
1. Aromatic rings keep their stability by electrophilic substitution, not addition.
2. Activating groups direct ortho and para, while strongly withdrawing groups direct meta.

Answer: (b) 1,3-Dinitrobenzene

Explanation:
1. The nitro group is a meta director
2. The second nitro group enters the meta position

Teacher's Note:
1. Work out the electrophile first (NO2+, SO3, R+, RCO+, Cl+) and then where it attacks.
2. Alkyl groups on the ring raise reactivity, while nitro and carbonyl groups lower it.

Answer: (a) Benzene triozonide

Explanation:
1. Three ozone molecules add across the three double bonds
2. The triozonide is formed first and hydrolysis gives three glyoxal

Teacher's Note:
1. Electrophilic addition goes through the more stable carbocation, so H adds to the carbon with more hydrogens.
2. With HBr and a peroxide the radical path reverses the orientation.