JEE Chemistry Hydrocarbons MCQs Set 23

Answer: (d) In alkylation, a polyalkylated product is formed

Explanation:
1. An alkyl group activates the ring, so the product reacts faster than benzene
2. This gives polyalkylation, while an acyl group deactivates the ring and stops at the mono product

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) Elimination

Explanation:
1. Benzene gives substitution, addition (hydrogenation, chlorination in light) and combustion
2. It has no leaving group or saturated carbon, so it cannot undergo elimination

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: (b) Aromatic compounds

Explanation:
1. Coal tar distillation gives benzene, toluene, phenol, naphthalene and anthracene
2. These are aromatic compounds

Teacher's Note:
1. Straight chain alkanes knock easily, while branched alkanes and aromatics resist knocking.
2. Octane number is the percentage of iso-octane in a mixture with n-heptane.

Answer: (c) 12 sigma and 3 pi

Explanation:
1. Benzene has six C-C and six C-H sigma bonds, 12 in all
2. Three pi bonds are formed from the six p orbitals

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) 4

Explanation:
1. Benzene has 12 sigma bonds and 3 pi bonds
2. The ratio is 12/3 = 4

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: (a) 120 degrees

Explanation:
1. All carbons are sp2 hybridised in a regular planar hexagon
2. The bond angle is 120 degrees

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) Hexachlorobenzene

Explanation:
1. Iodine acts as a halogen carrier and gives electrophilic substitution
2. With excess chlorine all six hydrogens are replaced, giving \( \ce{C6Cl6} \)
3. Benzene hexachloride forms by addition in sunlight

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) Benzene hexachloride

Explanation:
1. Gammexane (lindane) is the gamma isomer of 1,2,3,4,5,6-hexachlorocyclohexane
2. This is benzene hexachloride, \( \ce{C6H6Cl6} \)

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) \( \ce{C6H6 + C2H5Cl -> C6H5C2H5 + HCl} \) (anhydrous \( \ce{AlCl3} \))

Explanation:
1. The Friedel-Crafts reaction substitutes an alkyl or acyl group on the benzene ring using \( \ce{AlCl3} \)
2. Only (a) is balanced and correct

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) \( \ce{(C6H5)2CH^+} \)

Explanation:
1. Two phenyl rings delocalise the positive charge by delocalisation
2. So the diphenylmethyl cation is more stable than benzyl, ethyl or 2-phenylethyl cations

Teacher's Note:
1. Count sigma bonds and lone pairs to find hybridisation: 4 gives sp3, 3 gives sp2, 2 gives sp.
2. Bond length falls and bond strength rises with increasing s-character and bond order.

Answer: (b) o- and p-Chlorotoluenes

Explanation:
1. \( \ce{FeCl3} \) gives ring substitution by \( \ce{Cl+} \)
2. The methyl group directs to the ortho and para positions

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) Benzotrichloride

Explanation:
1. In light the side chain is substituted by radicals
2. \( \ce{C6H5CH3 -> C6H5CH2Cl -> C6H5CHCl2 -> C6H5CCl3} \)
3. With excess chlorine benzotrichloride forms

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: (d) \( \ce{NO2^+} \)

Explanation:
1. \( \ce{HNO3 + 2 H2SO4 -> NO2+ + H3O+ + 2 HSO4-} \)
2. The nitronium ion is the electrophile

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) Electrophilic substitution

Explanation:
1. The nitronium ion attacks the ring and a hydrogen is lost
2. The aromatic ring is kept, so it is electrophilic substitution

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) Both (a) and (b)

Explanation:
1. Halogenation needs a Lewis acid such as \( \ce{FeCl3} \) to make \( \ce{X+} \)
2. Cold and dark conditions avoid radical addition and side chain reactions

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: (d) Friedel-Crafts reaction

Explanation:
1. An alkyl halide alkylates the aromatic ring in the presence of \( \ce{AlCl3} \)
2. This is the Friedel-Crafts alkylation

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: (a) Chlorobenzene

Explanation:
1. Iron forms \( \ce{FeCl3} \) in situ, a halogen carrier
2. Ring substitution gives chlorobenzene

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) Faraday

Explanation:
1. Michael Faraday isolated benzene in 1825 from the oil obtained from illuminating gas
2. He called it bicarburet of hydrogen

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) Armstrong and Baeyer

Explanation:
1. Armstrong and Baeyer proposed the centric formula with the fourth valence of each carbon directed to the centre
2. Dewar and Ladenburg gave other formulae

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) Between one and two

Explanation:
1. Delocalisation of the pi electrons makes all six C-C bonds equal
2. Each has bond order 1.5, which is between one and two

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) One type

Explanation:
1. All six carbons are sp2 hybridised and equivalent
2. They are of one type

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) Benzene

Explanation:
1. Decarboxylation: \( \ce{C6H5COONa + NaOH -> C6H6 + Na2CO3} \)
2. Benzene is formed

Teacher's Note:
1. Write the balanced equation first and track every carbon.
2. Check whether the reagent gives elimination (alcoholic KOH) or substitution (aqueous KOH).

Answer: (b) Sulphonation

Explanation:
1. Sulphonation of benzene is reversible
2. Heating the sulphonic acid with dilute acid or steam removes the \( \ce{-SO3H} \) group

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) \( \ce{SO3} \)

Explanation:
1. Sulphur trioxide, present in oleum and formed from conc. \( \ce{H2SO4} \), is a strong electrophile
2. It attacks the ring

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: (c) Benzoic acid

Explanation:
1. A benzylic methyl group is fully oxidised to \( \ce{-COOH} \)
2. Benzoic acid is formed

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.