JEE Chemistry Hydrocarbons MCQs Set 25

Answer: (b) 3-Nitrochlorobenzene

Explanation:
1. Nitration gives nitrobenzene (X)
2. The nitro group directs chlorination to the meta position, giving 1-chloro-3-nitrobenzene

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) Aromatic hydrocarbons

Explanation:
1. Aromatic hydrocarbons burn smoothly and have high octane numbers
2. So their tendency to knock is lowest

Teacher's Note:
1. Remember the composition of common fuels: LPG is butane and propane, natural gas is mostly methane, producer gas is CO with N2.
2. Match each fuel to its main constituent.

Answer: (b) Benzene

Explanation:
1. Delocalisation in benzene makes all C-C bonds equal at 139 pm
2. The others have different single and multiple bonds

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) \( \ce{C6H5COONa} \)

Explanation:
1. Heating sodium benzoate with soda lime removes \( \ce{CO2} \)
2. Benzene is the product
3. Phenol can be reduced to benzene only with zinc dust, which is the less common route

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: (c) \( \ce{Cl+} \)

Explanation:
1. The Lewis acid polarises \( \ce{Cl-Cl} \)
2. The electrophile, effectively \( \ce{Cl+} \), attacks the ring

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) Planar

Explanation:
1. All twelve atoms of benzene lie in one plane
2. The carbons are sp2 hybridised

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) The nitrogen in pyridine is \( \ce{sp^2} \) hybridised

Explanation:
1. The lone pair of pyridine is in an sp2 orbital with 33% s-character
2. It is held more tightly than the sp3 lone pair of triethylamine
3. It is not part of the pi system

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) \( \ce{Cl+} \)

Explanation:
1. \( \ce{FeCl3} \) polarises \( \ce{Cl2} \) to \( \ce{Cl+} \) \( \ce{FeCl4-} \)
2. The electrophile is \( \ce{Cl+} \)

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) Ethyl benzoate

Explanation:
1. The ester group \( \ce{-COOEt} \) withdraws electrons and is meta directing
2. Chlorobenzene and phenol are o,p-directing, and ethoxyethane is not aromatic

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) Both o- and p-positions

Explanation:
1. The methyl group activates the ortho and para positions
2. Both nitrotoluenes are formed

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) \( \ce{HNO3} \)

Explanation:
1. Benzene is nitrated by nitrating mixture
2. It does not react with bromine water, water or methanol

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) p-Xylene

Explanation:
1. p-Xylene has only one type of ring hydrogen, so there is one mononitro product
2. The second nitro group can go to 2,3-, 2,5- or 2,6- relative to the methyls, giving three dinitro isomers
3. The other isomers give more mononitro products

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) Cyclohexane

Explanation:
1. Benzene adds three moles of \( \ce{H2} \) over Ni at high temperature
2. Cyclohexane forms

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: (b) Coaltar

Explanation:
1. The light oil fraction of coal tar has benzene
2. It is a major source

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: (b) Naphthalene

Explanation:
1. Light oil gives benzene, toluene and xylenes
2. Middle oil gives phenol and naphthalene, and anthracene comes later

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) Hexa-1,5-diyne

Explanation:
1. Unsaturation decolourises bromine
2. A precipitate with ammoniacal \( \ce{AgNO3} \) needs a terminal alkyne, \( \ce{HC#C-CH2CH2-C#CH} \)
3. Hexa-2,4-diyne has no terminal alkyne and cyclohexatriene is just benzene

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) \( \ce{C2H6 > C6H6 > C2H4 > C2H2} \)

Explanation:
1. The lengths are ethane 154 pm, benzene 139 pm, ethene 134 pm and ethyne 120 pm
2. The order is \( \ce{C2H6 > C6H6 > C2H4 > C2H2} \)

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) Soda lime and zinc dust

Explanation:
1. Benzoic acid is decarboxylated by soda lime: \( \ce{C6H5COOH -> C6H6} \)
2. Phenol is reduced to benzene by hot zinc dust: \( \ce{C6H5OH + Zn -> C6H6 + ZnO} \)

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: (b) Toluene

Explanation:
1. The seven-carbon chain cyclises and dehydrogenates
2. A six-membered ring with a methyl group forms, giving toluene

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) Benzylamine

Explanation:
1. The lone pair of benzylamine is not conjugated with the ring
2. In aniline it is delocalised, and in acetanilide and p-nitroaniline it is delocalised even more
3. Benzylamine is the strongest base

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) Benzoic acid

Explanation:
1. Excess chlorine in light gives benzotrichloride, \( \ce{C6H5CCl3} \)
2. Hydrolysis with aqueous NaOH converts the \( \ce{-CCl3} \) group into \( \ce{-COOH} \), which forms sodium benzoate

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) The members of the series are isomers of each other

Explanation:
1. Successive members differ by \( \ce{CH2} \) and therefore have different molecular formulae
2. So they cannot be isomers of each other

Teacher's Note:
1. Trends such as boiling point and stability depend on molecular size, branching and substituent effects.
2. Apply the trend directly to rank the options.

Answer: (a) 6, 2, 2 and 1

Explanation:
1. The molecule is 2,3,3,5-tetramethylheptane, \( \ce{C11H24} \)
2. Primary: six \( \ce{CH3} \) groups. Secondary: two \( \ce{CH2} \) groups. Tertiary: two CH carbons (C-2 and C-5). Quaternary: C-3
3. Total 6 + 2 + 2 + 1 = 11

Teacher's Note:
1. Recall the key fact or reaction first, then eliminate options that contradict it.
2. Link each answer to the underlying bonding or mechanism rather than memorising it.

Answer: (c) 60% n-heptane + 40% iso-octane

Explanation:
1. Octane number is the percentage of iso-octane in an iso-octane and n-heptane mixture
2. So 40 means 40% iso-octane and 60% n-heptane

Teacher's Note:
1. Remember the composition of common fuels: LPG is butane and propane, natural gas is mostly methane, producer gas is CO with N2.
2. Match each fuel to its main constituent.

Answer: (a) Saytzeff's rule

Explanation:
1. Dehydration of an alcohol gives the more substituted alkene
2. This is Saytzeff's rule, and but-2-ene is more substituted than but-1-ene

Teacher's Note:
1. Check the type of reaction involved and what it does to the carbon skeleton.
2. Compare the options one by one against that.