Q.(a) [3 marks] Hydrocarbon n-butane on treatment with a limited amount of bromine undergoes monobromination only to produce compound "X". Compound "X" on treatment with alcoholic potash gives a new compound "Y" which can show geometrical isomerism. Identify the compound "X" and the two geometrical isomers of "Y".
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Start your 14-day free trial to unlock the full solution →Monobromination of n-butane gives 2-bromobutane (X); its dehydrohalogenation with alcoholic KOH gives but-2-ene (Y), which shows cis/trans geometrical isomerism. Benzene converts to chlorobenzene via electrophilic aromatic substitution: Cl⁺ generated by FeCl3 attacks the ring, forming a resonance-stabilised arenium intermediate that loses H⁺ to restore aromaticity.
(a) Identifying X and the isomers of Y:
n-Butane is CH3–CH2–CH2–CH3. Free-radical monobromination with a limited amount of Br2 occurs preferentially at the more substituted (secondary) carbon, because the intermediate secondary radical is more stable than a primary radical. This gives:
Treating X with alcoholic KOH causes a dehydrohalogenation (elimination, E2/Saytzeff) reaction, removing H–Br to form the more substituted alkene:
Because but-2-ene has two different groups (CH3 and H) on each doubly-bonded carbon, it shows geometrical (cis-trans) isomerism:
- cis-but-2-ene: both CH3 groups on the same side of the double bond.
- trans-but-2-ene: the two CH3 groups on opposite sides of the double bond.
(b) Mechanism: benzene → chlorobenzene (electrophilic aromatic substitution):
- Generation of the electrophile: FeCl3 (a Lewis acid catalyst) polarises the Cl–Cl bond of Cl2, forming a complex that releases the electrophile Cl⁺ and the counter-ion FeCl4⁻: …
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