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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".

(b) [2 marks] Write the mechanism involved in the conversion of benzene to chlorobenzene. OR
(a) [3 marks] An alkyne "A" with molar mass 26, on passing through red hot iron tube at 600°C is converted to another hydrocarbon "B". Treatment with acetyl chloride in the presence of ferric chloride introduces a functional group in "B" to give compound "C". Identify the compounds "A", "B" and "C".
(b) [2 marks] Write the mechanism involved in the conversion of 2-methylpropene to 2-bromo-2-methylpropane.
Manipur CohsemCouncil of Higher Secondary Education, Manipur (Higher Secondary 1st Year) 2024Subjective· 5mImportance★★★★★
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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:

X=CH3−CHBr−CH2−CH3 (2-bromobutane)X = CH_3-CHBr-CH_2-CH_3\ (\text{2-bromobutane})

Treating X with alcoholic KOH causes a dehydrohalogenation (elimination, E2/Saytzeff) reaction, removing H–Br to form the more substituted alkene:

Y=CH3−CH=CH−CH3 (but-2-ene)Y = CH_3-CH=CH-CH_3\ (\text{but-2-ene})

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):

  1. 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⁻: Cl2+FeCl3→Cl++FeCl4−Cl_2 + FeCl_3 \rightarrow Cl^+ + FeCl_4^- …

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