Chemistry Hydrocarbons SET 2019
PMDC Verified Question 25 of 65
In the reaction sequence given here: \( \text{H}_3\text{C}-\text{CH}_3 + \text{Br}_2 \xrightarrow{h\nu} \text{A} \xrightarrow{\text{Alc KOH}} \text{B} \). The end product is an unsaturated hydrocarbon. Identify the nature of reaction in the two steps.
A
Step I is a nucleophilic substitution and step II is elimination
B
Step I is addition and step II is nucleophilic substitution
C
Step I is free radical substitution and step II nucleophilic substitution
D
Step I is free radical substitution and step II is elimination
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: Step I is free radical substitution and step II is elimination
Concept:

Halogenation of alkanes in UV light proceeds via radicals, and reacting the resulting alkyl halide with alcoholic KOH triggers an elimination.

Solution:

  • Step I: Ethane (\( \text{H}_3\text{C}-\text{CH}_3 \)) reacts with \( \text{Br}_2 \) under UV light (\( h\nu \)). UV light initiates a homolytic cleavage, resulting in a free radical substitution that yields ethyl bromide (Compound A).


  • Step II: Ethyl bromide is treated with alcoholic KOH. Alcoholic KOH is a strong base that removes a proton and the bromide ion from adjacent carbons. This process is elimination (specifically dehydrohalogenation), yielding ethene (Compound B), an unsaturated hydrocarbon.


Why other options are incorrect:

  • Step I is not nucleophilic substitution or addition, because alkanes only react via free radicals under these conditions.


  • Step II is not nucleophilic substitution; substitution would require aqueous KOH to produce an alcohol, not the unsaturated hydrocarbon mentioned in the stem.

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