JEE Mains · Chemistry · STD 12 - 6. Haloalkanes and Haloarenes
An optically active alkyl bromide C\(_4\)H\(_9\)Br, reacts with ethanolic KOH to form major compound [A] which reacts with bromine to give compound [B]. Compound [B] reacts with ethanolic KOH and sodamide to give compound [C]. One molecule of water adds to compound [C] on warming with mercuric sulphate and dilute sulphuric acid at \(333\) K to form compound [D]. The functional group in compound D will be confirmed by :
- A Haloform test
- B Lucas test
- C Silver mirror test
- D Benedict test
Answer & Solution
Correct Answer
(A) Haloform test
Step-by-step Solution
Detailed explanation
The optically active alkyl bromide \(\text{C}_4\text{H}_9\text{Br}\) is \(2\)-bromobutane, \(\text{CH}_3\text{CH}(\text{Br})\text{CH}_2\text{CH}_3\). Reaction with ethanolic \(\text{KOH}\) undergoes dehydrohalogenation to form but-\(2\)-ene as the major product [A] (Zaitsev's…
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