2 moles of iso-pentylene on reaction with compound $A$ gives compound $B$ , which on treatment with 1-phenyl-2-butyne followed by oxidation with $H_{2} O_{2}$ and $N a O H$ results in the compound $C$ . $A$ is in turn generated by treating $B C l_{3}$ , with $L i A l H_{4}$ . The compound $C$ gives:

Tollen’s reagent

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Iodoform test

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Fehling’s test

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a & c

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Solution

The correct option is B
Iodoform test

$4 B C l_{3} + 3 L i A l H_{4} ⟶ 2 B_{2} H_{6} + 3 L i A l C l_{4}$

So we have a dialkylborane where there is just the one hydrogen atom bonded to the boron atom for reduction. Plus, the alkyl group - the substituent attached to the boron is a bulky one. When this dialkylborane is made to react with 1-phenylbut-2-ene, the hydrogen attaches itself with the carbon 3 from the methyl group because there is a bulky alkyl group present in the reducing agent (dialkylborane).

$C$ is a methyl ketone, which means, it will give a positive haloform test!

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