Compound(A) $C_{6} H_{12} O$ is optically active. Compound(A) give negative Tollens test and positive test with $2 - 4$ -di-nitro phenyl hydrazine. Identify A.

C H_{3} - O | | C - C H_{2} - C | C H_{3} H - C H_{3}

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C H_{3} - O | | C - C | C H_{2} - C H_{3} H - C H_{3}

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H - O | | C - C H_{2} - C | C H_{2} - C H_{3} H - C H_{3}

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C H_{3} - O | | C - C H_{2} - C H_{2} - C H_{2} - C H_{3}

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Solution

The correct option is B $C H_{3} - O | | C - C | C H_{2} - C H_{3} H - C H_{3}$

Aldehyde gives $\Rightarrow$ positive Tollen's test
$\Rightarrow$ positive $2, 4$ -DNP-test
Ketone (Refer to Image 1) gives $\Rightarrow$ negative Tollen's test
$\Rightarrow$ positive $2, 4$ -DNP-test
So, the compound must have (Refer to Image 2) and as the compound is optically active, the compound must have chiral carbon.
(Refer to Image 3)

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C H_{3} - C H_{2} - C H_{2} - C H_{2} - C H | C H_{3} - C | C H_{3} - C H_{3} - C H_{2} - C H_{3} | C H - C H_{2} - C H_{2} - C H_{2} - C H_{3}

C H_{3} - C H_{3}, C H_{2} = C H_{2}

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S_{N} 2

K I

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C H_{3} - C H_{2} - C H_{2} - B r ∣ C H - C H_{3} a n d C H_{3} - C H_{3} ∣ C H - C H_{2} - B r ∣ C H - C H_{3}

(I) (I I)

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