underset{text{symmetrical Ketone}}{(A)}, xrig | vedclass

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(C) The reaction sequence is as follows:$1$. $(A)$ is a symmetrical ketone. Reduction of a ketone with $LiAlH_{4}$ yields a secondary alcohol $(B)$.$2

Vedclass · Accepted Answer

$CH_{3}-CH_{2}-C(=O)-CH_{2}-CH_{3}$

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(C) The reaction sequence is as follows:$1$. $(A)$ is a symmetrical ketone. Reduction of a ketone with $LiAlH_{4}$ yields a secondary alcohol $(B)$.$2$. Dehydration of the secondary alcohol $(B)$ in the presence of $H^{\oplus}$ and heat $(\Delta)$ produces an alkene.$3$. For the product to exhibit diastereomerism (specifically geometric isomerism),the alkene formed must have different groups attached to the double-bonded carbons.$4$. Let's analyze option $(C)$: $CH_{3}CH_{2}C(=O)CH_{2}CH_{3}$ (Pentan$-3-$one).$5$. $CH_{3}CH_{2}C(=O)CH_{2}CH_{3} \xrightarrow{LiAlH_{4}} CH_{3}CH_{2}CH(OH)CH_{2}CH_{3}$ (Pentan$-3-$ol).$6$. $CH_{3}CH_{2}CH(OH)CH_{2}CH_{3} \xrightarrow{H^{\oplus}, \Delta} CH_{3}CH=CHCH_{2}CH_{3}$ (Pent$-2-$ene).$7$. Pent$-2-$ene exists as cis and trans isomers,which are diastereomers.Therefore,the correct reactant $(A)$ is pentan$-3-$one.

$\underset{\text{symmetrical Ketone}}{(A)}\,$ $\xrightarrow{LiAlH_{4}}(B)$ $\xrightarrow[\Delta ]{H^{\oplus }}\text{Diastereomers}$
Reactant $(A)$ is

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$\underset{\text{symmetrical Ketone}}{(A)}\,$ $\xrightarrow{LiAlH_{4}}(B)$ $\xrightarrow[\Delta ]{H^{\oplus }}\text{Diastereomers}$Reactant $(A)$ is