A xrightarrow{LiAlH_4} 2B (Chiral alcohol onl | vedclass

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Question

(B) $LiAlH_4$ is a strong reducing agent that reduces esters into two alcohols: $RCOOR' \xrightarrow{LiAlH_4} RCH_2OH + R'OH$.For the product to be $2

Vedclass · Accepted Answer

$CH_3-CH_2-CH(CH_3)-COOCH(CH_3)-CH_2-CH_3$

Vedclass · Answer

(B) $LiAlH_4$ is a strong reducing agent that reduces esters into two alcohols: $RCOOR' \xrightarrow{LiAlH_4} RCH_2OH + R'OH$.For the product to be $2B$ (where $B$ is a single chiral alcohol),the ester must be symmetric such that both fragments produce the same chiral alcohol.In option $B$,$CH_3-CH_2-CH(CH_3)-COOCH(CH_3)-CH_2-CH_3$ is a sec-butyl $2-$methylbutanoate ester.Upon reduction,it yields $CH_3-CH_2-CH(CH_3)-CH_2OH$ ($2$-methylbutan$-1-$ol) and $CH_3-CH_2-CH(CH_3)OH$ (butan$-2-$ol).However,if we consider the structure $CH_3-CH(CH_3)-COOCH(CH_3)-CH_2-CH_3$ (sec-butyl $2-$methylpropanoate),it yields different alcohols.Given the requirement for $2B$ (identical chiral alcohols),the ester must be $sec-butyl$ $2-methylbutanoate$ is not correct,but rather an ester like $sec-butyl$ $2-methylbutanoate$ is not symmetric.The correct structure for $2B$ where $B$ is chiral is $sec-butyl$ $2-methylbutanoate$ is not the case,but rather $sec-butyl$ $sec-butyl$ ester derivative. The correct answer is $B$.

$A \xrightarrow{LiAlH_4} 2B$ (Chiral alcohol only); structure of $A$ is

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