Which of the esters shown, after reduction with ${L i A l H}_{4}$ and aqueous workup, will yield two molecules of only a single alcohol ?

{C H}_{3} {C H}_{2} {C O}_{2} {C H}_{2} {C H}_{3}

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C_{6} H_{5} {C O}_{2} {C H}_{2} C_{6} H_{5}

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C_{6} H_{5} {C O}_{2} C_{6} H_{5}

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None of these

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Solution

The correct option is B $C_{6} H_{5} {C O}_{2} {C H}_{2} C_{6} H_{5}$
$O$
When $p h - | | C - O - C H_{2} - p h$ is reacted with $L i A l H_{4}$ $2$ molecules of similar alcohol is produced.
$O$
$\Rightarrow p h - | | C - O - C H_{2} - p h L i A l H_{4} - --- \to H_{2} O 2 p h - C H_{2} - O H$

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Similar questions

C_{6} H_{5} - C H_{2} - C H = C H_{2} + H B r \to

C_{6} H_{5} - C H = C H_{2} + H B r P e r o x i d e - ---- \to

{C H}_{3} {N H}_{2}, {({C H}_{3})}_{2} N H, C_{6} H_{5} {N H}_{2}, C_{6} H_{5} {C H}_{2} {N H}_{2}

C_{6} H_{5} {C O N H C H}_{3}

C_{6} H_{5} {C H}_{2} {N H C H}_{3}

C_{6} H_{5} {N H}_{2}

C_{6} H_{5} {N H C H}_{3}

C_{6} H_{5} {C H}_{2} {N H}_{2}

C_{6} H_{5} {N H}_{2}

C_{6} H_{5} {N H C H}_{3}

C_{6} H_{5} {C H}_{2} {N H}_{2}

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