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The major product of the following reaction is :

CH_{3}CH=CHCO_{2}CH_{3}\xrightarrow{LiAlH_{4}}

  • Option 1)

     CH_{3}CH_{2}CH_{2}CO_{2}CH_{3}          

  • Option 2)

    CH_{3}CH=CHCH_{2}OH

  • Option 3)

    CH_{3}CH_{2}CH_{2}CH_{2}OH

  • Option 4)

     CH_{3}CH_{2}CH_{2}CHO

Answers (1)

 

Alcohol formation by reduction of esters -

Yields two alcohols

- wherein

RCOOR'+4[H]\rightarrow RCH_{2}OH+R'OH

 

 

CH_{3}CH=CHCO_{2}CH_{3}\xrightarrow{LiAlH_{4}}

LiAlH_{4} do not reduce alkene


Option 1)

 CH_{3}CH_{2}CH_{2}CO_{2}CH_{3}          

Option 2)

CH_{3}CH=CHCH_{2}OH

Option 3)

CH_{3}CH_{2}CH_{2}CH_{2}OH

Option 4)

 CH_{3}CH_{2}CH_{2}CHO

Posted by

Vakul

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