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JEE CHEMISTRY MCQ: ALDEHYDES & KETONES & CARBOXYLIC ACIDS

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Analyzing +4 | -1
NaAlH4NaAl{H_4} reduces an ester into HCHOHCHO and (CH3)2CHOH.\left({C{H_3}} \right){ _2}CHOH. This ester changes to ______ on reaction with CH3MgBr/H3O+.C{H_3}MgBr/{H_3}{O^ + }.
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Correct option: B
CH3CHOC{H_3}CHO
Option C (CH3CHOC{H_3}CHO) is correct because the ester is H−CO∣∣−O−CH(CH3)2H - \mathop C\limits^{\begin{array}{cccccccccccccccccccc} O \\ {||} \end{array}} - O - CH{\left( {C{H_3}} \right)_2}, and reaction with CH3MgBr/H3O+CH_3MgBr/H_3O^+ yields acetaldehyde.

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