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the reactant $\mathrm{A}$ is
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Verified Answer
The correct answer is:
$\mathrm{CH}_{3} \mathrm{CHOHCH}_{3}$
$\underset{\text { A }}{\mathrm{CH}_{3} \mathrm{CHOHCH}_{3}} \frac{\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}}{\text { dil.H }_{2} \mathrm{SO}_{4}} \underset{\mathrm{B}}{\mathrm{CH}_{3} \mathrm{COCH}_{3}}$
$\frac{\mathrm{CH}_{3} \mathrm{MgI}}{\mathrm{H}_{2} \mathrm{O}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}$
$\frac{\mathrm{CH}_{3} \mathrm{MgI}}{\mathrm{H}_{2} \mathrm{O}}\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}$
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