Question

Acetyl bromide reacts with excess of $C{H}_{3}MgI$ followed by treatment with a saturated solution of $N{H}_{4}C1$ gives:

• A. acetone
• B. acetamide
• C. 2-methyl-2-propanol
• D. acetyl iodide

Answer 1

The correct option is C 2-methyl-2-propanol

Acetyl bromide reacts with excess of
$C{H}_{3}MgI$ followed by
treatment with a saturated solution of
$N{H}_{4}Cl$ gives 2-methyl-2-propanol.