Question

Acetyl bromide reacts with an excess of CH3​MgI followed by treatment with a saturated solution of NH4​Cl gives:
A. acetone
B. acetamide
C. 2 - methyl - 2 – propanol
D. acetyl iodide

Answer 1

To answer this question, you should recall the concept of preparation of alcohols. The simplest method to produce all primary, secondary and tertiary alcohol is by using a Grignard reagent.

Complete step-by-step answer: We know that to produce primary alcohol, the Grignard reagent is reacted with formaldehyde.

Methyl Magnesium Formaldehyde Ethanol
Bromide
Further reacting a Grignard reagent with any other aldehyde will lead to a secondary alcohol.

Acetaldehyde 2-propanal
Ultimately, reacting a Grignard reagent with a ketone will generate tertiary alcohol.

tert-butyl alcohol
For the reaction given in the question: Acetyl bromide reacts with an excess of $C{H_{3}}MgI\;$ followed by treatment with a saturated solution of $N{H_4}Cl\;$gives tertiary alcohol which in the given option is 2-methyl-2-propanol. The mechanism can be visualized as follows:

Therefore, we can conclude that the correct answer to this question is option C.

Note: Grignard reagent refers to an organo-magnesium halide with the formula of ${\text{RMgX}}$, where X is a halogen, and R is an alkyl or aryl. Apart from preparation of alcohols, it is also used for determining the number of halogen atoms present in a halogen compound. Currently, another important use involves the chemical analysis of certain triacylglycerols.