Question

Acetyl bromide reacts with excess of CH$_3$MgI followed by treatment with a saturated solution of NH$_4$Cl gives:
a.) – acetone
b.) – acetamide
c.) – 2-methyl-2propanol
d.) – acetyl iodide

Answer 1

Hint: This question is based on the reactions of Grignard reagent. Write the formula of acetyl bromide, and then react it with CH$_3$MgI, then further it will react with a saturated solution to produce the end product. In the Grignard reagent, magnesium is electropositive metal.

Complete step by step answer:
Now, we have to attain the end product of the reaction between acetyl bromide, and Grignard reagent with a saturated solution. We will move step by step for this reaction.
Let us first write the chemical formula of acetyl bromide, i.e. CH$_3$COBr and as mentioned in the Grignard reagent CH$_3$MgI, the electropositive character of magnesium, so it can stabilize the partial positive charge.
Then, the methyl carbon atom of Grignard reagent will achieve a partial negative charge, so it can show nucleophilic nature. It means that it will donate an electron to the electrophilic atom.
If we see in the acetyl bromide the electrophilic atom is the carbon attached to that of oxygen, as oxygen is an electronegative element.
Thus, oxygen will have a partial negative charge, and carbon will have a partial positive charge on it. Now, we can say that carbon will accept an electron from a nucleophilic atom.
Therefore, there will be a formation of an intermediate product. The chemical reaction is CH$_3$COBr + CH$_3$MgI $\rightarrow$ (CH$_3$)$_2$CO + MgBrI

Now, we can see that acetone is formed as an intermediate product with the by-product magnesium bromoiodide.

Now, we can further react it with a saturated solution of ammonium chloride then, the chemical reaction is CH$_3$COBr + CH$_3$MgI $\rightarrow$ (CH$_3$)$_2$CO + MgBrI $\xrightarrow {NH_4Cl}$ (CH$_3$)$_3$COH + MgOHI

Here, we can say that the acetyl bromide reacts with the Grignard reagent, and acetone is formed. The reaction is further followed by ammonium chloride, and leads to the formation of alcohol.
Thus, we can conclude that there is a formation of alcohol i.e. 2-methyl-2-propanol.
So, the correct answer is “Option C”.

Note: Don’t get confused while performing the reaction. The confusion can arise why oxygen obtained partial negative charge while performing the reaction with the Grignard reagent. We know that oxygen is a highly electronegative element than carbon, so it will itself attain a partial negative charge making the carbon have a partial positive charge.