Acetyl bromide reacts with excess of $$CH_3MgI$$ followed by treatment with a saturated solution of $$NH_4Cl$$ given

The reaction takes place in three steps.

1. Reaction of acetyl bromide with methylmagnesium iodide

Acetyl bromide reacts with methylmagnesium iodide through nucleophilic acyl substitution. The methyl group from the Grignard reagent attacks the carbonyl carbon, and bromide ion is eliminated.

$$CH_3COBr + CH_3MgI \rightarrow CH_3COCH_3$$

Thus, the intermediate formed is acetone.

1. Reaction of acetone with excess $$CH_3MgI$$

Since excess Grignard reagent is present, the newly formed acetone further reacts with another molecule of $$CH_3MgI$$. The methyl group attacks the carbonyl carbon to form a tertiary alkoxide.

$$CH_3COCH_3 + CH_3MgI \rightarrow (CH_3)_3C-O^-MgI^+$$

1. Acidic workup with saturated $$NH_4Cl$$

The alkoxide ion is protonated by saturated ammonium chloride solution to give the corresponding alcohol.

$$ (CH_3)_3C-O^-MgI^+ + NH_4Cl \rightarrow (CH_3)_3COH $$

Hence, the final product formed is

$$\boxed{(CH_3)_3COH}$$

which is 2-methyl-2-propanol (tert-butyl alcohol).

Therefore, the correct answer is Option (C).