In the following reaction sequence:

Compound I is

We start with compound I, whose molecular formula is given as $$C_3H_6Cl_2$$. We need to decide which dichloropropane isomer it is. The whole decision will be based on the behaviour of the products formed in the subsequent steps.

First, I is treated with aqueous $$KOH$$. A well-known result of treating a gem-dichloride (both chlorines on the same carbon) with aqueous alkali is hydrolysis to a gem-diol, which then immediately loses water to give a carbonyl compound. The overall transformation is

$$R_2CCl_2 + 2\,KOH_{(aq)} \;\rightarrow\; R_2C(OH)_2 + 2\,KCl$$

followed by

$$R_2C(OH)_2 \;\rightarrow\; R_2C=O + H_2O$$

If I were the gem-dichloride $$CH_3-C(Cl)_2-CH_3$$ (2,2-dichloropropane), the reaction would be

$$CH_3-C(Cl)_2-CH_3 + 2\,KOH_{(aq)} \;\rightarrow\; CH_3-C(OH)_2-CH_3 + 2\,KCl$$

and immediately

$$CH_3-C(OH)_2-CH_3 \;\rightarrow\; CH_3-CO-CH_3 + H_2O$$

Thus compound II would be $$CH_3-CO-CH_3$$, that is, acetone (propan-2-one).

Now II is treated with methylmagnesium bromide and then hydrolysed. Before using the reagent, we recall the Grignard addition formula to a ketone:

$$R_2C=O + R'MgX \;\rightarrow\; R_2C(OMgX)R' \;\overset{H_2O/H^+}{\longrightarrow}\; R_2C(OH)R' + Mg(OH)X$$

Applying it to acetone:

$$CH_3-CO-CH_3 + CH_3MgBr \;\rightarrow\; (CH_3)_3C-O^-MgBr$$

and acid work-up gives

$$ (CH_3)_3C-O^-MgBr + H_2O/H^+ \;\rightarrow\; (CH_3)_3C-OH + MgBr(OH)$$

Therefore compound III is $$ (CH_3)_3C-OH$$, namely tert-butyl alcohol (2-methyl-2-propanol).

Finally, III is tested with Lucas reagent (anhydrous $$ZnCl_2$$ + concentrated $$HCl$$). The Lucas test relies on the following:

- Tertiary alcohols give turbidity immediately.
- Secondary alcohols turn turbid in about 5 min.
- Primary alcohols remain clear for a long time.

Because turbidity appears at once, III must be a tertiary alcohol. Our deduction already shows III is tert-butyl alcohol, perfectly matching the observation.

Hence compound I must be the gem-dichloride $$CH_3-C(Cl)(Cl)-CH_3$$, that is, 2,2-dichloropropane.

Hence, the correct answer is Option A.