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(D) $III$ gives immediate turbidity with Lucas reagent $(Anhy. ZnCl_2 + Conc. HCl)$,which indicates that $III$ is a tertiary $(3^\circ)$ alcohol.$A$ $

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$CH_3-C(Cl)_2-CH_3$

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(D) $III$ gives immediate turbidity with Lucas reagent $(Anhy. ZnCl_2 + Conc. HCl)$,which indicates that $III$ is a tertiary $(3^\circ)$ alcohol.$A$ $3^\circ$ alcohol is formed when a ketone reacts with a Grignard reagent $(CH_3MgBr)$. Therefore,$II$ must be a ketone.Since $II$ is formed from $I (C_3H_6Cl_2)$ by reaction with aqueous $KOH$,$I$ must be a geminal dihalide. Thus,$I$ is $2,2$-dichloropropane $(CH_3-C(Cl)_2-CH_3)$.The reaction sequence is:$CH_3-C(Cl)_2-CH_3 (I)$ $\xrightarrow{KOH_{(aq)}} [CH_3-C(OH)_2-CH_3]$ $\xrightarrow{-H_2O} CH_3-CO-CH_3 (II)$$CH_3-CO-CH_3 (II) \xrightarrow{(i) CH_3MgBr, (ii) H_2O/H^{+}} (CH_3)_3C-OH (III)$$(CH_3)_3C-OH (III) \xrightarrow{Lucas\,Reagent} 	ext{Immediate turbidity}$.

In the following reaction sequence,the compound $I$ is:
$I (C_3H_6Cl_2)$ $\xrightarrow{KOH_{(aq)}} II$ $\xrightarrow{(i) CH_3MgBr, (ii) H_2O/H^{+}} III$ $\xrightarrow{Anhy. ZnCl_2 + Conc. HCl}$ gives turbidity immediately.

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In the following reaction sequence,the compound $I$ is:$I (C_3H_6Cl_2)$ $\xrightarrow{KOH_{(aq)}} II$ $\xrightarrow{(i) CH_3MgBr, (ii) H_2O/H^{+}} III$ $\xrightarrow{Anhy. ZnCl_2 + Conc. HCl}$ gives turbidity immediately.