Question

In the following reaction sequence:

$\underset{\left(C_3{H}_{6}C{l}_2\right)}{I}\stackrel{KOH\left(aq\right)}{\to }II\underset{\left(ii\right)H_{2}O/H^{+}}{\overset{\left(i\right)C{H}_{3}MgBr}{\to }}III\stackrel{Anhy.ZnC{l}_2+Con.HCl}{\to }$ give trubidity immediately

• A. $Cl-\underset{Cl}{\underset{|}{CH}}-C{H}_2-C{H}_3$
• B. $\underset{Cl}{\underset{|}{C{H}_2}}-C{H}_2-\underset{Cl}{\underset{|}{C{H}_2}}$
• C. $C{H}_3-\stackrel{Cl}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_3$
• D. $\underset{Cl}{\underset{|}{C{H}_2}}-\underset{Cl}{\underset{|}{CH}}-C{H}_3$

Answer 1

The correct option is C $C{H}_3-\stackrel{Cl}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_3$

The compound (I) is represented by the option (C).
$\underset{\left(C_3{H}_{6}C{l}_2\right),I}{C{H}_3-\stackrel{Cl}{\stackrel{|}{\underset{Cl}{\underset{|}{C}}}}-C{H}_3}\stackrel{KOH\left(aq\right)}{\to }\underset{II}{C{H}_3-\stackrel{OH}{\stackrel{|}{\underset{OH}{\underset{|}{C}}}}-C{H}_3}\rightleftharpoons C{H}_3-\underset{O}{\underset{||}{C}}-C{H}_3\underset{\left(ii\right)H_{2}O/H^{+}}{\overset{\left(i\right)C{H}_{3}MgBr}{\to }}\underset{III}{C{H}_3-\stackrel{OH}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_3}\stackrel{Anhy.ZnC{l}_2+Con.HCl}{\to }C{H}_3-\stackrel{Cl}{\stackrel{|}{\underset{C{H}_3}{\underset{|}{C}}}}-C{H}_3$
Hydrolysis of compound (I), a geminal dichloride gives compound (II), a geminal diol. It loses a molecule of water to form acetone. Attack of grignard reagent gives tertiary alcohol.
Tertiary alcohols give rapid turbidity with Lucas reagent $Anhy.ZnC{l}_2+Con.HCl$

Hence, the correct option is C