JEE Main 2017 · 9 Apr · Shift 1 · Chemistry Q59

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ChemistrySingle correct+4 / -1Question 59 of 90

In the following reaction sequence :
$\underset{(\ce{C3H6Cl2})}{\text{I}}\ce{->[KOH(aq)]}\text{II}\xrightarrow[\text{(ii) }\ce{H2O/H+}]{\text{(i) }\ce{CH3MgBr}}\text{III}\xrightarrow{\text{Anhy. }\ce{ZnCl2}\text{ + Con. HCl}}\text{gives turbidity immediately}$
The compound I is :

\ce{CH2Cl-CHCl-CH3}

\ce{CH2Cl-CH2-CH2Cl}

\ce{CHCl2-CH2-CH3}

\ce{CH3-CCl2-CH3}

Answer and worked solution

Official answer: D

Turbidity at once with the Lucas reagent means III is a tertiary alcohol. 2,2-Dichloropropane is hydrolysed by aqueous KOH (via the gem-diol) to acetone (II), and $\ce{CH3MgBr}$ then gives tert-butyl alcohol (III). So I is $\ce{CH3-CCl2-CH3}$ .

Topics:Alcohols Nucleophilic addition Polyhalogen compounds

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JEE Main 2017 · 9 Apr · Shift 1, Chemistry question 59