In the following reaction sequence,the compou | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$CH_3-C(Cl)_2-CH_3$

Vedclass · Answer

(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.

Similar Questions

Absolute alcohol ($100\%$ alcohol) is prepared by distilling rectified spirit over:

$C_6H_6$ $\xrightarrow[(ii)\,\,Zn-Hg/HCl]{(i)\,\,CH_3COCl/AlCl_3}$ $(P)$ $\xrightarrow[(ii)\,\,H_2O/K_2CO_3]{(i)\,\,NBS}$ $(Q)$
Major product $(Q)$ of the following sequence is:

$HO-CH_2CH_2CH_2CH_2-Br \xrightarrow{KOH} \text{Product}$. The product is:

The reaction $CH_3-C(CH_3)(OH)-CH_3 \xrightarrow{H_2SO_4} CH_3-C(CH_3)=CH_2$ is an example of:

Ethyl alcohol on oxidation with $K_2Cr_2O_7$ gives

Vedclass Test Series

Exam Paper Generator

Online Exam Module

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.