Match the reactants given in Column-$I$ with the products formed in Column-$II$.
Column-$I$ (Reactants) Column-$II$ (Products) $A$. $ROH + SOCl_2$ $i$. $RBr + NaHSO_4 + H_2O$ $B$. $ROH + PCl_5$ $ii$. $3RX + H_3PO_3$ $C$. $3ROH + PX_3$ $iii$. $RCl + SO_2 + HCl$ $D$. $ROH + NaBr + H_2SO_4$ $iv$. $RCl + POCl_3 + HCl$
| Column-$I$ (Reactants) | Column-$II$ (Products) |
| $A$. $ROH + SOCl_2$ | $i$. $RBr + NaHSO_4 + H_2O$ |
| $B$. $ROH + PCl_5$ | $ii$. $3RX + H_3PO_3$ |
| $C$. $3ROH + PX_3$ | $iii$. $RCl + SO_2 + HCl$ |
| $D$. $ROH + NaBr + H_2SO_4$ | $iv$. $RCl + POCl_3 + HCl$ |
(A-III, B-IV, C-II, D-I) The reactions are as follows:
$A$. $ROH + SOCl_2 \rightarrow RCl + SO_2 + HCl$ (Darbzens process) $(A \rightarrow iii)$
$B$. $ROH + PCl_5 \rightarrow RCl + POCl_3 + HCl$ $(B \rightarrow iv)$
$C$. $3ROH + PX_3 \rightarrow 3RX + H_3PO_3$ $(C \rightarrow ii)$
$D$. $ROH + NaBr + H_2SO_4 \rightarrow RBr + NaHSO_4 + H_2O$ $(D \rightarrow i)$
Therefore,the correct matching is $A-iii, B-iv, C-ii, D-i$.
$A$. $ROH + SOCl_2 \rightarrow RCl + SO_2 + HCl$ (Darbzens process) $(A \rightarrow iii)$
$B$. $ROH + PCl_5 \rightarrow RCl + POCl_3 + HCl$ $(B \rightarrow iv)$
$C$. $3ROH + PX_3 \rightarrow 3RX + H_3PO_3$ $(C \rightarrow ii)$
$D$. $ROH + NaBr + H_2SO_4 \rightarrow RBr + NaHSO_4 + H_2O$ $(D \rightarrow i)$
Therefore,the correct matching is $A-iii, B-iv, C-ii, D-i$.
Explore More
Similar Questions
The boiling point of methanol $(CH_3OH)$ is higher than that of methyl thiol $(CH_3SH)$ because:
Medium
View SolutionOf the following alcohols,the one that would react fastest with conc. $HCl$ and anhydrous $ZnCl_{2}$ is,
MediumNEET 2017
View Solution$Ph-C(OH)(Ph)-C(OH)(CH_3)_2$ $(A)$ $+$ $Ph-C(OH)(Ph)-C(OH)(Et)_2$ $(B)$ $\xrightarrow{H_2SO_4}$
$(p)$ $Ph-C(Ph)(CH_3)-CO-CH_3$
$(q)$ $Ph-C(Ph)(Et)-CO-Et$
$(r)$ $Ph-C(Ph)(Et)-CO-CH_3$
$(s)$ $Ph-C(Ph)(CH_3)-CO-Et$
When $(A)$ and $(B)$ react with $H_2SO_4$,the products obtained are:
$(p)$ $Ph-C(Ph)(CH_3)-CO-CH_3$
$(q)$ $Ph-C(Ph)(Et)-CO-Et$
$(r)$ $Ph-C(Ph)(Et)-CO-CH_3$
$(s)$ $Ph-C(Ph)(CH_3)-CO-Et$
When $(A)$ and $(B)$ react with $H_2SO_4$,the products obtained are:
Difficult
View SolutionThe rose odour from an ester is formed by the action of $HCOOH$ on
Medium
View SolutionIn the following reaction sequence,$X$ and $Y$,respectively,are:
$CH_2=C(CH_3)_2$ $\xrightarrow{X} \text{Epoxide}$ $\xrightarrow{Y} (CH_3)_3C-OH$
$CH_2=C(CH_3)_2$ $\xrightarrow{X} \text{Epoxide}$ $\xrightarrow{Y} (CH_3)_3C-OH$
MediumKVPY 2011
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo