In the following reaction sequence, $X$ and $Z$, respectively, are:
$CH_{3}CH_{2}CH_{2}-OH + PCl_{5} \rightarrow CH_{3}CH_{2}CH_{2}Cl + X + HCl$
$CH_{3}CH_{2}CH_{2}Cl \xrightarrow{alc. KOH, \Delta} Y$
$Y \xrightarrow{HBr, (C_{6}H_{5}CO)_{2}O_{2}} Z$
$CH_{3}CH_{2}CH_{2}-OH + PCl_{5} \rightarrow CH_{3}CH_{2}CH_{2}Cl + X + HCl$
$CH_{3}CH_{2}CH_{2}Cl \xrightarrow{alc. KOH, \Delta} Y$
$Y \xrightarrow{HBr, (C_{6}H_{5}CO)_{2}O_{2}} Z$
- A$X = POCl_{3}$; $Z = CH_{3}-CH(Br)-CH_{3}$
- B$X = POCl_{3}$; $Z = CH_{3}CH_{2}CH_{2}-Br$
- C$X = H_{3}PO_{3}$; $Z = CH_{3}-CH(Br)-CH_{3}$
- D$X = H_{3}PO_{3}$; $Z = CH_{3}CH_{2}CH_{2}-Br$
Explore More
Similar Questions
Which of the following undergoes hydrolysis most easily?
Difficult
View SolutionMatch the reactions given in Column-$I$ with their names in Column-$II$:
Column-$I$ (Reactants) Column-$II$ (Name) $A$. $R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX$ $i$. Electrophilic addition $B$. $H_3C-Br + AgF \to H_3C-F + AgBr$ $ii$. Finkelstein reaction $C$. $H_3C-CH_3 + Cl_2 \xrightarrow{hv} H_3C-CH_2Cl + HCl$ $iii$. Swarts reaction $D$. $CH_2=CH_2 + Br_2 \to BrCH_2-CH_2Br$ $iv$. Free radical substitution
| Column-$I$ (Reactants) | Column-$II$ (Name) |
| $A$. $R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX$ | $i$. Electrophilic addition |
| $B$. $H_3C-Br + AgF \to H_3C-F + AgBr$ | $ii$. Finkelstein reaction |
| $C$. $H_3C-CH_3 + Cl_2 \xrightarrow{hv} H_3C-CH_2Cl + HCl$ | $iii$. Swarts reaction |
| $D$. $CH_2=CH_2 + Br_2 \to BrCH_2-CH_2Br$ | $iv$. Free radical substitution |
Medium
View SolutionGiven below are two statements:
Statement $I$: $3$-phenylpropene reacts with $HBr$ and gives secondary alkyl bromide having a chiral carbon atom as the major product.
Statement $II$: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.
Statement $I$: $3$-phenylpropene reacts with $HBr$ and gives secondary alkyl bromide having a chiral carbon atom as the major product.
Statement $II$: Aryl chlorides and aryl cyanides can be prepared by Sandmeyer reaction as well as Gattermann reaction.
DifficultJEE MAIN 2026
View SolutionChoose the correct option:
$A$. $ROH + SOCl_2 \rightarrow$ $P$. Darzen process $B$. $RX (Cl, Br) + NaI \rightarrow$ $Q$. Finkelstein reaction $C$. $RX (Cl, Br) + AgF \rightarrow$ $R$. Swarts reaction $D$. $ArX + RX + Na \rightarrow$ $S$. Wurtz-Fittig reaction
| $A$. $ROH + SOCl_2 \rightarrow$ | $P$. Darzen process |
| $B$. $RX (Cl, Br) + NaI \rightarrow$ | $Q$. Finkelstein reaction |
| $C$. $RX (Cl, Br) + AgF \rightarrow$ | $R$. Swarts reaction |
| $D$. $ArX + RX + Na \rightarrow$ | $S$. Wurtz-Fittig reaction |
Easy
View SolutionMatch the reactants in Column-$I$ with their products in Column-$II$.
Column-$I$ (Reactants) Column-$II$ (Products) $A$. $H_2C=CH_2 + Br_2 \xrightarrow{CCl_4}$ $i$. $CH_3CH_2CH_2Br$ $B$. $CH_3CH=CH_2 + HI \rightarrow$ $ii$. $BrCH_2-CH_2Br$ $C$. $C_6H_5N_2^+X^- + KI \rightarrow$ $iii$. $CH_3CHICH_3$ $D$. $CH_3CH=CH_2 + HBr \xrightarrow{peroxide}$ $iv$. $C_6H_5I$
| Column-$I$ (Reactants) | Column-$II$ (Products) |
| $A$. $H_2C=CH_2 + Br_2 \xrightarrow{CCl_4}$ | $i$. $CH_3CH_2CH_2Br$ |
| $B$. $CH_3CH=CH_2 + HI \rightarrow$ | $ii$. $BrCH_2-CH_2Br$ |
| $C$. $C_6H_5N_2^+X^- + KI \rightarrow$ | $iii$. $CH_3CHICH_3$ |
| $D$. $CH_3CH=CH_2 + HBr \xrightarrow{peroxide}$ | $iv$. $C_6H_5I$ |
Medium
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