H_2C=CH-CH_2-CH_2-CH(OH)-CH_3 xrightarrow[tex | vedclass

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

Question

(C) $1.$ Reaction of alcohol with $SOCl_2$ in pyridine replaces the $-OH$ group with $-Cl$ with retention of configuration (or simple substitution): $

Vedclass · Accepted Answer

$HO-CH_2-CH_2-CH_2-CH(Cl)-CH_3$

Vedclass · Answer

(C) $1.$ Reaction of alcohol with $SOCl_2$ in pyridine replaces the $-OH$ group with $-Cl$ with retention of configuration (or simple substitution): $H_2C=CH-CH_2-CH_2-CH(OH)-CH_3 + SOCl_2 \rightarrow H_2C=CH-CH_2-CH_2-CH(Cl)-CH_3 (A)$. $2.$ Ozonolysis of the terminal alkene $(A)$ using $O_3/Zn, H_2O$ cleaves the double bond to form an aldehyde: $H_2C=CH-CH_2-CH_2-CH(Cl)-CH_3 \rightarrow OHC-CH_2-CH_2-CH(Cl)-CH_3 (B) + HCHO$. $3.$ Reduction of the aldehyde $(B)$ with $NaBH_4$ yields the corresponding primary alcohol: $OHC-CH_2-CH_2-CH(Cl)-CH_3 \xrightarrow{NaBH_4} HO-CH_2-CH_2-CH_2-CH(Cl)-CH_3 (C)$.

$H_2C=CH-CH_2-CH_2-CH(OH)-CH_3$ $\xrightarrow[\text{Pyridine}]{SOCl_2} (A)$ $\xrightarrow[H_2O]{O_3/Zn} (B) (C_5H_9ClO)$ $\xrightarrow{NaBH_4} (C)$
Compound $(C)$ is:

Similar Questions

What are $X$ and $Y$ respectively in the following reaction sequence?

In the above chemical reaction sequence,"$A$" and "$B$" respectively are:

The major product $(Z)$ of the following chemical reaction is

Write about $(a)$ Tollen's Test and $(b)$ Fehling solution. $OR$ Write the oxidation reaction of only aldehyde. $OR$ Write tests which are utilized to differentiate aldehydes and ketones.

Which of the following compounds would undergo Cannizzaro's reaction?

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$H_2C=CH-CH_2-CH_2-CH(OH)-CH_3$ $\xrightarrow[\text{Pyridine}]{SOCl_2} (A)$ $\xrightarrow[H_2O]{O_3/Zn} (B) (C_5H_9ClO)$ $\xrightarrow{NaBH_4} (C)$ Compound $(C)$ is: