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

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

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

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(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:

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