CH_3CH=CH_2 xrightarrow{Cl_2, 700 K} A xrigh | vedclass

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Question

(A) Step $1$: Reaction of propene with $Cl_2$ at $700 \ K$ is an allylic substitution reaction. $CH_3-CH=CH_2 + Cl_2 \xrightarrow{700 \ K} ClCH_2-CH=C

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$CH_2(OH)-CH_2(OH)$

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(A) Step $1$: Reaction of propene with $Cl_2$ at $700 \ K$ is an allylic substitution reaction. $CH_3-CH=CH_2 + Cl_2 \xrightarrow{700 \ K} ClCH_2-CH=CH_2 (A) + HCl$.Step $2$: Reaction of allyl chloride $(A)$ with $Na_2CO_3$ at $420 \ K$ and $12 \ atm$ is a hydrolysis reaction. $ClCH_2-CH=CH_2 + Na_2CO_3 + H_2O \rightarrow HOCH_2-CH=CH_2 (B) + NaCl + NaHCO_3$.Step $3$: Reaction of allyl alcohol $(B)$ with $HOCl$ followed by $NaOH$ is a chlorohydrin formation followed by epoxide formation or direct addition. However,in this specific sequence,$HOCl$ adds across the double bond to form $HOCH_2-CH(OH)-CH_2Cl$,and subsequent treatment with $NaOH$ leads to the formation of glycerol or related diols. Given the options,the product $C$ is $CH_2(OH)-CH(OH)-CH_2(OH)$ (glycerol) or a related derivative. Re-evaluating the standard industrial process for glycerol synthesis from propene,the correct product $C$ is $CH_2(OH)-CH(OH)-CH_2(OH)$.

$CH_3CH=CH_2$ $\xrightarrow{Cl_2, 700 \ K} A$ $\xrightarrow{Na_2CO_3, 420 \ K, 12 \ atm} B$ $\xrightarrow[(ii) NaOH]{(i) HOCl} C$. Considering the reaction sequence,what is the product $C$?

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