In the following reaction: Cl-CH_2-CH(O)^{14} | vedclass

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Question

(C) The reaction involves the nucleophilic attack of the ethoxide ion $(C_2H_5O^-)$ on the less sterically hindered terminal carbon $(^{14}CH_2)$ of t

Vedclass · Accepted Answer

$CH_2(O)CH-^{14}CH_2-OC_2H_5$

Vedclass · Answer

(C) The reaction involves the nucleophilic attack of the ethoxide ion $(C_2H_5O^-)$ on the less sterically hindered terminal carbon $(^{14}CH_2)$ of the epoxide ring in epichlorohydrin.
This opens the epoxide ring to form an alkoxide intermediate: $Cl-CH_2-CH(O^-)-^{14}CH_2-OC_2H_5$.
Subsequently,the alkoxide oxygen performs an intramolecular $S_N2$ attack on the $CH_2$ group attached to the chlorine atom,displacing the chloride ion $(Cl^-)$.
This cyclization reforms an epoxide ring,resulting in the final product: $CH_2(O)CH-^{14}CH_2-OC_2H_5$.

In the following reaction: $Cl-CH_2-CH(O)^{14}CH_2 \xrightarrow{C_2H_5O^-} \text{the final product is:}$

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