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(C) The reaction involves the regioselective ring opening of an epoxide. $1$. For the formation of the $3^\circ$ alcohol,the nucleophilic hydride $(H^

Vedclass · Accepted Answer

$x = LiAlH_4, y = LiAlH_4 / AlCl_3$

Vedclass · Answer

(C) The reaction involves the regioselective ring opening of an epoxide. $1$. For the formation of the $3^\circ$ alcohol,the nucleophilic hydride $(H^-)$ from $LiAlH_4$ attacks the less sterically hindered carbon atom of the epoxide ring,followed by protonation to yield the $3^\circ$ alcohol. Thus,$x = LiAlH_4$. $2$. For the formation of the $2^\circ$ alcohol,the epoxide is treated with $LiAlH_4 / AlCl_3$. The $AlCl_3$ acts as a Lewis acid and coordinates with the oxygen atom of the epoxide,which increases the electrophilicity of the more substituted carbon. Consequently,the hydride attacks the more substituted carbon,leading to the formation of the $2^\circ$ alcohol. Thus,$y = LiAlH_4 / AlCl_3$. Therefore,the correct reagents are $x = LiAlH_4$ and $y = LiAlH_4 / AlCl_3$.

Find the missing reagents $(x)$ and $(y)$ for the following reaction sequence:

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