In the given reaction 3-Bromo-2,2-dimethylbut | vedclass

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(B) The reaction proceeds via an $S_N1$ mechanism involving the formation of a carbocation intermediate.$1$. The starting material $3-$Bromo$-2,2-$dim

Vedclass · Accepted Answer

$2-$Ethoxy$-2,3-$dimethylbutane.

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(B) The reaction proceeds via an $S_N1$ mechanism involving the formation of a carbocation intermediate.$1$. The starting material $3-$Bromo$-2,2-$dimethylbutane loses a bromide ion to form a secondary $(2^{\circ})$ carbocation.$2$. This $2^{\circ}$ carbocation undergoes a $1,2-$methyl shift to form a more stable tertiary $(3^{\circ})$ carbocation.$3$. The nucleophile $C_2H_5OH$ then attacks the stable $3^{\circ}$ carbocation.$4$. After deprotonation,the final major product formed is $2-$Ethoxy$-2,3-$dimethylbutane.

In the given reaction $3-$Bromo$-2,2-$dimethylbutane $\xrightarrow{C_2H_5OH}$ (Major product) $'A'$. Product $A$ is:

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