Question

Consider the following reactions:
$I.{Me}_{3}C-Br\underset{{SN}^1}{\overset{H_{2}O+NaBr}{\to }}Products$
$II.{Ph}_{3}C-Br\underset{{SN}^1}{\overset{H_{2}O+NaBr}{\to }}Products$
Which of the following statements are correct about the above reactions?

• A. The products in reactions $\left(I\right)$ and $\left(II\right)$ are mixture of $({Me}_3-OH+{Me}_{3}Br)$ and $({Ph}_{3}C-OH+{Ph}_{3}C-Br)$, respectively
• B. The product in $\left(I\right)$ is $({Me}_{3}C-OH)$ and in $\left(II\right)$ is $({Ph}_{3}C-OH+{Ph}_{3}C-Br)$
• C. The product in $\left(I\right)$ is $({Me}_{3}C-OH)$ and in $\left(II\right)$ is $({Ph}_{3}C-OH)$
• D. ${Ph}_3{C}^{\oplus }$ is more stable than ${Me}_3{C}^{\oplus }$

Answer 1

Answer: (B), (D)

The product in $\left(I\right)$ is $({Me}_{3}C-OH)$ and in $\left(II\right)$ is $({Ph}_{3}C-OH+{Ph}_{3}C-Br)$
${Ph}_3{C}^{\oplus }$ is more stable than ${Me}_3{C}^{\oplus }$

Tertiary alkyl halides in the presence of a mixture of $NaBr$ and water undergo hydrolysis in which halogen is replaced with hydroxide group. The reaction proceeds via $S_{N}1$ mechanism.

The product in $\left(I\right)$ is $({Me}_{3}C-OH)$ and in $\left(II\right)$ is $({Ph}_{3}C-OH+{Ph}_{3}C-Br)$

${Ph}_3{C}^{\oplus }$ is more stable than ${Me}_3{C}^{\oplus }$. Hence, in reaction $\left(1\right)$ only one product is formed and in reaction $\left(2\right)$ both the products are formed.