Question

Energy diagram of $S_{N}1$ and $S_{N}2$ reactions. The order of hydrolysis of RX by $S_{N}1$ is $3^o>2^o$ $1^o$ $RX$ and by ${SN}^2$ path is $1^o>2^o>3^o$ $RX$.
Which of the following statements are correct for the above energy diagras of $S_{N}1$ and $S_{N}2$ reactions?

• A. $\left(I\right)$ is Fig (a) and $\left(IV\right)$ in Fig. (b) represent $S_{N}1$ reaction
• B. $\left(II\right)$ is Fig (a) and $\left(III\right)$ in Fig. (b) represent $S_{N}2$ reaction
• C. Fig. (a) and Fig. (b) are the energy diagrams for $1^o$ $RX$ and $3^o$ $RX$ respectively
• D. Fig. (a) and Fig. (b) are the energy diagrams for $3^o$ RX and $1^o$ $RX$ respectively

Answer 1

Answer: (A), (B), (C)

The correct options are
A $\left(I\right)$ is Fig (a) and $\left(IV\right)$ in Fig. (b) represent $S_{N}1$ reaction
B $\left(II\right)$ is Fig (a) and $\left(III\right)$ in Fig. (b) represent $S_{N}2$ reaction
C Fig. (a) and Fig. (b) are the energy diagrams for $1^o$ $RX$ and $3^o$ $RX$ respectively

$S_{N}1$ has two steps and has two transition state intermediate carbocations. In the first, higher T.S $(C-X)$ bond stretches ($R_3{C}^{+\delta }....X^{-\delta })$.
In the second. lower T.S bond formation between carbocation intermediate and a nucleophile takes place $(R_3{C}^{+\delta }.....\stackrel{\stackrel{\cdot \cdot }{\ominus }}{Nu})$.
So in Figs (a), (b), $\left(I\right)$ and $\left(IV\right)$ represent $S_{N}1$ reactions respectively.
$S_{N}2$ has one step, has one T.S and no intermediate.
So in Figs. (a), (b), $\left(II\right)$ and $\left(III\right)$ represent $S_{N}2$ reactions, respectively.
The hydrolysis of $RX$ by $S_{N}1$ path is $3^o>2^o>1^o$ and by $S_{N}2$ path is $1^o>2^o>3^o$.
Therefore, Fig(a) represents hydrolysis of $1^o$ RX, since $S_{N}2$ path is of lower energy.
Fig. (b) represents hydrolysis of $3^o$ RX, since ${SN}^1$ path is of lower energy
So statements (A), (B) and (C) are correct.