Question

The order of reactivity towards \( \text{S}_\text{N}2 \) reaction among the following is
\[ (a)\ \text{CH}_3\text{CH}_2\text{Cl} \quad (b)\ \text{CH}_3\text{CH(Cl)}\text{CH}_3 \quad (c)\ \text{(CH}_3\text{)}_3\text{CCI} \quad (d)\ \text{CH}_3\text{CH}_2\text{Cl} \]

• A. \( a > d > b > c \)
• B. \( a > b > c > d \)
• C. \( a > c > b > d \)
• D. \( b > a > c > d \)
• E. \( c > b > d > a \)

Hint:

In \( \text{S}_\text{N}2 \) reactions, primary carbons are most reactive due to less steric hindrance, while tertiary carbons are least reactive.

Answer 1

The Correct Option is A

Step 1: Understand \( \text{S}_\text{N}2 \) reactivity.
The reactivity in \( \text{S}_\text{N}2 \) reactions follows the order: \[ \text{CH}_3\text{CH}_2\text{Cl} > \text{CH}_3\text{CH(Cl)}\text{CH}_3 > \text{(CH}_3\text{)}_3\text{CCI} > \text{CH}_3\text{CH}_2\text{Cl} \] This is based on steric hindrance and the nature of the leaving group.

Step 2: Recognize the best leaving groups.
In \( \text{S}_\text{N}2 \) reactions, the reactivity is enhanced by a good leaving group and less steric hindrance. Here, \( \text{CH}_3\text{CH}_2\text{Cl} \) has a primary carbon, which is less hindered and thus reacts more readily than secondary or tertiary carbons.

Step 3: Compare the options.
- Option \( a \): \( \text{CH}_3\text{CH}_2\text{Cl} \), with a less sterically hindered primary carbon, is the most reactive.
- Option \( b \): \( \text{CH}_3\text{CH(Cl)}\text{CH}_3 \) is a secondary carbon, which is less reactive than primary carbon.
- Option \( c \): \( \text{(CH}_3\text{)}_3\text{CCI} \) is a tertiary carbon, which is highly hindered and has less reactivity in \( \text{S}_\text{N}2 \) reactions.

Step 4: Final conclusion.
Thus, the order of reactivity is: \[ a > d > b > c \] Therefore, the correct option is: \[ \boxed{(1)\ a > d > b > c} \]