Which among the following compounds will undergo fastest SN2 reaction.

Step 1: Understand $S_N2$ mechanism

$S_N2$ (bimolecular nucleophilic substitution):

• Backside attack by nucleophile
• Requires accessible carbon (less steric hindrance)
• Transition state has partial bonds

Step 2: Analyze ring strain

Ring strain in cycloalkanes:

| Ring Size | Strain | Angle Deviation | |-----------|--------|------------------| | Cyclopropane | Highest | 109.5° → 60° | | Cyclobutane | High | 109.5° → 90° | | Cyclopentane | Low | ~108° | | Cyclohexane | Minimal | ~109.5° |

Step 3: Relate strain to reactivity

For $S_N2$ reactions with cycloalkyl halides:

Cyclopropyl bromide:

• High ring strain (60° bond angles)
• C-Br bond is weak (high s-character)
• Breaking C-Br relieves ring strain
• Transition state is stabilized
• Fastest $S_N2$ reaction ✓

Cyclobutane, cyclopentane, cyclohexane:

• Less ring strain
• Normal C-Br bond strength
• Slower $S_N2$ reactions

Step 4: Order of reactivity

Cyclopropyl > Cyclobutyl > Cyclopentyl ≈ Cyclohexyl

Answer: (d) Cyclopropyl bromide

Key Concepts:

• Ring strain increases reactivity in $S_N2$
• Cyclopropane: highest strain, most reactive
• Breaking C-X bond in strained ring → relief of strain → faster reaction
• Steric hindrance decreases $S_N2$ rate (but not the main factor here)

Note: This is an exception to the general rule that less hindered substrates react faster. Here, strain relief is the dominant factor.