Question:
Which of the following four compounds (I to IV) are correctly arranged in decreasing order of reactivity towards \(S_{N}2\) reaction?
I. 1-Bromobutane
II. 1-Bromo-2-methylbutane
III. 1-Bromo-2,2-dimethylpropane
IV. 1-Bromo-3-methylbutane
Which of the following four compounds (I to IV) are correctly arranged in decreasing order of reactivity towards \(S_{N}2\) reaction?
I. 1-Bromobutane
II. 1-Bromo-2-methylbutane
III. 1-Bromo-2,2-dimethylpropane
IV. 1-Bromo-3-methylbutane
I. 1-Bromobutane
II. 1-Bromo-2-methylbutane
III. 1-Bromo-2,2-dimethylpropane
IV. 1-Bromo-3-methylbutane
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\(S_N2\) reactions are extremely sensitive to steric hindrance. Branching near the reactive carbon slows the reaction drastically.
Updated On: Jun 7, 2026
- \(I > IV > III > II\)
- \(I > II > III > IV\)
- \(I > III > IV > II\)
- \(IV > III > II > I\)
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The Correct Option is A
Solution and Explanation
Concept:
\(S_N2\) reactions occur through backside attack. Greater steric hindrance near the reacting carbon decreases the reaction rate.
Step 1: Analyze steric hindrance in each compound.
• I: 1-Bromobutane is unbranched \(\rightarrow\) least hindered
• IV: Branching is far from the reaction center
• III: Greater steric crowding near the reactive site
• II: Maximum branching near the reacting carbon
Step 2: Arrange in decreasing \(S_N2\) reactivity.
Least hindered compounds react fastest: \[ I > IV > III > II \] Hence: \[ \boxed{(A)} \]
Step 1: Analyze steric hindrance in each compound.
• I: 1-Bromobutane is unbranched \(\rightarrow\) least hindered
• IV: Branching is far from the reaction center
• III: Greater steric crowding near the reactive site
• II: Maximum branching near the reacting carbon
Step 2: Arrange in decreasing \(S_N2\) reactivity.
Least hindered compounds react fastest: \[ I > IV > III > II \] Hence: \[ \boxed{(A)} \]
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