Question:medium
The order of reactivity of the following compounds towards \( S_N2 \) displacement reaction is (i) 2-Bromo-2-methylbutane (ii) 1-Bromopentane (iii) 2-Bromopentane
The order of reactivity of the following compounds towards \( S_N2 \) displacement reaction is (i) 2-Bromo-2-methylbutane (ii) 1-Bromopentane (iii) 2-Bromopentane
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Less steric hindrance $\Rightarrow$ faster \( S_N2 \).
Updated On: May 10, 2026
- (ii) > (i) > (iii)
- (iii) > (i) > (ii)
- (ii) > (iii) > (i)
- (i) > (ii) > (iii)
- (iii) > (ii) > (i)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the S\(_N\)2 Reaction Mechanism:
The S\(_N\)2 (bimolecular nucleophilic substitution) reaction is a single-step process where a nucleophile attacks the carbon atom bearing the leaving group from the backside. The rate of the S\(_N\)2 reaction is highly dependent on steric hindrance. Bulky groups around the reaction center block the nucleophile's path, slowing down or preventing the reaction.
Step 2: Classifying the Alkyl Halides:
We need to determine the type (primary, secondary, or tertiary) of each alkyl halide, as this is the primary factor determining steric hindrance.
(i) 2-Bromo-2-methylbutane: The bromine atom is attached to a carbon that is bonded to three other carbon atoms. This is a tertiary (3\(^\circ\)) alkyl halide.
(ii) 1-Bromopentane: The bromine atom is attached to a carbon that is bonded to only one other carbon atom. This is a primary (1\(^\circ\)) alkyl halide.
(iii) 2-Bromopentane: The bromine atom is attached to a carbon that is bonded to two other carbon atoms. This is a secondary (2\(^\circ\)) alkyl halide.
Step 3: Determining the Order of Reactivity:
The reactivity order for S\(_N\)2 reactions based on steric hindrance is:
Methyl halide > Primary (1\(^\circ\)) > Secondary (2\(^\circ\)) > Tertiary (3\(^\circ\))
Tertiary halides are the most sterically hindered and generally do not undergo S\(_N\)2 reactions. Primary halides are the least hindered (among these options) and react the fastest.
Applying this order to the given compounds:
1-Bromopentane (1\(^\circ\)) > 2-Bromopentane (2\(^\circ\)) > 2-Bromo-2-methylbutane (3\(^\circ\))
This corresponds to the order: (ii) > (iii) > (i).
Step 4: Final Answer:
The correct order of reactivity towards S\(_N\)2 displacement is (ii) > (iii) > (i).
The S\(_N\)2 (bimolecular nucleophilic substitution) reaction is a single-step process where a nucleophile attacks the carbon atom bearing the leaving group from the backside. The rate of the S\(_N\)2 reaction is highly dependent on steric hindrance. Bulky groups around the reaction center block the nucleophile's path, slowing down or preventing the reaction.
Step 2: Classifying the Alkyl Halides:
We need to determine the type (primary, secondary, or tertiary) of each alkyl halide, as this is the primary factor determining steric hindrance.
(i) 2-Bromo-2-methylbutane: The bromine atom is attached to a carbon that is bonded to three other carbon atoms. This is a tertiary (3\(^\circ\)) alkyl halide.
(ii) 1-Bromopentane: The bromine atom is attached to a carbon that is bonded to only one other carbon atom. This is a primary (1\(^\circ\)) alkyl halide.
(iii) 2-Bromopentane: The bromine atom is attached to a carbon that is bonded to two other carbon atoms. This is a secondary (2\(^\circ\)) alkyl halide.
Step 3: Determining the Order of Reactivity:
The reactivity order for S\(_N\)2 reactions based on steric hindrance is:
Methyl halide > Primary (1\(^\circ\)) > Secondary (2\(^\circ\)) > Tertiary (3\(^\circ\))
Tertiary halides are the most sterically hindered and generally do not undergo S\(_N\)2 reactions. Primary halides are the least hindered (among these options) and react the fastest.
Applying this order to the given compounds:
1-Bromopentane (1\(^\circ\)) > 2-Bromopentane (2\(^\circ\)) > 2-Bromo-2-methylbutane (3\(^\circ\))
This corresponds to the order: (ii) > (iii) > (i).
Step 4: Final Answer:
The correct order of reactivity towards S\(_N\)2 displacement is (ii) > (iii) > (i).
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