Question

The one giving maximum number of isomeric alkenes on dehydrohalogenation reaction is (excluding rearrangement):

• A. 1-Bromo-2-methylbutane
• B. 2-Bromopropane
• C. 2-Bromopentane
• D. 2-Bromo-3,3-dimethylpentane

Hint:

Always look for a \(\beta\)-carbon that allows for double bond formation in two different directions, and check if the resulting internal alkene is substituted enough to have E/Z isomers.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept: 
Dehydrohalogenation (an E2 elimination reaction) removes a halogen atom and a hydrogen atom from an adjacent ($\beta$) carbon, creating a carbon-carbon double bond. The number of isomeric alkenes depends on the number of different types of $\beta$-hydrogens and the possibility of geometrical (cis/trans or E/Z) isomerism in the resulting alkenes. 
Step 2: Detailed Explanation: 
Let's analyze each option independently: 
(a) 1-Bromo-2-methylbutane: $\text{Br-CH}_2\text{-CH(CH}_3\text{)-CH}_2\text{-CH}_3$ 
The halogen is at C1. The only adjacent carbon is C2. 
Removal of H from C2 gives: $\text{CH}_2\text{=C(CH}_3\text{)-CH}_2\text{-CH}_3$ (2-methylbut-1-ene). 
This terminal alkene does not exhibit geometrical isomerism. Total isomers = 1. 
(b) 2-Bromopropane: $\text{CH}_3\text{-CH(Br)-CH}_3$ 
The $\beta$-carbons are identical (both are methyl groups). 
Elimination gives: $\text{CH}_2\text{=CH-CH}_3$ (Propene). 
Propene has no geometrical isomerism. Total isomers = 1. 
(c) 2-Bromopentane: $\text{CH}_3\text{-CH(Br)-CH}_2\text{-CH}_2\text{-CH}_3$ 
There are two different adjacent carbons with hydrogens: C1 and C3. 
Elimination with C1 gives: $\text{CH}_2\text{=CH-CH}_2\text{-CH}_2\text{-CH}_3$ (Pent-1-ene) $\rightarrow$ No geometrical isomerism (1 isomer). 
Elimination with C3 gives: $\text{CH}_3\text{-CH=CH-CH}_2\text{-CH}_3$ (Pent-2-ene). 
Pent-2-ene can exist as both {cis}- and {trans}- isomers (2 isomers). 
Total isomers = 1 + 2 = 3. 
(d) 2-Bromo-3,3-dimethylpentane: $\text{CH}_3\text{-CH(Br)-C(CH}_3\text{)}_2\text{-CH}_2\text{-CH}_3$ 
The $\beta$-carbons are C1 and C3. However, C3 is a quaternary carbon (it has no hydrogens attached to it!). 
Therefore, elimination can only occur by removing a hydrogen from C1. 
Product: $\text{CH}_2\text{=CH-C(CH}_3\text{)}_2\text{-CH}_2\text{-CH}_3$ (3,3-dimethylpent-1-ene). 
No geometrical isomerism possible. Total isomers = 1.
 Step 3: Final Answer: 
2-Bromopentane gives the maximum number of isomeric alkenes (3 isomers).