Question

Which of the following haloalkanes will give more than one isomeric product, on being heated with alc. KOH?

• A. 1-Chloro-2-methylbutane
• B. 1-Chloropentane
• C. 2-Chloro-3,3-dimethylpentane
• D. 2-Chlorobutane

Hint:

In elimination reactions with alcoholic KOH, check all possible \(\beta\)-hydrogens. If different \(\beta\)-hydrogens lead to different alkenes, more than one isomeric product is formed.

Answer 1

The Correct Option is D

Step 1: Understand the reaction with alcoholic KOH.
When haloalkanes are heated with alcoholic KOH, they undergo \(\beta\)-elimination reaction.
In this reaction, hydrogen is removed from a \(\beta\)-carbon and halogen is removed from the \(\alpha\)-carbon, forming an alkene.
\[ R-CH_2-CHX-R' \xrightarrow{alc. \, KOH} Alkene \]

Step 2: Check option (A), 1-chloro-2-methylbutane.
In 1-chloro-2-methylbutane, chlorine is present on terminal carbon.
Elimination can occur mainly from the adjacent \(\beta\)-carbon only.
So, it gives only one major alkene product.
Therefore, option (A) is not correct.

Step 3: Check option (B), 1-chloropentane.
In 1-chloropentane, chlorine is present on carbon \(1\).
There is only one adjacent \(\beta\)-carbon available for elimination.
Hence, it forms only pent-1-ene as the elimination product.
\[ CH_3CH_2CH_2CH_2CH_2Cl \xrightarrow{alc. \, KOH} CH_2=CHCH_2CH_2CH_3 \] Therefore, option (B) is not correct.

Step 4: Check option (C), 2-chloro-3,3-dimethylpentane.
In 2-chloro-3,3-dimethylpentane, the \(\alpha\)-carbon is carbon \(2\).
The adjacent \(\beta\)-carbons are carbon \(1\) and carbon \(3\).
However, carbon \(3\) is a quaternary carbon and has no \(\beta\)-hydrogen.
Therefore, elimination can occur only from carbon \(1\).
So, only one alkene product is formed.

Step 5: Check option (D), 2-chlorobutane.
In 2-chlorobutane, chlorine is present on carbon \(2\).
There are two different \(\beta\)-carbons available, carbon \(1\) and carbon \(3\).
Removal of hydrogen from carbon \(1\) gives but-1-ene.
\[ CH_3CHClCH_2CH_3 \xrightarrow{alc. \, KOH} CH_2=CHCH_2CH_3 \] Removal of hydrogen from carbon \(3\) gives but-2-ene.
\[ CH_3CHClCH_2CH_3 \xrightarrow{alc. \, KOH} CH_3CH=CHCH_3 \]

Step 6: Identify more than one isomeric product.
2-Chlorobutane gives but-1-ene and but-2-ene.
Also, but-2-ene can exist as cis and trans geometrical isomers.
Thus, 2-chlorobutane gives more than one isomeric product on heating with alcoholic KOH.
Final Answer:
The haloalkane which gives more than one isomeric product is:
\[ \boxed{\text{2-Chlorobutane}} \]