Question

In a reaction sequence involving haloalkanes, the compound \( \text{CH}_3\text{CH}_2\text{Br} \) is treated with alcoholic KOH followed by ozonolysis. If the final product obtained is methanal, then identify the alkene formed in the intermediate step.

• A. \( \text{Ethene} \)
• B. \( \text{Propene} \)
• C. \( \text{But-1-ene} \)
• D. \( \text{2-Methylpropene} \)

Hint:

In ozonolysis:
• Terminal alkene carbon having two hydrogens produces methanal.
• Breaking the double bond helps reconstruct the original alkene.
• Alcoholic KOH always favors elimination in haloalkanes.

Answer 1

The Correct Option is A

Concept:
Haloalkanes undergo elimination reactions in the presence of alcoholic KOH to form alkenes. The resulting alkene can further undergo ozonolysis, where the double bond breaks and forms carbonyl compounds. Ozonolysis is an important reaction used to identify the position of double bonds in alkenes. The key concepts involved are:
• Dehydrohalogenation: Removal of HX from haloalkanes using alcoholic KOH.
• Ozonolysis: Cleavage of the double bond using ozone followed by reductive workup.
• Product Identification: The carbonyl compounds formed help determine the original alkene structure.

Step 1: Formation of alkene from haloalkane.
The given haloalkane is: \[ \text{CH}_3\text{CH}_2\text{Br} \] When treated with alcoholic KOH, elimination of HBr occurs: \[ \text{CH}_3\text{CH}_2\text{Br} \xrightarrow[\Delta]{alc.\ KOH} \text{CH}_2=\text{CH}_2 \] Thus, the alkene formed is ethene.

Step 2: Ozonolysis of the alkene formed.
Ethene undergoes ozonolysis as: \[ \text{CH}_2=\text{CH}_2 \xrightarrow[\text{Zn/H}_2\text{O}]{O_3} 2\text{HCHO} \] The product formed is methanal (formaldehyde).

Step 3: Verification using product analysis.
Since methanal is formed as the only product, the original alkene must contain two terminal hydrogen atoms around the double bond, which is only possible in ethene. Therefore, the intermediate alkene formed is: \[ \boxed{\text{Ethene}} \]