Question

An alkene X (C\(_4\)H\(_8\)) exhibits geometrical isomerism. Oxidation of X with KMnO\(_4\)/H\(^+\) gave Y. On heating sodium salt of Y with a mixture of NaOH and CaO gave Z. What is Z?

• A. CH\(_3\)CH\(_3\)
• B. CH\(_3\)CH\(_2\)CH\(_3\)
• C. CH\(_3\)CH\(_2\)CH\(_2\)CH\(_3\)
• D. CH\(_4\)

Hint:

The decarboxylation reaction removes a carbon atom from a carboxylic acid, producing an alkane.

Answer 1

The Correct Option is D

An alkene that exhibits geometrical isomerism must have a carbon-carbon double bond, with each of the two carbons in the double bond attached to two different groups. Of the alkenes with formula C\(_4\)H\(_8\), only 2-butene exhibits geometrical isomerism: \[\text{CH}_3 \text{CH} = \text{CHCH}_3.\]Oxidation of an alkene with KMnO\(_4\)/H\(^+\) cleaves the double bond, and converts each carbon in the double bond to a carbonyl group. Thus, Y is \[\text{CH}_3 \text{COOH}.\]Heating the sodium salt of a carboxylic acid with a mixture of NaOH and CaO produces an alkane with one fewer carbon atom. This reaction is called the decarboxylation reaction. Thus, Z is CH\(_4\).