Question

Match List-I with List-II. \begin{tabbing} \hspace{5cm} \= \hspace{3cm} \= \hspace{4cm} \= \hspace{5cm} \= \kill List-I \hspace{1cm} \= Isomers of \( C_{10}H_{14} \) \hspace{2cm} \= List-II \hspace{2cm} \= Ozonolysis product

• A. \= \hspace{1cm} \= \text{Cyclohexene derivative} \hspace{1cm} \= (I) \hspace{2cm} \= \text{Aldehyde product}
• B. \= \hspace{1cm} \= \text{1,2-Dimethylcyclohexene} \hspace{1cm} \= (II) \hspace{2cm} \= \text{Diketone product}
• C. \= \hspace{1cm} \= \text{1-Methylcyclohexene} \hspace{1cm} \= (III) \hspace{2cm} \= \text{Aldehyde and ketone product}
• D. \= \hspace{1cm} \= \text{1,4-Dimethylcyclohexene} \hspace{1cm} \= (IV) \hspace{2cm} \= \text{Aldehyde product} \end{tabbing}

Hint:

Ozonolysis of alkenes results in cleavage of the double bond and formation of carbonyl compounds (aldehydes or ketones), depending on the structure of the alkene.

Answer 1

The Correct Option is B

(A) Cyclohexene derivative: Ozonolysis leads to two aldehyde groups, hence it matches with (II).
- (B) 1,2-Dimethylcyclohexene: Ozonolysis leads to a diketone, hence it matches with (IV).
- (C) 1-Methylcyclohexene: Ozonolysis gives one aldehyde and one ketone, hence it matches with (I).
- (D) 1,4-Dimethylcyclohexene: Ozonolysis results in two aldehydes, hence it matches with (III). Thus, the correct matching is (A)-(II), (B)-(IV), (C)-(I), (D)-(III).