Question

An alkyl halide with molecular formula C$_5$H$_11$Br on dehydrohalogenation gives two isomeric alkenes X and Y with formula C$_5$H$_10$. On reductive ozonolysis, X and Y give four compounds CH$_3$CHO, (CH$_3$)$_2$CHO, CH$_3$CO, HCHO. The alkyl halide is:

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

Hint:

Ozonolysis of alkenes splits the carbon-carbon double bond, producing two carbonyl compounds (aldehydes or ketones). The structure of the products helps deduce the original alkene and thus the starting alkyl halide.

Answer 1

The Correct Option is B

Given the information about the alkyl halide and its ozonolysis products, let us first analyze the products of ozonolysis: - CH₃CHO (acetaldehyde), (CH₃)₂CHO (isopropanol), CH₃CO (acetone), and HCHO (formaldehyde) are typical products of ozonolysis when alkenes undergo oxidative cleavage. In the ozonolysis reaction, the carbon-carbon double bond is cleaved, forming two carbonyl compounds (aldehydes or ketones). The pattern of products provides insight into the structure of the original alkenes.
1. The compound CH₃CHO (acetaldehyde) indicates a terminal double bond (i.e., one alkene carbon is part of an aldehyde group).
2. The compound (CH₃)₂CHO (isopropanol) suggests a branching alkene where the central carbon is attached to two methyl groups.
3. CH₃CO (acetone) suggests another ketone product formed from cleavage.
4. HCHO (formaldehyde) indicates that one of the products is a smaller aldehyde. \begin{center} \includegraphics[width=0.4\textwidth]{q26_diagram.png} % Replace with actual path \end{center} Step-by-Step Reaction: - The alkyl halide likely has a structure that forms two isomeric alkenes (X and Y) with a carbon-carbon double bond in different positions or orientations. - Upon ozonolysis, the two alkenes produce acetaldehyde, acetone, isopropanol, and formaldehyde. - The only alkyl halide from the choices that could produce this combination of ozonolysis products is 2-Bromo-3-methylbutane, which after dehydrohalogenation, forms two alkenes: 1-butene and 2-butene. These two alkenes upon ozonolysis give the products as described above.
Thus, the alkyl halide is 2-Bromo-3-methylbutane. The ozonolysis reaction for 1-butene and 2-butene: \[ \text{CH}_2=\text{CH}_2 + O_3 \rightarrow \text{CH}_3\text{CHO} + \text{HCHO} \] \[ \text{CH}_3\text{CH}=\text{CH}_2 + O_3 \rightarrow \text{CH}_3\text{CO} + \text{(CH}_3\text{)}_2\text{CHO} \] This gives the expected four products.