Question

Identify X and Y in the following reaction $\text{CH}_{2} = \text{CH}_{2} \xrightarrow{\text{(i) X}} \text{CH}_{3}\text{CH}_{2}\text{I} \xrightarrow{\text{(ii) Y}}$

• A. \(\text{HBr, NaI/dry CH}_{3}\text{COCH}_{3}\)
• B. $\text{HBr},\ \text{I}_{2}/\text{dry CH}_{3}\text{COCH}_{3}$
• C. $\text{Br}_{2},\ \text{NaI}/\text{dry CH}_{3}\text{COCH}_{3}$
• D. $\text{Br}_{2},\ \text{I}_{2}/\text{dry CH}_{3}\text{COCH}_{3}$

Hint:

In reactions like these, alkyl halides are produced through the combination of alkene and halogen acids.

Answer 1

The Correct Option is A

To identify X and Y in the given reaction, we need to understand the chemical transformations taking place:

The reaction is: \(\text{CH}_{2} = \text{CH}_{2} \xrightarrow{\text{(i) X}} \text{CH}_{3}\text{CH}_{2}\text{I} \xrightarrow{\text{(ii) Y}}\)

Step 1: Convert ethene \(\text{(CH}_{2}=\text{CH}_{2})\) to ethyl iodide \(\text{(CH}_{3}\text{CH}_{2}\text{I})\).
This conversion suggests that ethene is first converted into ethyl bromide \(\text{(CH}_{3}\text{CH}_{2}\text{Br})\) using \(\text{HBr}\), a process that involves adding HBr across the double bond of ethene via a hydrohalogenation reaction.

Step 2: Convert ethyl bromide \(\text{(CH}_{3}\text{CH}_{2}\text{Br})\) to ethyl iodide \(\text{(CH}_{3}\text{CH}_{2}\text{I})\).
This transformation can be achieved using NaI in a dry acetone solvent (\(\text{NaI/dry CH}_{3}\text{COCH}_{3}\)), which is known as the Finkelstein reaction. In this reaction, the bromine in ethyl bromide is replaced by iodine, producing ethyl iodide.

Conclusion: Therefore, X is \(\text{HBr}\) and Y is \(\text{NaI/dry CH}_{3}\text{COCH}_{3}\).