Question

Species A, B, C, D formed in the following bond cleavages respectively are

• A. \[ \text{CH}_3\text{CH}_2^+ , \; \text{I}^- , \; \text{CH}_3\text{CH}_2^- , \; \text{Cu}^+ \]
• B. \[ \text{CH}_3\text{CH}_2^+ , \; \text{I}^- , \; \text{CH}_3\text{CH}_2^- , \; \text{Cu} \]
• C. \[ \text{CH}_3\text{CH}_2^- , \; \text{I}^+ , \; \text{CH}_3\text{CH}_2^+ , \; \text{Cu}^+ \]
• D. \[ \text{CH}_3\text{CH}_2^- , \; \text{I}^+ , \; \text{CH}_3\text{CH}_2^+ , \; \text{Cu} \]

Hint:

For heterolytic bond cleavage, consider the electronegativity of the elements involved. The more electronegative element typically takes the electrons, forming a negative ion.

Answer 1

The Correct Option is A

Step 1: Understanding the bond cleavage 
1. The first reaction: \[ \text{CH}_3\text{CH}_2 - I \quad \longrightarrow \quad \text{CH}_3\text{CH}_2^+ + I^- \] This is a typical ionization reaction where ethyl iodide (\( \text{CH}_3\text{CH}_2I \)) dissociates to form the ethyl carbocation (\( \text{CH}_3\text{CH}_2^+ \)) and iodide anion (\( I^- \)). 2. The second reaction: \[ \text{CH}_3\text{CH}_2 - Cu \quad \longrightarrow \quad \text{CH}_3\text{CH}_2^+ + Cu^+ \] This follows a similar dissociation mechanism where ethyl copper compound breaks into the ethyl carbocation (\( \text{CH}_3\text{CH}_2^+ \)) and copper ion (\( Cu^+ \)). Step 2: Verifying the correct answer 
From the reactions: - \( A = \text{CH}_3\text{CH}_2^+ \)
- \( B = I^- \)
- \( C = \text{CH}_3\text{CH}_2^+ \)
- \( D = Cu^+ \)
Thus, the correct answer is: 

\((A) \text{CH}_3\text{CH}_2^+, I^- , \text{CH}_3\text{CH}_2^+, Cu^+\)