See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Ionization of Alkyl Bromides **Step 1: Identify the structures** - **I.** $CH_2=CH-CHBr-CH_3$ (Br on C-3, a secondary allylic position) - **II.** $CH_2=CH-CHBr-CH_3$ (Br on C-1, a primary allylic position) Wait—both structures appear identical as written. Re-examining: Structure I has Br on the internal carbon of the double bond region, while II has Br on the terminal carbon. **Step 2: Ionization mechanism** Both compounds undergo heterolytic cleavage: $$R-Br \rightarrow R^+ + Br^-$$ Both produce the **same bromide anion**: $Br^-$ **Step 3: Analyze the carbocation formed** In both compounds, loss of Br⁻ generates an allylic carbocation that can be stabilized by resonance with the adjacent $C=C$ double bond. Due to resonance delocalization, **both form the same allylic carbocation**: $$CH_2=CH-\overset{+}{CH}-CH_3 \leftrightarrow CH_2^{\oplus}-CH=CH-CH_3$$ **Step 4: Conclusion** - Same anion produced: $Br^-$ ✓ - Same cation produced: Allylic carbocation (resonance stabilized) ✓ **Answer: (A) same cation** — Both compounds yield the same allylic cation and the same bromide anion upon ionization.