In which of the following pairs, indicated bond is of greater strength :
`CH_(3) - CH = CH underset(uarr" ")-Br" and "CH_(3) underset(rightarrow)-underset(Br)underset(|)CH - CH_(3)`

Answer

To determine which bond is stronger between the two given compounds, we need to analyze the stability of the resulting carbocations after the bond breaks. Let's break down the solution step by step.

### Step 1: Identify the Compounds
We have two compounds:
1. **Compound A**: \( CH_3 - CH = CH \uparrow Br \)
2. **Compound B**: \( CH_3 \uparrow Br - CH - CH_3 \)

### Step 2: Draw the Carbocations
When the bond with bromine (Br) breaks, it will leave behind a carbocation in both cases.

- For **Compound A**:
- The bond between the \( CH \) and \( Br \) breaks, resulting in:
\[ CH_3 - CH^+ - CH \]
Here, the positive charge is on the sp² hybridized carbon.

- For **Compound B**:
- The bond between the \( Br \) and \( CH \) breaks, resulting in:
\[ CH_3 - CH^+ - CH_3 \]
Here, the positive charge is on a carbon that is sp³ hybridized.

### Step 3: Analyze Carbocation Stability
The stability of carbocations can be influenced by several factors, including hybridization and inductive effects.

- **Compound A**: The carbocation formed has a positive charge on an sp² hybridized carbon. This carbon is more electronegative and less stable due to the lack of electron-donating groups around it. The sp² carbon has less hyperconjugation compared to sp³.

- **Compound B**: The carbocation formed has a positive charge on an sp³ hybridized carbon. The presence of two \( CH_3 \) groups provides a +I (inductive) effect, which stabilizes the positive charge through hyperconjugation and electron donation.

### Step 4: Compare Bond Strength
Since the stability of the carbocation formed from Compound B is greater than that from Compound A, we can conclude that the bond in Compound A is stronger. This is because a more stable carbocation means the bond is more likely to break easily.

### Conclusion
The bond in **Compound A** ( \( CH_3 - CH = CH \uparrow Br \) ) is of greater strength compared to the bond in **Compound B** ( \( CH_3 \uparrow Br - CH - CH_3 \) ).

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