`CH_(3)-underset((a))(C)-=C-CH_(2)-CHunderset((b))(=)CH_(2)`
Reactivity toward electrophilic addition reaction.

Options

A. `a gt b`
B. `b gt a`
C. `a = b`
D. Cannot predict

Answer

To determine the reactivity of the given compound toward electrophilic addition reactions, we need to analyze the structure and hybridization of the involved functional groups, which are an alkene and an alkyne. Let's break down the steps:

### Step 1: Identify the Functional Groups
The given compound is:
\[ CH_3-\overset{(a)}{C}=\overset{(b)}{C}-CH_2-CH=\overset{(b)}{CH_2} \]

Here, we have:
- An alkene (C=C) at position (b)
- An alkyne (C≡C) at position (a)

### Step 2: Determine the Hybridization
- For the alkene (C=C), each carbon is sp² hybridized. This is because there are one sigma bond and one pi bond between the two carbon atoms.
- For the alkyne (C≡C), each carbon is sp hybridized. This is due to the presence of one sigma bond and two pi bonds.

### Step 3: Analyze Electrophilic Addition
Electrophilic addition reactions involve the attack of an electrophile on a nucleophile. In this case:
- The electrophile (e.g., Br₂) will attack the site with higher electron density.
- The hybridization affects the electron density around the carbon atoms.

### Step 4: Compare Electronegativity and Electron Density
- The sp hybridized carbon (in the alkyne) has a higher s-character (50%) compared to the sp² hybridized carbon (in the alkene) which has 33% s-character.
- Higher s-character means that the electrons are held closer to the nucleus, making the carbon less electron-rich and more electronegative.

### Step 5: Conclusion on Reactivity
- Since the alkene (sp²) is less electronegative and has a higher electron density than the alkyne (sp), the electrophile will preferentially attack the alkene.
- Therefore, the reactivity toward electrophilic addition will be higher at the alkene site than at the alkyne site.

### Final Answer
The reactivity toward electrophilic addition is greater at the alkene (b) than at the alkyne (a).

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