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(I) Cl-underset(Cl)underset(|)overset(Cl...

(I) `Cl-underset(Cl)underset(|)overset(Cl)overset(|)(C^(3))-overset(2)(C)H=overset(1)(C)H_(2)`
(II) `H_(3)overset(3)(C)-overset(2)(C)H=overset(1)(C)H_(2)`
In addition of HOBr to (I) and (II)

A

Br is at `C_(2)` in both cases

B

Br is at `C_(2)` in II and at `C_(1)` in I

C

Br is at `C_(2)` in I and at `C_(1)` in II

D

Br is at `C_(2)` in both the cases.

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem of the addition of HOBr to the given alkenes (I) and (II), we will follow the mechanism of electrophilic addition. Let's break down the steps for each compound. ### Step-by-Step Solution: #### For Compound (I): `Cl3C=CH2` 1. **Identify the Alkene**: The given structure has a double bond between the carbon atoms (C=C) with three chlorine atoms attached to one carbon. 2. **Electrophilic Addition Mechanism**: HOBr exists in the form of OH⁻ (nucleophile) and Br⁺ (electrophile). The double bond will react with Br⁺ first. 3. **Formation of Carbocation**: When Br⁺ attacks, it will preferentially attack the carbon atom that leads to the more stable carbocation. In this case, the attack will occur at the terminal carbon (C=CH2), forming a secondary carbocation at the carbon with the chlorine substituents. 4. **Intermediate Formation**: The carbocation formed will be: - CCl3-CH⁺ (where the positive charge is on the carbon that was double-bonded to the other carbon). 5. **Nucleophilic Attack by OH⁻**: The hydroxide ion (OH⁻) will then attack the carbocation, leading to the final product: - CCl3-CH(OH)-CH2Br. #### For Compound (II): `H3C=CH2` 1. **Identify the Alkene**: The structure has a double bond between two carbon atoms, both of which are attached to hydrogen atoms. 2. **Electrophilic Addition Mechanism**: Similar to the first compound, the double bond will react with Br⁺ first. 3. **Formation of Carbocation**: The Br⁺ will attack one of the carbon atoms, forming a primary carbocation. 4. **Intermediate Formation**: The carbocation formed will be: - CH3-CH⁺. 5. **Nucleophilic Attack by OH⁻**: The hydroxide ion (OH⁻) will then attack the carbocation, leading to the final product: - CH3-CH(OH)-Br. ### Final Products: - For Compound (I): CCl3-CH(OH)-CH2Br. - For Compound (II): CH3-CH(OH)-Br.
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