Question

The major product of the following reaction is CH₂ = CH - CH₂ - OH product

• A. CH₃ - CHBr - CH₂Br
• B. CH₂ = CH - CH₂Br
• C. CH₃ - CHBr - CH₂ - OH
• D. CH₃ - CHOH - CH₂OH

Answer 1

The Correct Option is A

In the given reaction, the alkene reacts with excess HBr, leading to an electrophilic addition.

• The double bond in \( \text{CH}_2 = \text{CH} - \text{CH}_2 - \text{OH} \) breaks and the hydrogen from HBr adds to the most substituted carbon (according to Markovnikov's rule).
• The bromine atom (Br⁻) adds to the other carbon.
• Since there is excess HBr, both carbons of the double bond will be attached to bromine atoms, leading to the major product being a dibromide.

The correct answer is (A) : CH₃ - CHBr - CH₂Br.

Answer 2

To solve this problem, we need to determine the major product of the given reaction, which involves the addition of HBr to an alkene.

1. Analyzing the Reaction:
The given reaction is the addition of excess HBr to an alkene, specifically to propene (CH₂=CH-CH₃). In an alkene, the addition of HBr proceeds via an electrophilic addition mechanism, where the pi bond of the alkene reacts with the proton (H⁺) from HBr, generating a carbocation intermediate. The bromide ion (Br⁻) then attacks the carbocation, leading to the formation of the final product.

2. Markovnikov's Rule:
According to Markovnikov’s rule, when a protic acid (like HBr) adds to an asymmetrical alkene, the proton (H⁺) will add to the carbon of the double bond that has the greatest number of hydrogen atoms (the more substituted carbon), and the halide ion (Br⁻) will add to the carbon with fewer hydrogen atoms.

3. Reaction Mechanism:
The alkene in the given reaction is propene (CH₂=CH-CH₃). The double bond between the carbon atoms undergoes electrophilic addition:

• The proton (H⁺) from HBr adds to the carbon atom of the double bond with more hydrogens (the CH₂ group), resulting in the formation of a secondary carbocation at the C-2 position (CH₃-CH⁺-CH₃).
• The Br⁻ then attacks the carbocation, forming the major product, which is 2-bromopropane (CH₃-CHBr-CH₃).

4. Identifying the Major Product:
The major product of the reaction is 2-bromopropane (CH₃-CHBr-CH₃), as the bromine attaches to the carbon that was initially part of the double bond with fewer hydrogen atoms (C-2).

5. Reviewing the Options:
- Option (D) "CH₃-CHBr-CH₂Br" is incorrect because this product would imply the addition of two bromine atoms to the alkene, which is not the case in this reaction.
- Option (B) "CH₂=CH-CH₂Br" is incorrect because it suggests the formation of a terminal alkene, but the reaction with HBr leads to a saturated product.
- Option (C) "CH₃-CH₂-CH₂OH" is incorrect because the product should contain bromine, not a hydroxyl group.
- Option (A) "CH₃-CHBr-CH₃" is correct because it is the major product formed when excess HBr adds to the alkene under the given conditions.

Final Answer:
The major product of the reaction is (A) "CH₃-CHBr-CH₃".