Question

What is the major product in the following reaction? 

                   CH₃-CH₂-CH₂-CH=CH₂+HBr\(\to\)

• A. CH₃-CH₂-CH₂-CH₂-CH₂-Br
• B.
• C.
• D.
• E. CH₃-CH₂-CH₂-CH₂-CH₃

Answer 1

The Correct Option is B

Reaction Mechanism: Electrophilic Addition

• Protonation: The hydrogen from HBr (the electrophile) adds to one of the carbons of the double bond, forming a carbocation intermediate.
• Bromide Attack: The bromide ion (Br-) acts as a nucleophile and attacks the carbocation, forming the alkyl halide product.

Markovnikov's Rule

In the addition of HX (where X is a halogen) to an alkene, the hydrogen adds to the carbon with more hydrogens already attached (the less substituted carbon), and the halogen adds to the carbon with fewer hydrogens attached (the more substituted carbon). This rule is also stated as hydrogen will add to the carbon which will produce the most stable (most substituted) carbocation.

Applying Markovnikov's Rule to the Given Reaction

In the given reaction:

CH3-CH2-CH2-CH=CH2 + HBr

The double bond is between carbons 4 and 5 (numbering from the left).

Carbon 4 (CH=) has one hydrogen attached.

Carbon 5 (=CH2) has two hydrogens attached.

According to Markovnikov's rule, the hydrogen from HBr will add to carbon 5 (the carbon with more hydrogens), and the bromine will add to carbon 4 (the carbon with fewer hydrogens). This gives the following major product:

CH3-CH2-CH2-CHBr-CH3

Therefore, the correct answer is (B) CH3-CH2-CH2-CHBr-CH3

Answer 2

1. Reaction analysis:

The given reaction involves the addition of hydrogen bromide (\( \text{HBr} \)) to 1-pentene (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2 \)). This is an example of an electrophilic addition reaction, where \( \text{HBr} \) adds across the double bond. The major product depends on the regioselectivity of the reaction, which is governed by Markovnikov's rule.

2. Apply Markovnikov's Rule:

• Markovnikov's rule states that in the addition of \( \text{HX} \) (where \( \text{X} \) is a halogen) to an alkene:
  • The hydrogen atom (\( \text{H}^+ \)) adds to the carbon atom of the double bond that has the more hydrogen atoms (i.e., the less substituted carbon).
  • The halogen atom (\( \text{X}^- \)) adds to the carbon atom of the double bond that has the fewer hydrogen atoms (i.e., the more substituted carbon).
• In the given alkene (\( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2 \)):
  • The left carbon of the double bond is bonded to three hydrogen atoms.
  • The right carbon of the double bond is bonded to only one hydrogen atom.
• According to Markovnikov's rule:
  • The \( \text{H}^+ \) will add to the left carbon (more hydrogen atoms).
  • The \( \text{Br}^- \) will add to the right carbon (fewer hydrogen atoms).

3. Write the major product:

• Following Markovnikov's rule, the major product is:
  • \[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CHBr}\text{CH}_3 \]

4. Match the product with the given options:

• The major product is \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CHBr}\text{CH}_3 \), which corresponds to Option (B).