Question

Markovnikov's rule is applicable to which of the following reactions?

• A. Addition of HBr to propene
• B. Addition of HBr to propene in the presence of peroxides
• C. Addition of Br₂ to ethene
• D. Hydroboration-oxidation of alkenes

Hint:

Remember: "The rich get richer" - hydrogen adds to the carbon with more hydrogens.

Answer 1

The Correct Option is A

Step 1: Markovnikov's Rule

Markovnikov's rule states that in the electrophilic addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens, and the halogen to the carbon with fewer hydrogens, forming the more stable carbocation.

Step 2: Evaluate Options

Option (a): HBr to propene

Propene (\(CH_3-CH=CH_2\)) reacts with HBr:

\[ CH_3-CH=CH_2 + \text{HBr} \rightarrow CH_3-CHBr-CH_3 \]

H\(^+\) adds to the \(CH_2\) carbon (more hydrogens), and Br\(^-\) to the \(CH\) carbon, following Markovnikov's rule.

Result: Applicable.

Option (b): HBr to propene with peroxides

With peroxides, HBr addition to propene (\(CH_3-CH=CH_2\)) is free radical, yielding anti-Markovnikov product:

\[ CH_3-CH=CH_2 \rightarrow CH_3-CH_2-CH_2Br \]

Br adds to the \(CH_2\) carbon, opposite to Markovnikov's rule.

Result: Not applicable.

Option (c): Br\(_2\) to ethene

Ethene (\(CH_2=CH_2\)) is symmetrical, and Br\(_2\) addition forms:

\[ CH_2=CH_2 + \text{Br}_2 \rightarrow CH_2Br-CH_2Br \]

No regioselectivity is needed, so Markovnikov's rule does not apply.

Result: Not applicable.

Option (d): Hydroboration-oxidation

Hydroboration-oxidation of propene (\(CH_3-CH=CH_2\)) gives:

\[ CH_3-CH=CH_2 \xrightarrow{\text{BH}_3, \text{H}_2\text{O}_2, \text{OH}^-} CH_3-CH_2-CH_2OH \]

Boron adds to the less substituted carbon, yielding anti-Markovnikov addition.

Result: Not applicable.

Step 3: Final Answer

Only option (a) follows Markovnikov's rule.

\[ \boxed{\text{(a)}} \]