Question

Which one of the following compound will show addition of HBr to alkene according to markownikoff's rule?

• A. But-2-ene
• B. Hex-3-ene
• C. Ethene
• D. 2-methyl propene

Hint:

Look for the alkene whose two double-bond carbons are different.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Markovnikov rule: when HBr adds to an unsymmetrical alkene, the hydrogen goes to the carbon that already has more hydrogens, and bromine goes to the carbon with fewer hydrogens. The rule only has meaning if the two double-bond carbons are different.

Step 2: Check each alkene:
But-2-ene \(CH_3CH=CHCH_3\) is symmetrical. Hex-3-ene \(CH_3CH_2CH=CHCH_2CH_3\) is symmetrical. Ethene \(CH_2=CH_2\) is symmetrical. In each, both ways of adding give the same product, so there is no choice to make.

Step 3: The unsymmetrical one:
2-Methylpropene is \((CH_3)_2C=CH_2\). One carbon has two methyl groups and no H, the other has two H atoms. So the rule applies.

Step 4: Product:
H adds to \(CH_2\) and Br to the carbon with two methyls, forming the more stable tertiary carbocation:
\[ (CH_3)_2C=CH_2 + HBr \rightarrow (CH_3)_3C-Br \]
So the answer is option (D).

Final Answer:
Only 2-methylpropene is unsymmetrical, so Markovnikov addition applies. \[ \boxed{D:\ \text{2-methyl propene}} \]