Question

Which of the following alkenes does not undergo anti-Markovnikov addition of HBr?

• A. Propene
• B. 1-Butene
• C. 2-Butene
• D. 3-Methyl-2-Pentene

Hint:

In the presence of peroxides, the addition of HBr to alkenes follows the anti-Markovnikov rule, where the hydrogen atom attaches to the carbon with fewer alkyl groups.

Answer 1

The Correct Option is C

Anti-Markovnikov addition occurs when HBr adds to an alkene such that the hydrogen atom attaches to the carbon atom with the fewest alkyl groups, resulting in the formation of the less stable carbocation. This addition typically occurs with peroxides (as in the case of the anti-Markovnikov mechanism).
Step 1: Propene (C$_3$H$_6$) undergoes anti-Markovnikov addition of HBr in the presence of peroxides. In this reaction, the hydrogen adds to the carbon with fewer alkyl groups, and the bromine adds to the carbon with more alkyl groups.
Step 2: 1-Butene (C$_4$H$_8$) also undergoes anti-Markovnikov addition of HBr in the presence of peroxides, similar to propene. 
Step 3: 2-Butene (C$_4$H$_8$) undergoes Markovnikov addition of HBr, where the bromine atom attaches to the carbon with more alkyl groups, and the hydrogen attaches to the carbon with fewer alkyl groups. Therefore, 2-Butene does not undergo anti-Markovnikov addition. 
Step 4: 3-Methyl-2-pentene (C$_6$H$_{12}$) undergoes Markovnikov addition and does not follow the anti-Markovnikov addition rule. 
Thus, the correct answer is 2-Butene, which does not undergo anti-Markovnikov addition.

Answer 2

To determine which of the following alkenes does not undergo anti-Markovnikov addition of HBr, let's consider the mechanism and conditions required for the reaction. The anti-Markovnikov addition of HBr occurs in the presence of peroxides. This reaction is possible due to a free radical mechanism where the bromine atom adds to the less substituted carbon of the alkene, leading to the anti-Markovnikov product.

• For Propene: The presence of a terminal double bond allows for anti-Markovnikov addition with HBr.
• For 1-Butene: Similarly, as a terminal alkene, it can undergo anti-Markovnikov addition of HBr.
• For 2-Butene: This is an internal alkene, which does not have the necessary conditions (i.e., presence of a terminal hydrogen) to undergo anti-Markovnikov addition with HBr.
• For 3-Methyl-2-Pentene: As a terminally branched alkene, it can also undergo anti-Markovnikov addition.

Hence, the correct answer is 2-Butene, which does not undergo anti-Markovnikov addition of HBr due to its internal double bond nature.