Question:

Number of isomeric products formed by monochlorination of 2-methylbutane in presence of sunlight is _______.

Updated On: Nov 5, 2025
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Correct Answer: 6

Approach Solution - 1

To determine the number of isomeric products formed by monochlorination of 2-methylbutane in the presence of sunlight, we need to consider the possible different hydrogen atoms that can be replaced by a chlorine atom. 2-Methylbutane, also known as isopentane, has the structural formula:

           CH3
           |
CH3-CH-CH2-CH3
           |
           CH3

The molecule has a total of four types of hydrogen positions: the primary carbon hydrogens (1°) on both terminal methyl (CH3) groups, the secondary carbon hydrogens (2°) on the middle methylene (CH2) group, and the tertiary carbon hydrogen (3°) on the central carbon bonded to three other carbons. By analyzing these positions, we can determine the distinct monochlorinated products:

  1. Chlorination at a primary hydrogen on the terminal methyl group generates 1-chloro-2-methylbutane or 2-chloro-2-methylbutane (2 each due to symmetry).
  2. Chlorination at the secondary hydrogen on the methylene group forms 3-chloro-2-methylbutane.
  3. Chlorination at the tertiary hydrogen on the central carbon forms 2-chloro-2-methylbutane.

Overall, this results in 6 distinct constitutional isomers:

PositionIsomers
1° (external CH3)1-chloro-2-methylbutane (2 isomers)
1° (internal CH3)2-chloro-2-methylbutane
2° (CH2)3-chloro-2-methylbutane
3° (internal CH)2-chloro-2-methylbutane

The number of isomeric products is thus 6. This value is confirmed to fall within the expected range of 6,6.

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Approach Solution -2

In the monochlorination of 2-methylbutane, chlorine can replace a hydrogen atom at different positions, leading to the formation of structural and stereoisomeric products. The possible isomeric products are:

\[ \text{CH}_3 - \text{CH}(\text{Cl}) - \text{CH}_2 - \text{CH}(\text{CH}_3) - \text{CH}_3 \, (\pm) \] \[ \text{CH}_3 - \text{CH}(\text{CH}_3) - \text{CH}_2 - \text{CH}(\text{Cl}) - \text{CH}_3 \] \[ \text{CH}_3 - \text{C}(\text{Cl})(\text{CH}_3) - \text{CH}_2 - \text{CH}_3 \, (\pm) \]

Therefore, there are 6 isomeric products.

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