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

We consider 2-methylbutane of formula $$C_5H_{12}$$ and number its longest chain as C1-C4 so that the methyl substituent is at C2. The structure is:

$$ \text{C1: }CH_3\;-\;\text{C2: }CH(CH_3)\;-\;\text{C3: }CH_2\;-\;\text{C4: }CH_3 $$

Next we identify the distinct types of hydrogen atoms and count their number:

Type 1 (C1-H): three equivalent primary hydrogens at C1.

Type 2 (C4-H): three equivalent primary hydrogens at C4, which is not equivalent to C1.

Type 3 (C2-H): one secondary hydrogen at C2.

Type 4 (C3-H): two equivalent secondary hydrogens at C3.

On monochlorination each hydrogen type gives a distinct chlorinated product. We must also check whether the carbon bearing chlorine becomes a chiral center.

Case 1: Substitution at C1 or at C4 converts a $$CH_3$$ into $$CH_2Cl$$. In each case the carbon has two identical hydrogens left, so the product is achiral and gives only one isomer for C1 and one for C4.

Case 2: Substitution of the single hydrogen at C2 gives $$CHCl$$ at C2 bonded to -CH_3, -C_3H_7 (from C1), and -C_2H_5 (from C3-C4). These three substituents are all different, so C2 becomes a chiral center. Therefore this case yields two enantiomers (2 isomers).

Case 3: Substitution of one of the two equivalent hydrogens at C3 also creates $$CHCl$$ at C3 with four different groups, making C3 a chiral center. Replacement of either prochiral hydrogen gives an enantiomeric pair, so this case also yields two isomers.

Summing the isomers from all cases:

1 (from C1) + 1 (from C4) + 2 (from C2) + 2 (from C3) = 6.

Therefore, the total number of isomeric monochlorinated products of 2-methylbutane is 6.