Number of isomeric products formed by mono-chlorination of 2-methylbutane in presence of sunlight is ________

### Core Logic The structure of 2-methylbutane is CH_3-CH(CH_3)-CH_2-CH_3$CH_3-CH(CH_3)-CH_2-CH_3$. It has four different types of hydrogen atoms, which can be substituted to form structural isomers: 1) 1-chloro-2-methylbutane: Chlorination at terminal CH_3$CH_3$ near branch. Yields a chiral center at C2 (2 enantiomers). 2) 2-chloro-2-methylbutane: Chlorination at the tertiary carbon (1 achiral product). 3) 2-chloro-3-methylbutane: Chlorination at the CH_2$CH_2$ group. Yields a chiral center at C2 (2 enantiomers). 4) 1-chloro-3-methylbutane: Chlorination at the far terminal CH_3$CH_3$. No chiral center (1 achiral product).

### Step 1: Counting Stereoisomers Total isomeric products = 2 (from 1st) + 1 (from 2nd) + 2 (from 3rd) + 1 (from 4th) = 6$2 (from 1st) + 1 (from 2nd) + 2 (from 3rd) + 1 (from 4th) = 6$. ### Pattern Recognition When asked for "isomeric products" in halogenation without specifying "structural isomers", you must count stereoisomers (enantiomers) as well. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 12 Chemistry: Haloalkanes and Haloarenes