Isomeric hydrocarbons giving negative Baeyer's test have the molecular formula C_9H_12$C_{9}H_{12}$. The total number of isomers from above with exactly four different non-aliphatic substitution sites 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

### Core Logic Kolbe's electrolytic method generates free radicals at the anode, which then combine to form alkanes. The given mixture yields two types of carboxylate radicals which decarboxylate to form alkyl radicals: CH_3COONa rightarrow dotCH_3$CH_3COONa \rightarrow \dot{C}H_3$ C_2H_5COONa rightarrow dotC_2H_5$C_2H_5COONa \rightarrow \dot{C}_2H_5$ These radicals can couple in three different ways: 1. Cross coupling: dotCH_3 + dotC_2H_5 rightarrow CH_3-CH_2-CH_3$\dot{C}H_3 + \dot{C}_2H_5 \rightarrow CH_3-CH_2-CH_3$ (Propane) 2. Self-coupling 1: dotCH_3 + dotCH_3 rightarrow CH_3-CH_3$\dot{C}H_3 + \dot{C}H_3 \rightarrow CH_3-CH_3$ (Ethane) 3. Self-coupling 2: dotC_2H_5 + dotC_2H_5 rightarrow CH_3-CH_2-CH_2-CH_3$\dot{C}_2H_5 + \dot{C}_2H_5 \rightarrow CH_3-CH_2-CH_2-CH_3$ (Butane) Thus, a total of 3 different alkanes are formed. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons