See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Identify the starting material and products** Starting with $C_6H_5Cl$ (chlorobenzene), we need to find: - $m$ = number of $C_6H_3Cl_3$ isomers from o-isomers of $C_6H_4Cl_2$ - $n$ = number of $C_6H_3Cl_3$ isomers from p-isomer of $C_6H_4Cl_2$ - $q$ = number of $C_6H_3Cl_3$ isomers from m-isomer of $C_6H_4Cl_2$ **Step 2: Count isomers from o-dichlorobenzene** When o-$C_6H_4Cl_2$ undergoes further chlorination: The two chlorine atoms are **adjacent** (ortho). Adding a third chlorine can give positions that are: - 2,3-dichlorobenzene - 2,4-dichlorobenzene - 2,5-dichlorobenzene - 2,6-dichlorobenzene This yields **$m = 2$** distinct $C_6H_3Cl_3$ isomers (accounting for symmetry). **Step 3: Count isomers from p-dichlorobenzene** The two chlorines are **para** (opposite ends). The benzene ring has equivalent positions by symmetry. Adding one more chlorine gives only **$n = 1$** $C_6H_3Cl_3$ isomer. **Step 4: Count isomers from m-dichlorobenzene** The two chlorines are **meta** (one position apart). This gives: - 2,3-dichlorobenzene - 2,4-dichlorobenzene - 2,5-dichlorobenzene - 2,6-dichlorobenzene - 3,4-dichlorobenzene - 3,5-dichlorobenzene This yields **$q = 3$** distinct $C_6H_3Cl_3$ isomers. **Therefore: $m = 2$, $n = 1$, $q = 3$** **The answer is (B) 2, 1, 3** ✓