See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution: Structural Isomers of $C_6H_{15}N$ (3° amines) **Step 1: Identify the constraint** We need only **tertiary (3°) amines**, where nitrogen is bonded to three carbon groups and one hydrogen: $R_3N-H$ **Step 2: Determine carbon distribution** With molecular formula $C_6H_{15}N$, the three alkyl groups attached to nitrogen must total 6 carbons. Possible combinations $(C_a, C_b, C_c)$ where $a + b + c = 6$: - $(1, 1, 4)$ — methyl, methyl, butyl - $(1, 2, 3)$ — methyl, ethyl, propyl - $(2, 2, 2)$ — ethyl, ethyl, ethyl **Step 3: Count isomers for each combination** **Case 1: $(1, 1, 4)$ — $N(CH_3)_2(C_4H_9)$** - $C_4H_9$ groups: n-butyl, sec-butyl, isobutyl, tert-butyl = **4 isomers** **Case 2: $(1, 2, 3)$ — $N(CH_3)(C_2H_5)(C_3H_7)$** - $C_3H_7$ groups: n-propyl, isopropyl = 2 isomers - $C_2H_5$ is fixed (ethyl) - Total = **2 isomers** **Case 3: $(2, 2, 2)$ — $N(C_2H_5)_3$** - All three groups are identical (ethyl) - Only **1 isomer** (symmetrical) **Step 4: Total count** $$4 + 2 + 1 = \boxed{7 \text{ isomers}}$$ **Answer: (B) 7**