See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Structural Isomers of Tertiary Amines: $C_6H_{15}N$ **Step 1: Understand tertiary amines** Tertiary amines have the general formula $R_3N$ where three carbon-containing groups are bonded to nitrogen (no N–H bonds). **Step 2: Determine carbon distribution** For $C_6H_{15}N$, we distribute 6 carbons among three alkyl groups: $R_1$, $R_2$, $R_3$. The possible distributions are: - (6, 0, 0) → not valid for tertiary amine - (5, 1, 0) → not valid - (4, 2, 0) → not valid - (4, 1, 1) ✓ - (3, 3, 0) → not valid - (3, 2, 1) ✓ - (2, 2, 2) ✓ **Step 3: Count isomers for each distribution** **Case 1: (4, 1, 1)** — One butyl, two methyls: $N(CH_3)_2(C_4H_9)$ - $n$-butyl: $N(CH_3)_2(n\text{-}Bu)$ - Isobutyl: $N(CH_3)_2(i\text{-}Bu)$ - *sec*-butyl: $N(CH_3)_2(\text{sec-}Bu)$ - *tert*-butyl: $N(CH_3)_2(t\text{-}Bu)$ **Total: 4 isomers** **Case 2: (3, 2, 1)** — One propyl, one ethyl, one methyl: $N(CH_3)(C_2H_5)(C_3H_7)$ - $n$-propyl: $N(CH_3)(C_2H_5)(n\text{-}Pr)$ - Isopropyl: $N(CH_3)(C_2H_5)(i\text{-}Pr)$ **Total: 2 isomers** **Case 3: (2, 2, 2)** — Three ethyl groups - Triethylamine: $N(C_2H_5)_3$ (only one structure) **Total: 1 isomer** (