See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Solution for $C_6H_{14}O$ Isomerism Types **Step 1: Determine degree of unsaturation** $$\text{DBE} = \frac{2(6) + 2 - 14}{2} = 0$$ No rings or double bonds — only single bonds possible. **Step 2: Chain Isomerism** $C_6H_{14}O$ can form different carbon skeletons: - Straight chain hexanol: $CH_3CH_2CH_2CH_2CH_2CH_2OH$ - Branched: $(CH_3)_2CHCH_2CH_2CH_2OH$ (2-methylpentanol skeleton) - More branched isomers exist ✓ **Chain isomerism is present** **Step 3: Functional Group Isomerism** The same formula $C_6H_{14}O$ can represent: - **Alcohols**: hexanols ($-OH$ group) - **Ethers**: e.g., $CH_3OCH_2CH_2CH_2CH_3$ (methyl pentyl ether) ✓ **Functional isomerism is present** **Step 4: Positional Isomerism (Metammerism)** Even with the same functional group (alcohol), the $-OH$ can be attached at different positions: - 1-hexanol: $OH$ at C1 - 2-hexanol: $OH$ at C2 - 3-hexanol: $OH$ at C3 ✓ **Metammerism is present** **Answer: (D) All of these** — $C_6H_{14}O$ exhibits chain isomerism, functional isomerism, and positional isomerism simultaneously.