See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

# Structural Isomers of $C_5H_8$ with One Triple Bond ## Step 1: Determine the carbon skeleton With one triple bond ($C \equiv C$), we have: - Triple bond uses 2 carbons and accounts for 2 degrees of unsaturation - Remaining 3 carbons form a chain/branches - The triple bond can be positioned at different locations along the carbon framework ## Step 2: Identify possible positions of the triple bond For a 5-carbon chain with one triple bond: 1. **$HC \equiv C-CH_2-CH_2-CH_3$** (1-pentyne) 2. **$CH_3-C \equiv C-CH_2-CH_3$** (2-pentyne) 3. **$CH_2=C=CH-CH_2-CH_3$** (Not a triple bond—this is allene, excluded) For branched structures with triple bond: 4. **$CH_3-C \equiv C-CH(CH_3)_2$** (3-methyl-1-butyne) Wait—let me reconsider systematically by checking where the triple bond can be placed: - Position 1: $HC \equiv C-CH_2-CH_2-CH_3$ ✓ - Position 2: $CH_3-C \equiv C-CH_2-CH_3$ ✓ - Position 3: Same as position 2 (symmetry) - With branching: $HC \equiv C-CH(CH_3)_2$ ✓ ## Step 3: List all valid isomers 1. 1-pentyne 2. 2-pentyne 3. 3-methyl-1-butyne **Answer: (2) 3 isomers**