Question

For the alkyne with formula \( \text{C}_6\text{H}_{10} \), the number of alkynes with acidic hydrogens is \( x \) and number of alkynes with no acidic hydrogens is \( y \). \( x \) and \( y \) are respectively

• A. 2, 5
• B. 3, 4
• C. 4, 3
• D. 5, 2

Hint:

Draw carbon skeletons first: \( C_6 \), \( C_5(Me) \), \( C_4(Me_2) \). Then try placing the triple bond at terminal positions for acidic isomers and internal positions for non-acidic ones, ensuring valency rules are met.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

Alkynes have the general formula \( \text{C}_n\text{H}_{2n-2} \). For \( n=6 \), it is \( \text{C}_6\text{H}_{10} \). Acidic hydrogens are present in terminal alkynes (triple bond at the end of the chain, \( \text{R-C}\equiv\text{CH} \)). No acidic hydrogens are present in internal alkynes (triple bond in the middle, \( \text{R-C}\equiv\text{C-R'} \)).
Step 2: Detailed Explanation:

We draw the structural isomers of hexyne. 1. Straight Chain (6 carbons):

• \( \text{H-C}\equiv\text{C-CH}_2\text{CH}_2\text{CH}_2\text{CH}_3 \) (1-Hexyne) $\rightarrow$ Acidic
• \( \text{CH}_3\text{-C}\equiv\text{C-CH}_2\text{CH}_2\text{CH}_3 \) (2-Hexyne) $\rightarrow$ Non-acidic
• \( \text{CH}_3\text{CH}_2\text{-C}\equiv\text{C-CH}_2\text{CH}_3 \) (3-Hexyne) $\rightarrow$ Non-acidic

2. Branched Chain (Isopentane skeleton - 5 carbons main chain):

• \( \text{H-C}\equiv\text{C-CH(CH}_3\text{)-CH}_2\text{CH}_3 \) (3-Methyl-1-pentyne) $\rightarrow$ Acidic
• \( \text{H-C}\equiv\text{C-CH}_2\text{-CH(CH}_3\text{)_2} \) (4-Methyl-1-pentyne) $\rightarrow$ Acidic
• \( \text{CH}_3\text{-C}\equiv\text{C-CH(CH}_3\text{)_2} \) (4-Methyl-2-pentyne) $\rightarrow$ Non-acidic
• Note: We cannot place a triple bond at position 2 if position 3 has a methyl group in a 5-carbon chain because carbon valency would exceed 4 (e.g., \( \text{C}_2 \) forms 3 bonds to \( \text{C}_3 \), so \( \text{C}_3 \) can only have 1 more bond, but it is attached to Methyl and C4. Wait, 4-methyl-2-pentyne is valid. \( \text{CH}_3\text{-C}\equiv\text{C-CH(CH}_3\text{)_2} \). \( \text{C}_3 \) has bond to \( \text{C}_2 \), \( \text{C}_4 \), and \( \text{H} \)? No, \( \text{C}_3 \) is part of triple bond. \( \text{C-C}\equiv\text{C-C} \). \( \text{C}_3 \) is bonded to \( \text{C}_2 \) (triple) and \( \text{C}_4 \) (single). 4 valency satisfied. So 4-methyl-2-pentyne is valid.
• What about 3-methyl-2-pentyne? \( \text{CH}_3\text{-C}\equiv\text{C(CH}_3\text{)-CH}_2\text{CH}_3 \). The carbon at pos 3 would have 3 bonds to C2, 1 to C4, 1 to methyl = 5 bonds. Impossible.

3. Branched Chain (Neopentane skeleton - 4 carbons main chain):

• \( \text{H-C}\equiv\text{C-C(CH}_3\text{)_3} \) (3,3-Dimethyl-1-butyne) $\rightarrow$ Acidic

Counting:

• Acidic (Terminal): 1-Hexyne, 3-Methyl-1-pentyne, 4-Methyl-1-pentyne, 3,3-Dimethyl-1-butyne. Total \( x = 4 \).
• Non-acidic (Internal): 2-Hexyne, 3-Hexyne, 4-Methyl-2-pentyne. Total \( y = 3 \).

Step 3: Final Answer:

\( x = 4 \), \( y = 3 \).