Question

The total number of 'sigma' and 'Pi' bonds in 2-oxohex-4-ynoic acid is____.

Answer 1

Correct Answer: 18

To determine the total number of sigma (\(\sigma\)) and pi (\(\pi\)) bonds in 2-oxohex-4-ynoic acid, analyze the structure:
\[\text{HO-CH}_2 - \text{C(=O)} - \text{CH}_2 - \text{C} \equiv \text{C} - \text{CH}_3\]
Count the sigma bonds (\(\sigma\)-bonds):
\(6 \, \sigma\)-bonds in carbon-hydrogen (C-H) bonds.
\(5 \, \sigma\)-bonds in carbon-carbon (C-C) single bonds.
\(2 \, \sigma\)-bonds in carbon-oxygen (C=O and C-O) bonds.
\(1 \, \sigma\)-bond in the hydroxyl (O-H) group.
Total \(\sigma\)-bonds: \(6 + 5 + 2 + 1 = 14\)
Count the pi bonds (\(\pi\)-bonds):
\(1 \, \pi\)-bond in the C=O bond.
\(3 \, \pi\)-bonds in the C \(\equiv\) C triple bond (\(2 \, \pi\)-bonds in the triple bond).
Total \(\pi\)-bonds: \(1 + 3 = 4\)
Therefore, the total number of bonds (sigma and pi) is:
\[14 + 4 = 18\]

Answer 2

Step 1: Write the structure of 2-oxohex-4-ynoic acid
The name 2-oxohex-4-ynoic acid can be analyzed as follows:
- “Hex” → 6 carbon atoms in the main chain.
- “4-ynoic” → triple bond between C₄ and C₅.
- “2-oxo” → carbonyl group (C=O) at C₂.
- “oic acid” → carboxylic acid group (–COOH) at the terminal C₁.

Therefore, the structure is:
HOOC–CO–CH₂–C≡C–CH₃

Step 2: Count total sigma (σ) and pi (π) bonds
(a) Carboxylic acid group (–COOH):
- 1 C–C σ (between COOH and next carbon)
- 1 C=O → 1 σ + 1 π
- 1 C–O σ (single bond)
- 1 O–H σ
Total from –COOH = 3 σ (C–C, C–O, O–H) + 1 σ (from C=O) + 1 π = 4 σ + 1 π

(b) Carbonyl group at position 2 (C=O):
C=O → 1 σ + 1 π
Plus C–C σ bond to the next carbon
→ total = 2 σ + 1 π

(c) Two C–C single bonds (from CH₂–C≡C):
Between C₂–C₃ and C₃–C₄ → 2 σ

(d) One triple bond (C₄≡C₅):
Triple bond = 1 σ + 2 π
→ total = 1 σ + 2 π

(e) One C–C single bond between C₅–C₆:
→ 1 σ

(f) Hydrogen atoms:
Let’s assign hydrogens:
- C₁: part of –COOH → no hydrogens
- C₂: has one double bond (C=O) and two single bonds (C–C), so no H
- C₃: CH₂ → 2 C–H σ bonds
- C₄: part of triple bond → 0 H
- C₅: part of triple bond → 0 H
- C₆: CH₃ → 3 C–H σ bonds
Total C–H σ bonds = 2 + 3 = 5 σ

Step 3: Summarize all bonds
- σ bonds = (from step a to f):
(4) + (2) + (2) + (1) + (1) + (5) = 15 σ bonds
- π bonds = (from step a to d):
(1) + (1) + (0) + (2) = 4 π bonds

Wait — check carefully: total should count unique bonds only. Let's recount to verify.

Detailed check of each bond: 1. Between atoms:
C₁–C₂ (σ)
C₁=O (1 σ + 1 π)
C₁–O(H) (σ)
O–H (σ)
C₂=O (1 σ + 1 π)
C₂–C₃ (σ)
C₃–H (2 σ)
C₃–C₄ (σ)
C₄≡C₅ (1 σ + 2 π)
C₅–C₆ (σ)
C₆–H (3 σ)

Counting all:
σ bonds = 1(C₁–C₂) + 1(C₁=O) + 1(C₁–O) + 1(O–H) + 1(C₂=O) + 1(C₂–C₃) + 2(C₃–H) + 1(C₃–C₄) + 1(C₄≡C₅) + 1(C₅–C₆) + 3(C₆–H) = 14 σ
π bonds = 1(C₁=O) + 1(C₂=O) + 2(C₄≡C₅) = 4 π

Total bonds = 14 + 4 = 18 bonds.

Step 4: Final answer

18