Question

24 multiple choice 2 points how many carbon-carbon σ bonds are present in the molecule shown? h—c≡c—ch₂ch₃ 4 1 5 3 2

Explanation:

Step1: Identify carbon atoms

The molecule is \( \ce{H - C\equiv C - CH_2 - CH_3} \). Carbon atoms: \( \ce{C_1 - C_2 - C_3 - C_4} \) (numbering from left: \( \ce{C_1} \) (attached to H), \( \ce{C_2} \) (triple bonded to \( \ce{C_1} \)), \( \ce{C_3} \) (attached to \( \ce{C_2} \) and \( \ce{CH_3} \) via \( \ce{CH_2} \)), \( \ce{C_4} \) (in \( \ce{CH_3} \))).

Step2: Count C - C σ bonds

• \( \ce{C_1 - C_2} \): 1 σ (triple bond has 1 σ, 2 π)
• \( \ce{C_2 - C_3} \): 1 σ (single bond)
• \( \ce{C_3 - C_4} \): 1 σ (single bond)

Wait, no—wait, the structure is \( \ce{H - C\equiv C - CH_2CH_3} \), which is \( \ce{H - C\equiv C - CH_2 - CH_3} \). So carbon - carbon bonds: \( \ce{C1 - C2} \) (triple, 1 σ), \( \ce{C2 - C3} \) (single, 1 σ), \( \ce{C3 - C4} \) (single, 1 σ)? Wait, no, \( \ce{CH_2CH_3} \) is \( \ce{CH_2 - CH_3} \), so \( \ce{C3 - C4} \) is another single bond. Wait, no, let's re - number:

First carbon: \( \ce{C1} \) (H - C≡), second: \( \ce{C2} \) (C≡C - ), third: \( \ce{C3} \) ( - CH₂ - ), fourth: \( \ce{C4} \) ( - CH₃). So bonds between C1 - C2 (1 σ), C2 - C3 (1 σ), C3 - C4 (1 σ)? No, that's 3? Wait, no, wait the formula is \( \ce{H - C\equiv C - CH_2CH_3} \), which is \( \ce{H - C\equiv C - CH_2 - CH_3} \). So the carbon - carbon bonds are:

1. \( \ce{C1 - C2} \) (triple bond: 1 σ)
2. \( \ce{C2 - C3} \) (single bond: 1 σ)
3. \( \ce{C3 - C4} \) (single bond: 1 σ)

Wait, but that's 3? Wait, no, maybe I misread. Wait, \( \ce{CH_2CH_3} \) is ethyl group, so the chain is \( \ce{C\equiv C - CH_2 - CH_3} \), so carbon atoms: 4? Wait, no, \( \ce{H - C\equiv C - CH_2 - CH_3} \) has 4 carbon atoms? Wait, no: \( \ce{C1} \) (H - C), \( \ce{C2} \) (C≡C), \( \ce{C3} \) (CH₂), \( \ce{C4} \) (CH₃). So C - C bonds: C1 - C2, C2 - C3, C3 - C4. Wait, but that's 3. Wait, but let's check again.

Wait, the molecule is \( \ce{H - C\equiv C - CH_2CH_3} \), which is but - 1 - yne? Wait, no, \( \ce{H - C\equiv C - CH_2 - CH_3} \) is pent - 1 - yne? Wait, no, number of carbons: 1 (H - C), 2 (C≡C), 3 (CH₂), 4 (CH₃). So 4 carbons. So C - C bonds:

• Between C1 and C2: 1 σ (triple bond has 1 σ)
• Between C2 and C3: 1 σ (single bond)
• Between C3 and C4: 1 σ (single bond)

Wait, that's 3. But wait, maybe I made a mistake. Wait, \( \ce{CH_2CH_3} \) is \( \ce{CH_2 - CH_3} \), so C3 - C4 is a single bond. So total C - C σ bonds: 3? Wait, but let's count again.

Wait, the structure: \( \ce{H - C\equiv C - CH_2 - CH_3} \). So the carbon - carbon bonds are:

1. \( \ce{C1 - C2} \): 1 σ (triple bond)
2. \( \ce{C2 - C3} \): 1 σ (single bond)
3. \( \ce{C3 - C4} \): 1 σ (single bond)

So total 3.

Answer:

3 (corresponding to the option with 3)