Question

question 43 of 49
identify the optimum c-c-c bond angle for each of the indicated carbon atoms in the molecule.
(the image shows a molecular structure with carbon atoms and an answer bank with options 120°, 109.5°, 90°, 180°)

Explanation:

Step1: Assign hybridization to each C

Use VSEPR:

• $sp$ (2 electron domains): bond angle $180^\circ$
• $sp^2$ (3 electron domains): bond angle $120^\circ$
• $sp^3$ (4 electron domains): bond angle $109.5^\circ$

Step2: Analyze top alkyne carbons

Leftmost $C$ ($\text{CH}_3\text{C}\equiv$): $sp$ hybridized.
$\text{C-C-C}$ angle: $180^\circ$
Middle alkyne $C$ ($\equiv\text{C-CH}_2-$): $sp$ hybridized.
$\text{C-C-C}$ angle: $180^\circ$

Step3: Analyze $\text{CH}_2$ adjacent to alkyne

$\text{-C-CH}_2-$: $sp^3$ hybridized.
$\text{C-C-C}$ angle: $109.5^\circ$

Step4: Analyze central branched $C$

Central $C$ (connected to 4 groups): $sp^3$ hybridized.
$\text{C-C-C}$ angle: $109.5^\circ$

Step5: Analyze alkene carbons

Right alkene $C$ ($\text{-CH=CH-CH}_2-$): $sp^2$ hybridized.
$\text{C-C-C}$ angle: $120^\circ$
Left alkene $C$ ($\text{-C-CH=CH}-$): $sp^2$ hybridized.
$\text{C-C-C}$ angle: $120^\circ$

Step6: Analyze bottom alkene carbons

Bottom right alkene $C$ ($\text{-C=CH-CH}_3$): $sp^2$ hybridized.
$\text{C-C-C}$ angle: $120^\circ$
Bottom left alkene $C$ ($\text{-CH-C=CH}-$): $sp^2$ hybridized.
$\text{C-C-C}$ angle: $120^\circ$

Answer:

1. Topmost (alkyne $\text{CH}_3\text{C}\equiv$): $180^\circ$
2. Second top (middle alkyne $\equiv\text{C}-$): $180^\circ$
3. Third top ($\equiv\text{C-CH}_2-$): $109.5^\circ$
4. Central branched $C$: $109.5^\circ$
5. Bottom right alkene ($\text{-CH=CH-CH}_2-$): $120^\circ$
6. Bottom middle alkene ($\text{-C-CH=CH}-$): $120^\circ$
7. Bottom left alkene ($\text{-CH-C=CH}-$): $120^\circ$
8. Bottommost ($\text{-C=CH-CH}_3$): $120^\circ$