Question

The \( C-O-H \) bond angle in A is \( X \) and \( C-O-C \) bond angle in B is \( Y \). What are X and Y? 

• A. \( X > 109^\circ 28' , Y > 109^\circ 28' \)
• B. \( X < 109^\circ 28' , Y < 109^\circ 28' \)
• C. \( X > 109^\circ 28' , Y = 109^\circ 28' \)
• D. \( X < 109^\circ 28' , Y > 109^\circ 28' \) \

Hint:

In alcohols, the bond angle at oxygen is slightly greater than \( 109^\circ 28' \) due to lone pair repulsions, whereas in ethers, the \( C-O-C \) bond angle is close to \( 109^\circ 28' \).

Answer 1

The Correct Option is C

Step 1: Understanding Bond Angles in Alcohols and Ethers 
- Structure A: Methanol (\( CH_3OH \))
- The central oxygen atom in methanol is sp\(^3\) hybridized with a bent geometry.
- Due to the presence of lone pairs on oxygen, the actual bond angle is slightly greater than the tetrahedral angle \( 109^\circ 28' \).
- This increase is due to repulsion between the lone pairs and bond pairs.
- Structure B: Dimethyl Ether (\( CH_3OCH_3 \))
- Oxygen is sp\(^3\) hybridized and forms two sigma bonds with carbon.
- The bond angle in ethers is approximately \( 109^\circ 28' \) due to similar electron pair repulsions.
Step 2: Evaluating the Given Options 
- Option (1): Incorrect, as \( Y \) should be equal to \( 109^\circ 28' \), not greater.
- Option (2): Incorrect, as alcohols generally have bond angles slightly greater than \( 109^\circ 28' \).
- Option (3): Correct, as \( X>109^\circ 28' \) (methanol) and \( Y = 109^\circ 28' \) (ether).
- Option (4): Incorrect, as it misplaces the bond angle conditions.
Thus, the correct answer is 

Option (3).