Question

The ${\text{C - O - C}}$ angle in ether is about?
A.$180^\circ $
B.$180^\circ 28'$
C.$110^\circ $
D.$105^\circ $

Answer 1

Hint: The angle between the bonds and atoms of ${\text{C - O - C}}$ depend on the hybridization of the central atom which is oxygen. The alkyl groups and the lone pair present on oxygen atoms of ether also affects the structure of ether and bond angle ${\text{C - O - C}}$.

Complete step by step answer:
Ethers are the organic compounds that contain an Oxygen between the two alkyl or aryl groups. It is represented as ${\text{R - O - R'}}$ where ${\text{R and R'}}$ are the alkyl or aryl groups.
In the ether the oxygen atom present between the two alkyl or aryl groups is ${\text{s}}{{\text{p}}^3}$ hybridized and also contains two lone pairs that’s why its structure should be tetrahedral and according to structure the correspondent angle between the bonds should be $109^\circ 28'$ but due to the presence of lone pair the angle must decrease. But in actuality the angle increases from $109^\circ 28'$to approximate $110^\circ $. This happens because the two alkyl or aryl groups present are bulky, so they repel each other and counter balance the repulsion generated by the lone pairs, so instead of decreasing the angle between the bonds slightly increases.
Therefore the ${\text{C - O - C}}$ angle in ether is about $110^\circ $.
Hence option (C) is correct, $110^\circ $.

Additional information:
Ether molecules can’t form hydrogen bonds with each other which results in low boiling points of ether.
Ethers are slightly polar also, they have polarity because the angle between the bonds changes to $110^\circ $ which allows them to maintain their dipoles.


Note:
Ethers don’t show any interaction between the molecules easily but the lone pairs present on the oxygen atom of ether makes hydrogen bonding possible with water molecules. Cyclic ethers are found to be miscible in water.