Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices.
Assertion : Bond angle in ethers is slightly less than the tetrahedral angle.
Reason : There is a repulsion between the two bulky (—R) groups.
We know ethers are represented by R-O-R (where R is an alkyl group or even Ph group).
We know that the tetrahedral bond angle around oxygen atom is equal to 109°28’.
If we look at the structure, we can observe that the electron pairs(which includes the bond pairs as well as lone pairs) surrounding the Oxygen atom assume tetrahedral arrangement that is sp3 hybrid state. As we know, generally the bond angle observed for sp3 hybrid atom is 109°28’, but in ethers it is not so. In ethers, bond angle is larger. Why is that so?
This deviation in angle caused by the repulsive interactions due to the bulky alkyl groups due to their ‘electron pushing nature’ as we shown in the diagram. This can also be termed ‘steric crowding’ of the alkyl groups.
Bulkier the alkyl group, more would be the repulsion or repulsive interactions and therefore larger would be the deviation in C-O-C angle.
So bond angle comparison is :
ethoxyethane> methoxymethane> methane.
Note:
One would say that we would observe lone pair-lone pair repulsion so if we consider that case the bond angle should definitely be lower than the standard tetrahedral angle.
The explanation is that, yes, we do observe lone pair-lone pair repulsion (lp-lp) but the deviation caused by the steric, bulky alkyl groups are so massive that the lp-lp repulsion cannot overpower it. That is why bond angle is larger.
But if we compare the bond angle of water with that of the standard tetrahedral angle, it is lower due to the same lp-lp repulsion. The reason is same as discussed, that Hydrogen being lighter, experience the repulsions as a result of which the bond angle decreases to around 106°