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Why does dimethyl ether have a bond angle bigger than 109.5 degrees, in contradiction with VSEPR theory?

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    $\begingroup$ Due to replusion between the two methyl groups, according to Competition Science Vision, October 2008, page 1033. $\endgroup$
    – DavePhD
    Dec 13, 2014 at 14:56
  • $\begingroup$ Related: chemistry.stackexchange.com/q/18697 $\endgroup$
    – Jan
    Nov 20, 2016 at 19:38

2 Answers 2

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In case of $\ce{Me2O}$, the positive deviation of the bond angle can be explained on the basis of steric crowding of two methyl groups.

There are other contradiction too like this:

In case of $\ce{(SiH3)2O}$, the huge bond angle increase is mainly due to back bonding and partly due to steric repulsion.

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Let us compare the $\ce{C-O-C}$ bond angle in dimethyl ether (110.3°) with H-O-H bond angle in $\ce{H2O}$ (104.5°).

The bond angle in dimethyl ether is greater due to the following reason. The $\ce{CH3}$ functional group is many times heavier than the H atom (both act as side atoms). So the effect of lone pair - lone pair repulsion in $\ce{H2O}$ is pronounced in $\ce{H2O}$ whereas due to heavy molecular weight the repulsive forces aren't able to fully force $\ce{CH3}$ functional groups to compress them.

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