# What is the bond angle of the C-N-H bond angle in (CH3)2NH?

Jan 9, 2016

This compound is drawn like so, with the $\text{CNH}$ atoms highlighted:

This is called dimethylamine; amine, because of the $\text{NH}$, and dimethyl because of the two ${\text{CH}}_{3}$ groups on the nitrogen.

(You might also call this N,N-dimethylamine to emphasize the locations of the methyl groups.)

We can see the lone pair of electrons on the nitrogen, which immediately means that one of the $\text{CNH}$ bond angles will not be "standard", so to speak.

(This compound is NOT trigonal planar, by the way.)

You can see that it has four electron groups: two methyls, one hydrogen, and one lone pair. That corresponds to a tetrahedral electron geometry, and thus a trigonal pyramidal molecular geometry.

As a result, we can expect the bond angle to be close to ${109.5}^{\circ}$.

However, due to the lone pair of electrons, which take up quite a bit of space when they aren't bonding, the molecule "crunches" up a little bit, and the angle becomes LESS than ${109.5}^{\circ}$. We can see this better when we redraw the compound like so:

Looking up the actual bond angle, it is about $\textcolor{b l u e}{{108.9}^{\circ}}$. You can look it up yourself as well here:

http://cccbdb.nist.gov/geomcalc1x.asp

Simply enter the molecular formula, then choose the three atoms you want for your bond angle (four if it is a dihedral/torsion angle).