In gaseous state:

$$\ce{(CH3)3N} > \ce{(CH3)2NH} > \ce{CH3NH2} > \ce{NH3}$$

However, when dissolved in water:

$$\ce{(CH3)2NH} > \ce{CH3NH2} > \ce{(CH3)3N} > \ce{NH3}$$

My notes mentioned the reason for this is because of the increased steric hindrance from the greater number of alkyl group which makes the lone pair of electrons on the N atom less available. But, if this is the case, why doesn't steric hindrance affect the basicity of tertiary amines when in a gaseous state?

After some research, I found that the conjugate acid of an amine can be stabilised by solvation. But, shouldn't the stabilisation be the same, because of the attraction of electronegative O in a water molecule to the positive charge on $\ce{N}$ in an ammonium salt? (can ammonium salts still be stabilised by hydrogen bonding? I thought ions don't form hydrogen bonds with water, they form ion-dipole attractions)

Unless the effect of steric hindrance comes in here, where the alkyl groups reduce the strength of the ion-dipole attraction between the ammonium ion and thus the stability?

I'm really confused how everything links up; appreciate it if someone can kind of clear this up.


2 Answers 2


A possible answer from what I've searched on the internet so far..

The strength of basicity of amines depends on two factors: 1) the availability of the lone pair for donation to a proton, 2) the stability of the conjugate acid of the amine.

In methylamine, the conjugate acid, $CH_3NH_3^+$ will attract water molecules and ion-dipole interactions will form between the cation (the conjugate acid of methylamine) and the partial negative charge on O in the water molecule. This ion-dipole interaction or attractive force helps to stabilize the positive charge on the cation.

So, the more water molecules you have solvating your cation, the more stable the conjugate base. However, the more alkyl groups you have, the less stable the conjugate base, because these alkyl groups are hindering water molecules from coming and stabilizing the conjugate base.

[The N atom is "shielded" by the bulky alkyl groups which take up a lot of space and prevent any water molecules from stabilizing the conjugate acid. Water molecules can only come in from where the H atom is where the shape of the conjugate acid is that of a pyramid. Note that there is a positive charge on N.]

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As methylamine has the least number of alkyl groups as compared to dimethylamine and trimethylamine, at this point, it would be the strongest base. However, basicity also depends on the availability of the lone pair of electrons on the N atom of the amine itself. As methylamine has the least number of alkyl groups, the inductive effect by the alkyl group is the least out of methylamine, dimethylamine, and trimethylamine.

So, it really is a measure of which factor is more dominant or which factor outweighs the other. In this case, although trimethylamine has the most number of alkyl groups, the three bulky alkyl groups prevent a lot of water molecules from stabilizing the conjugate acid. And, for dimethylamine, although dimethylamine has a fewer number of alkyl groups than trimethylamine, the inductive effect of the alkyl groups outweighs the steric hindrance from the 2 alkyl groups as compared to the three in trimethylamine.

Hence, in an aqueous solution, $(CH_3)_2NH>CH_3NH_2>(CH_3)_3N>NH_3$

However, in the gaseous state, there are no water molecules to stabilize the conjugate acid and basicity depends on the inductive effect of alkyl groups; there are no water molecules to hinder. As trimethylamine is trigonal bipyramidal in shape, the factor of steric hindrance does not apply here. Hence, as trimethylamine has the most number of alkyl groups, the lone pair of electrons on trimethylamine is most available for donation.


In the gaseous state the basicity can simply be compared on the basis of availability of lone pair. In tertiary amine the nitrogen atom receives +I effect from all the three methyl groups which increases the electron density on the nitrogen atom ,thus its lone pair is readily available. Whereas in secondary and primary amines the nitrogen atom receives+I effect from two and one methyl group respectively. So there lone pair are not as readily available as of tertiary amines. When in aqueous phase the amines interact with water. This interaction also includes hydrogen bonding. This interaction causes steric repulsion. Thus it affects the basicity and causes the order to be so. Note : the shape of amines is Tetrahedral with three surrounding atoms and one lone pair. So there is minimal steric repulsion in gaseous state. The main reason for steric repulsion in aqueous phase is the interaction with water.

  • $\begingroup$ How does this interaction with water cause steric hindrance? Does the water molecules kind of form like a shield around the conjugate acid of the amine(cation)? But, if that's the case, would there also be steric hindrance for the other 2 molecules? $\endgroup$
    – xander
    May 19, 2019 at 11:33

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