How can I rationalise the different basicities of cyclohexylamine, piperidine and morpholine?

Question

Compare the basicity of the following compounds:

Comparing compounds I and II

Since there is no conjugated system, there is no resonance at play and we only have to discuss hyperconjucation and inductive effect.

The conjugate acid of compound I has one α-hydrogen while the conjugate acid of compound II has four α-hydrogens. So, according to hyperconjugation, the conjugate acid of compound II is more stable than the conjugate acid of compound I and hence the basicity of compound II is more than that of compound I.

Further, the conjugate acid of compound I has one α-carbon, two β-carbons, two γ-carbons and one δ-carbon. While the conjugate acid of compound II has two α-carbons, two β-carbons and one γ-carbon.

Alkyl groups are electron releasing groups and, since the conjugate acid of compound II has its carbon atoms closer to the amine than in the case of compound I, it is further stabilised, and hence compound II is more basic.

Comparing compounds II and III

The conjugate acid of compound III has an ether group attached to the amine while that of compound II has an alkyl group to the amine. Since ether is an electron withdrawing group and alkyl is an electron releasing group, the former will destabilise the conjugate acid compound III while the latter will stabilise the compound II, and hence compound II is more basic than compound III.

So, I know that compound II is the most basic, but I still don't know how to compare compounds I and III.

Can anyone help?

Answer 1

You are right that (I) < (II) and (II) > (III), but my explanation for (I) < (II) is different from yours: nitrogen atom in (II) has more alkyl group than (I), therefore the its conjugated acid is more stable.

As inductive effect decrease fast along the carbon chain, even though cyclohexyl group have more carbons, its effect is still weaker than two alkyl group (you can treat piperidine as ethylpropylamine, they're similar). Inductive effect is usually stronger than hyperconjugation.

In (III), both two branches have -C effect, as oxygen is an electronegative element, which destabilises the conjugated acid. In (I), the only alkyl group has +C effect, which stabilise the conjugated acid. Therefore, (I) is a stronger base than (III)

Hence, we have the sequence of basicity: (II) > (I) > (III).

We can look up pKb of each substance to check the result:

Cyclohexylamine (I): 3.36
Piperidine (II): 2.78
Morpholine (III): 5.64

So, my comparison is rational.

Answer 2

In this answer of mine, I outlined why hyperconjugation is helpful in stabilising carbocations but why it also does not contribute to the stabilisation of ammonium cations. Indeed, when comparing the inductive effects of neighbouring groups on ammonium cations, only the corresponding electronegativities should be considered.

Cyclohexylammonium has three hydrogens and one carbon connected to it; both morpholinium and piperidinium have two carbons and two hydrogens attached. Since carbon is more electronegative than hydrogen, nitrogen is polarised less negatively if it is attached to more carbon atoms. (The $\ce{N-H}$ bonds are polarised towards nitrogen more strongly than $\ce{N-C}$ bonds.) Therefore, cyclohexylammonium has the largest electron density on nitrogen, meaning the positive charge is best stabilised and therefore the cation is most stable; thus, it is the most basic of the three.

Between piperidinium and morpholinium, the only difference is the oxygen atom in the latter. This oxygen atom draws away electrons inductively across the entire σ bonded ring, meaning that nitrogen is slightly less negatively polarised in morpholine. Therefore, the morpholinium cation is less stable than piperidinium.

Answer 3

Here cyclohexamine (1) is more basic than piperidine (2) which is more basic than morpholine (3).

The $\mathrm{1^{st}}$ compound has more methyl groups than the $\mathrm{2^{nd}}$ compound. Alkyl groups are electron donating group and thus stabilizes the positive charge of nitrogen.

In the $\mathrm{3^{rd}}$ compound, oxygen is present which is an electron withdrawing group, and thus reduces the basicity of nitrogen as it attracts the electrons towards itself. Hence, the $\mathrm{1^{st}}$ is most basic and $\mathrm{3^{rd}}$ is least basic.