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How can I rationalise the different basicities of cyclohexylamine, piperidine and morpholine?
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Khari Slaughter
How can I rationalise the different basicities of cyclohexylamine, piperidine and morpholine?
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.
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.
@Srishti Inductive effect decrease fast in chains, therefore having more -CH2- groups makes barely any difference. (II) compound has two inductive effects from two alkyl groups, which means it is more basic.More
But piperidine is a secondary amine while cyclohexylamine is a primary amine. Inductive effect is distance dependent. Also cyclohexamine has only 1 alpha hydrogen while piperidine has 4. So hyperconjugation flavours piperidineMore
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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:
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:
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.
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.
Hyperconjugation certainly isnt relevant for the octet-complete ammonium cations. However, cyclohexylamine (or its conjugate acid) has a pKa of 10.64; piperidine 11.22. (More
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.
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.
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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.
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.
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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.
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.
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