Would the electropositive alkyl groups repel protons?
It's more to do with the bulky nature of the alkyl groups than the charge. Depending on the size of the alkyl groups then the slightly higher electronegativity of nitrogen will hardly have an effect on them and thus protons wouldn't really be repelled
Would the electropositive alkyl groups repel protons?
Heavens no.
Alkyl groups aren't really electropositive. They stabilise low electron density, and the method through which they do this is often simplified to "inductive effects", but it's really more accurate to think of it in terms of partial orbital donation. They certainly don't "repel protons".
Alkyl groups aren't really electropositive. They stabilise low electron density, and the method through which they do this is often simplified to "inductive effects", but it's really more accurate to think of it in terms of partial orbital donation. They certainly don't "repel protons".
Could you explain this in a way an A2 student could understand? I'm genuinely interested.
Could you explain this in a way an A2 student could understand? I'm genuinely interested.
No, not really, unless you understand what molecular orbitals are.
I'll give it a go though.
Imagine a tertiary carbocation CEt3+
Under the "inductive effects" explanation, we are to believe that the stability of the carbocation is due to the fact that, as the electronegativity of C is 2.5 and H is 2.2, that the carbon is electron rich, and then (for some reason) pushes electrons it steals from the hydrogen onto the cation, hence stabilising the charge. This isn't really a good explanation; the electronegativity difference isn't really high enough for that to be a viable thing to occur. In our tertiary carbocation, the central carbon is sp2 hybridised, that is to say, it is planar, and it has an empty p orbital perpendicular to the central C atom, and the 3 C atoms of the R groups (ethyl). In the 3 first C atoms of the Et groups, however, the geometry is tetrahedral, as the atoms are sp3 hybridised. This means that the sigma bonds are have a component in the direction defined by the p orbital of the central cation, and one of the bonds will (for these purposes) be coplanar with the p orbital. Hence, there will be an interaction between the empty p orbital and the filled sigma orbital of the bond from the first Et carbon to the second Et carbon (as will be the case in this example here), which can be though of for all intents and purposes as a "partial donation of electron density from the sigma bond which is in phase with the empty p orbital of the cation", hence a stabilising effect is present.
No, not really, unless you understand what molecular orbitals are.
I'll give it a go though.
Imagine a tertiary carbocation CEt3+
Under the "inductive effects" explanation, we are to believe that the stability of the carbocation is due to the fact that, as the electronegativity of C is 2.5 and H is 2.2, that the carbon is electron rich, and then (for some reason) pushes electrons it steals from the hydrogen onto the cation, hence stabilising the charge. This isn't really a good explanation; the electronegativity difference isn't really high enough for that to be a viable thing to occur. In our tertiary carbocation, the central carbon is sp2 hybridised, that is to say, it is planar, and it has an empty p orbital perpendicular to the central C atom, and the 3 C atoms of the R groups (ethyl). In the 3 first C atoms of the Et groups, however, the geometry is tetrahedral, as the atoms are sp3 hybridised. This means that the sigma bonds are have a component in the direction defined by the p orbital of the central cation, and one of the bonds will (for these purposes) be coplanar with the p orbital. Hence, there will be an interaction between the empty p orbital and the filled sigma orbital of the bond from the first Et carbon to the second Et carbon (as will be the case in this example here), which can be though of for all intents and purposes as a "partial donation of electron density from the sigma bond which is in phase with the empty p orbital of the cation", hence a stabilising effect is present.
What you're describing with MO theory is still an inductive effect though, just with a better description of its origins!
What you're describing with MO theory is still an inductive effect though, just with a better description of its origins!
It isn't inductive. Induction is the tendency for charge to be transmitted through a network of atoms, resulting in the existence of a dipole. That isn't the cause of the stabilisation; that is a quantum effect.
It isn't inductive. Induction is the tendency for charge to be transmitted through a network of atoms, resulting in the existence of a dipole. That isn't the cause of the stabilisation; that is a quantum effect.
Sorry, I misread your explanation. You're right - it's hyperconjugation for carbocations. I'd still say induction is still a reasonable explanation for the basicity of substituted amines, though since they can't be planar.
Sorry, I misread your explanation. You're right - it's hyperconjugation for carbocations. I'd still say induction is still a reasonable explanation for the basicity of substituted amines, though since they can't be planar.
Oh, of course for amines, nitrogen is far more electronegative than carbon.
(Sorry, my explanation was mainly geared for the "alkyl groups are electron releasing" comment, as opposed to the features of substituted amines- I tend to ignore the OP sometimes)