enter image description here According to the source of the question answer is c>a>b but shouldn't the order be c>b>a as (a) is attached to a benzene ring and the lone pair will be delocalized?

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    $\begingroup$ B is effectively non-basic as it is part of an amide. The given answer is correct. $\endgroup$ – Waylander Mar 24 '19 at 17:38

I think we can all agree that the aliphatic amine has the greatest basicity since the lone pair is not delocalised into any neighbouring $\pi$ network. You have mentioned about the delocalisation of the lone pair on the $\ce {-NH2}$ substituent into the aromatic ring. However, as Waylander commented, the the lone pair of the $\ce {N}$ in the amide is also delocalised into the adjacent carbonyl group and he even goes to say that it is "effectively non-basic". He is certainly justified in making this comment because the protonation of amides actually occur at the carbonyl $\ce {O}$ and not the $\ce {N}$. To see this better, we need to examine the extent of delocalisation of the lone pairs on $\ce {N}$.

This analysis I believe can be carried out in two ways.

Looking at the resonance structures

Firstly, we could take a look at the resonance structures which can be drawn to illustrate the delocalisation. Shown below the resonance structures that can be drawn for the delocalisation of the lone pair in aniline, taken from here.

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Shown below is the resonance strcutres which can be drawn for the delocalisation in an amide, taken from here.

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Yes, more resonance structures can be drawn for the delocalisation of the lone pair into the aromatic ring. But this does not imply that the extent of delocalisation is greater. We have to take a look at the extent of contribution of the resonance structures to the resonance hybrid. Observe that in the three resonance structures, we have $\ce {N}$ bearing a positive formal charge while another $\ce {C}$ is bearing a formal negative charge. Recall that $\ce {N}$ is more electronegative than $\ce {C}$, thus this distribution of charge is less likely to be the case. Hence, the three resonance structures are likely to have small contributions to the overall resonance hybrid.

For the amide, we have a resonance structure where there is $\ce {N}$ incurring a positive formal charge while an $\ce {O}$ takes up the formal negative charge. This is a much more favourable and hence, much more likely charge distribution in the amide. Hence, this resonance structure has quite significant contribution to the resonance hybrid, albeit still less than the neutral structure.

Looking at some data

For there to be maximum delocalisation, the $\ce {N}$ atom must be as close to the $\ce {sp^2}$ hybridised state as possible where it can have an unhybridised $\ce {p}$ orbital being parallel to the adjacent $\pi$ system. How can we assess its hybridisation state? Well... We can look at the deviation from planarity of the $\ce {NH2}$ group. We can look at the value of the pyramidalisation angle between the $\ce {C-N}$ bond and the bisector of the $\ce {H-N-H}$ bond angle. The following values are obtained from Wiki:

\begin{array}{|c|c|c|c|} \hline \text{Compound} & \text{Pyramidalisation angle} \\ \hline \ce{Formamide} & 180 ^\circ\\ \hline \ce{Aniline} & 142.5 ^\circ\ \\ \hline \ce{Methylamine} & 125 ^\circ\ \\ \hline \end{array}

Notice that the amide is fully planar! This suggests a hybridisation state of $\ce {N}$ in the amide that is almost fully $\ce {sp^2}$ Although the $\ce {NH2}$ in aniline may not be fully planar, it is still relatively more planar than the pyramidal methylamine, suggesting greater $\ce {sp^2}$ character in the $\ce {N}$ of aniline than the usual amine. More evidence regarding the deviation from planarity can also be obtained from the energetics data for pyramidal inversion given by Ron here.


Both the analysis based on resonance structures and that based on empirical data converge to give the same conclusion:

There is greater delocalisation in the amide than in aniline.

Thus, the given answer is fully correct.

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