# Decreasing order of basic strength

I need to find the decreasing basicity of the following compounds(I, II, III, IV, respectively):

I know that to compare the basic strengths, we need to find the stability of their conjugate acids. Following are the conjugate acids, I drew (I don't know if they're correct):

According to me, due to resonance of the benzyl groups or lack of resonance, the order of basic strength should be:

$$\mathrm{III > II > I > IV}$$

$$\mathrm{IV > I > II > III}$$

Where am I going wrong?

• Resonance doesn't occur here because N can't expand its octet. Now I am not sure thereon, but I think that due to a lack of resonance the conjugate acids are roughly the same stability so we look at the stabilities of the original bases. Now here the lone pair is in resonance in three structures and you can see the order of stability. Thus the most stable base will have poor basic strength and you will get the answer. However, I am not sure of this. Maybe someone can give a definitive explanation. Mar 12, 2017 at 6:27
• @Sawarnik , yes. That could be a method. But, why can't N expand its octet here? Mar 12, 2017 at 6:45
• It has no d orbitals, so with only the s and p orbitals it can accomodate a max. of 8 electrons. If you drew resonance structures, it will have 5 bonds aka 10 electrons which is not possible. Mar 12, 2017 at 6:47
• To be more precise: nitrogen does have virtual d orbitals, like sulphur, phosphorus or whatever. None of them, however, expand their octet because the d orbitals are energetically not accessible.
– Jan
Oct 19, 2017 at 12:25

Basicity is how well an atom can donate electrons. More is the electron density more is the basicity of the compound.

In Aniline the lone pair of electrons present on the nitrogen are involved in resonance, so the electron density on N is less than expected.

In compound II (Diphenylamine), since two benzene ring is present so the electron density will be less as now lone pair are involved in resonance of both the ring. Thus electron density on Nitrogen is less than as compare to Nitrogen atom of Aniline

Similarly, in III compound (Triphenylamine), since three benzene ring is present the lone pair are more localized, and, thus electron density will be much less (least) as compared to the other compounds

In 4th structure, there is no resonance so, lone pair are present with nitrogen itself, and the electron density will be high, also due to inductive effect of the alkyl groups the electron density is more, and thus it can easily donate the electrons as compared to all other compounds. So the basicity is the highest for compound IV.

Therefore, the order of basicity is IV>I>II>III.

• Exactly what i thought .. i have seen a few questions which were solved by the same logic (where N was acting as a base). Why we don't care about the conjugate acid is something i don't fully understand though. Mar 12, 2017 at 9:26
• You are thinking in advance, like conditions after the H+ is accepted, but in order to accept H+ atom, first there should be some electron density present around the donor atom. If there is no (less) electron density then H+ will not get attached, and this is where you go wrong. I suggest you to solve the basicity problems of this type by considering the electron density, and problems based on acidity by conjugate acid-base method. Mar 12, 2017 at 10:16

As we know if conjugate acid is more stable then their parent compound is least stable.so in 4th option their is no any resonance so it is least stable therefore it's parent compound is more stable.so correct option is iv>I>ii>iii

The aryl group is electron withdrawing in nature, so as the no.of of aryl groups increase, the acidity increase while $$\ce{CH3}$$ being electron donating in nature leads to +I effect and thus to basic strength.. That's why (IV) will come first then (I) then (II) and then after (III). So the order as per decreasing basic strength is (IV)> (I)>(II)> (III)

i guess i could help out on this one..as i have completed the book(for JEE) The reason for the abnormal order is that the presence of more than one phenyl groups dosent actually increase the stability of the conj acid...rather it just increases the steric crowding around 'N'....The reason why more phenyl groups dont add to stability is that the phenyl grups due to repulsions between themseles keep rotating about the axis and are not always in the same plane for conjugation...thus do not really help in stabilising the N+ ion.....and i guess youd know that cyclohexyl amine is more basic than anilene...so for that simple reason.....4>1>2>3

P.S:The answer was provided to me by a teacher at FIITJEE...so you can be sure its authentic

POINT 1. If you think resonance will play a role try drawing resonance structure. Got stuck hmmm, because nitrogen can not form 5 bonds or can not expand its octet. Hence, resonance will not play any role.

POINT 2. Now, coming to conjugate method, add $\ce{H+}$ to the structures and then try to compare stability by considering inductive effect ( phenyl group shows -I effect ). You can not use resonance effect for comparing because of reason already discussed in point 1, that is, -I effect will be maximum in the case of triphenyl then di phenyl then phenyl and so on so the correct order will be 4>1>2>3.

POINT3. Now consider steric factors. The greather the steric hindrance, the less will be the basicity because $\ce{H+}$ will face more difficulty in coming close. So again, according to steric factors, the order will be 4>1>2>3.

In the first, $\ce{NH3+}$ is in conjugation with the benzene and also it is a $+M$ group which increases the negative charge and decreases the $K_\mathrm a$ value and in turn increases the $\mathrm pK_\mathrm a$ value, meaning it increases the pH of the solution (this is the same for the 1st 2nd and the 3rd). The fourth one doesn't have conjugation with the ring so the fourth has the highest acidity.
Now you have to give the order according to the steric hindrance so the correct answer would be: $4>1>2>3$