Amongst the following the most basic compound is
(a) p-nitroaniline
(b) acetanilide
(c) aniline
(d)

Q: Why doesn't the lone pair in benzylamine resonate with the benzene ring?

Because there is a $CH_2$ group acting as a buffer between the nitrogen and the benzene ring. The nitrogen is attached to an $sp^3$ hybridized carbon, which breaks the continuous $p$-orbital overlap required for resonance.

Q: What is cross-conjugation in acetanilide?

In acetanilide, the nitrogen's lone pair is pulled in two opposite directions: towards the benzene ring and towards the highly electronegative oxygen of the carbonyl group. This 'tug-of-war' is called cross-conjugation, making the lone pair highly unavailable.

Q: Why is p-nitroaniline a weaker base than aniline?

The $NO_2$ group is a powerful electron-withdrawing group. It pulls electron density out of the benzene ring via both $-I$ (inductive) and $-R$ (resonance) effects, which in turn pulls the lone pair away from the nitrogen even more strongly than in plain aniline.

Q: Does the phenyl group always act as an electron-withdrawing group?

When attached directly to an atom with a lone pair (like the nitrogen in aniline), the phenyl group acts as an electron-withdrawing group via resonance ($-R$ effect), delocalizing the lone pair into the ring.

Q: How does the $+I$ effect work in benzylamine?

The benzyl group ($C_6H_5CH_2-$) as a whole acts as an electron-donating group through the sigma bonds (inductive effect, $+I$). It pushes electron density towards the nitrogen, making the lone pair even more ready to accept a proton.

Question

Amongst the following the most basic compound is

Options:

(A) p-nitroaniline

(B) acetanilide

(C) aniline

(D) benzylamine

Sigma Percentile · Accepted Answer

### The Battle of the Bases: Finding the Strongest Amine

When we talk about the basicity of amines, we are essentially talking about generosity. A base is a chemical species that is willing to donate its lone pair of electrons to an incoming proton ($H^+$). The more available this lone pair is, the stronger the base.

However, if the lone pair is distracted or pulled away by other parts of the molecule, the basicity plummets. Let's evaluate our four contenders to see who is the most generous with their electrons.

### The Delocalized Losers

Let's start with **aniline**. In aniline, the nitrogen atom is directly attached to a benzene ring. The benzene ring acts as an electron-withdrawing group via resonance ($-R$ effect). The lone pair on the nitrogen delocalizes into the $\pi$-system of the ring, making it less available for donation.

Things get even worse with **$p$-nitroaniline**. Here, we have a powerful $-NO_2$ group sitting opposite the amine. The nitro group exerts both a strong $-I$ (inductive) and $-R$ (resonance) effect, aggressively pulling electron density out of the ring and, consequently, away from the nitrogen atom. This makes $p$-nitroaniline an extremely weak base.

What about **acetanilide**? In this molecule, the nitrogen's lone pair is caught in a tug-of-war. It can resonate into the benzene ring, but it can also resonate into the highly electronegative oxygen of the adjacent carbonyl ($C=O$) group. This phenomenon is known as **cross-conjugation**. Because the lone pair is heavily involved in resonance with the carbonyl group, its availability for protonation is drastically reduced.

### The Localized Champion

Finally, we arrive at **benzylamine** ($C_6H_5CH_2NH_2$). Notice a crucial structural difference here: the nitrogen atom is *not* directly attached to the benzene ring. Instead, it is attached to an $sp^3$ hybridized $-CH_2-$ group.

This $-CH_2-$ group acts as an insulating buffer. It completely breaks the conjugation, meaning the lone pair on the nitrogen is strictly **localized**. It cannot delocalize into the benzene ring. Furthermore, the benzyl group as a whole exerts a slight $+I$ (electron-donating inductive) effect, which actually pushes a bit more electron density onto the nitrogen atom.

### Final Conclusion

Because the lone pair in benzylamine is fully localized and its electron density is enhanced by the $+I$ effect, it is the most readily available for protonation. Therefore, benzylamine stands victorious as the strongest base among the given options.

**Correct Answer: (d) benzylamine**

Animated Solution for Chemistry - Organic Chemistry: Amongst the following the most basic compound is

Select Answer:

Visualized Solution

$C o m p a r in g B a s i c i t y o f A min es$

$C o n ce pt o f B a s i c i t y$

$A na l y s i s o f A ni l in e$

$A na l y s i s o f p -nitroaniline$

$A na l y s i s o f A ce t ani l i d e$

$A na l y s i s o f B e n z y l amin e$

$C o n c l u s i o n$

The Sigma Insight: Amines

The Battle of the Bases

The Delocalized Losers

The Localized Champion

Final Conclusion

Similar Questions

A. phenyl methanamine B. N,N-dimethylaniline C. N-methyl aniline D. Benzenamine Choose the correct order of basic nature of the above amines.

The decreasing order of basicity of the following amines is

Arrange the following amines in the decreasing order of basicity :

Which one of the following is the strongest base in aqueous solution ?

The increasing basicity order of the following compounds is (A) $CH_{3} CH_{2} NH_{2}$ (B) $CH_{3} CH_{2} NH ∣ CH_{2} CH_{3}$ (C) $H_{3} C - N - CH_{3} ∣ CH_{3}$ (D) $Ph - N - H ∣ CH_{3}$

In the following compounds, the decreasing order of basic strength will be

Which of the following is least basic?

The correct order of increasing basic nature for the bases $NH_{3}, CH_{3} NH_{2}$ and $(CH_{3})_{2} NH$ is

Considering the basic strength of amines in aqueous solution, which one has the smallest $p K_{b}$ value?

Consider the following four compounds I, II, III, and IV. Choose the correct statement(s).