Why does presence of a nitro group make the benzene ring less reactive in comparison to the unsubstituted benz

NCERT Exemplar · Q29

Q. Why does presence of a nitro group make the benzene ring less reactive in comparison to the unsubstituted benzene ring. Explain.

Rajasthan RbseShort· 3mImportance★★★★★est

76% · 68/90 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — Resonance Effects Comparison

Resonance Effects Comparison: The First Meeting

Imagine you're pushing a child on a swing. If you push at random moments, the swing jerks but never goes high. But if you push exactly when the swing is coming back toward you — matching its natural rhythm — each push adds energy, and the swing soars. That matching of your push to the swing's natural frequency is resonance.

Now, resonance isn't just for swings. It happens in electrical circuits, sound waves, bridges, and even atoms. The key idea is always the same: when the driving frequency matches the natural frequency, the response becomes maximum.

The Intuition First

Think of a system that can vibrate or oscillate — a pendulum, a guitar string, an electrical circuit with a capacitor and inductor. Every such system has a natural frequency at which it "likes" to oscillate. If you apply an external force (a push, a voltage, a sound wave) at that exact frequency, the system absorbs energy most efficiently. The amplitude of oscillation grows.

If you push at a frequency slightly different from the natural one, the system still responds, but less strongly. Far away from the natural frequency, the response is tiny.

Note

Resonance is not about "breaking" things — it's about energy transfer efficiency. The dramatic effects (like a singer shattering a glass) happen because energy builds up over many cycles.

The Precise Statement

For any oscillatory system driven by a periodic force, the amplitude of the steady-state response depends on the driving frequency $\omega$ . The response is maximum when:

\omega_{\text{driving}} = \omega_{\text{natural}}

This is the resonance condition.

For a simple damped harmonic oscillator (mass $m$ , spring constant $k$ , damping coefficient $b$ ), the natural frequency is:

\omega_0 = \sqrt{\frac{k}{m}}

And the amplitude $A$ of the driven oscillation is:

A(\omega) = \frac{F_0 / m}{\sqrt{(\omega_0^2 - \omega^2)^2 + (b\omega / m)^2}}

where $F_0$ is the amplitude of the driving force.

A_{\text{max}} = \frac{F_0}{b \omega_0} \quad \text{(at resonance, } \omega = \omega_0\text{)}

Notice: the maximum amplitude is inversely proportional to damping $b$ . Less damping means sharper, taller resonance.

Comparing Resonance Effects: What Changes?

When you compare resonance effects across different systems, you look at three things:

Sharpness of the peak — How quickly does the response drop off as you move away from resonance? This is measured by the quality factor $Q$ :

Q = \frac{\omega_0}{\Delta \omega}

where $\Delta \omega$ is the width of the resonance curve at half the maximum amplitude. High $Q$ means a very sharp, narrow peak (low damping). Low $Q$ means a broad, flat peak (high damping).

Peak height — The maximum amplitude at resonance. Higher $Q$ gives a taller peak.
Resonance frequency shift — In some systems (like with strong damping), the actual peak occurs slightly below $\omega_0$ . This shift is usually small but matters in precision applications.

Tip

For exams: When comparing two oscillators, always check damping first. The one with less damping has a higher and sharper resonance peak. The natural frequency itself depends only on $k$ and $m$ (or equivalent), not on damping.

A Concrete Comparison

Consider two pendulums:

Pendulum A: light string, small bob, no air resistance (low damping)
Pendulum B: heavy bob, thick string, in oil (high damping)

Both have the same length, so same natural frequency. But: …

Start your 14-day free trial to unlock the full solution →

The nitro group is a strong electron-withdrawing group that pulls electron density away from the benzene ring via both inductive and resonance effects, making the ring electron-deficient and thus less reactive toward electrophilic substitution.

The key to understanding this lies in the electronic structure of the nitro group itself. A nitro group ( $-\text{NO}_2$ ) has a nitrogen atom doubly bonded to one oxygen and singly bonded to another, with the second oxygen carrying a formal negative charge and the nitrogen a formal positive charge. This internal polarity is the starting point.

When attached to a benzene ring, the nitro group exerts two simultaneous effects that drain electron density from the ring.

1. The inductive effect — The nitrogen in $-\text{NO}_2$ is highly electronegative and carries a formal positive charge. Through the sigma bonds, it pulls electron density away from the carbon it is attached to. This is a through-bond, distance-dependent effect. It makes the ring carbon slightly positive, and this electron deficiency propagates weakly to the ortho and para positions. The inductive effect alone would reduce reactivity, but it is not the dominant factor here.

2. The resonance effect — This is the real story. The nitro group has a vacant p-orbital on the nitrogen (since it is $sp^2$ hybridised and forms a $\pi$ bond with one oxygen). The benzene ring's $\pi$ electrons can be delocalised into this vacant orbital. Draw the resonance structures: you will see that the negative charge from the ring moves onto the oxygen atoms of the nitro group, leaving positive charges on the ortho and para carbons of the ring.

\text{C}_6\text{H}_5-\text{NO}_2 \longleftrightarrow \text{structures with } \delta^+ \text{ at ortho/para carbons}

This is the critical point. In electrophilic substitution, the electrophile attacks the ring at a position of high electron density. But here, the nitro group has removed electron density from the ring, especially from the ortho and para positions — exactly where attack normally occurs. The ring is now electron-poor, or deactivated. …

Start your 14-day free trial to unlock the full solution →

←Previous questionQ28 — Despite their -I effect, halogens are o- and p-directing in haloarenes…→Next questionQ30 — Suggest a route for the preparation of nitrobenzene starting from acet…

Q.Why does presence of a nitro group make the benzene ring less reactive in comparison to the unsubstituted benzene ring. Explain.

🎓Unlock everything free for 14 days

Unlock everything free for 14 days