Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why?

Exercises · 9.23

Q. Out of benzene, m-dinitrobenzene and toluene which will undergo nitration most easily and why?

Rajasthan RbseTextbookSubjective· 2mImportance★★★★★est

41% · 37/90 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 ease of nitration depends on how strongly the substituent already on the ring activates it toward electrophilic attack. Toluene has a strongly activating methyl group, benzene has no substituent, and m-dinitrobenzene has two strongly deactivating nitro groups. So toluene undergoes nitration most easily.

The core idea: activation vs. deactivation

Nitration is an electrophilic aromatic substitution reaction. The benzene ring acts as a nucleophile, attacking the nitronium ion ( $\text{NO}_2^+$ ). Anything that increases the electron density on the ring makes it a better nucleophile — that speeds up the reaction. Anything that pulls electron density away slows it down.

Substituents already on the ring do one of two things:

Activating groups (like $-\text{CH}_3$ ) push electrons into the ring, making it more reactive than benzene itself.
Deactivating groups (like $-\text{NO}_2$ ) pull electrons out of the ring, making it less reactive.

The question gives us three compounds: benzene (no substituent), toluene (methyl group), and m-dinitrobenzene (two nitro groups). Let's compare them.

Step-by-step reasoning

1. Identify the substituent effect in each compound

Benzene: No substituent. This is our baseline — its reactivity is the reference point.
Toluene: The methyl group ( $-\text{CH}_3$ ) is an activating group. It donates electron density to the ring through the hyperconjugation and inductive effect (it's weakly electron-releasing). This makes the ring more electron-rich than benzene.
m-Dinitrobenzene: Each nitro group ( $-\text{NO}_2$ ) is a strongly deactivating group. It pulls electron density away from the ring through both the inductive effect (electronegative nitrogen and oxygen) and the resonance effect (the nitro group has a positive nitrogen that can accept electron density from the ring). Two such groups make the ring extremely electron-poor.

Watch out

A common mistake is to think that because m-dinitrobenzene already has nitro groups, it must be "used to" nitration and therefore reacts easily. In fact, the opposite is true: the nitro groups make further substitution very difficult. The second nitration of benzene (to give m-dinitrobenzene) is already much slower than the first; a third nitration is even harder.

2. Compare the electron density on the ring

The rate of electrophilic substitution depends directly on how much electron density the ring can offer to the attacking $\text{NO}_2^+$ ion.

In toluene, the methyl group pushes electrons in, so the ring has higher electron density than benzene. The transition state for nitration is stabilised by the methyl group.
In benzene, there is no extra push or pull — it's neutral.
In m-dinitrobenzene, both nitro groups pull electrons out, so the ring has much lower electron density than benzene. The transition state is destabilised.

Tip

You can remember the order of reactivity for common substituents using the mnemonic: "Ortho-para directors are activators (except halogens), meta directors are deactivators." Methyl is an ortho-para director and an activator; nitro is a meta director and a deactivator. …

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

←Previous question · exercises9.22 — Arrange the following set of compounds in order of their decreasing re…→Next question · exercises9.24 — Suggest the name of a Lewis acid other than anhydrous aluminium chlori…

Q.Out of benzene, m–dinitrobenzene and toluene which will undergo nitration most easily and why?

The core idea: activation vs. deactivation

Step-by-step reasoning

🎓Unlock everything free for 14 days

Unlock everything free for 14 days