Assertion: tert-Butyl bromide undergoes Wurtz reaction to give 2,2,3,3-tetramethylbutane. Reason: In Wurtz rea

NCERT Exemplar · Q88

Q. Assertion: tert-Butyl bromide undergoes Wurtz reaction to give 2,2,3,3-tetramethylbutane. Reason: In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon containing double the number of carbon atoms present in the halide.

(i) Assertion and reason both are correct and reason is correct explanation of assertion.

(ii) Assertion and reason both are wrong statements.

(iii) Assertion is correct but reason is wrong statement.

(iv) Assertion is wrong but reason is correct statement.

(v) Assertion and reason both are correct statements but reason is not correct explanation of assertion.

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Concept understanding — Inductive Effect on Acidity

Inductive Effect on Acidity – From Intuition to Precision

Imagine you are holding a rope tied to a heavy box. If you pull the rope, the box moves toward you. Now imagine the rope is made of rubber bands — the pull still reaches the box, but it gets weaker the farther away you are. That is exactly how the inductive effect works inside a molecule.

The Core Intuition

An acid donates a proton ( $H^+$ ). After it does, the remaining part (the conjugate base) carries a negative charge. The stability of that negative charge determines how willing the molecule is to give up the proton. More stable conjugate base → stronger acid.

Now, some atoms or groups are electron-withdrawing — they pull electron density toward themselves through the sigma bonds. If such a group is attached near the acidic proton, it pulls some electron density away from the negative charge on the conjugate base. That spreads out (delocalises) the negative charge, making the conjugate base more stable. The acid becomes stronger.

Conversely, electron-donating groups push electron density toward the negative charge, concentrating it and making the conjugate base less stable. The acid becomes weaker.

Note

The inductive effect operates through sigma bonds only. It does not involve pi bonds or resonance. It is a permanent, through-bond polarisation.

The Precise Statement

Inductive effect on acidity: The acidity of a compound increases with the presence of electron-withdrawing groups (EWGs) near the acidic site, and decreases with electron-donating groups (EDGs). The effect is strongest when the group is closest to the acidic proton, and diminishes rapidly with distance.

Mathematically, for a series of substituted carboxylic acids:

\text{R-COOH} \quad \text{where R = substituent}

The acid dissociation constant $K_a$ changes as:

If R is electron-withdrawing (e.g., $-\text{Cl}$ , $-\text{NO}_2$ , $-\text{CF}_3$ ): $K_a$ increases → stronger acid.
If R is electron-donating (e.g., $-\text{CH}_3$ , $-\text{C}_2\text{H}_5$ ): $K_a$ decreases → weaker acid.

Why Distance Matters

The inductive effect falls off with distance because sigma bonds are localised. Each bond attenuates the effect by roughly a factor of 2–3. For example, compare:

Compound	$pK_a$	Explanation
$\text{CH}_3\text{COOH}$	4.76	Reference (no EWG)
$\text{ClCH}_2\text{COOH}$	2.86	Cl withdraws through one bond
$\text{Cl}_2\text{CHCOOH}$	1.29	Two Cl atoms, stronger withdrawal
$\text{Cl}_3\text{CCOOH}$	0.65	Three Cl atoms, strongest withdrawal
$\text{CH}_3\text{CH}_2\text{COOH}$	4.87	Ethyl group is electron-donating (slightly weaker acid)

Notice: $\text{ClCH}_2\text{COOH}$ is about 100 times stronger than acetic acid ( $\Delta pK_a \approx 1.9$ ). But if the Cl is moved further away:

Compound	$pK_a$
$\text{ClCH}_2\text{CH}_2\text{COOH}$	4.08
$\text{ClCH}_2\text{CH}_2\text{CH}_2\text{COOH}$	4.52

The effect fades as the chlorine moves farther from the carboxyl group. …

Why this formula?

Inductive Effect on Acidity: Why It Works

The inductive effect is a through-bond electron displacement caused by differences in electronegativity. When we ask why it affects acidity, we must first understand what acidity means at the molecular level.

The Core Idea: Stabilising the Conjugate Base

Acidity is governed by the equilibrium:

\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-

The stronger the acid, the more it favours the right side. This happens when the conjugate base $\text{A}^-$ is more stable. The inductive effect directly influences this stability.

Why Electron-Withdrawing Groups (EWG) Increase Acidity

Consider a carboxylic acid with an electronegative atom (like Cl) attached to the carbon chain:

\text{Cl} - \text{CH}_2 - \text{COOH}

The inductive pull: The Cl atom is more electronegative than carbon. It pulls electron density toward itself through the sigma bonds.
Effect on the O–H bond: This electron withdrawal travels along the carbon chain, reducing electron density around the O–H bond. The bond becomes more polarised, making the H⁺ easier to remove.
Stabilising the conjugate base: After losing H⁺, the negative charge on the carboxylate ion ( $\text{RCOO}^-$ ) is delocalised by resonance. But the inductive effect further stabilises this negative charge by pulling electron density away from the oxygen atoms. This makes the conjugate base less reactive (more stable), shifting equilibrium toward dissociation.

Key insight: The inductive effect doesn't just weaken the O–H bond — it stabilises the anion that forms after deprotonation.

The Quantitative Relationship: Hammett Equation

For substituted benzoic acids, the effect is quantified by the Hammett equation:

\log \left( \frac{K_a}{K_a^0} \right) = \sigma \rho

Where:

$K_a$ = acid dissociation constant of substituted acid
$K_a^0$ = acid dissociation constant of unsubstituted benzoic acid
$\sigma$ = substituent constant (measures inductive + resonance effect)
$\rho$ = reaction constant (sensitivity of the reaction to substituent effects)

Why This Formula Holds

The derivation comes from linear free-energy relationships:

Free energy change: For any acid dissociation:

\Delta G^\circ = -RT \ln K_a

Effect of substituent: A substituent changes $\Delta G^\circ$ by an amount proportional to its electronic effect:

\Delta (\Delta G^\circ) = -RT \ln \left( \frac{K_a}{K_a^0} \right)

Separability assumption: The total effect of a substituent on any reaction can be factored into:
- A substituent-specific term ( $\sigma$ ) — how strongly it pulls/pushes electrons
- A reaction-specific term ( $\rho$ ) — how sensitive the reaction is to electronic effects
Empirical validation: Hammett found that for meta and para substituted benzoic acids, plotting $\log(K_a/K_a^0)$ against $\sigma$ gives a straight line. This confirms the additive nature of inductive effects.

The Inductive Effect Constant ( $\sigma_I$ ) …

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The Wurtz reaction couples two molecules of an alkyl halide through a new C–C bond, so the hydrocarbon formed has double the halide's carbon count. Two tert-butyl bromide molecules ( $\mathrm{C_4}$ each) couple to $(\mathrm{CH_3})_3\mathrm{C{-}C}(\mathrm{CH_3})_3$ — which is exactly 2,2,3,3-tetramethylbutane ( $\mathrm{C_8}$ ). The assertion is correct, the reason is correct, and the reason (the doubling rule) is precisely what predicts the asserted product. The correct option is (i).

The Wurtz reaction

2\,\mathrm{R{-}X} + 2\,\mathrm{Na} \xrightarrow{\text{dry ether}} \mathrm{R{-}R} + 2\,\mathrm{NaX}

Two alkyl halide molecules join through a new carbon–carbon bond, giving a symmetrical alkane with twice the number of carbon atoms of the starting halide. This is exactly what the Reason states — as a general description of the Wurtz reaction it is correct.

Checking the assertion

tert-Butyl bromide is $(\mathrm{CH_3})_3\mathrm{C{-}Br}$ — a 4-carbon halide (three methyl carbons plus the central carbon). Applying the Wurtz coupling:

2\,(\mathrm{CH_3})_3\mathrm{C{-}Br} + 2\,\mathrm{Na} \xrightarrow{\text{dry ether}} (\mathrm{CH_3})_3\mathrm{C{-}C}(\mathrm{CH_3})_3 + 2\,\mathrm{NaBr}

Name the product: the longest chain through the two central carbons is a butane chain, and C2 and C3 each carry two methyl groups — so the product is 2,2,3,3-tetramethylbutane, an 8-carbon alkane. That is exactly the compound the assertion names, so the assertion is correct.

Is the reason the correct explanation?

Yes. The reason's doubling rule ( $\mathrm{C_4}$ halide → $\mathrm{C_8}$ hydrocarbon by coupling of the two alkyl groups) is precisely the logic that produces 2,2,3,3-tetramethylbutane from tert-butyl bromide. Both statements are correct and the reason explains the assertion — option (i). …

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Q.Assertion: tert-Butyl bromide undergoes Wurtz reaction to give 2,2,3,3-tetramethylbutane.
Reason: In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon containing double the number of carbon atoms present in the halide.

Concept understanding — Inductive Effect on Acidity

Inductive Effect on Acidity – From Intuition to Precision

The Core Intuition

The Precise Statement

Why Distance Matters

Why this formula?

Inductive Effect on Acidity: Why It Works

The Core Idea: Stabilising the Conjugate Base

Why Electron-Withdrawing Groups (EWG) Increase Acidity

The Quantitative Relationship: Hammett Equation

Why This Formula Holds

The Inductive Effect Constant ( $\sigma_I$ ) …

The Wurtz reaction

Checking the assertion

Is the reason the correct explanation?

Concept: Wurtz Reaction

Method: Step-by-Step Verification

Common Mistakes Students Make on This Question

1. Counting tert-butyl bromide as 3 carbons

2. Mis-naming the coupled product

3. Over-applying the "tertiary halides eliminate" caveat and calling the assertion wrong

Q.Assertion: tert-Butyl bromide undergoes Wurtz reaction to give 2,2,3,3-tetramethylbutane. Reason: In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon containing double the number of carbon atoms present in the halide.

The Wurtz reaction

Checking the assertion

Is the reason the correct explanation?

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Q.Assertion: tert-Butyl bromide undergoes Wurtz reaction to give 2,2,3,3-tetramethylbutane.
Reason: In Wurtz reaction, alkyl halides react with sodium in dry ether to give hydrocarbon containing double the number of carbon atoms present in the halide.

Unlock everything free for 14 days