The correct order of the ease of dehydration of the following alcohols by the action of conc. $H_2SO_4$ is: (A) $(CH_3)_3C-OH > (CH_3)_2CH-OH > CH_3CH_2-OH$ (B) $(CH_3)_2CH-OH > CH_3CH_2-OH > (CH_3)_3C-OH$ (C) $CH_3CH_2-OH > (CH_3)_2CH-OH > (CH_3)_3C-OH$ (D) $(CH_3)_2CH-OH > (CH_3)_3C-OH > CH_3CH_2-OH$

The correct order of the ease of dehydration of the following alcohols by the action of conc. H₂SO₄ is: (A) (C

Question

H_2SO_4

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Concept understanding — Hindrance By Sulfuric Acid

Concentrated Sulfuric Acid: Its Several (Sometimes Conflicting) Roles

Concentrated sulfuric acid isn't just "a strong acid" in organic chemistry — depending on what it's reacting with, it can act as an oxidising agent, a dehydrating agent, or simply a strong Brønsted acid/proton source. Which role it plays is exactly what decides whether it HELPS or HINDERS the reaction you actually wanted.

Role 1: Oxidising Agent — Why It Can't Be Used to Make HX from a Halide Salt

If you try to prepare an alkyl iodide from an alcohol using $NaI$ and concentrated $H_2SO_4$ (instead of a non-oxidising acid), the reaction fails to give a clean product. Concentrated $H_2SO_4$ oxidises the iodide ion (and, less readily, bromide) straight to the free halogen:

2\,I^- + H_2SO_4 \longrightarrow I_2 + SO_2 + 2\,H_2O + \text{(other side products)}

The iodine you actually wanted to install ends up as elemental $I_2$ vapour instead of forming $HI$ or reacting further, so the intended substitution is hindered rather than helped. This is exactly why phosphoric acid ( $H_3PO_4$ ), which is not oxidising, is used instead when converting an alcohol to an alkyl iodide via $NaI$ / $KI$ .

Watch out

This oxidising side-reaction is specific to $H_2SO_4$ (and to a lesser extent nitric acid) — it is not a general property of "strong acids." Hydrochloric and phosphoric acids are strong/moderately strong acids that are NOT oxidising, which is exactly why they're the correct choice whenever a halide ion needs to survive the reaction untouched.

Role 2: Dehydrating Agent — Alcohol to Alkene

Heated with concentrated $H_2SO_4$ , an alcohol loses a molecule of water to form an alkene ( $E1$ pathway via a carbocation intermediate for 2°/3° alcohols):

R{-}CH_2{-}CH_2{-}OH \xrightarrow{\text{conc. } H_2SO_4, \Delta} R{-}CH{=}CH_2 + H_2O

Ease of dehydration follows carbocation stability: $3^\circ > 2^\circ > 1^\circ$ (a tertiary alcohol dehydrates fastest and at the lowest temperature).

Role 3: Protonating/Cleaving Agent — Ethers

Cold concentrated $H_2SO_4$ protonates an ether's oxygen lone pair to form an oxonium salt (this is also the basis of the solubility test that distinguishes ethers from alkanes, since the ether dissolves in cold conc. $H_2SO_4$ while an alkane does not): …

Why this formula?

Great — let’s break down Hindrance by Sulfuric Acid from first principles. This concept typically appears in qualitative inorganic analysis (salt analysis), specifically when testing for chloride ( $Cl^-$ ) in the presence of other anions.

What is “Hindrance by Sulfuric Acid”?

In salt analysis, we often add dilute sulfuric acid ( $H_2SO_4$ ) to a salt solution to test for carbonate ( $CO_3^{2-}$ ) or sulfite ( $SO_3^{2-}$ ) — they produce $CO_2$ or $SO_2$ gas.

But if the salt also contains chloride ( $Cl^-$ ), the sulfuric acid can hinder the detection of chloride later. Why? Because concentrated $H_2SO_4$ reacts with chloride to produce HCl gas, which escapes — but in the process, it also interferes with tests for other anions.

The key formula here is the reaction of chloride with concentrated sulfuric acid:

2NaCl + H_2SO_4 \ (\text{conc.}) \xrightarrow{\Delta} Na_2SO_4 + 2HCl \uparrow

But that’s just the result. Let’s see why this happens.

Why does this reaction occur?

1. Acid strength and volatility

$H_2SO_4$ is a strong, non-volatile acid.
$HCl$ is a volatile acid (gas at room temperature).
When a non-volatile acid reacts with a salt of a volatile acid, the volatile acid is displaced and escapes as gas.

This is a general principle:

\text{Salt of volatile acid} + \text{Non-volatile acid} \rightarrow \text{New salt} + \text{Volatile acid} \uparrow

Here:

Salt of volatile acid = $NaCl$ (gives $HCl$ )
Non-volatile acid = $H_2SO_4$

So the reaction is thermodynamically favoured because the product $HCl$ leaves the system (gas), driving the equilibrium forward.

2. Why “hindrance” in analysis?

When you add dilute $H_2SO_4$ to test for carbonate, you get $CO_2$ gas. But if chloride is present, the concentrated $H_2SO_4$ (used later in a separate test) reacts with chloride to produce $HCl$ gas. This $HCl$ can:

Mask the test for other gases (e.g., $SO_2$ from sulfite)
Interfere with confirmatory tests (e.g., with $AgNO_3$ — but that’s a separate step)

Thus, the hindrance is that the presence of chloride consumes the sulfuric acid and produces a gas that confuses the analysis.

The key formula — with reasoning

The core reaction is:

\boxed{2Cl^- + H_2SO_4 \ (\text{conc.}) \rightarrow SO_4^{2-} + 2HCl \uparrow}

Why this formula holds: …

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Alcohol dehydration proceeds via carbocation intermediates; tertiary carbocations are most stable, so tertiary alcohols dehydrate fastest. The order is tertiary > secondary > primary.

Why Carbocation Stability Governs Dehydration

When concentrated sulfuric acid acts on an alcohol, it protonates the hydroxyl group to form $\ce{ROH2^+}$ , converting a poor leaving group ( $\ce{OH^-}$ ) into an excellent one ( $\ce{H2O}$ ). The mechanism then proceeds through loss of water to generate a carbocation intermediate, which rapidly loses a proton to form the alkene.

The rate-determining step is carbocation formation. Since tertiary carbocations are stabilized by hyperconjugation and inductive effects from three alkyl groups, they form much more readily than secondary (two alkyl groups) or primary (one alkyl group) carbocations. This stability difference directly translates into reaction rate: the more stable the carbocation, the faster the dehydration.

Let's identify each alcohol and rank them:

$(CH_3)_3C-OH$ is tert-butanol — a tertiary ( $3°$ ) alcohol. Upon protonation and loss of water, it forms the tert-butyl cation $(CH_3)_3C^+$ , which is stabilized by nine $\alpha$ -hydrogens through hyperconjugation and three electron-donating methyl groups. This is the most stable carbocation of the three.
$(CH_3)_2CH-OH$ is isopropanol — a secondary ( $2°$ ) alcohol. It forms the isopropyl cation $(CH_3)_2CH^+$ , stabilized by six $\alpha$ -hydrogens and two methyl groups. Moderately stable, but less so than tertiary. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2024Set 56/1/11 markMCQ
$H_2SO_4$
›Reveal solutionSolution
Alcohol dehydration proceeds via carbocation intermediates; tertiary carbocations are most stable, so tertiary alcohols dehydrate fastest. The order is tertiary > secondary > primary.

Why Carbocation Stability Governs Dehydration

When concentrated sulfuric acid acts on an alcohol, it protonates the hydroxyl group to form $\ce{ROH2^+}$ , converting a poor leaving group ( $\ce{OH^-}$ ) into an excellent one ( $\ce{H2O}$ ). The mechanism then proceeds through loss of water to generate a carbocation intermediate, which rapidly loses a proton to form the alkene.

The rate-determining step is carbocation formation. Since tertiary carbocations are stabilized by hyperconjugation and inductive effects from three alkyl groups, they form much more readily than secondary (two alkyl groups) or primary (one alkyl group) carbocations. This stability difference directly translates into reaction rate: the more stable the carbocation, the faster the dehydration.

Let's identify each alcohol and rank them:

$(CH_3)_3C-OH$ is tert-butanol — a tertiary ( $3°$ ) alcohol. Upon protonation and loss of water, it forms the tert-butyl cation $(CH_3)_3C^+$ , which is stabilized by nine $\alpha$ -hydrogens through hyperconjugation and three electron-donating methyl groups. This is the most stable carbocation of the three.

$(CH_3)_2CH-OH$ is isopropanol — a secondary ( $2°$ ) alcohol. It forms the isopropyl cation $(CH_3)_2CH^+$ , stabilized by six $\alpha$ -hydrogens and two methyl groups. Moderately stable, but less so than tertiary. …
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Why Carbocation Stability Governs Dehydration

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