Alcohol Dehydration

8 MCQs9-step worked example

Source: NCERT Alcohols, Phenols, Ethers and Carbonyl CompoundsOfficial key: NTA-verifiedLast updated: 25 Sep 2026

Alcohol Dehydration, explained for NEET

Alcohol dehydration is the acid-catalysed elimination of water from an alcohol to form an alkene. The reaction proceeds via an E1 mechanism for secondary and tertiary alcohols, and the product distribution follows Zaitsev's rule — the more substituted alkene is the major product.

The mechanism in three steps (for 2° and 3° alcohols):

Protonation — the –OH group is protonated by concentrated H₂SO₄ or H₃PO₄ to form an oxonium ion (–OH₂⁺), converting –OH into a good leaving group.
Loss of water — the leaving group departs, generating a carbocation intermediate.
Deprotonation — a base (HSO₄⁻ or water) abstracts a β-hydrogen, forming the C=C double bond.

Ease of dehydration: 3° > 2° > 1° alcohols. Tertiary alcohols dehydrate at lower temperatures (~443 K) because the tertiary carbocation intermediate is most stable. Primary alcohols require higher temperatures (~453 K) and may proceed via an E2-like pathway.

Zaitsev's rule applied: When multiple β-hydrogens are available, the alkene with the greater number of alkyl substituents on the double bond predominates. For example, butan-2-ol dehydration gives but-2-ene (major) over but-1-ene (minor).

Carbocation rearrangement — the trap NEET exploits: If the initial carbocation can rearrange via a 1,2-hydride or 1,2-methyl shift to a more stable carbocation, the rearranged product forms. Example: 3,3-dimethylbutan-2-ol → 2,3-dimethylbut-2-ene (via methyl shift from secondary to tertiary carbocation). NEET questions often give a substrate where the "obvious" product is the non-rearranged alkene — pick the rearranged one.

Key conditions (NCERT Class 12 Chemistry Chapter 7, page 208): concentrated H₂SO₄ at 443 K is the standard dehydration condition. Alumina (Al₂O₃) at 623 K is the alternative heterogeneous catalyst.

Can you answer these Alcohol Dehydration MCQs?

Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.

MCQ 1Easy RecallPractice

The ease of acid-catalysed dehydration of alcohols follows the order:

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Answer: B. Tertiary alcohols form the most stable carbocation intermediate, requiring the least activation energy for dehydration. The order is 3° > 2° > 1° (NCERT Class 12 Chemistry Chapter 7, page 208).

Why A is wrong: A reverses the actual order. The stability of the carbocation intermediate increases from primary to tertiary, making tertiary alcohols easiest to dehydrate, not hardest.

Why C is wrong: C incorrectly places secondary above tertiary. The tertiary carbocation is more stable than the secondary due to greater hyperconjugation and inductive stabilisation.

Why D is wrong: D scrambles the order. There is no mechanistic basis for primary being easier than tertiary — primary carbocations are the least stable.

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MCQ 2Easy RecallPractice

Which reagent/condition is the standard laboratory method for dehydration of alcohols as described in NCERT?

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Answer: C. Concentrated sulphuric acid at 443 K (170 °C) is the standard acid catalyst for alcohol dehydration (NCERT Class 12 Chemistry Chapter 7, page 208). Dilute acid and lower temperatures are insufficient.

Why A is wrong: A uses dilute acid at room temperature — insufficient acid concentration and temperature to protonate the hydroxyl group and drive elimination.

Why B is wrong: B uses HCl, which favours nucleophilic substitution (SN) to form alkyl halides rather than elimination. HCl is not a dehydrating agent.

Why D is wrong: D describes the Lucas reagent conditions (anhydrous ZnCl₂ + conc. HCl), which test for alcohol classification via substitution, not dehydration.

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MCQ 3Easy RecallPractice

What is the role of H₂SO₄ in the first step of the E1 dehydration mechanism of a secondary alcohol?

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Answer: B. In the first mechanistic step, the acid protonates –OH to form –OH₂⁺, converting a poor leaving group (OH⁻) into a good one (H₂O). This enables the subsequent C–O bond cleavage (NCERT Class 12 Chemistry Chapter 7, page 208).

Why A is wrong: A confuses dehydration with oxidation. Concentrated H₂SO₄ at 443 K causes elimination, not oxidation. Oxidation would require reagents like PCC or KMnO₄.

Why C is wrong: C describes a later step. β-hydrogen abstraction occurs AFTER the carbocation forms (step 3). In step 1, the acid's role is protonation of oxygen, not hydrogen abstraction.

Why D is wrong: D is chemically implausible. H₂SO₄ is an oxidising acid, not a reducing agent. Reduction to an alkane would require catalytic hydrogenation or LiAlH₄ under different conditions.

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MCQ 4Direct ApplicationPractice

Butan-2-ol undergoes acid-catalysed dehydration. According to Zaitsev's rule, the major product is:

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Answer: D. Zaitsev's rule states that the more substituted alkene is the major product. Butan-2-ol has β-hydrogens on C-1 and C-3. Elimination toward C-3 gives but-2-ene (disubstituted), while elimination toward C-1 gives but-1-ene (monosubstituted). But-2-ene is the major product.

Why A is wrong: A is the less substituted (Hofmann) product. But-1-ene is monosubstituted, while but-2-ene is disubstituted. Under acid-catalysed E1 conditions, Zaitsev's rule applies — the more substituted alkene predominates.

Why B is wrong: B is an alkane with no double bond. Dehydration is an elimination producing an alkene. Formation of butane would require reduction, not elimination.

Why C is wrong: C requires loss of two molecules of H₂O or a different mechanism entirely. Simple acid-catalysed dehydration removes one H₂O to form a single double bond, not a diene.

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MCQ 5Direct ApplicationPractice

3,3-Dimethylbutan-2-ol is heated with concentrated H₂SO₄. The major product is:

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Answer: C. The initial secondary carbocation at C-2 rearranges via a 1,2-methyl shift to a more stable tertiary carbocation at C-3 (now bearing three alkyl groups). Subsequent β-hydrogen elimination from this rearranged carbocation gives 2,3-dimethylbut-2-ene as the major product (most substituted alkene, Zaitsev's rule).

Why A is wrong: A is the product expected if no rearrangement occurs AND Hofmann elimination is followed. However, under E1 conditions, the secondary carbocation rearranges to a tertiary one via methyl shift before elimination.

Why B is wrong: B would form from the rearranged tertiary carbocation by eliminating toward the less substituted side. This is the minor (Hofmann) product — Zaitsev's rule favours the tetrasubstituted alkene instead.

Why D is wrong: D has the wrong carbon skeleton. No bond-breaking/forming sequence from 3,3-dimethylbutan-2-ol can generate a five-carbon chain. The methyl shift rearranges the branching pattern but preserves the six-carbon skeleton.

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MCQ 6Direct ApplicationPractice

Which of the following alcohols requires the highest temperature for acid-catalysed dehydration?

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Answer: D. Ethanol is a primary alcohol, and primary alcohols need the harshest conditions because primary carbocations are very unstable: NCERT gives ethanol with concentrated H₂SO₄ at 443 K, secondary alcohols with 85% H₃PO₄ at 440 K, and tertiary alcohols with 20% H₃PO₄ at 358 K. So 2-methylpropan-2-ol (3°) dehydrates most easily, cyclohexanol and propan-2-ol (2°) come next, and ethanol needs the most.

Why A is wrong: A is a tertiary alcohol — it forms the most stable carbocation and dehydrates at the LOWEST temperature among these options, not the highest.

Why B is wrong: B is a secondary alcohol. Secondary alcohols dehydrate with 85% H₃PO₄ at about 440 K, milder conditions than ethanol (primary) needs.

Why C is wrong: C is a secondary alcohol. Propan-2-ol dehydrates at a moderate temperature, between tertiary (lowest) and primary (highest).

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MCQ 7Concept TrapPractice

During the dehydration of neopentyl alcohol (2,2-dimethylpropan-1-ol) with H₂SO₄, a rearranged alkene is obtained. The rearrangement occurs because:

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Answer: A. The initial primary carbocation at C-1 is highly unstable. A methyl group on the adjacent quaternary carbon migrates (1,2-methyl shift) to C-1, converting the primary carbocation into a more stable tertiary carbocation at C-2. This rearranged cation then loses a proton to give 2-methylbut-2-ene.

Why B is wrong: B identifies the wrong type of shift. The adjacent carbon (C-2) in neopentyl alcohol is quaternary — it has no hydrogen to migrate. Only a methyl group can shift from C-2 to C-1.

Why C is wrong: C is irrelevant here. Ring expansion occurs in cyclic systems (e.g., cyclopentane → cyclohexane). Neopentyl alcohol is acyclic, so no ring expansion is possible.

Why D is wrong: D denies the rearrangement. A primary carbocation adjacent to a quaternary centre with available methyl groups will always rearrange. The unrearranged product (3,3-dimethylbut-1-ene from direct E2) is at best a very minor pathway under these conditions.

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MCQ 8CalculationPractice

When 2-methylcyclohexanol undergoes acid-catalysed dehydration, two alkene products are possible. Applying Zaitsev's rule, the major product is:

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Answer: A. The carbocation forms at C-1 after water loss. There are β-hydrogens on C-2 (giving 1-methylcyclohexene, trisubstituted) and on C-6 (giving 3-methylcyclohexene, disubstituted). Zaitsev's rule predicts the more substituted alkene: 1-methylcyclohexene (trisubstituted, with the double bond between C-1 and C-2) is the major product.

Why B is wrong: B is the less substituted (disubstituted) alkene, formed by removing a β-hydrogen from C-6. Zaitsev's rule dictates that the trisubstituted 1-methylcyclohexene is the major product.

Why C is wrong: C (methylenecyclohexane) would have an exocyclic double bond. This is the least substituted possibility (only disubstituted with one ring carbon and the exocyclic CH₂). Zaitsev's rule strongly disfavours this product — it would be minor at best.

Why D is wrong: D is cyclohexene without a methyl group. Loss of the methyl group does not occur during dehydration — only water is lost. The carbon skeleton is preserved.

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How do you solve a Alcohol Dehydration question? A worked example

1
Given
3,3-Dimethylbutan-2-ol is heated with concentrated H₂SO₄.
2
Required
Identify the major dehydration product.
3
Concept
Acid-catalysed dehydration follows E1 mechanism for secondary/tertiary alcohols. If a carbocation intermediate can rearrange to a more stable cation via a 1,2-shift, the rearranged product predominates. Zaitsev's rule then selects the most substituted alkene from the rearranged carbocation.
4
Formula / Principle
Zaitsev's rule: the more substituted alkene is the major product. Carbocation stability: 3° > 2° > 1°.
5
Substitution / Setup
• Substrate: CH₃–C(CH₃)₂–CH(OH)–CH₃ (3,3-dimethylbutan-2-ol)• Step 1: –OH protonated → –OH₂⁺• Step 2: H₂O leaves → secondary carbocation at C-2: CH₃–C(CH₃)₂–C⁺H–CH₃• Assess: Can C-2 (2°) rearrange? Adjacent C-3 has three methyl groups. A 1,2-methyl shift moves one methyl from C-3 to C-2.
6
Calculation / Reasoning
After 1,2-methyl shift:• C-2 now has two methyl groups and is bonded to C-3 → tertiary carbocation at C-3: CH₃–C(CH₃)=C(CH₃)–CH₃ path available.• The rearranged tertiary carbocation: (CH₃)₂C⁺–CH(CH₃)–CH₃ is equivalent to the 2,3-dimethylbutyl cation at C-2 (tertiary).• β-Hydrogen elimination by Zaitsev's rule: remove H from the most substituted side → tetrasubstituted alkene.
7
Final answer
Major product: 2,3-dimethylbut-2-ene (tetrasubstituted alkene from the rearranged tertiary carbocation).
8
Common trap
Selecting 3,3-dimethylbut-1-ene (the product from direct elimination of the unrearranged secondary carbocation). NEET distractors exploit students who forget to check for possible 1,2-shifts before applying Zaitsev's rule.
9
Similar NEET-style question
"When neopentyl alcohol (2,2-dimethylpropan-1-ol) is treated with H₂SO₄ at 443 K, the major product is: (a) 3,3-dimethylbut-1-ene, (b) 2-methylbut-2-ene, (c) 2-methylbut-1-ene, (d) 3-methylbut-1-ene." Answer: (b) — 1,2-methyl shift converts the primary carbocation to a tertiary one, then Zaitsev elimination gives the trisubstituted alkene.

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What to remember before solving Alcohol Dehydration questions

Key Fact

Anomalous properties of nitrogen

However, it does not form compounds in +5 oxidation state with halogens as nitrogen does not have d-orbitals to accommodate electrons from other elements to form bonds. Nitrogen is restricted to a maximum covalency of 4 since only four (one s and three p) orbitals are available for bonding. The heavier elements have vacant d orbitals in the outermost shell which can be used for bonding (covalency) and hence, expand their covalence as in PF6–. Anomalous properties of nitrogen Nitrogen differs from the rest of the members of this group due to its small size, high electronegativity, high ionisation enthalpy and non-availability of d orbitals. Nitrogen has unique ability to form pπ-pπ multiple bonds with itself and with other elements having small size and high electronegativity (e.g., C, O). Heavier elements of this group do not form pπ-pπ bonds as their atomic orbitals are so large and diffuse that they cannot have effective overlapping. Thus, nitrogen exists as a diatomic molecule with a triple bond (one s and two p) between the two atoms. Consequently, its bond enthalpy (941.4 kJ mol–1) is very high. On the contrary, phosphorus, arsenic and antimony form single bonds as P–P, As–As and Sb–Sb while bismuth forms metallic bonds in elemental state. However, the single N–N bond is weaker than the single P–P bond because of high interelectronic repulsion of the non-bonding electrons, owing to the small bond length. As a result the catenation tendency is weaker in nitrogen. Another factor which affects the chemistry of nitrogen is the absence of d orbitals in its valence shell. Besides restricting its covalency to four, nitrogen cannot form dπ –pπ bond as the heavier elements can e.g., R3P = O or R3P = CH2 (R = alkyl group).

-- NCERT Class 12 Chemistry (pre-2023 edition), Chapter 7, p. 172

More in Alcohols, Phenols, Ethers and Carbonyl Compounds: 3 exam traps and mistakes · 2 formulas · 3 question patterns from its other lessons.

Alcohol Dehydration questions from past NEET papers

No question in our NEET 2020–2025 set targets this topic directly.

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Sources

NCERT refs: Class 12 Chemistry Chapter 7, p.208

Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.

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