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Predicting the product of an alcohol dehydration looks like a one-step question: take out OH and H, draw the double bond. The marks are lost on the hidden step in the middle. The reaction passes through a carbocation, and a carbocation can rearrange before it loses a proton. This post sets out the mechanism NCERT gives, adds the rearrangement check, and works three alcohols to their major alkene.
What do the syllabi ask for?
The Syllabus for JEE (Main) 2026 and the NEET (UG) 2026 syllabus both list, under alcohols, "identification of primary, secondary and tertiary alcohols" and the "mechanism of dehydration". The JEE (Advanced) 2026 brochure lists "dehydration (formation of alkenes and ethers)". The topic sits in Unit 7 of the NCERT Class 12 book, Alcohols, Phenols and Ethers. Check the syllabus for your exam year.
What does NCERT say about the reaction?
Alcohols lose water to form alkenes when treated with a protic acid such as concentrated H₂SO₄ or H₃PO₄, or with catalysts such as anhydrous zinc chloride or alumina. Ethanol needs concentrated H₂SO₄ at 443 K. Secondary and tertiary alcohols react under milder conditions, and NCERT gives the ease of dehydration as tertiary > secondary > primary.
The mechanism has three steps:
Protonation. The OH takes an H⁺ from the acid and becomes –OH₂⁺, which can leave as water.
Carbocation formation. Water leaves. NCERT calls this "the slowest step and hence, the rate determining step".
Loss of a proton from a carbon next to the positive carbon, forming the C=C. The acid comes back, and the alkene is removed as it forms to drive the equilibrium.
For exam answers, add one check between steps 2 and 3: can the carbocation become more stable by a shift?
When does the carbocation rearrange?
OpenStax Organic Chemistry describes how F. C. Whitmore traced unexpected products to carbocation rearrangements. A hydrogen with its electron pair (a hydride shift) or an alkyl group with its electron pair (for example a methyl shift) moves to the positive carbon from the carbon next door, turning a secondary carbocation into a more stable tertiary one. OpenStax's chapter on reactions of alcohols adds that acid-catalysed dehydrations usually follow Zaitsev's rule, giving the more substituted alkene, and that they are E1 processes, so tertiary alcohols react fastest.
Worked example 1: butan-2-ol
CH₃–CH(OH)–CH₂–CH₃.
Carbocation at C2, secondary. A hydride shift from C3 would give another secondary carbocation, so there is no gain and no shift.
Losing H from C3 gives CH₃–CH=CH–CH₃, but-2-ene, with two alkyl groups on the double bond.
Losing H from C1 gives CH₂=CH–CH₂–CH₃, but-1-ene, with one.
Major product: but-2-ene.
Worked example 2: 3,3-dimethylbutan-2-ol, a methyl shift
(CH₃)₃C–CH(OH)–CH₃.
Carbocation at C2, secondary. C3 has no hydrogen, so without a shift the only proton to lose is on C1, giving 3,3-dimethylbut-1-ene.
But a methyl can move from C3 to C2. The positive charge lands on C3, which now carries two methyls and the rest of the chain: a tertiary carbocation, (CH₃)₂C⁺–CH(CH₃)₂.
Losing H from the neighbouring CH gives (CH₃)₂C=C(CH₃)₂, 2,3-dimethylbut-2-ene, with four alkyl groups on the double bond. Losing H from a methyl instead gives 2,3-dimethylbut-1-ene, with two.
Expected major product: 2,3-dimethylbut-2-ene. This is the same secondary carbocation OpenStax shows rearranging by a methyl shift.
A numbers check you might meet in a board paper: 8.8 g of this alcohol (C₆H₁₄O, M ≈ 102.2 g/mol) is 0.0861 mol. Every possible alkene here is C₆H₁₂ (M ≈ 84.2 g/mol), so the most alkene you could get is 0.0861 × 84.2 ≈ 7.25 g. Mass alone cannot tell you which isomer formed; the mechanism can.
Worked example 3: 1-methylcyclohexanol
This is NCERT's intext question 7.7(i).
The OH is on a ring carbon that also carries a methyl, so the carbocation is tertiary at once. No shift helps.
Losing H from a ring CH₂ gives 1-methylcyclohexene, with three alkyl groups on the double bond. Losing H from the methyl gives methylenecyclohexane, with two.
Major product: 1-methylcyclohexene, the product OpenStax reports for this alcohol.
Alkene or ether?
Temperature matters. NCERT notes that ethanol with sulphuric acid gives ethene at 443 K but mainly ethoxyethane at 413 K. It adds that this ether route suits only primary, unhindered alcohols; with secondary and tertiary alcohols "elimination competes over substitution" and alkenes form.
Why bother writing the mechanism?
Grove, Cooper and Rush (2012) followed second-year organic chemistry students through a year and found a large proportion choosing not to use mechanisms in their work. In dehydration that habit costs marks, because the shift in example 2 is invisible unless the carbocation is on the page.
Which mistakes cost marks?
Removing H from the carbon that carried the OH. The proton comes from a neighbouring carbon.
Skipping the shift check, or shifting when the new carbocation is no more stable.
Counting hydrogens instead of alkyl groups on the double bond.
Mixing this up with alcoholic KOH, which removes HX from haloalkanes. Our post on E1 and E2 elimination reactions covers that route, and the one on alcohols, phenols and ethers covers the rest of the chapter.
When a product does not match the key, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can find the step where your mechanism went another way.
A practice routine
Day 1: Write the three NCERT steps for ethanol from memory, with curved arrows.
Day 2: Five secondary alcohols. For each, draw the carbocation and decide yes or no on a shift.
Day 3: NCERT intext question 7.7(ii), butan-1-ol, then two ring alcohols.
Day 4: A mixed set of six, half needing a rearrangement.
Today, predict the major alkene from 2-methylpentan-2-ol. You should get 2-methylpent-2-ene.
