Dehydration of 1-Phenylpropan-1-ol – Alcohol Dehydration JEE Main 2021 PYQ
Quick Summary
- Question Type:
- Mechanism + product stability prediction (acid-catalysed dehydration)
- Chapter:
- Alcohols, Phenols and Ethers
- Difficulty:
- ⭐⭐⭐ Medium
- Time to Solve:
- 2-3 minutes
- Key Concept:
- E1 dehydration via a benzylic carbocation → Saytzeff alkene that is also ring-conjugated wins
- Correct Answer:
- (C) Compound A will be the major product
- Why:
- A (C6H5–CH=CH–CH3) is both the more substituted alkene and conjugated with the benzene ring; B (allylbenzene) has neither advantage
The Question
JEE Main 2021 (27 July – Shift 2) – Alcohols, Phenols and Ethers
Consider the above reaction, and choose the correct statement:
The question figure shows 1-phenylpropan-1-ol (the OH sits on the carbon directly attached
to the benzene ring) heated with concentrated sulfuric acid, giving two alkene products A and B:
–Conc. H2SO4, Δ→
A = C6H5–CH=CH–CH3
+
B = C6H5–CH2–CH=CH2
In the original figure, A is the alkene whose double bond touches the ring
(drawn in the trans orientation), while B is the alkene one carbon
farther away (allylbenzene). The four statements offered were:
Quick Answer
Correct Option: (C) Compound A will be the major product
Reasoning: Hot concentrated H2SO4 dehydrates this secondary
benzylic alcohol through a resonance-stabilized carbocation, and the alkene that survives in largest
amount is the most stable one:
- Step 1 of the mechanism: OH is protonated, water leaves → a secondary benzylic carbocation forms (C6H5–CH+–CH2–CH3)
- β-removal from the CH2 of the ethyl group gives A: C6H5–CH=CH–CH3 — disubstituted and conjugated with the ring
- The alternative alkene B (allylbenzene) is monosubstituted and not conjugated — two stability levels below A
- Statement (A) is self-contradictory — the acid is the reagent; (B) ignores the stability gap; (D) crowns the weaker product. Only (C) matches the chemistry ✓
Understanding the Concept
How Acid-Catalysed Dehydration of Alcohols Works
An –OH group is a terrible leaving group, but concentrated sulfuric acid fixes that.
Protonation converts it into water, a superb leaving group, and heat pushes the reaction
towards elimination instead of substitution. For secondary and tertiary alcohols the
mechanism is E1: the C–O bond breaks first to give a carbocation, and a weak
base (water or HSO4−) removes a β-hydrogen in a separate, fast step.
Primary alcohols, which cannot support a carbocation, dehydrate by the concerted E2 route
instead. So the instant you see “alcohol + conc. H2SO4 + Δ”, your brain
should say: carbocation chemistry incoming.
- Protonation: ROH + H+ → ROH2+ (turns –OH into H2O, a good leaving group)
- Ionisation: loss of H2O → carbocation (rate-determining step of E1)
- Deprotonation: β-H removed → alkene + H+ regenerated (acid is a catalyst)
- Reactivity order towards dehydration: 3° > 2° > 1° (follows carbocation stability)
The Benzylic Carbocation Advantage
Here the cation formed is C6H5–CH+–CH2–CH3.
Its positive charge sits on the carbon directly attached to the ring, so it is delocalized by
resonance into the aromatic π-system (positive charge appears at the ortho and
para positions of the ring). This “benzylic” stabilization lowers the energy barrier
of the ionisation step dramatically — which is exactly why the question promises a clean,
high-yielding dehydration rather than a sluggish one. One caution the same resonance teaches:
no rearrangement can improve on a benzylic cation, so any hydride or methyl
shift you are tempted to draw would only lose stabilization and is wrong.
Choosing the Major Alkene: Saytzeff + Conjugation
With a small, unconcentrated base like HSO4− or water, β-elimination
follows the Saytzeff rule: the more substituted alkene dominates, because the
product spread is thermodynamic. But substitution is only the first filter — the second,
stronger one here is conjugation:
- A — C6H5–CH=CH–CH3: the C=C is part of the
ring’s π-system (a styrene unit). Conjugation with a phenyl ring is worth roughly
12–15 kJ/mol of extra stability, on top of being disubstituted. Within A, the
trans (E) isomer beats the cis (Z) isomer — that is the one drawn in the paper. - B — C6H5–CH2–CH=CH2: a terminal,
monosubstituted double bond separated from the ring by an sp3 CH2.
No conjugation, less substitution → clearly the minor product.
The Key Principle
To solve any “which product is major?” dehydration question:
- Identify where the carbocation forms and confirm it is stabilized (benzylic/allylic/3°)
- Check whether any rearrangement genuinely improves stability — usually it does not
- List every feasible alkene and rank them: ring-conjugated > more substituted > terminal, and E > Z
- Match the ranking against the statements — the major product is the thermodynamic winner
Detailed Step-by-Step Solution
Step 1: Read the Substrate and the Conditions
- Substrate: C6H5–CH(OH)–CH2–CH3 = 1-phenylpropan-1-ol, a secondary benzylic alcohol
- Reagents: Conc. H2SO4 + Δ — the classic dehydrating pair; heat favours elimination over substitution
- Verdict before any mechanism: E1 dehydration, carbocation intermediate expected
Step 2: Build the Carbocation
- Protonation: the OH oxygen grabs H+ → C6H5–CH(OH2+)–CH2–CH3
- Water departs → C6H5–CH+–CH2–CH3, a secondary benzylic carbocation
- Resonance spreads the charge into the ring (ortho/para positions) — this ion is far more comfortable than a normal 2° cation, so the dehydration is smooth
- No rearrangement: shifting H− or CH3+ would migrate the charge away from the ring — stability would only drop
Step 3: Generate and Rank the Alkenes
- Direct β-removal from the CH2 next to the cation gives A: C6H5–CH=CH–CH3 (main pathway; the trans isomer dominates)
- A small amount of B (allylbenzene) appears later: in hot acid, A can be reprotonated to a secondary cation and deprotonated again at the other end — a double-bond “walk” (isomerization)
- Stability score: A = disubstituted plus conjugated with the phenyl ring (≈ 12–15 kJ/mol bonus); B = monosubstituted, isolated terminal C=C
- Thermodynamic sink = A → the equilibrium piles up on A, leaving only traces of B
Step 4: Test Each Statement
- (A) “Not possible in acidic medium”: ✗ backwards — the acid protonates the OH; without it the reaction cannot start
- (B) “Both formed equally”: ✗ A and B differ hugely in stability (conjugation + substitution), so their amounts differ too
- (C) “Compound A will be the major product”: ✓ exactly what the mechanism predicts
- (D) “Compound B will be the major product”: ✗ crowns the minor, less stable isomer
Final Answer
Option (C): Compound A will be the major product ✓
Dehydration proceeds through a resonance-stabilized benzylic carbocation, and the alkene that
carries the C=C directly on the ring-carbon — A, 1-phenylprop-1-ene — is both
the Saytzeff (more substituted) product and the ring-conjugated one. B, an isolated terminal
alkene, is formed only in minor amounts, so statement (C) is the correct one.
Essential Facts for This Topic
The Working Relationships
- Dehydration conditions:
- Conc. H2SO4 (or H3PO4) + Δ = dehydrating pair; Δ shifts the balance to elimination
- Reactivity: 3° > 2° > 1° — tracks carbocation stability
- Mechanism by alcohol class:
- 3° and 2° alcohols → E1 (carbocation; watch for rearrangements)
- 1° alcohols → E2 (no carbocation; no rearrangement)
- Saytzeff orientation:
- Small base (HSO4−, H2O) → most substituted alkene is major
- Bulky base (t-BuO−) → Hofmann (less substituted) alkene — the standard exception
- Alkene stability ladder: ring-conjugated (styrene-type) > more substituted > less substituted > terminal; within the same substitution, trans (E) > cis (Z)
Facts Worth Memorising
- C6H5–CH(OH)–CH2–CH3 + Conc. H2SO4, Δ → (E)-C6H5–CH=CH–CH3 (major) + C6H5–CH2–CH=CH2 (minor)
- Benzylic and allylic carbocations are resonance-stabilized — they form fastest and refuse to rearrange into less stabilized cations
- Phenyl conjugation adds roughly 12–15 kJ/mol of stability to an alkene — enough to dominate any substitution-count comparison
- Never propose an exocyclic C=C drawn into an aromatic ring — it would destroy aromaticity, so that “product” does not exist here
Common Mistakes to Avoid
❌ Mistake 1: Calling the Acidic Medium the Enemy
Wrong Thinking: “Alcohols react badly with acids, so this reaction is not
possible in acidic medium” — and option (A) suddenly looks attractive.
Correct Approach: The acid is not a spectator, it is the reagent.
Protonating the –OH group is step 1 of every acid-catalysed dehydration; without H2SO4
there is no reaction at all. The statement is the exact opposite of the truth.
❌ Mistake 2: Declaring a 50 : 50 Split
Wrong Thinking: “Two alkenes are possible, so both A and B must form in equal
amounts” — option (B).
Correct Approach: Product ratios follow product stability. A is disubstituted
and conjugated with the benzene ring; B is a lonely terminal alkene. In a thermodynamic
(E1) reaction that gap translates directly into an A ≫ B ratio — never equal.
❌ Mistake 3: Importing Hoffmann Logic Without a Bulky Base
Wrong Thinking: “B is less substituted, therefore less hindered, therefore it
forms faster” — pushing option (D).
Correct Approach: Hoffmann orientation needs a bulky base (t-BuO−)
attacking a hindered β-H. Here the bases are tiny (H2O, HSO4−) and
the pathway is E1, so Saytzeff rules — and the phenyl conjugation bonus makes A the runaway winner.
❌ Mistake 4: Rearranging Away From Resonance
Wrong Thinking: “Every carbocation rearranges — let me shift the hydride and put
the charge on the middle carbon before eliminating.”
Correct Approach: The benzylic cation is already resonance-stabilized; a 1,2-shift
would move the charge off the benzylic carbon and destroy that stabilization, so it never
happens. Also remember the final polish: within A itself, the trans (E) isomer drawn in
the paper is the real major product.
Key Concept Summary
What You Must Remember
- “Alcohol + conc. H2SO4 + Δ” = dehydration: protonate OH → lose water → carbocation → deprotonate
- Benzylic carbocations are premium real estate: resonance into the ring makes them form fast and resist rearrangement
- Saytzeff is the default for small bases: the most substituted alkene is the major product in E1 chemistry
- Conjugation outranks everything: a C=C attached to a benzene ring (styrene unit) beats any isolated alkene, every time
- E beats Z: among the same alkene’s stereoisomers, the trans form drawn in exam figures is the true thermodynamic product
The Golden Rule for Dehydration Questions
“Hot acid strips water, then lets stability vote — and an alkene conjugated with a benzene ring always wins the election.”
Frequently Asked Questions
Q1: How does acid-catalysed dehydration of 1-phenylpropan-1-ol proceed?
A: It follows the E1 mechanism: the OH group is first protonated by concentrated H2SO4, water leaves to give a resonance-stabilized secondary benzylic carbocation, and a base (HSO4- or water) then removes a beta-hydrogen to form the alkene. Heat makes elimination win over substitution.
Q2: Why is compound A (1-phenylprop-1-ene) the major product?
A: Because its C=C double bond is directly attached to the benzene ring, so it is conjugated with the ring, and it is also the more substituted (Saytzeff) alkene. Both factors make A thermodynamically more stable than the isolated terminal alkene B, so both the rate and the equilibrium favour A.
Q3: What is the Saytzeff rule in dehydration of alcohols?
A: In beta-elimination reactions that are not carried out with a bulky base, the major alkene is the one with the greater number of alkyl or aryl groups on the double-bond carbons — the more substituted alkene. In E1 dehydrations the product spread is thermodynamic, so the Saytzeff (most stable) alkene dominates.
Q4: Can compound B (allylbenzene) ever be the major product?
A: Practically not under these conditions. B is an isolated, monosubstituted terminal alkene, so it is less stable than A. It appears only in small amounts, formed when A is protonated again in the hot acid and loses a different beta-hydrogen (double-bond migration). With concentrated H2SO4 and heat the equilibrium always lies towards the ring-conjugated alkene A.
Q5: Why does the benzylic carbocation form so easily in this reaction?
A: The positive charge on the benzylic carbon is delocalized by resonance into the aromatic ring (ortho and para positions), which greatly lowers the energy of the ion. A stabilized carbocation means a low-energy, fast step, so secondary benzylic alcohols dehydrate readily with concentrated H2SO4 and heat.
Prerequisites to Solve This Question
Before attempting this problem, you should be comfortable with:
- Acid-base role of the –OH group: why protonation is needed to turn a poor leaving group into water
- Carbocation stability order: 3° > 2° > 1°, with benzylic and allylic ions elevated by resonance
- E1 vs E2 selection: substrate class, base strength and temperature deciding the pathway
- Saytzeff vs Hoffmann orientation: when the more substituted alkene wins and when it does not
- Alkene stability factors: substitution count, conjugation with aromatic rings, and E/Z geometry
After Solving This, You Can:
- Predict the major alkene of any 2° or 3° alcohol dehydration in under 30 seconds
- Reject rearrangement steps that would move a charge away from a benzylic or allylic position
- Rank a mixed set of alkenes using substitution, ring conjugation and E/Z geometry in one pass
- Handle the twin traps — “not possible in acidic medium” and “both formed equally” — in every future dehydration PYQ
Study Tips for This Topic
For JEE Main:
- Build a reagent reflex: whenever you see an alcohol beside “Conc. H2SO4, Δ”, auto-play the film — protonation, carbocation, β-elimination — before reading the options.
- Ask the two-question checklist: where does the cation form, and which β-removal gives the most substituted and best-conjugated C=C? This kills the entire question family.
- Budget 60–90 seconds: no calculation is needed — only mechanism recall plus one stability comparison.
Common JEE Variants:
- Rearrangement trap: dehydration of 3,3-dimethylbutan-2-ol — a methyl shift does operate here because the starting cation is not resonance-stabilized
- Primary alcohol dehydration: E2 route, Saytzeff still preferred, no rearrangement (e.g. butan-1-ol → but-1-ene/but-2-ene)
- Reagent contrast: POCl3/pyridine gives dehydration without rearrangement — a favourite “which reagent avoids shifts?” question
- Cyclic alcohols: dehydration of substituted cyclohexanols with ring-expansion or hydride-shift options
- Pinacol-pinacolone: the diol version of the same carbocation story (JEE Advanced favourite)
Difficulty Rating & Exam Frequency
Written by Nishant Kumar Gupta
Former Faculty · Allen · Aakash · Narayana — Quantum Chemistry Classes, Arrah (Bihar)
Last Updated: 31 Aug 2026 ·
Question Source: JEE Main 2021 (27 July – Shift 2) Previous Year Question ·
Topic: Alcohols, Phenols and Ethers – Acid-Catalysed Dehydration (Saytzeff Orientation)