Mechanism of Some Important Reactions

ChemistryAlcohols, Phenols And EthersFor JEE aspirants

The mechanism of some important reactions of alcohols and ethers comes down to four moves: protonating an oxygen, losing water to a carbocation, attack by a nucleophile, and losing a proton. Hydration of alkenes and dehydration of alcohols run through carbocations, so they follow Markovnikov and Saytzeff rules and can rearrange; ether formation, PBr $_{3}$ and SOCl $_{2}$ reactions are S $_{N}$ 2, so the skeleton is kept. Knowing the mechanism of some important reactions lets you predict products instead of memorising them, for JEE Main, JEE Advanced and NEET.

On this page1Four moves2Hydration3Dehydration4Alcohol + HX5

PB r_{3}

SOC l_{2}

6Epoxides7Which mechanism?8Examples

Key Formulas - Quick Reference

★ Must learnHydration: C=C + $H_{3} O^{+}$ $\to$ carbocation $\to$ ROH (Markovnikov; 3 steps)
★ Must learnDehydration: ROH + $H^{+}$ $\to$ $RO H_{2}^{+}$ $\to$ $R^{+}$ (slow) $\to$ alkene; ease 3° $>$ 2° $>$ 1°
Ether formation (413 K): ROH + $RO H_{2}^{+}$ $\to$ ROR (S $_{N}$ 2, 1° alcohols)
★ Must learnROH + HX: 3°, 2° by S $_{N}$ 1 (carbocation); $C H_{3} OH$ , 1° by S $_{N}$ 2
★ Must learn $PB r_{3}$ , $SOC l_{2}$ : O-P or O-S bond first, then S $_{N}$ 2: no rearrangement, inversion
★ Must learnEpoxide + Nu: acid at the more substituted C, base at the less hindered C
Carbocation shifts: H or $C H_{3}$ moves to give the more stable (3°) cation

1. Four Moves Behind Every Mechanism

The reactions of alcohols and ethers look numerous, but their mechanisms use the same four steps in different orders: (1) an oxygen lone pair takes a proton, which turns OH into $O H_{2}^{+}$ , a good leaving group; (2) water leaves, giving a carbocation, if that cation is stable enough; (3) a nucleophile attacks carbon, either the carbocation or, from the back, the carbon still bonded to the leaving group; (4) a proton is lost, from oxygen or from a $β$ -carbon.

Figure 1: Almost every reaction in this chapter uses one of four steps: protonation of oxygen, loss of water to a carbocation, backside attack by a nucleophile, or loss of a proton.

Reaction	Mechanism	Page where it is used
Acid hydration of alkenes	carbocation (reverse of E1)	Preparation of Alcohol
Dehydration of alcohols to alkenes	E1 (2°, 3°)	Properties of Alcohol
Dehydration of 1° alcohols to ethers	S $_{N}$ 2	Preparation of Ether
Alcohols with HX, Lucas test	S $_{N}$ 1 or S $_{N}$ 2	Properties of Alcohol
Williamson synthesis	S $_{N}$ 2	Preparation of Ether
Cleavage of ethers by HI	S $_{N}$ 2 or S $_{N}$ 1	Properties of Ether

2. Acid-Catalysed Hydration of Alkenes

Alkenes add water in the presence of an acid in three steps. The $π$ electrons take a proton from $H_{3} O^{+}$ to form the more stable carbocation (slow step, which fixes the Markovnikov orientation). Water attacks the carbocation to give an oxonium ion, and a second water molecule removes a proton, regenerating the acid.

C H_{3} CH = C H_{2} + H_{2} O H^{+} C H_{3} CH (OH) C H_{3}

Figure 2: Hydration of an alkene in three steps. Step 1 is slow and decides the regiochemistry: H

^{+}

goes to the carbon that gives the more stable carbocation, so OH ends up on the more substituted carbon (Markovnikov).

Every step is reversible: the same three steps run backwards are the dehydration of an alcohol. Dilute acid and excess water push the equilibrium towards the alcohol; concentrated acid and heat (which drives off the alkene) push it towards the alkene.

Key idea

Hydration: H

^{+}

goes where the more stable cation forms, then water follows; because a free carbocation forms, the skeleton may rearrange.

3. Dehydration of Alcohols

Heated with concentrated sulphuric acid (or phosphoric acid, or passed over hot alumina) alcohols lose water to give alkenes. The NCERT mechanism for ethanol has three steps: protonation of the OH, slow loss of water to give a carbocation, and removal of a $β$ -hydrogen by a base.

C_{2} H_{5} OH conc. H_{2} S O_{4}, 443 K C H_{2} = C H_{2} + H_{2} O

Figure 3: Dehydration of ethanol (NCERT E1 picture). Step 2, loss of water to the carbocation, is slow, which is why 3° alcohols (stable cations) dehydrate most easily: 3°

>

2°

>

1°.

Because step 2 forms a carbocation, dehydration becomes easier as the cation becomes more stable: tertiary alcohols dehydrate with 20% $H_{2} S O_{4}$ at 358 K, primary ones need concentrated acid at 443 K. The same carbocation can rearrange by a 1,2-hydride or methyl shift before losing a proton, and the more substituted alkene (Saytzeff) is the major product.

Figure 4: Because dehydration goes through a carbocation, the skeleton can change. A 1,2-methyl shift turns the 2° cation into a 3° cation, and the Saytzeff alkene, 2,3-dimethylbut-2-ene, is the major product.

Exam Trick

"Cation? Check the neighbour." Whenever a carbocation forms, look at the next carbon: if moving an H or $C H_{3}$ from it turns a 2° cation into a 3° one, the shift happens first.

JEE Advanced

For primary alcohols a free primary carbocation is too unstable; the loss of water and the removal of the $β$ -hydrogen are concerted (E2-like on the protonated alcohol). NCERT draws the E1 picture for ethanol, which is the accepted exam answer; the key point for both is that protonation must come first because $O H^{-}$ is a poor leaving group while $H_{2} O$ is a good one.

Quick Recall: tap to check

Rate-determining step of alcohol dehydration?

Loss of water from the protonated alcohol to form the carbocation.

Ease of dehydration: 1°, 2°, 3°?

3°

>

2°

>

1° (carbocation stability).

Major product from 3,3-dimethylbutan-2-ol with acid?

2,3-Dimethylbut-2-ene (after a methyl shift).

4. Dehydration to Ethers

At 413 K with excess of a primary alcohol, the protonated alcohol is attacked by a second alcohol molecule in an S $_{N}$ 2 step before it can lose water to a carbocation. The protonated ether then loses a proton. The full three-step drawing is on the Preparation of Ether page; the essential equations are:

C H_{3} C H_{2} OH + H^{+} ⇌ C H_{3} C H_{2} O H_{2}^{+}

C H_{3} C H_{2} OH + C H_{3} C H_{2} O H_{2}^{+} S_{N} 2 C H_{3} C H_{2} O H^{+} C H_{2} C H_{3} + H_{2} O

C H_{3} C H_{2} O H^{+} C H_{2} C H_{3} C H_{3} C H_{2} OC H_{2} C H_{3} + H^{+}

Key idea

413 K and excess alcohol: substitution (ether); 443 K and excess acid: elimination (alkene).

5. Alcohols with Hydrogen Halides; the Lucas Reagent

The alcohol is first protonated. A tertiary (or secondary) protonated alcohol loses water to a carbocation, which the halide captures (S $_{N}$ 1). Methanol and primary alcohols cannot form stable cations, so the halide attacks the protonated alcohol from the back (S $_{N}$ 2). Reactivity is 3° $>$ 2° $>$ 1° for the S $_{N}$ 1 group, with methanol faster than other primary alcohols (least hindered), and HI $>$ HBr $>$ HCl.

Chloride is a weak nucleophile, so HCl needs anhydrous zinc chloride with 1° and 2° alcohols. $ZnC l_{2}$ , a Lewis acid, binds the OH oxygen and makes a far better leaving group. The time taken for the insoluble alkyl chloride to cloud the solution is the Lucas test.

Figure 5: Why ZnCl

_{2}

is needed. Cl

^{-}

is a weak nucleophile, so the leaving group must be improved: Zn

^{2 +}

binds the OH oxygen, the C-O bond breaks (S

_{N}

1 for 2° and 3°), and chloride captures the cation.

_{N}

1 (3°, 2°)carbocation intermediate
rearrangement possible
rate depends on

[RO H_{2}^{+}

] only

_{N}

(C H_{3} OH

, 1°)one step, backside attack
no rearrangement, inversion
rate depends on [

X^{-}

] too

6. Phosphorus Halides and Thionyl Chloride

$PB r_{3}$ , $PC l_{3}$ and $SOC l_{2}$ convert 1° and 2° alcohols to alkyl halides without strong acid. The oxygen first bonds to phosphorus (or sulphur), displacing a halide ion; the halide then attacks the carbon from the back and pushes out the phosphorus (or sulphur) group. No carbocation forms, so there is no rearrangement.

3ROH + PB r_{3} 3RBr + H_{3} P O_{3}

Figure 6: PBr

_{3}

turns OH into a very good leaving group (OPBr

_{2}

) and bromide displaces it by S

_{N}

2. No carbocation forms, so there is no rearrangement, and a chiral carbon is inverted.

ROH + SOC l_{2} RCl + S O_{2} ↑ + HCl ↑

Figure 7: SOCl

_{2}

is the cleanest route to an alkyl chloride: both by-products, SO

_{2}

and HCl, are gases and leave the flask, so the product needs no separation.

Exam Trick

"No cation, no surprise." $PB r_{3}$ and $SOC l_{2}$ keep the carbon skeleton; HX with 2° or 3° alcohols can rearrange it. Choose $PB r_{3}$ when the question warns about rearrangement.

7. Ring Opening of Epoxides

The strained three-membered ring opens under milder conditions than ordinary ethers. In acid the oxygen is protonated; the C-O bond to the more substituted carbon is longer and weaker and that carbon carries more positive charge, so the nucleophile attacks it (S $_{N}$ 1-like). In base a strong nucleophile attacks the less hindered carbon in a clean S $_{N}$ 2 step. In both cases attack is from the side opposite the oxygen, so cyclic epoxides give trans products.

Figure 8: The same ring, two regiochemistries. In acid the more substituted carbon carries more positive charge (S

_{N}

1-like); in base steric hindrance decides (S

_{N}

2). Both attacks come from the side opposite the oxygen (anti opening).

Quick Recall: tap to check

Methyloxirane +

C H_{3} OH

H^{+}

: which carbon is attacked?

The more substituted CH carbon: 2-methoxypropan-1-ol.

Methyloxirane +

C H_{3} O^{-}

: which carbon?

The

C H_{2}

carbon: 1-methoxypropan-2-ol.

Why do epoxides open with nucleophiles that ordinary ethers ignore?

Ring strain of the three-membered ring.

8. Which Mechanism?

The flowchart asks one question in several forms: does a carbocation form? If yes, expect Markovnikov or Saytzeff selectivity and possible shifts; if no, expect S $_{N}$ 2 with the skeleton unchanged. The mind map places every mechanism of the chapter on one screen.

Figure 9: Flowchart: decide whether a carbocation forms. If it does (2°, 3°, strong acid, heat), expect rearrangement and Saytzeff products; if not (PBr

_{3}

, SOCl

_{2}

, 1° with HX), the skeleton is kept.

Figure 10: Mind map: every mechanism on one screen. The one question to ask each time is whether a carbocation forms.

9. Solved Examples

Solved Example 1

Why does

C H_{3} OH

react faster with HX than other primary alcohols?

Solution:

Methanol and primary alcohols react by S $_{N}$ 2 (the methyl cation is far too unstable for S $_{N}$ 1). In S $_{N}$ 2 the rate falls with crowding at the carbon; the methyl carbon has only hydrogens around it, so backside attack is easiest and methanol reacts fastest.

Solved Example 2

The predominant product of

(C H_{3})_{3} C - CH (OH) C H_{3}

with HBr is
(A)

(C H_{3})_{3} C - CHBrC H_{3}

(B)

(C H_{3})_{2} CBr - CH (C H_{3})_{2}

Solution:

Answer: (B). The protonated 2° alcohol loses water to a 2° carbocation; a methyl shifts from the neighbouring quaternary carbon to give a 3° cation, which bromide captures: 2-bromo-2,3-dimethylbutane.

Solved Example 3

Dehydration of cyclopentylmethanol with conc.

H_{2} S O_{4}

gives
(A) cyclopentene
(B) cyclohexene
(C) cyclohexane
(D) none

Solution:

Answer: (B). Loss of water would give an unstable primary cation; a ring C-C bond migrates instead (ring expansion), giving the less strained cyclohexyl cation, which loses $H^{+}$ to form cyclohexene.

Solved Example 4

HBr reacts fastest with
(A) 2-methylpropan-2-ol
(B) propan-2-ol
(C) propan-1-ol
(D) 2-methylpropan-1-ol

Solution:

Answer: (A). It is the tertiary alcohol: S $_{N}$ 1 through the stable tert-butyl cation.

Solved Example 5

Predict the major product of heating 2-methylbutan-2-ol with conc.

H_{2} S O_{4}

Solution:

The 3° cation $(C H_{3})_{2} C^{+} C H_{2} C H_{3}$ can lose H from a $C H_{3}$ (giving 2-methylbut-1-ene) or from the $C H_{2}$ (giving 2-methylbut-2-ene). The more substituted, trisubstituted alkene is major: 2-methylbut-2-ene (Saytzeff).

Solved Example 6

(R)-Butan-2-ol is treated with

PB r_{3}

. What is the configuration of the 2-bromobutane formed?

Solution:

(S)-2-Bromobutane. Bromide attacks the carbon from the side opposite the $- OPB r_{2}$ group (S $_{N}$ 2), so the configuration is inverted. With aqueous HBr a 2° alcohol partly racemises through the carbocation.

Solved Example 7

Write the mechanism of the conversion of 2-allylphenol into 2-methylchroman (2-methyl-3,4-dihydro-2H-1-benzopyran) with acid.

Solution:

(1) $H^{+}$ adds to the terminal $C H_{2}$ of the allyl group (Markovnikov), giving a 2° carbocation on the chain. (2) The phenolic oxygen, held nearby, attacks the cation intramolecularly, closing a six-membered ring. (3) The oxonium ion loses $H^{+}$ . The product is the cyclic ether with a methyl group on the carbon next to oxygen.

Solved Example 8

Methyloxirane is treated (a) with

C H_{3} OH

and a trace of

H_{2} S O_{4}

, (b) with

C H_{3} ONa

in methanol. Give the products.

Solution:

(a) Acid: attack at the more substituted carbon, 2-methoxypropan-1-ol, $C H_{3} CH (OC H_{3}) C H_{2} OH$ . (b) Base: attack at the $C H_{2}$ carbon, 1-methoxypropan-2-ol, $C H_{3} CH (OH) C H_{2} OC H_{3}$ .

Solved Example 9

Hydration of alkenes and dehydration of alcohols share one mechanism. What decides the direction?

Solution:

The equilibrium $alkene + H_{2} O ⇌ alcohol$ . Dilute acid and a large excess of water (low temperature) favour the alcohol; concentrated acid, high temperature and removal of the volatile alkene favour dehydration (Le Chatelier). Both directions pass through the same carbocation.

Practice Questions

Give the mechanism type of ethanol → diethyl ether at 413 K.Answer: S $_{N}$ 2 on the protonated alcohol.
Which step is rate-determining in acid hydration of propene?Answer: protonation of the alkene to the carbocation.
Product of 3-methylbutan-2-ol with HBr (major)?Answer: 2-bromo-2-methylbutane (after a hydride shift).
Why does $ZnC l_{2}$ help HCl react with 1° and 2° alcohols?Answer: it binds the OH oxygen and makes a better leaving group.
Name the gaseous by-products of ROH + $SOC l_{2}$ .Answer: $S O_{2}$ and HCl.
Suggest a mechanism for 1-phenylprop-2-yn-1-ol + $H_{3} O^{+}$ → cinnamaldehyde.Answer: protonation, loss of water to a propargyl/allenyl cation, water attack at the alkyne end, enol → aldehyde.
Which carbon of 2,2-dimethyloxirane does $H_{2} O$ attack in acid?Answer: the tertiary carbon.

Common Mistakes to Avoid

Watch out

Forgetting the protonation step. $O H^{-}$ is a poor leaving group; the alcohol must be protonated first.
Writing a primary carbocation freely. Primary alcohols react by SN2 (ether formation, HX).
Ignoring rearrangement in HX or $H_{2} S O_{4}$ reactions of 2° alcohols. Check for a hydride or methyl shift.
Expecting rearrangement with $PB r_{3}$ or $SOC l_{2}$ . No carbocation forms, so the skeleton is kept.
Writing the less substituted alkene as major. Dehydration gives the Saytzeff (more substituted) alkene.
Swapping acid and base regiochemistry for epoxides. Acid: more substituted C; base: less hindered C.
Treating hydration as irreversible. It is the exact reverse of dehydration; conditions decide the direction.
Putting the slow step of dehydration at protonation. The slow step is loss of water to the carbocation.

Frequently Asked Questions

What is the mechanism of acid-catalysed hydration of alkenes?

It has three steps. The alkene takes a proton from the hydronium ion to form the more stable carbocation, water attacks the carbocation to form an oxonium ion, and a second water molecule removes a proton to give the alcohol and regenerate the acid. The first step decides the Markovnikov orientation.

What is the mechanism of dehydration of ethanol?

Ethanol is protonated by sulphuric acid, the protonated alcohol loses water to form a carbocation in the slow step, and a base removes a hydrogen from the neighbouring carbon to form ethene. It is the reverse of acid catalysed hydration.

Why do tertiary alcohols dehydrate most easily?

The slow step of dehydration is the loss of water to form a carbocation. Tertiary carbocations are the most stable because three alkyl groups donate electron density, so the order of ease of dehydration is tertiary greater than secondary greater than primary.

How is diethyl ether formed from ethanol mechanistically?

At 413 K with excess ethanol, one ethanol molecule is protonated and a second ethanol molecule attacks its carbon from the back in an SN2 step, pushing out water. The protonated ether then loses a proton to give diethyl ether.

Why do PBr3 and SOCl2 not cause rearrangement?

They first convert the OH group into an excellent leaving group bonded to phosphorus or sulphur, and the halide then displaces it from the back in an SN2 step. No free carbocation forms, so there is nothing to rearrange.

Why does an epoxide open at different carbons in acid and base?

In acid the protonated epoxide has more positive charge on the more substituted carbon, so the nucleophile goes there. In base there is no charge, and a strong nucleophile attacks the less hindered carbon by SN2. Both attacks come from the side opposite the oxygen.

Which mechanisms of alcohols are asked in NEET?

NEET follows NCERT: the three-step mechanisms of acid hydration of alkenes, dehydration of ethanol to ethene and formation of diethyl ether, plus the order of reactivity of alcohols with HX. Knowing which step forms the carbocation answers most questions.

How does JEE test carbocation rearrangements in alcohols?

JEE Main and Advanced give a secondary alcohol next to a quaternary or tertiary carbon, or a cyclic carbinol, and ask for the product with HBr or hot acid. A hydride, methyl or ring-bond shift to the more stable cation comes first, then capture or Saytzeff elimination.

Previous year questions on Mechanism of Some Important Reactions

3 questions from past papers, each with a step-by-step solution.

Ready to master Alcohols, Phenols And Ethers?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test