Preparation of Alcohol

Q: Which method gives the anti-Markovnikov alcohol from an alkene?

Hydroboration oxidation. Diborane or BH₃ · THF adds with boron on the less substituted carbon, and alkaline hydrogen peroxide then replaces boron by -OH. Propene gives propan-1-ol, while acid hydration of the same alkene gives propan-2-ol.

Q: What is the difference between NaBH4 and LiAlH4?

NaBH₄ is a mild hydride donor that reduces only aldehydes and ketones and can even be used in water or alcohol. LiAlH₄ is far more powerful, reducing esters, carboxylic acids and amides as well, but it reacts violently with water so it must be used in dry ether.

Q: What is the difference between syn and anti hydroxylation?

Cold dilute KMnO₄ or OsO₄ delivers both oxygen atoms from one cyclic intermediate, so the two -OH groups add to the same face and give a cis diol. Forming an epoxide first and opening it with acidic water attacks from the opposite face, giving a trans diol.

ChemistryAlcohols, Phenols And EthersFor JEE aspirants

The preparation of alcohols comes down to six reliable routes: hydration of alkenes (acid catalysed, oxymercuration or hydroboration), hydrolysis of alkyl halides, reduction of aldehydes, ketones, esters and acids, and addition of Grignard reagents to carbonyl compounds. Each route is chosen for what it controls: acid hydration and oxymercuration give the Markovnikov alcohol, hydroboration the anti-Markovnikov one, and only the Grignard route builds a new $C - C$ bond. The preparation of alcohols appears in almost every NEET and JEE Main paper.

On this page1All routes2From alkenes3From halides4Reduction5Grignard6Industry7Choose a route8Examples

Key Formulas - Quick Reference

★ Must learnAcid hydration: $RCH = CH_{2} + H_{2} O H^{+} Markovnikov alcohol$ (carbocation, so rearrangement is possible).
★ Must learnOxymercuration: $Hg(OAc)_{2} / H_{2} O$ , then $NaBH_{4}$ . Markovnikov, no rearrangement.
★ Must learnHydroboration: $B_{2} H_{6}$ , then $H_{2} O_{2} / OH^{-}$ . Anti-Markovnikov and syn.
Hydrolysis: $R - X + aq. KOH \to R - OH + KX$ (moist $Ag_{2} O$ is cleaner).
★ Must learnReduction: aldehyde $\to 1^{\circ}$ alcohol, ketone $\to 2^{\circ}$ alcohol, ester or acid $\to 1^{\circ}$ alcohol with $LiAlH_{4}$ .
★ Must learnGrignard: $HCHO \to 1^{\circ}$ , $RCHO \to 2^{\circ}$ , $R_{2} CO \to 3^{\circ}$ , epoxide $\to 1^{\circ}$ with two extra carbons.
Every Grignard reaction ends with an $H_{3} O^{+}$ work-up, and every Grignard reaction needs perfectly dry apparatus.

1. The Full Map of Routes to an Alcohol

Before learning any single method it helps to see all of them at once, because most exam questions are really asking which route gives which alcohol, not how to run the reaction. The table lists all six; the flowchart and mind map in section 7 turn it into a decision you can make in seconds.

Start from	Reagent	Alcohol obtained
alkene	dil. $H_{2} S O_{4}$ (or $H_{3} P O_{4}$ )	Markovnikov; skeleton may rearrange
alkene	$Hg (OAc)_{2}$ / $H_{2} O$ , then $NaB H_{4}$	Markovnikov; no rearrangement
alkene	$B_{2} H_{6}$ , then $H_{2} O_{2}$ / $O H^{-}$	anti-Markovnikov; syn addition
alkyl halide	aq. $KOH$ or moist $A g_{2} O$	same skeleton, $- OH$ replaces X
aldehyde, ketone, ester, acid	$NaB H_{4}$ or $LiAl H_{4}$	$1^{\circ}$ or $2^{\circ}$ alcohol
carbonyl compound	RMgX in dry ether, then $H_{3} O^{+}$	$1^{\circ}$ , $2^{\circ}$ or $3^{\circ}$ , new C-C bond

2. From Alkenes

2.1 Acid catalysed hydration

Alkenes add water in the presence of a dilute acid such as $H_{2} SO_{4}$ or $H_{3} PO_{4}$ . The addition follows Markovnikov's rule, so the $- OH$ group goes to the more substituted carbon.

Figure 1: Acid catalysed hydration. Top: propene gives propan-2-ol, the Markovnikov alcohol. Bottom: the mechanism; protonation makes the more stable

2^{\circ}

carbocation, water attacks it and loses

H^{+}

, so carbocation stability alone decides where

- OH

goes.

Every step of this mechanism is reversible. Dilute acid with plenty of water pushes the equilibrium towards the alcohol; hot concentrated acid pulls it back and dehydrates the alcohol to the alkene. The same two arrows, read in opposite directions, are the whole of acid catalysed hydration and acid catalysed dehydration.

2.2 The limitation: carbocation rearrangement

Because a free carbocation is formed, a hydride or an alkyl group can migrate to give a more stable cation before water ever attacks. The alcohol you isolate then has a different carbon skeleton from the one you expected.

Figure 2: Why acid hydration can give the wrong skeleton. The

2^{\circ}

cation from 3,3-dimethylbut-1-ene becomes a

3^{\circ}

cation by a 1,2-methyl shift, so the major product is 2,3-dimethylbutan-2-ol, not the expected 3,3-dimethylbutan-2-ol.

2.3 Oxymercuration and demercuration

This two-step sequence gives the same Markovnikov alcohol but avoids the rearrangement problem completely, because the intermediate is a bridged mercurinium ion and never a free carbocation.

Figure 3: Oxymercuration gives the same Markovnikov alcohol as acid hydration, but the bridged mercurinium ion prevents any carbocation rearrangement.

Exam Trick

If a question gives you a skeleton that could rearrange (a carbon next to a quaternary centre) and asks for the unrearranged Markovnikov alcohol, the answer is oxymercuration and demercuration.

Net addition is of $H$ and $OH$ with Markovnikov orientation, and typical yields are around 90 per cent.

2.4 Hydroboration and oxidation

Diborane or a $BH_{3} \cdot THF$ complex adds across the double bond with boron going to the less substituted carbon. Alkaline hydrogen peroxide then replaces boron by $- OH$ with retention of configuration, so the overall result is anti-Markovnikov hydration.

Figure 4: Hydroboration oxidation. Top: boron takes the less hindered carbon and

H_{2} O_{2} / OH^{-}

swaps it for

- OH

with retention, giving propan-1-ol. Bottom: in the four-centre transition state

δ +

builds on the more substituted carbon (which receives H) and boron (

δ -

) bonds to the

CH_{2}

end, both on the same face (syn).

JEE Advanced Hydroboration controls two kinds of selectivity at once. Regioselectivity: boron lands on the less hindered carbon, so the

- OH

finally appears there. Stereoselectivity: the four-centre transition state forces

H

and

B

onto the same face, so the addition is syn. Since the oxidation step keeps the configuration at carbon, the stereochemistry set in the first step survives into the product.

Route	Reagents	Regiochemistry	Stereochemistry	Rearrangement
Acid hydration	$H_{2} O / H^{+}$	Markovnikov	not controlled	possible
Oxymercuration	$Hg(OAc)_{2}$ , then $NaBH_{4}$	Markovnikov	not controlled	none
Hydroboration	$B_{2} H_{6}$ , then $H_{2} O_{2} / OH^{-}$	anti-Markovnikov	syn addition	none

Oxymercuration-demercuration

Hg(OAc)_{2} / H_{2} O

, then

NaBH_{4}

bridged mercurinium ion
Markovnikov, no rearrangement
propene gives propan-2-ol

Hydroboration-oxidation

B_{2} H_{6}

, then

H_{2} O_{2} / OH^{-}

four-centre transition state
anti-Markovnikov, syn, no rearrangement
propene gives propan-1-ol

2.5 Hydroxylation: making two hydroxyl groups at once

Cold dilute alkaline $KMnO_{4}$ (Baeyer's reagent) or $OsO_{4}$ followed by $NaHSO_{3}$ adds two $- OH$ groups to the same face of the double bond, giving a cis diol. Opening an epoxide with acidic water instead gives the trans diol.

Figure 5: Two ways to put two

- OH

groups on one double bond.

KMnO_{4}

OsO_{4}

give the cis (syn) diol; the epoxide route gives the trans (anti) diol.

Key idea

Same alkene, three reagents: acid and mercury put

- OH

on the more substituted carbon, borane on the less substituted one; only acid lets the skeleton rearrange.

Quick Recall: tap to check

Propene with

B_{2} H_{6}

, then

H_{2} O_{2} / OH^{-}

Propan-1-ol (anti-Markovnikov).

Why does oxymercuration never rearrange?

Its intermediate is a bridged mercurinium ion, not a free carbocation.

Cyclopentene with cold dilute alkaline

KMnO_{4}

cis-Cyclopentane-1,2-diol (syn hydroxylation).

3. From Alkyl Halides

An alkyl halide heated with aqueous alkali gives an alcohol by nucleophilic substitution. The reaction is clean for primary halides but competes badly with elimination for tertiary ones.

Figure 6: Aqueous alkali converts an alkyl halide to an alcohol. For a

1^{\circ}

halide it is a clean

S_{N} 2

attack from the back, so the configuration is inverted.

Aqueous

KOH

is both a nucleophile and a base, so tertiary halides mostly give alkenes. Moist silver oxide,

Ag_{2} O / H_{2} O

, supplies

OH^{-}

in a much less basic environment and gives far better alcohol yields. This is why textbooks call the alkali method unsatisfactory for tertiary halides.

4. From Carbonyl Compounds by Reduction

4.1 Aldehydes and ketones

Figure 7: Reduction of aldehydes and ketones. Top: an aldehyde always gives a

1^{\circ}

alcohol and a ketone a

2^{\circ}

alcohol. Bottom:

H^{-}

adds to the

δ +

carbonyl carbon, then the alkoxide picks up

H^{+}

in the separate work-up.

Reagent	Reduces	Does not reduce	Note
$NaBH_{4}$	aldehydes, ketones	esters, acids, $C = C$	can be used in water or alcohol
$LiAlH_{4}$	aldehydes, ketones, esters, acids, amides	isolated $C = C$	violent with water, use dry ether
$H_{2}$ with Ni, Pd or Pt	$C = O$ and $C = C$	nothing much	not selective

NaBH_{4}

: gentlereduces aldehydes and ketones only
leaves esters, acids and

C = C

alone
safe in water or ethanol

LiAlH_{4}

: powerfulalso reduces esters, acids and amides
leaves an isolated

C = C

alone
reacts violently with water: dry ether, then work-up

Exam Trick

"Count the R's." Reduction only adds hydrogen, so the degree of the alcohol equals the number of carbon groups already on the carbonyl carbon: an aldehyde (one R) gives a $1^{\circ}$ alcohol, a ketone (two R) a $2^{\circ}$ alcohol. No reduction can ever give a $3^{\circ}$ alcohol.

4.2 Esters and carboxylic acids

Figure 8:

LiAlH_{4}

pushes esters and acids all the way down to primary alcohols.

NaBH_{4}

cannot do this.

Catalytic hydrogenolysis of an ester needs far harsher conditions than ordinary hydrogenation: about

250^{\circ} C

and very high pressure over copper chromite,

CuO \cdot CuCr_{2} O_{4}

. This is the industrial route to long chain primary alcohols used in detergents.

Key idea

Reduction keeps the carbon skeleton: aldehydes and esters end as

1^{\circ}

alcohols, ketones as

2^{\circ}

, and only

LiAlH_{4}

can reach esters and acids.

5. From Grignard Reagents

Victor Grignard discovered organomagnesium halides in 1900 and received the Nobel Prize in 1912. Their value is simple: they put a negative charge on carbon, and a carbon nucleophile can attack a carbonyl carbon to make a new carbon to carbon bond.

Figure 9: The Grignard reagent. Top:

R-X

and Mg in dry ether; reactivity

R-I > R-Br > R-Cl

, and the

C - Mg

bond puts a negative charge on carbon. Bottom: that carbanion attacks

C = O

to build a new

C - C

bond; only the final hydrolysis brings in water.

5.1 Which carbonyl gives which alcohol

Figure 10: The one table you must know for Grignard questions. Methanal gives

1^{\circ}

, any other aldehyde gives

2^{\circ}

, a ketone gives

3^{\circ}

, and ethylene oxide adds two carbons to give a

1^{\circ}

alcohol.

Exam Trick

Count the groups on the carbonyl carbon before the reaction. Methanal has none, so you get $1^{\circ}$ ; one group gives $2^{\circ}$ ; two groups give $3^{\circ}$ .

An ester with excess Grignard gives a tertiary alcohol with two identical alkyl groups, because the reagent adds twice.

An ester of formic acid with excess Grignard gives a secondary alcohol, again with two identical groups.

Ethylene oxide is the only common reagent that adds two carbons and still gives a primary alcohol.

5.2 Restrictions you must respect

Figure 11: A Grignard reagent is a strong base first and a nucleophile second. Any acidic hydrogen in the flask converts

R MgX

into the alkane

R H

and the synthesis fails.

5.3 Planning a Grignard synthesis

Work backwards. Look at the carbon carrying $- OH$ , break one of the bonds to it, and the two fragments tell you which carbonyl compound and which Grignard reagent to start from.

Figure 12: How to plan a Grignard synthesis: look at the carbon carrying

- OH

, cut one of its bonds, and the two pieces are your carbonyl compound and your Grignard reagent.

Key idea

Grignard synthesis is the only route that adds carbons: cut one bond at the

- OH

carbon, and the two pieces are the carbonyl compound and the Grignard reagent.

Quick Recall: tap to check

Which carbonyl compound gives a

1^{\circ}

alcohol with RMgX?

Methanal, HCHO (or ethylene oxide, which adds two carbons).

Ethyl ethanoate with excess

CH_{3} MgBr

, then

H_{3} O^{+}

2-Methylpropan-2-ol: the reagent adds twice.

Why must the ether be dry?

Water protonates RMgX to the alkane RH and destroys the reagent.

6. Industrial and Commercial Routes

Ethanol by fermentation. Starch or sugar is hydrolysed by the enzymes diastase and maltase, then zymase converts glucose into ethanol and $CO_{2}$ . Fractional distillation gives rectified spirit, about 95 per cent ethanol.
Ethanol by hydration of ethene. $CH_{2} = CH_{2} + H_{2} O$ over phosphoric acid at about $300^{\circ} C$ and 70 atm, the main modern route.
Methanol from synthesis gas. $CO + 2 H_{2}$ over $ZnO - Cr_{2} O_{3}$ at high pressure.
The oxo process. An alkene with $CO$ and $H_{2}$ gives an aldehyde, which is then hydrogenated to a primary alcohol.

7. Choosing the Right Route

A synthesis question gives you a target and asks for the reagent. Ask four questions in order: does the target need new carbons, is the start an alkene (and which carbon should get $- OH$ ), is it a carbonyl compound, or is it a halide?

Figure 13: Flowchart: pick the route from the product you need. Ask first whether the target has more carbons than the starting material, because only the Grignard route builds a new

C - C

bond.

Figure 14: Mind map: eight routes, each with its reagent and the alcohol it gives. The three alkene routes differ only in regiochemistry and rearrangement.

8. Solved Examples

Solved Example 1

Which reagent converts but-1-ene into butan-1-ol, and which converts it into butan-2-ol?

Solution:

Butan-1-ol is the anti-Markovnikov product, so use $B_{2} H_{6}$ followed by $H_{2} O_{2} / OH^{-}$ . Butan-2-ol is the Markovnikov product, so use $H_{2} O / H^{+}$ , or better $Hg(OAc)_{2} / H_{2} O$ followed by $NaBH_{4}$ , which avoids any rearrangement.

Solved Example 2

3,3-dimethylbut-1-ene is treated with dilute

H_{2} SO_{4}

. Predict the major product and explain.

Solution:

Protonation gives a secondary carbocation at C-2. A methyl group migrates from C-3 to C-2, converting it into a more stable tertiary carbocation at C-3. Water then attacks there, so the major product is 2,3-dimethylbutan-2-ol and not 3,3-dimethylbutan-2-ol. This rearrangement is exactly why acid hydration has limited synthetic value.

Solved Example 3

How would you prepare 2-phenylethanol,

C_{6} H_{5} CH_{2} CH_{2} OH

, from bromobenzene?

Solution:

Make phenylmagnesium bromide from bromobenzene with magnesium in dry ether. React it with ethylene oxide, which opens at the less substituted carbon and adds two carbons, then hydrolyse with $H_{3} O^{+}$ . The product is $C_{6} H_{5} CH_{2} CH_{2} OH$ , a primary alcohol two carbons longer than the reagent.

Solved Example 4

Suggest two different Grignard routes to 2-methylbutan-2-ol.

Solution:

The $- OH$ carbon carries a methyl, an ethyl and another methyl group. Cutting one methyl gives butan-2-one with $CH_{3} MgBr$ . Cutting the ethyl group gives propanone with $C_{2} H_{5} MgBr$ . Both are followed by $H_{3} O^{+}$ and both give the same tertiary alcohol, so the choice is decided by which starting material is cheaper.

Solved Example 5

Why does

CH_{3} MgBr

fail to convert

HOCH_{2} CH_{2} CHO

into the expected alcohol?

Solution:

The molecule already contains an $- OH$ group, whose hydrogen is acidic towards a Grignard reagent. The first mole of $CH_{3} MgBr$ is consumed as a base, giving methane and the magnesium alkoxide, so that first mole never reaches the $C = O$ group. Either protect the hydroxyl first, or sacrifice an extra mole of $CH_{3} MgBr$ : with two moles the diol product does form.

Solved Example 6

An alkene

C_{4} H_{8}

gives butan-2-ol on oxymercuration and demercuration, and butan-1-ol on hydroboration and oxidation. Identify the alkene.

Solution:

The two products differ only in the position of the $- OH$ group, so the alkene must be terminal and unbranched. That gives but-1-ene, $CH_{3} CH_{2} CH = CH_{2}$ . Markovnikov addition puts $- OH$ on C-2 and anti-Markovnikov addition puts it on C-1.

Solved Example 7

Complete: ethyl benzoate

+ 2 C_{2} H_{5} MgBr

, followed by

H_{3} O^{+}

, gives what?

Solution:

An ester reacts with two moles of Grignard reagent. The first addition expels the ethoxide group to give propiophenone, and the second adds to that ketone. After hydrolysis the product is 3-phenylpentan-3-ol, a tertiary alcohol carrying two identical ethyl groups, which is the signature of the ester route.

Solved Example 8

Propene can be converted into propan-1-ol by
(A)

H_{2} O / H^{+}

(B)

Hg(OAc)_{2} / H_{2} O

, then

NaBH_{4}

(C)

B_{2} H_{6}

, then

H_{2} O_{2} / OH^{-}

(D) aq.

KOH

Solution:

Answer: (C). Propan-1-ol has $- OH$ on the less substituted carbon, the anti-Markovnikov product, which only hydroboration-oxidation gives. (A) and (B) give propan-2-ol; (D) does not react with an alkene.

Solved Example 9

Which pair gives 2-methylpropan-2-ol after acid hydrolysis?
(A) HCHO and

(CH_{3})_{2} CHMgBr

(B)

CH_{3} CHO

and

C_{2} H_{5} MgBr

(C)

CH_{3} COCH_{3}

and

CH_{3} MgBr

(D) HCHO and

CH_{3} MgBr

Solution:

Answer: (C). A $3^{\circ}$ alcohol needs a ketone. Propanone plus a methyl group gives $(CH_{3})_{3} COH$ . (A) gives 2-methylpropan-1-ol, (B) butan-2-ol and (D) ethanol.

Solved Example 10

How many moles of

CH_{3} MgBr

does one mole of 4-hydroxybutan-2-one,

HOCH_{2} CH_{2} COCH_{3}

, consume? Name the product after hydrolysis.

Solution:

Two moles. The first is destroyed by the acidic $O - H$ hydrogen (giving methane); the second adds to the $C = O$ group. Hydrolysis gives $HOCH_{2} CH_{2} C(OH)(CH_{3})_{2}$ , 3-methylbutane-1,3-diol.

Practice Questions

Product of but-1-ene with $B_{2} H_{6}$ , then $H_{2} O_{2} / OH^{-}$ ?Answer: butan-1-ol.
Product of 2-methylpropene with dilute $H_{2} SO_{4}$ ?Answer: 2-methylpropan-2-ol.
Reagent that converts ethyl ethanoate into ethanol only.Answer: $LiAlH_{4}$ , then $H_{3} O^{+}$ (2 mol of ethanol per mole of ester).
$CH_{3} MgBr$ with ethylene oxide, then $H_{3} O^{+}$ , gives?Answer: propan-1-ol, two carbons longer than the reagent.
Give one carbonyl compound and Grignard reagent pair that makes butan-2-ol.Answer: $CH_{3} CHO + C_{2} H_{5} MgBr$ (or propanal $+ CH_{3} MgBr$ ).
Cyclohexene with cold dilute alkaline $KMnO_{4}$ gives?Answer: cis-cyclohexane-1,2-diol.
Which reagent turns $CH_{2} = CHCHO$ into $CH_{2} = CHCH_{2} OH$ , and why not $H_{2}$ /Ni?Answer: $NaBH_{4}$ , which reduces only the $C = O$ ; $H_{2}$ /Ni would also saturate the $C = C$ .

Common Mistakes to Avoid

Watch out

Writing the Markovnikov product for hydroboration. $B_{2} H_{6}$ followed by $H_{2} O_{2} / OH^{-}$ always gives the anti-Markovnikov alcohol.
Forgetting that acid hydration goes through a free carbocation, so the skeleton can rearrange.
Adding water along with $LiAlH_{4}$ or a Grignard reagent. Water is added only in the separate work-up step.
Using $NaBH_{4}$ to reduce an ester or a carboxylic acid. It is not strong enough.
Expecting a tertiary alkyl halide to give a good yield of alcohol with aqueous $KOH$ . Elimination wins; use moist $Ag_{2} O$ .
Preparing a Grignard reagent from a molecule that contains $- OH$ , $- NH_{2}$ , $- COOH$ or a terminal alkyne hydrogen.
Saying hydroboration gives anti addition. It is syn addition with anti-Markovnikov regiochemistry, two separate ideas.
Forgetting that methanal is the only aldehyde that gives a primary alcohol with a Grignard reagent.

Frequently Asked Questions

Which method gives the anti-Markovnikov alcohol from an alkene?

Hydroboration oxidation. Diborane or $BH_{3} \cdot THF$ adds with boron on the less substituted carbon, and alkaline hydrogen peroxide then replaces boron by $- OH$ . Propene gives propan-1-ol, while acid hydration of the same alkene gives propan-2-ol.

Why is oxymercuration preferred over acid catalysed hydration?

Acid hydration forms a free carbocation, which can rearrange and give the wrong carbon skeleton. Oxymercuration goes through a bridged mercurinium ion, so no rearrangement is possible, and it gives the same Markovnikov alcohol in about 90 per cent yield under mild conditions.

What is the difference between NaBH4 and LiAlH4?

$NaBH_{4}$ is a mild hydride donor that reduces only aldehydes and ketones and can even be used in water or alcohol. $LiAlH_{4}$ is far more powerful, reducing esters, carboxylic acids and amides as well, but it reacts violently with water so it must be used in dry ether.

Why must Grignard reactions be carried out in perfectly dry conditions?

The carbon of a Grignard reagent behaves like a carbanion, which is a very strong base. Any acidic hydrogen, including that of water, protonates it immediately to give the alkane $R - H$ , destroying the reagent before it can attack the carbonyl group.

Which carbonyl compound gives a tertiary alcohol with a Grignard reagent?

A ketone. The carbonyl carbon of a ketone already carries two alkyl groups, and the Grignard reagent adds a third, so after hydrolysis you get a tertiary alcohol. Esters with excess Grignard also give tertiary alcohols, but with two identical alkyl groups.

What is the difference between syn and anti hydroxylation?

Cold dilute $KMnO_{4}$ or $OsO_{4}$ delivers both oxygen atoms from one cyclic intermediate, so the two $- OH$ groups add to the same face and give a cis diol. Forming an epoxide first and opening it with acidic water attacks from the opposite face, giving a trans diol.

Which preparation of alcohols questions are most common in NEET?

NEET mostly asks the product of acid hydration or hydroboration of propene or but-1-ene, the product of reducing an aldehyde or ketone with NaBH4 or LiAlH4, and which carbonyl compound gives a primary, secondary or tertiary alcohol with a Grignard reagent. Markovnikov versus anti-Markovnikov settles most of them.

What kind of Grignard problems appear in JEE Main?

JEE Main uses Grignard reagents in multi-step problems: choose the carbonyl compound and reagent for a target alcohol, count the moles used up when the molecule has an acidic hydrogen, or predict the tertiary alcohol from an ester and excess reagent. Always work backwards from the carbon carrying the OH group.

Previous year questions on Preparation of Alcohol

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