Properties of Alcohol

Q: Why do alcohols have higher boiling points than ethers of the same molecular mass?

An alcohol has an O-H bond, so its molecules associate through hydrogen bonds and behave like much larger units. An ether has no O-H bond and cannot do this. Ethanol boils at 78 °C while its isomer dimethyl ether boils at about -24 °C.

Q: Why is the acidity order of alcohols 1 degree greater than 2 degree greater than 3 degree?

Alkyl groups release electrons by the +I effect. More alkyl groups mean more electron density on the oxygen, which makes the O-H bond less polar and the alkoxide ion less stable. A tertiary alcohol has three such groups, so it is the weakest acid of the three.

Q: How does the Lucas test distinguish alcohols?

Lucas reagent is conc. HCl with anhydrous ZnCl₂. It converts the alcohol into an alkyl chloride, which is insoluble and makes the solution cloudy. A tertiary alcohol turns cloudy at once, a secondary in about five minutes and a primary not at all at room temperature, because the test measures how easily a carbocation forms.

Q: Which alcohols give a positive iodoform test?

Only ethanol and secondary alcohols with the structure CH₃CH(OH)R, because they are oxidised to a methyl ketone that then reacts with I₂/NaOH to give a yellow precipitate of iodoform, CHI₃.

ChemistryAlcohols, Phenols And EthersFor NEET aspirants

The properties of alcohols all follow from one small group, $- OH$ . Hydrogen bonding through that group gives alcohols unusually high boiling points and good solubility in water. Chemically, only two bonds can break: the $O - H$ bond, which makes an alcohol a weak acid that reacts with sodium but not with $NaOH$ , and the $C - O$ bond, which lets $- OH$ be replaced by halogen or removed as water. Oxidation then separates $1^{\circ}$ , $2^{\circ}$ and $3^{\circ}$ alcohols cleanly, which is why the properties of alcohols carry marks in NEET and JEE Main.

On this page1Physical2Which bond?3O-H reactions4C-O reactions5Dehydration6Oxidation7Tests8Examples

Key Formulas - Quick Reference

★ Must learnBoiling point: alcohol $>$ ether $>$ alkane of similar mass, because only the alcohol hydrogen bonds. Among isomers, $1^{\circ} > 2^{\circ} > 3^{\circ}$ .
★ Must learnAcidity: $RCOOH > ArOH > H_{2} O > ROH$ , and within alcohols $1^{\circ} > 2^{\circ} > 3^{\circ}$ ( $+ I$ effect).
$2 R - OH + 2 Na \to 2 R - O^{-} Na^{+} + H_{2}$ , but no reaction at all with $NaOH$ .
★ Must learnReactivity towards $HX$ : $3^{\circ} > 2^{\circ} > 1^{\circ}$ , and $HI > HBr > HCl$ . $HCl$ needs $ZnCl_{2}$ for $1^{\circ}$ and $2^{\circ}$ .
★ Must learnDehydration: $3^{\circ} > 2^{\circ} > 1^{\circ}$ , $E1$ , Saytzeff product major. Lower temperature gives the ether, higher temperature gives the alkene.
★ Must learnOxidation: $1^{\circ} \to$ aldehyde $\to$ acid, $2^{\circ} \to$ ketone, $3^{\circ} \to$ no reaction.
Tests: Lucas ( $3^{\circ}$ at once, $2^{\circ}$ in 5 min, $1^{\circ}$ not in the cold), Victor Meyer (red, blue, colourless), $CrO_{3}$ (orange to blue green, negative for $3^{\circ}$ ).

1. Physical Properties

1.1 Hydrogen bonding is the whole story

An alcohol has a hydrogen attached to a strongly electronegative oxygen and two lone pairs on that same oxygen. Every molecule can therefore both donate and accept a hydrogen bond, and the molecules link into chains.

Figure 1: Hydrogen bonding is the whole story. Top: each

O - H

bonds to the next oxygen, so the molecules behave like one large unit. Bottom: ethanol boils at

78^{\circ} C

, its isomer dimethyl ether (no

O - H

) at

- 24^{\circ} C

: a

102^{\circ} C

gap at the same molecular mass.

Dimethyl ether and ethanol have exactly the same molecular formula,

C_{2} H_{6} O

, and almost the same molecular mass. The ether has no

O - H

bond, so it cannot hydrogen bond with itself, and it boils about

100^{\circ} C

lower. This single comparison is worth more than any list of boiling points.

1.2 Branching, chain length and solubility

Two shape effects follow. Branching makes a molecule compact, so its surface contact and its boiling point fall. A longer carbon chain adds a part that cannot hydrogen bond with water, so solubility falls.

Figure 2: Top: among isomers the boiling point falls with branching,

1^{\circ} > 2^{\circ} > 3^{\circ}

(118, 100,

83^{\circ} C

). Bottom: the

- OH

group hydrogen bonds to water but the chain does not; beyond about four carbons the chain wins and solubility collapses.

Methanol, ethanol and propan-1-ol are miscible with water in all proportions.
Solubility falls steadily as the carbon chain grows, because the hydrocarbon part cannot hydrogen bond.
More $- OH$ groups means more solubility, which is why ethane-1,2-diol and glycerol dissolve freely.
Branching raises solubility a little: tert-butanol is miscible with water while n-butanol is not, because the compact shape disturbs the water structure less.

Key idea

One

O - H

explains every physical property: high boiling point, solubility of the small alcohols, and the huge gap between an alcohol and its isomeric ether.

2. Chemical Properties: Which Bond Breaks?

Every reaction in this chapter is easier to remember once you ask a single question: does the reagent attack the hydrogen of $O - H$ , or does it replace the whole $- OH$ group?

Figure 3: Sort every reaction of an alcohol into one of two boxes. If the

- OH

group leaves, the

C - O

bond broke. If only the hydrogen leaves, the

O - H

bond broke.

3. Reactions Involving the O-H Bond

3.1 Acidity of alcohols

Figure 4: Top: the

O - H

bond pair stays on oxygen when the proton leaves, but the alkyl group (

+ I

) crowds more charge on to the alkoxide oxygen, so

RO^{-}

is less stable than

OH^{-}

. Bottom: each extra alkyl group adds to the effect, so acidity falls

1^{\circ} > 2^{\circ} > 3^{\circ}

Alcohols are very weak acids, with $K_{a}$ of the order of $1 0^{- 16}$ to $1 0^{- 18}$ , which is even weaker than water ( $p K_{a} = 15.7$ ). They do not turn blue litmus red.

JEE Advanced Two effects act together. In the alcohol, the electron releasing alkyl group reduces the polarity of the

O - H

bond, so the proton is harder to release. In the alkoxide ion that same group intensifies the negative charge on oxygen and destabilises the ion. Both effects grow as alkyl groups are added, which is why acidity falls in the order

1^{\circ} > 2^{\circ} > 3^{\circ}

. Note that this is the opposite of the stability order of carbocations, so the two must never be confused.

3.2 Reaction with active metals

Figure 5: Top: the

p K_{a}

ladder; an alcohol (

p K_{a}

16 to 18) sits below water (15.7). Bottom: so sodium metal releases

H_{2}

from an alcohol, but

NaOH

cannot deprotonate it: that would make the stronger acid (water) from the weaker one.

3.3 Esterification

Figure 6: Esterification is reversible, so it is driven forward by removing water. Isotope labelling shows the acid loses

- OH

while the alcohol loses only its

H

Exam Trick

In esterification the alcohol loses only its hydrogen, while the acid loses the whole $- OH$ . Isotopic labelling with $^{18} O$ proves it.

The reaction is reversible, so concentrated $H_{2} SO_{4}$ is used both as catalyst and as a dehydrating agent.

Acid anhydrides and acid chlorides esterify faster and irreversibly, which is why aspirin is made from acetic anhydride and not from acetic acid.

Key idea

O-H reactions follow acidity:

1^{\circ} > 2^{\circ} > 3^{\circ}

. The alcohol keeps its oxygen, even in esterification, where it loses only

H

Quick Recall: tap to check

Does ethanol react with

NaOH

No. Ethanol is a weaker acid than water, so the equilibrium lies on the left.

Which is the strongest acid: methanol, propan-2-ol or 2-methylpropan-2-ol?

Methanol: it has the fewest electron releasing alkyl groups.

In esterification, which molecule supplies the ester oxygen?

The alcohol (

^{18} O

labelling shows it keeps its oxygen).

4. Reactions Involving the C-O Bond

4.1 Reaction with hydrogen halides

An $- OH$ group is a very poor leaving group, so the first step is always protonation. Once it becomes $- OH_{2}^{+}$ , water can leave and a halide ion takes its place.

Figure 7: Top: with

HX

, protonation turns

- OH

into the good leaving group

H_{2} O

; for a

3^{\circ}

alcohol the slow step is loss of water. Bottom: the Lucas test times that slow step:

3^{\circ}

cloudy at once,

2^{\circ}

in about 5 min,

1^{\circ}

clear in the cold.

Order	Reason
Reactivity of alcohols: $3^{\circ} > 2^{\circ} > 1^{\circ} > CH_{3}$	a more substituted carbocation forms more easily
Reactivity of halides: $HI > HBr > HCl$	$HI$ is the strongest acid and $I^{-}$ the best nucleophile
$HCl$ needs anhydrous $ZnCl_{2}$ for $1^{\circ}$ and $2^{\circ}$ alcohols	$Cl^{-}$ is a weak nucleophile, so the Lewis acid must first make a better leaving group
$HF$ is generally unreactive	$HF$ is a weak acid and $F^{-}$ a poor nucleophile

Acidity (O-H breaks)

1^{\circ} > 2^{\circ} > 3^{\circ}

alkyl groups destabilise the alkoxide ion
shows in: reaction with Na, esterification rate

C-O cleavage (C-O breaks)

3^{\circ} > 2^{\circ} > 1^{\circ}

alkyl groups stabilise the carbocation
shows in: HX, Lucas test, dehydration

4.2 The Lucas test

Lucas reagent is concentrated $HCl$ with anhydrous $ZnCl_{2}$ . It converts the alcohol into an alkyl chloride, which is insoluble in the reagent and turns the solution cloudy. The time taken measures how fast the carbocation forms, so the test sorts the three classes (lower panel of the figure above).

Exam Trick

"T-S-P: Turbid, Slow, Plain." Tertiary turns turbid at once, Secondary is Slow (about five minutes), Primary stays Plain in the cold. Methanol and primary alcohols need heating.

4.3 With phosphorus halides and thionyl chloride

These reagents avoid carbocations altogether, so they convert an alcohol into a halide without rearrangement. That is exactly why they are preferred for primary and secondary alcohols.

Figure 8: Three ways to swap

- OH

for

- Cl

SOCl_{2}

is preferred because both by-products are gases, so the alkyl halide is left pure.

4.4 Dehydration

Figure 9: Top: dehydration is

E1

: protonate, lose water to a carbocation, strip a

β

hydrogen; it is the reverse of acid catalysed hydration. Bottom: Saytzeff's rule; butan-2-ol gives the more substituted but-2-ene as the major alkene.

Figure 10: Two products from one reaction. Lower temperature with excess alcohol favours the ether; higher temperature with excess acid favours the alkene.

Alcohol	Typical conditions	Product
Ethanol ( $1^{\circ}$ )	conc. $H_{2} SO_{4}$ , $443 K$	ethene
Cyclohexanol ( $2^{\circ}$ )	$85% H_{3} PO_{4}$ , $440 K$	cyclohexene
$(CH_{3})_{3} COH$ ( $3^{\circ}$ )	$20% H_{3} PO_{4}$ , $358 K$	2-methylpropene

Because dehydration goes through a carbocation, rearranged alkenes can appear whenever a hydride or alkyl shift produces a more stable cation. If a question gives a skeleton that can rearrange, check for the shifted product before answering.

Key idea

C-O reactions follow carbocation stability:

3^{\circ} > 2^{\circ} > 1^{\circ}

. Wherever a carbocation forms, check for a hydride or methyl shift before writing the product.

5. Oxidation and Dehydrogenation

Figure 11: Top: oxidation identifies the class;

1^{\circ}

gives an aldehyde then an acid,

2^{\circ}

a ketone,

3^{\circ}

nothing under normal conditions. Bottom: hot copper at

573 K

removes

H_{2}

from

1^{\circ}

and

2^{\circ}

alcohols, but a

3^{\circ}

alcohol dehydrates to an alkene.

Reagent	$1^{\circ}$ alcohol	$2^{\circ}$ alcohol	$3^{\circ}$ alcohol
PCC in $CH_{2} Cl_{2}$	aldehyde (stops there)	ketone	no reaction
Acidified $KMnO_{4}$ or $K_{2} Cr_{2} O_{7}$	carboxylic acid	ketone	no reaction (cleaves only in drastic acid)
Cu at $573 K$	aldehyde	ketone	alkene (dehydration)

PCC in

CH_{2} Cl_{2}

mild and anhydrous

1^{\circ}

alcohol stops at the aldehyde

2^{\circ}

gives the ketone

Acidified

KMnO_{4}

K_{2} Cr_{2} O_{7}

strong, aqueous

1^{\circ}

alcohol goes on to the carboxylic acid

2^{\circ}

gives the ketone

Exam Trick

"Al, One, None." Oxidise a $1^{\circ}$ alcohol and you get an aldehyde, a $2^{\circ}$ alcohol gives a ketone, a $3^{\circ}$ alcohol gives none, because it has no H on the carbinol carbon. The same pattern holds over hot copper, except that the $3^{\circ}$ alcohol dehydrates instead.

Key idea

The H on the carbinol carbon decides oxidation: two H (1°) go all the way to the acid, one H (2°) stops at the ketone, none (3°) means no reaction.

Quick Recall: tap to check

Propan-1-ol with PCC?

Propanal: PCC stops at the aldehyde.

Butan-2-ol with acidified

K_{2} Cr_{2} O_{7}

Butanone (a ketone).

2-Methylpropan-2-ol over Cu at

573 K

2-Methylpropene: a

3^{\circ}

alcohol dehydrates.

6. Tests for Alcohols

Sodium metal: brisk effervescence of $H_{2}$ shows an active hydrogen. A wet sample does the same, so the sample must be dry.
Chromic anhydride, $CrO_{3}$ in aqueous $H_{2} SO_{4}$ : the clear orange solution turns opaque blue green within about two seconds for $1^{\circ}$ and $2^{\circ}$ alcohols. Tertiary alcohols give no change.
Ester test: a sweet smelling ester on warming with a carboxylic acid and conc. $H_{2} SO_{4}$ confirms an $- OH$ group.
Alcohols do not decolourise bromine in carbon tetrachloride, which separates them from alkenes and alkynes.
Iodoform test: only ethanol and alcohols of the type $CH_{3} CH(OH)R$ give a yellow precipitate of $CHI_{3}$ with $I_{2} / NaOH$ .

Figure 12: The Victor Meyer or red-blue test. Three identical steps, then the colour identifies the class: red for

1^{\circ}

, blue for

2^{\circ}

, colourless for

3^{\circ}

Test	$1^{\circ}$	$2^{\circ}$	$3^{\circ}$
Lucas reagent	no turbidity in the cold	turbidity in about 5 min	turbidity at once
Victor Meyer	red	blue	colourless
Oxidation product	aldehyde then acid	ketone	no reaction
Cu at $573 K$	aldehyde	ketone	alkene

The flowchart turns these tests into a procedure: confirm the $- OH$ group, sort the class with Lucas reagent, then confirm with a second test. The mind map closes the page with every property on one screen.

Figure 13: Flowchart: one reagent sorts the three classes by how fast the carbocation forms, and a second test (oxidation or Victor Meyer) confirms the answer.

Figure 14: Mind map: two bonds, three orders. Acidity runs

1^{\circ} > 2^{\circ} > 3^{\circ}

; reaction with

HX

and dehydration run

3^{\circ} > 2^{\circ} > 1^{\circ}

7. Solved Examples

Solved Example 1

Arrange in increasing order of boiling point: butane, butan-1-ol, methoxypropane.

Solution:

All three have similar molecular masses. Butane has only weak dispersion forces, methoxypropane adds dipole interactions but has no $O - H$ bond, while butan-1-ol hydrogen bonds strongly. The order is butane $<$ methoxypropane $<$ butan-1-ol.

Solved Example 2

Why does ethanol react with sodium but not with sodium hydroxide?

Solution:

Sodium is a metal and reduces the $O - H$ hydrogen to $H_{2}$ , giving sodium ethoxide, so that reaction is thermodynamically downhill. With $NaOH$ the reaction would have to convert the weaker acid (ethanol) into the stronger acid (water), which the equilibrium refuses to do. Hence no reaction with $NaOH$ .

Solved Example 3

Three bottles contain butan-1-ol, butan-2-ol and 2-methylpropan-2-ol. How would you identify each using one reagent?

Solution:

Use Lucas reagent, conc. $HCl$ with anhydrous $ZnCl_{2}$ . The bottle that turns cloudy immediately is 2-methylpropan-2-ol ( $3^{\circ}$ ); the one that turns cloudy in about five minutes is butan-2-ol ( $2^{\circ}$ ); the one that stays clear at room temperature is butan-1-ol ( $1^{\circ}$ ).

Solved Example 4

Butan-2-ol is heated with concentrated

H_{2} SO_{4}

. Write the major product and justify it.

Solution:

Dehydration can remove a hydrogen from C-1 or from C-3. Saytzeff's rule says the hydrogen comes from the carbon with fewer hydrogens, giving the more substituted alkene. The major product is therefore but-2-ene, whose double bond carries two alkyl groups, with but-1-ene as the minor product.

Solved Example 5

An alcohol

C_{4} H_{10} O

reacts rapidly with sodium but hardly at all with Lucas reagent. On heating with hot conc.

H_{2} SO_{4}

it gives

C_{4} H_{8}

, which on hydration gives a new alcohol that is inert to sodium but reacts at once with Lucas reagent. Identify all three compounds.

Solution:

No reaction with Lucas reagent in the cold means the first alcohol is primary. It is 2-methylpropan-1-ol, $(CH_{3})_{2} CHCH_{2} OH$ . Dehydration gives 2-methylpropene, $(CH_{3})_{2} C = CH_{2}$ . Markovnikov hydration of that alkene gives 2-methylpropan-2-ol, a tertiary alcohol, which is why it reacts instantly with Lucas reagent. (The statement that it is inert to sodium simply reflects how very slowly a $3^{\circ}$ alcohol reacts.)

Solved Example 6

Why is

SOCl_{2}

preferred over conc.

HCl

for preparing an alkyl chloride from a primary alcohol?

Solution:

Conc. $HCl$ needs $ZnCl_{2}$ and heat for a primary alcohol, and the reaction can produce rearranged products because a carbocation like species is involved. Thionyl chloride reacts under mild conditions without a carbocation, so there is no rearrangement, and both by-products, $SO_{2}$ and $HCl$ , escape as gases leaving a pure product.

Solved Example 7

Predict the organic products: (a) propan-1-ol with PCC, (b) propan-1-ol with acidified

K_{2} Cr_{2} O_{7}

, (c) propan-2-ol over Cu at

573 K

, (d) 2-methylpropan-2-ol over Cu at

573 K

Solution:

(a) Propanal, since PCC stops at the aldehyde. (b) Propanoic acid, since the strong oxidant carries on. (c) Propanone, because a secondary alcohol is dehydrogenated to a ketone. (d) 2-Methylpropene, because a tertiary alcohol has no hydrogen on the carbinol carbon and therefore dehydrates instead of losing hydrogen.

Solved Example 8

Which alcohol turns Lucas reagent cloudy fastest at room temperature?
(A) butan-1-ol
(B) butan-2-ol
(C) 2-methylpropan-2-ol
(D) 2-methylpropan-1-ol

Solution:

Answer: (C). It is the only $3^{\circ}$ alcohol, so it forms the most stable carbocation and its chloride separates at once. (B) takes about five minutes; (A) and (D) are $1^{\circ}$ and stay clear in the cold.

Solved Example 9

The correct order of acid strength is
(A)

CH_{3} COOH > C_{6} H_{5} OH > H_{2} O > C_{2} H_{5} OH

(B)

C_{6} H_{5} OH > CH_{3} COOH > C_{2} H_{5} OH > H_{2} O

(C)

CH_{3} COOH > H_{2} O > C_{6} H_{5} OH > C_{2} H_{5} OH

(D)

H_{2} O > CH_{3} COOH > C_{2} H_{5} OH > C_{6} H_{5} OH

Solution:

Answer: (A). $p K_{a}$ values: acetic acid about 4.8, phenol about 10, water 15.7, ethanol about 16. Resonance stabilises the carboxylate and phenoxide ions; the ethyl group ( $+ I$ ) destabilises ethoxide.

Solved Example 10

Which of these give a yellow precipitate with

I_{2} / NaOH

: ethanol, propan-1-ol, propan-2-ol, butan-2-ol, 2-methylpropan-2-ol?

Solution:

The iodoform test needs a $CH_{3} CH(OH) -$ group, which $I_{2} / NaOH$ first oxidises to $CH_{3} CO -$ . Ethanol, propan-2-ol and butan-2-ol are positive. Propan-1-ol has no such group, and 2-methylpropan-2-ol has no H on its carbinol carbon, so both are negative.

Practice Questions

Arrange in increasing boiling point: ethanol, propane, dimethyl ether.Answer: propane $<$ dimethyl ether $<$ ethanol.
Why is 2-methylpropan-2-ol miscible with water while butan-1-ol is not?Answer: its compact shape disturbs the hydrogen-bonded water structure less.
Write the products of ethanol with $PCl_{5}$ .Answer: $C_{2} H_{5} Cl + POCl_{3} + HCl$ .
Major product when 2-methylbutan-2-ol is heated with conc. $H_{2} SO_{4}$ ?Answer: 2-methylbut-2-ene (Saytzeff product).
Ethanol with conc. $H_{2} SO_{4}$ at $413 K$ , and at $443 K$ ?Answer: ethoxyethane at 413 K; ethene at 443 K.
Product of cyclohexanol with PCC?Answer: cyclohexanone.
Which alcohol $C_{4} H_{10} O$ gives a ketone over Cu at $573 K$ ?Answer: butan-2-ol, giving butanone.

Common Mistakes to Avoid

Watch out

Saying alcohols react with $NaOH$ . They are weaker acids than water, so they do not.
Mixing up the two orders. Acidity is $1^{\circ} > 2^{\circ} > 3^{\circ}$ , but reactivity with $HX$ and ease of dehydration are $3^{\circ} > 2^{\circ} > 1^{\circ}$ .
Claiming that a tertiary alcohol is never oxidised. Under drastic acidic conditions it dehydrates first and the alkene is then cleaved; the usual statement means no reaction under normal conditions.
Forgetting $ZnCl_{2}$ when writing the Lucas reagent, or expecting a primary alcohol to respond in the cold.
Using acidified dichromate when the question asks for an aldehyde. That oxidises straight through to the acid; PCC is the reagent that stops at the aldehyde.
Writing but-1-ene as the major dehydration product of butan-2-ol. Saytzeff gives but-2-ene.
Assuming ethers boil higher than alcohols. Without an $O - H$ bond there is no hydrogen bonding, so ethers boil close to alkanes.
Saying the alcohol loses $- OH$ during esterification. The isotope experiment shows the alcohol loses only $H$ .

Frequently Asked Questions

Why do alcohols have higher boiling points than ethers of the same molecular mass?

An alcohol has an $O - H$ bond, so its molecules associate through hydrogen bonds and behave like much larger units. An ether has no $O - H$ bond and cannot do this. Ethanol boils at $78^{\circ} C$ while its isomer dimethyl ether boils at about $- 24^{\circ} C$ .

Why is the acidity order of alcohols 1 degree greater than 2 degree greater than 3 degree?

Alkyl groups release electrons by the $+ I$ effect. More alkyl groups mean more electron density on the oxygen, which makes the $O - H$ bond less polar and the alkoxide ion less stable. A tertiary alcohol has three such groups, so it is the weakest acid of the three.

How does the Lucas test distinguish alcohols?

Lucas reagent is conc. $HCl$ with anhydrous $ZnCl_{2}$ . It converts the alcohol into an alkyl chloride, which is insoluble and makes the solution cloudy. A tertiary alcohol turns cloudy at once, a secondary in about five minutes and a primary not at all at room temperature, because the test measures how easily a carbocation forms.

What is the difference between the products of dehydration at 413 K and at 443 K?

At the lower temperature with excess alcohol the reaction is bimolecular and two alcohol molecules combine to give an ether. At the higher temperature with excess acid the protonated alcohol loses water and then a beta hydrogen, giving an alkene.

Why are tertiary alcohols resistant to oxidation?

Oxidation of an alcohol removes the hydrogen on the carbon that carries the $- OH$ group so that a $C = O$ bond can form. A tertiary alcohol has no such hydrogen, so oxidation would require breaking a carbon to carbon bond, which happens only under drastic conditions.

Which alcohols give a positive iodoform test?

Only ethanol and secondary alcohols with the structure $CH_{3} CH(OH)R$ , because they are oxidised to a methyl ketone that then reacts with $I_{2} / NaOH$ to give a yellow precipitate of iodoform, $CHI_{3}$ .

Which properties of alcohols are asked most often in NEET?

NEET favours the Lucas test order, the oxidation products of primary, secondary and tertiary alcohols, dehydration of ethanol at 413 K and 443 K, the acidity order of acids, phenol, water and alcohols, and why alcohols boil far above isomeric ethers. The mind map on this page holds all five.

How does JEE Main test the chemical properties of alcohols?

JEE Main chains the steps: an alcohol is dehydrated, the alkene rehydrated or oxidised, and you identify each compound from Lucas, iodoform or oxidation results. Carbocation rearrangement during dehydration or reaction with HX is the favourite trap, so always check for a hydride or methyl shift.

Previous year questions on Properties of Alcohol

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

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