Alcohols, Phenols and Ethers: Class 12 Chemistry · Chapter 7

Chemistry Part 2 · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

100 min
1 simulation
48 practice questions
39 MCQs

What you will learn in Alcohols, Phenols and Ethers

Classify alcohols, phenols and ethers and name them by common and IUPAC rules
Relate the bonding and bond angles at oxygen to the structures of alcohols, phenols and ethers
Choose reagents to make alcohols from alkenes, carbonyl compounds and Grignard reagents, and phenol from benzene derivatives and cumene
Explain boiling points and solubility in terms of hydrogen bonding, chain length and branching
Compare the acidity of alcohols and phenols and predict how ring substituents change it
Predict the products of esterification, substitution, dehydration and oxidation of alcohols and of electrophilic substitution, Kolbe and Reimer–Tiemann reactions of phenol
Describe how methanol and ethanol are made and used, and why methanol is poisonous
Plan ether syntheses by dehydration or Williamson synthesis and predict the products of ether cleavage and of substitution in aryl ethers

Original companion text aligned to NCERT Chemistry Class 12, Chapter 7, pp. 193–226. Not affiliated with or endorsed by NCERT.

Squeeze a drop of hand sanitiser and you smell an alcohol, usually ethanol or propan-2-ol. Dab clove oil on an aching tooth and you are using eugenol, a phenol that also carries an ether group. Refiners have blended MTBE, an ether, into petrol to help it burn smoothly. All three share one feature: an oxygen atom with a single bond on each side.

An alcohol has an –OH group on an $s p^{3}$ carbon of an alkyl group, written $R - O H$ . A phenol has the –OH bonded directly to a carbon of an aromatic ring, $A r - O H$ . An ether has the oxygen between two carbon groups, $R - O - R^{'}$ , where each group may be alkyl or aryl. A neat way to hold the three families together is to start from water and replace its hydrogen atoms with carbon groups (Fig. 7.1).

4/4Carbon groups replace both H atoms: an ether, $R - O - R^{'}$ .

Fig. 7.1Replace one hydrogen of water by a carbon group for an alcohol or phenol; replace both for an ether.NCERT Unit opener · p. 193

A bent oxygen atom with two lone pairs sits at the top centre with one bond to each side. Step 1 shows two hydrogen atoms on the oxygen: water. Step 2 replaces the left hydrogen with an ethyl group, giving an alcohol (ethanol). Step 3 puts a benzene ring on the left instead, giving a phenol. Step 4 has an ethyl group on the left and a methyl group on the right, giving an ether, methoxyethane. The group that has changed is shaded each time.

This unit follows the three families through classification and naming, the shapes around the oxygen, methods of preparation, physical properties and reactions, and ends with two alcohols that industry makes by the million tonnes.

7.1 Classification

Two questions sort almost every compound in this unit. How many –OH groups does it carry? And what kind of carbon holds each one?

7.1.1 Alcohols — Mono, Di, Tri or Polyhydric alcohols

Count the –OH groups first. One makes a compound monohydric, two dihydric, three trihydric, and more than three polyhydric. Each –OH normally sits on a different carbon atom; a carbon holding two –OH groups usually loses water at once.

Table 7.1Counting the hydroxyl groups

OH groups	Example
one (monohydric)	$(C H_{3})_{2} C H O H$ , propan-2-ol, in sanitisers
two (dihydric)	$C H_{3} C H (O H) C H_{2} O H$ , propane-1,2-diol
three (trihydric)	$H O C H_{2} C H (O H) C H_{2} O H$ , glycerol
six (polyhydric)	$C_{6} H_{8} (O H)_{6}$ , sorbitol, a sweetener

Monohydric alcohols are sorted further by the carbon that carries the –OH. When that carbon is $s p^{3}$ hybridised, look at its neighbours (Fig. 7.2):

Primary (1°): the –OH carbon is bonded to one other carbon (or to none, in methanol), as in ethanol.
Secondary (2°): it is bonded to two carbons, as in propan-2-ol.
Tertiary (3°): it is bonded to three carbons, as in 2-methylpropan-2-ol.
Allylic: it is the $s p^{3}$ carbon next to a C=C double bond, as in but-2-en-1-ol, $C H_{3} C H = C H C H_{2} O H$ .
Benzylic: it is the $s p^{3}$ carbon next to a benzene ring, as in 1-phenylethanol, $C_{6} H_{5} C H (O H) C H_{3}$ .

Notice that allylic and benzylic name the neighbour, not the degree, so the two labels combine: but-2-en-1-ol is a primary allylic alcohol, and 1-phenylethanol is a secondary benzylic one.

3/62-Methylpropan-2-ol: three carbon neighbours.

Fig. 7.2The carbon that holds the –OH, and what it is bonded to, decides the class.NCERT p. 194

Each step shows one alcohol with its OH-bearing carbon circled and its carbon neighbours boxed. Step 1, ethanol: one carbon neighbour, primary. Step 2, propan-2-ol: two carbon neighbours, secondary. Step 3, 2-methylpropan-2-ol: three methyl neighbours, tertiary. Step 4, but-2-en-1-ol: the OH carbon sits next to a C=C, allylic. Step 5, 1-phenylethanol: the OH carbon sits next to a benzene ring, benzylic. Step 6, prop-1-en-2-ol: the OH is on a carbon of the C=C itself, vinylic.

When the –OH sits on an $s p^{2}$ carbon of a C=C bond, the compound is a vinylic alcohol, also called an enol. Simple enols do not last: the hydrogen moves from oxygen to the neighbouring carbon, so prop-1-en-2-ol rearranges at once to propanone. Phenols also carry –OH on an $s p^{2}$ carbon, but there the aromatic ring keeps the structure stable.

7.1.2 Phenols — Mono, Di and trihydric phenols

Phenols are counted the same way. Phenol itself and thymol, the phenol that gives ajwain seeds their sharp smell, are monohydric. The three benzenediols (catechol, resorcinol and hydroquinone) are dihydric, and benzene-1,2,3-triol (pyrogallol), once used to develop photographs, is trihydric.

7.1.3 Ethers

For ethers, look at the two groups on the oxygen. Identical groups make a symmetrical (simple) ether, such as methoxymethane, $C H_{3} O C H_{3}$ , or diphenyl ether, $C_{6} H_{5} O C_{6} H_{5}$ . Two different groups make an unsymmetrical (mixed) ether, such as methoxybenzene, $C_{6} H_{5} O C H_{3}$ , or MTBE, $C H_{3} O C (C H_{3})_{3}$ .

Quick check

1 of 3

Glycerol,

H O C H_{2} C H (O H) C H_{2} O H

, is best described as

a trihydric alcohol with two 1° and one 2° –OHa trihydric alcohol with three 1° –OH groupsa dihydric alcohola trihydric phenol

Show answer

Answer (A) a trihydric alcohol with two 1° and one 2° –OH

Three –OH groups make it trihydric. The two end carbons each have one carbon neighbour (1°); the middle carbon has two (2°). It has no ring, so it is not a phenol.

7.2 Nomenclature

Most compounds in this unit have two names worth knowing: a common name that you meet on labels and in older books, and the systematic IUPAC name, from which you can draw the structure.

Naming alcohols

Common names of alcohols are two words: the name of the alkyl group, then alcohol. So $(C H_{3})_{2} C H O H$ is isopropyl alcohol and $C_{6} H_{5} C H_{2} O H$ is benzyl alcohol.

In the IUPAC system an alcohol is named as a substituted alkane whose final -e becomes -ol. Four rules do most of the work:

Choose the longest carbon chain that contains the carbon carrying the –OH.
Number that chain from the end that gives the –OH carbon the lower locant, even if a branch then gets a higher number.
Add the other substituents as prefixes, in alphabetical order, each with its locant.
For two or more –OH groups keep the -e of the alkane and add -diol, -triol and so on, with every locant: $H O (C H_{2})_{4} O H$ is butane-1,4-diol.

In a ring, the carbon carrying the –OH is C1, and you number round the ring in the direction that gives the next substituent the lower locant: a methyl group two carbons further round makes 3-methylcyclohexan-1-ol, usually shortened to 3-methylcyclohexanol. Fig. 7.3 builds one name step by step.

4/4Prefix, parent, locant and suffix: 5-methylhexan-3-ol.

Fig. 7.3The –OH sets the direction of numbering.NCERT p. 196

A zig-zag chain of six carbon groups has a methyl branch on the second carbon from the left and an OH group on the fourth. Step 1 shades the longest chain that includes the OH carbon: six carbons, so the parent is hexane. Step 2 numbers the chain from the right-hand end, which gives the OH carbon number 3 rather than 4. Step 3 circles the methyl branch, which is now on carbon 5. Step 4 shows the full name, 5-methylhexan-3-ol.

Table 7.2Common and IUPAC names of some alcoholsNCERT Table 7.1 · p. 196

Formula	Common; IUPAC
$(C H_{3})_{2} C H$ – $C H_{2} C H_{2} O H$	isopentyl alcohol; 3-methylbutan-1-ol
$(C H_{3})_{3} C$ – $C H_{2} O H$	neopentyl alcohol; 2,2-dimethylpropan-1-ol
$(C H_{3})_{2} C (O H)$ – $C_{2} H_{5}$	tert-pentyl alcohol; 2-methylbutan-2-ol
$C H_{2} = C H C H_{2} O H$	allyl alcohol; prop-2-en-1-ol
$C_{6} H_{5} C H_{2} O H$	benzyl alcohol; phenylmethanol
$C H_{3} C H (O H)$ – $C H_{2} O H$	propylene glycol; propane-1,2-diol
$C_{5} H_{9} O H$ (ring)	cyclopentyl alcohol; cyclopentanol

Naming phenols

Phenol, $C_{6} H_{5} O H$ , is both the common and the IUPAC name of the parent compound. In a substituted phenol, the ring carbon carrying –OH is C1. For two groups on a ring, older names use the prefixes o-, m- and p- (ortho = 1,2; meta = 1,3; para = 1,4), and the methylphenols have their own name, the cresols.

Table 7.3Names of some phenolsNCERT p. 197

Structure	Common; IUPAC
$m$ - $C H_{3} C_{6} H_{4} O H$	m-cresol; 3-methylphenol
$o$ - $C_{6} H_{4} (O H)_{2}$	catechol; benzene-1,2-diol
$m$ - $C_{6} H_{4} (O H)_{2}$	resorcinol; benzene-1,3-diol
$p$ - $C_{6} H_{4} (O H)_{2}$	hydroquinone (quinol); benzene-1,4-diol
OH at C1, isopropyl at C2, methyl at C5	thymol; 2-isopropyl-5-methylphenol

Naming ethers

To give an ether its common name, write the two groups on the oxygen in alphabetical order and finish with ether: $C_{6} H_{5} O C H_{2} C H_{3}$ is ethyl phenyl ether. Two identical groups take the prefix di, as in dipropyl ether.

For the IUPAC name, choose the larger group, or the ring, as the parent hydrocarbon and name everything else, oxygen included, as an alkoxy (–OR) or aryloxy (–OAr) prefix: $C H_{3} O C H_{2} C H_{3}$ is methoxyethane and $C_{6} H_{5} O C H_{2} C H_{2} C H_{3}$ is propoxybenzene. If the molecule also has an –OH group, the –OH takes the suffix and the ether part becomes a prefix.

Table 7.4Common and IUPAC names of some ethersNCERT Table 7.2 · p. 197

Formula	Common; IUPAC
$C H_{3} O$ – $C (C H_{3})_{3}$	tert-butyl methyl ether (MTBE); 2-methoxy-2-methylpropane
$C_{2} H_{5} O$ – $(C H_{2})_{3} C H_{3}$	butyl ethyl ether; 1-ethoxybutane
$C_{6} H_{5} O$ – $C H_{2} C H_{2} C H_{3}$	phenyl propyl ether; propoxybenzene
$C_{6} H_{5} O C_{6} H_{5}$	diphenyl ether; phenoxybenzene

Example 7.1NCERT Example 7.1 · p. 198

Naming from structures

Give the IUPAC name of each compound.

$(C H_{3})_{2} C (O H) C H_{2}$ – $C H (C l) C H_{3}$
$C H_{3} C H_{2} O C H (C H_{3}) C H_{2} C H_{3}$
a phenol with Cl para to –OH and $C H_{3}$ next to the Cl
$C H_{3} O C H_{2} C H_{2} O H$

Try it yourself first. Then reveal the solution one step at a time.

Practice · 1 question

Practice: naming: questions with answers

1
Short answerMediumNCERT Intext Question 7.3
Write the IUPAC name of each compound.
1. $C H_{3} C H (O H) C H_{2}$ – $C H (C H_{3}) C H_{2} C H_{3}$
  Show solution
  4-Methylhexan-2-ol: the six-carbon chain runs through the ethyl end; numbering from the left gives –OH the locant 2.
2. $H O C H_{2} C H (C H_{3})$ – $C H (O H) C H_{3}$
  Show solution
  2-Methylbutane-1,3-diol: the –OH locants {1,3} beat {2,4}.
3. a cyclopentane ring with –OH on one carbon and Br on the next
  Show solution
  2-Bromocyclopentan-1-ol (2-bromocyclopentanol): the –OH carbon is C1.
4. $C H_{2} = C H C H_{2}$ – $C H (O H) C H_{3}$
  Show solution
  Pent-4-en-2-ol: number from the right so that –OH gets 2; the double bond then starts at C4.
5. $C H_{3} C \equiv C C H_{2} O H$
  Show solution
  But-2-yn-1-ol: –OH on C1, triple bond between C2 and C3.

7.3 Structures of Functional Groups

Start from water. Its oxygen is $s p^{3}$ hybridised: two hybrid orbitals bond to hydrogen atoms and two hold lone pairs. Lone pairs take up more room than bonding pairs, so they close the H–O–H angle to 104.5°, well below the tetrahedral 109.5° (Fig. 7.4).

Swap one hydrogen for a methyl group and you have methanol. The oxygen is still $s p^{3}$ , and the C–O–H angle opens to about 109°, still just under tetrahedral. Swap the second hydrogen too, giving methoxymethane, and the two methyl groups crowd each other enough to push the C–O–C angle to 111.7°, a little above tetrahedral. Each C–O bond of this ether, 141 pm long, matches the C–O bond of methanol to within a picometre or so.

Phenol differs at the carbon, not at the oxygen. Its –OH is attached to an $s p^{2}$ ring carbon, and the C–O bond shrinks to 136 pm (methanol: 142 pm). An $s p^{2}$ orbital has more s character and holds its bonding pair closer, and one lone pair of oxygen is shared with the ring, which gives the C–O bond a little double-bond character (acidity of phenols).

3/4C–O–C is 111.7°: two methyl groups crowd each other.

Fig. 7.4Lone pairs close the angle at oxygen; bulky groups open it.NCERT Fig. 7.1 · p. 198

An oxygen atom with two lone pairs is drawn with two bonds pointing down. Step 1, water: two hydrogens at 104.5 degrees, squeezed by the lone pairs. Step 2, methanol: a methyl group and a hydrogen at about 109 degrees. Step 3, methoxymethane: two methyl groups at 111.7 degrees, pushed apart by their bulk. Step 4, phenol: a benzene ring and a hydrogen on the oxygen, with the C–O bond labelled 136 pm, shorter than the 142 pm bond of methanol.

Quick check

1 of 3

Arrange by the bond angle at oxygen, smallest first.

C–O–C in methoxymethane
H–O–H in water
C–O–H in methanol

Show answer

Answer

H–O–H in water
C–O–H in methanol
C–O–C in methoxymethane

Lone-pair repulsion squeezes water most (104.5°); replacing one H by

C H_{3}

opens it to about 109°; two methyl groups push it to 111.7°.

7.4 Alcohols and Phenols

We now see how alcohols and phenols are made, why they boil and dissolve as they do, and how they react.

7.4.1 Preparation of Alcohols

From alkenes

Acid-catalysed hydration. Water adds across a C=C bond when an acid such as dilute sulphuric acid is present as catalyst. With an unsymmetrical alkene the hydrogen goes to the carbon that already has more hydrogens and the –OH to the other carbon (Markovnikov's rule), because that order passes through the more stable carbocation.

\begin{aligned} (C H_{3})_{2} C = C H_{2} \\ + H_{2} O \overset{H^{+}}{\to} \\ (C H_{3})_{3} C - O H \end{aligned}

(7.1)

The mechanism has three steps (Fig. 7.5). First, the π electrons of the C=C pick up a proton from $H_{3} O^{+}$ . The proton goes to the $C H_{2}$ end, so the positive charge lands on the carbon that carries more alkyl groups. Second, a water molecule bonds through a lone pair to the cationic carbon, giving a protonated alcohol (an oxonium ion). Third, another water molecule removes the extra proton; the alcohol forms and $H_{3} O^{+}$ is regenerated.

2/4A lone pair on water bonds to the positive carbon.

Fig. 7.5Protonate, add water, lose a proton: the acid is a catalyst.NCERT p. 199

Step 1 shows 2-methylpropene beside a hydronium ion, with a curved arrow from the C=C bond to a hydrogen of the hydronium ion and a second arrow showing the O–H bond electrons moving to oxygen. Step 2 shows the tertiary carbocation, (CH3)3C+, with a curved arrow from a lone pair of a water molecule to the positive carbon. Step 3 shows the oxonium ion, with a second water molecule pulling off one of the hydrogens on oxygen. Step 4 shows the products: 2-methylpropan-2-ol and a regenerated hydronium ion.

Hydroboration–oxidation. This two-stage route gives the alcohol that acid hydration cannot. First, borane, $B H_{3}$ (supplied as its dimer, diborane, $B_{2} H_{6}$ ), adds across the C=C. Boron bonds to the less substituted carbon, the one with more hydrogens and less crowding, while a hydrogen goes to the other carbon; each borane takes up three alkenes to form a trialkylborane. Second, alkaline hydrogen peroxide swaps every C–B bond for a C–OH bond. The net change puts H and OH across the double bond the anti-Markovnikov way, with few side products. This is hydroboration–oxidation.

\begin{aligned} 3 R C H = C H_{2} + B H_{3} \\ \overset{}{\to} (R C H_{2} C H_{2})_{3} B \\ \to_{O H^{-}}^{H_{2} O_{2}} 3 R C H_{2} C H_{2} O H \end{aligned}

(7.2)

Compare

One alkene, two different alcohols

Acid-catalysed hydration

$(C H_{3})_{2} C = C H_{2}$ gives 2-methylpropan-2-ol
–OH ends on the more substituted carbon
goes through a carbocation, which may rearrange

Hydroboration–oxidation

$(C H_{3})_{2} C = C H_{2}$ gives 2-methylpropan-1-ol
–OH ends on the less substituted carbon
no carbocation, so no rearrangement

From carbonyl compounds

Reducing aldehydes and ketones. Reduction adds two hydrogen atoms to a C=O group, one to carbon and one to oxygen. The starting compound decides the product: the –CHO of an aldehyde becomes $- C H_{2} O H$ , a primary alcohol, and the C=O of a ketone becomes CH–OH, a secondary alcohol. On a large scale, hydrogen gas with a metal catalyst (Ni, Pd or Pt) does the job; in the laboratory, the hydride donors sodium borohydride, $N a B H_{4}$ , and lithium aluminium hydride, $L i A l H_{4}$ , are more convenient.

\begin{aligned} R C H O \to_{N i}^{H_{2}} R C H_{2} O H \\ and \\ R_{2} C = O \\ \overset{N a B H_{4}}{\to} R_{2} C H O H \end{aligned}

(7.3)

Reducing carboxylic acids and esters. The –COOH group is harder to reduce: $N a B H_{4}$ leaves it untouched, and only the far stronger $L i A l H_{4}$ , followed by water, converts it to $- C H_{2} O H$ . The two numbers on the arrow below mean: add the first reagent, let that reaction finish, then add the second. $L i A l H_{4}$ costs too much for bulk chemicals, so manufacturers take a detour: they make an ester of the acid and hydrogenate the ester over a metal catalyst.

\begin{aligned} R C O O H \\ \to_{2 \cdot H_{2} O}^{1 \cdot L i A l H_{4}} R C H_{2} O H \end{aligned}

(7.4)

From Grignard reagents

A Grignard reagent, $R^{'} M g X$ , has a carbon with a partial negative charge bonded to magnesium. That carbon attacks the partly positive carbon of a carbonyl group (nucleophilic addition) and forms a new C–C bond, giving a magnesium alkoxide, the adduct. Water or dilute acid then hydrolyses the adduct to the alcohol.

\begin{aligned} R_{2} C = O + R^{'} M g X \\ \overset{}{\to} R_{2} C (R^{'}) O M g X \\ \overset{H_{2} O}{\to} R_{2} C (R^{'}) O H \\ + M g (O H) X \end{aligned}

(7.5)

The carbonyl partner decides the class of alcohol (Fig. 7.6): methanal gives a primary alcohol, any other aldehyde a secondary alcohol, and a ketone a tertiary alcohol. Because the chain also grows by the R′ group, this is one of the most useful ways to build a carbon skeleton.

3/3Propanone gives 2-methylbutan-2-ol, a 3° alcohol.

Fig. 7.6Methanal → 1°, other aldehydes → 2°, ketones → 3°.NCERT p. 201

Three rows show the same Grignard reagent, ethylmagnesium bromide followed by water, reacting with three carbonyl compounds. Row 1: methanal gives propan-1-ol, a primary alcohol. Row 2: ethanal gives butan-2-ol, a secondary alcohol. Row 3: propanone gives 2-methylbutan-2-ol, a tertiary alcohol. Each step shades one row.

Example 7.2NCERT Example 7.2 · p. 201

Predicting alcohols from three reactions

Give the structure and IUPAC name of the main product of each reaction.

Pentan-3-one with sodium borohydride.
2-Methylbut-2-ene, $(C H_{3})_{2} C = C H C H_{3}$ , with water and dilute sulphuric acid.
Cyclohexanone with methylmagnesium iodide, then water.

Try it yourself first. Then reveal the solution one step at a time.

Example

Planning a Grignard synthesis

Suggest two different pairs of a Grignard reagent and a carbonyl compound that give 2-methylbutan-2-ol, $(C H_{3})_{2} C (O H) C H_{2} C H_{3}$ .

Try it yourself first. Then reveal the solution one step at a time.

7.4.2 Preparation of Phenols

Coal tar was the first source of phenol, sold as carbolic acid. Modern phenol is made from benzene and its derivatives instead; Fig. 7.7 maps four routes.

From haloarenes. Sodium hydroxide displaces chlorine from chlorobenzene only under brute-force conditions: about 623 K and some 300 atm. That tells you how firmly an aryl C–Cl bond resists substitution. The result is sodium phenoxide; adding acid frees the phenol.
From benzenesulphonic acid. Here the group that leaves is $- S O_{3} H$ instead of chlorine. Benzene and oleum (fuming sulphuric acid) give benzenesulphonic acid; fusing it with solid NaOH swaps the sulphonate group for –O⁻, and acid turns the phenoxide into phenol.
From diazonium salts. Cold nitrous acid (from $N a N O_{2}$ and HCl, kept at 273–278 K) turns aniline into benzenediazonium chloride. The salt is stable only in the cold: warm the solution and nitrogen bubbles off while water takes its place on the ring, giving phenol. You will meet diazonium salts again in Unit 9.
From cumene. The industrial favourite starts from cumene (isopropylbenzene), itself made from benzene and propene. Blowing air through cumene forms cumene hydroperoxide; dilute acid then splits it into phenol and propanone. Two saleable products from one process keep costs low, and most phenol made today comes from this route.

\begin{aligned} C_{6} H_{5} C l + N a O H \\ \to_{h i g h P}^{623 K} C_{6} H_{5} O N a \\ \overset{H^{+}}{\to} C_{6} H_{5} O H \end{aligned}

(7.6)

\begin{aligned} C_{6} H_{6} \overset{o l e u m}{\to} \\ C_{6} H_{5} S O_{3} H \\ \to_{f u s e}^{N a O H} C_{6} H_{5} O N a \\ \overset{H^{+}}{\to} C_{6} H_{5} O H \end{aligned}

(7.7)

\begin{aligned} C_{6} H_{5} N H_{2} \to_{H C l}^{N a N O_{2}} \\ C_{6} H_{5} N_{2}^{+} C l^{-} \\ \to_{w a r m}^{H_{2} O} C_{6} H_{5} O H \\ + N_{2} ↑ + H C l \end{aligned}

(7.8)

\begin{aligned} C_{6} H_{5} C H (C H_{3})_{2} \overset{O_{2}}{\to} \\ C_{6} H_{5} C (C H_{3})_{2} O O H \\ \to_{H_{2} O}^{H^{+}} C_{6} H_{5} O H \\ + C H_{3} C O C H_{3} \end{aligned}

(7.9)

4/4Cumene with air, then dilute acid: phenol plus propanone.

Fig. 7.7Four routes to phenol; the cumene route also yields propanone.NCERT p. 202

A phenol ring sits in the centre. Arrows reach it from four corners: chlorobenzene (top left, with NaOH at 623 K then acid), benzenediazonium chloride made from aniline (top right, warmed with water), benzenesulphonic acid (bottom left, fused with NaOH then acid) and cumene (bottom right, air then dilute acid, with propanone as a second product). Each step shades one route.

Practice · 2 questions

Practice: preparing alcohols: questions with answers

1
Short answerMediumNCERT Intext Question 7.4
Which Grignard reagent, added to methanal and followed by water, gives each alcohol?
1. $(C H_{3})_{2} C H C H_{2} C H_{2} O H$
  Show solution
  Remove $C H_{2} O H$ : the Grignard reagent is $(C H_{3})_{2} C H C H_{2} M g B r$ , giving 3-methylbutan-1-ol.
2. $C_{6} H_{5} C H_{2} C H_{2} O H$
  Show solution
  Benzylmagnesium chloride, $C_{6} H_{5} C H_{2} M g C l$ , with methanal gives 2-phenylethanol.
2
Short answerMediumNCERT Intext Question 7.5
Write the main organic product of each reaction.
1. $C H_{3} C H_{2} C H = C H_{2}$ with $H_{2} O$ / $H^{+}$
  Show solution
  Butan-2-ol, $C H_{3} C H_{2}$ – $C H (O H) C H_{3}$ , because the –OH goes to the inner carbon.
2. $C H_{3} C O C H_{2}$ – $C H_{2} C O O C H_{3}$ with $N a B H_{4}$
  Show solution
  Only the ketone is reduced, giving methyl 4-hydroxypentanoate, $C H_{3} C H (O H)$ – $C H_{2} C H_{2}$ – $C O O C H_{3}$ . The ester survives.
3. $C H_{3} (C H_{2})_{2}$ – $C H (C H_{3}) C H O$ with $N a B H_{4}$
  Show solution
  The aldehyde becomes a primary alcohol, 2-methylpentan-1-ol: $C H_{3} (C H_{2})_{2}$ – $C H (C H_{3})$ – $C H_{2} O H$ .

7.4.3 Physical Properties

Picture each molecule as a water-like head, the –OH group, on a hydrocarbon tail. The head sets the pattern of physical behaviour; the tail fine-tunes it.

Boiling points. The four butanols make a neat test. All are $C_{4} H_{10} O$ , yet butan-1-ol boils at 391 K, 2-methylpropan-1-ol at 381 K, butan-2-ol at 373 K and the compact, nearly spherical 2-methylpropan-2-ol at 355 K. The more branched the molecule, the less surface it can press against its neighbours, and the weaker the van der Waals (dispersion) attraction. Lengthening the chain does the opposite, so boiling points climb steadily from methanol to ethanol, propanol and beyond.

The larger effect is hydrogen bonding (हाइड्रोजन आबंध). The O–H hydrogen of one molecule is attracted to a lone pair on the oxygen of another, so alcohol molecules cling together in chains and clusters (Fig. 7.8). Boiling has to break these links, which costs extra energy.

1/3The O–H hydrogen of each molecule bonds to the oxygen of the next.

Fig. 7.8Alcohols can both give and accept hydrogen bonds; ethers can only accept them.NCERT p. 203

Step 1 shows three alcohol molecules, R–O–H, arranged in a diagonal chain, with dotted lines joining the H of each molecule to the O of the next: hydrogen bonds between molecules. Step 2 shows three ether molecules, R–O–R prime, scattered with no dotted lines, because an ether has no O–H hydrogen. Step 3 shows one alcohol molecule with a water molecule giving a hydrogen bond to its oxygen and another water molecule accepting a hydrogen bond from its O–H.

Table 7.5Similar masses, very different boiling points

Compound (molar mass)	Boiling point
butane (58 g/mol)	273 K
methoxyethane (60 g/mol)	281 K
chloroethane (64.5 g/mol)	285 K
propan-1-ol (60 g/mol)	370 K

Butane, methoxyethane and propan-1-ol have almost the same molar mass, yet the alcohol boils about 90 K above the ether. Only the alcohol has an O–H hydrogen to share. Chloroethane is polar and a little heavier, but without hydrogen bonds it boils near the ether. Alcohols and phenols boil well above hydrocarbons, ethers and haloalkanes of similar mass for this reason.

Solubility. Water lets a molecule in only if it can join water's network of hydrogen bonds. The –OH head can, both giving and accepting hydrogen bonds; the hydrocarbon tail cannot, and it pushes water molecules aside. The contest between head and tail decides the result. With a short tail the head wins outright: methanol, ethanol and propan-1-ol mix with water in any proportion. As the tail grows the head loses ground: about 7 g of butan-1-ol dissolves in 100 mL of water, about 2 g of pentan-1-ol and only about 0.6 g of hexan-1-ol.

Try this

Where does the alcohol go?

You’ll need

ethanol (rectified spirit)
butan-1-ol
liquid paraffin
three test tubes with stoppers
a 10 mL measuring cylinder

Predict first

Which tubes will show two layers?

butan-1-ol and paraffinethanol onlyall three

Pour 5 mL of water into each of three labelled test tubes.
Add 2 mL of ethanol to the first tube, 2 mL of butan-1-ol to the second and 2 mL of liquid paraffin to the third.
Stopper each tube, shake for 20 seconds and let it stand for a minute.
Look for one clear layer or two layers, and note which layer is on top.

Table 7.6Observations

Liquid added	One layer or two?
Ethanol
Butan-1-ol
Liquid paraffin

Your entries are saved automatically.

Example 7.3NCERT Example 7.3 · p. 204

Ranking boiling points

Arrange each set in order of increasing boiling point and give the reason.

Hexan-1-ol, 2-methylbutan-2-ol, ethanol, pentan-1-ol, pentan-2-ol.
Butanal, propane-1,2-diol, ethoxyethane, butan-1-ol (molar masses 72–76 g/mol).

Try it yourself first. Then reveal the solution one step at a time.

7.4.4 Chemical Reactions

An alcohol can react at either of its two polar bonds. As a nucleophile it uses a lone pair on oxygen to attack an electrophile, and in the end the O–H bond breaks. As an electrophile it first takes a proton on its oxygen; the protonated alcohol then loses water and the C–O bond breaks.

Compare

Which bond breaks?

O–H breaks

the alcohol acts as an acid or a nucleophile
gives alkoxides and esters
phenols react this way too

C–O breaks

the protonated alcohol is the electrophile
gives alkyl halides and alkenes
phenols almost never (only with zinc dust)

\begin{aligned} R - O H + H^{+} \\ \overset{}{⇌} R - O H_{2}^{+} \\ R - O H_{2}^{+} + B r^{-} \\ \overset{}{\to} R - B r + H_{2} O \end{aligned}

(7.10)

(a) Reactions involving cleavage of the O–H bond

1. Acidity: reaction with active metals. Drop a small piece of sodium into ethanol and it fizzes: hydrogen escapes and sodium ethoxide stays in solution. Potassium and aluminium behave the same way, and so does phenol. Aluminium isopropoxide, made this way from propan-2-ol, is a useful laboratory reagent.

\begin{aligned} 2 R - O H + 2 N a \\ \overset{}{\to} 2 R - O^{-} N a^{+} \\ + H_{2} ↑ \end{aligned}

(7.11)

\begin{aligned} 6 (C H_{3})_{2} C H O H \\ + 2 A l \overset{}{\to} \\ 2 [(C H_{3})_{2} C H O]_{3} A l \\ + 3 H_{2} ↑ \end{aligned}

(7.12)

Phenol goes further: it also dissolves in aqueous sodium hydroxide, forming sodium phenoxide and water. Alcohols barely react with aqueous NaOH. Both families are therefore Brønsted acids (proton donors), but phenol is the stronger acid.

\begin{aligned} C_{6} H_{5} O H + N a O H \\ \overset{}{\to} C_{6} H_{5} O^{-} N a^{+} \\ + H_{2} O \end{aligned}

(7.13)

Acidity of alcohols. Why can an alcohol give up a proton at all? Oxygen pulls the shared electrons of the O–H bond towards itself, leaving the hydrogen slightly positive, so a strong enough base can take it. Alkyl groups on the carbon next to oxygen push electron density back towards the oxygen (a +I effect). That makes the O–H bond less polar and the resulting alkoxide ion less stable. In solution, acid strength therefore falls from methanol to primary, secondary and tertiary alcohols.

Put sodium ethoxide in water and it does not survive: it takes a proton from water, giving ethanol and hydroxide ions. Proton transfer runs mainly towards the weaker acid, so water must be the stronger acid of the pair. Most alcohols are indeed slightly weaker acids than water (methanol is about equal), and an alkoxide ion is a stronger base than hydroxide.

\begin{aligned} R - O^{-} + H_{2} O \\ \overset{}{\to} R - O H + O H^{-} \end{aligned}

(7.14)

The same oxygen can also take a proton. In concentrated acid an alcohol exists partly as its protonated form, $R O H_{2}^{+}$ , so alcohols are weak Brønsted bases too, and this protonation starts most reactions in which the C–O bond breaks.

Acidity of phenols. Phenol, with $p K_{a} \approx 10$ , is about a million times stronger an acid than ethanol ( $p K_{a} \approx 16$ ). $p K_{a} = - \log K_{a}$ . A lower $p K_{a}$ means a stronger acid; one unit of $p K_{a}$ is a factor of 10 in $K_{a}$ . Two features favour loss of its proton. First, the oxygen is bonded to an $s p^{2}$ ring carbon, which is more electronegative than an $s p^{3}$ carbon, so the O–H bond is more polar. Second, and more important, the negative charge of the phenoxide ion is spread over the oxygen and three ring carbons, while in an alkoxide ion it stays on the oxygen alone (Fig. 7.9).

\begin{aligned} C_{6} H_{5} O H + H_{2} O \\ \overset{}{⇌} C_{6} H_{5} O^{-} \\ + H_{3} O^{+} \end{aligned}

(7.15)

5/5The real ion is a blend: charge shared by O and three ring carbons.

Fig. 7.9Resonance spreads the phenoxide charge over four atoms; an alkoxide keeps it on one.NCERT p. 206

A phenoxide ion is drawn with oxygen at the top of a benzene ring. Step 1: the negative charge is on oxygen and the ring has three alternating double bonds. Step 2: oxygen forms a C=O double bond and the negative charge appears on the ortho carbon to the right. Step 3: the charge has moved to the para carbon at the bottom. Step 4: it is on the other ortho carbon, to the left. Step 5 shows the hybrid: a ring with a circle, a partial double bond to oxygen and partial negative charges on oxygen and the two ortho and one para carbons, beside ethoxide, whose charge stays on oxygen.

Phenol itself also has resonance structures, but they separate charge, with a positive oxygen and a negative ring carbon, which costs energy. The phenoxide ion's structures all carry one negative charge without separating anything. Resonance therefore stabilises the ion more than the acid, and that pushes equilibrium 7.15 to the right.

SimulationOne charge on one atom, or one charge shared

Two groups of fixed negative charges sit on the stage. On the left a single charge of −1 stands for the oxygen of an alkoxide ion. On the right four smaller charges, −0.55 at the top for the phenoxide oxygen and −0.15 at each of the two ortho carbons and the para carbon, add up to the same −1. Field lines converge on both groups. A positive test charge, standing for a proton, can be dragged anywhere, and an arrow with a readout shows the force on it.

Fig. 7.10Left: an alkoxide, with its whole −1 charge on oxygen. Right: a phenoxide, with the same −1 shared between oxygen (−0.55) and three ring carbons (illustrative shares). The positive test charge plays a proton.

Try this

Drag the proton to sit just above the single left-hand charge, then just above the top right-hand charge at about the same distance. Which pull is stronger?
Move the proton far away from both groups. Do the two pulls still differ? Why do they become similar?
Which ion would hold on to a proton more tightly, and so be the stronger base? What does that say about the strength of the matching acids?

Substituents on the ring. Read Table 7.7 from the top. A methyl group, as in 4-methylphenol (a cresol), nudges the pKa up: electron-releasing groups push charge into a ring that is already carrying the negative charge, so the acid gets slightly weaker. A nitro group pulls the pKa down, by almost 3 units at the para position but by only about 1.6 units at the meta position. The difference comes from resonance. With $- N O_{2}$ at an ortho or para carbon, one resonance structure of the anion places the negative charge on exactly that carbon, and it can spread onto the electronegative oxygens of the nitro group. At the meta position resonance never puts the charge there, so the nitro group helps only through the σ bonds (−I effect).

Table 7.7Acid strength in water at 25 °C (lower pKa = stronger acid)NCERT Table 7.3 · p. 207

Compound	pKa
ethanol	15.9
4-methylphenol	10.3
phenol	10.0
4-chlorophenol	9.4
3-nitrophenol	8.4
4-nitrophenol	7.2
2,4-dinitrophenol	4.1
2,4,6-trinitrophenol (picric acid)	0.4

InteractivePhenol acidity explorer

Buttons choose a substituent on the phenol ring (none, methyl, methoxy, chloro, cyano or nitro) and its position (ortho, meta or para). A horizontal pKa scale from 0 to 18 marks the chosen phenol against ethanol, phenol and carbonic acid, and a line states how many times stronger or weaker the acid is than phenol and whether it would release carbon dioxide from sodium hydrogencarbonate. A drawing of the phenoxide ion shows the substituent in place; the Step button moves the negative charge from oxygen to the ortho, para and other ortho carbons, and when a nitro or cyano group sits ortho or para an extra structure shows the charge on that group. A text line explains the effect of the group.

Fig. 7.11Choose a substituent and a ring position, then step through the resonance structures of the phenoxide ion.

Example 7.4NCERT Example 7.4 · p. 207

Ranking acid strength

Arrange in increasing order of acid strength: 4-nitrophenol, ethanol, 4-chlorophenol, phenol, 2,4-dinitrophenol, 4-methylphenol.

Try it yourself first. Then reveal the solution one step at a time.

2. Esterification. Swap the hydrogen of an –OH group for an acyl group, $R C O^{-}$ , and you have an ester. The acyl group can come from a carboxylic acid, an acid chloride or an acid anhydride, and both alcohols and phenols take part. This is esterification (एस्टरीकरण).

\begin{aligned} R^{'} O H + R C O O H \\ \overset{H^{+}}{⇌} R C O O R^{'} + H_{2} O \end{aligned}

(7.16)

\begin{aligned} R^{'} O H + R C O C l \\ \overset{p y r i d i n e}{\to} R C O O R^{'} \\ + H C l \end{aligned}

(7.17)

The three reagents need different handling. A carboxylic acid reacts slowly and only part of the way, so a few drops of concentrated sulphuric acid speed it up, and the water formed is taken away (or excess alcohol is used) to keep the equilibrium moving forward. An acid chloride reacts readily but releases HCl; a base such as pyridine mops up the HCl and drives the reaction to completion. Phenols react only sluggishly with free carboxylic acids, so they are esterified with acid chlorides or anhydrides.

When the acyl group is acetyl, $C H_{3} C O^{-}$ , the reaction is called acetylation. Aspirin, a medicine for pain, fever and inflammation, is made exactly this way: ethanoic anhydride acetylates the phenolic –OH of salicylic acid (2-hydroxybenzoic acid).

(b) Reactions involving cleavage of the C–O bond in alcohols

These reactions belong to alcohols. In a phenol the C–O bond has partial double-bond character, and a ring carbon cannot be attacked from behind, so phenols do not swap their –OH for a halogen. The one C–O cleavage of phenol you need is reduction by zinc dust ((c) Reactions of phenols).

1. With hydrogen halides. Alcohols give alkyl halides with HCl, HBr or HI, as you saw in Unit 6. Protonation turns –OH into a good leaving group, water. Tertiary alcohols react fastest, because they form stable carbocations.

\begin{aligned} R - O H + H X \\ \overset{}{\to} R - X + H_{2} O \end{aligned}

(7.18)

Because the three classes react with HCl at such different speeds, the reaction doubles as a test, the Lucas test. Lucas reagent is concentrated hydrochloric acid with anhydrous zinc chloride. The trick is solubility: a small alcohol dissolves in the reagent but its chloride does not, so the moment chloride forms, the clear liquid turns milky (turbid). Watch the clock: a tertiary alcohol clouds at once, a secondary alcohol within about five minutes, and a primary alcohol stays clear at room temperature.

Try thisNCERT Lucas test · p. 208

Racing three butanols

You’ll need

Lucas reagent (prepared and dispensed by the teacher)
butan-1-ol, butan-2-ol and 2-methylpropan-2-ol
three dry test tubes with stoppers
a stopwatch

Predict first

Which tube will turn cloudy first?

butan-1-olbutan-2-ol2-methylpropan-2-ol

Your teacher puts 2 mL of Lucas reagent into each of three labelled, dry test tubes.
Add five drops of a different butanol to each tube, stopper, shake once and start the stopwatch.
Hold the tubes against a dark background and note when each turns cloudy. Stop after ten minutes.

Table 7.8Observations

Alcohol	Time
Butan-1-ol
Butan-2-ol
2-Methylpropan-2-ol

Your entries are saved automatically.

2. With phosphorus halides. Phosphorus tribromide converts alcohols to alkyl bromides, and phosphorus trichloride, phosphorus pentachloride or thionyl chloride give alkyl chlorides.

\begin{aligned} 3 R - O H + P B r_{3} \\ \overset{}{\to} 3 R - B r \\ + H_{3} P O_{3} \end{aligned}

(7.19)

3. Dehydration. Take a molecule of water out of an alcohol and an alkene is left. Strong protic acids, concentrated sulphuric or phosphoric acid, do this in the liquid; for alcohol vapour, hot alumina or anhydrous zinc chloride serves as the catalyst. This is dehydration (निर्जलीकरण). The class of alcohol sets how hard you must push. Ethanol needs concentrated sulphuric acid at about 443 K; secondary alcohols lose water under milder conditions; tertiary alcohols need only fairly dilute acid a little above 350 K. The ease of dehydration, tertiary > secondary > primary, follows the stability of the carbocations involved.

\begin{aligned} C H_{3} C H_{2} O H \\ \to_{443 K}^{H_{2} S O_{4}} C H_{2} = C H_{2} \\ + H_{2} O \end{aligned}

(7.20)

The mechanism is acid-catalysed hydration run backwards (Fig. 7.12). First the –OH oxygen picks up a proton from the acid; this is fast and reversible. Next the protonated alcohol lets go of a water molecule, leaving a carbocation. That step is the slowest, so it controls the rate. Finally a base (water or $H S O_{4}^{-}$ ) plucks a proton from a carbon next to the cation, and the C=C forms. Count the protons: one goes in at the start and one comes out at the end, so the acid is not used up; it is a catalyst. Because every step can run backwards, the alkene is distilled away as soon as it forms.

2/4Water leaves; the carbocation forms (slow step).

Fig. 7.12Protonate, lose water (slow), lose a proton: the alkene forms and the acid is returned.NCERT Mechanism · p. 209

A cyclohexane ring carries an OH group at its upper right carbon. Step 1: a curved arrow from a lone pair of the OH oxygen to a proton, labelled fast. Step 2: the oxygen is now OH2 plus, and a curved arrow shows the C–O bond electrons leaving with water, labelled slow and rate-determining. Step 3: the ring carbon is now positive, and a water molecule pulls off a hydrogen from the next carbon while that C–H bond becomes a double bond. Step 4: the product, cyclohexene, with the acid catalyst returned and a note that the alkene is distilled off.

For primary alcohols such as ethanol, textbooks draw the same three steps. Strictly, a primary carbocation is too unstable to exist on its own; with a primary alcohol, steps 2 and 3 happen together as the base removes a β-hydrogen while water leaves. The product is the same.

Example

Which alkene from dehydration?

Predict the main product when 3-methylpentan-3-ol, $C H_{3} C H_{2} C (O H) (C H_{3}) C H_{2} C H_{3}$ , is warmed with acid.

Try it yourself first. Then reveal the solution one step at a time.

4. Oxidation. In oxidation an alcohol loses two hydrogen atoms, the one on oxygen and one on the carbon next to it, and a C=O bond takes their place; chemists therefore also call it dehydrogenation. Whether that is possible, and how far it goes, depends on how many hydrogens the carbon carrying –OH has (Fig. 7.13).

1/4Primary: 2 H; secondary: 1 H; tertiary: none.

Fig. 7.13The number of H atoms on the carbinol carbon decides the product.NCERT p. 209

Three rows show a primary, a secondary and a tertiary alcohol, each with the carbon carrying OH drawn with its four bonds. Step 1 circles the hydrogens on that carbon: two for primary, one for secondary, none for tertiary. Step 2 shows the primary alcohol oxidised to an aldehyde and then to a carboxylic acid, noting that PCC stops at the aldehyde. Step 3 shows the secondary alcohol oxidised to a ketone, which stops there. Step 4 notes that the tertiary alcohol has no hydrogen to remove, so no C=O forms.

Primary alcohols give aldehydes, which are easily oxidised further to carboxylic acids. Acidified potassium permanganate or dichromate, both strong oxidants, go straight to the acid. To stop at the aldehyde, use a milder reagent: pyridinium chlorochromate (PCC, made from chromium trioxide, pyridine and HCl), or $C r O_{3}$ in an anhydrous solvent.
Secondary alcohols give ketones, for example with $C r O_{3}$ or acidified dichromate. A ketone has no hydrogen on its carbonyl carbon, so it resists further oxidation.
Tertiary alcohols have no hydrogen on the carbon carrying –OH, so ordinary oxidising agents leave them alone. Force the issue with hot, concentrated permanganate and the carbon skeleton itself breaks, leaving a mixture of smaller acids.

\begin{aligned} R C H_{2} O H \\ \overset{P C C}{\to} R C H O \\ or \\ \to_{H^{+}}^{K M n O_{4}} R C O O H \end{aligned}

(7.21)

\begin{aligned} R_{2} C H O H \\ \overset{C r O_{3}}{\to} R_{2} C = O \end{aligned}

(7.22)

Over hot copper. Copper at 573 K offers a reagent-free version. Alcohol vapour streams over the hot metal and hydrogen gas comes off, leaving an aldehyde (from a 1° alcohol) or a ketone (from a 2° alcohol). A tertiary alcohol has no hydrogen to give, so over hot copper it loses water instead and becomes an alkene.

\begin{aligned} (C H_{3})_{2} C H O H \to_{573 K}^{C u} \\ C H_{3} C O C H_{3} + H_{2} \\ (C H_{3})_{3} C O H \to_{573 K}^{C u} \\ (C H_{3})_{2} C = C H_{2} \\ + H_{2} O \end{aligned}

(7.23)

(c) Reactions of phenols

The –OH group makes the ring of phenol far more reactive towards electrophiles than benzene is. A lone pair on oxygen is delocalised into the ring, by the same resonance as in the phenoxide ion (Fig. 7.9), so the ortho and para carbons become electron-rich. The –OH group is therefore an activating, ortho/para-directing group, and the reactions below take place at those positions.

1. Nitration. Phenol is so reactive that dilute nitric acid alone, at about 298 K, nitrates it; benzene needs a hot mixture of concentrated nitric and sulphuric acids. The product is a mixture of 2-nitrophenol and 4-nitrophenol, which steam distillation separates: 2-nitrophenol is carried over with the steam and 4-nitrophenol stays behind. In 2-nitrophenol the O–H hydrogen bonds to an oxygen of the nitro group right next to it, inside the same molecule (intramolecular), so the molecules hardly cling to each other or to water. In 4-nitrophenol the two groups are too far apart, so hydrogen bonds join different molecules (intermolecular) and hold them back (Fig. 7.14).

\begin{aligned} C_{6} H_{5} O H \to_{298 K}^{d i l \cdot H N O_{3}} \\ H O C_{6} H_{4} N O_{2} \\ (2- and 4-isomers) \end{aligned}

(7.24)

Fig. 7.14A hydrogen bond inside one molecule (left) leaves nothing to bind the neighbours; between molecules (right) it holds them together.NCERT p. 211

On the left, 2-nitrophenol: the OH group and the nitro group sit on neighbouring ring carbons, and a dotted line joins the OH hydrogen to one nitro oxygen, closing a six-membered ring inside the molecule. On the right, 4-nitrophenol: the OH group and the nitro group are on opposite sides of the ring, and dotted lines join its OH hydrogen to the nitro group of one neighbouring molecule and its nitro oxygen to the OH of another, linking separate molecules.

Pushing on to three nitro groups is harder. Concentrated nitric acid alone oxidises much of the phenol, so the yield of 2,4,6-trinitrophenol (picric acid) is poor. Sulphonating first protects the ring: concentrated sulphuric acid gives phenol-2,4-disulphonic acid, and concentrated nitric acid then replaces both $- S O_{3} H$ groups by nitro groups and adds a third. Picric acid is a strong acid (pKa 0.4), because three nitro groups draw charge out of its anion.

2. Halogenation. Benzene needs a Lewis acid such as $F e B r_{3}$ to polarise bromine; phenol's activated ring does the polarising by itself. How far bromination goes depends on the solvent. In carbon disulphide or chloroform at 273 K, where phenol stays un-ionised, one bromine enters, mostly para to –OH. In water, some phenol ionises to the far more reactive phenoxide ion, and all three ortho and para positions are brominated at once: 2,4,6-tribromophenol separates as a white solid. Bromine colour vanishing as a white precipitate appears is a quick test for phenol.

\begin{aligned} C_{6} H_{5} O H + B r_{2} \\ \to_{273 K}^{C S_{2}} B r C_{6} H_{4} O H \\ + H B r \\ (mostly para) \end{aligned}

(7.25)

\begin{aligned} C_{6} H_{5} O H + 3 B r_{2} \overset{H_{2} O}{\to} \\ C_{6} H_{2} B r_{3} O H ↓ \\ + 3 H B r \end{aligned}

(7.26)

Try this

Two quick tests for a phenol

You’ll need

dilute (about 1 %) phenol solution, handled by the teacher
ethanol
neutral iron(III) chloride solution
bromine water
four test tubes

Predict first

Which liquid will turn violet with neutral iron(III) chloride?

ethanolboththe phenol solutionneither

Your teacher pours 2 mL of dilute phenol solution into two tubes and 2 mL of ethanol into two more.
Add two drops of neutral iron(III) chloride solution to one phenol tube and one ethanol tube. Note the colour.
Add bromine water drop by drop, shaking, to the other two tubes. Note the colour and any solid.

Table 7.9Observations

Tube	What you see
Phenol + iron(III) chloride
Ethanol + iron(III) chloride
Phenol + bromine water
Ethanol + bromine water

Your entries are saved automatically.

Example 7.5NCERT Example 7.5 · p. 212

Where do the new groups go?

Give the major product of each reaction.

Mononitration of 4-methylphenol.
Excess bromine water with 3-methylphenol.
Mononitration of 4-methoxyphenol.

Try it yourself first. Then reveal the solution one step at a time.

3. Kolbe's reaction. Carbon dioxide is far too feeble an electrophile to attack benzene, or even phenol. The phenoxide ion, with a full negative charge feeding the ring, is reactive enough. Heat dry sodium phenoxide with $C O_{2}$ at about 400 K and 4–7 atm, and a carboxylate group joins the ring next to the oxygen; acidify, and 2-hydroxybenzoic acid (salicylic acid), the starting material for aspirin, is obtained. This is the Kolbe reaction.

\begin{aligned} C_{6} H_{5} O N a + C O_{2} \\ \to_{4 - 7 a t m}^{400 K} \\ C_{6} H_{4} (O H) C O O N a \\ \overset{H^{+}}{\to} C_{6} H_{4} (O H) C O O H \end{aligned}

(7.27)

4. Reimer–Tiemann reaction. Chloroform and aqueous alkali supply a different one-carbon electrophile. The base strips HCl from $C H C l_{3}$ to make dichlorocarbene, $: C C l_{2}$ , which attacks the phenoxide ring next to the oxygen. The $- C H C l_{2}$ group that results (making the intermediate a substituted benzal chloride) is hydrolysed by the same alkali to –CHO, and acid work-up gives 2-hydroxybenzaldehyde (salicylaldehyde). This is the Reimer–Tiemann reaction.

\begin{aligned} C_{6} H_{5} O H \to_{N a O H}^{C H C l_{3}} \\ C_{6} H_{4} (O^{-}) C H C l_{2} \\ \overset{O H^{-}}{\to} \overset{H^{+}}{\to} \\ C_{6} H_{4} (O H) C H O \end{aligned}

(7.28)

5. Reduction with zinc dust. Heated with zinc dust, phenol loses its oxygen and gives benzene, while the zinc becomes zinc oxide. This is the one reaction here in which the C–O bond of a phenol breaks.

\begin{aligned} C_{6} H_{5} O H + Z n \\ \overset{h e a t}{\to} C_{6} H_{6} + Z n O \end{aligned}

(7.29)

6. Oxidation. Phenol's ring is easily oxidised. Chromic acid (acidified dichromate) converts it into 1,4-benzoquinone, a yellow compound with two C=O groups in conjugation. Even air manages this slowly: a bottle of phenol left for months turns pink and then brown as quinone-type products build up.

\begin{aligned} C_{6} H_{5} O H \\ \to_{H_{2} S O_{4}}^{N a_{2} C r_{2} O_{7}} O = C_{6} H_{4} = O \end{aligned}

(7.30)

Practice · 4 questions

Practice: reactions of alcohols and phenols: questions with answers

1
Short answerMediumNCERT Intext Question 7.6
Give the main organic product of each alcohol with (i) HCl and $Z n C l_{2}$ , (ii) HBr, (iii) $S O C l_{2}$ .
1. Pentan-1-ol
  Show solution
  (i) 1-chloropentane (slowly, on heating); (ii) 1-bromopentane; (iii) 1-chloropentane, with $S O_{2}$ and HCl as gases.
2. 1-Methylcyclopentan-1-ol
  Show solution
  (i) 1-chloro-1-methylcyclopentane (at once: a tertiary alcohol); (ii) 1-bromo-1-methylcyclopentane; (iii) 1-chloro-1-methylcyclopentane.
2
Short answerMediumNCERT Intext Question 7.7
Predict the major alkene formed when each alcohol is heated with concentrated phosphoric acid.
1. 2-Methylbutan-2-ol
  2-methylbut-2-ene2-methylbut-1-ene3-methylbut-1-ene
  Show answer
  Answer (A) 2-methylbut-2-ene
  The tertiary cation on C2 loses H from C3, giving the trisubstituted alkene $(C H_{3})_{2} C = C H C H_{3}$ .
2. Pentan-2-ol
  pent-1-enepentan-2-onepent-2-ene
  Show answer
  Answer (C) pent-2-ene
  Losing H from C3 gives the disubstituted pent-2-ene, more stable than pent-1-ene. A ketone would need an oxidising agent.
Long answerHardNCERT Intext Question 7.8
Use resonance structures of the anions to explain why 4-nitrophenol (pKa 7.2) is a much stronger acid than 3-nitrophenol (pKa 8.4).
Show solution
In the phenoxide ion the negative charge appears on oxygen and on C2, C4 and C6.
In 4-nitrophenoxide one structure puts the charge on C4, the carbon carrying $- N O_{2}$ . The charge then moves onto the nitro oxygens, giving a structure with C=N and two O atoms sharing the charge. Delocalising onto these electronegative atoms stabilises the anion strongly.
In 3-nitrophenoxide the charge never reaches C3, so the nitro group can only pull electrons through σ bonds (−I). The anion is less stabilised, and the acid is weaker, though still stronger than phenol (pKa 10.0).
4
Short answerMediumNCERT Intext Question 7.9
Write the reactions that convert 4-methylphenol into (a) 2-hydroxy-5-methylbenzoic acid and (b) 2-hydroxy-5-methylbenzaldehyde. Name each reaction.
1. 2-Hydroxy-5-methylbenzoic acid
  Show solution
  Kolbe reaction: NaOH gives sodium 4-methylphenoxide; heat with $C O_{2}$ at about 400 K and 4–7 atm; acidify. –COOH enters ortho to –OH, which is C5 relative to the acid group.
2. 2-Hydroxy-5-methylbenzaldehyde
  Show solution
  Reimer–Tiemann reaction: warm with $C H C l_{3}$ and aqueous NaOH, then acidify. –CHO enters ortho to –OH.

InteractiveReaction bench: alcohols and phenol

Four buttons choose the compound: ethanol (primary), propan-2-ol (secondary), 2-methylpropan-2-ol (tertiary) or phenol. Nine buttons choose the reagent: sodium metal, aqueous sodium hydroxide, Lucas reagent, PCC, acidified dichromate, hot copper at 573 K, concentrated sulphuric acid with heat, bromine water and neutral iron(III) chloride. Pressing Add shows a test tube whose contents change colour, cloud over, bubble or form a precipitate as appropriate, together with the product formula and name and a sentence explaining the result. Reagents that are not used with phenol say so.

Fig. 7.15Pick a compound and a reagent, then press Add to see the product and what you would observe.

7.5 Some Commercially Important Alcohols

Two alcohols are made on a huge scale: methanol and ethanol.

Methanol

The old name wood spirit recalls where methanol, $C H_{3} O H$ , first came from: the liquid given off when wood is heated in a closed vessel (destructive distillation). Modern plants build it from synthesis gas instead, combining carbon monoxide and hydrogen over a catalyst at high pressure and temperature. The original process used a zinc oxide–chromium(III) oxide catalyst at a few hundred atmospheres; newer copper–zinc oxide catalysts work at lower pressure.

\begin{aligned} C O + 2 H_{2} \\ \to_{h i g h P, T}^{Z n O - C r_{2} O_{3}} C H_{3} O H \end{aligned}

(7.31)

Methanol (b.p. 338 K) looks and smells much like ethanol, which is exactly what makes it dangerous: swallowing a few millilitres can blind a person, and a little more can kill (why methanol is a poison). Its biggest single use is as the raw material for methanal (formaldehyde); it also serves as a solvent in paints and varnishes and, increasingly, as a fuel.

Ethanol

Yeast has been making ethanol, $C_{2} H_{5} O H$ , for people for thousands of years by fermentation (किण्वन) of sugars. In India most fermentation ethanol comes from molasses, the dark syrup left after sugar is crystallised from cane juice. Two enzymes do the chemistry: invertase splits sucrose into glucose and fructose, and zymase, from yeast, converts both into ethanol and carbon dioxide (Fig. 7.16).

\begin{aligned} C_{12} H_{22} O_{11} + H_{2} O \\ \overset{i n v e r t a s e}{\to} \\ C_{6} H_{12} O_{6} + C_{6} H_{12} O_{6} \\ (glucose + fructose) \end{aligned}

(7.32)

\begin{aligned} C_{6} H_{12} O_{6} \overset{z y m a s e}{\to} \\ 2 C_{2} H_{5} O H + 2 C O_{2} ↑ \end{aligned}

(7.33)

3/4Zymase turns the sugars into ethanol; $C O_{2}$ bubbles away.

Fig. 7.16Two enzymes, no air: sucrose → glucose and fructose → ethanol and carbon dioxide.NCERT p. 214

A closed vessel with no air sits in the centre. Step 1: it holds sucrose from molasses, and yeast is added. Step 2: the enzyme invertase has split the sucrose into glucose and fructose, C6H12O6. Step 3: the enzyme zymase from yeast turns the sugars into ethanol, and bubbles of carbon dioxide rise and escape from the top. Step 4: at about 14 per cent ethanol the yeast stops working; a side note warns that if air gets in, the ethanol is oxidised to ethanoic acid.

Two limits shape the process. Yeast ferments sugar only without air (anaerobic conditions), so fermenters are sealed and the carbon dioxide escapes through a valve. And yeast cannot tolerate its own product: at roughly 14 % ethanol the zymase stops working, so anything stronger has to be distilled. Let air in and other microbes take over, oxidising ethanol to ethanoic acid; that is how vinegar is made on purpose, and how a carelessly stored drink spoils.

Ethanol meant for industry is deliberately spoiled so that nobody drinks it. This denaturation adds a little copper sulphate, which tints it blue, and pyridine, which makes it smell foul. Besides fermentation, industry also makes ethanol by hydrating ethene (From alkenes). With a boiling point of 351 K and the power to dissolve both polar and non-polar substances, ethanol is a workhorse solvent for paints, medicines and sanitisers, a fuel, and a starting material for many other carbon compounds.

7.6 Ethers

With no O–H bond, ethers are much less reactive than alcohols. That makes them useful solvents: diethyl ether dissolves fats, oils and many organic compounds without reacting with them.

7.6.1 Preparation of Ethers

1. Dehydration of alcohols. Heat ethanol with concentrated sulphuric acid to 443 K and ethene comes off ((b) Reactions involving cleavage of the C–O bond in alcohols). Keep the mixture cooler, near 413 K, with ethanol in excess, and two ethanol molecules join through one oxygen instead: the product is ethoxyethane (diethyl ether).

\begin{aligned} 2 C_{2} H_{5} O H \to_{413 K}^{H_{2} S O_{4}} \\ C_{2} H_{5} O C_{2} H_{5} + H_{2} O \end{aligned}

(7.34)

The ether forms by an SN2 reaction. One ethanol molecule is protonated. A second ethanol molecule, acting as the nucleophile, attacks the $C H_{2}$ carbon of the first from the back and pushes out water. Loss of a proton then gives the ether. In the equations, R stands for the ethyl group, $C_{2} H_{5}$ .

\begin{aligned} R O H + H^{+} \\ \overset{}{⇌} R O H_{2}^{+} \\ R O H + R O H_{2}^{+} \\ \overset{}{\to} R - O^{+} (H) - R \\ + H_{2} O \\ R - O^{+} (H) - R \\ \overset{}{\to} R - O - R + H^{+} \end{aligned}

(7.35)

Three conditions must hold. An SN2 step needs an open, primary carbon to attack. The temperature must stay low, because heat favours elimination. And both halves of the ether should come from the same alcohol. Secondary and tertiary alcohols fail the first test: in acid they form carbocations, which shed a proton to give alkenes much faster than they pick up a second alcohol molecule. Two different alcohols fail the last test: heating ethanol with propan-1-ol in acid gives diethyl, dipropyl and ethyl propyl ethers together, which then have to be separated.

2. Williamson synthesis. In the laboratory, the dependable way to join two groups through an oxygen is to build the C–O bond by SN2: an alkoxide ion, made from an alcohol and sodium, attacks an alkyl halide and pushes out the halide ion. It works for symmetrical and unsymmetrical ethers alike. This is the Williamson synthesis.

\begin{aligned} R - X + R^{'} - O^{-} N a^{+} \\ \overset{}{\to} R - O - R^{'} + N a X \end{aligned}

(7.36)

Every rule for this reaction follows from that SN2 step. The carbon under attack must be open: methyl or primary. An alkoxide is also a strong base, so a crowded secondary halide gives a mixture of ether and alkene, and a tertiary halide gives essentially nothing but alkene. Hence the planning rule: a tertiary or other bulky group must arrive as the alkoxide, never as the halide (Fig. 7.17).

3/3Methoxide pulls off a β-H from the 3° halide: an alkene forms.

Fig. 7.17Put the bulky group on the alkoxide and the small group on the halide.NCERT p. 216

The target ether, tert-butyl methyl ether (CH3)3C–O–CH3, is drawn at the top. Step 1 marks its two C–O bonds, a on the methyl side and b on the tert-butyl side. Step 2 shows route a: sodium tert-butoxide plus bromomethane gives the ether by SN2 at the unhindered methyl carbon, marked with a tick. Step 3 adds route b: sodium methoxide plus 2-bromo-2-methylpropane gives 2-methylpropene instead, because the methoxide acts as a base and removes a β-hydrogen; no ether forms, marked with a cross.

Phenols give alkyl aryl ethers the same way. Sodium hydroxide turns the phenol into sodium phenoxide, which then attacks a primary alkyl halide. The reverse pairing fails: an alkoxide cannot attack the $s p^{2}$ carbon of an ordinary aryl halide from behind, so no ether forms.

\begin{aligned} C_{6} H_{5} O H \overset{N a O H}{\to} \\ C_{6} H_{5} O^{-} N a^{+} \\ \overset{C H_{3} I}{\to} C_{6} H_{5} O C H_{3} \\ + N a I \end{aligned}

(7.37)

Example 7.6NCERT Example 7.6 · p. 216

Choosing the right Williamson pair

A student tries to make 2-methoxy-2-methylbutane, $(C H_{3})_{2} C (O C H_{3}) C H_{2} C H_{3}$ , a petrol additive, by heating sodium methoxide with 2-chloro-2-methylbutane.

What is the main product?
Suggest a better pair of reactants.

Try it yourself first. Then reveal the solution one step at a time.

7.6.2 Physical Properties

An ether molecule is bent, so its two polar C–O bonds leave it with a small dipole moment. Small is the key word. Table 7.10 shows 1-ethoxypropane boiling within a few kelvin of hexane, while pentan-1-ol, of the same mass, boils about 70 K higher. The alcohol's molecules are tied together by hydrogen bonds; the ether, with no O–H hydrogen, cannot form them between its own molecules.

Table 7.10An alkane, an ether and an alcohol of similar mass

Compound (molar mass)	Boiling point
hexane (86 g/mol)	342 K
1-ethoxypropane (88 g/mol)	about 337 K
pentan-1-ol (88 g/mol)	411 K

With water the picture changes. An ether oxygen still has lone pairs, so it can accept a hydrogen bond from a water molecule. Ethers therefore dissolve far better than hydrocarbons of similar size, and an ether is usually about as soluble as an alcohol of similar mass. A striking case is tetrahydrofuran, a five-membered ring containing one oxygen, which mixes with water in any proportion, while cyclopentane, the same ring with $C H_{2}$ in place of the oxygen, hardly dissolves at all.

7.6.3 Chemical Reactions

Apart from burning, ethers do very little at room temperature, which is why chemists trust them as solvents. Two reactions matter here.

1. Cleavage of the C–O bond. Hot concentrated HI or HBr breaks an ether into an alcohol and an alkyl halide; with excess acid the alcohol goes on to a second alkyl halide. The order of reactivity is HI > HBr > HCl, because HI is the strongest acid and iodide the best nucleophile; HCl hardly works.

\begin{aligned} R - O - R + H X \\ \overset{}{\to} R - X + R - O H \\ excess HX: \\ R - O H + H X \\ \overset{}{\to} R - X + H_{2} O \end{aligned}

(7.38)

The mechanism (Fig. 7.18) starts by protonating the ether oxygen, which turns an alkoxy group into a neutral alcohol that can leave. Then the halide ion attacks. If both groups are primary or secondary, iodide attacks by SN2 at the less hindered carbon, so the smaller group becomes the alkyl iodide. If one group is tertiary, its C–O bond breaks first (SN1), because that gives a stable tertiary carbocation; the tertiary group then ends up as the iodide.

2/4 $I^{-}$ attacks the methyl carbon, the less crowded side.

Fig. 7.18Protonate first; then SN2 at the smaller group, or SN1 if one group is tertiary. The aryl–O bond survives.NCERT Mechanism · p. 218

Step 1: 2-methoxypropane and HI give the protonated ether and an iodide ion. Step 2: iodide attacks the methyl carbon, the less crowded side, giving iodomethane and propan-2-ol. Step 3: with tert-butyl methyl ether the protonated ether splits into methanol and a tert-butyl carbocation, which then takes iodide to give 2-iodo-2-methylpropane. Step 4: protonated methoxybenzene gives phenol and iodomethane; the ring–oxygen bond is never broken.

Alkyl aryl ethers always break at the alkyl–oxygen bond and give a phenol and an alkyl halide. The aryl–oxygen bond is stronger, with some double-bond character, and a ring carbon cannot undergo SN2. For the same reason the phenol does not go on to an aryl halide, even with excess HI.

\begin{aligned} C_{6} H_{5} O C H_{3} + H I \\ \overset{}{\to} C_{6} H_{5} O H + C H_{3} I \end{aligned}

(7.39)

Example 7.7NCERT Example 7.7 · p. 219

Which bond does HI break?

Give the main products when each ether is heated with one equivalent of HI.

$C H_{3} C H_{2} O C H_{2} C H (C H_{3})_{2}$
$(C H_{3})_{2} C (O C_{2} H_{5}) C H_{2} C H_{3}$
4-methylmethoxybenzene, $p$ - $C H_{3} C_{6} H_{4} O C H_{3}$

Try it yourself first. Then reveal the solution one step at a time.

2. Electrophilic substitution in aryl alkyl ethers. Like –OH, an alkoxy group gives a lone pair to the ring. It activates the ring and directs incoming groups to the ortho and para positions; the para product usually dominates because the alkoxy group crowds the ortho positions.

Halogenation. Bromine dissolved in ethanoic acid brominates methoxybenzene (anisole) with no Lewis-acid catalyst at all; nearly all of the product is 4-bromoanisole.
Friedel–Crafts reactions. Alkyl halides and acyl chlorides attack too, with anhydrous $A l C l_{3}$ as the catalyst, and again the para product dominates: chloroethane gives mainly 4-ethylanisole, and propanoyl chloride gives mainly 1-(4-methoxyphenyl)propan-1-one.
Nitration. A mixture of concentrated nitric and sulphuric acids gives 2- and 4-nitroanisole, mostly the 4-isomer.

\begin{aligned} C_{6} H_{5} O C H_{3} + B r_{2} \\ \overset{C H_{3} C O O H}{\to} \\ B r C_{6} H_{4} O C H_{3} \\ + H B r \end{aligned}

(7.40)

\begin{aligned} C_{6} H_{5} O C H_{3} \\ + C_{2} H_{5} C O C l \overset{A l C l_{3}}{\to} \\ C_{2} H_{5} C O C_{6} H_{4} O C H_{3} \\ + H C l \end{aligned}

(7.41)

\begin{aligned} C_{6} H_{5} O C H_{3} \to_{H_{2} S O_{4}}^{H N O_{3}} \\ O_{2} N C_{6} H_{4} O C H_{3} \\ + H_{2} O \end{aligned}

(7.42)

Practice · 3 questions

Practice: ethers: questions with answers

Short answerMediumNCERT Intext Question 7.10
Show how 2-methoxy-4-methylpentane, $C H_{3} C H (O C H_{3})$ – $C H_{2} C H (C H_{3})_{2}$ , can be made by a Williamson synthesis starting from methanol and 4-methylpentan-2-ol.
Show solution
Turn methanol into a methyl halide with HBr (or $P B r_{3}$ ): $C H_{3} O H + H B r$ → $C H_{3} B r + H_{2} O$ .
Add sodium metal to 4-methylpentan-2-ol: hydrogen is given off and the sodium alkoxide forms.
SN2 of the alkoxide on bromomethane gives the ether and NaBr. Using sodium methoxide with 2-bromo-4-methylpentane instead would give mostly alkene by elimination.
Quick answerMediumNCERT Intext Question 7.11
Which pair is the better choice for making 1-ethoxy-4-methylbenzene, and why?
4-bromotoluene + $C_{2} H_{5} O N a$ 4-methylphenol + ethanol with acidsodium 4-methylphenoxide + $C_{2} H_{5} B r$
Show answer
Answer (C) sodium 4-methylphenoxide + $C_{2} H_{5} B r$
The phenoxide attacks the primary carbon of bromoethane by SN2. An alkoxide cannot attack the $s p^{2}$ carbon of 4-bromotoluene, and acid dehydration does not make aryl ethers.
3
Short answerMediumNCERT Intext Question 7.12
Predict the products.
1. $C H_{3} O C H_{2} C H_{2} C H_{2} C H_{3}$ + HBr
  Show solution
  Bromomethane + butan-1-ol (SN2 at the methyl carbon).
2. $C_{6} H_{5} O C H_{3}$ + HBr
  Show solution
  Phenol + bromomethane.
3. $C_{6} H_{5} O C_{2} H_{5}$ + $C_{2} H_{5} C l$ , $A l C l_{3}$
  Show solution
  Mainly 1-ethoxy-4-ethylbenzene, with some of the 2-isomer: the ethoxy group directs the new ethyl group para.
4. $(C H_{3})_{3} C O C H_{3}$ + HI
  Show solution
  2-Iodo-2-methylpropane + methanol (SN1 through the tertiary carbocation).

Recap

Alcohols, Phenols and Ethers: key points and summary

Alcohols ( $R - O H$ ), phenols ( $A r - O H$ ) and ethers ( $R - O - R^{'}$ ) are classified by the number of –OH groups, by the carbon that holds the –OH (1°, 2°, 3°, allylic, benzylic, vinylic) and, for ethers, by whether the two groups match (Section 7.1).
IUPAC names: alkane -e → -ol, with the lowest locant to –OH; diols keep the -e; ethers are alkoxy derivatives of the larger group (Section 7.2).
The oxygen is $s p^{3}$ : lone pairs close the angle below 109.5° in alcohols, bulky groups open it to about 112° in ethers, and phenol's C–O bond is short (136 pm) (Section 7.3).
Alcohols come from alkenes (acid hydration, Markovnikov; hydroboration–oxidation, anti-Markovnikov), from reduction of aldehydes, ketones, acids and esters, and from Grignard reagents with methanal (1°), other aldehydes (2°) or ketones (3°) (Section 7.4.1).
Phenol is made from chlorobenzene, benzenesulphonic acid, benzenediazonium salts or, mostly, cumene (Section 7.4.2).
Hydrogen bonding gives alcohols and phenols high boiling points and water solubility; boiling points rise with chain length and fall with branching (Section 7.4.3).
Phenol (pKa 10) is a much stronger acid than alcohols (pKa about 16) because the phenoxide charge is delocalised; ortho/para electron-withdrawing groups strengthen it, alkyl groups weaken it ((a) Reactions involving cleavage of the O–H bond).
Alcohols form esters, give alkyl halides with HX (Lucas test: 3° fast, 1° slow), dehydrate to alkenes (3° > 2° > 1°), and oxidise: 1° → aldehyde → acid, 2° → ketone, 3° resist ((b) Reactions involving cleavage of the C–O bond in alcohols).
The –OH group activates phenol's ring for ortho/para substitution: nitration, bromination, Kolbe (→ salicylic acid) and Reimer–Tiemann (→ salicylaldehyde) ((c) Reactions of phenols).
Methanol comes from CO and $H_{2}$ and is highly toxic; ethanol comes from fermenting sugars and from ethene (Section 7.5).
Ethers are made by acid dehydration of primary alcohols at 413 K or by Williamson synthesis (alkoxide + primary halide) (Section 7.6.1).
Ethers boil like alkanes but dissolve like alcohols; HI cleaves them by SN2 at the smaller group or SN1 at a tertiary one, and alkoxy groups direct ortho/para (Section 7.6.3).

Flash cards · 8

Named reactions and tests

1 / 8

Practice · 36 questions

Practice questions with answers (NCERT “Exercises”)

1
Short answerMediumNCERT Exercises Q7.1
Write the IUPAC name of each compound.
1. $(C H_{3})_{2} C H$ – $C H (O H) C H_{2} C H_{3}$
  Show solution
  2-Methylpentan-3-ol: –OH is on C3 from either end, and the methyl group then takes 2 rather than 4.
2. $H O C H_{2} C H_{2}$ – $C H (O H) C H_{3}$
  Show solution
  Butane-1,3-diol: locants {1,3} beat {2,4}.
3. $(C H_{3})_{3} C C H_{2} C H_{2} O H$
  Show solution
  3,3-Dimethylbutan-1-ol.
4. $o$ - $C_{2} H_{5} C_{6} H_{4} O H$
  Show solution
  2-Ethylphenol.
5. a phenol with $C H_{3}$ groups at both meta positions
  Show solution
  3,5-Dimethylphenol.
6. thymol: –OH at C1, isopropyl at C2, methyl at C5
  Show solution
  5-Methyl-2-(propan-2-yl)phenol, also written 2-isopropyl-5-methylphenol: locants {2,5} beat {3,6}.
7. $C H_{3} O (C H_{2})_{4} C H_{3}$
  Show solution
  1-Methoxypentane.
8. $C_{6} H_{5} O C H_{2} C H_{2} C H_{3}$
  Show solution
  Propoxybenzene.
2
Short answerMediumNCERT Exercises Q7.2
Write the structure of each compound.
1. 3-Methylpentan-3-ol
  Show solution
  $C H_{3} C H_{2}$ – $C (O H) (C H_{3})$ – $C H_{2} C H_{3}$
2. 2-Phenylethan-1-ol
  Show solution
  $C_{6} H_{5} C H_{2} C H_{2} O H$
3. 2-Methylbutane-1,4-diol
  Show solution
  $H O C H_{2} C H (C H_{3})$ – $C H_{2} C H_{2} O H$
4. 2,4-Dichlorophenol
  Show solution
  A phenol ring with Cl on C2 (next to –OH) and on C4 (opposite –OH).
5. 1-Methoxybutane
  Show solution
  $C H_{3} O (C H_{2})_{3} C H_{3}$
6. 3-Ethoxy-2-methylhexane
  Show solution
  $(C H_{3})_{2} C H$ – $C H (O C_{2} H_{5})$ – $C H_{2} C H_{2} C H_{3}$
7. 2-Cyclohexylpropan-2-ol
  Show solution
  $C_{6} H_{11} C (C H_{3})_{2} O H$ : a cyclohexane ring on the C–OH carbon, which also has two methyl groups.
8. Cyclohex-2-en-1-ol
  Show solution
  A cyclohexene ring with –OH on C1 and the double bond between C2 and C3.
Long answerMediumNCERT Exercises Q7.3
Seven compounds have the formula $C_{4} H_{10} O$ . Draw and name all of them, and classify the alcohols as 1°, 2° or 3°.
Show solution
Alcohols: butan-1-ol, $C H_{3} (C H_{2})_{3} O H$ (1°); butan-2-ol, $C H_{3} C H_{2} C H (O H) C H_{3}$ (2°); 2-methylpropan-1-ol, $(C H_{3})_{2} C H C H_{2} O H$ (1°); 2-methylpropan-2-ol, $(C H_{3})_{3} C O H$ (3°).
Ethers: 1-methoxypropane, $C H_{3} O C H_{2} C H_{2} C H_{3}$ ; 2-methoxypropane, $C H_{3} O C H (C H_{3})_{2}$ ; ethoxyethane, $C_{2} H_{5} O C_{2} H_{5}$ .
Short answerEasyNCERT Exercises Q7.4
Butan-1-ol ( $M = 74$ ) boils at 391 K but pentane ( $M = 72$ ) boils at 309 K. Explain.
Show solution
The masses, and so the dispersion forces, are similar. Butan-1-ol molecules are also held together by hydrogen bonds between the O–H hydrogen of one molecule and the oxygen of the next, and these must be broken to boil the liquid. Pentane has only dispersion forces.
Short answerEasyNCERT Exercises Q7.5
Methanol mixes with water in all proportions, but only about 0.6 g of hexan-1-ol dissolves in 100 mL of water. Explain.
Show solution
Both have one –OH group that hydrogen-bonds with water. In methanol it dominates a tiny $C H_{3}$ group. In hexan-1-ol a long $C_{6} H_{13}$ chain cannot hydrogen-bond and would disrupt the hydrogen-bonded network of water, so very little dissolves.
Short answerMediumNCERT Exercises Q7.6
What product does hydroboration–oxidation give from 1-methylcyclohexene, and how does it differ from the product of acid-catalysed hydration?
Show solution
Hydroboration–oxidation: boron adds to C2 (the CH of the C=C) and is replaced by –OH, giving 2-methylcyclohexan-1-ol (anti-Markovnikov).
Acid hydration: $H^{+}$ adds to C2, the tertiary cation forms at C1, and water adds there, giving 1-methylcyclohexan-1-ol (Markovnikov).
Short answerHardNCERT Exercises Q7.7
How many monohydric phenols have the formula $C_{8} H_{10} O$ ? List them.
Show solution
Ethylphenols: 2-, 3- and 4-ethylphenol (3).
Dimethylphenols: 2,3-, 2,4-, 2,5-, 2,6-, 3,4- and 3,5-dimethylphenol (6). Total 9.
Short answerMediumNCERT Exercises Q7.8
2-Hydroxybenzaldehyde distils with steam; 4-hydroxybenzaldehyde does not. Explain.
Show solution
In the 2-isomer the O–H hydrogen forms a hydrogen bond with the neighbouring C=O oxygen inside the same molecule, so molecules are only weakly attracted to each other and to water, and the compound is volatile. In the 4-isomer the groups are far apart, so hydrogen bonds link different molecules and the compound is much less volatile.
Short answerMediumNCERT Exercises Q7.9
Starting from benzene and propene, write the steps that make phenol by the cumene route. Name the second product.
Show solution
Benzene + propene with an acid catalyst ( $H_{3} P O_{4}$ or $A l C l_{3}$ ) gives cumene, $C_{6} H_{5} C H (C H_{3})_{2}$ .
Air oxidises cumene to cumene hydroperoxide, $C_{6} H_{5} C (C H_{3})_{2} O O H$ .
Dilute acid splits the hydroperoxide into phenol and propanone (acetone), the second product.
Short answerEasyNCERT Exercises Q7.10
Write the reactions for making phenol from chlorobenzene, with conditions, and explain why the last step needs an acid.
Show solution
Chlorobenzene with NaOH at about 623 K and high pressure gives sodium phenoxide, $C_{6} H_{5} O^{-} N a^{+}$ .
In alkali, phenol exists as the phenoxide salt because it is stronger than water as an acid. Adding a strong acid such as HCl protonates the phenoxide and releases phenol.
Long answerMediumNCERT Exercises Q7.11
Show, step by step, how propene and water give propan-2-ol in the presence of an acid, and explain why propan-1-ol is not the main product.
Show solution
Step 1: $H_{3} O^{+}$ gives a proton to the $C H_{2}$ end of propene. This forms the secondary carbocation $(C H_{3})_{2} C H^{+}$ and water.
Step 2: a water molecule bonds to the positive carbon, giving the oxonium ion $(C H_{3})_{2} C H O H_{2}^{+}$ .
Step 3: another water molecule removes a proton, giving propan-2-ol and $H_{3} O^{+}$ .
Adding the proton to the middle carbon would give a primary carbocation, which is much less stable, so the –OH ends on C2.
Short answerMediumNCERT Exercises Q7.12
You have toluene, concentrated sulphuric acid, sodium hydroxide and dilute hydrochloric acid. Write the steps for making 4-methylphenol.
Show solution
Sulphonate toluene with concentrated $H_{2} S O_{4}$ to give mainly 4-methylbenzenesulphonic acid.
Fuse it with solid NaOH at high temperature to give sodium 4-methylphenoxide.
Acidify with dilute HCl to release 4-methylphenol.
13
Short answerMediumNCERT Exercises Q7.13
Show how you would make each alcohol.
1. 2-Phenylpropan-2-ol from an alkene
  Show solution
  2-Phenylpropene, $C_{6} H_{5} C (C H_{3}) = C H_{2}$ , with water and dilute acid: $H^{+}$ adds to the $C H_{2}$ end and the benzylic tertiary cation takes water, giving $C_{6} H_{5} C (C H_{3})_{2} O H$ .
2. 2-Cyclohexylethan-1-ol from an alkyl halide by SN2
  Show solution
  (2-Bromoethyl)cyclohexane, $C_{6} H_{11} C H_{2} C H_{2} B r$ , warmed with aqueous NaOH: hydroxide replaces bromide at the primary carbon.
3. 3-Methylbutan-1-ol from an alkyl halide
  Show solution
  1-Bromo-3-methylbutane with aqueous NaOH or KOH.
Short answerMediumNCERT Exercises Q7.14
Describe two experiments that reveal phenol to be an acid, and one that shows it is weaker than carbonic acid. Roughly how many times stronger an acid is phenol than ethanol?
Show solution
Phenol gives hydrogen with sodium: $2 C_{6} H_{5} O H + 2 N a$ → $2 C_{6} H_{5} O N a + H_{2}$ .
It dissolves in aqueous NaOH: $C_{6} H_{5} O H + N a O H$ → $C_{6} H_{5} O N a + H_{2} O$ .
Phenol does not release $C O_{2}$ from sodium hydrogencarbonate solution, so it is weaker than carbonic acid (pKa 6.4).
Ethanol does not react with aqueous NaOH. Phenol (pKa 10) is about $10^{6}$ times stronger, because the phenoxide charge is delocalised over the ring.
Short answerMediumNCERT Exercises Q7.15
Explain why 4-cyanophenol (pKa 7.9) is a stronger acid than 4-methoxyphenol (pKa 10.2).
Show solution
In 4-cyanophenoxide the negative charge reaches C4 and can spread onto the nitrogen of the –C≡N group, which is electronegative; the anion is stabilised and the acid is stronger.
The methoxy group donates a lone pair into the ring (+R), which adds electron density where the phenoxide charge already sits, destabilising the anion. 4-Methoxyphenol is even slightly weaker than phenol.
Short answerMediumNCERT Exercises Q7.16
Bromine water brominates phenol at once with no catalyst, but benzene needs liquid bromine and $F e B r_{3}$ . Explain, and say where the bromine atoms go.
Show solution
A lone pair on the oxygen of –OH is delocalised into the ring, so resonance structures place negative charge on C2, C4 and C6. The ring is far richer in electrons than benzene and polarises $B r_{2}$ without a catalyst.
The electrophile attacks the electron-rich ortho and para carbons, so all three positions (2, 4 and 6) are brominated in water.
17
Short answerMediumNCERT Exercises Q7.17
Write the main organic product of each reaction.
1. Butan-1-ol heated under reflux with excess acidified potassium dichromate
  Show solution
  Butanoic acid, $C H_{3} C H_{2} C H_{2} C O O H$ .
2. 4-Methylphenol with $B r_{2}$ in $C S_{2}$ at 273 K
  Show solution
  2-Bromo-4-methylphenol: the para position is blocked, so bromine goes ortho to –OH.
3. 2-Methylphenol with dilute nitric acid
  Show solution
  2-Methyl-4-nitrophenol and 2-methyl-6-nitrophenol (para and ortho to –OH).
4. Phenol heated with zinc dust
  Show solution
  Benzene (and zinc oxide).
18
Long answerMediumNCERT Exercises Q7.18
Explain each with an example.
1. Kolbe reaction
  Show solution
  Sodium 2-methylphenoxide heated with $C O_{2}$ at about 400 K and 4–7 atm, then acidified, gives 2-hydroxy-3-methylbenzoic acid: –COOH enters the free position ortho to –OH.
2. Reimer–Tiemann reaction
  Show solution
  Phenol warmed with chloroform and aqueous NaOH forms an ortho dichloromethyl intermediate, hydrolysed by the alkali; acidification gives 2-hydroxybenzaldehyde.
3. Williamson synthesis
  Show solution
  Sodium propoxide with bromoethane gives 1-ethoxypropane and NaBr by SN2.
4. Unsymmetrical ether
  Show solution
  An ether with two different groups on oxygen, for example methoxybenzene, $C_{6} H_{5} O C H_{3}$ , or 2-methoxypropane.
Long answerMediumNCERT Exercises Q7.19
Propan-2-ol heated with concentrated sulphuric acid gives propene. Draw the mechanism in three steps and mark the slow step.
Show solution
Step 1 (fast): $(C H_{3})_{2} C H O H + H^{+}$ ⇌ $(C H_{3})_{2} C H O H_{2}^{+}$ .
Step 2 (slow): the protonated alcohol loses water to give the secondary carbocation $(C H_{3})_{2} C H^{+}$ .
Step 3: a base removes a proton from a methyl group, giving $C H_{3} C H = C H_{2}$ and returning $H^{+}$ .
20
Short answerMediumNCERT Exercises Q7.20
How would you carry out each conversion?
1. But-1-ene → butan-1-ol
  Show solution
  Hydroboration–oxidation: $B H_{3}$ (from $B_{2} H_{6}$ ), then $H_{2} O_{2}$ and NaOH.
2. 4-Methylbenzyl chloride → (4-methylphenyl)methanol
  Show solution
  Warm with aqueous NaOH (SN2 at the benzylic carbon).
3. Methylmagnesium bromide → ethanol
  Show solution
  Add methanal, then water or dilute acid: $C H_{3} M g B r + H C H O$ gives $C H_{3} C H_{2} O M g B r$ , hydrolysed to $C H_{3} C H_{2} O H$ .
4. Ethylmagnesium bromide → 3-methylpentan-3-ol
  Show solution
  Add butanone, $C H_{3} C O C H_{2} C H_{3}$ , then water: the ethyl group bonds to the carbonyl carbon, which ends with $C H_{3}$ , two ethyl groups and –OH.
21
Short answerEasyNCERT Exercises Q7.21
Name a reagent for each change.
1. Propan-2-ol → propanone
  Show solution
  $C r O_{3}$ , PCC or acidified dichromate.
2. Ethanol → ethanal (stopping there)
  Show solution
  Pyridinium chlorochromate (PCC).
3. Phenol → 4-bromophenol
  Show solution
  Bromine in $C S_{2}$ (or $C H C l_{3}$ ) at 273 K.
4. Ethanol → ethanoic acid
  Show solution
  Acidified potassium dichromate or permanganate.
5. Butan-2-ol → but-2-ene
  Show solution
  Concentrated sulphuric or phosphoric acid, heated.
6. Propanal → propan-1-ol
  Show solution
  $N a B H_{4}$ , $L i A l H_{4}$ or $H_{2}$ over Ni.
Short answerEasyNCERT Exercises Q7.22
Propan-1-ol and methoxyethane have the same molar mass (60 g/mol), but they boil at 370 K and 281 K. Explain.
Show solution
Propan-1-ol molecules hydrogen-bond to each other through their O–H groups, which takes extra energy to break. Methoxyethane has no O–H hydrogen, so only weak dipole and dispersion forces hold its molecules together.
23
Short answerMediumNCERT Exercises Q7.23
Give the IUPAC name of each ether.
1. $C H_{3} O C H_{2} C H_{2} C H_{2} C l$
  Show solution
  1-Chloro-3-methoxypropane (prefixes in alphabetical order).
2. $p$ - $C H_{3} C_{6} H_{4} O C H_{3}$
  Show solution
  1-Methoxy-4-methylbenzene.
3. $(C H_{3} C H_{2} C H_{2})_{2} O$
  Show solution
  1-Propoxypropane.
4. $C_{6} H_{5} O C H_{2} C_{6} H_{5}$
  Show solution
  (Benzyloxy)benzene; common name benzyl phenyl ether.
5. $C_{2} H_{5} O C_{5} H_{9}$ (a cyclopentane ring)
  Show solution
  Ethoxycyclopentane.
6. $(C H_{3})_{3} C O C_{2} H_{5}$
  Show solution
  2-Ethoxy-2-methylpropane.
24
Short answerMediumNCERT Exercises Q7.24
Give the reagents for a Williamson synthesis of each ether.
1. 1-Butoxybutane
  Show solution
  Sodium butoxide + 1-bromobutane.
2. Methoxybenzene
  Show solution
  Sodium phenoxide + iodomethane (not bromobenzene + methoxide).
3. 2-Ethoxy-2-methylbutane
  Show solution
  Sodium 2-methylbutan-2-oxide + bromoethane.
4. 1-Methoxypropane
  Show solution
  Sodium methoxide + 1-bromopropane, or sodium propoxide + bromomethane: both halides are unhindered.
Short answerHardNCERT Exercises Q7.25
A student tried to make di-tert-butyl ether and diphenyl ether by Williamson synthesis. Explain why both attempts fail.
Show solution
Di-tert-butyl ether needs a tertiary halide whichever way it is split. tert-Butoxide is a strong base, so it removes a β-hydrogen from 2-bromo-2-methylpropane and gives 2-methylpropene.
Diphenyl ether needs phenoxide to attack an aryl halide, but an $s p^{2}$ ring carbon cannot be attacked from behind by SN2, so no ether forms under Williamson conditions.
Long answerMediumNCERT Exercises Q7.26
Dibutyl ether (1-butoxybutane) can be made from butan-1-ol alone. Give the conditions and show the mechanism.
Show solution
Heat butan-1-ol with concentrated $H_{2} S O_{4}$ at a moderate temperature, with the alcohol in excess.
Step 1: $C_{4} H_{9} O H + H^{+}$ ⇌ $C_{4} H_{9} O H_{2}^{+}$ .
Step 2 (SN2): a second molecule of butan-1-ol attacks the $C H_{2}$ next to $O H_{2}^{+}$ , displacing water to give $C_{4} H_{9} O^{+} (H) C_{4} H_{9}$ .
Step 3: loss of $H^{+}$ gives $C_{4} H_{9} O C_{4} H_{9}$ .
Short answerMediumNCERT Exercises Q7.27
Heating 2-methylpropan-2-ol with concentrated sulphuric acid gives 2-methylpropene, not di-tert-butyl ether. Why?
Show solution
The protonated tertiary alcohol loses water at once to a stable tertiary carbocation. The cation loses a β-hydrogen faster than a bulky second alcohol molecule can reach it, so elimination wins. The SN2 ether route needs an unhindered primary carbon.
28
Short answerMediumNCERT Exercises Q7.28
Write the products of each ether with hot concentrated HI.
1. 1-Butoxybutane with excess HI
  Show solution
  Two molecules of 1-iodobutane (and water).
2. Ethoxybenzene
  Show solution
  Phenol + iodoethane.
3. 2-Methoxy-2-methylpropane
  Show solution
  2-Iodo-2-methylpropane + methanol.
Short answerMediumNCERT Exercises Q7.29
Explain why nitration of ethoxybenzene is faster than nitration of benzene and gives mainly ortho and para products.
Show solution
A lone pair on the ether oxygen is delocalised into the ring, placing extra electron density on C2, C4 and C6. The ring attracts the electrophile $N O_{2}^{+}$ faster than benzene does, and attack at those carbons gives intermediates in which oxygen can share the positive charge. The para product dominates because the ethoxy group crowds the ortho positions.
Long answerMediumNCERT Exercises Q7.30
Methoxyethane is heated with one equivalent of HI. Show each step of the mechanism and decide which alkyl iodide forms.
Show solution
Step 1: $C H_{3} O C H_{2} C H_{3} + H I$ ⇌ $C H_{3} O^{+} (H)$ – $C H_{2} C H_{3}$ + $I^{-}$ .
Step 2 (SN2): $I^{-}$ attacks the methyl carbon, which is less hindered than the ethyl $C H_{2}$ , giving $C H_{3} I$ and ethanol.
With excess HI, ethanol is converted further to iodoethane.
31
Short answerMediumNCERT Exercises Q7.31
Write the main product of each reaction of ethoxybenzene.
1. Chloromethane, anhydrous $A l C l_{3}$
  Show solution
  1-Ethoxy-4-methylbenzene (with some 2-isomer).
2. Concentrated nitric and sulphuric acids
  Show solution
  1-Ethoxy-4-nitrobenzene (with some 2-isomer).
3. Bromine in ethanoic acid
  Show solution
  1-Bromo-4-ethoxybenzene.
4. Ethanoyl chloride, anhydrous $A l C l_{3}$
  Show solution
  1-(4-Ethoxyphenyl)ethan-1-one.
32
Short answerHardNCERT Exercises Q7.32
Which alkene, and which method, would you use to make each alcohol?
1. 2-Methylpentan-2-ol
  Show solution
  2-Methylpent-1-ene (or 2-methylpent-2-ene) with water and dilute acid.
2. 1-Ethylcyclopentan-1-ol
  Show solution
  1-Ethylcyclopentene with water and dilute acid.
3. 2-Methylpropan-1-ol
  Show solution
  2-Methylpropene by hydroboration–oxidation.
4. Cyclohexylmethanol
  Show solution
  Methylenecyclohexane, $C_{6} H_{10} = C H_{2}$ , by hydroboration–oxidation.
Long answerHardNCERT Exercises Q7.33
2-Methylpentan-3-ol reacts with HBr, but the product is 2-bromo-2-methylpentane: the bromine ends up on C2, not C3. Explain with a mechanism.
Show solution
Step 1: the –OH on C3 takes a proton and becomes $- O H_{2}^{+}$ , a good leaving group.
Step 2: water leaves, giving a secondary carbocation on C3.
Step 3: a hydrogen on C2 moves, with its bonding pair, to C3 (1,2-hydride shift). The positive charge is now on C2, which carries two methyl groups: a more stable tertiary carbocation.
Step 4: $B r^{-}$ bonds to C2, giving $(C H_{3})_{2} C B r C H_{2} C H_{2} C H_{3}$ , 2-bromo-2-methylpentane.
Quick answerEasy
Which reagent distinguishes phenol from cyclohexanol?
sodium metalneutral iron(III) chlorideconcentrated HClwater
Show answer
Answer (B) neutral iron(III) chloride
Both compounds give hydrogen with sodium; only the phenol gives a violet colour with neutral iron(III) chloride.
35
Quick answerEasy
Arrange in increasing ease of dehydration with acid.
1. 2-methylpropan-2-ol
2. butan-1-ol
3. butan-2-ol
Show answer
Answer
butan-1-ol
butan-2-ol
2-methylpropan-2-ol
The easier the carbocation forms (3° > 2° > 1°), the easier the dehydration.
Quick answerMedium
Picric acid, 2,4,6-trinitrophenol, releases $C O_{2}$ from sodium hydrogencarbonate solution; phenol does not.
TrueFalse
Show answer
Answer True
Picric acid (pKa 0.4) is far stronger than carbonic acid (pKa 6.4); phenol (pKa 10) is weaker.

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Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbook	Practice question here
NCERT Intext Question 7.1	Practice: classifying alcohols · Q1
NCERT Intext Question 7.2	Practice: classifying alcohols · Q2
NCERT Intext Question 7.3	Practice: naming · Q1
NCERT Intext Question 7.4	Practice: preparing alcohols · Q1
NCERT Intext Question 7.5	Practice: preparing alcohols · Q2
NCERT Intext Question 7.6	Practice: reactions of alcohols and phenols · Q1
NCERT Intext Question 7.7	Practice: reactions of alcohols and phenols · Q2
NCERT Intext Question 7.8	Practice: reactions of alcohols and phenols · Q3
NCERT Intext Question 7.9	Practice: reactions of alcohols and phenols · Q4
NCERT Intext Question 7.10	Practice: ethers · Q1
NCERT Intext Question 7.11	Practice: ethers · Q2
NCERT Intext Question 7.12	Practice: ethers · Q3
NCERT Exercises Q7.1	Exercises · Q1
NCERT Exercises Q7.2	Exercises · Q2
NCERT Exercises Q7.3	Exercises · Q3
NCERT Exercises Q7.4	Exercises · Q4
NCERT Exercises Q7.5	Exercises · Q5
NCERT Exercises Q7.6	Exercises · Q6
NCERT Exercises Q7.7	Exercises · Q7
NCERT Exercises Q7.8	Exercises · Q8
NCERT Exercises Q7.9	Exercises · Q9
NCERT Exercises Q7.10	Exercises · Q10
NCERT Exercises Q7.11	Exercises · Q11
NCERT Exercises Q7.12	Exercises · Q12
NCERT Exercises Q7.13	Exercises · Q13
NCERT Exercises Q7.14	Exercises · Q14
NCERT Exercises Q7.15	Exercises · Q15
NCERT Exercises Q7.16	Exercises · Q16
NCERT Exercises Q7.17	Exercises · Q17
NCERT Exercises Q7.18	Exercises · Q18
NCERT Exercises Q7.19	Exercises · Q19
NCERT Exercises Q7.20	Exercises · Q20
NCERT Exercises Q7.21	Exercises · Q21
NCERT Exercises Q7.22	Exercises · Q22
NCERT Exercises Q7.23	Exercises · Q23
NCERT Exercises Q7.24	Exercises · Q24
NCERT Exercises Q7.25	Exercises · Q25
NCERT Exercises Q7.26	Exercises · Q26
NCERT Exercises Q7.27	Exercises · Q27
NCERT Exercises Q7.28	Exercises · Q28
NCERT Exercises Q7.29	Exercises · Q29
NCERT Exercises Q7.30	Exercises · Q30
NCERT Exercises Q7.31	Exercises · Q31
NCERT Exercises Q7.32	Exercises · Q32
NCERT Exercises Q7.33	Exercises · Q33

Quick check: MCQs on Alcohols, Phenols and Ethers

7.1.3 Ethers3 questions
Naming ethers4 questions
7.3 Structures of Functional Groups3 questions
7.4.2 Preparation of Phenols4 questions
7.4.3 Physical Properties4 questions
(a) Reactions involving cleavage of the O–H bond5 questions
(b) Reactions involving cleavage of the C–O bond in alcohols4 questions
(c) Reactions of phenols4 questions
Ethanol4 questions
7.6.3 Chemical Reactions4 questions

Glossary · 15

Key terms (English–Hindi)

alcoholऐल्कोहॉलaelkohol: A compound with an –OH group bonded to an $s p^{3}$ carbon, $R - O H$ .First seen in the chapter →

dehydrationनिर्जलीकरणnirjaleekaran: Removal of a water molecule, for example from an alcohol to give an alkene or an ether.First seen in the chapter →
denaturation of alcohol: Making ethanol unfit to drink by adding substances such as copper sulphate and pyridine.First seen in the chapter →

esterificationएस्टरीकरणesteereekaran: Forming an ester from an alcohol or phenol and a carboxylic acid, acid chloride or anhydride.First seen in the chapter →
etherईथरeethar: A compound with an oxygen atom bonded to two carbon groups, $R - O - R^{'}$ .First seen in the chapter →

fermentationकिण्वनkinvan: Breakdown of sugars by enzymes from microbes such as yeast, without air, giving ethanol and carbon dioxide.First seen in the chapter →

hydroboration–oxidation: Adding $B H_{3}$ to an alkene, then $H_{2} O_{2}$ and NaOH, to give an alcohol with –OH on the less substituted carbon.First seen in the chapter →
hydrogen bondहाइड्रोजन आबंधhaaidrojan aabandh: An attraction between a hydrogen bonded to O, N or F and a lone pair on another such atom.First seen in the chapter →

Kolbe reaction: Sodium phenoxide with $C O_{2}$ under pressure, then acid, giving 2-hydroxybenzoic (salicylic) acid.First seen in the chapter →

Lucas test: Conc. HCl with $Z n C l_{2}$ : a 3° alcohol turns it cloudy at once, a 2° alcohol within minutes, a 1° alcohol not at room temperature.First seen in the chapter →

phenolफ़ीनॉलpheenol: A compound with an –OH group bonded directly to a carbon of an aromatic ring, $A r - O H$ ; also the simplest one, $C_{6} H_{5} O H$ .First seen in the chapter →
phenoxide ion: The anion $C_{6} H_{5} O^{-}$ left when phenol loses a proton; its charge is spread over oxygen and three ring carbons.First seen in the chapter →

Reimer–Tiemann reaction: Phenol with $C H C l_{3}$ and aqueous NaOH, then acid, giving 2-hydroxybenzaldehyde.First seen in the chapter →

symmetrical ether: An ether whose two groups on oxygen are identical; if they differ, the ether is unsymmetrical (mixed).First seen in the chapter →

Williamson synthesis: Making an ether by SN2 reaction of a sodium alkoxide or phenoxide with a primary alkyl halide.First seen in the chapter →

Alcohols, Phenols and Ethers: Class 12 Chemistry · Chapter 7

What you will learn in Alcohols, Phenols and Ethers

7.1 Classification

7.1.1 Alcohols — Mono, Di, Tri or Polyhydric alcohols

7.1.2 Phenols — Mono, Di and trihydric phenols

7.1.3 Ethers

Practice: classifying alcohols: questions with answers

7.2 Nomenclature

Naming alcohols

Naming phenols

Naming ethers

Practice: naming: questions with answers

7.3 Structures of Functional Groups

7.4 Alcohols and Phenols

7.4.1 Preparation of Alcohols

From alkenes

From carbonyl compounds

From Grignard reagents

7.4.2 Preparation of Phenols

Practice: preparing alcohols: questions with answers

7.4.3 Physical Properties

7.4.4 Chemical Reactions

(a) Reactions involving cleavage of the O–H bond

(b) Reactions involving cleavage of the C–O bond in alcohols

(c) Reactions of phenols

Practice: reactions of alcohols and phenols: questions with answers

7.5 Some Commercially Important Alcohols

Methanol

Ethanol

7.6 Ethers

7.6.1 Preparation of Ethers

7.6.2 Physical Properties

7.6.3 Chemical Reactions

Practice: ethers: questions with answers

Alcohols, Phenols and Ethers: key points and summary

Practice questions with answers (NCERT “Exercises”)

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Quick check: MCQs on Alcohols, Phenols and Ethers

Key terms (English–Hindi)

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