Classification

Organic compounds containing hydroxyl (–OH) groups attached to carbon atoms are fundamental. When the –OH group replaces a hydrogen atom in an aliphatic hydrocarbon, the resulting compounds are called alcohols (e.g., methanol, $\textsf{CH}_3\text{OH}$). If the –OH group replaces a hydrogen atom in an aromatic hydrocarbon, they are called phenols (e.g., phenol, $\textsf{C}_6\text{H}_5\text{OH}$).

Replacing a hydrogen atom of the hydroxyl group in an alcohol or phenol by an alkyl (R–) or aryl (Ar–) group leads to another class of compounds known as ethers (e.g., dimethyl ether, $\textsf{CH}_3\text{OCH}_3$). Ethers can also be seen as derivatives of hydrocarbons where a hydrogen is substituted by an alkoxy (R–O–) or aryloxy (Ar–O–) group.

Alcohols— Mono, Di, Tri Or Polyhydric Alcohols

Alcohols are classified based on the number of hydroxyl groups present in their structure:

• Monohydric Alcohols: Contain one –OH group.
• Dihydric Alcohols: Contain two –OH groups (e.g., ethane-1,2-diol or ethylene glycol, $\textsf{HO–CH}_2\text{–CH}_2\text{–OH}$).
• Trihydric Alcohols: Contain three –OH groups (e.g., propane-1,2,3-triol or glycerol, $\textsf{HO–CH}_2\text{–CH(OH)–CH}_2\text{–OH}$).
• Polyhydric Alcohols: Contain more than three –OH groups.

Monohydric alcohols can be further classified based on the hybridisation of the carbon atom to which the –OH group is directly attached:

(i) Compounds containing $\textsf{C(sp}^3\text{)–OH}$ bond: The hydroxyl group is attached to an $\textsf{sp}^3$ hybridised carbon atom of an alkyl group.

• Primary ($1^\circ$) Alcohols: The –OH group is attached to a primary carbon atom (a carbon bonded to only one other carbon atom).
• Secondary ($2^\circ$) Alcohols: The –OH group is attached to a secondary carbon atom (a carbon bonded to two other carbon atoms).
• Tertiary ($3^\circ$) Alcohols: The –OH group is attached to a tertiary carbon atom (a carbon bonded to three other carbon atoms).
• Allylic Alcohols: The –OH group is attached to an $\textsf{sp}^3$ hybridised carbon atom that is adjacent to a carbon-carbon double bond (an allylic carbon).
  Allylic alcohols can be primary, secondary, or tertiary depending on the substitution on the allylic carbon.
• Benzylic Alcohols: The –OH group is attached to an $\textsf{sp}^3$ hybridised carbon atom that is next to an aromatic ring.
  Benzylic alcohols can also be primary, secondary, or tertiary.

(ii) Compounds containing $\textsf{C(sp}^2\text{)–OH}$ bond: These alcohols have the –OH group directly bonded to an $\textsf{sp}^2$ hybridised carbon atom.

• Vinylic Alcohols: The –OH group is attached to a vinylic carbon (one of the carbons of a carbon-carbon double bond, e.g., $\textsf{CH}_2 \text{= CH – OH}$).
• Phenols: The –OH group is attached to an aryl carbon (one of the carbons of an aromatic ring).

Phenols— Mono, Di And Trihydric Phenols

Phenols are classified based on the number of hydroxyl groups directly attached to the aromatic ring:

• Monohydric Phenols: Contain one –OH group attached to the aromatic ring (e.g., phenol, $\textsf{C}_6\text{H}_5\text{OH}$).
• Dihydric Phenols: Contain two –OH groups attached to the aromatic ring. They can be *ortho* (1,2-), *meta* (1,3-), or *para* (1,4-) substituted.
• Trihydric Phenols: Contain three –OH groups attached to the aromatic ring.

Ethers

Ethers are classified based on the nature of the alkyl or aryl groups attached to the oxygen atom:

• Simple or Symmetrical Ethers: The two groups attached to the oxygen atom are identical (e.g., diethyl ether, $\textsf{C}_2\text{H}_5\text{OC}_2\text{H}_5$).
• Mixed or Unsymmetrical Ethers: The two groups attached to the oxygen atom are different (e.g., ethyl methyl ether, $\textsf{C}_2\text{H}_5\text{OCH}_3$).

Nomenclature

Naming alcohols, phenols, and ethers follows both common and IUPAC systems.

Alcohols

Common Names: Derived from the common name of the alkyl group followed by the word "alcohol" (e.g., $\textsf{CH}_3\text{OH}$ is methyl alcohol).

IUPAC Names: Derived from the name of the parent alkane by replacing the final 'e' with 'ol'. The longest carbon chain containing the –OH group is the parent chain. Numbering starts from the end nearest to the –OH group. The position of the –OH group and other substituents are indicated by numbers. For polyhydric alcohols, the 'e' of the alkane is kept, and 'diol', 'triol', etc., is added with positional locants (e.g., $\textsf{HO–CH}_2\text{–CH}_2\text{–OH}$ is ethane-1,2-diol). Cyclic alcohols use the prefix 'cyclo-'.

Examples of alcohol nomenclature:

Phenols

The simplest hydroxy derivative of benzene ($\textsf{C}_6\text{H}_5\text{OH}$) is called phenol, which is both its common and accepted IUPAC name. For substituted phenols, common names often use *ortho* (1,2-), *meta* (1,3-), and *para* (1,4-) prefixes. IUPAC names treat phenol as the parent structure, with substituents numbered relative to the carbon bearing the –OH group (C-1).

Examples of phenol nomenclature:

Ethers

Common Names: Derived by naming the alkyl/aryl groups attached to the oxygen in alphabetical order, followed by the word "ether" (e.g., $\textsf{CH}_3\text{OC}_2\text{H}_5$ is ethyl methyl ether). If the groups are the same, 'di-' is used (e.g., $\textsf{C}_2\text{H}_5\text{OC}_2\text{H}_5$ is diethyl ether).

IUPAC Names: Treated as alkoxy or aryloxy derivatives of the larger hydrocarbon chain. The smaller group becomes part of the alkoxy/aryloxy substituent (e.g., $\textsf{CH}_3\text{OCH}_3$ is methoxymethane, $\textsf{C}_6\text{H}_5\text{OCH}_3$ is methoxybenzene or anisole).

Examples of ether nomenclature:

Compound	Common name	IUPAC name
$\textsf{CH}_3\text{OCH}_3$	Dimethyl ether	Methoxymethane
$\textsf{C}_2\text{H}_5\text{OC}_2\text{H}_5$	Diethyl ether	Ethoxyethane
$\textsf{CH}_3\text{OCH}_2\text{CH}_2\text{CH}_3$	Methyl n-propyl ether	1-Methoxypropane
$\textsf{C}_6\text{H}_5\text{OCH}_3$	Methyl phenyl ether	Methoxybenzene (Anisole)
$\textsf{C}_6\text{H}_5\text{OCH}_2\text{CH}_3$	Ethyl phenyl ether	Ethoxybenzene (Phenetole)
$\textsf{C}_6\text{H}_5\text{O(CH}_2)_6 \text{ – CH}_3$	Heptyl phenyl ether	1-Phenoxyheptane
$\textsf{(CH}_3)_2\text{CH O CH}_3$	Methyl isopropyl ether	2-Methoxypropane
$\textsf{CH}_3\text{– O – CH}_2\text{ – CH}_2\text{ – OCH}_3$	—	1,2-Dimethoxyethane
	—	2-Ethoxy-1,1-dimethylcyclohexane

Example 11.1. Give IUPAC names of the following compounds:

(i)

(ii)

(iii)

(iv)

Answer:

Structures Of Functional Groups

The structure around the oxygen atom in alcohols, phenols, and ethers can be understood in terms of hybridisation and electron pair repulsion.

In alcohols ($\textsf{R–OH}$), the oxygen atom is $\textsf{sp}^3$ hybridised. Two $\textsf{sp}^3$ hybrid orbitals form sigma bonds with the carbon atom (of the R group) and the hydrogen atom. The other two $\textsf{sp}^3$ hybrid orbitals contain lone pairs of electrons. The C–O–H bond angle is typically slightly less than the ideal tetrahedral angle of 109.5°, around 108.9° in methanol. This is due to the greater repulsion between the two lone pairs on oxygen compared to the repulsion between bond pairs.

In phenols ($\textsf{Ar–OH}$), the oxygen atom is attached to an $\textsf{sp}^2$ hybridised carbon atom of the aromatic ring. The C–O bond length in phenol (around 136 pm) is slightly shorter than in methanol (around 142 pm). This shorter length is attributed to two factors: (i) the partial double bond character of the C–O bond due to the conjugation of the oxygen's lone pair electrons with the aromatic ring's $\pi$ system (resonance), and (ii) the $\textsf{sp}^2$ hybridisation of the carbon atom, which has more s character and forms a shorter bond compared to an $\textsf{sp}^3$ carbon.

In ethers ($\textsf{R–O–R'}$), the oxygen atom is also $\textsf{sp}^3$ hybridised, bonded to two alkyl/aryl groups. The four electron pairs (two bond pairs and two lone pairs) around oxygen are arranged approximately tetrahedrally. However, the bond angle (C–O–C) is typically slightly larger than the tetrahedral angle, for example, around 111.7° in methoxymethane. This is due to the repulsive interaction between the two bulky alkyl/aryl groups attached to the oxygen. The C–O bond length in ethers (around 141 pm) is similar to that in alcohols.

Alcohols And Phenols

Alcohols and phenols contain the hydroxyl group and exhibit characteristic reactions primarily due to this functional group, though the attached alkyl or aryl group influences reactivity.

Preparation Of Alcohols

Alcohols can be prepared by several methods:

1. From Alkenes:

• (i) By Acid Catalysed Hydration: Alkenes react with water in the presence of an acid catalyst ($\textsf{H}_2\text{SO}_4$ or $\textsf{H}_3\text{PO}_4$) to form alcohols. The addition follows Markovnikov's rule, where the hydrogen atom of water adds to the carbon with more hydrogen atoms, and the –OH group adds to the carbon with fewer hydrogen atoms.

  Mechanism involves:

  Step 1: Protonation of the alkene by $\textsf{H}_3\text{O}^+$ to form a carbocation (electrophilic attack).

  Step 2: Nucleophilic attack by water on the carbocation.

  Step 3: Deprotonation of the protonated alcohol by water to form the alcohol.

• (ii) By Hydroboration–Oxidation: Diborane ($\textsf{(BH}_3)_2$) reacts with alkenes forming trialkyl boranes. These are then oxidised by hydrogen peroxide ($\textsf{H}_2\text{O}_2$) in the presence of aqueous sodium hydroxide ($\textsf{NaOH}$). Borane adds such that boron attaches to the less substituted carbon of the double bond. Subsequent oxidation replaces the boron-carbon bond with a hydroxyl group, resulting in overall addition of water in an anti-Markovnikov fashion. This method provides alcohols in high yield.

2. From Carbonyl Compounds:

• (i) By Reduction of Aldehydes and Ketones: Aldehydes and ketones are reduced to primary and secondary alcohols, respectively. This can be done by catalytic hydrogenation (using $\textsf{H}_2$ with catalysts like Pt, Pd, or Ni) or by using reducing agents like sodium borohydride ($\textsf{NaBH}_4$) or lithium aluminium hydride ($\textsf{LiAlH}_4$).

  Aldehydes yield primary alcohols:

  Ketones yield secondary alcohols:

• (ii) By Reduction of Carboxylic Acids and Esters: Carboxylic acids can be reduced to primary alcohols using a strong reducing agent like lithium aluminium hydride ($\textsf{LiAlH}_4$).

  However, $\textsf{LiAlH}_4$ is expensive. Commercially, carboxylic acids are first converted to esters, which are then reduced to primary alcohols by catalytic hydrogenation.

3. From Grignard Reagents:

Grignard reagents ($\textsf{RMgX}$) react with aldehydes and ketones through nucleophilic addition to the carbonyl carbon. The initial adduct is then hydrolysed to yield alcohols.

Step 1: Nucleophilic addition to carbonyl:

Step 2: Hydrolysis of the adduct:

Specific reactions with different carbonyl compounds:

• Reaction with methanal (formaldehyde, $\textsf{HCHO}$) produces a primary alcohol.
• Reaction with other aldehydes produces a secondary alcohol.
• Reaction with ketones produces a tertiary alcohol.

Answer:

Preparation Of Phenols

Phenols are mainly produced synthetically. Laboratory methods include:

• 1. From Haloarenes: Chlorobenzene is heated with aqueous sodium hydroxide ($\textsf{NaOH}$) at high temperature (623 K) and pressure (320 atm). This produces sodium phenoxide, which upon acidification (with dilute acid) yields phenol. This is known as Dow's process.
• 2. From Benzenesulphonic Acid: Benzene is first sulphonated with oleum ($\textsf{H}_2\text{S}_2\text{O}_7$) to form benzenesulphonic acid. This is then heated with molten sodium hydroxide, converting it to sodium phenoxide. Subsequent acidification gives phenol.
• 3. From Diazonium Salts: Aromatic primary amines (like aniline) are treated with nitrous acid ($\textsf{NaNO}_2$ + $\textsf{HCl}$) at low temperature (273-278 K) to form diazonium salts (e.g., benzenediazonium chloride). Diazonium salts are hydrolysed to phenols by warming with water or dilute acid.
• 4. From Cumene: Most commercial phenol is produced from cumene (isopropylbenzene). Cumene is oxidised by air to form cumene hydroperoxide, which is then treated with dilute acid to give phenol and acetone. Acetone is a valuable co-product.

Physical Properties

The physical properties of alcohols and phenols are significantly influenced by the presence of the polar hydroxyl (–OH) group, which can form hydrogen bonds.

Boiling Points:

• Boiling points of alcohols and phenols increase with increasing number of carbon atoms due to increased van der Waals forces.
• In isomeric alcohols, boiling points decrease with increasing branching because the surface area decreases, leading to weaker van der Waals forces.
• Alcohols and phenols have significantly higher boiling points than hydrocarbons, ethers, or haloalkanes of comparable molecular mass. This is because the –OH group allows for strong **intermolecular hydrogen bonding**.
  Ethers lack the hydrogen atom directly bonded to oxygen, preventing them from forming such strong intermolecular hydrogen bonds with each other.

Comparison of boiling points for molecules of similar molecular mass:

Solubility:

Alcohols and phenols are soluble in water because they can form **hydrogen bonds with water molecules**.

Answer:

Chemical Reactions

Alcohols are versatile reactants, capable of acting as both nucleophiles (breaking O–H bond) and electrophiles (breaking C–O bond). Phenols primarily react via O–H bond cleavage (acidity, reactions of the hydroxyl group) and electrophilic substitution on the aromatic ring.

Alcohols as nucleophiles (O–H cleavage): The electron pair on oxygen can attack an electrophile.

Protonated alcohols as electrophiles (C–O cleavage): After protonation, the C–O bond is weakened, and the alkyl group can be attacked by a nucleophile.

Reactions can be broadly categorised by bond cleavage:

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

• 1. Acidity of Alcohols and Phenols:
  • (i) Reaction with Metals: Both alcohols and phenols react with active metals (Na, K, Al) to form alkoxides or phenoxides and release hydrogen gas, demonstrating their acidic nature.
    Phenols are sufficiently acidic to also react with aqueous sodium hydroxide (a strong base) to form sodium phenoxides.
    These reactions confirm that alcohols and phenols act as Brönsted acids, donating a proton ($\textsf{H}^+$).
  • (ii) Acidity of Alcohols: The acidity of alcohols is due to the polarity of the O–H bond. Electron-releasing groups (like alkyl groups) attached to the carbon bonded to –OH increase electron density on the oxygen, reducing O–H bond polarity and thus decreasing acidity. The acidic strength of alcohols decreases in the order: Primary > Secondary > Tertiary.
    Alcohols are generally weaker acids than water. This is shown by the reaction of an alkoxide with water; water readily donates a proton to the alkoxide, forming hydroxide and the alcohol.
    This implies water is a stronger acid than alcohol, and alkoxides are stronger bases than hydroxides.
  • (iii) Acidity of Phenols: Phenols are significantly more acidic than alcohols and water. This is because the –OH group is attached to an $\textsf{sp}^2$ hybridised carbon of the benzene ring, which is electron-withdrawing. More importantly, the phenoxide ion formed after deprotonation is stabilised by resonance (delocalisation of the negative charge into the aromatic ring).
    The resonance structures of phenol itself involve charge separation, making the phenoxide ion relatively more stable than the neutral phenol molecule, thus favouring deprotonation. Alkoxide ions, in contrast, lack such resonance stabilisation, with the negative charge localised on the oxygen.

  Effect of Substituents on Phenol Acidity:

  • Electron-withdrawing groups (e.g., –$\textsf{NO}_2$) on the aromatic ring enhance acidity, especially at *ortho* and *para* positions, by further stabilising the phenoxide ion through delocalisation. For instance, 2,4,6-trinitrophenol (picric acid) is a very strong acid.
  • Electron-releasing groups (e.g., alkyl groups) decrease acidity by destabilising the phenoxide ion (or making the negative charge more concentrated on oxygen). Cresols (methylphenols) are less acidic than phenol.

  Acidity is often expressed using the $\textsf{p}K_\text{a}$ value. A lower $\textsf{p}K_\text{a}$ indicates a stronger acid.

  Compound	Formula	$\textsf{p}K_\text{a}$
  Phenol	$\textsf{C}_6\text{H}_5\text{–OH}$	10.0
  o-Nitrophenol	o–$\textsf{O}_2\text{N–C}_6\text{H}_4\text{–OH}$	7.2
  m-Nitrophenol	m–$\textsf{O}_2\text{N–C}_6\text{H}_4\text{–OH}$	8.3
  p-Nitrophenol	p–$\textsf{O}_2\text{N–C}_6\text{H}_4\text{–OH}$	7.1
  o-Cresol	o–$\textsf{CH}_3\text{–C}_6\text{H}_4\text{–OH}$	10.2
  m-Cresol	m–$\textsf{CH}_3\text{C}_6\text{H}_4\text{–OH}$	10.1
  p-Cresol	p–$\textsf{CH}_3\text{–C}_6\text{H}_4\text{–OH}$	10.2
  Ethanol	$\textsf{C}_2\text{H}_5\text{OH}$	15.9

  The $\textsf{p}K_\text{a}$ data confirms phenols ($\textsf{p}K_\text{a} \approx 10$) are much more acidic than ethanol ($\textsf{p}K_\text{a} \approx 15.9$). Nitro groups decrease $\textsf{p}K_\text{a}$ (increase acidity), while methyl groups slightly increase $\textsf{p}K_\text{a}$ (decrease acidity) compared to phenol.

• 2. Esterification: Alcohols and phenols react with carboxylic acids, acid chlorides, or acid anhydrides to form esters, involving the cleavage of the O–H bond.
  • With carboxylic acids (acid catalyst, reversible):
    Water is removed to shift equilibrium.
  • With acid chlorides (in presence of base like pyridine):
    Pyridine neutralises $\textsf{HCl}$ formed, favouring product formation.
  • With acid anhydrides:

  Introduction of the acetyl ($\textsf{CH}_3\text{CO}$) group is called acetylation. Acetylation of salicylic acid produces aspirin (acetylsalicylic acid).

Answer:

(b) Reactions involving cleavage of carbon – oxygen (C–O) bond in alcohols: These reactions typically occur only in alcohols, not phenols (except reaction with Zn).

• 1. Reaction with Hydrogen Halides: Alcohols react with hydrogen halides ($\textsf{HCl}$, $\textsf{HBr}$, $\textsf{HI}$) to form alkyl halides.
  The reactivity order of alcohols with $\textsf{HCl}$ (in presence of $\textsf{ZnCl}_2$) is $3^\circ > 2^\circ > 1^\circ$. This difference in reactivity is used in the **Lucas test** to distinguish primary, secondary, and tertiary alcohols. Tertiary alcohols react immediately (turbidity), secondary react slowly, and primary react only upon heating.
• 2. Reaction with Phosphorus Trihalides: Alcohols react with phosphorus trihalides ($\textsf{PX}_3$, where $\textsf{X = Cl, Br}$) to form alkyl halides.
• 3. Dehydration: Alcohols undergo dehydration (removal of water) upon heating with protic acids ($\textsf{conc. H}_2\text{SO}_4$, $\textsf{H}_3\text{PO}_4$) or catalysts ($\textsf{anhydrous ZnCl}_2$, alumina) to form alkenes.

  Ethanol dehydration with $\textsf{conc. H}_2\text{SO}_4$ at 443 K yields ethene:

  Secondary and tertiary alcohols dehydrate under milder conditions. The ease of dehydration follows the order: Tertiary > Secondary > Primary. This is related to the stability of carbocation intermediates formed during the mechanism.

  Mechanism of ethanol dehydration:

  Step 1: Protonation of the alcohol by the acid.

  Step 2: Formation of a carbocation by the removal of a water molecule. This is the slow, rate-determining step.

  Step 3: Elimination of a proton from the carbocation to form the alkene. The acid catalyst is regenerated.

• 4. Oxidation: Oxidation of alcohols involves breaking O–H and C–H bonds on the carbon bearing the –OH group, forming a carbon-oxygen double bond. This is also called dehydrogenation (removal of $\textsf{H}_2$).

Primary alcohols can be oxidised to aldehydes or carboxylic acids depending on the oxidising agent and conditions:

• With strong oxidising agents (acidified $\textsf{KMnO}_4$), primary alcohols are oxidised directly to carboxylic acids.
• With milder oxidising agents (e.g., $\textsf{CrO}_3$ in anhydrous medium), primary alcohols are oxidised to aldehydes.
  Pyridinium chlorochromate (PCC) is a better reagent for obtaining aldehydes in good yield from primary alcohols.

Secondary alcohols are oxidised to ketones using reagents like chromic anhydride ($\textsf{CrO}_3$).

Tertiary alcohols are generally resistant to oxidation under normal conditions as they lack a hydrogen atom on the carbon bearing the –OH group. Strong conditions cause C–C bond cleavage, yielding mixtures of carboxylic acids with fewer carbons.

Passing alcohol vapours over heated copper at 573 K causes dehydrogenation: primary alcohols yield aldehydes, secondary alcohols yield ketones, and tertiary alcohols undergo dehydration to alkenes.

Note on biological oxidation: Methanol oxidation in the body produces toxic formaldehyde and formic acid, causing blindness/death. Ethanol produces acetaldehyde and acetic acid. Ethanol is used to treat methanol poisoning by competing for the enzyme responsible for oxidation, allowing methanol to be excreted.

(c) Reactions of phenols: Phenols undergo reactions primarily on the aromatic ring (electrophilic substitution) and some involving the hydroxyl group, distinct from alcohols.

• 1. Electrophilic Aromatic Substitution (EAS): The –OH group on the benzene ring is strongly activating towards EAS and directs incoming electrophiles to the *ortho* and *para* positions. This is because the –OH group's lone pair can be donated into the ring via resonance, increasing electron density, especially at the *ortho* and *para* carbons. (See resonance structures under Acidity of Phenols).
• (i) Nitration:
  • With dilute nitric acid at 298 K, phenol gives a mixture of *ortho* and *para* nitrophenols.
    The *ortho* and *para* isomers can be separated by steam distillation: *o*-nitrophenol is steam volatile (intramolecular H-bonding), while *p*-nitrophenol is less volatile (intermolecular H-bonding).
  • With concentrated nitric acid, phenol is converted to 2,4,6-trinitrophenol (picric acid).
    Modern synthesis of picric acid involves sulphonating phenol first to protect the ring and activate it further, followed by nitration.
• (ii) Halogenation: Phenol reacts with halogens, with products depending on conditions.
  • In solvents of low polarity ($\textsf{CHCl}_3$, $\textsf{CS}_2$) at low temperature, monobromophenols are formed (*ortho* and *para* isomers).
    The –OH group's strong activating effect polarises the bromine molecule even without a Lewis acid catalyst like $\textsf{FeBr}_3$.
  • With bromine water, rapid polysubstitution occurs, forming 2,4,6-tribromophenol as a white precipitate.
• 2. Kolbe’s reaction: Treating phenol with sodium hydroxide produces phenoxide ion, which is even more reactive towards EAS. It reacts with carbon dioxide (a weak electrophile) at high temperature and pressure, followed by acidification, to produce *ortho*-hydroxybenzoic acid (salicylic acid) as the main product.
• 3. Reimer-Tiemann reaction: Reaction of phenol with chloroform ($\textsf{CHCl}_3$) in the presence of aqueous sodium hydroxide introduces a –CHO group at the *ortho* position, forming salicylaldehyde. The reaction involves dichlorocarbene (:$\textsf{CCl}_2$) as the electrophile.
  An intermediate dichloromethyl substituted compound is formed, which is hydrolysed by alkali to the aldehyde.
• 4. Reaction of phenol with zinc dust: Heating phenol with zinc dust causes reduction, converting phenol to benzene.
• 5. Oxidation: Oxidation of phenol with chromic acid ($\textsf{H}_2\text{CrO}_4$, from $\textsf{Na}_2\text{Cr}_2\text{O}_7/\textsf{H}_2\text{SO}_4$) produces a conjugated diketone called benzoquinone.
  Phenols can also be slowly oxidised by air, often leading to coloured quinone mixtures.

Answer:

(a) 3-methylphenol (m-cresol) has two activating groups: –OH (strong *ortho*, *para* director) and –$\textsf{CH}_3$ (weak *ortho*, *para* director). The –OH group is a stronger activator and determines the main positions. Positions *ortho* and *para* to –OH are C2, C4, and C6. The methyl group is on C3. The possible nitration sites are C2 (*ortho* to –OH, *ortho* to –$\textsf{CH}_3$), C4 (*para* to –OH, *meta* to –$\textsf{CH}_3$), and C6 (*ortho* to –OH, *para* to –$\textsf{CH}_3$). Steric hindrance at C6 (between nitronium ion and methyl group) might slightly disfavour C6. The major products are usually *ortho* and *para* to the strongest activator, considering steric effects. The most likely major products are nitration at C2 and C4.

Major products: 2-nitro-3-methylphenol and 4-nitro-3-methylphenol.

(b) Dinitration will add two nitro groups. Following the strongest directing effect of –OH, the most activated positions are C2, C4, and C6. With two nitro groups, expect substitution at two of these positions. Nitration at C2 and C4, C2 and C6, or C4 and C6 are possible. Given the position of the methyl group, C2 and C6 are both *ortho* to –OH and *ortho*/*para* to –$\textsf{CH}_3$. C4 is *para* to –OH and *meta* to –$\textsf{CH}_3$. The strongest activation is at C2, C4, and C6. Considering possible combinations for dinitration, 2,4- and 2,6-dinitrophenols are likely substituted forms relative to the hydroxyl group, so the methyl group would remain at position 3. The nitro groups would add to the available *ortho* and *para* positions relative to the –OH, which are C2, C4, and C6. Given two substitutions, the products would likely be on C2 and C4, or C2 and C6, or C4 and C6. Since position 3 is blocked by a methyl group, and the strongest activation by –OH is at 2, 4, and 6, dinitration will fill two of these positions. The methyl group on C3 is a weaker activator. The preferred positions are usually 2 and 4, or 2 and 6, or 4 and 6 relative to the –OH. The methyl group is a substituent on C3. The positions available for substitution are C2, C4, C5, C6. Strongest activation is at C2, C4, C6 by –OH. Substitutions are directed primarily to these. With methyl on C3, positions 2, 4, 6 relative to –OH are still activated. So nitro groups will likely go to two of these positions. Possible dinitration products are 2,4-dinitro-3-methylphenol, 2,6-dinitro-3-methylphenol, and 4,6-dinitro-3-methylphenol.

Major products: 2,4-dinitro-3-methylphenol and 2,6-dinitro-3-methylphenol.

(c) Phenyl methanoate is

It consists of a benzene ring with a –O–CO–H group attached. This group contains an ester linkage. The oxygen atom directly attached to the ring is part of the –OCOH group. Due to resonance involving the carbonyl oxygen, the electron-donating effect of the ether-like oxygen is significantly reduced, and the electron-withdrawing effect of the carbonyl dominates. The –O–CO–H group is actually deactivating (due to electron withdrawal by resonance and induction) and *meta*-directing for EAS on the benzene ring. However, the text focuses on the activation by –OH/alkoxy groups. Let's re-read the prompt carefully. The question asks for the product of mononitration of phenyl methanoate. Phenyl methanoate is an ester, not a phenol or ether discussed in the main text's EAS section. The example might be out of scope or expects application of general EAS rules. The group attached to the benzene ring is a formyloxy group, -OCHO. The oxygen directly attached to the ring has lone pairs that could potentially activate, but the adjacent carbonyl is electron withdrawing. Esters like phenyl acetate where the oxygen is directly attached to the ring and bonded to a carbonyl, are typically weakly activating and ortho/para directing, due to the balance between oxygen's resonance donation and the carbonyl's electron withdrawal. Assuming it is weakly activating and ortho/para directing like other aryl esters, nitration would occur at the ortho and para positions relative to the ester group. Given mononitration, we expect substitution at the activated positions.

Major products: ortho-nitrophenyl methanoate and para-nitrophenyl methanoate.

Some Commercially Important Alcohols

Methanol and ethanol are two alcohols produced and used on a large scale in industry.

Methanol

Methanol ($\textsf{CH}_3\text{OH}$), historically known as 'wood spirit', was first produced by the destructive distillation of wood. Currently, it is mainly synthesised from carbon monoxide and hydrogen by catalytic hydrogenation under high pressure and temperature, using a catalyst like $\textsf{ZnO} – \textsf{Cr}_2\text{O}_3$.

$\textsf{CO (g)} + 2\textsf{H}_2\text{ (g)} \xrightarrow{\textsf{ZnO–Cr}_2\text{O}_3, 573-673 K, 200-300 atm} \textsf{CH}_3\text{OH (l)}$

Methanol is a colourless liquid with a boiling point of 337 K. It is highly toxic; even small amounts ingested can cause blindness, and larger amounts can be fatal. Methanol is widely used as a solvent for paints and varnishes and as a raw material, particularly for producing formaldehyde ($\textsf{HCHO}$).

Ethanol

Ethanol ($\textsf{C}_2\text{H}_5\text{OH}$) is commercially produced via fermentation of sugars, an ancient method. Sugars from sources like molasses, sugarcane, or fruits (grapes) are converted into glucose and fructose ($\textsf{C}_6\text{H}_{12}\text{O}_6$) by the enzyme invertase.

$\textsf{C}_{12}\text{H}_{22}\text{O}_{11} \text{ (Sucrose)} + \textsf{H}_2\text{O} \xrightarrow{\textsf{Invertase}} \textsf{C}_6\text{H}_{12}\text{O}_6 \text{ (Glucose)} + \textsf{C}_6\text{H}_{12}\text{O}_6 \text{ (Fructose)}$

Glucose and fructose then undergo fermentation catalyzed by the enzyme zymase (found in yeast), producing ethanol and carbon dioxide.

$\textsf{C}_6\text{H}_{12}\text{O}_6 \text{ (Glucose/Fructose)} \xrightarrow{\textsf{Zymase}} 2\textsf{C}_2\text{H}_5\text{OH} \text{ (Ethanol)} + 2\textsf{CO}_2\text{ (g)}$

Fermentation is carried out under anaerobic (absence of air) conditions. The zymase enzyme activity is inhibited if the ethanol concentration exceeds about 14%. If air is present, ethanol can be oxidised by oxygen to ethanoic acid ($\textsf{CH}_3\text{COOH}$), spoiling the product.

Ethanol is a colourless liquid boiling at 351 K. It serves as a solvent (e.g., in paint industry) and a precursor for many organic compounds. Commercial alcohol is often denatured (made unfit for drinking) by adding small amounts of toxic substances like copper sulfate (for colour) and pyridine (for foul smell).

Large-scale industrial ethanol production also occurs via the hydration of ethene, which was discussed earlier as a method of alcohol preparation.

Ingestion of ethanol affects the central nervous system. Moderate consumption impairs judgment; higher doses can cause unconsciousness or death by interfering with respiration. Ethanol is used medically as an antidote for methanol poisoning by inhibiting the enzyme that metabolises methanol into toxic products, allowing methanol to be excreted.

Ethers

Ethers ($\textsf{R–O–R'}$) are organic compounds characterised by an oxygen atom bonded to two alkyl or aryl groups.

Preparation Of Ethers

Ethers can be prepared by:

• 1. By Dehydration of Alcohols: Alcohols can undergo dehydration in the presence of protic acids ($\textsf{H}_2\text{SO}_4$, $\textsf{H}_3\text{PO}_4$). The reaction product (alkene or ether) depends on the reaction conditions, especially temperature and the specific alcohol. For example, ethanol with concentrated $\textsf{H}_2\text{SO}_4$ at 443 K yields ethene, but at a lower temperature of 413 K, ethoxyethane (diethyl ether) is the main product.

  Dehydration to alkene: $\textsf{CH}_3\text{CH}_2\text{OH} \xrightarrow{\textsf{H}_2\text{SO}_4, 443 K} \textsf{CH}_2\text{=CH}_2 + \textsf{H}_2\text{O}$

  Dehydration to ether: $2\textsf{CH}_3\text{CH}_2\text{OH} \xrightarrow{\textsf{H}_2\text{SO}_4, 413 K} \textsf{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 + \textsf{H}_2\text{O}$

  The formation of symmetrical ethers from primary alcohols typically follows a nucleophilic bimolecular ($\textsf{S}_{\text{N}}2$) mechanism:

  Step 1: Protonation of the alcohol.

  Step 2: Nucleophilic attack by a second alcohol molecule on the protonated alcohol, displacing water.

  This method is suitable mainly for symmetrical ethers from primary alcohols. If the alkyl group is sterically hindered or the temperature is too high, elimination to form alkenes becomes the predominant reaction. Secondary and tertiary alcohols readily form alkenes under these acidic dehydration conditions via carbocation intermediates ($\textsf{S}_{\text{N}}1$ pathway), making this method unsuitable for preparing ethers from them.

• 2. Williamson Synthesis: This is a widely used laboratory method for preparing both symmetrical and unsymmetrical ethers. It involves the reaction of an alkyl halide with a sodium alkoxide or a sodium phenoxide.

  General reaction: $\textsf{R–X} + \textsf{R'–ONa}^+ \rightarrow \textsf{R–O–R'} + \textsf{Na}^+\textsf{X}^-$

  The reaction proceeds via an $\textsf{S}_{\text{N}}2$ mechanism, where the alkoxide/phenoxide ion acts as a nucleophile attacking the alkyl halide.

  For good yields of ethers, the alkyl halide should be **primary**. If a secondary or tertiary alkyl halide is used, the alkoxide/phenoxide ion acts as a strong base and elimination reaction predominates, leading to the formation of an alkene instead of an ether. For example, reacting sodium methoxide ($\textsf{CH}_3\text{ONa}$) with *tert*-butyl bromide ($\textsf{(CH}_3)_3\text{C–Br}$) exclusively produces 2-methylpropene (an alkene).

  Phenols can also be converted to ethers using this method by reacting a sodium phenoxide with an alkyl halide (preferably primary).

Example 11.6. The following is not an appropriate reaction for the preparation of t-butyl ethyl ether.

(i) What would be the major product of this reaction ?

(ii) Write a suitable reaction for the preparation of t-butylethyl ether.

Answer:

(i) The reaction uses sodium ethoxide ($\textsf{CH}_3\text{CH}_2\text{ONa}$), which is a strong base and a strong nucleophile, and *tert*-butyl bromide ($\textsf{(CH}_3)_3\text{C–Br}$), which is a tertiary alkyl halide. Due to the bulk of the tertiary carbon and the strong basicity of the ethoxide, the predominant reaction is elimination ($\textsf{E}2$) rather than $\textsf{S}_{\text{N}}2$ substitution. The ethoxide removes a $\beta$-hydrogen, leading to the formation of an alkene.

The major product is 2-methylpropene.

(ii) To prepare *tert*-butyl ethyl ether using Williamson synthesis, we need to use a primary alkyl halide and the alkoxide derived from the tertiary alcohol. This ensures an $\textsf{S}_{\text{N}}2$ reaction where the nucleophile attacks the primary carbon. We should react sodium *tert*-butoxide ($\textsf{(CH}_3)_3\text{CONa}$) with ethyl bromide ($\textsf{CH}_3\text{CH}_2\text{Br}$).

Suitable reaction: $\textsf{(CH}_3)_3\text{CONa} + \textsf{CH}_3\text{CH}_2\text{Br} \rightarrow \textsf{(CH}_3)_3\text{COCH}_2\text{CH}_3 + \textsf{NaBr}$.

Physical Properties

Ethers are polar molecules due to the polar C–O bonds, resulting in a net dipole moment. However, they lack the ability to form strong intermolecular hydrogen bonds with each other (as they don't have an O–H bond).

Boiling Points: The boiling points of ethers are comparable to those of alkanes with similar molecular masses. They are significantly lower than the boiling points of alcohols with comparable molecular masses because alcohols exhibit strong intermolecular hydrogen bonding. For example, butan-1-ol ($\sim 74$) boils much higher than ethoxyethane ($\sim 74$) or n-pentane ($\sim 72$).

Solubility: Although ethers cannot form hydrogen bonds with themselves, the oxygen atom can form hydrogen bonds with water molecules.

Chemical Reactions

Ethers are generally less reactive compared to alcohols and phenols, mainly due to the stable C–O–C linkage. Their reactions primarily involve the cleavage of the C–O bond or electrophilic substitution on the aromatic ring if an aryl group is present.

1. Cleavage of C–O bond in Ethers: The C–O bond in ethers can be cleaved under drastic conditions using excess hydrogen halides ($\textsf{HX}$). The reactivity of hydrogen halides follows the order $\textsf{HI > HBr > HCl}$. Concentrated $\textsf{HI}$ or $\textsf{HBr}$ at high temperature is typically required.

Reaction of dialkyl ethers with excess $\textsf{HX}$ yields two molecules of alkyl halide:

$\textsf{R–O–R} + 2\textsf{HX} \xrightarrow{Heat} 2\textsf{R–X} + \textsf{H}_2\text{O}$

Alkyl aryl ethers are cleaved at the alkyl-oxygen bond, yielding phenol and alkyl halide, because the bond between oxygen and the aryl group is stronger (due to partial double bond character from resonance with the ring).

$\textsf{Ar–O–R} + \textsf{HX} \xrightarrow{Heat} \textsf{Ar–OH} + \textsf{R–X}$

Cleavage mechanism for dialkyl ethers with HI (excess) involves protonation followed by nucleophilic attack:

Step 1: Protonation of the ether oxygen by the strong acid HI.

Step 2: Nucleophilic attack by iodide ion ($\textsf{I}^-$) on the less substituted carbon of the protonated ether ($\textsf{S}_{\text{N}}2$ pathway), displacing an alcohol molecule.

Step 3: If excess HI is present and heated, the alcohol formed in Step 2 reacts further with HI to form another molecule of alkyl iodide and water.

For mixed ethers with primary or secondary alkyl groups, the iodide attacks the less hindered carbon, leading to cleavage that forms the alkyl iodide from the smaller alkyl group and alcohol from the larger group (following $\textsf{S}_{\text{N}}2$ at the less substituted carbon). However, if one alkyl group is tertiary, the reaction proceeds via an $\textsf{S}_{\text{N}}1$ mechanism. Step 2 involves dissociation of the protonated ether to form a stable tertiary carbocation and an alcohol molecule, followed by rapid attack of iodide on the carbocation, yielding a tertiary alkyl halide and an alcohol from the other group.

For alkyl aryl ethers (like anisole, $\textsf{C}_6\text{H}_5\text{OCH}_3$), protonation occurs on the oxygen. The subsequent nucleophilic attack by $\textsf{I}^-$ cleaves the weaker oxygen-alkyl bond (O–$\textsf{CH}_3$) because the bond between oxygen and the $\textsf{sp}^2$ hybridised phenyl carbon (O–$\textsf{C}_6\text{H}_5$) has partial double bond character and is much stronger. This results in phenol and alkyl iodide ($\textsf{CH}_3\text{I}$). Phenol does not react further with HI under these conditions because the $\textsf{sp}^2$ hybridised carbon of the benzene ring cannot undergo nucleophilic substitution to form a halide.

Example 11.7. Give the major products that are formed by heating each of the following ethers with HI.

(i) $\textsf{CH}_3\text{CH}_2\text{CH}_2\text{OCH}_3$

(ii)

(iii)

Answer:

Heating ethers with HI causes C–O bond cleavage. The reaction mechanism depends on the groups attached to oxygen. In general, the weaker C–O bond breaks, and iodide attacks the carbon that can best support either an $\textsf{S}_{\text{N}}2$ displacement or an $\textsf{S}_{\text{N}}1$ carbocation.

(i) $\textsf{CH}_3\text{CH}_2\text{CH}_2\text{OCH}_3$ (Methyl n-propyl ether). This ether has two primary alkyl groups, but one is methyl, the least substituted. Cleavage follows an $\textsf{S}_{\text{N}}2$ pathway, with iodide attacking the carbon that is least sterically hindered and can better accommodate the nucleophilic attack from the backside. The methyl carbon is less hindered than the primary carbon of the n-propyl group.

Major products: Propan-1-ol and Methyl iodide ($\textsf{CH}_3\text{I}$).

(ii) Methoxybenzene (anisole). This is an alkyl aryl ether. The phenyl-oxygen bond is strong due to resonance. Cleavage occurs at the oxygen-alkyl bond. Iodide attacks the methyl group via $\textsf{S}_{\text{N}}2$ mechanism, as it's a primary carbon.

Major products: Phenol and Methyl iodide ($\textsf{CH}_3\text{I}$).

(iii) *tert*-Butyl ethyl ether. This ether contains a tertiary alkyl group (*tert*-butyl) and a primary alkyl group (ethyl). The cleavage proceeds via an $\textsf{S}_{\text{N}}1$ mechanism because the tertiary carbon can form a stable tertiary carbocation. Protonation of the ether is followed by dissociation to form the tertiary carbocation and ethanol. Iodide then attacks the carbocation.

Major products: **tert-Butyl iodide** ($\textsf{(CH}_3)_3\text{CI}$) and **Ethanol** ($\textsf{CH}_3\text{CH}_2\text{OH}$).

2. Electrophilic Substitution: Alkyl aryl ethers, like anisole, undergo EAS on the aromatic ring. The alkoxy group (–OR) is activating and *ortho*, *para* directing because the oxygen atom can donate electron density to the ring through resonance. This increases the electron density on the ring, particularly at the *ortho* and *para* positions, making them more attractive to electrophiles.

• (i) Halogenation: Anisole undergoes bromination with bromine in ethanoic acid without a Lewis acid catalyst because the methoxy group is highly activating. The major product is the *para* isomer due to steric hindrance at the *ortho* positions.
• (ii) Friedel-Crafts reaction: Anisole undergoes Friedel-Crafts alkylation and acetylation with alkyl or acyl halides in the presence of anhydrous $\textsf{AlCl}_3$ (Lewis acid catalyst). The reactions occur at the *ortho* and *para* positions. The *para* isomer is usually the major product.
• (iii) Nitration: Anisole reacts with a mixture of concentrated sulphuric acid and concentrated nitric acid (nitrating mixture) to yield a mixture of *ortho* and *para* nitroanisole. The *para* isomer is the major product.

Summary

• Alcohols, phenols, and ethers are classified by the number of –OH groups (alcohols/phenols) or the groups attached to oxygen (ethers), and by the hybridisation ($\textsf{sp}^3$ or $\textsf{sp}^2$) of the carbon bonded to –OH.
• Alcohols are prepared by alkene hydration (acid-catalysed or hydroboration-oxidation), reduction of carbonyl compounds (aldehydes, ketones, carboxylic acids, esters), or reaction with Grignard reagents.
• Phenols are prepared from haloarenes, benzenesulphonic acid, diazonium salts, or industrially from cumene.
• Alcohols and phenols have higher boiling points than comparable hydrocarbons, ethers, and haloalkanes due to intermolecular hydrogen bonding. Their solubility in water is also due to H-bonding with water.
• Alcohols and phenols are acidic. Phenols are more acidic than alcohols due to resonance stabilisation of the phenoxide ion. Electron-withdrawing groups increase phenol acidity, while electron-releasing groups decrease it.
• Alcohols undergo O–H cleavage (e.g., esterification, acidity) and C–O cleavage (e.g., reaction with HX, PX3, dehydration, oxidation). Primary alcohols oxidise to aldehydes/acids, secondary to ketones, tertiary are resistant.
• Phenols undergo electrophilic substitution on the activated, *ortho*, *para*-directing aromatic ring (e.g., nitration, halogenation, Kolbe's reaction, Reimer-Tiemann reaction). They can be reduced to benzene with zinc dust or oxidised to quinones.
• Ethers are prepared by alcohol dehydration (suitable for primary alcohols) or Williamson synthesis (alkyl halide + alkoxide/phenoxide, best with primary alkyl halides).
• Ethers have boiling points similar to alkanes but are water soluble (H-bonding with water).
• Ethers are relatively unreactive. Their C–O bond can be cleaved by strong hydrogen halides. Alkyl aryl ethers cleave at the alkyl-oxygen bond.
• Alkoxy groups in alkyl aryl ethers activate the aromatic ring towards EAS and direct to *ortho*, *para* positions.

Points to Ponder

Key takeaways and further thoughts:

• The hydroxyl group (–OH) dictates much of the chemistry of alcohols and phenols due to its polarity and ability to form hydrogen bonds.
• The nature of the attached alkyl or aryl group significantly modifies the reactivity of the –OH group and influences the overall properties (e.g., acidity of phenols vs alcohols, ease of dehydration, $\textsf{S}_{\text{N}}1$ vs $\textsf{S}_{\text{N}}2$ reactions).
• Phenols' enhanced acidity and aromatic reactivity compared to alcohols highlight the impact of conjugation and resonance with the aromatic ring.
• Ethers' relative inertness stems from the stable C–O–C bond and lack of an O–H group, limiting intermolecular hydrogen bonding with themselves and reducing reactivity towards many reagents that target the –OH group.
• Understanding reaction mechanisms (e.g., carbocation stability in dehydration or $\textsf{S}_{\text{N}}1$ ether cleavage, steric effects in $\textsf{S}_{\text{N}}2$ Williamson synthesis) is crucial for predicting products and choosing appropriate synthetic methods.

Question 11.1. Classify the following as primary, secondary and tertiary alcohols:

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Question 11.3. Name the following compounds according to IUPAC system.

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Question 11.4. Show how are the following alcohols prepared by the reaction of a suitable Grignard reagent on methanal ?

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Question 11.5. Write structures of the products of the following reactions:

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Question 11.11. Which of the following is an appropriate set of reactants for the preparation of 1-methoxy-4-nitrobenzene and why?

(i)

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Question 11.12. Predict the products of the following reactions:

(i) $CH_3-CH_2-CH_2-O-CH_3 + HBr \rightarrow$

(ii) $(CH_3)_3C-OC_2H_5 + HI \rightarrow$

(iii)

(iv)

Answer:

Question 11.1. Write IUPAC names of the following compounds:

(i)

(ii)

(iii)

(iv)

(v)

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(ix)

(x) $C_6H_5–O–C_2H_5$

(xi) $C_6H_5–O–C_7H_{15}(n–)$

(xii)

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Question 11.23. Give IUPAC names of the following ethers:

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Question 11.32. Show how would you synthesise the following alcohols from appropriate alkenes?

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Question 11.33. When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:

Give a mechanism for this reaction.

(Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom.)

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