Classification

(i) Alcohols and phenols are classified on the basis of the number of hydroxyl groups present in the molecule as monohydric, dihydric, trihydric or polyhydric depending on whether they contain one, two, three or many - $\mathrm{OH}$ groups

(ii) Monohydric alcohols are further classified on the basis of hybridisation of the carbon atom to which - $\mathrm{OH}$ group is attached,

(a) $-\mathrm{OH}$ group attached to $-\mathrm{sp}^{3}$ hybridised $\mathrm{C}$ atom are classified as primary $\left(1^{\circ}\right)$, secondary $\left(2^{\circ}\right)$ or tertiary $\left(3^{\circ}\right)$ alcohols depending on whether hydroxyl is attached to primary, secondary or tertiary carbon atom

Allylic alcohol

• When the - $\mathrm{OH}$ group attached $\mathrm{sp}^{3}$ hybridised carbon is next to a $\mathrm{C}=\mathrm{C}$ system, then such alcohols are known as allylic alcohols. Allylic alcohols can also be $1^{\circ}, 2^{\circ}$ or $3^{\circ}$.

• When the $\mathrm{sp}^{3}$ hybridised carbon atom containing - $\mathrm{OH}$ group is attached to a benzene ring, then such alcohols are classified as benzylic alcohols. Benzylic alcohols can also be $1^{\circ}, 2^{\circ}$ or $3^{\circ}$.

(b) Vinyl alcohol - $-\mathrm{OH}$ group attached to $\mathrm{a}-\mathrm{sp}^{2}$ hybridised $\mathrm{C}$ atom are classified as vinylic alcohols, $\mathrm{CH} _{2}=\mathrm{CH}-\mathrm{OH}$ or phenols (when attached to an aryl carbon)

Preparation of Alcohols

1. From alkenes

2. Reduction of carbonyl compounds

aldehydes give primary alcohol, ketones give secondary alcohol.

1. $\mathrm{NaBH} _{4}$ does not affect the double bond.

$\mathrm{CH} _{2}=\mathrm{CH}-\mathrm{CH} _{2}-\mathrm{CHO} \xrightarrow{\mathrm{NaBH} _{4}} \mathrm{CH} _{2}=\mathrm{CH}-\mathrm{CH} _{2}-\mathrm{CH} _{2} \mathrm{OH}$

2. $\mathrm{LiAlH} _{4}$ also does not affect the double bond but if the compound is $\alpha, \beta$ - unsaturated carbonyl compound then double bond also undergoes reduction.

$\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CH}=\mathrm{CH}-\mathrm{CHO} \xrightarrow{\mathrm{LiAlH} _{4}} \mathrm{C} _{6} \mathrm{H} _{5} \mathrm{CH} _{2}-\mathrm{CH} _{2}-\mathrm{CH} _{2} \mathrm{OH}$

3. From grignard reagents

Mechanism

• $\quad$ With formaldehyde, $1^{\circ}$ alcohol is formed

• With ethylene oxide, $1^{\circ}$ alcohol is formed

• With any other aldehyde except formaldehyde, $2^{\circ}$ alcohol is formed

• $\quad$ With ketones, $3^{\circ}$ alcohol is formed

4. From alkyl halides

$\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{Br}+\mathrm{aqKOH} \xrightarrow{\mathrm{S} _{\mathrm{N}} 2} \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}+\mathrm{KBr}$

Not suitable for preparation of tertiary alcohols because tertiary halides with strong base prefer to undergo E-2 elimination.

Other bases can also be used like aqueous sodium carbonate or moist silver oxide.

5. By reduction of carboxylic acids and its derivatives

By reduction of carboxylic acids

By reduction of ester

By reduction of acid chloride

Other reducing agents like $\mathrm{Na} / \mathrm{C} _{2} \mathrm{H} _{5} \mathrm{OH}$ or hydrogen in presence of a catalyst can also be used for esters

6. From amines (via diazonium salts)

$$ \begin{aligned} & \underset{\text{ethylamine}}{\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{NH} _{2}}+\underset{\text { nitrous acid }}{\mathrm{HNO} _{2}} \longrightarrow \underset{\text {ethanol} }{\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}}+\mathrm{N} _{2}+\mathrm{H} _{2} \mathrm{O} \\ & \end{aligned} $$

Preparation of Phenols

Phenol also known as carbolic acid, was earlier obtained from coal tar. Other commercial methods are,

1. From benzene sulphonic acid

2. From diazonium salts

3. From aryl halides (Dow’s process)

4. From cumene

5. By decarboxylation of salicylic acid

Physical Properties

1. Boiling points

Due to the presence of intermolecular hydrogen bonding in alcohols and phenols they have higher boiling points than corresponding alkanes and alkyl halides.

(a) Boiling points increase with increase in molecular mass due to increase in van der Waals forces.

(b) Boiling points decrease with increase in branching due to decrease in van der Waals forces with decrease in surface area.

2. Solubility in water is due to the ability of alcohols and phenols to form hydrogen bonds with water

(a) It decreases with increase in molecular mass, since as the hydrocarbon part increases tendency to form hydrogen bonds decreases.

(b) Solubility increases with branching because as the branching increases surface area of non polar hydrocarbon part decreases hence solubility increases.

t-butyl alcohol>sec- butyl alcohol>isobutyl alcohol>n-butyl alcohol

Chemical Properties

I. Reactions involving cleavage of oxygen-hydrogen bond $[\mathrm{C}-\mathrm{O}+\mathrm{H}]$ with substitution or removal of hydrogen as proton.

When 0 - $\mathrm{H}$ bond is broken, alcohols and phenols act as nucleophiles

1. Reaction with metals

Alcohols and phenols are acidic in nature.

$2 \mathrm{CH} _{3} \mathrm{OH}+2 \mathrm{Na} \longrightarrow 2 \mathrm{CH} _{3} \mathrm{O}^{-} \mathrm{Na}^{+}+\mathrm{H} _{2}$

$2 \mathrm{CH} _{3} \mathrm{OH}+\mathrm{Mg} \longrightarrow\left(\mathrm{CH} _{3} \mathrm{O}\right) _{2} \mathrm{Mg}+\mathrm{H} _{2}$

Phenols are stronger acids than alcohol. Thus they even react with aqueous sodium hydroxide.

2. Esterification

Reaction with carboxylic acids, acid chlorides and anhydrides

When $\mathrm{HCl}$ gas is used as catalyst, reaction is known as Fischer-Speier esterification.

Acetylation

$\underset{\text { acetyl chloride }}{\mathrm{CH}_3 \mathrm{COCl}}+\underset{\text { methanol }}{\mathrm{CH}_3 \mathrm{OH}} \xrightarrow{\text { pyridine }} \underset{\text { methyl ethanoate }}{\mathrm{CH}_3 \mathrm{COOCH}_3}+\mathrm{HCl}$

$\underset{\text { acetic anhydride }}{\left(\mathrm{CH}_3 \mathrm{CO}\right)_2 \mathrm{O}}+\underset{\text { methanol }}{\mathrm{CH}_3 \mathrm{OH}} \xrightarrow[]{\text { pyridine }} \underset{\text { methyl ethanoate }}{\mathrm{CH}_3 \mathrm{COOCH}_3}+\underset{\text { acetic acid }}{\mathrm{CH}_3 \mathrm{COOH}}$

Benzoylation

Schotten Baumann reaction

$ \begin{aligned} & \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}+\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{COCl} \xrightarrow{\mathrm{NaOH}} \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OCOC} _{6} \mathrm{H} _{5}+\mathrm{HCl} \\ & \text { ethanol benzoyl chloride } \quad \text { ethyl benzoate } \end{aligned} $

Order of reactivity of alcohols

$\mathrm{CH} _{3} \mathrm{OH}>\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}>\left(\mathrm{CH} _{3}\right) _{2} \mathrm{CHOH}>\left(\mathrm{CH} _{3}\right) _{3} \mathrm{COH}$

Order of reactivity of acids

$\mathrm{HCOOH}>\mathrm{CH} _{3} \mathrm{COOH}>\left(\mathrm{CH} _{3}\right) _{2} \mathrm{CHCOOH}>\left(\mathrm{CH} _{3}\right) _{3} \mathrm{CCOOH}$

As the size of $\mathrm{R}$ group in acids and alcohols increases, rate of esterification decreases

3. Reaction with grignard reagents

$\underset{\text { methanol }}{\mathrm{CH}_3 \mathrm{OH}}+\underset{\substack{\text { ethyl magnesium } \ \text { bromide }}}{\mathrm{C}_2 \mathrm{H}_5 \mathrm{MgBr}} \longrightarrow \underset{\text { ethane }}{\mathrm{C}_2 \mathrm{H}_6}+\mathrm{Mg}\left(\mathrm{OCH}_3\right) \mathrm{Br}$

II. Reactions involving cleavage of carbon oxygen bond $\left[\mathrm{C}^{\prime}+\mathrm{OH}\right]$ with substitution or removal of $\mathrm{OH}$ group.

When the bond between $\mathrm{C}-0$ is broken, alcohols act as electrophiles.

1. Reaction with halogen acids

$\mathrm{R}-\mathrm{OH}+\mathrm{HX} \longrightarrow$ $\mathrm{RX}+\mathrm{H} _{2} \mathrm{O}$

Order of reactivity of halogen acids

$\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$

Order of reactivity of alcohols : $3^{0}>2^{\circ}>1^{\circ}$

a) With $\mathrm{HCl}$

Due to low reactivity of $1^{\circ}$ and $2^{\circ}$ alcohols, they require a lewis acid catalyst, anhydrous $\mathrm{ZnCl} _{2}$

This reaction is used as Lucas Test to differentiate between $1^{\circ}, 2^{\circ}$ and $3^{\circ}$ alcohols. A solution of conc. hydrochloric acid and anhydrous zinc chloride is known as Lucas Reagent. Tertiary alcohols react with this reagent immediately to give cloudiness because of the formation of insoluble alkyl chloride. Secondary alcohols take about five minutes to react and cloudiness to appear whereas primary alcohols do not react and thus no turbidity appears at room temperature.

b) With $\mathrm{HBr}$

$\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}+\mathrm{HBr} \xrightarrow[\text { reflux }]{\text { conc. } \mathrm{H} _{2} \mathrm{~S} _{4}} \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{Br}+\mathrm{H} _{2} \mathrm{O}$

No catalyst is required for $2^{\circ}$ and $3^{\circ}$ alcohols as they undergo dehydration in presence of conc. $\mathrm{H} _{2} \mathrm{SO} _{4}$ to form alkenes

c) With $\mathrm{HI}$

$\mathrm{CH} _{3} \mathrm{OH}+\mathrm{HI} \xrightarrow{\text { reflux }} \mathrm{CH} _{3} \mathrm{I}+\mathrm{H} _{2} \mathrm{O}$

Mechanism

$1^{\circ}$ alcohols react by $\mathrm{S} _{\mathrm{N}} 2$ mechanism

$2^{\circ}$ and $3^{\circ}$ alcohols react by $S _{N} 1$ mechanism

2. Reaction with phosphorus halides

$\mathrm{ROH}+\mathrm{PCl} _{5} \longrightarrow \mathrm{R}-\mathrm{Cl}+\mathrm{POCl} _{3}+\mathrm{HCl}$

$3 \mathrm{ROH}+\mathrm{PX} _{3} \longrightarrow 3 \mathrm{R}-\mathrm{X}+\mathrm{H} _{3} \mathrm{PO} _{3}(\mathrm{X}=\mathrm{Cl}, \mathrm{Br}, \mathrm{I})$

3. Reaction with thionyl chloride $\left(\mathrm{SOCl} _{2}\right)$

$\mathrm{ROH}+\mathrm{SOCl} _{2} \xrightarrow{\text { pyridine }} \mathrm{R}-\mathrm{Cl}+\mathrm{SO} _{2} \uparrow+\mathrm{HCl} \uparrow$

$\mathrm{SOCl} _{2}$ is better than $\mathrm{PCl} _{5}$ and $\mathrm{PCl} _{3}$ because by products are gaseous

4. Reaction with $\mathrm{NH} _{3}$

$\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}+\mathrm{NH} _{3} \xrightarrow[633 \mathrm{~K}]{\mathrm{Al} _{2} \mathrm{O} _{3}} \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{NH} _{2}+\mathrm{H} _{2} \mathrm{O}$

5. Reaction with Zn dust

III. Other reactions of alcohol

1. Reaction with Conc. $\mathrm{H} _{2} \mathrm{SO} _{4}$

a) At $383 \mathrm{~K}$, ethyl hydrogen sulphate is formed

$ \underset{\text { ethanol }}{\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}+\text { conc } \mathrm{H} _{2} \mathrm{SO} _{4} \xrightarrow{383 \mathrm{~K}}} \underset{\text { ethyl hydrogen sulphate }}{\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OSO} _{3} \mathrm{H}+\mathrm{H} _{2} \mathrm{O}} $

b) At 433-443K, dehydration of alcohol to alkene takes place

The rate of dehydration of alcohols to alkenes follows the order $3^{\circ}>2^{\circ}>1^{0}$.

Dehydration of alcohols to alkenes may also be accomplished when alcohol vapours are passed over activated alumina $\left(\mathrm{Al} _{2} \mathrm{O} _{3}\right)$ at high temperature.

c) At $413 K$, ether is formed

$ \begin{aligned} & \underset{\text { ethanol }}{2 \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}} \xrightarrow[413 \mathrm{~K}]{\mathrm{Conc} \mathrm{H} _{2} \mathrm{SO} _{4}} \underset{ \text { ethoxyethane }}{\mathrm{CH} _{3} \mathrm{CH} _{2}-\mathrm{O}-\mathrm{CH} _{2} \mathrm{CH} _{3}} \\ \end{aligned} $

2. Oxidation -

Involves formation of carbon-oxygen double bond with cleavage of $\mathrm{O}-\mathrm{H}$ and $\mathrm{C}-\mathrm{H}$ bond

$1^{\circ}$ alcohol

$ \mathrm{RCH}_2 \mathrm{OH} \underset{\text{(mild oxidising agent)}}{\xrightarrow[\text { PDC }]{\text { PCC or }}} \mathrm{RCHO} $

$\mathrm{PCC} \longrightarrow$ pyridinium chlorochromate

$\mathrm{PDC} \longrightarrow$ pyridinium dichromate

$\underset{\text{alcohol}}{\mathrm{RCH}_2 \mathrm{OH}} \underset{\text{strong oxidising agents}}{\xrightarrow{\substack{\mathrm{K}_2 \mathrm{Cr}_2, 0, \mathrm{H}_2 \mathrm{SO}_4 \text { or } \\ \mathrm{KMnO}_4 / \mathrm{OH}^{-}}}} \mathrm{RCOOH}$

$2^{\circ}$ alcohol

3. Action of heated copper

1. $1^{\circ} \text { alcohol } \xrightarrow{\text { dehydrogenation }} \text { aldehyde } $

$\underset{\text { ethanol }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}} \xrightarrow{\text { Cu/573K }} \underset{\text { ethanal }}{\mathrm{CH}_3 \mathrm{CHO}}+\mathrm{H}_2 $

2. $2^{\circ}$ alcohol $\xrightarrow{\text { dehydrogenation }}$ ketone

3. $3^{0}$ alcohol $\xrightarrow{\text { dehydration }}$ alkene

IV Other reactions of Phenols

Electrophilic Substitution in Phenols

1. Sulphonation

2. Halogenation

2, 4, 6-Tribromophenol

(white ppt.)

In aqueous solution, phenol ionizes to form phenoxide ion which activates the benzene ring leading to trisubstitution

In non polar solvents like $\mathrm{CS} _{2}, \mathrm{CCl} _{4}$ or $\mathrm{CHCl} _{3}$ ionization of phenol is suppressed. Ring gets activated only slightly leading to monosubstitution.

3. Nitration

The 0 -and p-isomers can be separated by steam distillation since 0-isomer is steam volatile due to chelation (intramolecular $\mathrm{H}$-bonding) while $\mathrm{p}$-isomer is not steam volatile due to association of molecules by intermolecular $\mathrm{H}$-bonding .

Poor yield is obtained because conc. $\mathrm{HNO} _{3}$ acts as an oxidizing agent so some oxidation products are also formed.

4. Friedel Crafts alkylation

5. Kolbe’s reaction

It is an electrophilic substitution in which $\mathrm{CO} _{2}$ behaves as electrophile.

Mechanism :

Salicylic acid is the starting material for many useful compounds.

6. Reimer Tiemann reaction :

Mechanism :

Step 1: Generation of electrophile

$ \mathrm{HO}^{-}+\mathrm{CHCl} _{3} \rightleftarrows \mathrm{H} _{2} \mathrm{O}+\mathrm{CCl} _{3}^{-} \longrightarrow \underset{\underset{\text{(Electrophile)}}{\text { Dichlorocarbene }}}{: \mathrm{CCl} _{2}+\mathrm{Cl}^{-}} $

Step 2 : Electrophilic substitution

7. Oxidation:

8. Reaction with $\mathrm{FeCl} _{3}$

It can be used as a test of distinction for phenol.

9. Coupling reaction - It can be used as a test of distinction for phenol

10. Fries rearrangement : Phenolic esters are converted into 0-and p-hydroxyketones in presence of anhydrous aluminum chloride.

11. Reduction

Test of Distinction

Distinction between $1^{\circ}, 2^{\circ}, 3^{\circ}$ alcohols

1. Victor Meyer Test

2. Lucas Test

$ \begin{array}{ccll} \text{Alcohol} & \xrightarrow[\text{(Lucas reagent)}]{\text{anhyd }\mathrm{ZnCl}_2+\text{conc. HCl}} & a. \quad \text{No turbidity} \rightarrow 1^{\circ} \text{ alcohol} \\ & & b. \quad \text{Turbidity appears after 5 minutes} \rightarrow 2^{\circ} \text{ alcohol} \\ & & c.\quad \text{ Turbidity appears immediately} \rightarrow 3^\circ \text{ alcohol} \\ \end{array} $

3. lodoform Test

Ethers

Classification : Ethers are classified on the basis of groups attached to oxygen. If the alkyl or aryl groups attached on either side of oxygen are same, the ether is known as simple or symmetrical, whereas if the two groups are different, ether is mixed or unsymmetrical.

Preparation

1. By acid catalysed dehydration of alcohols

Mechanism :

i. Protonation of alcohol

ii. Nucleophilic attack by unprotonated alcohol molecule on protonated alcohol molecule

iii. Loss of proton to form ethoxy ethane

At high temperature (443K) alkene is formed with the same reactants. To produce a high yield of ether the alcohol is added in excess. This method is suitable for producing only symmetrical ethers. Reaction involves nucleophilic bimolecular mechanism thus it is suitable for ethers having primary alkyl groups only.

2. By catalytic dehydration of alcohols

3. By the action of diazomethane on alcohols

$$ \begin{aligned} & \mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{OH}+\mathrm{CH} _{2} \mathrm{~N} _{2} \xrightarrow{\mathrm{HBF} _{4}} \mathrm{CH} _{3} \mathrm{CH} _{2}-\mathrm{O}-\mathrm{CH} _{3}+\mathrm{N} _{2} \\ & \text { Ethyl alcohol Diazomethane Ethyl methyl ether } \end{aligned} $$

4. From alkyl halides

Williamson synthesis

$$ \underset{\text { alkyl halide }}{\mathrm{R}-\mathrm{O}^{-} \mathrm{Na}^{+}+\mathrm{R}-\mathrm{X}} \xrightarrow{\mathrm{S} _{\mathrm{N}} 2} \underset{\text { Ether }}{\mathrm{R}-\mathrm{O}-\mathrm{R}}+\mathrm{Na}^{+}-\mathrm{X}^{-} $$

This method can be used for the preparation of both symmetrical and unsymmetrical ethers.

$$ \begin{aligned} & \underset{\text { Sod. ethoxide }}{\mathrm{CH} _{3} \mathrm{CH} _{2}-\mathrm{O}^{-} \mathrm{Na}^{+}}+\underset{\text { Ethyl bromide }}{\mathrm{CH} _{3} \mathrm{CH} _{2}-\mathrm{Br}} \xrightarrow[\mathrm{S_N}{2}]{330 \mathrm{~K}} \underset{\text { Diethyl ether }}{\mathrm{CH} _{3} \mathrm{CH} _{2}-\mathrm{O}-\mathrm{CH} _{2} \mathrm{CH} _{3}}+\mathrm{Na}^{+} \mathrm{Br} \\ \end{aligned} $$

i. Alkyl halide should have smaller alkyl group.

ii. Aryl halides are unreactive towards nucleophilic substitution.

Limitations

i. Diaryl ethers cannot be prepared since aryl halides are unreactive towards nucleophilic substitution.

ii. Di-tert butyl ether cannot be prepared because tert butyl halide undergoes elimination to form alkene.

5. From Grignard reagents

$\begin{aligned} & \underset{\substack{\text { Chloromethoxy } \\ \text { methane }}} {\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}_2 \mathrm{Cl}}+\underset{\substack{\text { Methylmagnesium } \\ \text { iodide }}}{\mathrm{CH}_3 \mathrm{Mgl}}\xrightarrow[\text { ether }]{\text { Dry }} \underset{\substack{\text { Ethyl methyl } \\ \text { ether }}}{\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}_2 \mathrm{CH}_3}+\underset{\substack{\text { Chloromagnesium } \\ \text { iodide }}}{\mathrm{Mg}[\mathrm{Cl}] \mathrm{l}} \\ & \end{aligned}$

PHYSICAL PROPERTIES

1. Dipolar nature

Because of the greater electronegativity of oxygen than carbon, the $\mathrm{C}-\mathrm{O}$ bonds are slightly polar and thus have a dipole moment. Since the two $\mathrm{C}-0$ bonds in ethers are inclined to each other at an angle of $110^{\circ}$, the two dipoles do not cancel each other.

2. Boiling points

Their boiling point are much lower than those of the isomeric alcohols. This is due to the reason that unlike alcohols, ethers do not form hydrogen bonds. Due to presence of weak dipole dipole interactions their boiling points are higher than alkanes which have vander Waals forces of attraction only.

3. Solubility

The solubility of lower ethers in water is due to the formation of hydrogen bonds between water and ether molecules.

As the molecular mass increases, the solubility of ethers in water decreases due to the corresponding increase in the hydrocarbon portion of the molecule.

CHEMICAL PROPERTIES

Ethers are relatively inert since the functional group of ethers does not contain any active site in their molecules.

I. Reaction of ethereal oxygen

1. Action of concentrated acids

Oxygen atom in ethers has two lone pairs. So, ethers behave as lewis bases and dissolve in concentrated inorganic acids to form stable oxonium salts.

2. Formation of coordination complexes

Being lewis bases, ethers form coordinate complexes with lewis acids such as $\mathrm{BF} _{3}, \mathrm{AlCl} _{3}$, $\mathrm{FeCl} _{3}$

II. Reactions involving cleavage of carbon-oxygen bond

Cleavage of $\mathrm{C}-0$ bond takes place only under very drastic conditions since ethers are comparatively unreactive

1. With halogen acids

$\mathrm{R}-\mathrm{O}-\mathrm{R}+\mathrm{HX} \xrightarrow{373 \mathrm{~K}} \mathrm{ROH}+\mathrm{R}-\mathrm{X} \quad(\mathrm{X}=\mathrm{Br}$ or $\mathrm{l})$

$\mathrm{R}-\mathrm{O}-\mathrm{R}+$ excess $\mathrm{HX} \xrightarrow{373 \mathrm{~K}} 2 \mathrm{R}-\mathrm{X}+\mathrm{H} _{2} \mathrm{O}$

Mechanism

Reactivity of halogen acids

$\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$

• In case of unsymmetrical ethers having two different alkyl groups, alcohol and alkyl halide formed depend on the nature of alkyl groups.

  • Looking at the mechanism, once the protonation of the ether has occured in the first step, the structure of alkyl groups decide the products.
    
    

• If one of the alkyl groups is tertiary, the alkyl halide is formed from tertiary alkyl groups. Since the reaction occurs by $\mathrm{S} _{\mathrm{N}} 1$ mechanism and the formation of products is controlled by stability of carbocation formed.

• If no tertiary alkyl group is present and a stable carbocation can’t be formed, $\mathrm{S} _{\mathrm{N}} 2$ mechanism is followed.

Due to steric hinderance, halide ion always attacks the smaller alkyl group.

• In case of alkyl aryl ethers, products formed are always phenol and alkyl halide and never aryl halide and alcohol because it is difficult to break ph- 0 bond.

• In case of benzyl alkyl ethers, reaction follows $\mathrm{S} _{\mathrm{N}} 1$ mechanism, since benzyl carbocation is more stable than methyl carbocation therefore benzyl halide is formed.

• Diaryl ethers cannot be cleaved by $\mathrm{HI}$ because due to resonance $\mathrm{C}-0$ has some double bond character.

2. With sulphuric acid

$\begin{aligned} &\underset{\text { diethyl ether }}{\mathrm{C}_2 \mathrm{H}_5-\mathrm{O}-\mathrm{C}_2 \mathrm{H}_5}+\mathrm{H}_2 \mathrm{O} \xrightarrow[\Delta]{\text { dil } \mathrm{H}_2 \mathrm{SO}_4} \underset{\text { ethanol }}{2 \mathrm{CH}_3 \mathrm{CH}_2 \mathrm{OH}} \end{aligned}$

ELECTROPHILIC SUBSTITUTION REACTIONS

In aryl alkyl ethers, the $+\mathrm{R}$ effect of the alkoxy group (OR) increases the electron density in the benzene ring (refer to resonance structures, I to V) thereby activating the benzene ring towards electrophilic substitution reactions.

Since the electron density increases more at ortho and para positions as compared to m-positions, therefore, electrophilic substitution reactions mainly occur at 0 -and $p$-positions.

1. Bromination

2. Nitration

3. Friedel crafts Alkylation

4. Friedel crafts acylation

PRACTICE QUESTIONS

Question 1- Which of the following is soluble in water?

(a) $\mathrm{CS} _{2}$

(b) $\mathrm{C} _{2} \mathrm{H} _{5} \mathrm{OH}$

(c) $\mathrm{CCl} _{4}$

(d) $\mathrm{CHCl} _{3}$

Question 2- Ethyl alcohol is heated with cone. $\mathrm{H} _{2} \mathrm{SO} _{4}$. The product formed is

(a) $\mathrm{CH} _{3} \mathrm{COOC} _{2} \mathrm{H} _{5}$

(b) $\mathrm{C} _{2} \mathrm{H} _{2}$

(c) $\mathrm{C} _{2} \mathrm{H} _{4}$

(d) $\mathrm{C} _{2} \mathrm{H} _{6}$

Question 3- $\mathrm{HBr}$ reacts fastest with

(a) 2-methylpropan-2-ol

(b) propan-1-0l

(c) propan-2-0l

(d) 2-methylpropan-1-0l

Question 4- 1-propanol and 2-propanol can be best distinguished by

(a) oxidation with alkaline $\mathrm{KMnO} _{4}$ followed by reaction with Fehling solution

(b) oxidation with acidic dichromate followed by reaction with Fehling solution

(c) oxidation by heating with copper followed by reaction with Fehling solution

(d) oxidation with concentrated $\mathrm{H} _{2} \mathrm{SO} _{4}$ followed by reaction with Fehling solution

Question 5- When phenyl magnesium bromide reacts with tert. butanol, which of the following is formed?

(a) Tert. butyl methyl ether

(b) Benzene

(c) Tert. butyl benzene

(d) Phenol

Question 6- The compound which reacts fastest with Lucas reagent at room temperature is

(a) butan-2-0l

(b) butan-1-0l

(c) 2-methyl propan-1-0l

(d) 2-methyl propan-2-0l

Question 7- The best method to prepare cyclohexene from cyclohexanol is by using

(a) conc. $\mathrm{HCl}+\mathrm{ZnCl} _{2}$

(b) conc. $\mathrm{H} _{3} \mathrm{PO} _{4}$

(c) $\mathrm{HBr}$

(d) conc. $\mathrm{HCl}$

Question 8- Which of the following is most acidic?

Question 9- In the given reaction:

$\mathrm{H} _{2} \mathrm{C}=\mathrm{CH}-\mathrm{CHO} \xrightarrow{[x]} \mathrm{H} _{2} \mathrm{C}=\mathrm{CH}-\mathrm{CH} _{2} \mathrm{OH}$

(a) $\mathrm{H} _{2} / \mathrm{Ni} / \Delta$

(b) $\mathrm{Pt} / \mathrm{H} _{2}$

(c) $\mathrm{NaBH} _{4}$

(d) $\mathrm{H} _{2} /$ Wilkinson Catalyst

Question 10- In the given reaction

$[X]$ will be :

Question 11- In the given reaction:

$[X]$ will be :

Question 12- Which of the following will most readily be dehydrated in acidic condition

Question 13- In the given reaction,

$[X]$ will be :

(a) Salicylic acid

(b) p-hydroxybenzoic acid

(c) Mixture of (a) and (b)

(d) Salicylaldehyde

Question 14- Phenol can be distinguished from alcohol by which reagent:

(a) $\mathrm{Na}$

(b) Alcoholic $\mathrm{FeCl} _{3}$

(c) $\mathrm{Br} _{2} / \mathrm{HOH}$

(d) $\mathrm{NaOH}$

Question 15- In the given reaction,

the reaction species is

(a) $\stackrel{+}{\mathrm{C}} \mathrm{HCl} _{2}$

(b) $\ddot{\mathrm{CCl}} _{2}$

(c) $\mathrm{CHCl} _{2}$

(d) $\overline{\mathrm{C}} \mathrm{HCl} _{2}$

Question 16- Phenol reacts with bromine in carbon disulphide at low temperature to give?

(a) m-bromophenol

(b) $0$-and p-bromophenol

(c) p-bromophenol

(d) 2,4,6-tribromophenol

Question 17- Phenol is least reactive for aromatic nucleophilic substitution because:

(a) Carbon-oxygen bond has some double bond character due to resonance.

(b) Oxygen is present in $\mathrm{sp}^{2}$ hybrid carbon which makes carbon-oxygen bond stronger.

(c) Oxygen is highly electronegative which decreases bond length between carbon and oxygen.

(d) All are correct.

Question 18- In the given reaction

$[Y]$ is

Question 19- Phenol reacts with benzenediazonium cation at $\mathrm{pH} 7.5$ to give :

(a) aniline

(b) chlorobenzene

(c) benzene

(d) azo dye

Question 20- When phenol is reacted with $\mathrm{CHCl} _{3}$ and $\mathrm{NaOH}$ followed by acidification, salicylaldehyde is formed. Which of the following species are involved in the above mentioned reaction as intermediates?

Question 21- In the following reaction, the products formed is (are)

(a) $\mathrm{P}$ (major)

(b) $\mathrm{Q}$ (minor)

(c) $\mathrm{R}$ (minor)

(d) $\mathrm{S}$ (major)

Question 22- The major product(s) of the following reaction is (are)

(a) $\mathrm{P}$

(b) $\mathrm{Q}$

(c) $\mathrm{R}$

(d) $\mathrm{S}$

Question 23- In the reaction the products are

Question 24- What are the products of the following reaction? $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{OCH} _{2} \mathrm{CH} _{2} \mathrm{OH} \xrightarrow[\text { Heat }]{\text { excess } \mathrm{HBr}}$ ?

(a) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{OH}+\mathrm{BrCH} _{2} \mathrm{CH} _{2} \mathrm{Br}$

(b) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{OH}+\mathrm{HOCH} _{2} \mathrm{CH} _{2} \mathrm{OH}$

(c) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{Br}+\mathrm{HOCH} _{2} \mathrm{CH} _{2} \mathrm{OH}$

(d) $\mathrm{C} _{6} \mathrm{H} _{5} \mathrm{OH}+\mathrm{BrCH} _{2} \mathrm{CH} _{2} \mathrm{OH}$

Question 25- Decreasing order of reactivity in Williamson’s ether synthesis of the following is

I. $\mathrm{Me} _{3} \mathrm{CCH} _{2} \mathrm{Br}$

II. $\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{CH} _{2} \mathrm{Br}$

III. $\mathrm{CH} _{2}=\mathrm{CHCH} _{2} \mathrm{Cl}$

IV. $\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{CH} _{2} \mathrm{Cl}$

(a) III > II > IV > I

(b) I > II > IV > III

(c) II > III > IV > I

(d) I > III > II > IV

Question 26- In the following reaction, $\mathrm{C} _{2} \mathrm{H} _{5} \mathrm{OC} _{2} \mathrm{H} _{5}+4 \mathrm{H} \xrightarrow{\text { Red } \mathrm{P}+\mathrm{HI}} 2 \mathrm{X}+\mathrm{H} _{2} \mathrm{O}, \mathrm{X}$ is

(a) Ethane

(b) Ethylene

(c) Butane

(d) Propane

Question 27- The product of the reaction

Question 28- In the reaction:

Which of the following compounds will be formed?

Question 29-

(a) $\left(\mathrm{CH} _{3}\right) _{2} \mathrm{CHCH} _{2} \mathrm{OH}$

(b) $\mathrm{CH} _{3} \mathrm{CH} _{2} \mathrm{CH} _{2} \mathrm{OH}$

(c) $\left(\mathrm{CH} _{3}\right) _{3} \mathrm{COH}$

(d) $\mathrm{CH} _{3} \mathrm{CHOHCH} _{2} \mathrm{CH} _{3}$

Question 30- Which of the following is most reactive towards electrophilic attack?

Question 31- tert-Butyl methyl ether on heating with $\mathrm{HI}$ gives a mixture of

(a) tert-Butyl alcohol and methyl iodide

(b) tert-Butyl iodide and methanol

(c) Isobutylene and methyl iodide

(d) Isobutylene and methanol