Alcohol - 5 Important questions with Notes | Class 12 chemistry | STEMFACT

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Aldehydes, ketone important questions with answers

🔥 Aldehyde, Ketone Important Questions - Class 12 Boards 2026

Class 12 Chemistry: Aldehydes, Ketones Important Questions for Board 2026

Here are Top 10 Important Questions of Aldehydes, Ketones and Carboxylic Acids. These are most expected for CBSE & Bihar Board 2026 [5 Marks Fixed].

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Q1. What are Aldehydes and Ketones? General Formula?

Ans: Aldehydes contain -CHO group (R-CHO) and Ketones contain >C=O group (R-CO-R'). Both have carbonyl functional group. Example: Formaldehyde (HCHO), Acetone (CH3COCH3).

Q2. Why aldehydes & ketones have higher boiling point than hydrocarbons?

Ans: Due to polar carbonyl group, they have dipole-dipole attraction, which is stronger than Van der Waals forces in non-polar hydrocarbons.

Q3. What is Cannizzaro Reaction?

Ans: Aldehydes having no alpha-H, on treatment with conc. NaOH give alcohol and salt of acid.
2HCHO + NaOH → CH3OH + HCOONa

Q4. What is Aldol Condensation?

Ans: Aldehydes/Ketones with alpha-H in presence of dil. NaOH give β-hydroxy carbonyl compound.
2CH3CHO → CH3CH(OH)CH2CHO (Aldol)

Q5. Tollen's Test - How to distinguish aldehyde and ketone?

Ans: Aldehyde gives silver mirror with Tollen's reagent [Ag(NH3)2]+ OH-, ketone does not react.
Reaction: R-CHO + 2[Ag(NH3)2]+ + 3OH- → RCOO- + 2Ag (Silver Mirror) + 4NH3 + 2H2O
Result: Silver mirror at the inner wall of test tube confirms aldehyde.

Q6. What is Rosenmund Reduction? Give equation.

Ans: It is a method to prepare aldehyde from acid chloride. Acid chloride is reduced with H2 gas in presence of Pd catalyst poisoned with BaSO4.
Equation: R-COCl + H2 --(Pd/BaSO4)--> R-CHO + HCl
Example: CH3COCl + H2 → CH3CHO + HCl

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Q7. What is Stephen Reaction?

Ans: It is a method to prepare aldehyde from alkyl cyanide. Nitrile is reduced with SnCl2 and HCl to aldimine, which on hydrolysis gives aldehyde.
Equation: R-C≡N + SnCl2 + 2HCl → R-CH=NH.HCl → [H2O] → R-CHO
Example: CH3CN → CH3CHO (Acetaldehyde)

Q8. Why carboxylic acid has higher boiling point than aldehyde and ketone?

Ans: Carboxylic acids have -OH group, so they form intermolecular Hydrogen Bonding. They exist as dimer (two molecules joined). Due to this strong H-bonding, more heat is required to break it, hence higher boiling point than aldehydes/ketones which have only dipole-dipole forces.

Q9. What is Hell-Volhard-Zelinsky (HVZ) Reaction?

Ans: Carboxylic acids having alpha-Hydrogen, when treated with Chlorine or Bromine in presence of Red Phosphorus, give alpha-halo carboxylic acid.
Equation: CH3COOH + Cl2 --(Red P)--> ClCH2COOH + HCl
Use: To prepare alpha-halo acids.

Q10. What is Haloform Reaction? How to convert Acetone to Chloroform?

Ans: Compounds containing CH3CO- group (methyl ketones) react with halogen and NaOH to give Haloform (CHX3).
Reaction (Acetone to Chloroform):
CH3COCH3 + 3Cl2 + 4NaOH → CHCl3 + CH3COONa + 3NaCl + 3H2O
CHCl3 is Chloroform. Iodoform test is also an example of this reaction.

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🔥 Frequently Asked Questions - Board 2026

Q1. Aldehyde ka general formula kya hai?

Ans: Aldehyde ka general formula R-CHO hota hai, jaha R = alkyl ya aryl group.

Q2. PCC ka use kya hai?

Ans: PCC (Pyridinium Chlorochromate) ka use Primary alcohol ko bina acid banne diye Aldehyde me badalne ke liye hota hai.

Q3. DIBAL-H kya karta hai?

Ans: DIBAL-H, Ester aur Nitrile ko low temperature (-78°C) par Aldehyde me convert karta hai. Board me isse 2 marks ka question pakka aata hai.

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Preparation of Aldehydes and Ketones: Methods, Reactions and Examples

Aldehydes and ketones are important classes of organic compounds containing the carbonyl group (>C=O). These compounds are widely studied in organic chemistry and are important for Class 12 Chemistry, CBSE, Bihar Board, NEET, and other competitive examinations.

In this article, we will discuss the important methods of preparation of aldehydes and ketones, their chemical equations, examples, and key points for examinations.

Preparation of Aldehydes - Board 2026 Complete Guide | Class 12 Chemistry

1. What Are Aldehydes and Ketones?

Aldehydes

Aldehydes are organic compounds in which the carbonyl group is attached to at least one hydrogen atom. Their general formula is R–CHO.

Example: Ethanal (CH₃CHO).

Ketones

Ketones are organic compounds in which the carbonyl group is attached to two carbon-containing groups. Their general formula is R–CO–R′.

Example: Propanone (CH₃COCH₃), commonly known as acetone.

2. Preparation of Aldehydes

2.1 Oxidation of Primary Alcohols

Primary alcohols can be oxidised to aldehydes under controlled conditions. Pyridinium chlorochromate (PCC) is commonly used to prevent further oxidation to carboxylic acids.

General reaction:

R–CH₂OH + [O] → R–CHO + H₂O

Example:

CH₃CH₂OH + [O] → CH₃CHO + H₂O

Ethanol is oxidised to ethanal under suitable controlled conditions.

Important point: Strong oxidising agents may further oxidise aldehydes to carboxylic acids.

2.2 Dehydrogenation of Primary Alcohols

When primary alcohol vapours are passed over heated copper at about 573 K, they undergo dehydrogenation to form aldehydes.

General reaction:

R–CH₂OH → R–CHO + H₂

Example:

CH₃CH₂OH → CH₃CHO + H₂

Condition: Heated copper, approximately 573 K.

2.3 Rosenmund Reduction

Rosenmund reduction is used to prepare aldehydes from acid chlorides. The acid chloride is reduced using hydrogen in the presence of palladium supported on barium sulfate. The catalyst is poisoned to prevent further reduction of the aldehyde.

General reaction:

R–COCl + H₂ → R–CHO + HCl

Reagents: H₂, Pd/BaSO₄.

Example:

CH₃COCl + H₂ → CH₃CHO + HCl

Exam tip: Rosenmund reduction converts an acid chloride into an aldehyde.

2.4 Stephen Reaction

The Stephen reaction is a method of preparing aldehydes from nitriles. A nitrile is reduced using stannous chloride and hydrochloric acid, followed by hydrolysis of the intermediate.

General reaction:

R–C≡N → R–CHO

Reagents: SnCl₂/HCl, followed by hydrolysis.

Example: Ethanenitrile can be converted into ethanal by the Stephen reaction.

Important point: The carbon atom of the nitrile group becomes the carbonyl carbon of the aldehyde.

2.5 Etard Reaction

The Etard reaction is used to prepare aromatic aldehydes by oxidising a suitable methyl group attached to an aromatic ring. Chromyl chloride (CrO₂Cl₂) is the commonly used reagent.

Example:

Toluene → Benzaldehyde

Reagent: Chromyl chloride (CrO₂Cl₂), followed by hydrolysis.

Exam tip: Etard reaction is associated with the conversion of toluene into benzaldehyde.

2.6 Gattermann–Koch Reaction

The Gattermann–Koch reaction is used to introduce a formyl group (–CHO) into an aromatic ring.

Example:

C₆H₆ + CO → C₆H₅CHO

Reagents: CO, HCl, anhydrous AlCl₃, and CuCl.

Benzene is converted into benzaldehyde under suitable reaction conditions.

2.7 Preparation of Aldehydes from Esters Using DIBAL-H

Diisobutylaluminium hydride (DIBAL-H) can reduce suitable esters to aldehydes under controlled conditions.

General reaction:

R–COOR′ → R–CHO

Reagent: DIBAL-H, generally at low temperature, followed by hydrolysis.

Important point: Careful control of reaction conditions is required to avoid further reduction to an alcohol.

3. Preparation of Ketones

3.1 Oxidation of Secondary Alcohols

Secondary alcohols are oxidised to ketones using suitable oxidising agents.

General reaction:

R–CH(OH)–R′ + [O] → R–CO–R′ + H₂O

Example:

CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O

Propan-2-ol is oxidised to propanone (acetone).

3.2 Dehydrogenation of Secondary Alcohols

When secondary alcohol vapours are passed over heated copper at approximately 573 K, ketones are formed.

General reaction:

R–CH(OH)–R′ → R–CO–R′ + H₂

Example:

CH₃CH(OH)CH₃ → CH₃COCH₃ + H₂

3.3 Ozonolysis of Alkenes

Ozonolysis involves the cleavage of a carbon-carbon double bond using ozone, followed by a suitable work-up. Depending on the structure of the alkene and the work-up conditions, aldehydes and/or ketones may be formed.

Example:

CH₃CH=CHCH₃ → 2 CH₃CHO

But-2-ene gives two molecules of ethanal on reductive ozonolysis.

Another example:

(CH₃)₂C=CH₂ → (CH₃)₂CO + HCHO

2-Methylpropene gives propanone and methanal on reductive ozonolysis.

3.4 Friedel–Crafts Acylation

Friedel–Crafts acylation is used to prepare aromatic ketones by introducing an acyl group (RCO–) into an aromatic ring.

General reaction:

Ar–H + RCOCl → Ar–CO–R

Reagents: An acyl chloride and anhydrous AlCl₃.

Example:

C₆H₆ + CH₃COCl → C₆H₅COCH₃

Benzene reacts with acetyl chloride to form acetophenone under suitable conditions.

3.5 Hydration of Alkynes

Hydration of alkynes in the presence of suitable catalysts generally produces carbonyl compounds after tautomerisation of the initially formed enol.

Example: Hydration of ethyne

HC≡CH + H₂O → CH₃CHO

In the presence of HgSO₄ and dilute H₂SO₄, ethyne forms ethanal.

Example: Hydration of propyne

CH₃C≡CH + H₂O → CH₃COCH₃

Under the usual mercury(II)-catalysed acidic conditions, propyne forms propanone.

3.6 Reaction of Acid Chlorides with Dialkylcadmium

Acid chlorides react with dialkylcadmium reagents to form ketones.

General reaction:

2 RCOCl + R′₂Cd → 2 R–CO–R′ + CdCl₂

Important point: This is a classical method of preparing ketones. Organocadmium compounds are highly toxic and are not suitable for routine laboratory use.

4. Difference Between Preparation of Aldehydes and Ketones

Method	Aldehydes	Ketones
Oxidation of alcohols	From primary alcohols under controlled conditions	From secondary alcohols
Dehydrogenation	Primary alcohols over heated copper	Secondary alcohols over heated copper
Reduction of acid chlorides	Rosenmund reduction	Not the usual Rosenmund product
Ozonolysis of alkenes	May be formed, depending on alkene structure	May be formed, depending on alkene structure
Aromatic compounds	Etard and Gattermann–Koch reactions	Friedel–Crafts acylation
Hydration of alkynes	Ethyne produces ethanal	Many other terminal alkynes produce methyl ketones under HgSO4/H2SO4 conditions

5. Important Named Reactions for Examinations

• Rosenmund reduction: Acid chloride to aldehyde.
• Stephen reaction: Nitrile to aldehyde.
• Etard reaction: Toluene to benzaldehyde.
• Gattermann–Koch reaction: Introduction of the –CHO group into an aromatic ring.
• Friedel–Crafts acylation: Preparation of aromatic ketones.
• Ozonolysis: Cleavage of a carbon-carbon double bond to form carbonyl compounds under suitable work-up conditions.

6. Important MCQs on Preparation of Aldehydes and Ketones

Question 1

Which reagent is commonly used in Rosenmund reduction?

• A. NaBH₄
• B. H₂/Pd-BaSO₄
• C. LiAlH₄
• D. KMnO₄

Correct answer: B. H₂/Pd-BaSO₄

Question 2

Which compound is obtained by oxidation of propan-2-ol?

• A. Propanal
• B. Propanoic acid
• C. Propanone
• D. Ethanal

Correct answer: C. Propanone

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Question 3

Which reaction converts toluene into benzaldehyde?

• A. Cannizzaro reaction
• B. Etard reaction
• C. Aldol condensation
• D. Haloform reaction

Correct answer: B. Etard reaction

Question 4

Which reaction is used to prepare aromatic ketones from benzene and an acyl chloride?

• A. Wurtz reaction
• B. Stephen reaction
• C. Friedel–Crafts acylation
• D. Rosenmund reduction

Correct answer: C. Friedel–Crafts acylation

Question 5

What is the product of oxidation of a primary alcohol under controlled conditions?

• A. Ketone
• B. Aldehyde
• C. Ether only
• D. Alkene only

Correct answer: B. Aldehyde

7. Conclusion

The preparation of aldehydes and ketones is an important topic in organic chemistry. Students should understand the reactions of primary and secondary alcohols, Rosenmund reduction, Stephen reaction, Etard reaction, Gattermann–Koch reaction, Friedel–Crafts acylation, and ozonolysis.

Learning the reagents, reaction conditions, and products of these reactions will help students solve organic chemistry problems in board examinations and competitive tests such as NEET.

Revision tip: Practise writing every general reaction and one example for each named reaction to improve your understanding and recall.

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Ethers: Classification, Preparation and Methods of Preparation

Ethers are organic compounds in which an oxygen atom is bonded to two alkyl or aryl groups. The general structure of an ether is represented as R–O–R′.

What are Ethers?

In ethers, the oxygen atom is connected to two carbon-containing groups. These groups may be alkyl groups, aryl groups, or one alkyl and one aryl group.

General formula: R–O–R′

Examples:

• CH₃–O–CH₃ — Dimethyl ether
• C₂H₅–O–C₂H₅ — Diethyl ether
• CH₃–O–C₂H₅ — Methyl ethyl ether
• C₆H₅–O–CH₃ — Anisole

Classification of Ethers

1. Simple or Symmetrical Ethers

In symmetrical ethers, the two groups attached to the oxygen atom are identical.

General structure: R–O–R

Example:

C₂H₅–O–C₂H₅ — Diethyl ether

2. Mixed or Unsymmetrical Ethers

In unsymmetrical ethers, the two groups attached to oxygen are different.

General structure: R–O–R′

Example:

CH₃–O–C₂H₅ — Methyl ethyl ether

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3. Aromatic Ethers

Aromatic ethers contain an aryl group attached to the oxygen atom.

Example:

C₆H₅–O–CH₃ — Anisole

Methods of Preparation of Ethers

1. Williamson Ether Synthesis

Williamson ether synthesis is one of the most important methods for preparing ethers. In this method, a sodium alkoxide reacts with an alkyl halide to form an ether.

General reaction:

R–O⁻Na⁺ + R′–X → R–O–R′ + NaX

For example, sodium ethoxide reacts with bromoethane to produce diethyl ether.

C₂H₅ONa + C₂H₅Br → C₂H₅–O–C₂H₅ + NaBr

Important point: Primary alkyl halides are generally preferred in Williamson synthesis because the reaction proceeds mainly through the S_N2 mechanism.

2. Dehydration of Alcohols

Symmetrical ethers can be prepared by heating a primary alcohol with concentrated sulphuric acid at about 413 K.

General reaction:

2R–OH → R–O–R + H₂O

For example, ethanol gives diethyl ether on heating with concentrated H₂SO₄ at about 413 K.

2C₂H₅OH → C₂H₅–O–C₂H₅ + H₂O

Conditions: Concentrated H₂SO₄, approximately 413 K.

Note: This method is mainly useful for preparing symmetrical ethers from primary alcohols.

3. Preparation of Aromatic Ethers

Aromatic ethers such as anisole can be prepared by Williamson ether synthesis using sodium phenoxide and a suitable alkyl halide.

First, phenol reacts with sodium hydroxide to form sodium phenoxide:

C₆H₅OH + NaOH → C₆H₅ONa + H₂O

The sodium phenoxide then reacts with methyl iodide:

C₆H₅ONa + CH₃I → C₆H₅OCH₃ + NaI

The product, C₆H₅OCH₃, is called anisole.

Important Exam Points

• Williamson ether synthesis is an important method for preparing ethers.
• Primary alkyl halides are preferred in Williamson synthesis.
• Secondary and tertiary alkyl halides may undergo elimination reactions instead of giving the desired ether efficiently.
• Primary alcohols can form symmetrical ethers by dehydration at about 413 K.
• At higher temperatures, alcohols may undergo dehydration to form alkenes.
• Anisole is an example of an aromatic ether.

Quick Revision

Method	Reactants	Product
Williamson synthesis	Sodium alkoxide + alkyl halide	Ether
Dehydration of alcohol	Alcohol + conc. H2SO4, 413 K	Symmetrical ether
Aromatic ether preparation	Sodium phenoxide + alkyl halide	Aromatic ether

Frequently Asked Questions

What is the general formula of ethers?

The general structure of ethers is R–O–R′.

What is Williamson ether synthesis?

It is a method of preparing ethers by reacting a sodium alkoxide with an alkyl halide.

How is diethyl ether prepared?

Diethyl ether can be prepared by heating ethanol with concentrated H₂SO₄ at about 413 K.

How is anisole prepared?

Anisole is prepared by reacting sodium phenoxide with methyl iodide.

Conclusion

Ethers are important oxygen-containing organic compounds. The two major methods of preparation are Williamson ether synthesis and dehydration of alcohols. Williamson synthesis is particularly useful for preparing both symmetrical and unsymmetrical ethers, while dehydration of primary alcohols is commonly used for symmetrical ethers.

Quantum Numbers: Definition, Types, Rules and Examples

Quantum numbers are a set of numbers used in quantum mechanics to describe the state of an electron in an atom. They provide important information about the energy level, subshell, orbital and spin of an electron. Every electron in an atom is described by four quantum numbers.

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What Are Quantum Numbers?

In the modern quantum mechanical model of the atom, electrons do not move around the nucleus in fixed circular paths. Instead, an electron is described by a wave function and has a certain probability of being found in a particular region of space. Quantum numbers are used to describe the properties and state of an electron.

There are four quantum numbers:

1. Principal quantum number
2. Azimuthal quantum number
3. Magnetic quantum number
4. Spin quantum number

Four Quantum Numbers at a Glance

Quantum Number	Symbol	Describes	Allowed Values
Principal	n	Shell or energy level	1, 2, 3, 4, ...
Azimuthal	l	Subshell and orbital shape	0 to n - 1
Magnetic	ml	Orientation of orbital	-l to +l
Spin	ms	Spin of electron	+1/2 or -1/2

1. Principal Quantum Number

The principal quantum number is represented by n. It indicates the principal shell or main energy level in which an electron is present.

The possible values of the principal quantum number are:

n = 1, 2, 3, 4, 5, ...

The shells are commonly represented by the following letters:

Principal Quantum Number	Shell
n = 1	K shell
n = 2	L shell
n = 3	M shell
n = 4	N shell

As the value of n increases, the electron generally belongs to a shell farther from the nucleus and has a higher principal energy level.

Example

For an electron present in the fourth shell:

n = 4

2. Azimuthal Quantum Number

The azimuthal quantum number is represented by l. It is also called the angular momentum quantum number. It determines the subshell and is related to the shape of the orbital.

For a particular value of n, the possible values of l are:

l = 0 to n - 1

The values of l correspond to subshells as follows:

l	Subshell	General Shape
0	s	Spherical
1	p	Dumbbell-shaped
2	d	Complex shapes
3	f	More complex shapes

Example

For the third shell, n = 3. Therefore:

l = 0, 1, 2

Thus, the third shell contains the 3s, 3p and 3d subshells.

3. Magnetic Quantum Number

The magnetic quantum number is represented by m_l. It describes the orientation of an orbital with respect to a chosen axis.

For a particular value of l, the possible values of the magnetic quantum number are:

m_l = -l, ..., 0, ..., +l

Therefore, the total number of possible values of m_l is:

Number of orbitals = 2l + 1

For s Subshell

For an s subshell:

l = 0

Therefore:

m_l = 0

Thus, an s subshell contains 1 orbital.

For p Subshell

For a p subshell:

l = 1

Therefore:

m_l = -1, 0, +1

Thus, a p subshell contains 3 orbitals.

For d Subshell

For a d subshell:

l = 2

Therefore:

m_l = -2, -1, 0, +1, +2

Thus, a d subshell contains 5 orbitals.

For f Subshell

For an f subshell:

l = 3

Therefore:

m_l = -3, -2, -1, 0, +1, +2, +3

Thus, an f subshell contains 7 orbitals.

Number of Orbitals in Different Subshells

Subshell	l	Number of Orbitals	Maximum Number of Electrons
s	0	1	2
p	1	3	6
d	2	5	10
f	3	7	14

4. Spin Quantum Number

The spin quantum number is represented by m_s. It describes the spin state of an electron.

An electron can have one of two possible spin quantum numbers:

m_s = +1/2

or

m_s = -1/2

An orbital can accommodate a maximum of two electrons. If two electrons occupy the same orbital, they must have opposite spins.

Relationship Between Quantum Numbers

The four quantum numbers are related to each other. The principal quantum number determines the possible values of the azimuthal quantum number. The azimuthal quantum number determines the possible values of the magnetic quantum number. The spin quantum number describes the spin state of the electron.

The sequence can be remembered as:

Shell → Subshell → Orbital → Spin

n → l → m_l → m_s

Example: Quantum Numbers of a 3p Electron

Consider an electron present in the 3p subshell.

Since the electron is in the third shell:

n = 3

Since it belongs to the p subshell:

l = 1

For l = 1, the possible values of m_l are:

m_l = -1, 0, +1

The possible spin values are:

m_s = +1/2 or -1/2

Therefore, one possible set of quantum numbers for a 3p electron is:

(3, 1, 0, +1/2)

Another possible set is:

(3, 1, +1, -1/2)

Example: Quantum Numbers of a 4d Electron

For a 4d electron:

n = 4

Since d corresponds to l = 2:

l = 2

Therefore:

m_l = -2, -1, 0, +1, +2

The spin quantum number can be:

m_s = +1/2 or -1/2

Hence, one possible set is:

(4, 2, -2, +1/2)

Quantum Numbers and Pauli Exclusion Principle

The Pauli exclusion principle states that no two electrons in the same atom can have the same set of all four quantum numbers.

For example, two electrons can occupy the same orbital because they may have the same values of n, l and m_l, but their spin quantum numbers must be different.

Thus, if one electron has:

(2, 0, 0, +1/2)

the other electron in the same orbital must have:

(2, 0, 0, -1/2)

Important Rules of Quantum Numbers

• The principal quantum number n can have values 1, 2, 3, 4, ...
• The azimuthal quantum number l ranges from 0 to n - 1.
• The magnetic quantum number m_l ranges from -l to +l.
• The spin quantum number m_s can only be +1/2 or -1/2.
• The number of orbitals in a subshell is given by 2l + 1.
• Each orbital can contain a maximum of two electrons.
• No two electrons in an atom can have identical values of all four quantum numbers.

Common Mistakes in Quantum Numbers

• For n = 1, l can only be 0.
• For n = 2, l can be 0 or 1, but not 2.
• For l = 1, m_l cannot be +2.
• The spin quantum number cannot be 0.
• An s subshell has one orbital, not two.
• A p subshell has three orbitals and can accommodate six electrons.
• A d subshell has five orbitals and can accommodate ten electrons.
• An f subshell has seven orbitals and can accommodate fourteen electrons.

Quick Revision Table

Quantity	Symbol	Values
Principal quantum number	n	1, 2, 3, ...
Azimuthal quantum number	l	0 to n - 1
Magnetic quantum number	ml	-l to +l
Spin quantum number	ms	+1/2 or -1/2

Conclusion

Quantum numbers are fundamental to understanding the electronic structure of atoms. The four quantum numbers provide information about an electron's shell, subshell, orbital orientation and spin.

The most important relationships to remember are:

l = 0 to n - 1

m_l = -l to +l

m_s = +1/2 or -1/2

Understanding these rules makes it easier to solve questions related to electronic configuration, orbitals, atomic structure and the Pauli exclusion principle.

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Preparation of Phenol: Methods, Reactions and Mechanisms

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Phenol, also known as hydroxybenzene, is an important aromatic organic compound with the molecular formula C₆H₅OH. It consists of a hydroxyl group (-OH) directly attached to a benzene ring. Phenol is widely used in the manufacture of resins, plastics, dyes, medicines and many other industrial chemicals.

There are several methods for the preparation of phenol. Some methods are particularly important for academic and competitive examinations, while the cumene process is an important industrial method.

1. Preparation of Phenol from Chlorobenzene: Dow Process

Phenol can be prepared from chlorobenzene by heating it with sodium hydroxide under high temperature and pressure. This method is known as the Dow process.

Step 1: Formation of Sodium Phenoxide

Chlorobenzene is treated with aqueous sodium hydroxide at about 623 K and high pressure. The reaction produces sodium phenoxide.

Reaction:

C₆H₅Cl + NaOH → C₆H₅ONa + HCl

Under industrial conditions, the reaction is commonly represented as:

C₆H₅Cl + NaOH → C₆H₅ONa + NaCl

Step 2: Acidification of Sodium Phenoxide

The sodium phenoxide formed in the first step is treated with an acid such as hydrochloric acid. Phenol is obtained as the final product.

C₆H₅ONa + HCl → C₆H₅OH + NaCl

Thus, chlorobenzene is converted into phenol through the intermediate sodium phenoxide.

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2. Preparation of Phenol from Benzene Sulphonic Acid

Phenol can also be prepared from benzene sulphonic acid or its sodium salt. The process involves alkaline fusion followed by acidification.

Step 1: Formation of Sodium Phenoxide

Sodium benzene sulphonate is heated with molten sodium hydroxide. The sulphonate group is replaced by the hydroxyl group, producing sodium phenoxide.

C₆H₅SO₃Na + 2NaOH → C₆H₅ONa + Na₂SO₃ + H₂O

Step 2: Acidification

The sodium phenoxide is then acidified with hydrochloric acid to obtain phenol.

C₆H₅ONa + HCl → C₆H₅OH + NaCl

3. Preparation of Phenol from Benzene Diazonium Chloride

Benzene diazonium chloride is an important intermediate in aromatic chemistry. When it is heated with water, the diazonium group is replaced by a hydroxyl group and phenol is formed. Nitrogen gas is released during the reaction.

Reaction:

C₆H₅N₂⁺Cl^- + H₂O → C₆H₅OH + N₂ + HCl

The reaction is carried out by warming the aqueous solution of benzene diazonium chloride. Because nitrogen gas is evolved, the reaction proceeds readily.

Why is this method important?

This method demonstrates the replacement of the diazonium group by a hydroxyl group. It is an important example of the chemistry of diazonium salts.

4. Industrial Preparation of Phenol: Cumene Process

The cumene process is an important industrial method for manufacturing phenol. It is particularly significant because it produces phenol and acetone simultaneously.

Cumene is also known as isopropylbenzene.

Step 1: Formation of Cumene

Benzene reacts with propene in the presence of an acid catalyst to form cumene.

C₆H₆ + CH₃CH=CH₂ → C₆H₅CH(CH₃)₂

The product is called cumene.

Step 2: Oxidation of Cumene

Cumene is oxidized by air to form cumene hydroperoxide.

Cumene + O₂ → Cumene hydroperoxide

Step 3: Acidic Cleavage

Cumene hydroperoxide undergoes acid-catalysed cleavage to produce phenol and acetone.

Cumene hydroperoxide → Phenol + Acetone

In simplified form:

C₆H₅C(CH₃)₂OOH → C₆H₅OH + CH₃COCH₃

The formation of acetone as a valuable co-product makes the cumene process commercially important.

5. Comparison of Important Methods

Method	Starting Material	Important Intermediate	Final Product
Dow Process	Chlorobenzene	Sodium phenoxide	Phenol
Alkaline Fusion	Sodium benzene sulphonate	Sodium phenoxide	Phenol
Diazonium Salt Hydrolysis	Benzene diazonium chloride	Diazonium salt	Phenol
Cumene Process	Cumene	Cumene hydroperoxide	Phenol + Acetone

6. Important Reactions to Remember

From chlorobenzene:

C₆H₅Cl → C₆H₅ONa → C₆H₅OH

From benzene sulphonic acid:

C₆H₅SO₃Na → C₆H₅ONa → C₆H₅OH

From diazonium salt:

C₆H₅N₂⁺Cl^- → C₆H₅OH

By cumene process:

Cumene → Cumene hydroperoxide → Phenol + Acetone

7. Exam-Oriented Points

• The molecular formula of phenol is C₆H₅OH.
• The Dow process uses chlorobenzene and sodium hydroxide.
• Sodium phenoxide is an important intermediate in the Dow process.
• Acidification of sodium phenoxide gives phenol.
• Benzene diazonium chloride gives phenol on warming with water.
• The cumene process is an important industrial method for phenol preparation.
• The cumene process produces acetone as a valuable co-product.
• Cumene is also called isopropylbenzene.

Conclusion

Phenol can be prepared by several chemical routes. The preparation from chlorobenzene through the Dow process, alkaline fusion of benzene sulphonate, hydrolysis of benzene diazonium chloride, and the industrial cumene process are the most important methods to study. Among these, the cumene process is especially important industrially because it produces both phenol and acetone.

Understanding the starting material, reaction conditions, intermediate and final product makes these preparation methods much easier to remember and apply in organic chemistry problems.

Mechanism of Breaking C–O and O–H Bonds in Alcohols

Alcohols contain two important bonds around the hydroxyl group: the C–O bond and the O–H bond. Depending on the reagent and reaction conditions, either of these bonds can undergo cleavage. Understanding how these bonds break is very important for studying reactions such as the formation of alkoxides, substitution reactions, and dehydration of alcohols.

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1. Structure of an Alcohol

The general structure of an alcohol is:

R–OH

Here, R represents an alkyl group. The hydroxyl group contains two different bonds:

• C–O bond between carbon and oxygen
• O–H bond between oxygen and hydrogen

Both bonds can participate in chemical reactions, but the way in which they break depends strongly on the reaction conditions.

2. Breaking of the O–H Bond

The O–H bond can break when an alcohol reacts with an active metal such as sodium or potassium. In this type of reaction, the alcohol behaves as a weak acid.

Reaction with Sodium

For example:

2ROH + 2Na → 2RONa + H₂↑

The sodium atom reacts with the hydrogen of the hydroxyl group. The bonding electrons of the O–H bond remain with oxygen, producing an alkoxide ion.

The simplified representation is:

R–O–H → R–O⁻ + H⁺

In the actual reaction with sodium, the alkoxide combines with sodium to form sodium alkoxide, while hydrogen gas is released.

Why Does the O–H Bond Break?

Oxygen is more electronegative than hydrogen. Therefore, the O–H bond is polar:

R–O^δ−–H^δ+

The hydrogen carries a partial positive charge, making it possible for the alcohol to lose hydrogen as a proton under suitable conditions.

However, alcohols are weak acids, so their O–H bond does not normally break completely on its own.

3. Breaking of the C–O Bond

The C–O bond can be broken during reactions in which the hydroxyl group is replaced by another group. A common example is the conversion of an alcohol into an alkyl halide.

For example:

ROH + HBr → RBr + H₂O

The important problem in this reaction is that the hydroxyl group itself is a poor leaving group. Therefore, the alcohol is first protonated.

4. Protonation of the Alcohol

The oxygen atom of an alcohol has lone pairs of electrons. It can use one of these lone pairs to accept a proton from an acid.

ROH + H⁺ → ROH₂⁺

The alcohol is therefore converted into a protonated alcohol, commonly represented as R–OH₂^+.

This step is important because it converts the poor leaving group OH⁻ into the much better leaving group H₂O.

5. Why Is OH⁻ a Poor Leaving Group?

A good leaving group is generally a species that can exist relatively comfortably after leaving the molecule. Hydroxide ion is a strong base and is therefore a relatively poor leaving group.

In acidic medium, however, the hydroxyl group is protonated:

R–OH + H⁺ → R–OH₂⁺

Water can then leave:

R–OH₂⁺ → R⁺ + H₂O

Thus, protonation makes C–O bond cleavage much easier under appropriate conditions.

6. C–O Bond Breaking by the SN1 Mechanism

Tertiary alcohols can undergo substitution through an SN1 mechanism under suitable acidic conditions.

The first important step is protonation:

ROH + H⁺ → ROH₂⁺

Water then leaves and a carbocation is formed:

ROH₂⁺ → R⁺ + H₂O

The nucleophile then attacks the carbocation:

R⁺ + Br⁻ → R–Br

Therefore, the overall process can be summarized as:

1. Protonation of the alcohol
2. Loss of water
3. Formation of a carbocation
4. Attack by the nucleophile

7. C–O Bond Breaking by the SN2 Mechanism

Primary alcohols generally do not form stable carbocations. Therefore, under suitable conditions, their conversion into alkyl halides can proceed through an SN2 mechanism.

The alcohol is first activated by protonation. The nucleophile then attacks the carbon atom while the C–O bond breaks in the same concerted step.

In simplified form:

R–OH₂⁺ + Br⁻ → R–Br + H₂O

Unlike SN1, there is no freely existing carbocation intermediate in the SN2 pathway.

8. Comparison Between O–H and C–O Bond Breaking

Feature	O–H Bond Breaking	C–O Bond Breaking
Bond involved	O–H	C–O
Common reaction	Reaction with Na or K	Conversion into alkyl halides
Important species	Alkoxide ion	Carbocation or substitution transition state
Role of acid	Not necessarily required	Often used to activate the –OH group
Leaving group	H+ equivalent	Usually H2O after protonation

9. Important Concept: Activation of the –OH Group

One of the most important concepts in alcohol chemistry is that the hydroxyl group is normally a poor leaving group.

Therefore, before C–O bond cleavage in many acid-catalyzed reactions, the oxygen is protonated:

R–OH → R–OH₂⁺

After protonation, water can leave much more easily:

R–OH₂⁺ → R⁺ + H₂O

This simple idea helps explain many reactions of alcohols, including substitution and dehydration.

10. Summary

Alcohols can undergo cleavage of either the O–H bond or the C–O bond, depending on the reaction conditions.

• O–H bond cleavage: Alcohol can react with active metals such as sodium to form an alkoxide and hydrogen gas.
• C–O bond cleavage: The hydroxyl group is usually activated first by protonation.
• Protonation: Converts the poor leaving group OH⁻ into the better leaving group H₂O.
• SN1 pathway: C–O bond cleavage can produce a carbocation before nucleophilic attack.
• SN2 pathway: Nucleophilic attack and C–O bond breaking occur in the same step.

Thus, understanding whether the O–H bond or the C–O bond is being broken is essential for predicting the mechanism and products of alcohol reactions.

Try MCQ of Alcohols learn more chemical properties of alcohol