Alcohols, Phenols and Ethers: Class 12 NCERT Chemistry Guide

Core Concepts: Defining Alcohols, Phenols, and Ethers

Alcohols

Organic compounds formed when a hydrogen atom in an aliphatic hydrocarbon (alkane) is replaced by a hydroxyl (–OH) group. The functional group is -OH attached to a saturated sp³ hybridized carbon atom. Their general formula is R–OH.

Phenols

Organic compounds where a hydroxyl (–OH) group is directly attached to an sp² hybridized carbon atom of an aromatic ring (like a benzene ring). The simplest example is phenol itself, C₆H₅OH.

Ethers

Organic compounds represented by the general formula R–O–R′, where R and R′ can be alkyl or aryl groups. They are considered derivatives of water or alcohols where both hydrogen atoms (in water) or the hydroxyl hydrogen (in alcohol) are replaced by alkyl/aryl groups.

Understanding the Structure and Classification

The properties of alcohols, phenols, and ethers are directly linked to their structures. In alcohols, the oxygen of the –OH group is attached to an sp³ hybridized carbon, and the C–O–H bond angle is approximately 108.9° (in methanol), close to the tetrahedral angle. In phenols, the –OH group is attached to an sp² hybridized carbon of the aromatic ring. The C–O bond length in phenol is shorter than in methanol because of the partial double bond character due to resonance. Ethers have a C–O–C bond angle of about 111.7° in dimethyl ether, which is larger than the H-O-H angle in water due to steric repulsion between the bulky alkyl groups.

Alcohols are classified based on the number of hydroxyl groups (monohydric, dihydric, etc.) and the nature of the carbon atom bonded to the -OH group:

• Primary (1°) Alcohol: The –OH group is attached to a primary carbon (a carbon bonded to only one other carbon), e.g., Ethanol (CH₃CH₂OH).
• Secondary (2°) Alcohol: The –OH group is attached to a secondary carbon (bonded to two other carbons), e.g., Propan-2-ol (CH₃CH(OH)CH₃).
• Tertiary (3°) Alcohol: The –OH group is attached to a tertiary carbon (bonded to three other carbons), e.g., 2-Methylpropan-2-ol ((CH₃)₃COH).

Worked Examples: Mechanism and Nomenclature

• Example 1: Mechanism of Dehydration of Ethanol to Ethene This reaction is an acid-catalyzed elimination. Let's see the mechanism for the conversion of ethanol to ethene at 443 K with concentrated H₂SO₄. Step 1: Protonation of Alcohol The lone pair of electrons on the oxygen atom of ethanol attacks a proton (H⁺) from the acid to form a protonated alcohol (ethyloxonium ion). This is a fast, reversible step. CH₃CH₂–Ö–H + H⁺ ⇌ CH₃CH₂–O⁺H₂ Step 2: Formation of a Carbocation The C–O bond in the protonated alcohol is weak. It breaks heterolytically, and the stable water molecule is eliminated, leaving behind a primary carbocation. This is the slowest step and hence, the rate-determining step of the reaction. CH₃CH₂–O⁺H₂ → CH₃–C⁺H₂ + H₂O * Step 3: Elimination of a Proton A proton is eliminated from the adjacent carbon atom of the carbocation by a base (like HSO₄⁻ or H₂O) to form the alkene, ethene. This regenerates the acid catalyst. CH₂(H)–C⁺H₂ + H₂O → CH₂=CH₂ + H₃O⁺ This mechanism is crucial for understanding elimination reactions in organic chemistry.
• Example 2: IUPAC Nomenclature Let's name the following compound: CH₃–CH(OH)–CH₂–CH(CH₃)–CHO Step 1: Identify the Principal Functional Group and Parent Chain. The principal functional group is the aldehyde (-CHO), which has higher priority than the hydroxyl (-OH) group. The longest carbon chain containing the aldehyde group has 5 carbons. So, the parent alkane is pentane. The suffix will be '-al'. Step 2: Number the Carbon Chain. Numbering starts from the aldehyde carbon, giving it position 1. ¹CHO–²CH(CH₃)–³CH₂–⁴CH(OH)–⁵CH₃ Wait, let's recheck the structure. The question has it as CH₃–CH(OH)–CH₂–CH(CH₃)–CHO. Let's number from the right: ⁵CH₃–⁴CH(OH)–³CH₂–²CH(CH₃)–¹CHO Step 3: Identify and Name the Substituents. We have a hydroxyl group at carbon 4 and a methyl group at carbon 2. - At C-4: Hydroxyl group (named 'hydroxy' as a prefix). - At C-2: Methyl group. Step 4: Assemble the Full Name. List the substituents alphabetically. The name is 4-Hydroxy-2-methylpentanal.

Exam Trap: Acidity of Phenols vs. Alcohols

A very common and high-yield question in board exams is: 'Why are phenols more acidic than alcohols?' Simply stating 'resonance' is not enough for full marks. You must explain the mechanism clearly.

The Correct Explanation:
Acidity is the ability to donate a proton (H⁺). The stability of the conjugate base formed after donating the proton determines the acidic strength.

1. In Alcohols: When an alcohol (R-OH) loses a proton, it forms an alkoxide ion (R-O⁻). The alkyl group (R) has a +I (positive inductive) effect, which means it pushes electron density towards the oxygen atom. This intensifies the negative charge on the oxygen, making the alkoxide ion less stable and less likely to form. Thus, alcohols are weak acids.

1. In Phenols: When phenol (C₆H₅OH) loses a proton, it forms a phenoxide ion (C₆H₅O⁻). The negative charge on the oxygen atom is delocalized over the entire benzene ring through resonance. The charge is spread across the ortho and para positions of the ring. This delocalization stabilizes the phenoxide ion significantly. Because the resulting conjugate base is very stable, phenol has a much greater tendency to donate a proton compared to an alcohol.

To score full marks: Draw the resonance structures of the phenoxide ion to show the delocalization of the negative charge.

Practice Questions with Solutions

• Q: Give the IUPAC name of the compound: CH₃-O-CH(CH₃)₂. A: Step 1: Identify the parent alkane. The longer alkyl chain attached to the oxygen atom is the isopropyl group (propane chain). So, the parent alkane is propane. Step 2: Identify the alkoxy group. The smaller group is CH₃-O-, which is named 'methoxy'. Step 3: Number the parent chain to give the alkoxy group the lowest possible number. Numbering from the end closer to the substituent gives the methoxy group position 2 on the propane chain. Final answer: The IUPAC name is 2-Methoxypropane.
• Q: Arrange the following in increasing order of their boiling points: Propan-1-ol, Butan-1-ol, Butan-2-ol, Pentan-1-ol. A: Step 1: Analyze the intermolecular forces. All given compounds are alcohols and exhibit strong intermolecular hydrogen bonding. Step 2: Relate boiling point to molecular mass. For compounds with similar functional groups, the boiling point increases with an increase in molecular mass due to increased van der Waals forces. Thus, Propan-1-ol < Butan-1-ol/Butan-2-ol < Pentan-1-ol. Step 3: Compare isomers. Among isomeric alcohols (Butan-1-ol and Butan-2-ol), boiling point decreases with an increase in branching. Branching reduces the surface area, leading to weaker van der Waals forces. Butan-1-ol (a straight chain) has a higher boiling point than Butan-2-ol (branched). Final answer: The increasing order of boiling points is: Propan-1-ol < Butan-2-ol < Butan-1-ol < Pentan-1-ol.
• Q: How would you prepare propan-2-ol from propene? A: Step 1: Identify the reaction type. This is the addition of water across a double bond (hydration of an alkene). Step 2: Choose the appropriate reagent and rule. According to Markovnikov's rule, when an unsymmetrical reagent (like H₂O) adds to an unsymmetrical alkene (propene), the negative part of the reagent (OH⁻) attaches to the carbon atom having a lesser number of hydrogen atoms. This is achieved by acid-catalyzed hydration. Step 3: Write the reaction. Propene is passed through dilute sulfuric acid (or water in the presence of an acid catalyst). CH₃-CH=CH₂ + H₂O (in presence of H⁺) → CH₃-CH(OH)-CH₃ Final answer: Propan-2-ol can be prepared by the acid-catalyzed hydration of propene.
• Q: Explain Williamson Synthesis for the preparation of ethers with an example. A: Step 1: Define the reaction. Williamson synthesis is a laboratory method to prepare both symmetrical and unsymmetrical ethers. It involves the reaction of an alkyl halide with a sodium alkoxide or sodium phenoxide. Step 2: Describe the mechanism. The reaction proceeds via an Sₙ2 mechanism where the alkoxide/phenoxide ion acts as a nucleophile and attacks the alkyl halide, displacing the halide ion. Step 3: Provide an example and a key limitation. To prepare methoxyethane, sodium ethoxide is reacted with methyl bromide. C₂H₅O⁻Na⁺ (Sodium ethoxide) + CH₃Br (Methyl bromide) → C₂H₅-O-CH₃ (Methoxyethane) + NaBr. A key limitation is that for good yields, the alkyl halide must be primary. If a tertiary alkyl halide is used, elimination becomes the major reaction pathway, producing an alkene instead of an ether. Final answer: Williamson synthesis is the reaction of a sodium alkoxide with a primary alkyl halide to form an ether via an Sₙ2 pathway. For example, sodium ethoxide reacts with methyl bromide to give methoxyethane.

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