Comprehension
Alcohols, phenols and ethers are oxygen-containing organic compounds. Primary alcohols yield aldehydes with mild oxidising agents and carboxylic acids with strong oxidizing agents. Secondary alcohols yield ketones on oxidation while tertiary alcohols are resistant to oxidation. Ethers may be prepared by dehydration of alcohols and Williamson synthesis. The C – O bond in ethers can be cleaved by hydrogen halides. The presence of – OH group in phenols activates the aromatic ring towards electrophilic substitution and directs the incoming group to ortho and para positions due to resonance effect. In presence of NaOH, phenol generates phenoxide ion which is even more reactive than phenol. Thus, in alkaline medium, phenol undergoes Kolbe's reaction.
Question: 1

Name the reagents used in the following reactions :
(i) Oxidation of a primary alcohol to an aldehyde
(ii) Oxidation of a primary alcohol to a carboxylic acid

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Mild oxidant (PCC) → aldehyde; strong oxidant (KMnO₄) → acid.
Updated On: Jun 16, 2026
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Solution and Explanation

Concept: Proteins and nucleic acids are big molecules (polymers) made by joining many small repeating units (monomers). Each has its own kind of building unit. Also, proteins come in two shapes, fibrous and globular, and the difference is mostly about how the chains are arranged.

Step 1: The basic building units
Proteins are built from $\mathrm{\alpha}$-amino acids joined by peptide bonds. Nucleic acids (DNA and RNA) are built from nucleotides joined together.

Step 2: Fibrous proteins
In a fibrous protein, the long polypeptide chains lie side by side, running roughly parallel, and are held firmly together by hydrogen bonds and disulphide bonds. This makes them tough, thread-like and not soluble in water. An example is keratin, found in hair and nails.

Step 3: Globular proteins
In a globular protein, the chain folds and coils up on itself into a compact, roughly ball-like (spherical) shape. These are usually soluble in water. An example is insulin.

Answer: Building units: $\mathrm{\alpha}$-amino acids for proteins, nucleotides for nucleic acids. Fibrous proteins have parallel chains and are tough and water-insoluble; globular proteins are folded into compact spheres and are water-soluble.
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Question: 2

Write the reaction involved in Kolbe's reaction.

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Phenoxide + CO₂ → salicylic acid.
Updated On: Jun 16, 2026
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Solution and Explanation

Concept: A single nucleotide of DNA is built from three parts joined together: a nitrogen base, a sugar, and a phosphate (phosphoric acid). When we hydrolyse it (break it apart with water), it simply falls back into those three pieces.

Step 1: Identify the three parts
For a DNA nucleotide that has thymine, the base is thymine, the sugar in DNA is 2-deoxy-D-ribose, and the third part is phosphoric acid.

Step 2: Write the hydrolysis products
So on complete hydrolysis this nucleotide gives three products: thymine, 2-deoxy-D-ribose and phosphoric acid.

Step 3: The OR part, DNA versus RNA
If the alternative is asked, one clear structural difference is the sugar: DNA contains 2-deoxyribose while RNA contains ribose. A second difference is in the bases: DNA uses thymine, whereas RNA uses uracil in its place.

Answer: (i) Hydrolysis gives thymine, 2-deoxy-D-ribose and phosphoric acid. OR (ii) DNA has 2-deoxyribose sugar and the base thymine, while RNA has ribose sugar and the base uracil.
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Question: 3

(i) Why are tertiary alcohols resistant to oxidation ?
(ii) Write the products of the following reaction : $\mathrm{(CH_3)_3C\text{-}O\text{-}C_2H_5 + HI \rightarrow}$

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3° alcohol has no α-H to remove; ether + HI → tert-butyl iodide + ethanol.
Updated On: Jun 16, 2026
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Solution and Explanation

(i) answer: Tertiary alcohols are hard to oxidise, and the reason is simple. To oxidise an alcohol, the oxidising agent has to pull off a hydrogen atom that sits on the very carbon that carries the $\mathrm{-OH}$ group. In a tertiary alcohol that carbon is already joined to three other carbon atoms, so there is no hydrogen left on it. Since there is no such hydrogen to remove, normal oxidation just cannot happen. The only way to oxidise it is to break a strong carbon to carbon bond, and that needs very harsh conditions. So under ordinary conditions a tertiary alcohol stays as it is and resists oxidation.

(ii) answer: When an ether is broken by HI, the iodide always goes to the side that can form the more stable carbocation. Here one side is the tertiary $\mathrm{(CH_3)_3C}$ group and the other side is the small ethyl group. The tertiary group makes a very stable tertiary carbocation, so the iodide attaches there and the ethyl side leaves as an alcohol. So the products are tert-butyl iodide and ethanol:\[ (CH_3)_3C\text{-}O\text{-}C_2H_5 + HI \rightarrow (CH_3)_3C\text{-}I + C_2H_5OH \]
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