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Unit 23: Carbonyl Compounds and Carboxylic Acids — Short Questions

12th Class Chemistry · Unit 23: Carbonyl Compounds and Carboxylic Acids

Q2.a.How is ethanol converted to acetic acid?

Ethanol is oxidized by refluxing with excess acidified K₂Cr₂O₇ (or acidified KMnO₄). Since excess oxidizing agent is used and the product is not removed, oxidation proceeds all the way to the carboxylic acid:
CH₃CH₂OH + 2[O] →(K₂Cr₂O₇/dil. H₂SO₄, Heat/reflux)→ CH₃COOH + H₂O

Q2.b.Give the mechanism for the reductions of ethanal with NaBH₄.

NaBH₄ supplies a hydride ion (H⁻), which acts as a nucleophile.
Step 1: H⁻ attacks the electrophilic (δ+) carbonyl carbon of ethanal, CH₃–CHO; the π electrons of C=O shift onto oxygen, giving a tetrahedral alkoxide ion, CH₃–CH₂–O⁻.
Step 2: The alkoxide ion is protonated by water/the alcoholic medium (H–OH), giving the alcohol and regenerating OH⁻.
Overall: CH₃CHO + 2[H] →(NaBH₄, H₂O/OH⁻)→ CH₃CH₂OH

Q2.c.What is iodofrom reaction?

The iodoform reaction is a substitution reaction in which aldehydes/ketones containing the CH₃CO– (acetyl) group — i.e. ethanal, or any methyl ketone — react with iodine in the presence of sodium hydroxide to give a yellow crystalline precipitate of triiodomethane (iodoform, CHI₃) and the sodium salt of a carboxylic acid. Since it gives a positive result only for compounds with the CH₃CO– group (and CH₃CH(OH)– alcohols, via oxidation to that group), it is used to identify such compounds.
CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃↓ + CH₃COONa + 3NaI + 3H₂O

Q2.d.Methanoic acid can be oxidized but other carboxylic acids cannot. Why?

Methanoic acid, H–COOH, uniquely has a hydrogen atom directly bonded to the carboxyl carbon, giving it an aldehyde-like (–CHO) portion within its structure in addition to the acidic –COOH portion. This aldehydic H can be lost on oxidation, so methanoic acid behaves as a reducing agent and is readily oxidized (e.g. by Tollens' or Fehling's reagent) to CO₂ and H₂O. Other carboxylic acids (R–COOH, R ≠ H) have no such H on the carboxyl carbon — the carbon is already fully oxidized — so they cannot be oxidized further.

Q2.e.Explain the influence of chlorine substitution on the acidity of carboxylic acids.

Chlorine is electronegative and exerts an electron-withdrawing inductive (–I) effect. When Cl replaces an H on the alkyl chain of a carboxylic acid, it pulls electron density away from the carboxylate ion formed on ionization, dispersing and stabilizing its negative charge. A more stable conjugate base means the acid ionizes more readily, so chlorine substitution increases acidity. The effect is additive: more chlorine atoms give a stronger acid. For example, pKa: ethanoic acid 4.74 > chloroacetic acid 2.86 > dichloroacetic acid 1.26 > trichloroacetic acid 0.64 (lower pKa = stronger acid).

Q2.f.Give the reaction of acyl chloride with phenol.

An acyl chloride reacts with phenol (on warming) by an addition–elimination mechanism to give a phenyl ester and HCl:
CH₃COCl + C₆H₅OH → CH₃COOC₆H₅ + HCl
(ethanoyl chloride) (phenol) (phenyl ethanoate)

Q2.g.Explain why carboxylic acid is more acidic than ethanol.

When a carboxylic acid ionizes, the resulting carboxylate ion has its negative charge delocalized equally over two oxygen atoms by resonance, which greatly stabilizes the ion and makes it reluctant to re-accept a proton. When ethanol ionizes, the ethoxide ion has its negative charge localized entirely on one oxygen atom with no resonance stabilization, so it is far less stable and readily re-accepts a proton. Because the conjugate base of the carboxylic acid is much more stable, carboxylic acids (Ka ≈ 10⁻⁵) are far more acidic than alcohols (Ka ≈ 10⁻¹⁶).

Q2.h.Give reaction of methanal with the Tollen's reagent.

Tollens' reagent, ammoniacal silver nitrate [Ag(NH₃)₂]OH, oxidizes methanal (an aldehyde) to the carboxylate, while the silver(I) ions are reduced to metallic silver, depositing as a silver mirror on the test-tube wall:
HCHO + 2[Ag(NH₃)₂]OH → 2Ag↓ + HCOONH₄ + 3NH₃ + H₂O

Q2.i.Write down structural formulae of the following compounds: i) 2-methylhexanal ii) 2-methyl-3-pentanone iii) 3-hydroxypentanal iv) 4-nitro-2-pentanone v) 2-chlorobutanol
Q3.a.Write down the equation for the reaction of CH₃CH₂-CHO with LiAlH₄ (dry ether).

LiAlH₄ reduces the aldehyde to a primary alcohol:
CH₃CH₂CHO + 2[H] →(LiAlH₄, dry ether)→ CH₃CH₂CH₂OH
(propanal) (propan-1-ol)

Q3.b.An alcohol, C₄H₁₀O undergoes oxidation to form a carboxylic acid, C₄H₈O₂. Give a chemical equation along with conditions.

A primary alcohol is required, since only primary alcohols oxidize all the way to carboxylic acids. Taking butan-1-ol as the C₄H₁₀O alcohol:
CH₃CH₂CH₂CH₂OH + 2[O] →(K₂Cr₂O₇/dil. H₂SO₄, Heat/reflux)→ CH₃CH₂CH₂COOH + H₂O
(butan-1-ol) (butanoic acid, C₄H₈O₂)

Q3.c.Write the structures of all the isomeric alcohols with formula C₄H₁₀O. Give their systematic names. Which one of them forms the above carboxylic acid? Explain why the others do not.
Q3.d.Propanoic acid can be reduced to propan-1-ol by reduction with LiAlH₄. Give a balanced chemical equation. Can we use NaBH₄ for this reaction? If not, give reason.

CH₃CH₂COOH + 4[H] →(LiAlH₄, dry ether)→ CH₃CH₂CH₂OH + H₂O

No, NaBH₄ cannot be used for this reduction. NaBH₄ is a comparatively weak/mild reducing agent that only reduces the more electrophilic carbonyl carbon of aldehydes and ketones; it is not powerful enough to reduce the less electrophilic carbonyl carbon of a carboxylic acid (stabilized by the –OH group). Only the powerful reducing agent LiAlH₄ can reduce carboxylic acids to primary alcohols.

Q6.How does ethanoic acid reacts with: I) NaOH ii) ethanol iii) SOCl₂ iv) PCl₅