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Difficulty: Very hardAlkanoic Acids, Esterification, Saponification, Fats, and Oils

An organic compound PP with the molecular formula C4H8O2\text{C}_4\text{H}_8\text{O}_2 reacts with aqueous sodium hydroxide upon heating to produce a sodium salt QQ and an alkanol RR. Complete oxidation of alkanol RR with acidified potassium dichromate(VI) yields an alkanoic acid identical to the acid produced when salt QQ is acidified with dilute hydrochloric acid. What is the IUPAC name of compound PP?

  1. A
    Methyl propanoate
  2. Ethyl ethanoateAnswer
  3. C
    Propyl methanoate
  4. D
    Butanoic acid

Answer

Ethyl ethanoate
Ethyl ethanoate has the molecular formula C4H8O2\text{C}_4\text{H}_8\text{O}_2. Alkaline hydrolysis of ethyl ethanoate with sodium hydroxide yields sodium ethanoate (salt QQ) and ethanol (alkanol RR). Acidification of sodium ethanoate yields ethanoic acid (2 carbons). Complete oxidation of ethanol with acidified potassium dichromate(VI) also yields ethanoic acid (2 carbons). Because both pathways yield the exact same acid (ethanoic acid), ethyl ethanoate satisfies all conditions.

Step-by-Step Solution

1
Determine the functional group of compound P.
Compound P (C4H8O2\text{C}_4\text{H}_8\text{O}_2) reacts with NaOH\text{NaOH} to yield a salt and an alkanol, identifying PP as an ester with general formula R1COOR2\text{R}^1\text{COOR}^2.
Alkaline hydrolysis (saponification) of esters produces a carboxylate salt and an alkanol.
2
Analyze the carbon distribution from the reaction products.
Acidifying salt QQ (R1COONa\text{R}^1\text{COONa}) gives alkanoic acid R1COOH\text{R}^1\text{COOH} (containing n1+1n_1 + 1 carbon atoms). Oxidation of primary alkanol RR (R2OH\text{R}^2\text{OH}) gives alkanoic acid R′COOH\text{R}'\text{COOH} (containing n2n_2 carbon atoms).
Primary alkanols undergo complete oxidation with acidified K2Cr2O7\text{K}_2\text{Cr}_2\text{O}_7 to produce alkanoic acids with the same number of carbon atoms as the alkanol.
3
Equate the carbon counts of the two acids formed.
Since both processes yield the identical acid, the acid must have 2 carbon atoms (ethanoic acid). Thus, n1+1=2  ⟹  n1=1n_1 + 1 = 2 \implies n_1 = 1 (methyl group CH3−\text{CH}_3-) and n2=2n_2 = 2 (ethyl group −C2H5-\text{C}_2\text{H}_5).
The total number of carbon atoms in ester PP is 4 (1+1+2=41 + 1 + 2 = 4). Dividing 4 total carbons equally between the acyl and alkoxy portions yields ethanoic acid derivative and ethanol derivative.
4
Deduce the structure and IUPAC name of ester P.
Ester PP is CH3COOCH2CH3\text{CH}_3\text{COOCH}_2\text{CH}_3, which has the IUPAC name ethyl ethanoate.
The IUPAC name of an ester consists of the alkyl group attached to the oxygen followed by the alkanoate chain.

Key Concept

Ester Saponification and Oxidation of Alkanols
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Topics

Alkanoic Acids, Esterification, Saponification, Fats, and Oils

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