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Chapter 5: Problem 10

An organic compound \(\mathrm{X}\left(\mathrm{MF}: \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{2}\right)\) on acid hydrolysis gave a carboxylic acid \(\mathrm{Y}\) and an alcohol \(\mathrm{Z}\). Oxidation of \(Z\) with chromic acid produced Y. Identify the compound (X).

Short Answer

Expert verified
Oxidation of Z with chromic acid gave Y, and both Y and Z were found to have the same number of carbon atoms. Answer: The structure of the unknown organic compound X is propyl propanoate (C2H5COOCH2CH2CH3).

Step by step solution

01

Identify the functional group of compound X

Since hydrolysis of compound X gives a carboxylic acid Y and an alcohol Z, it indicates that compound X contains an ester functional group. Esters have the general formula RCOOR'. We'll proceed by finding the structure of the ester.
02

Determine the structures of Y and Z

The molecular formula of X is C6H12O2. Upon acid hydrolysis, it forms a carboxylic acid Y and an alcohol Z. The sum of carbon atoms in Y and Z should be equal to 6. Let the number of carbon atoms in Y be m and the number of carbon atoms in Z be n, then m + n = 6. Oxidation of Z with chromic acid gave Y, which means Z should be a primary alcohol (RCH2OH). Also, since the carboxylic acid and alcohol have the same number of carbons, it leads to m = n. Thus, m = n = 3. Now we know that both Y and Z have 3 carbon atoms each. The carboxylic acid Y is then propanoic acid (C2H5COOH) and the alcohol Z is propan-1-ol (C3H7OH).
03

Deduce the structure of compound X

With the knowledge of the structures of Y and Z (propanoic acid and propan-1-ol), we can deduce the structure of the ester compound X. The general structure of an ester is RCOOR', and in our case, RCOO part belongs to propanoic acid which is C2H5COO and R' part belongs to propan-1-ol which is CH2CH2CH3. Therefore, the structure of the ester compound X is C2H5COOCH2CH2CH3 (propyl propanoate).

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Most popular questions from this chapter

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