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Chapter 24: Problem 50

Draw the condensed structures of the compounds formed from (a) butanoic acid and methanol, (b) benzoic acid and 2-propanol, (c) propanoic acid and dimethylamine. Name the compound in each case.

Short Answer

Expert verified
The condensed structures and names of the compounds formed are: (a) Methyl butanoate: \(CH_3CH_2CH_2COOCH_3\) (b) Isopropyl benzoate: \(C_6H_5COOCH(CH_3)_2\) (c) N,N-dimethylpropanamide: \(CH_3CH_2CON(CH_3)_2\)

Step by step solution

01

(a) Reaction of butanoic acid and methanol

: 1. Write down the reactants: \(CH_3CH_2CH_2COOH\) (butanoic acid) and \(CH_3OH\) (methanol). 2. Identify the functional group undergoing reaction: The carboxylic acid group (-COOH) in butanoic acid and the hydroxyl group (-OH) in methanol. 3. Form the ester by removing a water molecule: The carboxylic acid and the hydroxyl group will lose a water molecule (\(H_2O\)), resulting in a bond between the remaining oxygen and carbon. 4. Draw the condensed structure of the ester: The final structure is \(CH_3CH_2CH_2COOCH_3\). 5. Name the compound: The IUPAC name of the ester is "methyl butanoate."
02

(b) Reaction of benzoic acid and 2-propanol

: 1. Write down the reactants: \(C_6H_5COOH\) (benzoic acid) and \(CH_3CH(OH)CH_3\) (2-propanol). 2. Identify the functional group undergoing reaction: The carboxylic acid group (-COOH) in benzoic acid and the hydroxyl group (-OH) in 2-propanol. 3. Form the ester by removing a water molecule: The carboxylic acid and the hydroxyl group will lose a water molecule (\(H_2O\)), resulting in a bond between the remaining oxygen and carbon. 4. Draw the condensed structure of the ester: The final structure is \(C_6H_5COOCH(CH_3)_2\). 5. Name the compound: The IUPAC name of the ester is "isopropyl benzoate."
03

(c) Reaction of propanoic acid and dimethylamine

: 1. Write down the reactants: \(CH_3CH_2COOH\) (propanoic acid) and \(H_3CN(CH_3)_2\) (dimethylamine). 2. Identify the functional group undergoing reaction: The carboxylic acid group (-COOH) in propanoic acid and the amine group (-NH) in dimethylamine. 3. Form the amide by removing a water molecule: The carboxylic acid and the amine group will lose a water molecule (\(H_2O\)), resulting in a bond between the remaining nitrogen and carbon. 4. Draw the condensed structure of the amide: The final structure is \(CH_3CH_2CON(CH_3)_2\). 5. Name the compound: The IUPAC name of the amide is "N,N-dimethylpropanamide."

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