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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 from the given reactions are: (a) Butanoic acid and methanol: Product: Methyl butanoate, CH3CH2CH2COOCH3 (b) Benzoic acid and 2-propanol: Product: Isopropyl benzoate, C6H5COOCH(CH3)2 (c) Propanoic acid and dimethylamine: Product: N,N-dimethylpropanamide, CH3CH2CON(CH3)2

Step by step solution

01

(a) Butanoic acid and methanol reaction

1. Identify reactants and product type: Butanoic acid and methanol are reacting together to form an ester. 2. Draw the structures of the reactants: - Butanoic acid: CH3CH2CH2COOH - Methanol: CH3OH 3. Connect reactants to form the ester: The ester is formed by the reaction of the carboxylic acid and the alcohol, with the loss of water. The oxygen of the alcohol connects to the carbonyl carbon of the carboxylic acid. 4. Draw the condensed structure of the product: The product formed is CH3CH2CH2COOCH3 5. Name the compound: The IUPAC name for this compound is methyl butanoate.
02

(b) Benzoic acid and 2-propanol reaction

1. Identify reactants and product type: Benzoic acid and 2-propanol are reacting together to form an ester. 2. Draw the structures of the reactants: - Benzoic acid: C6H5COOH - 2-propanol: CH3CHOHCH3 3. Connect reactants to form the ester: The ester is formed by the reaction of the carboxylic acid and the alcohol, with the loss of water. The oxygen of the alcohol connects to the carbonyl carbon of the carboxylic acid. 4. Draw the condensed structure of the product: The product formed is C6H5COOCH(CH3)2 5. Name the compound: The IUPAC name for this compound is isopropyl benzoate.
03

(c) Propanoic acid and dimethylamine reaction

1. Identify reactants and product type: Propanoic acid and dimethylamine are reacting together to form an amide. 2. Draw the structures of the reactants: - Propanoic acid: CH3CH2COOH - Dimethylamine: (CH3)2NH 3. Connect reactants to form the amide: The amide is formed by the reaction of the carboxylic acid and the amine, with the loss of water. The nitrogen of the amine connects to the carbonyl carbon of the carboxylic acid. 4. Draw the condensed structure of the product: The product formed is CH3CH2CON(CH3)2 5. Name the compound: The IUPAC name for this compound is N,N-dimethylpropanamide.

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