This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the names for the given compounds: I) Step 1: Identify the functional group. The compound contains a carboxyl group (-COOH), so it is a carboxylic acid. The carbon of the carboxyl group is C1. Step 2: Identify the longest carbon chain that includes the carboxyl carbon. Numbering starts from the carboxyl carbon (C1). The main chain is: CH_3CH(Br)CH_2-COOH. This chain has 4 carbons, making the parent acid butanoic acid. Step 3: Identify and number the substituents. There is a bromine atom (Br) on C3. Step 4: Combine the information to form the IUPAC name. The name is 3-bromobutanoic acid. The name of compound (I) is: 3-bromobutanoic acid II) Step 1: Identify the functional group. The compound contains a carboxyl group (-COOH), so it is a carboxylic acid. The carbon of the carboxyl group is C1. Step 2: Identify the longest carbon chain that includes the carboxyl carbon. Numbering starts from the carboxyl carbon (C1). The main chain is: CH_3CH_2CH(CH_3)-COOH. This chain has 4 carbons, making the parent acid butanoic acid. Step 3: Identify and number the substituents. There is a methyl group (CH_3) on C2. Step 4: Combine the information to form the IUPAC name. The name is 2-methylbutanoic acid. The name of compound (II) is: 2-methylbutanoic acid III) Step 1: Identify the functional group. The compound contains a carboxyl group (-COOH), so it is a carboxylic acid. The carbon of the carboxyl group is C1. Step 2: Identify the longest carbon chain that includes the carboxyl carbon. Numbering starts from the carboxyl carbon (C1). The structure is: c CH_3 \\ | \\ CH_3-CH-CH_2-CH-COOH \\ | \\ CH_2CH_2CH_3 The longest continuous carbon chain that includes the carboxyl carbon (C1) is 5 carbons long: C1(COOH)-C2(CH)-C3(CH_2)-C4(CH)-C5(CH_3). This makes the parent acid pentanoic acid. Step 3: Identify and number the substituents. There is a propyl group (CH_2CH_2CH_3) on C2. There is a methyl group (CH_3) on C4. Step 4: Combine the information to form the IUPAC name, listing substituents alphabetically. The name is 4-methyl-2-propylpentanoic acid. The name of compound (III) is: 4-methyl-2-propylpentanoic acid That's 2 down. 3 left today — send the next one.