Answer:
The correct answer is c. 3.
Explanation:
To determine the total number of isomeric monocarboxylic acids that, upon decarboxylation, yield n-pentane, we need to consider the different possible structures and arrangements of atoms.
Decarboxylation is the process of removing a carboxyl group (-COOH) from a molecule, resulting in the formation of a smaller molecule. In this case, we are interested in the formation of n-pentane (C5H12) upon decarboxylation.
Monocarboxylic acids have the general formula CnH2n+1COOH, where n represents the number of carbon atoms in the carbon chain. In this case, we are interested in n-pentane, which has a carbon chain of 5.
Now, let's consider the possible structural isomers of monocarboxylic acids with a carbon chain of 5 (n = 5):
1. n-Pentanoic acid: CH3(CH2)3COOH
2. Isopentanoic acid: CH3CH(CH3)CH2COOH
3. Neopentanoic acid: (CH3)4CCOOH
These three structural isomers have different arrangements of atoms and are considered isomers because their chemical structures are different.
Therefore, the total number of isomeric monocarboxylic acids that, upon decarboxylation, give n-pentane is 3.
The correct answer is c. 3.
