Possible explanation:

To solve this problem, we need to consider the general reaction that occurs when a monocarboxylic acid is heated with soda lime (sodium hydroxide and calcium oxide) under reflux:

RCOOH + NaOH → RCOONa + H2O
RCOONa + CaO → RCOOCa + Na2O
RCOOCa + 2H2 → RH + CaCO3

This reaction sequence involves the conversion of the acid to its sodium salt, followed by its thermal decarboxylation to the corresponding hydrocarbon, which is then reduced by hydrogen to remove any functional groups and form an alkane. The carbon chain length of the alkane depends on the number of carbons in the acid, and the position of the methyl group can vary.

To obtain 3-methylpentane, we need an acid that has five carbons and a methyl group on the third carbon. There are several ways to achieve this, as shown by the following examples:

1. Pentanoic acid (CH3(CH2)3COOH) gives 3-methylpentane by losing CO2 and reducing the resulting pentan-3-ol:
CH3(CH2)3COOH + NaOH → CH3(CH2)3COONa + H2O
CH3(CH2)3COONa + CaO → CH3(CH2)3COOCa + Na2O
CH3(CH2)3COOCa + 2H2 → CH3(CH2)3CH3 + CaCO3

2. 4-Methylpentanoic acid (CH3(CH2)2CH(CH3)COOH) gives 3-methylpentane by losing CO2 and reducing the resulting 4-methylpentan-3-ol:
CH3(CH2)2CH(CH3)COOH + NaOH → CH3(CH2)2CH(CH3)COONa + H2O
CH3(CH2)2CH(CH3)COONa + CaO → CH3(CH2)2CH(CH3)COOCa + Na2O
CH3(CH2)2CH(CH3)COOCa + 2H2 → CH3(CH2)2CH(CH3)CH3 + CaCO3

3. 3-Methylbutanoic acid (CH3CH2CH(CH3)COOH) gives 3-methylpentane by losing CO2 and reducing the resulting 3-methylbutan-2-ol:
CH3CH2CH(CH3)COOH + NaOH → CH3CH2CH(CH3)COONa + H2O
CH3CH2CH(CH3)COONa + CaO → CH3CH2CH(CH3)COOCa + Na2O
CH3CH2CH(CH3)COOCa + 2H2 → CH3CH2CH(CH3)CH3 + CaCO3

4. 2-Methylbutanoic acid (CH3CH(CH3)CH2COOH) gives 3-methylpentane by losing CO2 and reducing the resulting 2-methylbutan-3-ol:
CH3CH(CH3)CH2COOH + NaOH → CH3CH(CH3)CH2COONa + H2O

Ans.is 6