Single Choice

Decreasing order of boiling point of $$I$$ to $$IV$$ follow $$\underset{I}{\text{Methylformate}}$$ $$\underset{II}{\text{Ethylformate}}$$ $$\underset{III}{\text{Iso-propylformate}}$$ $$\underset{IV}{\text{n-Propylformtae}}$$

AI > II > III > IV
BIII > IV > II > I
CIV > III > II > I
Correct Answer
DI > II > IV > III

Solution

As the molecular weight increases, the boiling point increases.
Also branched chain esters have lower boiling point than normal chain esters.
On going from methyl formate to ethyl formate and to npropyl fomate, the number of carbon atoms and the molecular weight increases. Also, isopropyl formate has lower boiling point than n propyl format as it is branched chain ester.
Hence, the decreasing order of boiling point of I to IV is IV > III > II > I.

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