Question

The decreasing order of melting points of the following compounds is :

A.) $(i) > (ii) > (iii) > (iv)$
B.) $(iv) > (iii) > (ii) > (i)$
C.) $(ii) > (i) > (iv) > (iii)$
D.) $(iii) > (iv) > (i) > (ii)$

Answer 1

Hint: In this question, we can see that all given compounds are alkanes and for alkanes the melting point is directly proportional to molecular weight of compounds.

Complete step by step answer:
As we know all that the given compounds have only single bonds without any functional group in them so they all are alkanes.
The melting point of alkanes depends on the molecular weight of alkanes. We can say that if a compound has more molecular weight then its melting point is more.
Also we know that molecular weight is dependent on the number of carbon atoms present in that compound that is more the number of carbons in a compound then more will be the melting point.
But here comes a confusion when we see option $(i)$and option $(ii)$. Both options have the same number of carbon that is six carbons but the first option shows cyclohexane and the second shows hexane. But we need to know that cycloalkanes have a higher melting point than corresponding alkanes. This is because of the compact nature of cycloalkanes. Hence, we can say that option $(i)$ (cyclohexane) has more melting point than option $(ii)$ - (hexane).
We can also conclude from above theory that option$(iii)$ – cyclopentane has higher melting point than option $(iv)$ – pentane.
Hence, the order can be explained as:
$(i)$ cyclohexane – (six carbons and cyclic compound)
$(ii)$ hexane – (six carbons)
$(iii)$ cyclopentane – (five carbons and cyclic alkane)
$(iv)$ pentane – (five carbons)
Therefore, the correct decreasing order of melting point is :
$(i) > (ii) > (iii) > (iv)$
Hence, option A.) is the correct option.

Note:
In such types of questions we need to remember that if two alkanes have the same number of carbons then their molecular mass will depend on whether they are cyclic or not. The cyclic one will have a higher melting point.