Question

Correct order for boiling point for the given compound is :
a)1-chloropropane
b)Isopropyl chloride
c)1-chlorobutane
A. $a < b < c$
B. $b > a > c$
C. $b < a < c$
D. $a > b > c$

Answer 1

Hint: In alkyl halides with the same halogen group, the boiling point decreases with an increase in branching of the carbon chain and increases with the increases in the length of the carbon chain of the alkyl group.

Complete step by step answer:
In the given three options mentioned above, 1-chloropropane is an alkyl halide with three carbon atom chain and one chlorine group attached to 1st carbon atom of the alkyl group, Isopropyl chloride which is 2-chloropropane has three carbon atom chain and one chlorine group attached to 2nd carbon atom of the alkyl group and 1-chlorobutane is an alkyl halide with four carbon atom chain and one chlorine group attached to 1st carbon atom of the alkyl group. Now, after discussing the structure of the given isomers, we can conclude that 1-chlorobutane has the longest carbon chain as the other two has three carbon atom chain and 1-chlorobutane has a alkyl group with four carbon atoms chain and hence, has the it has the highest boiling point. Now, Isopropyl chloride which is 2-chloropropane has three carbon atom chain and one chlorine group attached to 2nd carbon atom of the alkyl group which gives branching to the carbon chain and as already discussed, boiling point decreases with an increase in branching of the carbon chain and thus Isopropyl chloride has the lowest boiling point. This discussion brings us to the conclusion that 1-chlorobutane has the highest whereas Isopropyl chloride has the lowest boiling point. As per which, $b < a < c$ is the correct option.

Hence, the option C is correct

Note: In the above question, the compound with longer alkyl group chain and less branching will have the highest boiling point. Isopropyl chloride is also known as 2-propyl chloride and sec-propyl chloride. However, the IUPAC for Isopropyl chloride is 2-chloropropane.