Question

Rank the following substances in order of decreasing heat of combustion (maximum – minimum):

A. $1 > 2 > 3 > 4$
B. $3 > 4 > 2 > 1$
C. $2 > 4 > 1 > 3$
D. $1 > 3 > 2 > 4$

Answer 1

Hint: The heat of combustion is inversely proportional to the stability of the ring. Thus,
${\text{Heat of combustion}} \propto \dfrac{1}{{{\text{Stability of the ring}}}}$.

Complete step by step answer:
Step 1:
Determine the order of stability as follows:
The stability of the ring increases as the size of the ring increases. Substance 2 and substance 4 have the same ring size. The size of substance 1 is greater than substance 2 and 4 and the size of substance 3 is the greatest.
Thus, the order of stability is $3 > 1 > 2 = 4$.
The substances 2 and 4 are isomers. The trans isomer is more stable than the cis isomer. This is because in a trans isomer, the substituent groups are attached on the opposite sides of the carbon – carbon bond which decreases the strain in the ring resulting in an increased stability. Thus, the substance 4 (trans isomer) is more stable than substance 2 (cis isomer).
Thus, the order of stability is $3 > 1 > 4 > 2$.
Step 2:
Determine the order of heat of combustion as follows:
The heat of combustion is inversely proportional to the stability of the ring. Thus,
${\text{Heat of combustion}} \propto \dfrac{1}{{{\text{Stability of the ring}}}}$
Thus, the order of heat of combustion is the opposite of the order of the stability of the ring.
Thus, the order of heat of combustion is $2 > 4 > 1 > 3$.
Thus, the correct option is option (C).

Note: In a trans isomer, the substituent groups are attached on the opposite side of the carbon – carbon bond. In a cis isomer, the substituent groups are attached on the same side of the carbon – carbon bond. As a result, the steric hindrance due to the substituent groups decreases in a trans isomer. Thus, the trans isomer is always more stable than the cis isomer.