The hydrocarbon $\ce{C17H36}$ can be cracked. Which compound is the least likely to be produced in this reaction?
A: $\ce{C3H8}$
B: $\ce{C4H8}$
C: $\ce{C8H16}$
D: $\ce{C16H34}$
Is it D because cracking's purpose is the breaking of long hydrocarbon chains into smaller ones?
There are two ways to look at this question. One is simple carbon atom counts. Another is the saturation of the products.
Carbon atom counts. The question doesn't specify which of the 24894 isomers of heptadecane is to be cracked. Suppose it is "normal" or n-heptadecane. Cracking usually begins with the homolytic cleavage of a carbon-carbon bond. To make a $\ce{C16}$ product from n-heptadecane by cleaving carbon-carbon bonds, there are only two potential bonds to cleave, one at one end of the molecule, another at the other end. In either case, to obtain a $\ce{C16}$ product, only a single cleavage would be allowed, because further reactions would break more C-C bonds and change the product. Cracking is a high temperature reaction that is entropically driven. The system is increasing the disorder. Imagine rolling four 6-sided dice, and the total comes up 17. Then you take away either one or two of the dice (whether its one or two is random). What is the probability that the remaining dice sum to 16, compared to say 8 or 4?
Saturation. Choices B and C are unsaturated hydrocarbons. Choices A and D are saturated hydrocarbons. Both saturated and unsaturated compounds can be formed by cracking, but most modern cracking processes tend to focus on making unsaturated hydrocarbons.
Thus, primarily because of simple statistical unlikelihood, but also because of saturation considerations, D is the correct answer.
Wikipedia's article on cracking is very helpful. See here.$%edit$
