The hydrocarbon A, with a vapor density of 36 and forming only one monochloro substitution product, corresponds to methyl-cyclohexane. Here option D is correct.
To determine the structure of hydrocarbon A based on its vapor density (36) and its ability to form only one monochloro substitution product, we need to consider the possible structures and their corresponding vapor densities.
The vapor density of a hydrocarbon is related to its molecular weight and is given by the ratio of the molecular weight to the molar mass of hydrogen (2 g/mol). For hydrocarbons containing only carbon and hydrogen, the vapor density is often a multiple of 12.
Analyze the options:
A. Iso-pentane: C5H12 (vapor density = 36/3 = 12)
B. Neo-butane: C4H10 (vapor density = 36/2 = 18)
C. Cyclopentane: C5H10 (vapor density = 36/2 = 18)
D. Methyl-cyclohexane: C7H14 (vapor density = 36/3 = 12)
Considering the vapor density requirement of 36, we can eliminate options B and C, which have vapor densities of 18. Now, comparing A and D, both have vapor densities of 12, so we need to consider the fact that A forms only one monochloro substitution product.
The only hydrocarbon from the remaining options that forms only one monochloro substitution product is methyl-cyclohexane (Option D).
Complete question:
A hydrocarbon A (vapor denidy = 36) forms only one monochloro substitution product. A will be:
A - Iso-pantane
B - neo-butane
C - cyclopentane
D - methyl-cyclohexane