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Which would have one monochlorination possibility?
2,2-dimethylpropane:
or 2-methylpropane:
It seems I'm missing something because I thought both would only have one monochlorination possibility. (unless I am allowed to substitute the additional H on the 2-methylpropane with a Cl?)
Replacing Hydrogen is the only way how chlorination (or halogenation) works.
So in 2,2-dimethylpropane the central carbon (carbon-number 2) is quaternary, meaning all four of its valencies are satisfied by carbon atoms, which implies that there are no hydrogen atoms which can be substituted by chlorine (or any halogen), whereas all the remaining 4 carbon atoms are identical, so substitution on any one of those will result in the same compound (1-chloro-2,2-dimethylpropane).
Whereas in 2-methylpropane(commonly known as isobutane) the carbon-number 2 is only tertiary meaning only 3 of its valencies are satisfied by carbon-atoms and it has one hydrogen atom on it. So on substitution of this hydrogen atom by chlorine the product is 2-chloro-2-methylpropane and on substitution of any one of the terminal (last) carbon atoms would all result in 1-chloro-2-methylpropane. That is two different products.
“unless I am allowed to substitute the additional H on the 2-methylpropane with a Cl”
Why wouldn't you be? I think that's the key to the answer.
“unless I am allowed to substitute the additional H on the 2-methylpropane with a Cl” I assume your confusion lies on the fact that Cl substitutes preferably the H that bonds with a primary carbon, whereas Br would substitute preferably the H that bonds with a tertiary carbon. This is of course correct and this would be the main product. However, a small amount of the second product will indeed be produced and since the question of the exercise is so clear you would pick 2,2-dimethylpropane as your answer.
