Just a quick one: Why does cyclobutane have a less heat of combustion than methylcyclopropane?
Because the cyclopropane ring is more strained than the cyclobutane ring? Bayer strain Theory?
I know the strain theory, but I am not sure how to apply it here. My guess: because cyclopropane is less stable, it has a bigger energy content, so more energy is released in its combustion. Then, this could be applied to all isomers that have different stabilities.
Are the δ given?
Since the products of these reactions are the same; look at the stability of the starting material (any reason for destabilization?).
They aren't given. I actually shortened the problem because I thought that the other info won't help to distinguish between these two compounds. Here is the full version:
An unknown non-ionic compound A contains carbon, hydrogen and one more element.
When 1 mole of A reacts with 0.5 mole of HCl, the solution has pH 11.
When 1 mole of A reacts with 1 mole of HCl, the solution is slightly acidic.
When 1 mole of A reacts with MeI, a new compound is obtained. When 1 mole of the new compound reacts with 0.5 mole of HCl, the obtained solution is acidic. In H-NMR, A has 4 signals, one from 3 and one from 2 protons and two signals from 4 protons. Write its structure.
Did I miss something here, so my structure isn't okay, or are both structures concluded from the data given?