Why does there need to be a planar intermediate in 2,2'-dibromobiphenyl when one of the phenyl groups rotates? I was in class and my professor brought up the example of 2,2'-dibromobiphenyl:

In her example, one of the bromines was actually pointing into the page and the two phenyl groups were not actually on the same plane of the page. Let's call the phenyl with the bromine that points into the page group 1 and the one that is on the page group 2. She then said that in order for the group 1 phenyl to point out from the page (a conformer) there will be a planar intermediate where the bromines are very close to each other during the rotation (just like in the picture), and this will require a large amount of energy (20 kcal/mol) which will again drop off once the group 1 rotates till it's pointing out of the plane. My question is, why doesn't it just rotate the other way where the bromines will not come into close contact and there will not be a planar intermediate? Am I overthinking this?
I think that your professor was just mentioning a rotational angle vs relative energy plot. It is really important to understand relative energy difference between different points in such plots. The situation when two bromines are in the same plane and close to each other will represent the highest energy spot, this actually means that the probability of the molecule to be in such spot is extremely low, but technically not zero, that is why some reaction that require such spot may happen. In the picture below you can see the plot for biphenyl, in case of the dibromobiphenyl one of the peaks will be much higher then the others. But you are actually right, molecule itself would prefer to rotate the other way.