How does the effect of heat and effect of substituents affect racemization of biphenyl compounds?
Preparation: Take the parent compound per se, biphenyl. The bond keeping the two phenyl rings together leads to one carbon atom we call C1, the other C1'. Now if you label the other carbon atoms in each phenyl ring seperately in clockwise direction, you'll get C2, C3, C4, C5, C6 in one cycle; and C2', C3', C4', C5', C6' in the second. (I haven't a credit to upload a picture yet.)
Now: The larger, the more sterically demanding substitutents on C2, C6, C2' and C6' are (condition), the higher the energetical barrier is to interconvert the conformers into each other. Eventually, this may lead to atropisomerism, and the isomers are atropisomers.
Note, please, sometimes there may be already a preference for one or the other conformation. If, for example, you substitute C2 and C2' of biphenyl by nitrogen, free molecules of this compound 2,2'-bipyridine energetically favour a conformer with the nitrogens as far as possible separated (similar to a s-trans conformation). Offering a cation to bind, to complex to, however may lead to a preference towards the s-cis-like conformation.