So I have 2-methyl-1,3-cyclohexanone and I react it with LDA in THF. When working with LDA typically the trend is that a proton is removed at the less substituted carbon. However one of the answers in the my text says that the proton will be removed at the more substitued 2 carbon of the ring, where the methyl group is also bonded. Does this occur because the protons at that carbon are very acidic? Just wanted some conformation as this goes against my understanding of LDA/THF.
Thanks!