[jubba]

For the question below is the difference Because when the base (LDA) concentration isn't high enough it attacks the carbon of highest degree.
p.s LDA is lithium diisopropyl amide

Question 7 (10 minutes)
When 2-methylcyclohexanone is treated with 1.2 equivalents of lithium diisopropylamide
followed by bromomethane, 2,6-dimethylcyclohexanone is formed. However when 2-
methylcyclohexanone is treated with 0.5 equivalents of lithium diisopropylamide followed by
bromomethane, 2,2-dimethylcyclohexanone is formed.

[plu]

Remember, LDA is this big hulking base.  In high relative concentrations, it will always go for the more kinetically favourable proton.  In your case, another factor that would cause the secondary carbon to be deprotonated is that there are two protons on that carbon versus only one on the tertiary carbon (again, kinetic favourability).

[FeLiXe]

it's probably something about stereochemistry

I would say:
LDA because of its big substituents is only able to deprotonate the secondary C-atom and form the 6-anion. If you take excess LDA the secondary 6-C is completely deprotonated (because LDA is much stronger) and that's it. Add MeBr and get the 2,6-dimethyl-product.

if you only take .5 equivalents, you only deprotonate half of you Methyl-c-hexanone. so you have the best possibilities for an equilibrium. It seems that at least slowly the 6-anion is able to deprotonate the 2-C of another molecule and form it's 2-anion. Apparently the 2-anion is more stable because it's enolate has a higher substituted double bond. (generally carbon-substituted anions would be less stable).
if you wait long enough you'll have an equilibrium with pretty much only the 2-anion. add MeBr and get the 2,2-dimethyl-procuct.

[FeLiXe]

I'm sorry if I said too much, but I am happy that I found it out

[jubba]

i've been thinking about the question myself and this is my own idea.
The two equilibrium's we have to consider are the deprotonation of the cyclohexanone and the breaking of the MeBr

Cyclohexanone will be easily deprotonated and the equilibrium position will lie in the product side at when the concentration is greater than 1eq a second carbon can be deprotonated so at equilibrium there will be a significant amount where the carbon (2,6 anion will form)

The MeBr bond will break slowly as the Me cation is not stable. So there will always be enought (2,6 cyclohexane anion). The Me will bond to the form 2,2 dimethyl cyclohexane as it sterically favourable.

[jubba]

can anyone tell me if im right