What is the product of reacting #"Br"_2# in methanol with methylenecyclohexane?

For determining the product in these type of questions, we have to look at the functional groups of the reactants. In this case, we have a #Br_2# molecule and a cyclohexane ring with a double bond side group. As we know, double bonds are pretty reactive, so we expect something to happen there.

Now the #Br_2# molecule can line up with this double bond, creating partial charges. This is indicated in the figure below. Because of the partial positive charge (created on the side of the double bond), the electrons will attack there and create a cyclic structure with one of the #Br# atoms.

The other #Br# will be released.

Now we add #CH_3OH#. The free electrons on the #O# can react with the structure. This is showed in the mechanism below.

The carbon atom of the cyclic bromine group is partial positively charged, as indicated in the image above. This is created by hyperconjugation. This carbon atom is attached to 3 other carbon atoms (tertiary carbon atom), which stabilises a charge on the carbon atom more easily. Therefore the electrons from methanol will attack at that carbon atom, 'pushing away' the electrons from the bromine bond.

In the last step, the #H# atom will be released. This causes the oxygen to be neutral instead of positively charged.

And there you have your product!