"Methylcyclohexane reacts with molecular bromine (Br2) in the presence of light and heat. The major product obtained from this reaction is reacted with sodium ethoxide in ethanol at 55°C. Draw the skeletal structure of the major organic product resulting from the second reaction."

 

So for the first reaction, I'm assuming that methylcyclohexane with Br2 (and light) will form 1-bromo-1-methylcyclohexane (because there's a tertiary hydrogen). Taking that through the 2nd rxn with yeild 1-ethoxide-1-bromocyclohexane (ethoxide=strong base in protic solvent)

 

Hopefully I am right, i have no way to check the answer. I tried to search but could not find an answer and want to know if my logic is flawed or not. Also, is there going to be an E1 (minor) rxn after adding sodium ethoxide? or not?

 

I appreciate anyones help.

You are correct in saying that the tertiary bromide will form in the presence of light and heat via a radical mechanism. However NaOEt will likely not do a substitution reaction on a tertiary haloalkane, but would instead do an elimination (E2) of HBr by removing a proton beta to the bromide. This would furnish a tri-substituted alkene (1-methyl-cyclohexene).

 

If this was a primary haloalkane, you would likely get substitution unless you had a sterically hindered base such as tBuONa.