To begin, I always look for similarities in the products and find the longest carbon chain. In this case, the longest chain contains 5 carbons, pentane. We can also see that there are two methyl groups attached to the carbon chain at carbons 2 and 3 in the top reaction, and both at carbon 2 in the bottom reaction. Immediately this makes me think there will be some sort of methyl shift occurring that we should look out for. And finally, we know that a bromine will be attaching to the carbon chain, but need to consider the reaction conditions to determine where the bromine will be attached.
Okay, now to put all the pieces together and figure this puzzle out. In bottom reaction we see that a strong base is present, therefore a SN2 (one step) reaction is favorable. In the top reaction we see that a weak base (methanol) is present, therefore a SN1 (two step) reaction is favorable.
I used this information to determine my initial guess that bromine will replace the methoxy group currently attached to carbon 3 (lower reaction), resulting in the compound below. Since the lower reaction is an SN2 reaction there will not be a carbocation formed or any rearrangements. I also took this into consideration when forming my initial answer.


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