Monday, December 8, 2014

Sapling Ch 8: Problem 24

My favorite types of problems are ones like above, when you are given the product(s) and asked to find the starting material. In this particular case, not only do we need to find the starting material, but this particular starting compound needs to undergo two different reactions, yielding two different products.

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.

 
To verify that the correct alkyl bromide has been drawn, we need to perform the top reaction as a double check. The drawn alkyl bromide will undergo the top reaction (SN1) forming a carbocation on carbon 3 where the bromine is currently attached. The carbocation will undergo a methyl shift with carbon 2 to achieve greater stability. After the shift, the methoxy group attaches and the resulting ether is formed. So the drawn alkyl bromide satisfies both reactions! :)

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