Organic Chemistry / Leaving Group Conversions - Using HX
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Solution: Each of the following reactions has been described in the ch...

Question

Each of the following reactions has been described in the chemical literature and involves an organic starting material somewhat more complex than those we have encountered so far. Nevertheless, on the basis of the topics covered in this chapter, you should be able to write the structure of the principal organic product of each reaction. 

Video Transcript

Hey everyone's, let's take a look at this question here we have a benzene with two substituents coming off of it reacting with HBr and heat, so I want you to pay attention to is the two alcohol groups that we have, we have one right here and another one located down here. Now, alcohol is not a good leaving group but we know when we subjected to HX we can convert it into a good leaving group and then our reaction can go forward from then. Now, here these alcohols are, are they primary, secondary or tertiary because that's going to predict what kind of mechanism we're going to undergo. Well, both of these alcohols are in fact primary, this one's primary and this one's primary, because all this looks like is if we draw in without our atoms being shown is we simply have a CH2, CH2, OH and down here another CH2, CH2, OH, right? now using HBr that's going to be a strong acid and what that's going to do is it's going to first protonate both of our alcohols and turn them into good leaving groups, so this one's going to become a water and this one will become a water as well, with a positive charge. Now, we would actually undergo an SN2 reaction because what we have is a primary leaving group, so it's going to have a great back side. if this was a tertiary leaving group we would be more likely to do an SN1 because we form a stable carbocation. So, in our next step all is going to happen if we're simply going to have bromine around and what that's going to do is simply do an SN2 reaction on both of these waters, so we expect to get as our final product if we redraw this, if you would get 2 different substitutions occurring, here you have a CH2, CH2 and then Br, right? And same thing down here, here we would have another CH2, CH2 and then Br, okay? And this right here is going to be our final product, alright? and just remember that this proceeded through an SN2 mechanism. Alright because we had a primary alcohol to begin with. Alright, so I hope this made sense and let me know if you guys have any questions.