Select the compound in each of the following pairs that will be converted to the corresponding alkyl bromide more rapidly on being treated with hydrogen bromide. Explain the reason for your choice.
(g) 1-Cyclopentylethanol or 1-ethylcyclopentanol
Hey everyone. So, in this question we;re going to be looking at these two compounds and figure out which one's going to be more reactive towards hydrogen bromide to get the corresponding alkyl bromide, so that means, just think of both of these compounds reacting with HBr and note that you may have been confused on how to draw them because their names almost look like they were incorrect but what they were trying to indicate is that for the first one it says, 1-cyclo penta ethanol, so the route is ethanol, which is this group right here and then coming off our position number one we have a cyclo penta group. Now, for the next one we have cyclopentanol as our group, so that's going to be this right here, and then coming off position one, which is this one, we have an ethyl group, okay? So I hope that cleared things up. Now, if we were able to draw them correctly you would see that we're comparing 2 different alcohol groups, we have one right here coming off of this carbon and we have one right here coming off of this carbon. Now, the alcohol reactivity has a trend that we should definitely commit to memory, we know that tertiary alcohols are going to react the fastest, secondary alcohols will come next and then we have our primary. So, really we're actually comparing 2 different types of alcohols, one of them secondary and the other ones tertiary, which one is which? well this one's secondary, right? Because the two carbons are at this one in this one and the one on the right is actually tertiary because the carbons we're looking at is this one, this one and the ethyl group carbon, okay? So you'd see that our 1-ethyl cyclopentanol group is actually going to be the compound that's going to be more reactive towards our hydrogen bromine, okay? Because its tertiary. Alright, so I hope that makes sense and now let's take a look at our next question.