Two stereoisomers of 1-bromo-4-methylcyclohexane are formed when trans-4-methylcyclohexanol reacts with hydrogen bromide. Write structural formulas of:
(c) The two stereoisomers of 1-bromo-4-methylcyclohexane
Hey everyone, so now we're at Part C and we already saw how we can show our trans 4-methyl cyclohexanol up top where we have our methyl group on wedge, the OH on the dash, we then show that we have a carbocation intermediate but now we need to show how we can have 2 different stereoisomers isomers of our 1-bromo 4-methyl cyclohexane, so I want you to try and think what's going to happen after we now react with our br negative, what you expect is that it's going to come in and attack our carbocation, okay? and it only happen once, I'm just showing you it in a six membered ring and of course the chair, so the stereoisomers that is talking about is, let's start with our chair conformation, is if we redraw this, you can draw two of them that look like this, we're now going to have our methyl group here, in both, but now our bromine could either attack from the top or from the bottom, so when our planar structure is easy to see that our stereoisomers would be what you guys think? we have a methyl group here, we could have a bromine also on a wedge, or excuse me, yeah a wedge or a bromine on a dash and we can illustrate that the same over here, so they can either be cis or trans to each other, so we would have one bromine here and another bromine right here, okay? So this would be the one that they're trans to each other just like this one and the other one would be the one where they're cis each other. So, just like this one, okay? So, just remember that when we have that carbocation our bromine can either choose to attack from the top or from the bottom and that's how we come up with the 2 different stereoisomers, okay? So I hope this made sense and let me know if you have any questions on anything did.