(a) The reaction of butan-2-ol with concentrated aqueous HBr&n...

Question

(a) The reaction of butan-2-ol with concentrated aqueous HBr goes with partial racemization, giving more inversion than retention of configuration. Propose a mechanism that accounts for racemization with excess inversion.

Accepted Answer

All right. Hello everyone. So this question says that when Hexane 30 is treated with concentrated HBR solution, the reaction proceeds with partial reim with excess inversion provide the possible mechanism accounting for resim organization resulting in greater inversion than retention of configuration. So here our starting material is hexane three all. That's a six carbon parrot chain with a hydroxy group on carbon number three. So this alcohol is reacting with concentrated H pr solution. Now we have secondary alcohol and it's worth mentioning that bromide coming from H pr is a decently strong nuc file. So without other information to specify the type of reaction that's going to take place here, we actually have two possible mechanisms to consider. One is SN one and the other is SN two. Now, because we do have to consider what's happening to serial chemistry. What's happening to configuration for the purposes of this mechanism were going to use Rhn 30, which means that the hydroxy group on carbon river three is going to be drawn on a wedge though the same mechanism will apply even if we start with S hexane three all. But with that being said, let's go ahead and talk about the two possible mechanisms. The first of which is going to be our SN one mechanism, which will result in 50% inversion of stereo chemistry and 50% retention of configuration. This is because the SN one mechanism is the one that's going to result in a reca mixture. So let's go ahead and talk about how that happens. All right. So here I've gone ahead and redrawn the structure of R hexane 30, so I can go ahead and scroll down. So in the first step of the sn one mechanism, the oxygen of the hydroxy group is first going to be pronated with HBR. So one of the lone pairs of electrons on oxygen is going to take hydrogen from H PR which is going to break the bond between hydrogen and brome, resulting in those electrons going towards bromine to create a bromide ion. Now, in addition to this bromide ion, oxygen is now going to have a positive charge after this protonation step. Now, the reason why this happens is because hydroxy groups are not suitable living groups due to the fact that hydroxide is a strong base. So the pro nation step is going to make oxygen a more suitable leaving group for actual substitution. Now, at this point, because this is the acid one mechanism, our leaving group is going to go ahead and detach itself from carbon. So this creates our carbo cion intermediate, which is precisely why a reca mixture is formed. After the essen one mech mechanism, we also have water as a by-product. Now, after the formation of this carbel cion, the carpal cion itself is P two hybridized, which means that it is planar. So therefore, the next step of the reaction is going to be our nucleo file. In this case, the bromide ion attacking this planar carbel cion intermediate. So here nucleic attack can take place on either face of the carbo cion, which is what results in the reca mixture. So at this point, the hydroxy group is replaced with bromine. And so one half of our products is going to have bromine depicted on a wedge and the other is going to have bromine depicted on a dash. So now we've completed our first mechanism which means we can move on to our second mechanism. Now, the second mechanism is what ultimately explains why there is excess inversion because even though the first mechanism results in a reca mixture with a 5050 ratio of inversion and retention. The second mechanism which is SN two results in 100% inversion of configuration. This explains why there is an overall excess of inversion. So for the SN two mechanism, I'm going to go ahead and re draw our starting material. And our first step is actually going to proceed the same way as the ESSEN one mechanism because the formation of a more suitable leaving group is still going to be necessary before nucleic attack can take place. So first, the hydroxy oxygen is going to be pronated by HBR which is going to generate our nuclei the bromide ion. So in our next step, oxygen is going to have a positive charge which makes it a more suitable living group. However, this time around in the S and two mechanism, the bromide ion which is our nucleo is actually going to attack before the leaving group detaches itself from carbon. This is because the SN two mechanism is concerted, which means that everything is happening at the same time because the leaving group is still attached to carbon at the time that the nucleo attacks. The SN two mechanism is referred to as a backside attack. This results in inversion of stereo chemistry because the Nile is going to attack carbon from the opposite side relative to the leaving group. So the hydroxy group is ultimately still replaced with bromine. But now bromine is always going to be on a dash since our leaving group was on a wedge though we still get water as a by product and there you have it. So here are the two possible mechanisms. The first of which was SN one, which resulted in 50% inversion and 50% retention of configuration. And the second being en two which resulted in 100% inversion of configuration. And so with that being said, thank you so very much for watching And I hope you found this helpful.

(a) The reaction of butan-2-ol with concentrated aqueous HBr goes with partial racemization, giving more inversion than retention of configuration. Propose a mechanism that accounts for racemization with excess inversion.

Key Concepts

Racemization

Nucleophilic Substitution Mechanisms

Carbocation Stability

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