a. Optically active 2-bromobutane undergoes racemization on tr...

Question

a. Optically active 2-bromobutane undergoes racemization on treatment with a solution of KBr. Give a mechanism for this racemization.

b. In contrast, optically active butan-2-ol does not racemize on treatment with a solution of KOH. Explain why a reaction like that in part (a) does not occur.

Accepted Answer

All right. Hi, everyone. So this question says that when optically active two iodent is treated with sodium iodide solution, it undergoes rim organization show the mechanism for the racim organization. On the other hand, optically active pent 20 does not undergo reim organization when treated with sodium hydroxide solution. Explain why. So first lets go ahead and start out by drawing two Ayo pente, we have a five carbon chain with iodine attached to carbon too. Now KB two is a chiral center and because configuration is not specified here, let's go ahead and use the R configuration in which iodine is drawn on a wedge. Now, when this compound is treated with sodium iodide solution, sodium iodide is going to dissociate and form iodide anions with sodium cat ions. However, it's the iodide anion that's especially important for this discussion because here we have to talk about substitution, specifically nucleic substitution. Recall it reim organization is when the products of the reaction show both retention and inversion of the original stereo chemistry and rim is a consequence of nucleic substitution. So let's talk about some differences between the SN one and the SN two mechanisms. The SN one mechanism is characterized by the formation of a carbo cion intermediate. Because of this leaving groups that are attached to tertiary carbons are best suited for SN one because tertiary carbons make excellent carbo cats. Now, on the other hand, ascent two reactions are a single step process. The nuclear file is going to approach or is going to commit a backside attack on carbon which is going to go ahead and cause the leaving group to leave at the same time. Because of this, we're more concerned with hysteric hindrance when it comes to S and two reactions. That's why methyl and primary carbons that contain leaving groups are best suited for SN two reactions because of how hysterical the unhindered they happen to be. So in this context, we can also distinguish between the role of the Nile S and one reactions tend to favor quote unquote weak nucleo files, oftentimes neutral nucleo files. Whereas SN two reactions tend to favor quote unquote strong nucleo files generally with a negative charge. We can also talk about the role of the solvit because S and one reactions tend to favor polar product solvents like alcohols. Whereas SN two reactions tend to favor polar a product solvents such as acetone or DMSO. So with this in mind, let's consider this the given reaction, our leaving group iodine is attached to a secondary carbon on our starting material. And iodide, our nuclei is considered strong due to its negative charge because of this, this particular substitution reaction is going to take place via the SN two mechanism. So the first thing that happens is that the iodide, an ion is going to go ahead and attack the carbon atom that it contains the leaving group. Because the Nile approaches carbon, while the leaving group is still attached, it's going to do so from the opposite side. Hence the term backside attack, regardless though this causes iodine or N or leaving group, excuse me to go ahead and detach itself. So because this backside attack is so characteristic of the SN two mechanism, the stereo chemistry of the product is going to be inverted, it's going to be opposite that of the starting material. So because the product or the starting material shown is in the R configuration, the product must be in the S configuration. So we have S two iodopen with iodide as a by-product. So even though we're showing here an inversion of stereo chemistry, why is this still considered wasim organization? Well, recall that not all of the R two Ioane is going to be reacted with because of this S two Ioane can undergo substitution again because notice how our nucleo file has been regenerated. So after the S two Ioane undergoes substitution, again, R two iodent is regenerated which still results in a ramic mixture or a mixture of two tumors. So now let's compare this with Penan to all you have the same five carbon chain. But this time, a hydroxy group is on carbon too. And once again, I'm drawing in the R configuration. So the oh group is on a wedge. So the reason why the the active pent to old does not undergo reim organization is because it does not undergo substitution. But why is that well recalled to force substitution to take place. There has to be a suitable leaving group attached to carbon. When the O group detaches itself from carbon, it does. So as a hydroxide ion, because hydroxide is such a strong base, it is not a good leaving group as a result. This is in contrast to something like iodide, which is a weak base and therefore a more suitable leaving group. So because there is no good leaving group present substitution and therefore recreation does not occur in the case of pent tool and there you have it so very briefly, I'm going to pause this recording so we can write out our final answer. All right. So here I've resumed the recording and now we can see both parts of our final answer on the top of the screen is the mechanism for reim organization. And as for the explanation, it reads as follows, optically active pent tool does not undergo reim organization because the hydroxy group is a bad leaving group and there you have it. So with that being said, if you watch this video all the way through, thank you. So very much and I hope you found this helpful.

a. Optically active 2-bromobutane undergoes racemization on treatment with a solution of KBr. Give a mechanism for this racemization.
b. In contrast, optically active butan-2-ol does not racemize on treatment with a solution of KOH. Explain why a reaction like that in part (a) does not occur.

Key Concepts

Racemization

S_N1 Mechanism

S_N2 Mechanism

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