Optically active 2-bromobutane undergoes racemization on treat...

Question

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

Accepted Answer

Welcome back everyone. S. three Bromo but one is optically active but loses its optical activity. When treated with a solution of potassium bromide. We need to explain using a mechanism. Our first step is to draw out our structure of S. Three Bmo. But one in so our ending butte one in tells us that we have a four carbon chain. So we have 1234. Where at carbon number one. The in ending tells us we have our double bond. Next we have at carbon # three which will say is here a Bromo group. Now we need to recognize that we have a Cairo center here. We know that by assigning priority based on atomic mass and because bromine has the highest atomic mass, we would assign it to be on a wedge where we also want to draw in the implied hydrogen at this location. Since carbon should have four bonds, it would be bonded to this carbon, this bromine, this carbon and then this hydrogen for its fourth bond. We want the hydrogen to be on a dash Because it's our lowest priority group. So numbering priority based on atomic mass roaming would be priority number one Ethel would be priority # two. Metal priority number three and hydrogen priority number four And so following our priority numbers one through four, we have a counterclockwise rotation which confirms that we definitely do have us as our stereo chemistry. And so we know we have the proper structure. The prompt tells us that this is reacting with a treatment of potassium bromide. Recall that potassium bromide is an ionic compound an assault. And so it's going to dissociate into our bromide. An ion which will actually be our strong nuclear file in this case. The next thing we want to make note of is that our substrate, which is our s. three is an alcohol. Hey lead we know that because we have a broom a group which is a halogen attached to our alcohol chain in which the carbon that this promo group is attached to is surrounded by 12 other carbon atoms. So this is particularly a secondary alcohol. Hey lead. And so we have the potential of either undergoing an SN one or sN two mechanism. So how do we know which one will follow? We should note that because we have the presence of this double bond when our leaving group roaming leaves and we know that it will leave because bruning is weakly basic since it's the an ion of our strong base hydroponic acid, it's a good leaving group. And when romaine leaves our structure it's going to leave behind a car, bo katan in which because it's next to this double bond, we have the potential of resonance stabilization. And so we would say that due to our Sorry, this says residents down here due to our residents stabilized carbo cat ion. We would therefore undergo an sN one mechanism. So this is the mechanism we will show and show. And so we'll begin by depicting our leaving group which exits our structure. Since it's a good leaving group it leaves on its own which will result in the following structure where we have this, we have our double bond and a carbo Catalan at this position and this carbo Catalan is surrounded by two other carbon atoms, meaning that this is a secondary carbo catalon which is residents stabilized because our carbo Catalan is a lilic being that it's next to this double bond. And so this double bond will be able to shift over to this position, creating this second structure where we have a carbon copy on now at this primary position and a double bond in this position and here again we have a primary carbo can ion. Note that we also have our bromide, an ion in solution as our nuclear file. So let's not label that again, since we labeled it earlier. But being our nuclear foul in an SN one mechanism, we then have our nuclear filic attack where we should note that are carbo cad ions do not have stereo centers any longer. And so we therefore have planer carbo cat ions, meaning that our nuclear file can attack from either direction on both structures on both of these residents forms. We can have a nuclear filic attack from either the front side or back side and we should add in the resonance stabilization here of this car broken into. So in consideration of our nuclear filic attack, we should note that a secondary carbo catalon is much more stable than a primary carbo catalon, so are an ion or rather our nuclear foul is most likely to attack at the secondary carbo Catalan position again because it's more stable. And as we stated, it can do so in either direction, either a front side or a backside attack. Let's also note that our planer primary carbo katan is unstable and so it will not undergo a nuclear filic attack. And so as a result, because only our secondary Krivo Catalan can be or can undergo the nuclear filic attack. We will have the following two products where we'll have this structure with a roaming group now on a wedge at this position, which brings us back to our s structure being our S and anti armor in this case. And then we will have a second product in which we have bromine on a dash and we still have our double bond here and this is our our an anti armor. And so these two and anti armor products form a race emmick mixture because we'll have 50% of each of these products due to the fact that we had that nuclear filic attack on both sides of our secondary carbo catalon, as we noted above. And so as a result, we can understand that if we were to have had the nuclear filic attack to our planer primary carbo catty on which would be the structure, we would not have been able to have formed racism Ized products since this primary position is not a Cairo center. And so going back to our race, surmised products that form from the attack of our nuclear file giving us us and our and anti immerse. We should recall that a race emmick mixture does not have optical activity or no net optical rotation. Because the rotations of the two and anti MERS cancel one another out. And so because our an anti MERS cancel out optical activity. We can confirm that to explain why our S. Three bromo bute one in loses its optical activity when treated with potassium bromide is due to the formation of a planer carbo catalon, which has no stereo center in this reaction, which results in race um Ization. And we understand that racism. Ization cancels out optical activity. And so this statement highlighted here in yellow would represent our final answer to complete this example, which will correspond to choice be in the multiple choice. So, I hope this made sense. And let us know if you have any questions

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

Key Concepts

Optical Activity

Racemization

Nucleophilic Substitution Mechanism

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