Give the expected products of the following reactions. Include...

Question

Give the expected products of the following reactions. Include a protonation step where necessary.

(a) 2,2-dimethyloxirane + isopropylmagnesium bromide

(b) propylene oxide + n-butyllithium

(c) cyclopentyloxirane + ethyllithium

Accepted Answer

Hey everyone and welcome back to another video. In today's lesson, we are going to learn how to solve the following problem which asks us to determine the identity of products obtained in each of the reactions shown below if necessary include a proto nation stop number 12 to the metal oxide rain plus magnesium bromide. Number two propelling oxide plus ethel lithium and number three cycle pencil oxy rain plus third. Beautiful lithium. If we start with the first reaction, we just want to look at the structures of the two regions to to the metal oxide rain, we may draw oxygen rain. That's our three member drink with two carbon atoms and oxygen. And we understand that position, number two will have to method groups. So that would be two to the method. So this is our starting material and according to the problem we are going to use magnesium bromide. Right? So we control ethel magnesium bromide. We understand that this is our grain yard region, right? Because it has magnesium bromide, bonnets and alcohol chain. We call that in greener reactions. We first of all want to make sure that we identified the less substituted position in Oxford rain. Whenever we are thinking about a pox sides in general, graveyard regions attack a lass substituted position. That's basically because growing regions are basic and as we know it up oxides react in basic conditions by a ring opening at a less substituted position. So let's see what's happening. We're essentially having an attack at the less substituted position. We're opening our ring and when we open our ring. Of course we want to use that protein nation stopped just as the problem suggests to form our alcohol. So this is reaction number one and number two would be addition of hydro knee. Um some strong asset that would allow us to perform acidic work up. Now, if you think about our carbon atoms, we may essentially label our carbons in the crack sequence. So that would be carbon number one, number two, number three. And now we are essentially using this as our carbon number four and carbon number five. So what we're going to do here, we're going to draw a five member chain. So that would be 12345. And we understand that in particular, if we look at all of our carbon atoms carbon number 454321. So position number two has two substations, right? It has an alcohol group and a metal group. So let's draw it out. If we choose our position to to be over here, we're going to add a methyl group as well as an alcohol group. So that would be our final alcohol produced in this reaction. Right. So once again, don't forget that we definitely need to use acidic work up to produce the final alcohol because in the first step, when the gradient region attacks were forming an alc oxide and iron, Let's label this is our final product for reaction number one. Looking at reaction number two, we have propylene oxide reacting with ethyl lithium. So let's draw a properly in oxide. It's basically the structure which has once again a three member ring with oxygen. Again, it's in a box side. And we're going to add a metal group over here. So that would be properly an oxide. The reaction takes place with S. A lithium. So we are going to draw an ethyl group Bonner to lithium. This is a very similar reaction. Right? We're using an organic lithium reagent and because this region is once again basic, we're going to attack a less substituted position. Notice that there's an alcohol substation here. So this position is more substituted. So between the two electrical carbon atoms, this is a less substituted position. Once again a less substituted position similarly to what we have for the green region. For organa lithium regions, we have exactly the same regional selectivity. So we're going to attack that carbon atom followed by the loss of the leaving group. And of course we went to protein eight that the alcohol oxides were going to say that reaction number two would be acidic World Cup and then we will form our alcohol If we label the carbon atoms we may start with this as position number 123, four and five. Right? So again we have five carbon atoms And at position number two we now have our alcohol group. Well done. So this is our structure for Reaction # two. And finally we are going to look at reaction number three. For reaction number three, we're going to analyze the reaction between cyclo pencil ox rain. So let's draw ox rain and let's bond a psycho pencil group. So that would be a five member ring. Now this is psycho pencil ox rain. And we're using third beetle lithium now third beetle, lithium essential implies that we have a third beetle group. Let's throw it out. So that would be terra beetle group and then we have lithium bonnet tip. This is our number one. Once again, let's keep that in mind. But we may need to use acidic conditions once again, let's see what's happening here. So, similarly to the previous problem, we understand that this is a basic region. So between the two positions, we're going to choose a less substituted position. Right? Both organa lithium regions and grain yield regions attack that less substituted position because they exhibit basic character. So we're going to perform an attack at that less substituted position, open the ring and we will form our elk oxide, meaning indeed we will need to use acidic conditions. So reaction number two would be addition of an asset to protein eight elk oxide. And if we label our carbon atoms supposing that this is carbon number one, this will be carbon number two, then this would be carbon number three. This be carbon number four. So we may draw cycle. Benzene 1st and let's draw four carbon atoms 123, four, right 1234. Now, position number one will have an alcohol group because there was a loss of oxygen. Now it would be negatively charged to form Alcock side, then it gets protein ated. So we're going to draw old age position number two in particular. If you look at position number two will now have a third video group. So position number two or actually based on our numbering, let's do it another way. We can just say that position number two is bonded to carbon number three and then we have carbon number four. Carbon number number three has two metal substitutions. Right based on our numbering, this would be a more logical way to do that. So that's carbon number 123. Carbon number three has two metal substitue ints. So we may see that basically we're bonding that video group to position number two. You can think about it this way as well and that will be our found product. Let's label this is our final structure for reaction number three. And let's summarize our findings now in each of these reactions, we formed alcohols whether we used the Green yard region or the organ a lithium region, it didn't matter. We had three box sides and in each case we wanted to make sure that we were attacking that less substituted carbon which essentially gave us these three alcohols. We may simply conclude that the correct answer to this multiple choice question is a however, if you have any questions, don't have States, leave your comment down below in the comment section and thank you for watching.

Give the expected products of the following reactions. Include a protonation step where necessary.
(a) 2,2-dimethyloxirane + isopropylmagnesium bromide
(b) propylene oxide + n-butyllithium
(c) cyclopentyloxirane + ethyllithium

Key Concepts

Nucleophilic Addition to Epoxides

Organometallic Reagents

Protonation Steps

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