Predict the major products of the following reactions.
(...

Question

Predict the major products of the following reactions.

(b)

Accepted Answer

Hey everyone and welcome back. What made your product is expected from the given reaction. Given a reaction involves Esther because we have our our C. O. R group. Right? So we can clearly see that this fragment corresponds and Esther and now the reaction conditions are dies will be the aluminum hydrate at a low temperature negative 78 degrees Celsius, followed by the second reaction, hydrolyzed. Stop addition of water. Now, first of all, let's recall. The di belle age is a reducing agent. So let's label that that's how reducing agent. And now things start to make sense. If this is a reducing agent, we may have a natural question, why don't we use lithium aluminum hydride that we use for reduction of esters? It's basically because if we think about the comparison between the two regions, lithium aluminum hydride is used to reduce esters, two alcohols. But diable age is more selective. It reduces esters to al the heights and we don't reduce the result on al hide further as long as the temperature is low and there's only one equivalent of diable age. So now we understand that the product would be an al hide because our starting material as an extra. But let's look at the rationale behind our assumption or basically this theoretical statement that we wrote right, we want to have some sort of evidence that this is what we expect to get and to do that we are going to look at a partial mechanism of this reaction which is not too complicated. So not don't worry about it. So first of all what we're going to do, we're going to remember that there are two long parents in oxygen and we want to draw the structure of dia bel age. Diable age essential implies that you have aluminum which is bonded to hydrogen and then you have two eyes. A beetle groups. Okay, so you're going to throw three carbon atoms and then branch the chain so that you can actually get your is a beetle substitutions. Okay, so that's our di ISA beetle aluminum hydrate. You may recall that the lone pair in oxygen can preferably attack an electro positive center or basically our electric elec center. Right, aluminum is electoral positive. So it can be easily attacked. And in this step we are forming our first intermediate, we can through our five member ring. We can change the position of OC three for simplicity to make sure that our drawing is not too bulky. And then we have our double bonded auction over here, but actually it forms three bonds. So it forms a formal positive charge because now it's bonded to aluminum, aluminum in turn, that will form four bonds. So it will have a negative charge because it has hydrogen right basically has hydrogen attached to it and also to Aiso beetle groups. So 123, 123, we're going to add one more carbon atom and that's our complex. Now this is where the proton transfer takes place essentially. We want to notice that this aluminum hydrogen bond is strongly polarized, right? Essentially, hygiene is more electro negative than aluminum? And therefore there is a high electron density adjacent to hydrogen, meaning it can easily be transferred to an electra Felix center, such as our carbon group. In other words, it's a good electrolytic center because oxygen is positively charged, performing that carbon hydrogen bond. And because we are doing that, we have to break the adjacent pi bond and then the stuff we are forming the hemi acid I'll complex. Now let's see how it looks like we're going to draw Our five numbers rain. Now we clearly see that there must be hydrogen attached to this carbon and then oxygen is still bonded to aluminum with two eyes of beetle groups. We're going to use a shorthand notation instead of drawing that bulky structure, This is our hemi aside, I'll complex and this is where di belle age has done its job. Now we have to think about the hydrolyzed this stuff and we're going to use water for that during this step to make it simpler for us to understand, we just want to realize that this oxygen bond that is polarized, can actually be used in the transfer of these electrons into the formation of a pi bond, which results in the loss of the leaving group. And this is how we get our al to hide. So the following structures is obtained which corresponds to our final product, our five member ring. And now, if we draw it in the same perspective, that would be double bonded oxygen, right? And then we have the loss of the leaving group, then we only end up with hydrogen. That's indeed an AL hide. So going back to the reaction scheme, we can say that the expected product indeed, is an AL to hide. Just as we stated, we are going to turn this estrogen to an AL to hide by replacing that O. C three group with hydrogen. So that's it we got our AL to hide. We can label this is our final answer. And based on the structure of this alga hide, conclude that the correct answer to the multiple choice question is a thank you for watching.

Predict the major products of the following reactions.
(b)

Key Concepts

DIBAL-H Reduction

Mechanism of Carbonyl Reduction

Hydrolysis of Aldehydes

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