Predict the major products of the following reactions, and giv...

Question

Predict the major products of the following reactions, and give the structures of any intermediates. Include stereochemistry where appropriate.

(l)

Accepted Answer

everyone and welcome back in today's video, we are going to tackle the following problem which reads through the reaction intermediates and major products for the following reactions. Show the stereo chemistry where applicable. We understand that we are starting this reaction by using an L. Key in which is one muscle cycle, opened one in and we're treating it with number one ozone at low temperature and number two dimethyl sulfide, which as you recall implies that this reaction can be classified as also know is this in the first step we are going to draw the starting L. King And we are going to treat that with ozone which is 03. And the structure of ozone would be if formal negative charge on the first one. Formal positive charge and the second one because it forms three bonds in this auction is double bonded. So there's no charge on it. The addition of ozone across the triple bond is as follows. In particular, the liberal bond attacks this oxygen. There's a transfer of bonding electrons onto this oxygen in the form of a lone pair. And the negatively charged oxygen attacks back to form a ring. And that would correspond to our first intermediate. Now we may start drawing that intermediate by adding a five member drink the original methyl group. And now we need to add three oxygen atoms. So there will be two additional bonds and then we would have three oxygen atoms. This is our first intermediate. Remember that? Its name is small ozone. I and we may label the structure because we are interested in getting our intermediate right in this problem. However, more melissa night will form another intermediate after rearrangements. So let's see how that happens. The carbon carbon bond will be broken over here. Enviably forming a carbon auction double bond. There would be transfer of these bonding electrons in the form of a lone pair and this lone pair on oxygen would form another double bond at this position. Notice that we have a ring opening so to draw our final structure, it would make sense to label our carbon atoms from here up to this sixth carbon atom which also has hydrogen. So we may now draw A six member chain and also that hydrogen Which is located at position number six Position #2 will now have a double bomb with T. Oxygen's and one of them will have a formal positive charge, while another one has a formal negative charge Position # six now has a double bond and oxygen. Now these are our fragments after the C C bond cleavage. And now the fragments will recombine to form another intermediate, which would be our second intermediate in this problem. Now, let's see how that happens. This would require us to draw three different arrows. So first of all these pi electrons would attack the carbonell group. There will be transfer of the pi electrons onto oxygen and the negative charge. What attack the carbon atom to restore that ring. Now let's draw our second intermediate. We made through a fragment of a five member ring. But now there is one oxygen atom in particular at this position and they're two adjacent oxygen atoms, one of them here. Another one over here and they're bonded bye. A single bomb. Now we have this intermediate, there's an additional methyl group of this position and the name of this intermediate is also night. So that will be our second intermediate. Now the difference between two intermediate is if you carefully look at it, is that in particular there's the cleavage of carbon carbon bond. And instead of having this bond, we actually have oxygen now bonded to these two carbon atoms while the two other oxygen stay on the same side. So that's how you can think of it. We will also label this as our second intermediate. So these are our two intermediates. And now the world final step is reaction number two. We are done with reaction number one. We got our ozone nine. We essentially want to use a reduction reactions. We may label now that here we're using they metal sulfide. If we draw their muscle sulfide, there would be Sulfur with two method groups. There are two lone pairs in particular because this oxygen is partially positive. The lone pair would attack it. This bond, which is the carbon oxygen bond would form a double bond at this position. We would have the cleavage of this auction auction bond and the formation of a new double bond at this position, and then these bonding electrons would be transferred onto oxygen. This again, opens the ring. So you may think about it as the removal of this oxygen in forming a similar looking structure. But now we would not have that oxygen because now it has been captured by sulfur. So to draw our final product, we may simply redraw that intermediate without that additional oxygen. That said, our final product would look like this. That would be our final product. And in case you're wondering, we are also forming sulfur. We're two methyl groups. Now it has oxygen that is negatively charged, sulfur has a positive charge and it actually produces die methos sulfur dioxide D. M. S. O. However, we may label this as our final product because we wanted to predict our final product for this problem. So to conclude, we have two different intermediates in this reaction. One of them is small. Ozo night. The other one is the rearranged form of it, which is called ozone Aight and the final product that we are forming contains a kitchen group and an AL to hide group After we treat Osa night with dimethyl sulfide, that's it. Now, our final answer for this problem is be. So, if you chose B. You got it right. However, if you have any questions, don't have states. Leave your comment down below in the comment section and thank you for watching

Predict the major products of the following reactions, and give the structures of any intermediates. Include stereochemistry where appropriate.
(l)

Key Concepts

Ozonolysis

Stereochemistry

Reaction Mechanism

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