Identify the alkene that would give each of the following products upon ozonolysis followed by t

Identify the alkene that would give each of the following...

Key Concepts

Ozonolysis

Ozonolysis is a chemical reaction in which ozone reacts with alkenes to cleave the carbon–carbon double bond. This process leads to the formation of ozonide intermediates that, upon further treatment with a workup reagent, produce carbonyl compounds. It is a key method for determining the structure of alkenes and for synthetic transformations in organic chemistry.

Oxidative Workup

The oxidative workup step, often employing hydrogen peroxide, is used after the initial ozonolysis reaction to convert the intermediate ozonides into more stable products such as carboxylic acids, aldehydes, or ketones. This step ensures that the oxidation of the intermediate is complete, providing a predictable and useful set of products.

Alkene Structure and Cleavage Mechanism

Understanding the structure of the starting alkene is crucial because the cleavage during ozonolysis is governed by the positions of the double bonds. The reaction mechanism involves the formation of an initial cyclic ozonide that rearranges and cleaves to yield two fragments, with each fragment's fate determined by substituent effects and reaction conditions.

Carbonyl Compound Formation

The transformation of alkenes into carbonyl compounds (aldehydes, ketones, or carboxylic acids) is central to the reaction. The specific carbonyl products obtained depend on the nature of the substituents attached to the original alkene and the oxidative conditions used during the workup step.

Transcript

00:01 Okay, so we want to give the alkyen used to create the following products.

00:03 In the first one, we have a four -carbon, carbonsylchic acid, and acetone.

00:20 And so if we count the number of carbons that we have so far, we have four in this chain, and seven right here.

00:29 I mean, three right there, so until we have seven.

00:36 And if we draw the four -carbon, or create the four -carbon carbon carbon carbon dioxide first on carbon -fourbon, we would have an al -key to get the product on the left.

00:48 And actually if we take off one methyl group and add it right here, we'll get our products.

01:00 So we'll add an oxygen to this carbon, which would be a ketone, and we would get structure.

01:11 We have a 9 carbon chain.

01:24 And the fact that we have one product means we've started with a ring -like structure.

01:30 So if we draw a 6 carbon ring and we add so we have ketones which means you would have had r groups attached to these sp2 carbons so the one carbon has a methyl group and the other has an ethyl so we started with that structure the next one is this product and we have a 6 carbon carboxyl acid so we get the fact that we have two products means we started from a straight chain alken, i mean alken.

02:27 So if we add, we have six carbons here and three in that chain, instead of we have nine.

02:37 And if we create the product on the right first, carbon six is right here, so we'll add a key tune.

02:45 And i mean, we'll add an acine.

02:48 And if it cleaves this bond, it'll give us the product to the left.

02:55 But we need to add a key tune to this carbon, because it's the same carbon is right here.

03:00 So this is 1, 2, and 3.

03:04 1, 2, and 3.

03:09 So we started with that structure.

03:13 I'm sorry, we also have a ketone right here.

03:16 So as a result, we're going to add a ketone of that carbon.

03:20 This will be 6, 5, 4, 3, 2, and 1.

03:27 So this would be carbon 2.

03:35 And for d, we have cyclohexanine and propinolc acid.

03:55 So the fact that we have a ketone means that, we had some r group attached to the ketone and it would have been the akin would have been directly attached to the ring in order to get a ketone attached to the ring...

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