(a) The following questions relate to the formation of lactone 7 from keto-ester 6. (Sections 24

(a) The following questions relate to the formation of...

Key Concepts

Lactonization

Lactonization is the intramolecular esterification process by which a compound containing both an alcohol (or a nucleophilic site) and a carboxylic acid (or an electrophilic carbonyl) forms a cyclic ester known as a lactone. This transformation is common in organic synthesis and is influenced by factors such as ring strain, the proximity of reactive groups, and reaction conditions.

Keto-Ester Reactivity

Keto-esters contain both a ketone and an ester functional group, each of which can engage in different types of chemical reactivity. The presence of these functional groups allows for enolization and subsequent cyclization reactions, as the keto group can act as an electrophile while neighboring nucleophilic sites (such as an ester oxygen) can initiate ring closure under the proper conditions.

Intramolecular Cyclization

Intramolecular cyclization refers to the process in which reactive sites within the same molecule react with each other to form a ring structure. This approach is often kinetically favored over intermolecular reactions due to the proximity of the reactive centers, leading to the preferential formation of five- or six-membered rings which are generally more stable.

Transcript

00:01 Okay, so we have to answer the following questions, and we're starting with the structure.

00:07 So we have a 6 carbon ester with a ketone on carbons 2 and 6, and we're reacting with sodium borhydrod, which is a mild reducing agent.

00:18 So it's only going to reduce one of these ketones into an alcohol, but it's not strong enough to reduce a ketone and an ester.

00:26 And it's pretty much all or nothing.

00:27 It's not going to reduce the ketone and not the ester.

00:30 So therefore it's only going to reduce this ketone.

00:38 So we still have the ketone over here.

00:40 And this is now an alcohol.

00:44 So the ketone is going to depredate and acid.

01:07 So the alcohol is going to attack the carbonyl carbon.

01:11 And we're going to get a five carbon chain.

01:14 So this is one, two, three, four, and five.

01:24 Let me try that.

01:36 So on carbon one, we have this methyl right here.

01:41 And then two, three, four, and then five was turned into an alcohol, but it's also bound to an ester.

02:03 So the ester is going to depred an acid to turn itself into a better leaving group because the alcohol is not strong enough to kick out the ester.

02:17 If we had an alcoxod ion, it would be able to do so.

02:29 So we get the alcohol right here, and then this has a hydrogen with positive charge.

02:41 So it's going to reform a bond rate here and expel the ester.

02:45 So we're going to get the structure.

02:57 So if we react the same thing with lithium aluminum hydride, it's strong enough to reduce the ester and the ketone into an alcohol.

03:09 In this case, we'll have a primary alcohol.

03:17 So we get, but it's going to leave the one ketone as is.

03:32 So therefore, it's chemo selective.

03:34 It's not going to reduce both ketones.

03:39 And reacting this with so if we repeat the same steps the first one's going to be pertination so we'll get this the oxygen is going to attack the carbonil carbon we'll get a five a five carbon ring but the um it's not going to regenerate the ketone because we don't have an ester to kick out so it's going to remain as an alcohol after pertinating the acid so that's the only difference just the alcohol and the ketone so that's for for 3i or 2 this and this is 2 i okay so for b it's to the mechanism we have this structure reacting with this with um with peroxide which is a strong base so methoxite the alpha hydrogen but because we don't have any alpha hydrogens in this structure it's going to deprotinate one of these alpha hydrogens and we'll get a carbane ion.

05:28 So this is going to attack the carbonyl carbon on either one of these carbonyl carbons.

05:48 So this is going to add, it's going to be r for now.

05:52 This should be an oxide.

05:59 So this is strong enough to get rid of the ester...

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