Write the structure of the Dieckmann cyclization product formed on treatment of each of the foll

Write the structure of the Dieckmann cyclization product...

Key Concepts

Dieckmann Cyclization

Dieckmann cyclization is an intramolecular variant of the Claisen condensation where a diester, upon deprotonation of an ?-hydrogen by a base, forms an enolate that attacks the carbonyl carbon of the same or another ester group within the molecule. This results in the formation of a cyclic structure, usually a ?-keto ester, with ring sizes typically dictated by the distance between the reactive centers in the molecule.

Claisen Condensation

Claisen condensation is a fundamental carbon–carbon bond forming reaction in organic chemistry. It involves the nucleophilic addition of an enolate ion, derived from a carbonyl compound with ?-hydrogens, to another carbonyl compound, followed by elimination to form a ?-keto ester. The reaction can occur both inter- and intramolecularly, with the intramolecular version leading to cyclized products such as in the Dieckmann reaction.

Enolate Formation

Enolate formation is the process by which a base removes an ?-hydrogen from a carbonyl compound, producing a resonance-stabilized carbanion known as an enolate. This enolate serves as a powerful nucleophile, initiating carbon–carbon bond formation in reactions such as the Claisen condensation and its intramolecular counterpart, the Dieckmann cyclization.

?-Keto Ester Formation

A ?-keto ester is a compound containing both a ketone and an ester functional group separated by one carbon atom (the ?-position relative to the ester). These compounds are the common products formed from both Claisen-type condensations and Dieckmann cyclizations, and they serve as versatile intermediates in organic synthesis due to their reactivity and ability to undergo further transformations.

Transcript

00:01 Okay, this problem is asking us what would happen if we perform a dekman cyclization on diethyl -ethyl -thyl -threthyl -hephtane diowate.

00:06 Okay, so i'll write that out.

00:07 We have diethyl -ethyl -thre -methyl -heptane diowate.

00:14 Okay, and the reason i'm doing this is because we don't have the structure, right? it is not given to us.

00:19 So just for these purposes, i'm going to do one by one how to figure out the structure of the compound.

00:24 Okay, so diethyl -ethyl -thyl -threthane diolate.

00:27 Okay, so every time i'm given a name in organic chemistry and i don't know the structure, i start in the very back.

00:32 So on the very end.

00:34 So i'm going to start with dye -o -8.

00:35 Okay, so the eight part corresponds to an ester.

00:39 So die -o -8, we know that there's two because i have dye.

00:41 So die -o -8, i know i have two esters.

00:44 Okay, so if there are esters, i know that they're going to be on the very ends of my alkyal chain.

00:51 And i know that because if i have an auction like this connected to an auction, that auction is already the end of my carbon chain.

00:59 So right, so this will be the very end of my carbon chain because i can only go, this is, way, i can't go this way because i'm already connected to this oxygen.

01:05 So i know my diowates are going to be on the very ends of whatever alkyal chain i have.

01:10 Okay, so i have heptane.

01:12 So heptane corresponds to seven carbons.

01:14 So i'm now going to have seven carbons with two esters on the very end.

01:17 So one, two, three, four, five, six, seven.

01:20 Let's just confirm that.

01:22 One, two, three, four, five, six, seven.

01:23 Oops, missed one.

01:26 All right, added one.

01:28 Okay, so right here, erase that.

01:30 Okay, so that is seven carbons.

01:32 Okay.

01:33 And then the dialate, represents the ester.

01:35 So right here i have my carbonyl, carbonyl, and then my ester.

01:40 Okay, and we don't know yet what is going to be connected to the oxygen until we look at the rest of my compound.

01:45 But first, we have this three methyl.

01:48 That three methyl corresponds to an addition of a methyl group on carbon number three.

01:52 So carbon number three, it doesn't matter, we can either start with this way or this way.

01:56 Let's just do this way because it's considered symmetrical.

01:58 So one, two, three.

02:00 On carbon number three, we have a three methyl group.

02:03 So methyl.

02:04 Okay, next up, we have my diethyl.

02:07 So the diethyl in the very front corresponds to whatever is attached to my oxygen in my ester.

02:14 So diethyl, we know that it is an ethyl, and because we have two of them, those both correspond onto the di -o -weight nature.

02:20 So ethyl group, ethyl.

02:23 Okay, so that is diethyl -ethyl -ethyl -threthreathypth.

02:27 Okay, so that is diethyl -ethyl -ethyl -threathypth.

02:29 How to make a dekman cyclization of this.

02:33 So we're going to react this with, i'll do this part in green, we're going to react this with sodium ethoxide.

02:41 And the reason we do with sodium ethoxide is for a couple reasons.

02:45 One, sodium ethoxide is a very good base.

02:47 It's going to deprotonate a potential hydrogen to be deprotonated.

02:50 And two, to avoid transesterification.

02:52 So transesterification is when we have an ester, like we do, and we have an alcohol or an ethytide.

02:59 Oxide.

03:00 So, sorry, oxygen with a negative charge that is connected to a carbon chain.

03:06 So if we have that, we are prone to possible transesterification in which my oxygen connected to the carbon can potentially replace this group.

03:15 So because i have a thoxide, it is potential, i have the potential to move this compound onto its place.

03:22 But because they're the same, they both have the same number of carbons, we're not going to have that outcome.

03:27 So that's why we use sodium ethoxide.

03:30 But we don't really have to worry about that much because my sodium methoxide is primarily going to react as a base.

03:35 And my base is going to depotinate the most acidic hydrogen.

03:39 So on this molecule, we have two options for the most acidic hydrogen because we have two alpha carbons, right? we have an alpha carbon here, right? and we have an alpha carbon here.

03:50 So because this molecule is not symmetrical anymore because we just added this methyl group, that's why we have two different products, right? because we're going to either deprotate this one, and if we had the other choice, we're going to depronate this one.

04:03 So i'll demonstrate both possibilities, but the reason we don't have, or the reason we do have two products is because we have an unsymmetrical diester.

04:13 Okay, so let's do this hydrogen first.

04:17 So i'm going to deprotate this, move the electrons onto that carbon, and then we should get the following, in which we have this compound.

04:28 I'll just duplicate it to save time.

04:30 We have this compound, and then in the place of that former hydrogen, we have lump pairs.

04:37 Okay, so those lone pairs are going to be considered to be nucleophilic or basic.

04:42 In this case, with these deekman or kleinseng condensation reactions, they're considered to be nucleophilic.

04:48 Okay, so that is going to behave as a nucleophile, and now we just need an electrophile.

04:52 So we have two options.

04:53 We can either react this whole compound with another molecule of the certain material, or, which is actually more favorable, we can undergo a reaction with itself, an intramolecular reaction.

05:06 And the reason we can undergo an intramolecular reaction is because close by, we have this carbon.

05:12 And even more so, we have that carbon only one, two, three, four, five, six carbons away.

05:20 So in general, every time we have the potential to make an intramolecular reaction to make a ring structure, and we have the potential to make a six -membered ring, such as this.

05:32 This six -membered ring has virtually no angle strain, right? so every time we make a ring, we have the vulnerabilities of creating a bond that has a lot of angle strain, right? because with tetrahedral compounds, we would like to have all bonds be approximately 109 .5 degrees, degrees, right? but for this one, with my six carbon compound, we have virtually no angle strain, right? they're already at approximately 109 .5 so that's why it's so favorable to make a six -membered rain okay so we can also see the possibility of making a five -membered drain in another compound or a seven -membered drain right those are all very close to 109 .5 so if we have a six -membered drain or a possibility to make a six -membered rain that's why we're going to favor that outcome okay so i'm going to use this carbon this one right here to attack that carbineel and then i already labeled my one two three four five six we know it's going to be a six -membered rain.

06:29 So every time i'm making an interim molecular reaction to make a rain, i would recommend, highly recommend, to label or number your carbons.

06:36 Okay, so just go along the bond order until i get to my carbonyl.

06:39 So i have a six -membered rain.

06:41 So i'm going to draw out my six -membered rain.

06:43 One, two, three, four, five, six.

06:45 And then i had a teacher that just said to decorate the rain, right? so at number one, so here's my number one carbon.

06:51 I have one, two, three, four, five, six...