Benzyl bromide is converted into benzaldehyde by heating in dimethyl sulfoxide. Propose a struct

Benzyl bromide is converted into benzaldehyde by heating in...

Key Concepts

Nucleophilic Substitution (SN2) Reaction

This concept explains how a nucleophile (in this case, dimethyl sulfoxide) attacks an electrophilic carbon bearing a leaving group (the benzyl bromide). The SN2 mechanism involves a backside attack leading to inversion of configuration at the carbon center and the formation of a new bond while displacing the leaving group.

Formation of Sulfonium Ion Intermediate

When the sulfur atom of dimethyl sulfoxide attacks the benzyl bromide, a sulfonium salt intermediate is formed. This intermediate features the benzyl group attached to the sulfur center, thereby generating a positively charged species that is key in subsequent reaction steps.

Kornblum Oxidation

This oxidation process involves converting an alkyl halide into a carbonyl compound using a polar aprotic solvent like DMSO. It proceeds via the formation of a sulfonium ion intermediate, followed by deprotonation and rearrangement to yield the corresponding aldehyde or ketone, along with the release of a sulfide byproduct.

Ylide Formation and Rearrangement

After the sulfonium ion is formed, deprotonation at the carbon adjacent to sulfur generates a sulfur ylide. This ylide is an intermediate that undergoes rearrangement, facilitating the transfer of the alkyl group to the oxygen atom. This rearrangement ultimately leads to the formation of a carbonyl group (in this case, an aldehyde) while eliminating a neutral sulfide molecule.

Transcript

0:00 All right.

00:01 In this mechanism, benzobromide in the presence of dmso, dimethyl sulfoxide, you can heat that and you undergo an sn2 reaction to form some kind of intermediate that then undergoes an e2 reaction to form benzaldehyde.

00:20 This is a pretty wacky, wacky mechanism, right? so it's two steps.

00:26 They give you all the information you need.

00:28 So i would recommend just, and they also help you out by showing you the resonance structure of, of this, of dmso, of dimethyl sulfoxide, dmso.

00:39 Right.

00:40 So the same way you can push an electron density in a carbonyl to show a carbonyl and a carbonyl and a negatively charged oxygen, you could do the same thing with, with dmso, with these sulfoxides, to show that the sulfur is partially positive and to show that the oxygen is slightly.

00:58 Negative.

00:59 So in terms of sn2, right, you need a nucleophile.

01:04 So where is that nucleophile going to be? so really, the only thing that's a nucleophile in this case is going to be your negatively charged oxygen in the resident structure of dmso.

01:16 So i'm going to start by doing that.

01:21 So once i've established that that's probably going to end up being my nucleophile, what i want to do now is, is just try the mechanism, right? so you know it's sn2, which means that the br is your leaving group.

01:39 So if i take dimethyl sulfoxide and i use the sulfur oxygen bonds to attack the carbon and push the bromine out, that is a totally fine sn2 mechanism.

01:52 If you're uncomfortable drawing arrows from bonds, what you can do is push the loan pair onto oxygen and then use the oxygen loan pair.

02:01 It's the exact same thing.

02:03 Right.

02:03 So now you're going to make br minus plus benzene that has one carbon that is now bound to the oxygen in dmso.

02:24 So by extension, sulfur is going to have a positive charge still.

02:30 And your bromine is going to stabilize that.

02:37 All right.

02:38 So what's the next step, right? so i think when you have all these heteroatoms and you don't necessarily know, might not know what to do next, i think it's important to draw out everything, right? so we're doing a reaction that is going from having two carbon hydrogen bonds on this benzobromide.

03:01 And we only have one in the product, right? so we're going to have to deprotonate one of the protons on that carbon.

03:10 And we're told it's only two steps, right? and we've already done one step.

03:14 So all of this has to, and it has to be an e2.

03:17 So we have to eliminate something while maintaining the oxygen in the structure and losing a proton.

03:25 It's not really a proton, but it is in this case.

03:28 It's going to be an acid, right? so it's important to draw all your lone pairs as well because that will also tell you where the electron density is, is, right? so i drew a crappy lone pair in oxygen, right? when hetero atoms are positive, they are electron withdrawing.

03:47 And the reason for that, i mean, when anything is positive, it's electron withdrawing, because it needs electron density to stabilize that charge.

03:54 The positive charge on an atom means that there's an orbital that is lacking an electron...

Please give Ace some feedback