The herbicide oxyfluorfen can be prepared by reaction between a phenol and an aryl fluoride. Pro

The herbicide oxyfluorfen can be prepared by reaction...

Key Concepts

Activation by Electron-Withdrawing Groups

Electron-withdrawing substituents on the aromatic fluoride enhance its susceptibility to nucleophilic attack by stabilizing the negative charge developed in the Meisenheimer complex. This activation is crucial for efficient nucleophilic aromatic substitution, enabling the replacement of the fluoride leaving group with the nucleophile.

Phenol Deprotonation

In alkaline conditions, a phenol loses a proton to form a phenoxide ion. This ion is a stronger nucleophile because the negative charge on oxygen is better able to initiate a reaction with an electrophilic aromatic system. This step is key to activating the phenolic component for further reactions.

Nucleophilic Aromatic Substitution (S?Ar)

This mechanism involves the attack of the nucleophilic phenoxide ion on an activated aromatic ring bearing a good leaving group. The reaction proceeds through the formation of a negatively charged intermediate, often referred to as a Meisenheimer complex, which is stabilized by electron-withdrawing groups on the aromatic ring. The departure of the leaving group (in this case, a fluoride ion) finalizes the formation of the new aryl ether bond.

Transcript

00:01 Okay, so nucleophilic aromatic substitution, we have to propose a mechanism for these two molecules coming together to make this herbicide oxyphorfen that seems to look very similar to literally our two products push together somehow.

00:21 A lot of the substituents that are on those rings don't do anything.

00:24 They're not, they're unreactive.

00:26 Realistically, each is a tri -substituted benzene ring, and only two of those sites are reacting.

00:33 Right.

00:33 So in nucleophilic amarymatic substitution, you need a halide leaving group, right? on one side of the ring, fluorine is the best one to have.

00:44 And then in conjunction with that, you also need a strong electron withdrawing group that's either ortho or para to that, to that halide leaving group to optimize.

00:56 The intermediate that's formed.

00:59 So you have a para -nitro group, which is exactly what you need in that situation.

01:03 And now you just need to identify your nucleophile.

01:06 So if you notice the only bond that's being made here in your product is this new co bond, which comes from the cleavage of that oh bond to generate your phenoxy or your phenoxide anion, which will be your bait, which will be your nucleophile in this case.

01:28 How do you generate that under these conditions? that's why we're provided the k -o -h.

01:33 That hydroxide's going to pluck off that proton to make your phenoxide, and that's going to be your nucleophile.

01:38 So it looks like the three -step mechanism.

01:41 The first step is going to be that nucleophile formation.

01:44 So just for the sake of shorthand, i'm going to leave all the groups off of the ring because they're not doing any chemistry.

01:52 I'm only going to draw the things that are relevant to the mechanism.

01:56 So you have this hydroxide.

01:58 What i should do is make this bond a little more clear here.

02:03 There you go.

02:04 Right.

02:04 So here you go.

02:05 Acid -based chemistry.

02:06 You're going to do a very classic proton transfer.

02:08 You've probably did a million of these between your whole semester and all the chemistry you've done so far in this chapter.

02:14 So you're going to make your phenoxide, o minus, and the equivalent of water.

02:19 Plus, don't forget, there's groups here and here.

02:23 I'll go to the next page.

02:25 So i drew out the nitro group because you're able to do, you're able to do some, sorry, i have to pause for a second.

02:37 So i drew out the nitro group because we're going to push our electrons onto that group because the nitro group gives us some resident stabilization.

02:47 And i want to show what that looks like in the nitro group.

02:50 Right.

02:51 So what you're going to do is you're going to second step is nucleophilic attack at that site.

02:55 We're going to push our electrons accordingly.

02:57 So you're going to make this, you're going to basically push electrons across, so your del bonds are across from each other in the ring.

03:07 These electrons are going to go onto the nitro group, and then you're going to push these electrons up...

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