Write the mechanism for each of the following reactions: a....

Key Concepts

Nucleophilic Acyl Substitution

This mechanism is fundamental in organic chemistry where a nucleophile attacks a carbonyl carbon, resulting in the displacement of a leaving group. The process involves the formation of a tetrahedral intermediate that subsequently reforms the carbonyl bond while expelling the leaving group, thereby replacing it with the nucleophile. It is commonly observed in reactions involving acyl chlorides due to their high electrophilicity and good leaving group ability of chloride.

Tetrahedral Intermediate

The formation of a tetrahedral intermediate is a key step in many carbonyl reactions. Once a nucleophile attacks the electrophilic carbon of a carbonyl compound, the planar carbonyl group is temporarily converted into a tetrahedral structure. This intermediate is crucial as it allows for the reorganization of bonds, leading to the elimination of the leaving group and the formation of the final substituted product.

Leaving Group Ability

The concept of leaving group ability is vital in understanding substitution reactions. A good leaving group, such as a chloride ion in acyl chlorides, is one that can stabilize the negative charge once it departs from the substrate. This stabilization facilitates the breakdown of the tetrahedral intermediate, driving the reaction toward completion.

Proton Transfer Steps

Proton transfer steps often accompany nucleophilic substitution reactions, particularly in aqueous or amine-based environments. Protonation and deprotonation processes help stabilize intermediates and transition states, adjust acid-base properties, and ultimately assist in the formation of neutral, stable final products. These steps play an essential role in refining the overall mechanism, ensuring that the reaction proceeds efficiently.

Transcript

00:01 Okay, this problem is asking us to show the mechanisms of each of these reactions.

00:04 So first up, we have my acetyl chloride reacting with water in order to form my acetic acid.

00:08 Okay, so in order to show the mechanism, let's proceed.

00:11 So first up, we know that my acetyl chloride is particularly electrophilic, right? because if we have this carbon right there, there's a lot of electron density being pulled away from that carbon, both by resonance, if i are to move those electrons onto the oxygen, and also by the inductive effect because i can move electrons onto the oxygen, because auction is electroneg and also i can move them onto chlorine, because all chlorine is electronegenective.

00:32 Okay, so again, a lot of electron density being pulled away from that carbon.

00:35 Okay, so that is an electric phyllicite, and that is in relation to my nucleophilicite, which is my oxygen in this case of water.

00:41 Okay, so using those electrons on oxygen of water, i'm going to take them and attack this carbon of my carbonyl, and as i do that, i'm going to have to move the electrons onto this oxygen, because otherwise this carbon will have exceeded its octet.

00:52 Okay, so that will form this, my intermediate, my tetrahedral intermediate, in which i have my ch3, my o minus, my former water, which is now oh2, but i'm just going to separate those hydrogen, so just so we can kind of separate them for now.

01:09 Okay, and then that will be my tetrahedral intermediate.

01:13 Okay, so as we can see, we have a neutral molecule.

01:15 Right, we have a negative charge right there, and we have a positive charge right there.

01:18 They cancel out, they eventually form a neutral molecule.

01:21 So this is not necessarily unfavorable, but we can make it even more favorable by simply getting rid of both charges entirely.

01:28 And right, we can do that by simply doing a proton transfer.

01:31 And that is simply moving this hydrogen onto this oxygen, which will both relieve this oxygen of its positive charge and also relieve this oxygen of its negative charge.

01:40 So what we're going to do is simply depict that with proton transfer.

01:44 Okay, so the reason i'm just doing this as proton transfer is one, because proton transfer is depicted like this, are pretty standard within organic chemistry, and two, because a lot of steps are contained within this proton transfer, it is simply taking an outside molecule, an intermolecular reaction, and it's going to deprotinate this hydrogen, which is eventually going to transfer onto this oxygen.

02:04 Okay, so a lot of steps, which we can just depict with a proton transfer with pt.

02:08 Okay, so that is going to form this compound, in which i have my ch3, my oxygen up here, which is now protonated.

02:15 So it has the alcohol.

02:17 I have my other alcohol, which was formerly my water with a positive charge, but now it is neutral because we got rid of that positive charge.

02:24 And we also have my chlorine right there.

02:26 Okay, so this is the state that i'm at right now.

02:28 Obviously, i want to make this acetic acid and not this.

02:31 So what do i have to do? as i can see, i don't have any chlorine on that molecule, but i do have a chlorine right there.

02:36 So how do we get rid of that chlorine? all i need to do is simply move the electrons down from one of these two alcohols.

02:41 I'm going to move them down onto the single bond to make a double bond.

02:44 And as i do that, i'm going to have to kick off this chlorine.

02:47 Okay, so the reason i kick off that chlorine as opposed to another alcohol is simply because we have a difference in basicity.

02:53 Right, because i have chloride.

02:54 If that were by itself, it would be a weak base.

02:56 In relation to this hydroxide ion by itself, right? okay, because chloride by itself is a weaker base than hydroxide.

03:03 So that's the reason why we kick off chloride as opposed to hydroxide.

03:07 Okay, so with that mind, let's proceed.

03:09 Okay, so if i proceed with that reaction, i will form this product in which i have my ch3, my protonated carbonyl, just like that, and i'll have my alcohol down here with my chloride.

03:22 Okay, so as we can see, we have protonated acetic acid, and what we want to form is, just regular old acetic acid.

03:28 So what do i have to do? i simply have to deprotonate this molecule.

03:32 Okay, so i'll have a couple options.

03:33 I can either use the chloride ion to deprotonate this molecule, or because i started off with water, i can just use water again.

03:40 Okay, and i'm going to use water again simply because if i were to use chloride and i would protonate that, we'd form hcl, which has a very low pca, meaning it is very acidic.

03:49 If i just use water, we're going to form a strong acid, but not as strong as hcl, which is the reason that we want to use water instead of chloride...

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