What alkyl bromide(s) should be used in the malonic ester...

Key Concepts

Decarboxylation

After alkylation and ester hydrolysis, heating the intermediate leads to decarboxylation. In this process, one of the carboxyl groups is lost as carbon dioxide, resulting in the introduction of a methylene group between the alkyl substituent and the carboxyl group. This predictable outcome means that the final carboxylic acids have the structure R–CH2COOH, so one must work backwards from the target acid’s structure to determine the required R group and hence the appropriate alkyl halide.

Retrosynthetic Analysis of Carboxylic Acids

Understanding the final structure as R–CH2COOH allows for a retrosynthetic disconnection where the malonic ester synthesis has effectively inserted a CH2 unit between R and the carboxyl group. Thus, to prepare a desired acid, one must remove the –CH2COOH fragment to identify R. This analysis is critical for selecting the proper alkyl bromide because the carbon framework of R determines whether a target acid is accessible by this method or not.

Malonic Ester Synthesis

This is a classic method for preparing substituted acetic acids. It involves the alkylation of diethyl malonate (or a related malonate ester) at the methylene group with an alkyl halide, followed by hydrolysis and decarboxylation. The key feature is that after decarboxylation the product has a –CH2COOH group, meaning that the alkyl group introduced by the alkyl halide ultimately appears one carbon removed from the carboxyl group.

Alkylation Reaction

The alkylation step in the malonic ester synthesis is typically carried out via an SN2 mechanism. The active methylene group of the malonate is deprotonated to form a resonance?stabilized enolate anion, which then undergoes nucleophilic substitution with an alkyl bromide (or similar halide). Choosing the correct alkyl bromide is essential because it determines the carbon framework of the final acid via subsequent transformations.

Transcript

00:01 Okay, this problem is asking us, what is the alkyo bromide that was used for these particular synthesis? so first up, i have the transformation of this malonic ester into my propanoic acid.

00:09 Okay, so for each of these, i would recommend just drawing out the structure of my carboxylic acid, just so we can see the reactions a little bit more clearly.

00:17 So propanoic acid, i have propane that corresponds to three carbons, and then oic acid corresponds to a carboxylic acid.

00:24 So i know that carboxylic acid is always going to be present on my first or last carbon, so like this.

00:30 Acid there.

00:31 Okay, i know that it's not going to be two proponok acid, for example, because if i'd have this carbonyl on the second carbon, at that point it wouldn't be a carbonyl acid.

00:39 It would be a ketone.

00:40 So i don't have to draw out the numbering system.

00:43 I just have to draw out propanolc acid.

00:45 Okay, so that is my molecule, and i have to get that from my melanic ester.

00:49 So as far as malonic ester goes, i have this acidic hydrogen right there.

00:53 So as far as synthesis goes, i'm going to react it with a base, such as sodium o .c2h5, that would be my ethoxide.

01:04 That's going to take off this acidic proton.

01:06 Okay, and then we can proceed with our reaction.

01:09 So as far as what this molecule is going to be, it's going to look like this.

01:12 We're going to have my melanic ester in which that middle hydrogen is depronated.

01:18 I'm going to have that nucleophiloc carbon.

01:20 Okay, so this is where we're going to have the addition of my occiobromide.

01:24 So in my propinic acid, i see that i have this carbon, this carbon right here, i'll do this part in blue actually.

01:30 That carbon right there here corresponds to this carbon right there, and then this carbon right here corresponds to this carbon right there.

01:39 Everything else to the left of that carbon is not going to be involved.

01:42 That's just going to be split apart after we undergo the decarboxylation reaction.

01:47 So basically we have to figure out what are we going to add to this carbon.

01:51 And as we can see by this proponeog acid, we only added this extra carbon.

01:55 So that means i'm going to perform an s &2 reaction in which only i'm only adding ch3 br.

02:01 So this is going to behave in an s -n -2 reaction in which this carbon nucleophile attacks this carbon, and i release that bromine as a leaving group.

02:08 Okay, so that first one is going to be ch3br.

02:11 Okay, what about this one? so again, start off with the structure of my carboxylic acid.

02:16 So i'm going to have my oic acid.

02:19 That's going to be a carboxylic acid like that.

02:22 And that's going to be associated with three carbons.

02:25 So one, two, three.

02:26 And then on my second carbon, so right here, i have the presence of a methyl group.

02:30 So just like that.

02:32 Okay, so i'm forming a two -methal propanoic acid from my melanic ester.

02:36 So similar to before, we have to recognize that in order to get anything at all on this carbon, we're going to have to undergo a base reaction.

02:43 So using sodium -eth oxide, let's go ahead and do that.

02:46 So i'm going to have this structure in which i have my melanic ester.

02:51 Same thing, except it's going to be deprotonated at that central carbon.

02:55 Okay, so this is where we add our corresponding substituents, because as we can see, attached to this carbon, which again i'll draw in blue, attached to this carbon, that is what we're adding to our melanic ester.

03:07 Because again, this carbon right here corresponds to that carbon right there.

03:11 Okay, so what do we add to that carbon? well, as we can see, we have these two methyl groups.

03:16 Okay, but we can't add those two method groups at the same time.

03:18 We have to do them individually, one at a time.

03:20 So similar to before, i'm going to react one of those methyl groups.

03:23 I'm going to put one of those methyl groups on first using ch3 br...

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