Propose methods for preparing butanoic acid from each of the following: (a) 1-Butanol (b) Butana

Propose methods for preparing butanoic acid from each of the...

Key Concepts

Oxidation Reactions

Oxidation reactions involve transforming a functional group such as a primary alcohol or aldehyde into a carboxylic acid by increasing its oxidation state. In organic synthesis, reagents such as KMnO4, CrO3, or other oxidizing agents are used to achieve these conversions, making oxidation a fundamental method for preparing carboxylic acids from less oxidized starting materials.

Aldol Condensation

The aldol condensation is a carbon–carbon bond-forming reaction where aldehydes or ketones combine under basic conditions to create larger ?-hydroxy carbonyl compounds. This method is particularly useful for extending carbon chains when the starting material has fewer carbons than the target molecule, and the resulting aldol product can often be further oxidized to yield a carboxylic acid.

Grignard Reaction and Carboxylation

The Grignard reaction involves transforming an alkyl halide into an organomagnesium reagent, which can then react with carbon dioxide. Upon acidic workup, this forms a carboxylic acid. This process is critical for carbon–carbon bond formation and enables the extension of the carbon chain when synthesizing larger acids from smaller precursors.

Hydroboration-Oxidation

Hydroboration-oxidation is a two-step reaction that converts alkenes into alcohols. Initially, borane adds across the alkene in a syn addition, and the resulting organoborane is then oxidized with hydrogen peroxide to form an alcohol. This alcohol can subsequently undergo oxidation to yield a carboxylic acid, providing a pathway for transforming unsaturated compounds into acids.

Decarboxylation

Decarboxylation is the process by which a carboxyl group is removed from an organic molecule, typically releasing carbon dioxide. This reaction is particularly useful in synthesis for modifying the carbon skeleton, such as converting dicarboxylic acids into monocarboxylic acids to achieve the desired carbon count and functionality.

Transcript

00:01 Okay, this problem is asking us to make transformations starting with butinoic acid.

00:05 So, butynoic acid looks like this, where we have four carbons and then our carboxylic acid.

00:13 Okay, so that is butinic acid, and the first transformation they want us to complete is to complete a transformation into one butanol.

00:22 So one butanol looks like this.

00:24 We have four carbons, and on the fourth carbon, or the first carbon, we have an alcohol.

00:30 So how do we make this transformation? okay, so we need to get rid of an alcohol and then reduce this one.

00:39 So the best reducing agent that we are familiar with is lithium aluminum hydride.

00:46 Lithium aluminum hydride will reduce my carboxylic acid down into a primary alcohol.

00:52 And that is exactly what we have here because this carbon is connected to one other carbon.

00:57 Okay, so the first transformation is used with lithium aluminum hydride.

01:00 The second one, we want to make a butanal.

01:05 So that looks like this, where we have an aldehyde.

01:10 Okay, that is butanol.

01:12 And in order to do that, the reagent we need is called dibal.

01:19 Dybal is also a reducing agent, and it reduces it down to an aldehyde.

01:23 It also reduces down a variety of other compounds, a variety of other carboxylacid derivatives, but specifically for carboxic acid, it breaks it down into an aldehyde.

01:32 Okay, so that's that one.

01:35 The next one is the production of one bromobutane.

01:39 One bromobutane looks like this, where we have our four carbons and then attached to the one position is a bromine.

01:47 That is one bromobutane.

01:49 So how do we do that? we see that in our starter material, we also have four carbons, so we need to reduce down our carboxylic acid and then replace our alcohol with a bromine.

02:00 So we should use lithium aluminum hydride to first make it into this compound right here.

02:10 Okay, so once we get this compound, then we need to replace this alcohol with a bromine.

02:17 And in order to do that, we will need something like a pbr3.

02:24 Okay, so that's that.

02:27 For the next one, we need to transform our butanilic acid into.

02:32 To pentane nitrile.

02:35 So pentane nitrile looks like this, where we have five carbons, one, two, three, four, five, and on the fifth carbon, we are connected to a nitrogen that is triple bonded.

02:48 Okay, so one, two, three, four, five carbons, and the fifth carbon is triple bonded to the nitrogen.

02:55 So how do we make this transformation? well, if we look back to our previous examples, we had this compound right here.

03:02 This compound, if we were to react it with nicn, that would produce this compound right here.

03:10 So we would have the same compounds as the previous problem, where we have lithium -aluminum hydride, and then pbr3.

03:20 After that, we would have this reactant right here.

03:22 If we use an s -sn2 reaction and then react it with nicn, that cn, the nitral can come in and replace that bromine, and that bromine will leave as a leaving group.

03:32 So we would end up with our desired product, which is our pantane nitral.

03:39 Okay, for the next one, we need to produce one butene.

03:45 One butene looks like this, where we have our four carbons, and on the one position, so one, two, three, four.

03:52 On the one position, we have our alkyne.

03:55 So how do we make this transformation? well, if we look back to our previous examples, so obviously with the structure of this problem, we can look back and see which of these reactants would work better as an intermediate to our eventual product.

04:11 So with this one, i can see that with this compound right here, my bromobutine, if i were just to eliminate that bromine and replace it with a base to make an alken, i would end up with my desired product.

04:30 So i'm just going to use those same reactants, lithium -aluminum hydride, 2, pbr3, and step three, i need a base.

04:43 Because if i had a bromine right here, if i replace with a base, sorry, if i deprotonate a hydrogen, such as right here, i'll kick off that bromine, and for my alken, then i'll get my desired product.

04:59 So i'm going to use a base, specifically a bulky base such as turd butoxide.

05:07 Okay.

05:08 Next up, we want to make the compound n -methyl -pananamide.

05:14 So n -methyl pantanamide looks like this.

05:23 Okay, so we know it is n -methyl pantomide because the n -methel corresponds to this methyl group right here, and the panthenamide corresponds to the 1, 2, 3, 4, 5 carbons.

05:36 Okay, so how do we make this transformation? well, if we look back to our previous examples, this is a very good intermediate.

05:45 We see that.

05:46 We need to, if we have this compound right here, and we react that with magnesium, then we can create a greenard because in this compound we have five carbons and with our starting material we only have four carbon so we want to lengthen our carbon chain the best way we know how to do that is with greenards so if we start with this compound and we react that with magnesium we have our greenard reagent which we can further react with another carbon to lengthen our carbon chain so i'm going to take those same reactants and at this point i'm just going to highlight these reactants right here duplicate them and move them over here and then then i'm going to do a third step of reacting it with magnesium.

06:30 Okay, so with magnesium, i will end up with this intermediate right here, where i have my four carbons and then my magnesium bromide.

06:40 Okay, so this is my greenerid reagent.

06:42 That greenerid agent can react with a carbon dioxide molecule to lengthen my carbon chain by one, because i started out with butane, and i want to end up with a pentane, because four carbons to five carbons.

06:54 So i'm going to react that with co2.

06:56 2.

06:57 Okay, and because this whole chapter is about carboxylac acid and their derivatives, i'm going to transform that carboxylate into carboxylac acid using hydronium ion.

07:09 Okay, so after that, after steps 4 and 5, i should end up with this, where i lengthen my carbon chain by 1.

07:17 Okay, so now i have my carboxylac acid...

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