How can the following compounds be synthesized, starting...

Key Concepts

Carbon Backbone Conservation

This concept involves selecting a starting hydrocarbon that contains the same number of carbon atoms as the final target molecule. By preserving the carbon chain, the synthesis avoids additional steps needed to lengthen or shorten the molecule, thereby streamlining the synthetic design and reducing potential complications from rearrangements or side reactions.

Free Radical Halogenation

Free radical halogenation is a method used to introduce a halogen atom into an alkane by initiating a radical chain reaction. The process typically employs light or heat to generate radicals, which then abstract hydrogen atoms from the hydrocarbon. This transformation provides a functional handle that can be subsequently substituted with other groups, and it is especially useful when other direct functionalization routes are not available due to the inert nature of alkanes.

Nucleophilic Substitution Reactions

Nucleophilic substitution reactions, typically categorized as SN1 or SN2, enable the conversion of a haloalkane into a functionalized product by replacing a halogen atom with a nucleophile, such as hydroxide, cyanide, or ammonia. This mechanism is key in synthesizing compounds like alcohols where the substitution of the halogen with an –OH group occurs, allowing for the controlled installation of new functional groups on a preserved carbon skeleton.

Alcohol Synthesis from Alkyl Halides

Alcohol synthesis via alkyl halides is a strategy where an initially non-reactive alkane is first transformed into a more reactive haloalkane and then converted into an alcohol through nucleophilic substitution. This two-step process is particularly significant when starting from a saturated hydrocarbon, as the first step introduces the necessary functional group for further transformation while keeping the carbon framework intact.

Reaction Selectivity and Control

Achieving selectivity in reactions involving alkanes is critical because alkanes are typically unreactive and can form multiple products under uncontrolled conditions. By carefully choosing reagents, conditions (such as light, temperature, and solvent), and catalysts, chemists ensure that the halogenation and subsequent nucleophilic substitution proceed at the desired site. This control is essential to obtain high yields of the target product without contamination by side products.

Transcript

00:06 This question asks us how we can synthesize each of these molecules, starting with an alkyne, with the same number of carbons as the product.

00:14 So starting with a, we have one, two, three, four, five carbons here.

00:19 So we need a five carbon alkyne.

00:21 And because our aldehyde, we end up with an aldehyde, which is on the terminal carbon, we need to put the alkyne on the terminal carbon.

00:29 So we'll start with this.

00:31 Two, three, oops, four, five.

00:37 Carbons, right? that's what i said? five.

00:39 Okay.

00:40 And then to go from an alkyne to an aldehyde, we first have to add an o -h on, and then it willotomorize into the aldehyde.

00:47 So to do that, we need to add the oh onto the terminal carbon, which is an antimicobin addition.

00:55 The way that we do that is with hydroboration oxidation.

00:58 So we'll use bh3 with thf, and then we'll add hydroxide, peroxide, and water and that will give us the enol with the oh on that carbon and that will atomize into our product and b we have a four carbon alcohol so to do that we will actually use a hydrocarbon so it doesn't have to be an alkyne it can be an alken so that's what we'll use here so we'll do a four carbon alken again with the alken at the end of the chain because that's where the alcohol is.

01:44 And then to go from this alkyne to an alcohol on the terminal carbon, again, we're going to use hydroboration oxidation because that will add the oh onto the terminal carbon.

01:58 And because we're starting with an alkyne this time instead of an alkyne, we won't get an enol, we'll just get a regular alcohol so we don't have to worry about the tautomerization happening.

02:11 And then for c, we have this really long chain.

02:15 Let's see...

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