Draw the organic products formed in each reaction. |...

Name: Problem 52, Chapter 23: Organic Chemistry
Uploaded: 2019-06-10T19:07:00-08:00
Duration: 9 min 54 s
Channel: Nima Gharibi
Description: Draw the organic products formed in each reaction.

Key Concepts

Glucosamine

Glucosamine is an amino sugar commonly found in nature that plays a key role in the formation of various structural and functional biomolecules. It serves as a precursor for the synthesis of complex carbohydrates and glycosaminoglycans, which are important for maintaining the structural integrity of cartilage and connective tissues.

N-Acetylation

N-Acetylation is a biochemical modification process wherein an acetyl group is transferred to an amino group of a molecule. This modification changes the molecule’s properties, such as its reactivity and solubility, and in the context of glucosamine, results in the formation of N-acetylglucosamine.

N-Acetylglucosamine (NAG)

N-Acetylglucosamine (NAG) is the acetylated derivative of glucosamine and an important monomer in the biosynthesis of structural polysaccharides. It features a six-membered ring structure with multiple hydroxyl groups and an acetamido group, making it a critical building block in polymers like chitin.

Acetyl-CoA

Acetyl-CoA is a central metabolic intermediate that acts as an acetyl donor in many reactions, including the acetylation of glucosamine. Its role is pivotal in various biosynthetic and energy-producing pathways, facilitating the modification and integration of molecules within cells.

Chitin

Chitin is a natural polysaccharide comprised primarily of N-acetylglucosamine units. It forms the rigid exoskeletons of arthropods and cell walls of fungi, providing structural support and protection, and is a key example of how simple monomers can organize into complex, biologically significant polymers.

Transcript

0:00 Hello everyone.

00:01 Today we're doing chapter 23 problem 52 and this problem asks to draw the product in each of these reactions.

00:06 So let's look at a first.

00:09 So in a we have some sort of carbonyl group with an alpha halide and then we're adding some sort of reagents that will promote elimination.

00:17 So we know lithium bromide will dissociate in solution to lithium plus and br minus and br minus can act as a nucleophile to cause an elimination reaction to occur in which you'll pick up one of these protons making a bond and then the sigma bond will break to form a new pi bond.

00:36 So for a, we could say that we're going to get an elimination product and when you make a bond, you break a bone, that bromin will fall off and we will form a new elimination product in which you will form both the e and the z isomer.

00:56 So here we have we drew the e isomer, so we would also have the z with these two groups on the same side.

01:04 So for b, we're adding heat.

01:06 So what we know about heat is that it promotes spontaneous decarboxylations of alpha carboxylic acid group.

01:16 So if we see here, like this is a carbonylic acid.

01:18 So let me actually just blow this up a little bit.

01:22 Let's give it over this.

01:24 And then just blow this one up.

01:27 Yeah, that's better.

01:30 So now we see that this carboxylic acid is actually alpha to a carbonyl group.

01:36 So when you add heat, the lone parallels in this oxygen will go to form co2, making a bond.

01:44 You make a bond, you break a bond.

01:45 So this bond is going to break, go to this carbon nucleus and pick up this proton to neutralize it.

01:50 So for b, you can say that you're going to form a spontaneous decarboxylation product using heat.

01:57 And only for carbonyl, because we have another carbacilic acid here, but there's nothing alpha to it to promote the spontaneous decarboxylation.

02:04 Nothing is driving this reaction on this carbonic acid.

02:07 It's just happy to stay as is, but here because it's alpha to a carbonyl group it causes it promotes it to go through that spontaneous decarboxylation to form two products and increase the entropy of the system so now you have products for b now and see what we have well let's draw c out so you can actually visualize it so for c we're starting with one two three so one two three we're starting with this group and we're using lda, which is a strong base, to deprotonate this alpha proton, and then we're adding in alkali group.

02:56 So we're going to add those two carbons off this alpha position because we're forming a nucleophile, and these are an electrophil.

03:03 So we'll have an s &2 nucleophilic attack to add those two carbons here to form product c.

03:14 Now for d, what we have is an alpha halide, and we're reacting with a strong nucleophile, being this aiming group.

03:23 So what we can actually have is just a simple s &2 backside attack with inversion of stereochemistry here.

03:32 So for d, you just have a simple nucleophilic attack, s and 2 nucleophilic attack with inversion of stereochemistry...

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