Predict qualitatively the isomer ratio for nitration of each of the following compounds:

Name: Problem 1, Chapter 9: Advanced Organic Chemistry. Part A. Structure and Mechanisms
Uploaded: 2020-07-06T19:09:57-08:00
Duration: 5 min 54 s
Channel: Tom Rutherford
Description: Predict qualitatively the isomer ratio for nitration of each of the following compounds:

step 1 Okay, this problem is asking us to show the aldol products. So these are the reactants that they

Predict qualitatively the isomer ratio for nitration of each...

Key Concepts

Electrophilic Aromatic Substitution (EAS)

Electrophilic Aromatic Substitution is a fundamental reaction mechanism in aromatic chemistry where an electrophile replaces a hydrogen atom on an aromatic ring. This process proceeds through a resonance-stabilized arenium ion intermediate and is key to predicting where substitution will occur on the ring.

Directing Effects

Directing effects arise from substituents already attached to the aromatic ring. Activating groups, which donate electrons by resonance or inductively, tend to direct incoming electrophiles to the ortho and para positions. In contrast, deactivating groups, which withdraw electrons, usually direct substitution to the meta position, as they destabilize the arenium ion intermediate at ortho and para positions.

Resonance and Inductive Effects

Resonance effects involve the delocalization of electrons across the aromatic system that can stabilize or destabilize reaction intermediates, thereby influencing regioselectivity. Inductive effects pertain to the electron-withdrawing or -donating power of substituents through sigma bonds, which also affects the electron density distribution on the aromatic ring and guides the site of substitution.

Steric Factors

Steric factors refer to the physical hindrance that may affect the approach of the electrophile to certain positions on the aromatic ring. Bulky substituents can block access to nearby positions, often favoring substitution at less hindered sites, which can modify the predicted isomer ratios determined by electronic factors alone.

Transcript

00:01 Okay, this problem is asking us to show the aldol products.

00:03 So these are the reactants that they're giving us.

00:06 So we have this aldehyde right here.

00:10 Okay, we'll just start with this one.

00:12 So if this aldehyde were to undergo an aldeol reaction, my product would be a case in which i reacted with, n -a -o -h, for example, and water.

00:25 So first things first is my n -a -o -h, my sodium hydroxide, is considered to be a base.

00:29 So that base is going to deprotonate this alpha -carbon, this alpha -hydrogen, because that alpha -hydrogen is considered the most acidic hydrogen on this molecule.

00:39 So that's going to deprotonate it, leaving the electrons onto this carbon.

00:44 Okay, so we should get this after that immediate step, in which we have my carbonyl, and then instead of a proton, we have lone pairs there.

00:52 Okay, so those lone pairs are going to react with another molecule of this alvehyde.

00:57 So after that reaction, we should get the movement of these electrons onto this carbonyl to form this product in which we have this aldehyde, which i'll represent in red.

01:13 And then in the alpha position, so this position right here, we're going to draw its connection to this carbon, which i'll represent in blue.

01:22 So it will be connected to this carbon, which is connected to an oxygen, which is connected to a hydrogen, and then it's connected to the rest of my chain.

01:35 Okay, and then after propination by the water that we just created it in the previous step, then we should get the protonation of the oxygen to get this.

01:45 Okay, so representing that in a more understandable form, we should get this as my final product.

02:02 So this is where i'll draw in red.

02:05 So this right here, this compound here, is the starting material that we reacted our base with, and then this part of my molecule is the second carbonyl, the second aldehyde.

02:16 Okay, moving.

02:17 On to part b, we have a benzene.

02:24 So this benzene has a ketone, and that ketone is going to undergo my alpha, or sorry, my aldol reaction.

02:35 So same thing, sodium hydroxide.

02:37 That sodium hydroxide is going to deprotinate the most acidic hydrogen on this molecule.

02:42 And if we look to this carbon, there's no hydrogen attached to that because this is an sp2 hybridized carbon right here.

02:48 We can only have the one other connection that is already connected to the ketone.

02:52 So we don't have any hydrogens over here.

02:54 Instead, we're going to left it right here.

02:56 And on that carbon, we have three hydrogens.

02:59 So we only need one.

03:00 So i'm going to deprotinate the hydrogen, move the electrons up to this carbon, and then we should get the deprotent version of this molecule.

03:10 Okay, so just lone pairs on this carbon.

03:14 And that is the carbon in which a nucleophilic attack will occur and attack another molecule of my starting material...

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