Explain why alkylation of an α-carbon works best if the...

Key Concepts

Steric Hindrance

Steric hindrance refers to the crowding around the reactive center due to bulky groups attached to the substrate. In the context of alkylation, primary alkyl halides have minimal steric bulk, allowing the nucleophile to effectively perform a backside attack. Tertiary alkyl halides, however, possess significant steric bulk that blocks the nucleophile's approach, thereby suppressing the SN2 pathway and rendering the alkylation reaction ineffective.

Alkyl Halide Structure and Reactivity

The structure of the alkyl halide plays a crucial role in determining its reactivity. Primary alkyl halides, with less steric bulk, are well-suited for nucleophilic substitution via the SN2 mechanism, ensuring efficient alkylation of the ?-carbon. In contrast, tertiary alkyl halides are too hindered, which not only slows the reaction rate dramatically but often diverts the reaction pathway away from SN2, thereby preventing the intended alkylation from occurring.

SN2 Reaction Mechanism

The SN2 reaction mechanism involves a one-step bimolecular nucleophilic substitution where the nucleophile attacks the electrophilic carbon, leading to the simultaneous displacement of the leaving group. This mechanism is highly sensitive to steric effects because the nucleophile must approach the reactive center from the backside, making unhindered substrates, such as primary alkyl halides, ideal for these reactions.

Transcript

00:01 Okay, this problem is asking us to predict the product of this reaction, if this reaction occurred by shaking for several hours.

00:05 So the important thing about this problem is that we have a symmetrical ketone.

00:09 Okay, that's important because if we didn't have a symmetrical ketone, we could have a potential discrepancy in the hydrogen that we're going to depronate with my base.

00:16 But because it is symmetrical, we're not going to have that discrepancy.

00:18 So using my base, my n -a -o -d, that's going to go ahead and take off the most acidic hydrogen on this molecule, which again, because it is symmetrical, we're not going to have any situation in which the difference between this carbon and this carbon.

00:30 Is different.

00:30 So we're going to take off this hydrogen.

00:33 It doesn't matter, of course, using my od minus.

00:36 Okay, so that od minus is going to take off that hydrogen, meaning that the carbon hydrogen bond, it's going to move its electrons onto this single bond, making a double bond here, and after it does that, i'm going to have to move the electrons onto this oxygen.

00:47 Okay, so i'm going to have this as my product in which i have my cyclohexane, and then i have my oxygen with a negative charge, and then my double bond right there.

00:56 So that is my enolate ion.

00:58 Okay, and then after i do that i will have formed my h -o -d.

01:02 Okay, so this is the part where we could have a little bit of difference in what's going to occur next because in reality, what i probably should do is move the electrons from this oxygen to get protonated by my hydrogen or d, but that's just going to be basically gotten rid of because we're going to have to go through the mechanism again.

01:20 So basically, if we're to protonate this, i would have the formation of an alcohol, but because i still am in basic conditions, i'm going to have to reform this exact compound.

01:28 So i'm just going to skip those couple steps because i want to get to the important stuff.

01:32 So if i form this enolate ion, i'm going to go ahead and move the electrons back over here, which again, i would if i usually have undergone that protonation step, but because it's going to be the same thing regardless, i'm just moving them back onto here.

01:43 So i'm basically just doing this.

01:44 Okay, if i move them back onto this single bond to make a double bond, i'm going to have to move the electrons onto this particular carbon.

01:51 Okay, and that will form basically back where i started.

01:55 Okay, so i'm going to have my oxygen right there, and then my set of loan pair is right here.

02:00 Okay, so this is where things get interesting because if we didn't shake, then we could get a situation, in fact several situations in which we just go ahead and get protein by that hydrogen.

02:09 Okay, but if we shake this for several hours, shaking will help us get the most stable product.

02:14 And by most stable, i mean that we're going to form a carbon deuterium bond.

02:17 Carbon deuterium bonds are stronger than carbon hydrogen bonds because they're a little bit harder to break.

02:21 Okay, so by using this carbon nucleophile, i'm going to go ahead and do an acid -based reaction.

02:27 In which i get pronated by my deuterium...

Please give Ace some feedback