Ethers can often be prepared by SN 2 reaction of alkoxide...

Ethers can often be prepared by $S_{N} 2$ reaction of alkoxide ions, $R O^{-},$ with alkyl halides. Suppose you wanted to prepare cyclohexyl methyl ether. Which of the two possible routes shown below would you choose? Explain. (EQUATION CAN'T COPY)

Key Concepts

Ether Synthesis via Alkoxide Alkylation

Ether synthesis through the reaction of alkoxide ions with alkyl halides is a common method in organic chemistry. The mechanism typically follows an SN2 pathway, where an alkoxide ion attacks an alkyl halide to form an ether. The choice of which alkoxide ion and alkyl halide to pair is critical to avoid issues related to steric hindrance and to ensure a high-yield reaction.

Nucleophile and Electrophile Selection

In substitution reactions, selecting the appropriate pairing of nucleophile and electrophile is crucial. An effective strategy involves using a strong nucleophile with an electrophile that has minimal steric hindrance. This pairing enhances the likelihood of a successful backside attack in an SN2 process, leading to the desired product with optimal yield.

SN2 Reaction Mechanism

The SN2 reaction is a bimolecular nucleophilic substitution where the rate-determining step involves the simultaneous attack of a nucleophile on an electrophilic carbon and the departure of the leaving group. This mechanism requires a backside attack, making less sterically hindered electrophiles more reactive, and proceeds with inversion of configuration at the carbon center.

Steric Hindrance

Steric hindrance refers to the physical crowding around a reactive center in a molecule. In the context of SN2 reactions, less hindered substrates are preferred because bulky groups can block the approach of the nucleophile, thus slowing down or inhibiting the reaction. Choosing an electrophile with minimal steric bulk can lead to a more efficient reaction.

Transcript

00:01 So we have two different reactions here, and we would like to know which is the better reaction at synthesizing cyclohexyl methyl methyl ether.

00:10 So looking at the top reaction here, we see that we have a cyclohex oxide with our methyl iodide.

00:18 So we can go ahead and write which one's going to be our substrate and which one's going to be our nucleophile.

00:23 So our cyclohex oxide is going to be our nucleophile here.

00:28 So we go ahead and write that there.

00:35 And what we see here is that our methyl iodide, since it contains our alkylohalide, is going to be our substrate.

00:47 And we see that it is a primary alkylohalide.

00:51 So we can make that observation.

00:56 And for the second reaction here, we see that we have an iotocyclohexate.

01:01 And since that contains our alkylohalide, we're going to go ahead and say that is our substrate.

01:05 And it is a secondary substrate.

01:08 And for our nucleophile, we have a methoxide.

01:13 So we can go ahead and label that as our nucleophile.

01:20 Now, right off the bat, we can see that we are comparing a, the top reaction has a primary alkohalide with a second reaction has an iodo, i'm sorry, it has a secondary alkaliad.

01:30 We can't really make any conclusions yet from that.

01:32 So i want to go ahead and look at the reactions a little bit more in depth.

01:38 So looking at the first reaction here, we know that our substrate is going to be our methyliodide.

01:49 And for our nucleophile, we're going to have our cyclohex oxide, which i'll go ahead and draw there...

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