In light of your answer to Problem 11-31, what product might...

In light of your answer to Problem $11-31,$ what product might you expect from treatment of 4 -bromo-1-butanol with base?
$$
\mathrm{BrCH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \quad \stackrel{\text { Base }}{\longrightarrow} ?
$$

Key Concepts

Intramolecular Nucleophilic Substitution

This concept involves a nucleophile within a molecule attacking an electrophilic center elsewhere on the same molecule to form a ring. In such reactions, the proximity of the reactive groups can lead to cyclization via a substitution mechanism, often favoring the formation of rings with minimal strain.

Alkoxide Formation and S?N2 Mechanism

A base can deprotonate an alcohol to form an alkoxide ion, which is a strong nucleophile. This alkoxide can then participate in an S_N2 reaction, attacking an electrophilic carbon that bears a leaving group. The intramolecular nature of the reaction often enhances its rate and selectivity.

Williamson Ether Synthesis

The Williamson ether synthesis is a method to produce ethers by reacting an alkoxide with an alkyl halide. In intramolecular variants of this synthesis, the alkoxide attacks a tethered alkyl halide, leading to cyclization and the formation of cyclic ethers.

Ring Closure and Ring Size Preference

Formation of cyclic ethers depends on the size of the ring being formed and the associated ring strain. Five?membered rings, such as tetrahydrofuran, are commonly favored due to their low ring strain, making them especially stable and synthetically accessible products in cyclization reactions.

Transcript

00:01 So we essentially need to figure out the product that we expect from the treatment of the 4 -bromo 1 butanol with a base.

00:11 So let's just draw this out.

00:14 And i'm going to draw it as like an actual molecule instead of just the letters to help me really visualize what's happening and where the attacks and the s &2 reactions and eliminations are occurring.

00:29 So we have ch2, bonded to ch2, bonded to an oxygen, which is bonded to a hydrogen.

00:43 The ch2 is also bonded to another ch2, which is bonded to another ch2, which is bonded to a bromine.

00:53 And the bromine has six, and the oxygen over here has four.

00:59 So if we're going to start a reaction, we're going to have the double arrows, and it's going to react with o -c -h -3, where the oxygen has 1, 2, 3, 4, 5, 6, and a negative charge.

01:18 So these atoms right here are going to attack the hydrogen, which is then going to detach the hydrogen, which is then going to detach the hydrogen from the oxygen, pushing more charges onto the oxygen.

01:31 And this gives us we have the original ch2, bonded to the ch2, bonded to the ch2, bonded to the ch2, bonded to the ch2 as well.

01:50 We saw the bromine here.

01:52 The bromine is just the same as what it was.

01:55 However, this oxygen right here then has six atoms and a negative charge.

02:00 So the second thing here would be where the oxygen right here is going to attack this carbon.

02:16 And so that means that the bromine is then going to have another extra bond...

Please give Ace some feedback