When a single compound contains both a nucleophile and a leaving group, an intramolecular reacti

When a single compound contains both a nucleophile and a...

Key Concepts

Intramolecular Reactions

These are reactions where the reactive functional groups are within the same molecule, resulting in bond formation between different parts of the molecule. The close proximity of the reactive sites often leads to faster reaction rates and can result in the formation of cyclic structures, which is a common outcome in intramolecular processes.

Nucleophiles and Leaving Groups

A nucleophile is an electron-rich species that seeks an electrophilic center to donate electrons, while a leaving group is a substituent that can depart with a pair of electrons. In reactions where these groups are present within the same molecule, the nucleophile can attack the electrophilic site, displacing the leaving group and leading to the formation of a new intramolecular bond.

Cyclization

Cyclization refers to the process by which an acyclic molecule forms a ring structure. In intramolecular reactions, cyclization commonly occurs when a nucleophile attacks an electrophilic center within the same molecule. The size of the ring formed is often influenced by the spatial arrangement and the distance between the reactive centers, with five- and six-membered rings generally being the most favored due to their lower ring strain.

Reaction Mechanisms

Mechanistic pathways, such as nucleophilic substitution (for example, SN2), are crucial in understanding how the reaction progresses. In intramolecular nucleophilic substitutions, the mechanism can involve a backside attack by the nucleophile on the electrophilic carbon, leading directly to ring closure and the displacement of the leaving group. The detailed mechanism often depends on the nature of the substrate and the reaction conditions.

Transcript

00:04 Let's say we wanted to find the product of the following reaction with the formula c7h1202 as shown in the box to the right.

00:15 There are a few things we need to note before we get into the actual reaction.

00:19 The first thing to note is that this is a two -step reaction.

00:23 We're able to tell this because there are two arrows between our starting material on the far left and our product on the box on the right.

00:30 So i'm just going to clearly label the steps, step one, and step two here.

00:37 Now notice that step one has already been solved for us, and we're just here to solve step two.

00:43 Before we do that, let's go back and review step one.

00:47 And in order to do that, we need to first fill in any missing lone pairs on atoms, because often in organic chemistry, the loan pairs are implied but not stated.

00:56 Recall that every atom needs four electron group surrounding it in order to fill its octet.

01:04 So let's start with the compound on the far left.

01:08 The atoms that are missing the lone pairs.

01:10 First of all, this bromine here has one bond, therefore it is missing three sets of lone pairs in order to have a total of the eight valence electrons.

01:20 Moving up to the oxygen here, this one has two bonds, therefore it's missing two sets of lone pairs.

01:26 The oxygen up top has two bonds as it's double bonded, therefore it's only missing two sets of lone pairs.

01:35 And then the hydroxide group above the arrow only has one bond, therefore it is missing three sets of lone pairs.

01:44 Moving on to the intermediate, it looks fairly similar in that the bromine still has only one bond, therefore it's missing those three sets of lone pairs.

01:53 The double bonded oxygen still has two sets of lone pairs.

01:59 And now this oxygen, which previously had a hydrogen attached, now is negatively charged.

02:05 So that tells us that it now obtained an additional set of lone pairs, as it must have lost that hydrogen atom.

02:12 And moving down below to this water molecule, we know that oxygen having two bonds to hydrogen is only missing two sets of lone pairs.

02:21 So just to review step one, which has already been solved for us, this appears to be an acid -based reaction.

02:29 We were able to tell this because a hydrogen was exchanged between two compounds.

02:34 Just to highlight where that occurred, the alcohol group, like we noted, is missing a hydrogen atom.

02:41 Therefore, this must have been our acid as it donated its hydrogen.

02:46 And what did it donate its hydrogen to? well, that must have been water.

02:50 Therefore, hydroxide here is going to be our base that must have grabbed that hydrogen atom off of our acid.

02:59 So now we've reached our intermediate here, and this is where we want to understand how do we get to our product.

03:07 So notice that this oxygen atom here is negatively charged.

03:11 And in this chapter, we learned that negatively charged species are nucleophilic, meaning that they are electron -rich.

03:19 So what i'm going to do is label this species up above as nuc minus, which stands for nucleophile, and minus reminds me that nucleophiles tend to be electron -rich, and oftentimes they can be negatively charged.

03:35 And we also learned in this chapter that nucleophiles react with electrophiles, which are electron -poor species.

03:43 And generally, electrons ' poor species can show up as alkaliads.

03:50 In other words, a carbon halogen bond.

03:54 The reason being that carbons are electron -poor, partially positive, when attached to a halogen, which is electron -rich, so we can label that as partially negative, due to differences in electronegativity.

04:10 And notice that we do see that trend within this same...

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