PM1B - Autumn Functional Group Chemistry

Question 1

Why do we need to know about functional groups

Answer 1

We need to understand where the electrons are, where they moving and where we have areas of high electron density and low density because that’s what will form a reaction – it will be the movement of these electrons from high to low density that will form reactions

A good example of why we need to know about functional groups is to improve solubility- we can use the functional groups to change the solubility within the molecule and make it more soluble.

Question 2

Give examples of Carbon Reactive functional groups

Answer 2

1. Alkene
2. Ketone/Aldehyde
3. Ester
4. Amide
5. Carboxylic acid
6. Aromatic ring
7. Urea

Question 3

Give examples of Oxygen reactive functional groups

Answer 3

1. Ether
2. Alcohol / Hydroxyl
3. Phenol

Question 4

Give examples of Nitrogen reacive functional groups

Answer 4

1. Primary / Secondry / Tertiary amine
2. Aniline
3. Nitrile

We will see oxygen and nitrogen as nucleophiles

Question 6

Show the functional group of Halogens and what there role is in reactions

Answer 6

Halogens normally have F/CL/BR

There main role is being a leaving group

Question 7

Example of addition of Halogens to alkenes

Answer 7

1 - Polarisation between the bromines - Electrons from double bond will react with the bromine to form a C-Br bond and at the same time Br-Br is broken

2. C-Br sharing with bromine - the other carbon lost an electron so it has a positive charge ( gave electrons away to form the other C-Br bond)
3. Bromine will donate electrons to the +ve charge which forms our 3 membered ring with the positive bromine at the top.

At the same time the other bromine anion negative charge recognises carbon is lacking electrons so it moves its electrons towards that

4 - The Br attacks the carbon, we break the C-BR bond and we have added our neutral bromine across that bond

NB key points -1. halogens are very electronegative ) Holds electrons close and tight doesnt like donating )
2. Polarisability - when 2 areas of high electron density come close together the electrons repell one another. So in this scenario what we find is when the bromine gets close to the alkene, the electrons in the bromine-bromine bond will shift towards one bromine over the other and they will be repelled away from the alkene.As a result the bromine becomes electron deficient, because the bromine at the top is taking those repelled electrons towards itself.
3. The alkene is the nucleophile
4. Delta positive bromine lacks electrons
5. The BR positive charge is being shared with the carbon, but the electrons pushed more towards the BR as its electronegative. so - We then have another area of high electron density in this bromine anion with a negative charge. That then recognises where that carbon is lacking electrons, it will move its extra electrons towards that then we finally end up with a neutral species where both are happy.

Question 8

Exam Question Example 1

Answer 8

Question 9

Exam question example 1/2

Answer 9

Question 10

In Regards to electrophillic addition reactions for alkenes, what is the electrophile and Nucleophile for each of these reactions

Answer 10

the alkene is where the electrons are so that’s what we will identify as being the nucleophile.

So where we have HBR or water with acid in it, the acid has protons, and the protons are electro deficient, so the proton is going to act as the electrophile, and this will attract the electrons from the alkene as well.
When we look at an electrophile we see what they have in common. What they both have in common is they have both got this H plus. It might be bonded to the bromine but we know bromine is more electronegative, so we know hydrogen will be more electro deficient, so that’s whats going to act as our electrophile.

Question 11

Reaction mechanism for an addition of halogen to a symmetrical Alkene

Answer 11

1. We have a proton that lacks electrons ( full +ve charge) which will attract electrons in the alkene
2. Alkene reacts with and attacks the H+
3. the hydrogen will add to one of those carbons, the other carbon just like in the bromine example will have a positive charge.
4. So in the BR minus case, we would attack and add HBR across the double bond. In the water example, we would add the OH group. ( OH2 to start with then we lose a proton).

Question 12

Reaction mechanism for an addition of halogen to a unsymmetrical Alkene

Answer 12

In the example we have a Methyl group and Hydrogens
1. Alkene would come out and attack the hydrogen.
2. Break the Hydrogen Bromine =
3. This will give us our H+ and BR-
4. In this example the hydrogen is adding to the carbon at the top of the alkene, so the carbon at the bottom of the alkene gets a positive charge.
5. Br- will come and react with our product to give us our product

NB : 1 - we do have 2 different products
2. On the top we have a methyl group and 2 hydrogens, and here we have a methyl group bromine and 2 hydrogens.
3.What we can identify is, there are both carbocations. The first one is a primary
carbocation, its only bonded to the carbon there, everything else is hydrogens. The 2nd example we have a methyl group and carbon here, so this is a secondary carbocation, and we know these are generally more stable.
4. The most stable is the most favourable

Question 13

What is Markovnikovs rule in regards to helping us predict which product will result from electrophillic additions

Answer 13

we are going to form the most stable and favourable carbocation.

We add a hydrogen to the carbon that has already more hydrogens, we automatically make sure the positive charge ends up on the other carbon. That carbon will have less hydrogens which means its got more carbons.

Question 14

Homolytic vs Heterolytic

Answer 14

We also have homolytic. The difference is

Hetrolytic. Movement of 2 electrons to generate charged species. We do this with the curved arrow.

Homolytic. – movement of just electron to generate neutral species. What we actually do is produce radicals. A homolytic reaction is signalled by what is called a fish hook as per arrow above

Question 15

Using Markovnikovs rule, which reaction favours the rule

Answer 15

We are going to make the most stable carbocation and therefore make that particular product.

Homolytic reaction, similar process but the end result is total opposite.

The way we would see this is, heres our double bond in the middle. The heterolytic reaction is where we took our HBR, we call this an ionic process because we are saying this is H+ OR BR-, and this is what gives us our markovnikovs addition where we added the hydrogen to the carbon at the bottom, we form the most stable carbocation at the top, and then the bromine (the nucleophile ) came and reacted. So this gave us addition of hydrogen to carbon at the bottom, and addition of bromine to the carbon at the top.

When we do it by a radical process or homolytic process, what we get is the hydrogen attached to the carbon at the top and the bromine attached to the carbon at the bottom. We can also consider this to be anti-markovnikovs.

These 2 processes can give us different regional isomers.

Question 16

Give an Example of a symmetrical Homolytic Reaction

Answer 16

All we need to know is, if we see something like a peroxide in a reaction mixture, we have to think it’s a radical initiator, what that means is we have a BR Minus.

SO we have an oxygen bonded to an oxygen, if we do this in the oresence of light, shining a light will break this bond. One electron goes to one oxygen the other electron goes back to the other oxygen. This makes sense because we know oxygen is electronegative, it doesn’t like to give up its electrons, shining a light is essentially puting energy in the system. So the system is splitting that energy equally, the electrons are split and the system is happy. So we wnd of forming 2 oxygen radicals

Now if this is in the presence of our HBR, the hydrogen is electron difficiant, and we know we can use a pair of electrons to donate to that hydrogen so its still considered H+?.

Here we have electrons, and these radicals floating around. So essentially one electron is donated to the hydrogen, and in this H/BR bond, of the electrons that were helped to form this bond, the other one will go to the bromine.

(2nd diagram )
What we have done then is made a BR radical. The BR radical is a neutral species but it doesn’t have a full shell of electrons. We know BR minus is a nice stable Anion. In this case we have the electrons in this alkene. We also have the BR radical which is quite reactive so if we pair these electrons off it will essentially make it more stable.

What happens is the double bond breakes, one electron goes to the bromine radical ( so we can form the BR minus). The other goes to the Carbon atom (middle). This forms a carbon radical.

Radicals are reactive and we want to have our electrons paired up if we can. A radical is an unpaired electron so it would want to find a pair. Just like with the example of HBR, the carbon radical can attack the hydrogen, this will break the hydrogen bromine bond, one of the electrons will stay to form that new bond, the other will go to the bromine.

(last pic) |So what we have done is we have added the HBR across our double bond using a radical process and we have ended up with BR radical.

The reaction is similar, but instead of adding the H first, we form the bromine radical so we add the bromine first.