CD 13.10 Azo Compounds

An important aspect of the CD topic is the production of dyes. Azo compounds are used as dyes, and are more fast than diazonium salt dyes because they are more stable. The video above is made for the OCR specification, so go take a look and enjoy the jazzy music, check out Ash Z's channel for more revision videos.

Azo compounds contain an -N=N- group. It is highly stable because it becomes part of the delocalised system involving the arenes.

They generally look like this with different groups coming off eg. The OH group would come from reacting a diazonium ion with phenol as a coupling reagent, and the NH2 group would come from reacting with phenylamine.

The first thing we need is a diazonium salt. The book gives the example of the benzenediazonium ion. For the exam we need to know how this is made.

A cold solution of sodium nitrite (sodium nitrate (III)) is added to a solution of an arylamine (an arene with an NH2 group) in dilute acid. It's kept below 5 degrees because it'll decompose at higher temperatures.

First the sodium nitrate reacts with the acid (usually hydrochloric or sulfuric) to form nitrous acid.

The nitrous acid reacts with the arylamine (in this case phenylamine.)

Azo dyes are made in coupling reactions. The diazonium ion acts as an electrophile and reacts with the benzene ring of the coupling agent.
Examples in the book are with phenol and amines.

The different functional groups on the azo compounds causes a slight difference in colour as it affects the chromophore.

CD 13.6 Oils and Fats

Yo. 13.6 is all about the oils and the fats. The video above is pretty helpful, just don't get sidetracked watching all the weight loss videos it recommends after watching, not that that happened to me...

Anyways, most fats and oils are esters of glycerol:



Condensation reactions join carboxylic acids with long hydrocarbon chains to the glycerol molecule.
For example, palmitic acid can be joined onto the glycerol molecule to form a fat.
Natural triesters are often mixed triesters, meaning they have different fatty acids.

Oils and fats can be spit apart by hydrolysis. This is usually one by heating the oil/fat with concentrated sodium hydroxide, which forms glycerol and an acid salt.

Fats tend to be saturated molecules (no double bonds) which can be packed tightly together by intermolecular bonds. This means that they are harder to break apart, and so they are solid.

Oils tend to be unsaturated, with double bonds, causing it to have a kink in the chain. This means that the intermolecular bonds will be weaker, and it's more likely to be a liquid.

You can convert oils to fats by removing the double bonds. The reaction that does this is called hydrogenation, where hydrogen is passed through heated oil with a nickel catalyst. You can do this reaction to different extents, you can do it until is hardens, or even until it hardens so much that it is brittle. So depending on what you want, you can control how much hydrogenation occurs. In Chemical Ideas they give the example of margarine.

Answers to the past exam questions are here. Looking through the past exam papers, fats and oils does come up, but not in much detail. It's mainly the 'draw the structure' questions and hydrolysis of fats and oils.

CD 12.4 Reactions of Arenes: Friedel-Crafts Reactions

Finally, I've reached the end of the chapter! The final electrophilic substitution reactions to look at are the Friedel-Crafts reactions. These are used to introduce side chains to the benzene ring.

Alkylation is the addition of an alkyl group (think of it as an alkane) and Acylation is the addition of an acyl group (R-CO-) The video above explains alkylation only, but don't start worrying! Acylation isn't that different.



The example we're given in the book for alkylation is with chloromethane, but the reaction can be done with any halogen containing organic molecule.

The chloromethane and benzene are warmed with an aluminium chloride catalyst:

As with the other reactions, when the chloromethane approaches the benzene, it becomes polarised. The aluminium chloride polarises it further:

It becomes polarised so much that the molecule splits up:

The chlorine molecule binds to the catalyst, aluminium chloride, leaving a carbocation, CH3+. The carbocation is an electrophile (loves those little negative electrons!) It attacks the benzene as benzene is electron rich. In doing so, the benzene releases a hydrogen ion to remain stable and form methylbenzene.

The hydrogen ion, as seen in the other reactions, forms HCl by reacting with the AlCl4-, this regenerates the aluminium chloride catalyst.



Acylation is the same process, but instead of using chloromethane, an acyl chloride is used such as ethanoyl chloride.

Ethanoic anhydride can also be used:

Ionic liquids can be used as solvents/catalysts. They are liquids at room temperature and contain only ions. They are seen s being more 'green' because of the following reasons:

1. They have low volatility (doesn't evaporate easily at room temperature) so reduce emissions.
2. Low flammability and toxicity so are safer to use.
3. Reactions can be carries out at lower temperatures with the ionic liquid than by conventional methods.
4. The ionic liquid can be recycled.

Friedel-Crafts come up quite a lot in the exams so make sure you've got them up to scratch. Howdy doo dah, I'm glad this chapter is over, my paint skills were really being stretched.

Answers to the exam questions are here.

CD 12.4 Reactions of Arenes: Nitration of Benzene

The nitration of benzene requires a nitrating mixture composed of concentration sulfuric acid and concentrated nitric acid.
The overall equation for the reaction is:

This nitrating mixture provides the electrophile for the electrophilic substitution reaction:

The electrophile is NO₂⁺, as a positively charged molecule it is attracted towards the negative charge of the electrons. 

Therefore the  NO₂⁺ attacks the benzene and binds to a carbon atom in the ring. A carbocation is formed (a positive carbon atom) and the delocalised system of electrons is momentarily disrupted. To fix this, the carbon pinches the electrons from the bond with hydrogen, and a hydrogen ion is released. 

If the temperature is over 55 degrees, further substitution will occur to produce compounds like the ones below.

The video above goes through the whole process in more detail, but its just like the other reactions in terms of it's mechanism.

It has appeared a couple of times in past papers, and they like talk about the nitrating mixture (ie. which acid acts as a base, which acid is stronger etc.)

Answers for the question below can be found here.

CD 12.4 Reactions of Arenes: Chlorination of Benzene

The chlorination of benzene is very much similar to the bromination of benzene. The video above does a very good job serving as a visual aid, thanks to dnseducation. The only difference is that Chemical Ideas gives the catalyst to be aluminium chloride rather than iron (III) chloride (however they both have the same mechanism.) So in the exam I'd use aluminium chloride as the catalyst.

Anyway, the reaction must take place in anhydrous conditions as the aluminium chloride reacts with water, which we do not want do we? Noooo, we do not my precious. (Sorry about the movie quotes, I need to cheer myself up since I just wrote this out 10 minutes ago then accidentally logged out of google and lost it all, rant over, let the learning begin.)

The overall reaction equation is shown below:

The first step in the reaction is the polarisation of the chlorine molecule, caused by approaching the benzene. Benzene is electron rich, so causes the electrons in the bond between the chlorines to repel away. This makes it have a slightly positive and negative end.

Aluminium chloride also has areas of slightly positive and negative charge. The chlorine atoms are highly electronegative and so the aluminium atom becomes positive. To fix this, the aluminium wants to gain electrons, and so accepts a lone pair of electrons from the chlorine. This in turn, makes the chlorine even more polarised, to the point where it splits apart. This forms the electrophile, a chloride ion, and the other chloride ion binds to the catalyst to form AlCl₄^-.

An electrophile loves electrons, it's like a gold digger... for electrons. So the electron rich benzene is a prime 'gold digging' target. The chloride ion therefore 'attacks' the benzene ring, and binds the to carbon.

This upsets the delocalised system of electrons momentarily, because it loses one to the bond with the chloride ion. That bitch of a chloride ion stole your electrons and your delocalised system is in trouble, so what are you gonna do?! You're gonna get them back by kicking off a hydrogen ion.

So the benzene takes the electrons from the bond with hydrogen, releasing it into the wild. It doesn't have any electrons of it's own so is positively charged. So now chlorobenzene has been formed - it lost two electrons to form a bond with the chloride ion, and gained two from kicking off the hydrogen, it is now stable.

The hydrogen ion goes on to react with the AlCl₄^- to form HCl and regenerate the catalyst AlCl₃.

And that is the story of the gold digging, chloride electrophile and benzene. It is a classic tale of electrophilic substitution.

I've not managed to find any past exam questions on it, so be prepared for it to come up. But by the looks of things, the exams focus on bromination, nitration and the Friedel-Crafts reactions in this chapter.

CD 12.4 Reactions of Arenes: Sulfonation of Benzene

This chemical ideas chapter is a little bigger than most, hence why I've broken it down into the separate reactions. The video above (Thanks to T Holbrook, go give him a thumbs up if you like his video) shows the reaction in a little more detail than needed, but if you want a thorough understanding go and watch it.

For those that want the simplest explanation that will get you the marks in the exam, read on.

The general equation for the reaction is:

The sulfuric acid provides the electrophile for the reaction, which is, SO₃. The video above explains how that happens but for the exam we just need to know that the electrophile is SO₃ and that it comes from the sulfuric acid.

It's an electrophile because the oxygen atoms surrounding the sulfur atom are electronegative, leaving the sulfur atom with a slightly positive charge.

Hence it's attracted towards the electron dense benzene.
It's the sulfur that attacks the benzene ring and binds to the carbon. With the addition of a hydrogen ion (formed from the sulfuric acid), benzenesulfonic acid is formed.

Benzenesulfonic acid can form salts in alkaline solutions. This can be particularly helpful as the ionic groups allow the molecule to be soluble.

That's it for sulfonation anyway, it's a relatively short part of the chapter. There's not been any past exam questions on sulfonation, but that doesn't mean it's not going to come up! Just remember that it's an electrophilic substitution reaction!

CD - 12.4 Reactions of Arenes: Bromination of Benzene

All reactions of arenes in this topic are electrophilic substitution reactions. Implant that in your head for the exam, when they ask the 'Underline two words that describe this reaction'. Easy two marks for you there!

The first magical arene reaction we'll look at is the Bromination of benzene. You'll see above a video that goes through it, it is great. It's a little differet to how the book explains it but watch it, it helps because you end up feeling more intelligent than the book. Kudos to Khan Academy, go give him a thumbs up.

The reaction involves benzene (funnily enough), bromine and iron (III) chloride. The overall reaction is shown below:

(excuse my poor paint skills)

It's actually a little more complicated than that though.

First, the bromine approaches the benzene and becomes polarised. This is because the bromine has a high electron density, remember those delocalised electrons? So the electrons in the bond between the bromines, repel the electron density in the bromine, so one end is more negative than the other:

The addition of iron (III) bromide causes the bromine molecule to split apart. This is because iron (III) bromide has groups of different electronegativity, you can imagine it as seen on the below: 

The bromine atoms have a greater 'pull' on the electrons so are slightly more negative than the iron. So the iron atom is slightly positive and wants to gain electrons. It can accept a lone pair of electrons from the bromine:

This causes the bromine molecule to become attached to the iron (III) bromide molecule.

The iron (III) bromide and bromine molecule become polarised. The attached bromine atom wants to gain more electrons, which is does by stealing the bonding electron and splitting apart from the other bromine atom. This forms FeBr₄^-

This leaves a positively charged bromide ion (lost it's electron so becomes positive.) As an electrophile (electron loving bundle of joy) it is attracted towards the electron density that the benzene conveniently has. So it binds to the carbon ring.

This causes a bit of a problem, as the carbon it joins to becomes positively charged due to the addition of the bromide ion. Therefore it expels a hydrogen ion:

Things are starting to look a little better now. The positively charged hydrogen ion can grab hold of that bromine on the FeBr₄^- , so that the iron (III) bromide can reform and return to it's neutral, stable state.

And there we have it, our products, bromobenzene, hydrogen bromide and the reformed iron (III) bromide.

For the exam, you don't need as much detail, but understanding it will make it a lot easier to remember the steps, plus if you sound like you know what you're talking about you're more likely to get BOD marks! A summary of the steps more suitable for the exam is listed below:

1. Bromine becomes polarised as it approaches the benzene.
2. Iron (III) bromide accepts a lone pair of electrons from the bromine causing it to become so polarised it splits apart and  FeBr₄^- is formed.
3. The remaining bromide ion acts as an electrophile and binds to a ring carbon atom and a hydrogen ion is lost from the ring.
4. The hydrogen ion reacts with  FeBr₄^- to regenerate the iron (III) bromide catalyst and form hydrogen bromide.

If you're going to be asked a question on the bromination of bromine, it is likely to be about it's the process of the reaction. There's little mention of it in Storylines so I think we're safe on that front!

For those superb students, I've added a past exam question below, answers can be found here.

Any corrections to what I've said above is most welcome, I'm learning as well as you!