Content

Chains, Rings and Spectroscopy

Note: From 2008, this will be renamed 'Rings, Polymers and Analysis'.

This is an extension of the things from AS Chains and Rings. So while you revise this, it'll also be useful to take a quick look through both Foundation and Chains and Rings... a lot escapes the mind in the long sleep of the summer...

Benzene

Benzene is another type of hydrocarbon. Its structure is important in understanding its chemistry... and it puzzled chemists for a long time.

Benzene's basic formula is C6H6, but how can that be possible when each carbon needs to form four bonds? A scientist called Kekule had an idea that it contained bonds like this:

The alternating double and single bonds were accepted for a long time and in some textbooks today, you'll still see it for representing benzene. Unfortunately, though, it's wrong.

Using more sophisticated machinery, chemists were able to find that the lengths of the bonds were not long, short, long, short... but actually somewhere in between long and short, all the same length. How is this possible? The electrons found in the p orbital become delocalised and form a ring above and below the carbon chain. It actually looks like this:

  • This is a planar molecule.
  • The angle between the sigma bonds is 120o.
  • The delocalised π system contains six electrons, which move above and below the carbons.
  • The electron ring is formed by the overlapping of p orbitals.

In skeletal formula, it tends to look like this:

Naming

Benzene can form position isomerism, depending on where groups are found attached to the ring. Naming is relatively simple.

First we need to label each carbon with a number:

Next it's just a case of drawing on any alkyl or halogen groups attached, according to the number in their name.

Example: 1,3-dibromobenzene.

From the name, we know bromine will be found on carbon 1 and carbon 3.

No number means that it is attached to the first carbon.

Example: cloro-4-methlybenzene:

From the name, we know there is a chlorine attached to the first carbon and a methyl group attached to the 4th carbon.

Reactions

Though benzene is unsaturated, it doesn't undergo electrophilic addition like alkenes. The delocalised electron ring makes it much more stable, because the electrons are being shared by 6 carbons, rather than just two in a normal bond.

Instead, it undergoes electrophilic substitution. There are three reactions that you need to know: Nitration, Halogenation and Alkylation.

Nitration

This is the mononitration of benzene (adding a NO2+ nitronium ion), using cH2SO4 and HNO3.

Overall equation: C6H6 + HNO3 -----> C6H5NO2 + H2O
Catalyst: cH2SO4
Conditions: Reflux at 50oC.

We also need to know the mechanism, which is in three steps:

Step one... react the two "acids" together. H2SO4 is a stronger acid, and will therefore cause HNO3 to accept a proton and act as a base:

H2SO4 + HNO3 -----> HSO4- + H2NO3+

H2NO3+ isn't very stable, though and will break up to produce water and the nitronium ion:

H2NO3+ ------> H2O + NO2+

Step two... the nitronium ion reacts with benzene as follows:

Step three... To restore the benzene ring, the bond between benzene and H is broken, and the electrons rejoin the ring. H+ reactions with HSO4-, to regenerate the catalyst:


HSO4- + H+ -----> H2SO4

That's all there is to nitration... make sure you learn the curly arrows!

Halogenation

Because the benzene ring isn't very reactive, halogens can't react and substitute the hydrogens without the help of a halogen carrier. These cause a positive dipole on the halogen, enough to allow them to react.

Catalyst halogen carriers: Fe, FeBr3
Conditions: Room temp
Equation: C6H6 + Br2 -----> C6H5Br + HBr

No mechanisms to learn here, so it's pretty straight forward!

Alkylation

Alkyl groups are things like -CH3 and also require a halogen carrier (same reason as above!). This is called the Friedel-Crafts reaction.

Catalyst halogen carriers: AlCl3
Conditions: Reflux at 60oC
Equation: C6H6 + CH3Cl -----> C6H5CH3 + HCl

Benzene Uses

Benzene is useful in industry for three things:

  1. Explosives
    Converting benzene to methylbenzene in the Friedel-Crafts reaction is the first stage of making explosives. The methylbenzene is then nitrated to trinitromethylbenzene (TNT).
  2. Polystyrene
    Styrene (below) is formed from a series of reactions starting with benzene and can be polymerised to form polystyrene.
  3. Dyes
    Nitrobenzene can be converted into phenylamine, a starting material for dyes. We look at dyes later in this page...

And that's everything you need to know about benzene. Not too bad, hmm?

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Phenol

Phenol is a benzene ring with a OH group attached:

Naturally, the formula is: C6H5OH.

Reactions

Phenol is more reactive than benzene. This is because the 2 pairs of electrons in the p orbitals of the O become delocalised and join the ring of delocalised electrons. There are 10 electrons in the ring, rather than 6, and this means more electron density -- the benzene ring is said to be activated. The extra electron density can induce dipoles in nearby reagents, and so they can react without a halogen carrier.

Adding Br2 is a much easier electrophilic substitution:

+ 3 Br2 -----> + 3 HBr

The product is called 2,4,6-tribromophenol. Simple enough if you remember that you need 3 mol of Br2. Be careful if phenol already has groups attached to 2, 4 or 6... They won't be replaced!

Phenol is also a weak acid. (pH=3.4)

This occurs because, unlike alcohols, the negative charge on the phenoxide ion is spread around the benzene ring, thus giving the ion extra stability.

Reacting with sodium would produce fizzing (H2 gas), the metal would dissolve and sodium phenoxide would be produced:

+ Na -----> + H2

Reacting with sodium hydroxide would produce water, the metal would dissolve and sodium phenoxide would be formed again:

+ NaOH -----> + H2O

Phenol and its compounds are used as disinfectants and antiseptics.

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Carbonyl Compounds

Oooh... this ought to ring some bells... remember the old carbonyl functional group from Chains and Rings? Aldehydes and Ketones? Well, we've got to look at them again now...

The carbonyl functional group looks like this:

...and their general formula is CnH2nO.

Because of the C=O, they are all naturally unsaturated and are isomeric (same formula can produce different structures).

If the C in the C=O bond has a H attached, the carbonyl compound is an aldehyde. If not, it is a ketone. Example:

butanal (aldehyde)
butanone (ketone)

For carbon chains longer than butan-, the C=O needs to be identified on ketones. (e.g. hexan-2-one). This isn't the case with aldehydes because the C=O is always at the end of the carbon chain.

Reduction

As we found out at AS, carbonyls are formed from either oxidising a primary or secondary alcohol.

It stands to reason, then, that carbonyls can be reduced back into alcohols...

The reagent for reduction is sodium borohydride (NaBH4) (although other reducing agents like LiAlH4 can be used). In the equation, we just represent it with [H].

Simply, an aldehyde is reduced to a primary alcohol, and a ketone is reduced to a secondary alcohol.

aldehyde
 
primary alcohol
+ 2[H] ----->
CH3CHO
 
CH3CH2OH

ketone
 
secondary alcohol
+ 2[H] ----->
CH3CHOCH3
 
CH3CH(OH)CH3

Nucleophilic addition

Because of the electronegative O in the C=O, a dipole occurs in the aldehyde or ketone. δ+C=Oδ-. This means it will attract nucleophiles (electron pair doners) and nucleophilic addition occurs. The one reaction we need to know is the mechanism for hydrogen cyanide (HCN) with potassium cyanide as a catalyst.

Example with ethanal:

Step One

HCN -----> H+ + CN-
(throw in catalyst KCN to increase concentration).

Step Two

Step Three

The reaction is really useful because it increases the carbon chain by one carbon. The product is called a nitrile due to the -CN group. In this example, the product formed is ethanenitrile.

Nitriles can then be converted to carboxylic acids by reflux with dilute acid (hydrolysis).

This mechanism, like every other, is very important. You may be asked to draw it in the exam, so make sure you learn it!

2,4-dinitrophenylhydrazine

(try saying that word three times in a row... actually, do, it might help you learn it! 2,4-di...ni...tro...phen...yl...hy...dra...zine)

Or 2,4-DNP for short (but you need the full version in the exams) is a great test for the carbonyl group.

Chuck in 2,4-dinitrophenylhydrazine and orange crystals will quickly form.

If we filter these, recrystalise, find the melting point, we can then identify the carbonyl compound because each aldehyde and ketone has a slightly different melting point. Just compare your results with findings recorded in a book! Cool, huh?

Classic tests

Then we come onto the classic tests for distinguishing aldehydes from ketones. Not as cool as 2,4-dinitrophenylhydrazine because they don't tell you exactly what compound you have, but useful enough for telling you whether it's aldehyde or ketone...

Test Observation for aldehyde Observation for ketone
Tollen's Reagent
Warm with ammonical silver nitrate (producing the ion: [Ag(NH3)2]+ a mild oxidiser).
Aldehyde is oxidised to a carboxylic acid. The Ag is displaced and a silver mirror forms. No change because the ketone doesn't oxidise.
Potassium Dichromate
Warm acidified (cH2SO4) solution.
Aldehyde oxides to carboxylic acid again. Solution turns from orange to green. No change because the ketone doesn't oxidise.

So, if it's boring and doesn't do anything, you know you have a ketone!

That's it for carbonyls... make sure you learn it!

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Carboxylic acids

Carboxylic acids are a starting material in the manufacture of esters (which we look at later). The functional group is R-COOH, which we saw when we oxidised a primary alcohol twice back in AS...

The 120o bond angle means that it is a trigonal plainer.

General formula is CnCH2nO2.

Unless it involves benzene, naming is comprised of three stages:

  1. Work out the alkane carbon chain backbone and knock off the 'e' : propan.
    (N.B. If there's a C=C double bond, it would be propen, based on the alkene.)
  2. Work out what numbered carbon(s) the -COOH occurs on. For more than one -COOH group, remember di or tri. propan-1,3-di
  3. Finally stick on 'oic acid': propan-1,3-dioic acid.

Benzene-1,3,5-tricarboxylic acid would look like this:

Carboxylic acids, as the name suggests, are weak acids, about pH=4-5, and split like this:

CH3CH2COOH -----> CH3CH2COO- + H+

This occurs because of the electronegativity of the two oxygens. They pull electrons along the -OH bond, thus weakening the bond and making it more likely that H+ is formed.

In aqueous solution, water acts as a base to form:

CH3CH2COOH(aq) + H2O(l) CH3CH2COO- + H3O+

The equilibrium lies on the LHS.

Because of their acidic nature, they react with alkalis, bases and metals, to form salts. The anion (negative charge) is called a carboxylate ion, and so the salt CH3COO-Na+ would be called sodium ethanolate.

Some examples:

Reaction Equation
Zinc and ethanoic acid Zn + 2 CH3COOH -----> Zn(CH3COO)2 + H2
Sodium carbonate and methanoic acid

Na2CO3 + 2 HCOOH -----> 2 HCOO-Na+ + H2O + 2 CO2

Magnesium oxide and ethanoic acid MgO + 2 CH3COOH -----> Mg(CH3COO)2 + H2O
Benzoic acid and sodium hydroxide + NaOH -----> + H2O

And that's really all there is to carboxylic acids!

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Esters

Esters are typically defined by their "sweet smelling" nature (unlike most organic chemistry). They are formed by the reaction (reflux) of an alcohol with a carboxylic acid with cH2SO4 as a catalyst and the functional group is:

The red part is the carboxylic acid part of the ester and the blue is the alcohol part of the ester.

To name them:

  1. Deduce which carboxylic acid has been used: propanoic acid. Exchange 'oic' with 'oate': propanoate.
  2. Deduce which alcohol has been used: propanol. Exchange 'ol' with 'yl': propanyl.
  3. Put these together: propanyl propanoate.

Uses

  • Esters are good solvents and are used for dissolving substances like drugs and antibiotics.
  • Because of their pleasant smell and volatility, they are used in flavourings and perfumes.
  • Waxes are esters comprising of long chains of carboxylic acids or alcohols.

Hydrolysis

The opposite of esterification (formation of an ester) is a hydrolysis reaction (addition of water to split; from hydro=water and -lysis=splitting). When an ester is hydrolysised, the following reaction occurs:

Hydrolysis occurs when the ester is refluxed with a weak acid (HCl) as a catalyst.

If a base is used as a catalyst, however, (like NaOH) the hydrolysis occurs to form the acid and the alcohol, but then neutralisation occurs when the acid and base react... forming a salt and water.

Those are basic facts for esters!

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Amines

This is another important group you need to learn about. Useful for making dyes and we wouldn't be here if we didn't have amino acids in our bodies!

Okay, so what are amines? They are derived from ammonia (NH3) and have a functional group called the amonio... -NH2 bond.

Their general formula is CnH2n+1NH2.

They can be primary, secondary, tertiary or even quanternary:

R-NH2
Primary amine
Secondary amine
Tertiary amine
Quanternary amine

They are named depending on what carbon the amine group is attached. For example:

There is a methly group attached and so it is methyl amine. Although there is a benzene ring, it's referred to as phenol, so phenylamine.

Primary amines are bases. Because of the lone pair on the N, they will accept H+ ions. (Remember, the definition of a base is a proton acceptor).

They form ammonium ions -NH3+.

For example:

CH3CH2NH2 + H+ -----> CH3CH2NH3+
Ethylamine Ethylammonium ion

How basic the amine is depends on the electron density of the lone pair on N... the greater the density, the more basic.

The alkyl group has a positive inductive effect, which means that electrons from surrounding bonds can be pushed onto the N. So, the more carbon branches attached to the N, the more electrons, the greater the electron density, and the more basic.

A benzene ring pushes electrons in the opposite direction, allowing them to delocalise into the ring, and so has negative inductive effect. From this, we can figure out that:

Least basic.
NH3 More basic.
CH3NH2 More basic.
Very basic.

Acting as bases means that they'll react with acids to produce salt.

For example:

CH3NH2 + HCl -----> CH3NH3+Cl-
methylamine methylammonium chloride

Naming is based on the ammonium ion.

C6H5NH3+Cl- is called phenylammonium chloride, because the amine with a benzene ring is called phenlyamine.

Dyes

A good thing about phenylamine is that it can be used to make dyes. The bad thing is, you have to learn how it's done... all the conditions and steps. So take a deep breath, and let's begin. At least when you know it, you'll get lots of easy marks in the exam!

Overview of the process:

Step one

Just use a mixture tin and concentrated hydrochloric acid and reflux to ensure reduction reaction goes to completion:

Step two

Sodium nitrate (NaNO2) and hydrochloric acid form nitrous acid (HNO2) at less than 10oC. This then reacts with the phenlyamine to produce benzene diazonium chloride:

Step three

Finally, if phenol is added in the presence of an alkali, an azo dye is formed:

 

This reaction is very important economically because azo-dye compounds are the starting point for the dye industry. Azo-dyes are also used for acid-base indicators like methly orange.

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Amino Acids

All you biologists ought to be pretty knowledgeable about this one already, but for us mere chemists, here's what you need to know...

The general formula of the α amino acid is: RCH(NH2)COOH. You'll notice right away that it has the carboxylic acid functional group and the amine group. To draw this in display formula would look like:

There are two amino acids we need to learn:

Glycine
Alanine

Because of the two functional groups, amino acids can be either acids or bases, depending on the conditions they are subjected to:

  1. basic amino group (NH2) can accept protons (H+)
  2. carboxylic acid group (-COOH) can donate protons (H+)

This ability is given the posh name: amphoteric. Amino acids are amphoteric.

To show this, look at glycine and alanine reacting with either acidic or alkali solutions:

Amino acid With HCl With NaOH
Glycine
Alanine

Zwitterions

An unusual feature of amino acids is their much higher melting point than expected for their molecular masses. They exist as white crystalline solids. An explanation for these properties is that in intermediate pHs, they form a type of "inner salts" called zwitterions where the H is transferred from COOH to NH2 to form: NH3+ and COO-.

These two charges cancel out the overall charge of the molecule, but the charges attract opposite charges on neighbouring zwitterions and ionic bonds are formed. These are stronger than the usual hydrogen bonds that would exist on uncharged amino acids and explains the higher melting points.

The general formula of the zwitterion is:

Depending on the pH, however, this temporary formation changes. At a high alkali pH, the proton on NH3+ will be lost, leaving an overall negative charge from COO-. At a low acidic pH, a proton will be donated to COO-, leaving an overall positive charge from the NH3+. When there is no overall charge, the pH is called the isoelectric point. This is pH 7.2.

Example:

Substance 2-phenyl-2-aminoethanoic acid
pH = 2.1
pH = 7.2
pH = 13.6

Proteins

Proteins are chains of amino acids linked by peptide bonds.

Peptide bonds look like this:

When the amino acids react together to form bonds, H2O is lost. The proper term for this elimination of a small molecule is a condensation reaction.

Depending on the orientation of the amino acids, two different products can be formed. For example, when glycine and alanine reaction, the following reactions can take place:

 

+

----->
+ H2O

+

----->
+ H2O

To get back the individual amino acids, we need to add water -- hydrolysis. To achieve this, the protein had to be refluxed in concentrated (6M) hydrochloric acid for several hours.

That's just about it for amino acids!

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Acyl Chlorides

Acyl chlorides are speedy compounds... or they readily let other reactions occur.

They are based on the carboxylic acid and their functional group looks like this:

There are two reactions to form them:

CH3COOH(l) + PCl5(s) -----> CH3COOCl(l) + PCl3(l) + HCl(g)
ethanoic acid phosphorus (V) chloride   ethanoyl chloride  

and

C6H5COOH(l) + SOCl2(l) ------> C6H5COOCl(l) + SO2(g) + HCl(g)
benzoic acid sulphure dichloride oxide   benzoyl chloride  

The second reaction is less messy, because you don't have to separate two liquids - both by-products effervesce.

They are useful because they are more reactive than carboxylic acids and can be used in the preparation of esters, carboxylic acids and the forming of a peptide bond...

Preparation of esters

For example:

CH3COCl(l) + CH3OH(l) -----> CH3COCH3(l) + HCl(g)
ethanoyl chloride methanol   methyl ethanoate  

NaOH is often added to the reaction to neutralise the HCl produced and prevent the ester hydrolysing back to the carboxylic acid and alcohol.

Preparation of carboxylic acid

Though not soluble in water, they hydrolise quickly into a carboxylic acid and HCl. For example:

CH3COCl(l) + H2O(l) -----> CH3COOH(aq) + HCl(aq)
ethanoyl chloride     ethanoic acid  

Preparation of a peptide link

 

 

 

 

 

 

 

 

CH3COCl(l) + CH3NH2(l) -----> CH3CONHCH3(l) + HCl(g)
ethanoyl chloride methyl amine   amide  

This reaction takes place at room temperature and is an example of a condensation reaction.

That's all there is to acyl chlorides!

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Stereoism

Stereoisomers, remember, have the same structural formula, but a different display formula. They occupy a different 3D space.

You probably remember geometric stereoism from AS Foundation Chemistry. That's the type that involves cis or trans groups, across a C=C double bond, which has no free rotation.

The classic example is with but-2-ene:

Cis Trans
The CH3 groups are on the same side of the C=C double bond. The CH3 groups are across the C=C double bond, on opposite sides.

Another type of stereoism is called optical isomerism. For this, there has to be a chiral centre. A chiral centre, if you're wondering, is a carbon surrounded by four unique groups.

For example, an amino acid like alanine has a chiral centre - the middle carbon, with all the branches coming from it:

So, when there's a chiral centre, something called optical isomerism occurs. This is where two molecules with the same formula are mirror images of each other:

|
|
|

They can't be superimposed on each other... just try it out with a modelling kit if you don't believe me!

Chirality with Organic Synthesis

When an organic compound is synthesised to contain a chyral centre, two products are often formed in the lab, especially when planar molecules are used. For example, when ethanal reacts with CN- in nucleophilic addition, to increase the carbon chain by one, the nucleophile can either be added to the top of CH3CHO or the bottom:

This results in a mixture of products, to the ratio 1:1. The mixture is called a racemic mixture or racemate and only half will be optically active.

In nature, however, just one optical isomer is made, and the body, being natural, will only react to one type of the optical isomer. The product is therefore optically active.

For things made in the lab, with a mixture of optical isomers, (like medicines) there are a number of considerations:

  1. Requires a larger dose because half of it won't work...
  2. ...and therefore costs more to produce.
  3. What nasty side effects/unpredictable consequences might be caused by the optical isomer that doesn't work?
  4. How toxic is it to the population? (some people might be affected by the optical isomer, while others aren't).
  5. How effective is it?

That's it for this little topic. Nothing too nasty, huh?

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Polymerisation

If it was in GCSE and AS, what were the chances of escaping it at A2?

Well, remember that addition polymerisation is the joining of a long chain of monomers (alkenes) using the Ziegler catalyst, heat and pressure, and it can be drawn like this:

Or:

And to remind you of the problems:

  • Because of the non-polar bonds (C-H) or very strong polar bonds (C-F), they are non-biodegradable. They persist in the environment as litter.
  • When they're burned, they often produce toxic gases. Scientists are trying now to develop polymers that can be burned... and used as a fuel.
  • The best way to deal with them is to recycle them.
  • New processes are also being developed for cracking polymers back to alkenes.

Depending on the arrangement of the monomers within the polymer, the plastic can either be either a crystalline or soft, elastic, rubbery material. Take polypropene for example:

Polymer Type
Structure
Skeletal Formula
Properties
Isotactic
*crystalline
*packed closely together
*strong Van der Waals
*hard and strong
Syndio
-
tactic
*crystalline
*
packed closely together
*
strong Van der Waals
*
hard and strong
Atactic
*low crystallity
*
random structure
*
weak Van der Waals
*
soft and rubbery

Condensation Polymerisation

We've already met condensation reactions with amino acids, which means elimination of a small molecule when two molecules react. Condensation polymerisation is just the same, but is when lots of monomers react together and kick out lots of small molecules.

For it to take place, the two monomers must have two fuctional groups, meaning each one can react twice, and thus form a long chain.

The reacting functional groups are:

  1. -OH and -COOH to form an ester link (-CO-O-)
  2. -COOH and -NH2 to form a peptide link (-CO-NH-)

The molecule kicked out in both examples is water.

For example when the following monomers react:


ethan-1,2-diol

and
benzene-1,4-dicarboxylic acid
they eliminate H2O to form:

This is called PET (poly(ethylene terephthalate)).

And when the following monomers react:

H2N(CH2)6NH2

1,6-diamino-hexane

and
hexane-1,6-dicarboxylic acid
they eliminate H2O to form:

This is called Kevlar.

It takes a while to get used to drawing them... but you need to make sure you can do it!

Plastics formed in condensation reactions are used for fibres and bottles...though ester links can start to decompose if NaOH is transported in such a bottle (forming -O-Na+).

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Mass Spectrometry

Remember the old mass spec from AS foundation? Ionisation, acceleration, deflection, detection... If you can't, click here!

So far, we've only encountered elements going through the mass spectrometre, but organic molecules can go through too.

When ionisation takes place for organic molecules, the molecules are broken up into a range of ions. For example, ethanol might break up like this:

----->
OH+ + CH3+ +

Some of the ethanol, though, will only be hit by a few electrons, and will form CH3CH2OH+. This is called the parent ion or molecular ion and is the one that we're really interested in. It gives the relative molecular mass and is usually the largest peak (the most stable compound) and found to the right of the reading.

So, the mass spec of ethanol would look something like this:

Reading off the m/z - the mass charge density, we see 46 for ethanol. This corrisponds to CH3CH2OH... (12 x 2) + (6 x 1) + (16 x 1) = 46.

The question might ask you to deduce the relative molecular mass, and so it's simply a case of locating the parent ion and reading off the mass charge density... nothing to it!

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Infrared Spectroscopy

You'll no doubt remember this one from AS Chains and Rings too... So, after you've had a look through that page by clicking here, to recap:

Functional Group Wavenumber (cm-1)
Hydroxyl group (-OH) in carboxylic acids (-COOH). Very broad, between 2,500-3,300.
O-H hydrogen bonded in alcohols and phenols. Less broad, between 3,230-3,550.
Carbonyl group (C=O) in aldehydes, ketones and carboxylic acids. Narrow, 1,680-1,750.
  1. One broad peak (2,500-3,550) means you have an alcohol.
  2. No broad peak, but a narrow (1,680-1,750) peak is either a ketone or aldehydes.
  3. A reading containing both a broad peak and a narrow peak is a carboxylic acid.
  4. The only new compound you might need to identify is an ester. This has no broad peak, but a C=O (1,680-1,750) and also C-O narrow peaks (1,000-1,300).

When looking at the infrared, you might see the broad peak (2,500-3,550) which means you have an alcohol. If this is oxidised in reflux and the peak disappears, but a narrow (1,680-1,750) peak appears, you know the original was a secondary alcohol, oxidising to a ketone. If the broad peak remains and the narrow peak appears, you know the original was a primary alcohol oxidising twice to a carboxylic acid.

All the figures are given in the data sheet, so you don't even have to memorise them! Just make sure you copy them out accurately!

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NMR

Nuclear Magnetic Resonance spectroscopy is very useful for identifying compounds. Like the other spectroscopies, it draws out a graph, and this one shows the different hydrogen environments in a molecule.

What are hydrogen environments, you might wonder... take a look at the following three molecules:

Notice the symmetry of the molecule. All the hydrogens - 3 on each carbon - are in the same environment.

The Hs attached to the middle carbon are different to those on the end, so there are two environments.

Here, there are three different environments.

The area of the peaks gives a ratio of the hydrogens. For example for CH3CH2CH3 above, the ratio would be 3:1.

Each different environment has a slightly different δ shift value along the x-axis of the graph. This means, by comparing the peaks' positions with data, we can figure out exactly what each peak refers to.

Even better, like infrared, all the data is found in the data book!

Type of Proton
Chemical Shift, δ
R-CH3
0.7-1.6
R-CH2-R
1.2-1.4
R3CH
1.6-2.0
2.0-2.9
2.3-2.7
-O-CH3
-CH2-R
3.3-4.3
R-OH
3.5-5.5
6.5-7.0
7.1-7.7
9.5-10
11.0-11.7

Another interesting thing about n.m.r. is called high density n.m.r. The peaks of the graph sometimes slip up... and this depends on a rule called the n + 1 rule.

It uses the number of hydrogens on an adjacent carbon. For example:

There are two environments here, one containing OH and one containing CH2. For the OH group, there are no hydrogens on the adjacent carbon... and so therefore, using n + 1, the peak would split into one. For the CH2, however, there are two hydrogens on the adjacent carbon, and so, the peak would split into three.

So if you're given something like this, and asked to work out what the compound is:

First, we note that there are three peaks, with a ratio of 1:2:3.

Next, we look in on the data sheet for the similar δ values.

R-CH3 has δ = 0.6-1.1 so, fits the range of 3 peak.
R-CH2-R
has δ = 3.5 -3.9 so, fits the range of 2 peak.
R-OH
has δ = 3.5-5.5 and fits the range of 1 peak.

From this, we can deduce the molecule is:

This matches the ratio.

We can also check with high density n.m.r.

For -CH3, there are two H attached to the adjacent carbon, so the peak splits into three. This matches the graph.
For -CH2-, there are three hydrogens on the adjacent carbon, so it splits into four - as seen on the graph.
And finally, the hydrogen in -OH is only attached to a O, (no carbons and no other hydrogens), so it just splits into 1 peak.

An isotope of hydrogen, known as D, isn't detected by the n.m.r. reading. Therefore, if you mix an alcohol/carboxylic acid with D2O (isotopic water), something called proton shuffle will occur and the -OH in the alcohol/carboxylic acid will be replaced by -OD. When reading the n.m.r. of the alcohol again, the peak for the -OH will disappear.

That's all there is to n.m.r. It's awesome, doncha think?

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Synthetic Routes

Synthetic routes is here to make sure you didn't just sleep through Chains and Rings... It drags out all the equations that you might have ever learned and requires you to put a few in a logical order to get from one starting molecule to a finishing product. Included in this are the conditions and equations, so you really need to know it. On the plus, if you get it right, it's worth lots of marks!

So, just learn all these equations, and I mean, REALLY learn them, and you'll have no difficulty:

Reagents Equation Conditions
Combustion of propane C3H8 + 5O2 -----> 4H2O + 3CO2  
Chlorination of methane CH4 + Cl2 -----> CH3Cl + HCl

UV light
(free radical substitution)

Hydrogenation of ethene CH2=CH2 + H2 -----> CH3CH3 Ni catalyst/400oC
(electrophilic addition)
Addition of bromine to ethene CH2=CH2 + Br2 -----> CH2BrCH2Br Dark
(electrophilic addition)
Addition of hydrogen bromide to ethene CH2=CH2 + HBr -----> CH3CH2Br Room temperature
(electrophilic addition -- remember TWO products can form here).
Ethene with steam CH2=CH2 + H2O(g) -----> CH3CH2OH cH3PO4 cat/300oC/70 atmos
(electrophilic addition -- remember TWO products can form here).
Polymerisation of ethene

Ziegler cat/heat/pressure
(addition polymerisation)

Combustion of ethanol CH3CH2OH + 3O2 -----> 3H2O + 2CO2  
propan-2-ol and HBr CH3CH(OH)CH3 + HBr -----> CH3CHBrCH3 + H2O Remember HBr is made from NaBr + H2SO4.
ethanol and sodium CH3CH2OH + Na -----> CH3CH2O-Na+ + ½H2  
ethanol with concentrated sulphuric acid CH3CH2OH CH2=CH2 + H2O (dehydration)
ethanol with aluminium oxide CH3CH2OH CH2=CH2 + H2O (dehydration using pumice)
ethanoic acid and methanol

CH3COOH + CH3OH -----> + H2O

H2SO4 cat/reflux
(esterification)
Ethanol with K2Cr2O7 and H2SO4 to make ethanal. (Using [O]). CH3CH2OH + [O] -----> + H2O (oxidation)
Propan-2-ol with K2Cr2O7 and H2SO4 to make propanone. (Using [O]). CH3CH(OH)CH3 + [O] -----> + H2O (oxidation)
Propan-1-ol with K2Cr2O7 and H2SO4 to make propanoic acid. (Using [O]). CH3CH2CH2OH + 2[O] -----> + H2O (oxidation, twice)
bromoethane and aqueous sodium hydroxide CH3CH2Br + NaOH -----> CH3CH2OH + NaBr heat
(nucleophilic substitution)
ethanolic ammonium and bromoethane CH3CH2Br + NH3 -----> CH3CH2NH2 + HBr heat
(nucleophilic substitution)
Bromoethane with ethanolic sodium hydroxide CH3CH2Br + NaOH -----> CH2=CH2 + H2O + NaBr heat
(elimination)

For example, a question might ask you to convert cyclohexanol into cyclohexan-1,2-diol.

The first thing to do is to draw out your starting and ending products:

Start:
End:

Right, so thinking cap on here... If you have an alcohol, what can you do to it? Well, you can oxidise it, but that won't do much to it...you can change it to a salt, still not very useful, brominate it, and dehydrate it. Of those, dehyration is the only one which could act as an "intermediate" reaction:

It's no use hydrolysing our new alkene, because we'd be back to where we started, but we could halogenate it...

Finally, if we reflux with aqueous NaOH, we can substitue the Brs to OHs and get our end product!

That's the kind of mental process you need to go through... Keep looking at the end product and seeing if the step you're about to take will help you to get there...

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That's about everything for Chains, Rings and Spectroscopy! Hasn't it been fun? Let me know if I've missed anything!