6.2

Question 1

6.2.1 a) Explain the different types of amines, and outline the IUPAC system of naming them.

Answer 1

Amines are organic derivatives of ammonia (where one or more hydrogen atoms on the ammonia have been replaced by an alkyl chain or ring):
NH3 = ammonia
RNH2 = primary amine
RNHR’ = secondary amine
RNHR’R” = tertiary amine
When naming amines, the root is determined by the longest hydrocarbon chain, and any additional (alkyl) groups are added as prefixes. The suffix is amine, and ‘N-‘ or ‘N,N-‘ is given as a prefix for secondary and tertiary amines, respectively.
e.g. CH3(CH2)3NH2 = butylamine
e.g. CH3(CH2)2N(H)CH3 = N-methylpropylamine
e.g. CH3(CH2)2N(CH3CH2)2 = N,N-diethylpropylamine

Question 2

6.2.1 a) Explain why amines are considered to be bases.

Answer 2

Similar to ammonia, amines are weak bases. The lone pair of electrons on the nitrogen atom allows them to accept a proton: a dative covalent (or coordinate) bond is formed when the nitrogen atom donates its lone pair of electrons to the proton, producing an alkylammonium ion:
RNH2 + H+ → [RNH3]+

Question 3

6.2.1 a) Explain how amines react with dilute acids.

Answer 3

When amines react with acids, alkylammonium salts are produced: this is formed by the proton in the acid being replaced by an alkylammonium ion.
Strong acids fully ionise in solution:
HA(aq) + H2O(l) → H3O+(aq) + A-(aq)
The amine accepts a proton from the hydronium ion to form an alkylammonium ion :
RNH2(aq) + H3O+(aq) → [RNH3]+(aq) + H2O(l)
The overall equation shows the production of an alkylammonium salt:
HA(aq) + RNH2(aq) → [RNH3]+(aq) + A-(aq)

Question 4

6.2.1 a) Outline the overall equations for making alkylammonium salts with hydrochloric acid, nitric acid, and sulfuric acid.

Answer 4

1) Alkylammonium chloride:
RNH2 + HCl → RNH3+Cl-

e.g. propylamine reacts with hydrochloric acid to form the salt propylammonium chloride:
CH3CH2CH2NH2 + HCl → CH3CH2CH2NH3+Cl-

2) Alkylammonium nitrate:
RNH2 + HNO3 → RNH3+NO3-

e.g. butylamine
CH3CH2CH2CH2NH2 + HNO3 → CH3CH2CH2CH2NH3+NO3-

3) Alkylammonium sulfate:
2RNH2 + H2SO4 → (RNH3+)2(SO4)2-

e.g. ethylamine reacts with sulfuric acid to form ethylammonium sulfate:
2CH3CH2NH2 + H2SO4 → (CH3CH2NH3+)2(SO4)2-

Question 5

.6.2.1 b) Outline the preparation of aliphatic amines.

Answer 5

Aliphatic amines can be prepared by nucleophilic substitution of haloalkanes. Excess ethanolic ammonia (ammonia dissolved in ethanol) is added to the haloalkane (primary amines are added if the desired product is a secondary amine, and secondary amines are added if the desired product is a tertiary amine). The ethanol is used as a solvent instead of water - this is to prevent substitution of the haloalkane with water, where an alcohol would form rather than the intended amine.

Consider the formation of primary amines - this is a two step reaction:
1) The lone pair of electrons on ammonia allow it to act as a nucleophile. It substitutes the halogen in the haloalkane and forms an (alkyl)ammonium salt:
RX + NH3 → RNH3X
2) An additional ammonia molecule then reacts with the (alkyl)ammonium salt, forming both the primary amine and an ammonium halide salt:
RNH3X + NH3 → RNH2 + NH4X
The overall reaction is:
RX + 2NH3 → RNH2 + NH4X
[Note: alternatively, aqueous alkali (e.g. NaOH) can be added after the first step - it reacts with the (alkyl)ammonium salt to form a primary amine:
RNH3X + NaOH → RNH2 + NaX + H2O ]

Additional substitution can occur to produce a secondary amine:
RNH2 + RX → RNHR’ + HX
or
RNH2 + RX + NH3 → RNHR’ + NH4X
And further substitution can occur to produce a tertiary amine:
RNHR’ + RX → RNR’R” + HX
or
RNHR’ + RX + NH3 → RNR’R” + NH4X
These further substitutions are a result of the lone pair of electrons on the nitrogen atom that allow it to act as a nucleophile.

Question 6

6.2.1 b) ii) Outline the preparation of aromatic arenes.

Answer 6

Aromatic arenes are produced by the reduction of nitroarenes, using tin and concentrated hydrochloric acid. The mixture should be heated under reflux. A strong alkali, such as NaOH, is added to remove the excess HCl (in a neutralisation reaction).
Consider the reaction of nitrobenzene to phenylamine:
C6H5NO2 + 6[H] → C6H5NH2 + 2H2O

Question 7

6.2.2 a) What is an α-amino acid?

Answer 7

• An amino acid is an organic compound containing both the amine, NH2, and carboxylic acid, COOH, functional groups.
• When both functional groups are attached to the same carbon atom, the compound is called an α-amino acid.
• The general formula for an α-amino acid is: RCH(NH2)COOH.
  Amino acids are amphoteric (they can act as both an acid and a base):
• The carboxylic acid functional group partially dissociates in water, acting as a weak acid and donating a proton.
• The amine functional group acts as a base, with the lone pair of electrons on nitrogen allowing it to accept a proton.

Question 8

6.2.2 a) i) How can an amino acid react with an alkali?

Answer 8

An amino acid will react with an alkali to form a salt and water. This is because the carboxylic acid functional group donates a proton, forming the conjugate base, RCH(NH2)COO-. The conjugate base combines with a positive ion to form a salt, with water as a by-product:
RCH(NH2)COOH + AOH → RCH(NH2)COO-A+ + H2O

Question 9

6.2.2 a) i) Outline the reaction between amino acids and alcohols.

Answer 9

The carboxylic acid functional group in an amino acid will react with an alcohol, in the presence of a strong, concentrated acid (e.g. H2SO4), to form an ester:
RCH(NH2)COOH + ROH → RCH(NH2)COOR + H2O

Question 10

6.2.2 a) ii) How can an amino acid react with an acid?

Answer 10

The amine functional group can act as a base due to the lone pair of electrons on the nitrogen atom. It reacts with an acid by accepting a proton, forming a conjugate acid. This can combine with a negative ion to form an ammonium salt:
RCH(NH2)COOH + HA → RC(COOH)NH3+A-

Question 11

6.2.2 a) Explain what the following terms mean: zwitterion and isoelectric point.

Answer 11

In an α-amino acid, the amine group can accept a proton from the carboxylic acid group, to form an ion containing both a positive and negative charge. This ion is known as a zwitterion: RCH(NH3+)COO-. A zwitterion has no overall charge (the positive and negative charges cancel each other out).
The isoelectric point is the pH at which a zwitterion is formed - each amino acid has its own unique isoelectric point.

Question 12

6.2.2 a) Outline how changing the pH can affect an amino acid?

Answer 12

By changing the pH, you can alter the structure of an amino acid and its overall charge:

1) If an amino acid is added to a solution with a pH lower than its isoelectric point:
- where there is a high concentration of H+ ions
- the amino acid behaves as a base, and accepts a proton from the acid
- the carboxylic acid becomes -COOH and only the amine group is charged: -NH3+
- a positive ion forms in acidic solution
2) If an amino acid is added to a solution with a pH higher than its isoelectric point:
- where there is a low concentration of H+ ions
- the amino acid behaves as a acid, and donates a proton to the alkali
- the amine group becomes -NH2 and only the carboxylic acid is charged: -COO-
- a negative ion forms in alkaline solution

Question 13

6.2.2 b) Outline the structure of amides and the IUPAC system of naming them.

Answer 13

Amides contain the functional group -CONH2.
RCONH2 = primary amide
RCONHR’ = secondary amide
RCONHR’R” = tertiary amide
When naming amides, the root is determined by the longest hydrocarbon chain, and any additional (alkyl) groups are added as prefixes. The suffix is amide, and ‘N-‘ or ‘N,N-‘ is given as a prefix for secondary and tertiary amides, respectively.
e.g. CH3CH2CONH2 = propanamide
e.g. CH3CONHCH3 = N-methylethanamide
e.g. HCONCH3CH3 = N,N-dimethylmethanamide

Question 14

6.2.2 c) Explain what optical isomerism is.

Answer 14

Optical isomerism is an example of steroisomerism. It is found in molecules that contain a chiral centre. In organic chemistry, a chiral centre (usually a carbon atom) has four different groups attached to it. For each chiral centre, there is always one pair of optical isomers - although, some molecules have more than one chiral centre.

The presence of a chiral centre in a molecule leads to two optical isomers that are non-superimposable mirror images of each other. Optical isomers are otherwise known as enantiomers.

Question 15

6.2.3 a) i) Explain how a polyester can be formed.

Answer 15

Polyesters are a class of condensation polymers. They are formed in reactions between dicarboxylic acids (containing two carboxyl groups) and diols (containing two hydroxyl group).
Each carboxyl group (-COOH) on one monomer reacts with a hydroxyl group (-OH) on another monomer to form an ester link, RCOOR’ (or R-O-C(=O)-R). A water molecule is released in each condensation reaction.
HO-R-OH + HOOC-R-COOH → HO-RO-C(=O)R-COOH + H2O
This happens numerous times until a polyester has formed:
nHO-R-OH + nHOOC-R-COOH → [-O-RO-C(=O)R-C(=O)-]n + (2n - 1)H2O
Alternatively, a polyester might be made from one monomer containing both functional groups, e.g a hydroxycarboxylic acid monomer:
nHO-R-COOH → [-O-R-C(=O)-]n + (2n - 1)H2O
[Note: a water molecule is formed when the -OH is removed from the carboxy group, and the -H is removed from the hydroxyl group]

Question 16

6.2.3 a) ii) Explain how a polyamide can be formed.

Answer 16

Polyamides are a class of condensation polymers. They are formed in reactions between dicarboxylic acids (containing two carboxyl groups) and diamines (containing two amine groups).
The carboxyl group (-COOH) on one monomer reacts with an amine group (-NH2) on another monomer to form an amide link, RCONHR’ (or R-(O=)C-N(H)-R). A water molecule is released in each condensation reaction.
H2N-R-NH2 + HOOC-R-COOH → H2N-RN(H)-C(=O)R-COOH + H2O
This happens numerous times until a polyamide has formed:
nH2N-R-NH2 + nHOOC-R-COOH → [-(H)N-RN(H)-C(=O)R-CO-]n + (2n - 1)H2O
Alternatively, a polyamide might be made from one monomer containing both functional groups, e.g an amino acid monomer:
nH2N-R-COOH → [-N(H)-R-C(=O)-]n + (2n - 1)H2O
[Note: a water molecule is formed when the -OH is removed from the carboxy group, and the -H is removed from the amine group]

Alternatively, polyamides can be formed in reactions between diacyl chlorides and diamines. The acyl chloride group (-COCl) on one monomer reacts with an amine group (-NH2) on another monomer to form an amide link, RCONHR’ (or R-(O=)C-N(H)-R). A HCl molecule is released in each condensation reaction.
H2N-R-NH2 + ClOC-R-COCl → H2N-RN(H)-C(=O)R-COOH + (2n - 1)HCl

Question 17

6.2.3 b) i) Outline the acid hydrolysis of ester groups in polyesters.

Answer 17

When polyesters are hydrolysed using a hot aqueous acid, the reaction produces a diol and a dicarboxylic acid:
[-O-RO-C(=O)R-C(=O)-]n + 2nH2O ⇌ nHO-R-OH + nHOOC-R-COOH
{Remember to include the acid, or H+(aq), over the arrow}
The acid hydrolysis of polyesters tends to be relatively slow.

Question 18

6.2.3 b) i) Outline the base hydrolysis of ester groups in polyesters.

Answer 18

When polyesters are hydrolysed using a hot aqueous base (e.g. NaOH(aq)), the reaction produces a diol and a dicarboxylate salt:
[-O-RO-C(=O)R-C(=O)-]n + 2nNaOH(aq) ⇌ nHO-R-OH + n+Na-OOC-R-COO-Na+
The base hydrolysis of polyesters tends to be relatively fast.

Question 19

6.2.3 b) ii) Outline the acid hydrolysis of amide groups in polyamides.

Answer 19

When polyamides are hydrolysed using a hot aqueous acid, the reaction produces a dicarboxylic acid and a diammonium salt:
[-(H)N-RN(H)-C(=O)R-CO-]n + 2H2O + 2H+(aq)→ n+H3N-R-NH3+ + nHOOC-R-COOH
The acid hydrolysis of polyamides tends to be relatively fast.

Question 20

6.2.3 b) ii) Outline the base hydrolysis of amide groups in polyamides.

Answer 20

When polyamides are hydrolysed using a hot aqueous base, the reaction produces a diamine and a dicarboxylate salt:
[-(H)N-RN(H)-C(=O)R-CO-]n + 2NaOH → nH2N-R-NH2 + n+Na-OOC-R-COO-Na+
The base hydrolysis of polyamides tends to relatively slow.

Question 21

6.2.4 a) List four methods where the length of a carbon chain can be increased.

Answer 21

In organic formation, a C-C bond is formed to increase the length of a carbon chain. For example:

1. The formation of a substituted aromatic C-C by Friedel-Crafts:
   - alkylation, using a haloalkane and a halogen carrier (nucleophilic substitution)
   - acylation, using an acyl chloride and a halogen carrier (nucleophilic substitution)
2. The formation of C-C≡N bond by reaction of:
   - haloalkanes with CN- and ethanol (nucleopilic substitution)
   - carbonyl compounds with HCN (nucleophilic addition)

Question 22

6.2.4 b) i) Outline how a carbon chain length can be increased by reaction of haloalkanes.

Answer 22

Haloalkanes can be reacted with potassium cyanide/sodium cyanide (in an ethanol solution), and heated under reflux, to form a nitrile. This is a nucleophilic substitution reaction, where the cyanide ion (:C-≡N) acts as the nucleophile and is exchanged with the halogen.
Haloalkanes contain a polar C-X bond (due to the large difference in electronegativity). The cyanide ion donates a pair of electrons to the electron-deficient carbon atom and forms a dative bond (represented in an arrow going from the :C≡N to the Cδ+). A pair of electrons is then donated to the slightly negative halogen atom, and the C-X bond is split by heteolyctic fission (represented in an arrow going from the C-X bond to the Xδ- atom). The final profucts include a nitrile and a X- ion.
R-X + KCN/NaCN → R-CN + KX/NaX

Question 23

6.2.4 b) i) Outline how a carbon chain length can be increased by reaction of carbonyl compounds.

Answer 23

Hydrogen cyanide is a weak acid - it will partially dissociate in water:
HCN + H2O ⇄ CN- + H3O+
or HCN ⇄ CN- + H+
Hydrogen cyanide is a weak acid - it will partially dissociate in water:
HCN + H2O ⇄ CN- + H3O+
or HCN ⇄ CN- + H+
The cyanide ion, :C-≡N, acts as a nucleophile, reacting with carbonyl compounds in the presence of an acid to form hydroxynitrile. This is a nucleophilic addition reaction:
1. Carbonyl compounds have a dipole on the C=O group: this makes them susceptible to nucleophilic attack on the δ+ carbon atom. The lone pair of electrons on the cyanide ion, :CN-, are attracted and donated to the electron-deficient carbon atom (represented in a curly arrow going from the lone pair on :CN- to the Cδ+ atom), This forms a dative covalent bond between the cyanide ion and the carbon atom.
2. Simultaneously, the π-bond in the C=O breaks by heterolytic fission (represented in a curly arrow going from the double bond to the Oδ-). A reactive intermediate is formed (where there is an extra lone pair of electrons on the oxygen, :O-).
3. The lone pair of electrons on the oxygen is donated to either a hydrogen ion or a slightly positive hydrogen atom (on another molecule) - this is the protonation stage (and can be represented in a curly arrow going from the lone pair on :O- to H+/Hδ+). This creates a dative covalent bond between the oxygen and the hydrogen, forming a hydroxyl group.
The final product is a hydroxynitrile.

Question 24

6.2.4 c) i) Outline how amines are formed from nitriles.

Answer 24

A nitrile, or a hydroxynitril, can be reduced to a primary amine using a strong reducing agent:
In catalytic hydrogenation, nitriles are reduced using hydrogen gas and a nickel catalyst:
R-C≡N + 2H2 → R-CH2-NH2
Nitriles can also be reduced using LiAlH4 and a dilute acid catalyst:
R-C≡N + 4[H] → R-CH2-NH2

Question 25

6.2.4 c) ii) Outline how carboxylic acids are formed from nitriles.

Answer 25

Nitriles can be reduced using acid hydrolysis. They are heated under reflux , with a diluted strong acid catalyst, to form a carboxylic acid:
R-C≡N + 2H2O + HCl → R-COOH + NH4Cl

Question 26

6.2.5 a) ii) Outline how an organic solid can be separated from a solvent or liquid reaction mixture.

Answer 26

Organic solids often crystallise out of solution - filtration under reduced pressure can be used to separate a solid product from a solvent/liquid reaction mixture.

• Connect one end of some think-walled, rubber tubing (otherwise known as pressure tubing) to a vacuum line or outlet. Attach the other end of the rubber tubing to the sidearm of a Büchner flask.
• Then place a Büchner funnel (containing a piece of filter paper) into the flask.
• When you pour the reaction mixture into the Büchner funnel, the reduced pressure from the vacuum line will result in suction through the funnel. This causes the liquid to pass quickly into the flask, leaving behind dry crystals of your product.
• To ensure maximum yield - and prevent loss of product - rinse out the beaker containing the reaction mixture (using a solvent), and add this to the Büchner funnel.

[Note: a Büchner flask is also known as a vacuum flask, filter flask, suction flask or side-arm flask]

Question 27

6.2.5 a) ii) Outline how an organic solid can be purified.

Answer 27

A solid product obtained in organic synthesis often contains impurities. These impurities can be removed - and the solid purified - through a technique called recrystallisation:

1) Add the minimum volume of hot solvent needed to only just dissolve the impure solid. This produces a saturated solution of the impure product (a saturated solution contains the maximum possible amount of solid dissolved in solution).
2) Leave the solution to cool. As the temperature drops, the solubility of the product falls, and crystals start to form. The impurities stay in solution - they’re present in much smaller quantities that the product, and take longer to crystallise out of solution.
3) The crystals are removed by filtration under reduced pressure.

[Note: the choice of solvent for recrystallisation is very important:

1) The solid has to be very soluble in the hot solvent, but nearly insoluble when the solvent is cold
2) Both the desired product and the impurities need to have different solubilities in the chosen solvent]

Question 28

6.2.5 a) ii) Explain the importance in measuring the melting point of a solid.

Answer 28

A pure organic substance usually has a very specific melting and boiling point - in fact, the melting range (i.e. the difference at which a sample starts to melt and the temperature at which melting is complete) is very sharp, with a range of only one or two degrees. If the compound contains impurities, the melting point is lowered and boiling point is raised - in fact, the solid also melts over a wider range of temperatures.
Once you’ve measured either the melting point or the melting range of your sample, you can compare it to known values to determine its purity. If the melting point is lower or the melting range is wider than standard values, the sample contains impurities.

Question 29

6.2.5 a) ii) Outline how a melting point apparatus is used to determine the melting point of an organic solid.

Answer 29

Method 1:

1) Add a few grains of the organic solid to a sealed capillary tube. Gently insert the tube into an electrically heated melting point apparatus - ensure you also add an accurate thermometer with suitable range to the apparatus. Start to heat up the sample.
2) Look though the lens/magnifying window and observe the crystals. Record the temperature: once, when the solid has begun melting, and twice, when the sample has liquefied. The first value provides your melting point, and the two values together provide your melting range.

Question 30

6.2.5 a) ii) Outline how a Thiele tube is used to determine the melting point of an organic solid.

Answer 30

1) Add a few grains of the organic solid to a sealed capillary tube. Using a small rubber band, attach the tube to a thermometer. Submerge the thermometer into the oil of the Thiele tube, ensuring that the rubber band is above the oil line.
2) Using a micro-burner, heat the side arm of the Thiele tube while observing the solid. Record the temperature: once, when the solid has begun melting, and twice, when the sample has liquefied. The first value provides your melting point, and the two values together provide your melting range.