18 - Organic chemistry III Flashcards
(76 cards)
1
Q
What is the structure of a benzene ring?
A
- C6H6
- Each C atom is bonded to 2 other C atoms and 1 H atom by single sigma bonds.
- This leaves one unused electron on each C atom in a p orbital. These six p electrons are delocalised in a ring structure above and below the plane of carbon atoms.
2
Q
What does delocalised mean when describing delocalised electrons?
A
- delocalised means not attached to a particular atom.
3
Q
Problem 1 with kekule’s structure?
A
- Kekule’s structure shows that benzene has C=C double bonds.
- We would expect benzene to decolourise bromine water.
- However, bromine water is not decolourised.
4
Q
Problem 2 with kekule’s structure?
A
- isomerism
- If kekule’s structure was correct, dibromobenzene would have 4 isomers.
- However only 3 exist.
- The isomers with bromine on adjacent carbon atoms are identical.
- This suggests that the bonds between the carbon atoms in the benzene ring are identical, not different.
5
Q
Problem 3 with kekule’s structure?
A
- bond length of the C-C bonds in benzene ring were found to be inbetween bond length of C=C in cyclohexene and C-C in cyclohexene.
- This suggests that the carbon-carbon bonds in benzene are all the same and have intermediate character between C-C and C=C bonds.
6
Q
Problem 4 with kekule’s structure?
A
Enthalpies of hydrogenation
- Enthalpy of hydrogenation of cyclohexene is -120 and cyclohexa-1,4-diene is -239.
- These values suggest that the enthalpy change for adding one mol of H2 to 1 mol of C=C bonds is -120.
- We predict that the enthalpy of hydrogenation is -360 (since kekule’s structure suggests 3 C=C bonds).
- However actual value is -208 (152 lower than theoretical).
- This suggests that benzene does not have 3 distinct C=C double bonds.
7
Q
What is the new structure of benzene?
A
- C6H6
- Each carbon atom forms 3 sigma bonds: 2 with 2 other C atoms, 1 with a hydrogen.
- Each carbon atom has one electron in a p-orbital that remains unused.
- These p orbitals overlap sideways forming a pi-electron cloud above and below the plane.
- The 6 p electrons are delocalised above and below the plane of carbon atoms, in the pi-electron cloud.
8
Q
What does aromatic mean?
A
Aromatic refers to a hydrocarbon ring containing delocalised electrons (benzene ring).
9
Q
What does aromatic mean?
A
Aromatic refers to a hydrocarbon ring containing delocalised electrons (benzene ring).
10
Q
Methylbenzene vs benzene toxicity and reactivity?
A
- methylbenzene is less toxic
- more reactive because the methyl group releases electrons into the delocalised system making it more attractive to electrophiles.
11
Q
Methylbenzene vs benzene toxicity and reactivity?
A
- methylbenzene is less toxic
- more reactive because the methyl group releases electrons into the delocalised system making it more attractive to electrophiles.
12
Q
Hydrogenation of Benzene?
A
- reacts with 3 moles of H2 (to make it saturated).
- nickel catalyst
- heat under pressure
- this reaction is addition and reduction.
- cyclohexane is formed.
13
Q
Combustion of benzene?
A
- burns in the the air like any other hydrocarbon.
- forms 6CO2 and 3H2O.
- burns with a smoky flame.
14
Q
Halogenation of Benzene (e.g bromination)?
A
- reagent is bromine, Br2.
- catalyst of aluminium chloride AlCl3 (these are called halogen carriers).
- heated under reflux
- products are bromobenzene and hydrogen bromide.
- electrophillic substitution mechanism.
15
Q
In bromination of benzene, how is the hydrogen bromide formed?
A
- After the electrophilic substitution has occurred, an there is an H+ ion that is free which has been substituted.
- Remember the other compound formed when the electrophile was formed. AlCl3Br-.
- This compound reacts with the H+ ion to form HBr: the Br- is removed from AlCl3Br- and reacts with H+ to make HBr.
- Now you have your original AlCl3 compound. It is chemically unchanged at the end of the reaction -> catalyst.
16
Q
Nitration of benzene?
A
- reagent is concentrated nitric acid (source of NO2) in the presence of concentrated sulfuric acid (catalyst).
- heated at 60C (at higher temps, a second nitro group can be substituted onto different positions on the ring).
- products are nitrobenzene and water
- electrophilic substitution
17
Q
In the nitration of benzene, how do we get the electrophile NO2+?
A
1 mol of nitric acid and 1 mol of sulfuric acid reacts to form:
HSO4-, H2NO3+
The H2NO3- splits into:
H2O, NO2+
So essentially, HNO3 + H2SO4 gives
HSO4-
H2O
NO2+ (electrophile).
18
Q
In nitration of benzene, how is the H2SO4 catalyst reformed?
A
- After the electrophilic addition, there is a free H+ ion that has just been substituted by NO2.
- Remember the HSO4- from the formation of the electrophile.
- The H+ reacts with the HSO4- to reform H2SO4, the catalyst.
- You can see how the H2SO4 is chemically unchanged at the end of the reaction -> catalyst.
19
Q
What are friedel-crafts reactions?
A
- alkylation and acylation
- if there is an XY reagent, 1 H atom on benzene is substituted by Y and HX is also produced.
- catalyst required: aluminium chloride (or iron (III) bromide, iron (III) chloride).
- anhydrous conditions needed.
20
Q
Why are anhydrous conditions needed in friedel-crafts reactions?
A
Water would react with the catalyst and sometimes also with the organic product.
21
Q
Friedel-crafts alkylation?
A
- substitution of 1 H atom of benzene by an alkyl group.
- halogenoalkane (e.g chloroalkane)
- aluminium chloride catalyst
- anhydrous conditions
- heat under reflux.
- electrophilic substitution.
- alkylbenzene is the product.
22
Q
In friedel-crafts akylation, how is the electrophile (e.g CH3CH2+) formed?
A
- chloroethane reacts with the catalyst, e.g aluminium chloride (AlCl3)
- The chlorine atom moves to the catalyst, forming AlCl4- and CH3CH2+
23
Q
In friedel-crafts akylation, how does HCl get produced when reacting benzene with chloroethane?
A
- After the electrophilic substitution, there is a a free H+ ion that has just been substituted by the CH3CH2+.
- Remember the AlCl4- from the formation of the electrophile.
- This reacts with the H+ ion to form AlCl3 and HCl.
- notice how the AlCl3 is chemically unchanged at the end of the reaction -> catalyst.
24
Q
Friedel-crafts acylation?
A
- substitution of 1 H atom of benzene by an acyl group.
- acyl chloride (e.g ethanoyl chloride).
- aluminium chloride catalyst.
- anhydrous conditions (water can react with the catalyst and sometimes the organic product).
- heat under reflux.
- electrophilic substitution.
25
In friedel-crafts acylation, how is the electrophile CH3CO+ formed?
- reaction between ethanoyl chloride and aluminium chloride catalyst.
- Cl of ethanoyl chloride transfers to aluminium chloride to form AlCl4- and CH3CO+
26
In friedel-crafts acylation, how is HCl produced in the reaction between benzene and ethanoyl chloride?
- after the electrophilic substitution, there is a free H+ ion that has been substituted by the CH3CO+ electrophile.
- This H+ reacts with the AlCl4- (formed during formation of the electrophile), to form AlCl3 and HCl.
- notice how the AlCl3 is reformed and is chemically unchanged at the end of the reaction -> catalyst.
27
What is the acidity of phenol?
- very weakly acidic
- can react with sodium metal and sodium hydroxide and forms salt. (H of OH group on phenol is removed and replaced by Na+).
- cannot react with sodium carbonate as phenol is not strong enough of an acid to react.
28
Bromination of phenol?
- Bromine supplied by bromine water.
- 3 Br2 molecules react with phenol.
- forms 2,4,6-tribromophenol and 3HBr. (3 Br go onto phenol, 3 Br react with H+ ions from phenol ring to form 3HBr.
- this reaction does not need a catalyst and does. not need to be heated under reflux. Occurs at room temperature and multiple substitutions occur.
- bromine water is decolourised.
- product is a white solid.
29
Why is phenol able to be brominated without being needed to be heated under reflux and a catalyst?
- oxygen in the OH group has lone pairs of electrons are partially delocalised into the benzene ring.
- electron density above and below the ring of carbon atoms increases. This makes the molecule more reactive towards electrophiles.
- bromine molecules are originally non-polar. But when they approach the benzene ring, they are polarised and eventually the Br-Br bond breaks and the Br+ electrophile attacks the benzene ring/molecule.
30
What are some of the uses of phenol?
- production of plastics,
- antiseptics
- disinfectants
- resins for paints.
31
What kind of smells do amines have?
- fishy smell
32
Can amines dissolve in water?
- small amines can form hydrogen bonds with water and therefore dissolve readily in water.
33
Which amines act as bronsted-lowry bases and why? What does this mean?
- primary aliphatic amines
- the lone pair of electrons on the nitrogen is readily available for forming a dative covalent bond with a H+ ion.
- It can do this with water forming OH- too.
34
Why are primary aliphatic amines stronger bases than ammonia?
- alkyl groups are electron releasing
- this means that there is a greater electron density on the nitrogen atom.
- The lone pair of electrons on the nitrogen atom is more readily available to form a dative covalent bond with H+ ion.
- therefore stronger base.
35
Why are secondary aliphatic amines stronger bases than primary aliphatic amines?
- they have more alkyl groups that are substituted onto the N atom in place of H atoms.
- alkyl groups are electron releasing.
- More electron density is pushed onto the N atom.
- This means that the lone pair is more readily available to form dative covalent bond with H+ ions.
36
Describe and explain the basicity of aromatic amines?
primary aromatic amines such as phenylamine do not form basic solutions.
- benzene ring is electron withdrawing.
- this causes the lone pair on the N to delocalise with the benzene pi electrons.
- This means that N is less able to accept H+ ions.
37
amines and acids
- amines act as bases
- so they can react with acids
- to form ammonium salts
38
How do you convert ammonium salt back to amine?
- react with NaOH.
- Cl reacts with Na to form NaCl
- H reacts with OH to form H2O.
- and then you are left with amine.
39
How do you make a basic buffer?
- react a weak base with a salt of that weak base.
- e.g ammonia and ammonium chloride
- e.g methylamine and methylammonium chloride.
40
Amines and ethanoyl chloride?
- addition-elimination reaction. Two molecules join together, one small molecule is eliminated like condensation polymerisation.
- HCl is eliminated
- secondary amide is formed.
41
What is a way to make a secondary amide?
react a primary amine with an acyl chloride.
42
How do you go from alkane to a halogenoalkane?
- react with halogen molecule, e.g Cl2
- UV light
- free radical substitution.
43
primary amines and halogenoalkanes?
- secondary amine formed
| - inorganic product also formed such as HCl.
44
secondary amines and halogenoalkanes?
- tertiary amine
| - inorganic product also formed such as HCl.
45
tertiary amines and halogenoalkanes?
- quaternary ammonium salt is formed.
| - because of the lone pair on the nitrogen.
46
Preparing primary amine from halogenoalkane and ammonia?
- one H removed from NH3 to become NH2 and joins substitutes halogen on halogenoalkane.
- H and halogen X react to form HX.
- under pressure
- in a sealed container
- or react with concentrated aqueous ammonia.
47
Preparing secondary amine from halogenoalkane and primary amine.
- forms secondary amine
| - inorganic product formed e.g HCl.
48
Preparation of primary amines from reduction of nitriles?
- LiAlH4
- dry ether
or hydrogenation:
- 2H2
- nickel catalyst
- heat
49
Preparation of aromatic amines from reduction of nitrobenzene?
- react with Sn (tin) and concentrated HCl.
| - heat
50
Preparation of amides?
- mix acyl chloride with concentrated aqueous ammonia.
51
what is a zwitterion?
A zwitterion is an overall neutral molecule with both a positive and negative charge.
- in a zwitterion, H of OH of COOH group is transferred to NH2 group to form NH3+ and COO-
52
What is isoelectric point?
- The pH of an aqueous solution, in which the amino acid is neutral and exists as zwitterions.
53
What does a low isoelectric point indicate?
- Indicates that the molecule is predominantly acidic
54
What does a high isoelectric point indicate?
- Indicates that the molecule is predominantly basic.
55
What are isoelectric points linked to?
- the numbers of NH2 and COOH groups in the molecule.
- sometimes.
- more COOH makes the molecule predominantly acidic and have a low isoelectric point.
- more NH2 makes the molecule predominantly basic and have a high isoelectric point.
56
Do amino acids show optical activity?
- Almost all amino acids contain a chiral centre.
- except glycine.
- the enantiomers rotate plane polarised light.
- amino acids synthesised in a lab leads to the formation of a racemic mixture with no optical activity.
57
amino acids at low pH?
- not a zwitterion.
- amino acid is able to act as a base and gain a H atom to the COO- group.
- this forms an alkaline solution (since H+ ions are taken in by the amino acids).
58
amino acids at high pH?
- not a zwitterion.
- amino acid is able to act as an acid and donate a H atom from NH3+ group.
- this forms an acidic solution (since H+ ions are donated into the solution from the amino acid).
59
What is a peptide?
- when two amino acids react together, they join by a condensation reaction of the COOH and NH2 groups.
- H2O is eliminated.
- the two amino acids join by an amide group CONH, which is known as a peptide bond (in compounds where an amide bonds occurs that did not form between two amino acids, it is called an amide link).
60
How many unique combinations can dipeptides have?
2
61
how many unique combinations can tripeptides have?
6
62
What is a protein?
- a very long polypeptide.
- difference between a long-chain polypeptide and a protein is that proteins have further levels to their structures.
- The polypeptide chains in proteins interact with each other in 3d to give secondary, tertiary and quaternary structures.
63
How do you hydrolyse proteins?
- react with concentrated hydrochloric acid.
- prolonged heating.
dipeptide + H2O + 2H+
(water to reverse condensation reaction, H+ to result in protonated NH3 groups in amino acids only).
(number of H+ atoms depends on number of COOH groups present in amino acids).
- results in amino acids with protonated NH3 groups due to acidic conditions (compared to zwitterions, the COO- group gains H ion).
64
How do you use chromatography on amino acids?
- pencil line 1.5cm from bottom.
- with a capillary tube, put a small concentrated drop of mixture of amino acids on pencil line.
- stand the paper in a large beaker.
- the solvent in beaker should be below the pencil line.
leave for 20 mins.
- spray with ninhydrin and put in oven.
65
Why do we sometimes use ninhydrin as a developing agent when using chromatography on mixture of amino acids?
- amino acids are colourless.
- ninhydrin allows the positions of the amino acids to be seen with a vibrant colour.
- however ninhydrin is not normally used now due to its toxic nature.
66
How do you calculate Rf values of amino acids and what can you do with them?
Rf = distance moved by amino acid / distance moved by solvent.
- essentially a ratio.
- The calculated Rf values can be compared to known Rf values in a data book to identify the amino acids.
67
How do you prepare a Grignard reagent?
- react halogenoalkane with magnesium.
- dry ether (since Grignard reagents react with water).
- Mg slots itself between R group and halogen.
68
Grignard + CO2?
carboxylic acid
69
Grignard + methanal
primary alcohol
70
Grignard + aldehyde
secondary alcohol
71
Grignard + ketone
tertiary alcohol
72
What are the steps in Grignard reactions?
1. formation of Grignard reagent
2. Reaction with chosen reagent.
3. Hydrolysis using dilute acid (for this step on the arrow, write H2O/H+ on top).
73
Paper chromatography?
Stationary phase: water trapped in the fibres of the chromatography paper
Mobile phase: suitable solvent
74
Thin layer chromatography (TLC)?
stationary phase: silica gel or alumina (polar)
mobile phase: suitable solvent
- sheet of glass/plastic
75
Paper chromatography vs TLC
- TLC gives better separations
- TLC is faster to run
- TLC has better selectivity, has the ability to use different kinds of stationary points instead of just water in paper.
- TLC plates can withstand stronger solvents and colour forming agents than paper.
76
Column chromatography?
stationary phase: silica gel or alumina packed into the column and soaked in the solvent.
mobile phase: the solvent (eluent).
