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Chemistry organic speedrun v2 + other topics > 18 - Organic chemistry III > Flashcards

18 - Organic chemistry III Flashcards

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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.
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2
Q

What does delocalised mean when describing delocalised electrons?

A
  • delocalised means not attached to a particular atom.
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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.
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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.
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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.
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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.
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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.
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8
Q

What does aromatic mean?

A

Aromatic refers to a hydrocarbon ring containing delocalised electrons (benzene ring).

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9
Q

What does aromatic mean?

A

Aromatic refers to a hydrocarbon ring containing delocalised electrons (benzene ring).

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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.
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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.
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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.
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13
Q

Combustion of benzene?

A
  • burns in the the air like any other hydrocarbon.
  • forms 6CO2 and 3H2O.
  • burns with a smoky flame.
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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.
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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.
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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
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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).

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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.
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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.
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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.

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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.
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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+
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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.
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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.
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25
Q

In friedel-crafts acylation, how is the electrophile CH3CO+ formed?

A
  • reaction between ethanoyl chloride and aluminium chloride catalyst.
  • Cl of ethanoyl chloride transfers to aluminium chloride to form AlCl4- and CH3CO+
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26
Q

In friedel-crafts acylation, how is HCl produced in the reaction between benzene and ethanoyl chloride?

A
  • 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.
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27
Q

What is the acidity of phenol?

A
  • 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.
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28
Q

Bromination of phenol?

A
  • 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.
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29
Q

Why is phenol able to be brominated without being needed to be heated under reflux and a catalyst?

A
  • 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.
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30
Q

What are some of the uses of phenol?

A
  • production of plastics,
  • antiseptics
  • disinfectants
  • resins for paints.
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31
Q

What kind of smells do amines have?

A
  • fishy smell
32
Q

Can amines dissolve in water?

A
  • small amines can form hydrogen bonds with water and therefore dissolve readily in water.
33
Q

Which amines act as bronsted-lowry bases and why? What does this mean?

A
  • 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
Q

Why are primary aliphatic amines stronger bases than ammonia?

A
  • 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
Q

Why are secondary aliphatic amines stronger bases than primary aliphatic amines?

A
  • 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
Q

Describe and explain the basicity of aromatic amines?

A

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
Q

amines and acids

A
  • amines act as bases
  • so they can react with acids
  • to form ammonium salts
38
Q

How do you convert ammonium salt back to amine?

A
  • 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
Q

How do you make a basic buffer?

A
  • react a weak base with a salt of that weak base.
  • e.g ammonia and ammonium chloride
  • e.g methylamine and methylammonium chloride.
40
Q

Amines and ethanoyl chloride?

A
  • addition-elimination reaction. Two molecules join together, one small molecule is eliminated like condensation polymerisation.
  • HCl is eliminated
  • secondary amide is formed.
41
Q

What is a way to make a secondary amide?

A

react a primary amine with an acyl chloride.

42
Q

How do you go from alkane to a halogenoalkane?

A
  • react with halogen molecule, e.g Cl2
  • UV light
  • free radical substitution.
43
Q

primary amines and halogenoalkanes?

A
  • secondary amine formed

- inorganic product also formed such as HCl.

44
Q

secondary amines and halogenoalkanes?

A
  • tertiary amine

- inorganic product also formed such as HCl.

45
Q

tertiary amines and halogenoalkanes?

A
  • quaternary ammonium salt is formed.

- because of the lone pair on the nitrogen.

46
Q

Preparing primary amine from halogenoalkane and ammonia?

A
  • 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
Q

Preparing secondary amine from halogenoalkane and primary amine.

A
  • forms secondary amine

- inorganic product formed e.g HCl.

48
Q

Preparation of primary amines from reduction of nitriles?

A
  • LiAlH4
  • dry ether

or hydrogenation:

  • 2H2
  • nickel catalyst
  • heat
49
Q

Preparation of aromatic amines from reduction of nitrobenzene?

A
  • react with Sn (tin) and concentrated HCl.

- heat

50
Q

Preparation of amides?

A
  • mix acyl chloride with concentrated aqueous ammonia.
51
Q

what is a zwitterion?

A

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
Q

What is isoelectric point?

A
  • The pH of an aqueous solution, in which the amino acid is neutral and exists as zwitterions.
53
Q

What does a low isoelectric point indicate?

A
  • Indicates that the molecule is predominantly acidic
54
Q

What does a high isoelectric point indicate?

A
  • Indicates that the molecule is predominantly basic.
55
Q

What are isoelectric points linked to?

A
  • 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
Q

Do amino acids show optical activity?

A
  • 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
Q

amino acids at low pH?

A
  • 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
Q

amino acids at high pH?

A
  • 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
Q

What is a peptide?

A
  • 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
Q

How many unique combinations can dipeptides have?

A

2

61
Q

how many unique combinations can tripeptides have?

A

6

62
Q

What is a protein?

A
  • 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
Q

How do you hydrolyse proteins?

A
  • 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
Q

How do you use chromatography on amino acids?

A
  • 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
Q

Why do we sometimes use ninhydrin as a developing agent when using chromatography on mixture of amino acids?

A
  • 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
Q

How do you calculate Rf values of amino acids and what can you do with them?

A

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
Q

How do you prepare a Grignard reagent?

A
  • react halogenoalkane with magnesium.
  • dry ether (since Grignard reagents react with water).
  • Mg slots itself between R group and halogen.
68
Q

Grignard + CO2?

A

carboxylic acid

69
Q

Grignard + methanal

A

primary alcohol

70
Q

Grignard + aldehyde

A

secondary alcohol

71
Q

Grignard + ketone

A

tertiary alcohol

72
Q

What are the steps in Grignard reactions?

A
  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
Q

Paper chromatography?

A

Stationary phase: water trapped in the fibres of the chromatography paper

Mobile phase: suitable solvent

74
Q

Thin layer chromatography (TLC)?

A

stationary phase: silica gel or alumina (polar)

mobile phase: suitable solvent

  • sheet of glass/plastic
75
Q

Paper chromatography vs TLC

A
  • 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
Q

Column chromatography?

A

stationary phase: silica gel or alumina packed into the column and soaked in the solvent.

mobile phase: the solvent (eluent).

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