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A LEVEL - Chemistry (Both) > Topic 18 - Organic Chemistry III > Flashcards

Topic 18 - Organic Chemistry III Flashcards

1
Q

What is the formula of benzene?

A

C₆H₆

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2
Q

How can the structure of benzene be described?

A

It is cyclic.

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3
Q

What are the two ways of representing the structure of benzene?

A
  • Kekulé model

* Delocalised model

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4
Q

Which of the models of benzene structure came first?

A

Kekulé model

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5
Q

Describe the Kekulé model of benzene structure.

A
  • Planar ring of 6 carbon atoms
  • Alternating single and double bonds between carbons
  • Each carbon is bonded to one hydrogen atom
  • Single and double bonds are constantly switching over
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6
Q

Describe how the Kekulé model can be drawn using skeletal formulae.

A

Hexagon with alternating single and double lines.

See diagram pg 205 of revision guide

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7
Q

What are the two different structures of benzene in the Kekulé model referred to as?

A

Isomers

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8
Q

Remember to practice drawing out the structure of benzene in the Kekulé model.

A

Pg 205 of revision guide

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9
Q

Why was the Kekulé model for the structure of benzene shown to be incorrect?

A
  • The model predicts that 3 of the bonds would be shorter (C=C) and 3 would be longer (C-C)
  • X-ray diffraction showed that all the carbon-carbon bonds were of the same length, suggesting delocalisation
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10
Q

Describe and explain the delocalised model of benzene structure.

A
  • Planar ring of 6 carbon atoms
  • Each carbon atom forms three σ-bonds -> 1 to a hydrogen atom, 1 to each of its neighbouring carbon atoms (these are due to head-on overlap of orbitals)
  • Each carbon has one p-orbital that sticks out above and below the plane -> These overlap sideways to form a ring of π-bonds that are delocalised around the carbon ring
  • The delocalised π-bonds are made of two ring-shaped clouds of electrons (above and below the plane)
  • All the bonds in the ring are the same length
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11
Q

In the delocalised model of benzene structure, how does the delocalised ring of electrons form?

A
  • Each carbon has one remaining p-orbital that sticks out above and below the ring
  • These p-orbitals overlap sideways to form two rings of π-bonds
  • This is the delocalised ring of electrons
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12
Q

Why are the electrons in the benzene ring said to be delocalised?

A

They don’t belong to a specific carbon atom.

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13
Q

How is the delocalised ring of electrons represented in benzene?

A

A circle within the hexagon (although sometimes the structure is drawn as the Kekulé model)

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14
Q

Describe how the delocalised model of benzene structure can be drawn using skeletal formulae.

A

Hexagon with a circle inside it.

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15
Q

Remember to practise drawing out the structure of benzene in the delocalised model.

A

Pg 205 of revision guide

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16
Q

Remember to revise the formation of a delocalised ring of electrons in benzene.

A

Diagram at bottom of pg 205 of revision guide.

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17
Q

What is hydrogenation in alkenes?

A

When an alkenes reacts with hydrogen, so two hydrogen atoms add across the double bond.

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18
Q

What is the enthalpy change of hydrogenation?

A

The enthalpy change when hydrogen reacts with an alkene.

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19
Q

How many C=C bonds does cyclohexane have?

A

1

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20
Q

What is the enthalpy of hydrogenation of cyclohexane?

A

-120kJ/mol

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21
Q

Given that the enthalpy of hydrogenation of cyclohexane is -120kJ/mol, what would you expect the enthalpy be for the Kekulé structure of benzene?

A

3 x -120 = -360kJ/mol

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22
Q

How does the expected enthalpy of hydrogenation of the Kekulé structure of benzene compare to the actual enthalpy of hydrogenation of benzene?

A
  • The expected enthalpy for Kekulé structure is -360kJ/mol
  • The actual enthalpy is -208kJ/mol
  • This means that the structure must be more stable than predicted by Kekulé’s model
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23
Q

Why is the expected enthalpy of hydrogenation of the Kekulé model of benzene more exothermic than the actual value?

A

The delocalised model of benzene is more stable, so more energy is required to overcome this, making the overall enthalpy less exothermic.

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24
Q

What are the conditions in hydrogenation of benzene?

A
  • Nickel catalyst

* 200°C

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25
Q

Does benzene easily undergo addition reactions? Why?

A

No, because:
• The delocalised ring of electrons is very stable
• The charge is so spread out in the delocalised ring

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26
Q

Why is benzene so stable?

A

The delocalised ring of electrons.

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27
Q

When will benzene undergo addition reactions?

A

When there is:
• Hot benzene
• UV light

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28
Q

Will bromine water be decolourised by cyclohexane and benzene?

A
  • Cyclohexane - Yes

* Benzene - No

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29
Q

What is the preferred reaction type of benzene?

A

Electrophilic substitution

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30
Q

What are some pieces of evidence for the delocalised model of benzene over the Kekulé model?

A
  • Bond lengths (X-ray diffraction studies)
  • Enthalpy changes of hydrogenation
  • Reluctance to undergo addition reactions
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31
Q

Give the equation of benzene burning in air.

A

2C₆H₆ + 15O₂ -> 12CO₂ + 6H₂O

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32
Q

What is the type of flame when benzene is burnt in air?

A

Smoky, because there is insufficient oxygen to burn the benzene completely.

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33
Q

Why does combustion of benzene give a smokey flame?

A

Due to the high ratio of C to H in benzene.

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34
Q

What are aromatic compounds?

A

Compounds derived from benzene (i.e. that contain a benzene ring).

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35
Q

What are arenes?

A

Another name for aromatic compounds (those derived from a benzene ring).

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36
Q

What are the two ways of naming aromatic compounds?

A

1) Substituted benzene ring (e.g. NITRObenzene)
2) Compounds with a phenyl group attached (e.g. PHENol)

There is no easy way to tell which to use.

(See pg 207 of revision guide)

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37
Q

What is a phenyl group?

A
  • C₆H₅

* It is the name sometimes used when a benzene ring is attached onto a compound.

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38
Q

What are the rules for naming aromatic compounds with more than one functional group?

A

Number the carbons using these rules:
• If all the functional groups are the same, pick the group to start on that gives the smallest numbers when you count round
• If the functional groups are different, start from the functional group that gives the molecule its suffix and count the way round that gives the smallest numbers

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39
Q

When naming aromatic compounds, what are the rules used when all of the functional groups attached to the benzene are the same?

A

When numbering the carbons, start on the carbon that gives the smallest numbers when you count round.

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40
Q

When naming aromatic compounds, what are the rules used when the functional groups attached to the benzene are different?

A

When numbering the carbons, start from the functional group that gives the molecule its suffix and count the way round that gives the smallest numbers.

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41
Q

What is the name for chlorine attached to a benzene ring?

A

Chlorobenzene

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42
Q

What is the name for a nitro (NO₂) group attached to a benzene ring?

A

NITRObenzene

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43
Q

What is the name for a 2 methyl groups attached to a benzene ring (two carbons apart)?

A

1,3-dimethylbenzene

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44
Q

What is the name for a hydroxide group attached to a benzene ring?

A

Phenol

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45
Q

What is the name for an NH₂ group attached to a benzene ring?

A

Phenylamine

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46
Q

What is a the name for a benzene with these functional groups:
• CH₃
• 2 x NO₂ (1 and 3 carbons away from the CH₃)

A

2,4-dinitromethylbenzene

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47
Q

What is a the name for a benzene with these functional groups:
• CH₃
• OH (on adjacent carbons)

A

2-methylphenol

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48
Q

Remember to practise naming aromatic compounds.

A

Pg 207 of revision guide

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49
Q

When benzene undergoes electrophilic substitution, what is replaced by what?

A

A hydrogen is replaced by the electrophile.

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50
Q

What are the two steps of the mechanism for electrophilic substitution in benzene?

A

1) Addition of electrophile to form positively charged intermediate
2) Loss of H⁺ from the carbon atom attached to the electrophile

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51
Q

How is an electrophile symbolised in mechanisms?

A

E

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52
Q

Describe how to draw the mechanism for electrophilic substitution in benzene.

A

(NOTE: Electrophile must be generated first)

First step:
• Arrow goes from circle in benzene to E⁺
• Now there is an E and a H bonded to a single carbon. The circle has been replaced by a horseshoe that does not go beyond the adjacent carbons. There is a + charge in the middle.
Second step:
• Arrow goes from C-H bond to the positive charge in the ring
• The product is a benzene ring with E bonded to it, plus a H⁺ ion

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53
Q

Remember to practise drawing out the mechanism for electrophilic substitution in benzene.

A

See diagram pg 208 of revision guide

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54
Q

With arenes undergoing electrophilic substitution, what must you remember about the intermediate?

A

The horseshoe in the benzene ring cannot go beyond the two adjacent carbons to where to electrophile has attached.

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55
Q

What sort of electrophile is required to react with benzene and why?

A

A strong electrophile, because the negative charge density in benzene is spread out across the whole ring.

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56
Q

What can be used to make a stronger electrophile (for a reaction with a benzene ring)?

A

Halogen carriers

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57
Q

What are halogen carriers?

A
  • Molecules that increase the polarisation in an electrophile, so much that sometimes a carbonation forms.
  • This allows for the electrophile to be strong enough to react with a benzene ring
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58
Q

How does a halogen carrier work?

A
  • Halogen carrier accepts a lone pair of electrons from the electrophile, polarising it more
  • Sometimes a carbocation forms
  • This makes the electrophile stronger
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59
Q

Give an example of a halogen carrier.

A

AlCl₃

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60
Q

An what molecules do halogen carriers work?

A
  • Halogens
  • Acyl chlorides
  • Halogenoalkanes
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61
Q

Practice drawing out the mechanism by which a halogen carrier works.

A

Pg 208 of revision guide

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62
Q

What are some of the different halogen carriers?

A
  • Aluminium halides
  • Iron halides
  • Iron
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63
Q

What reactions of benzene do you need to know about?

A
  • Combustion
  • Bromination
  • Nitration (H₂SO₄ + HNO₃)
  • Sulphonation (Fuming H₂SO₄)
  • Alkylation (Friedel Crafts)
  • Acylation (Friedel Crafts)
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64
Q

What is bromination of benzene?

A

The electrophilic substitution of a halogen for a hydrogen in a benzene ring

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65
Q

What is the catalyst for the bromination of benzene?

A

Iron(III) Bromide (FeBr₃)

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66
Q

What are the reactants and products of the bromination of benzene?

A 
REACTANTS:
• Benzene
• Bromine
CATALYST:
• Iron(III) bromide (Halogen carrier catalyst)
PRODUCTS:
• Bromobenzene
• HBr
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67
Q

Describe the preparation of the electrophile in the bromination of benzene.

A
  • Fe + 3/2Br₂ -> FeBr₃
  • FeBr₃ + Br₂ -> [FeBr₄]⁻ + Br⁺

(The Br⁺ is the electrophile)

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68
Q

What are the conditions for the bromination of benzene?

A
  • FeBr₃ catalyst (Hydrogen carrier)

* RTP

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69
Q

Remember to practise drawing out the mechanism for bromination of benzene.

A

See diagram pg 208 of revision guide

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70
Q

What is the general name for the substitution of a halogen atom for a hydrogen atom in benzene?

A

Halogenation

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71
Q

What is alkylation of benzene?

A

The substitution of an alkyl group (e.g. CH₃CH₂) onto a benzene ring.

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72
Q

What are the two types of Friedel-Crafts reaction?

A
  • Alkylation

* Acylation

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73
Q

What is the catalyst in the alkylation of benzene?

A

Aluminium chloride (AlCl₃)

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74
Q

What are the reactants and products of the alkylation of benzene (with a normal alkyl electrophile)?

A 
REACTANTS:
• Benzene
• Halogenoalkane
CATALYST:
• AlCl₃ catalyst (Hydrogen carrier catalyst)
PRODUCTS:
• Alkylbenzene
• HCl
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75
Q

Give the equation for the preparation of the electrophile in the alkylation of benzene.

A

R-Cl + AlCl₃ -> R⁺ + [AlCl₄]⁻

Where R⁺ is the electrophile

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76
Q

Describe the mechanism for the generation of the electrophile in the alkylation of benzene.

A 
1-step reaction:
• At the start, there is R-Cl and AlCl₃
• Arrow goes from R-Cl bond to the Cl
• Arrow goes from the lone pair on the Cl to the Al in the AlCl₃
• R⁺ and [AlCl₄]⁺ are produced
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77
Q

In the alkylation of benzene, what is the electrophile?

A

The carbocation generated when the halogen is lost from a halogenoalkane.

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78
Q

What are the conditions for the alkylation of benzene?

A
  • AlCl₃ catalyst (Halogen carrier catalyst)

* Heating under reflux

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79
Q

What is the general equation for alkylation of benzene?

A

C₆H₆ + R-X -> C₆H₅R + HX

AlCl₃ catalyst + Heating under reflux

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80
Q

Remember to practise drawing out the mechanism for the alkylation of benzene with an electrophile containing an OAlCl₃⁻ group.

A

See diagram pg 209 of revision guide.

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81
Q

Remember to practise drawing out the mechanism for alkylation of benzene.

A

See diagram pg 209 of revision guide

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82
Q

Does Friedel-Crafts alkylation only happen with halogenoalkanes?

A

No, it can also happen with electrophiles containing OAlCl₃⁻ groups.

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83
Q

What other group do you need to know about that can be the electrophile in the alkylation of bromine?

A

The electrophile can be an alkyl chain containing an OAlCl₃⁻ group.

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84
Q

What is produced when the electrophile in alkylation of benzene contains an OAlCl₃⁻?

A

An alcohol containing a benzene ring.

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85
Q

What are the reactants and products of the acylation of benzene?

A 
REACTANTS:
• Benzene
• Acyl chloride
CATALYST:
• AlCl₃ catalyst (Hydrogen carrier catalyst)
PRODUCTS:
• Phenylketones (or benzaldehyde)
• HCl
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86
Q

What is the exception to the products of the acylation of benzene?

A
  • Most of the time, the organic product is a phenylketone

* However, if R = H in RCOCl, then the organic product is an aldehyde called benzaldehyde

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87
Q

What is acylation of benzene?

A

The substitution of an acyl group (e.g. COCl) onto a benzene ring.

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88
Q

What is the catalyst in the acylation of benzene?

A

Aluminium chloride (AlCl₃)

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89
Q

Remember to practise drawing out the mechanism for acylation of benzene.

A

See diagram pg 209 of revision guide

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90
Q

In the acylation of benzene, what is the electrophile?

A

The carbocation generated when the chlorine is lost from the acyl chloride.

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91
Q

Describe the preparation of the electrophile in the acylation of benzene.

A

AlCl₃ + CH₃COCl -> CH₃C⁺O + [AlCl₄]⁻

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92
Q

Give the general equation for the preparation of the electrophile in the acylation of benzene.

A

R-COCl + AlCl₃ -> RCO⁺ + [AlCl₄]⁻

Where R⁺ is the electrophile

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93
Q

Describe the mechanism for the generation of the electrophile in the acylation of benzene.

A

1-step reaction:
• At the start, there is R-COCl and AlCl₃
• Arrow goes from C-Cl bond to the Cl
• Arrow goes from the lone pair on the Cl to the Al in the AlCl₃
• RCO⁺ and [AlCl₄]⁺ are produced

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94
Q

In the acylation of benzene, what is the electrophile?

A

The carbocation generated when the chlorine is lost from an acyl chloride.

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95
Q

What are the conditions for the acylation of benzene?

A
  • AlCl₃ catalyst (Halogen carrier catalyst)

* Heating under reflux in dry ether

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96
Q

What is the general equation for acylation of benzene?

A

C₆H₆ + RCOCl -> C₆H₅COR + HCl

AlCl₃ catalyst + Heating under reflux in dry ether

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97
Q

What is nitration of benzene?

A

Substitution of an NO₂ into the benzene ring.

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98
Q

What is the catalyst in the nitration of benzene?

A

Concentrated sulphuric acid

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99
Q

What are the reactants and products of the nitration of benzene?

A 
REACTANTS:
• Benzene
• Concentrated sulphuric acid
CATALYST:
• Concentrated nitric acid
PRODUCTS:
• Nitrobenzene
• Water
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100
Q

In the nitration of benzene, what is the electrophile?

A

Nitronium ion (NO₂⁺)

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101
Q

Give the equation for the preparation of the electrophile in the nitration of benzene.

A
  • HNO₃ + H₂SO₄ -> H₂NO₃⁺ + HSO₄⁻
  • H₂NO₃⁺ -> NO₂⁺ + H₂O

(NO₂⁺ is the electrophile)

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102
Q

Describe the mechanism for the preparation of the electrophile in the nitration of benzene.

A

First step:
• There is the HO-NO-O (nitric acid) with a positive charge on the N, negative charge on the right O and 2 lone pairs on the left O
• There is the H-OSO₃H (sulphuric acid)
• Arrow goes from a lone pair on the O in the nitric acid to the H on the sulphuric acid
• Arrow goes from the same H-O bond to the S
• This produces H₂O⁺-NO-O and HSO₄⁻
Second step:
• Arrow goes from left O-N bond in the nitric acid to the positive O on the left
• Arrow goes from the right O to the same N-O bond on the right
• This produces O=N=O, H₂O and HSO₄⁻

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103
Q

What are the conditions for the nitration of benzene?

A
  • Sulphuric acid catalyst

* 50°C for benzene (lower temperatures for other benzenes)

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104
Q

What is the equation for the nitration of benzene?

A

C₆H₆ + HNO₃ -> C₆H₅NO₂ + H₂O

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105
Q

Remember to practise drawing out the mechanism for the nitration of benzene.

A

Pg 210 of revision guide

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106
Q

What bearing does temperature have on the nitration of benzene?

A
  • At 50°C -> 1 NO₂ group is added

* Above 50°C -> 2 NO₂ groups are added

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107
Q

When nitration of benzene happens above 50°C, at what position is the extra NO₂ group added and why?

A
  • In the 3rd position (2 carbons away from the first NO₂)

* Due to the electron withdrawal due to the extra NO₂ group

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108
Q

How many times can nitration of benzene occur?

A

Twice - the ring is too stable for more NO₂ groups.

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109
Q

What is the name for adding just one nitro group to a benzene?

A

Mononotration

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110
Q

What is sulphonation of benzene?

A

Substitution of the SO₃H group onto the benzene ring.

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111
Q

What are the reactants and products in the sulphonation of benzene?

A 
REACTANTS:
• Benzene
• SO₃
• Sulphuric acid
PRODUCTS:
• Benzesulphonic acid (C₆H₅SO₃H)
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112
Q

What is fuming sulphuric acid?

A

Mixture of:
• Concentrated H₂SO₄
• SO₃

It is used in the sulphonation of benzene.

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113
Q

In the sulphonation of benzene, what is the electrophile?

A

SO₃

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114
Q

Why can SO₃ act as an electrophile in electrophilic substitution in benzene?

A

Because the S is so positive due to the oxygens.

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115
Q

What are the conditions for the sulphonation of benzene?

A
  • Fuming sulphuric acid (H₂SO₄ + SO₃)

* Heat

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116
Q

What is the equation for the sulphonation of benzene?

A

C₆H₆ + H₂SO₄ -> C₆H₅SO₃H + H₂O

OR

C₆H₆ + SO₃ -> C₆H₅SO₃H

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117
Q

What is the name of the product of the sulphonation of benzene?

A

Benzesulphonic acid

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118
Q

In sulphonation of benzene, where does the extra H on the OH of the benzesulphonic acid product come from?

A
  • It is the hydrogen that is being substituted

* This is because the hydrogen does not have to be used to regenerate the catalyst like in other reactions

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119
Q

Remember to practise drawing out the reaction mechanism for the sulphonation of benzene.

A

See diagram online or pg 11 of booklet 5.4

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120
Q

What is a phenol?

A

A benzene ring with an OH group attached

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121
Q

What is the formula of phenol?

A

C₆H₅OH

122
Q

What is the name for a benzene ring with these functional groups:
• OH
• Cl (opposite the OH group)

A

4-chlorophenol

123
Q

What is the name for a benzene ring with these functional groups:
• OH
• NO₂ (opposite the OH)

A

4-nitrophenol

124
Q

What is the result of phenol being more reactive than benzene?

A

It is more likely to undergo electrophilic substitution.

125
Q

What is the name for a benzene ring with these functional groups:
• OH
• 2 x CH₃ (1 and 3 carbons away from the OH)

A

2,4-dimethylphenol

126
Q

What happens when the product of the sulphonation of benzene is reacted with sodium hydroxide?

A

It forms a salt and water.

-SO₃H -> -SO₃Na + H₂O

127
Q

What is more reactive, benzene or phenol? Why?

A

Phenol, because:
• One of the lone pairs of the O in the OH overlaps with the delocalised π-bonds in the benzene
• So the lone pair of electrons from the oxygen is partially delocalised into the π-system
• This increases the electron density of the ring, making it more likely to be attacked by electrophiles

128
Q

When naming a phenol, where does numbering of the carbons start?

A

The -OH carbon is number 1.

129
Q

Describe the two reactants in the production of aspirin.

A
  • Salicylic acid

* Ethanoic anhydride

130
Q

Describe the structure of salicylic acid.

A
  • Phenol ring
  • -COOH group is attached to the carbon next to the -OH on the phenol

This is a phenol derivative.

(See diagram pg 211 of revision guide)

131
Q

Describe the structure of ethanoic anhydride.

A

CH₃CO-O-COCH₃

This is a symmetrical molecule.

(See diagram pg 211 of revision guide)

132
Q

Write the word equation for the synthesis of aspirin.

A

Salicylic acid + Ethanoic anhydride -> Aspirin + Ethanoic acid

133
Q

What type of reaction is the production of aspirin?

A

Esterification -> The reaction is essentially between an alcohol (the salicylic acid) and a carboxylic acid (ethanoic anhydride is similar).

134
Q

Describe how aspirin can be synthesised.

A

1) Add some ethanoic anhydride and a few drops of phosphoric acid to salicylic acid in a test tube.
2) Warm the reaction mixture to 50°C and leave for 15 minutes.
3) Filter the crystals under reduced pressure.
4) Recrystallise the aspirin in a mixture of water and ethanol.

135
Q

What are the conditions for the synthesis of aspirin?

A
  • 50°C

* Acid (e.g. phosphoric acid)

136
Q

Remember to practise drawing out the displayed reaction for the formation of aspirin.

A

Pg 211 of revision guide

137
Q

What is an amine?

A

A ammonia molecule (NH₃) where one or more of the hydrogen’s is replaced with an organic group.

138
Q

What is the functional group for an anime?

A

-NR₂

Where R is an alkyl group or H.

139
Q

What is a primary amine?

A

A molecule of NH₃ where only 1 of the H’s is replaced by an organic group.

140
Q

What is a secondary amine?

A

A molecule of NH₃ where 2 of the H’s are replaced by organic groups.

141
Q

What is a tertiary amine?

A

A molecule of NH₃ where all 3 of the H’s are replaced by organic groups.

142
Q

What is the name for a nitrogen atom bonded to four alkyl groups?

A

Quaternary ammonium ion

Positively charged

143
Q

What are the two types of amine in terms of the attached groups?

A
  • Aliphatic amines

* Aromatic amines

144
Q

What is the name and type of this molecule:
• Central N atom
• 2 x H attached to the N
• CH₃ attached to the N

A

Methylamine

Primary amine

145
Q

What is the name and type of this molecule:
• Central N atom
• H attached to the N
• 2 x CH₃ attached to the N

A

Dimethylamine

Secondary amine

146
Q

What is the name and type of this molecule:
• Central N atom
• 3 x CH₃ attached to the N

A

Trimethylamine

Tertiary amine

147
Q

What is the name and type of this molecule:
• Central N atom
• 4 x CH₃ attached to the N

A

Tetramethylamine ion

Quaternary ammonium ion

148
Q

What is the name and type of this molecule:
• Central N atom
• 2 x H attached to the N
• Benzene ring attached to the N

A

Phenylamine
(Primary amine)

(Note: This is called an aromatic amine)

149
Q

When synthesising aspirin from salicylic acid and ethanoic anhydride, what is the by-product?

A

Ethanoic anhydride

150
Q

What are the different ways in which an aliphatic amine can be produced?

A

1) Halogenoalkane + Ethanolic ammonia
2) Reducing a nitrile

(Note: These give a primary amine at first, but more halogenoalkane can be used to create other amines.)

151
Q

Describe how an aliphatic amine can be produced from a halogenoalkane.

A

The halogenoalkane is made to react with an excess of ethanolic ammonia.

152
Q

What is the problem with producing aliphatic amines from halogenoalkanes? How can this be overcome?

A
  • It produces a mixture of primary, secondary and tertiary amines and quaternary ammonium salts
  • This is because the N has a lone pair so it can take part in multiple nucleophilic substitution reactions, so more than one hydrogen is replaced
  • This can be overcome by using an excess of ethanolic ammonia, which produces mostly primary amines. If a different type is needed, more halogenoalkanes can be added.
153
Q

Describe how an aliphatic amine can be produced from a nitrile.

A

The nitrile can be reduced in 2 ways:

1) Using LiAlH₄ in dry ether, then dilute acid
2) Using H₂ gas with a Pt/Ni catalyst at high temperature and pressure

154
Q

Describe how a nitrile can be reduced using lithium aluminium hydride.

A

It is reduced:
• In dry ether
• Followed by some dilute acid

This produces a primary aliphatic amine.

155
Q

Give the general equation for reducing a nitrile using aluminium hydride.

A

R-CH₂-CN + 4[H] -> R-CH₂-CH₂-N-H₂

156
Q

Describe how a nitrile can be reduced using hydrogen gas.

A

It is reduced:
• With a nickel/platinum catalyst
• At a high temperature and pressure

This produces a primary aliphatic amine.

157
Q

Give the general equation for reducing a nitrile using hydrogen gas.

A

R-CH₂-CN + 2H₂ -> R-CH₂-CH₂-N-H₂

158
Q

What is the name for reducing a nitrile using hydrogen and a catalyst at high temperature and pressure?

A

Catalytic hydrogenation

159
Q

Which method of reducing a nitrile to a primary amine is better industrially?

A
  • Catalytic hydrogenation

* Because LiAlH₄ is too expensive to use in industry

160
Q

Describe how an aromatic amine can be made.

A
  • Heat a nitro compound, tin metal and concentrated HCl under reflux -> This produces a salt
  • Add NaOH -> This produces the aromatic amine
161
Q

What type of reaction is the production of amines?

A

Reduction

162
Q

What is the most basic aromatic amine?

A

Phenylamine

163
Q

Describe the structure of phenylamine.

A

Benzene ring with NH₂ group attached to it

164
Q

When producing phenylamine from nitrobenzene, what is the by-product?

A

Water

165
Q

Remember to practise drawing out the formation of phenylamine from nitrobenzene.

A

Pg 212 of revision guide

166
Q

Describe briefly how you can produce phenylamine from benzene.

A 
Benzene to nitrobenzene:
• Conc. HNO₃
\+ Conc. H₂SO₄
• Under 55°
Nitrobenzene to phenylamine:
a) Sn + Conc. HCl + Heat
b) NaOH solution
167
Q

When producing an aromatic amine, why is NaIH added?

A

This converts the product from a salt to an amine.

168
Q

Are amines acids or bases? Why?

A

Weak bases because they accept protons.

169
Q

How can amines accept protons?

A
  • The N in the middle has a lone pair of electrons

* This allows it to make a dative covalent bond with an H⁺ ion

170
Q

What determines how strong of a base an amine is?

A
  • It depends on the availability of the electrons to form dative covalent bonds with a H⁺ ion
  • This depends on the electron density around the N, which depends on the attached groups
171
Q

Describe and explain the order of amines in terms of their strength as bases.

A

Weakest bases: Primary aliphatic bases
• The benzene ring draws electrons towards itself
• The nitrogen lone pair gets partially delocalised onto the ring, so the electron density on the nitrogen decreases
• So the lone pair is less available to form a dative covalent bond with a H⁺

Middle base: Ammonia

Strongest bases: Primary aliphatic amines
• The alkyl group pushes electrons towards the nitrogen
• The electron density on the nitrogen increases
• So the lone pair is more available to form a dative covalent bond with a H⁺

172
Q

What type of molecule are amines in terms of reactions?

A

Nucleophiles (due to the lone pair on the N)

173
Q

Can amines react with acids or bases?

A

Acids

174
Q

What happens when an acid reacts with an amine?

A

An ammonium salt is produced.

175
Q

CH₃CH₂CH₂CH₂NH₂ + HCl ->

A

CH₃CH₂CH₂CH₂NH₂ + HCl -> CH₃CH₂CH₂CH₂NH₃⁺Cl⁻

176
Q

Describe and explain the solubility of amines.

A

Small amines:
• Soluble
• Because the amine group can form hydrogen bonds with the water molecules
Larger amines:
• Less soluble
• The London forces between the amine molecules are greater and require more energy to overcome, so dissolving is not energetically viable

177
Q

How does the solubility of amines change as they increase in size?

A

The larger the amine, the less soluble.

178
Q

Describe what happens when an amine dissolves.

A

An alkaline solution is formed, containing:
• Alkyl ammonium ions
• Hydroxide ions

179
Q

CH₃CH₂CH₂CH₂NH₂(aq) + H₂O(l) ->

A

CH₃CH₂CH₂CH₂NH₂(aq) + H₂O(l) -> CH₃CH₂CH₂CH₂NH₃⁺(aq) + OH⁻ (aq)

180
Q

What complex ions are formed in copper(II) sulfate solution?

A

[Cu(H₂O)₆]²⁺

181
Q

What is the colour of copper(II) sulfate solution?

A

Blue

182
Q

Describe and explain what happens when butylamine is added to copper(II) sulfate solution.

A

Small amount of butylamine:
• The amine acts as a base and takes two H⁺ ions from the [Cu(H₂O)₆]²⁺ complexes
• This gives a pale blue precipitate of copper hydroxide [Cu(OH)₂(H₂O)₄]

With excess butylamine:
• The precipitate dissolves to form a deep blue solution
• Some of the ligands are replaced by the butylamine to give [Cu(CH₃(CH₂)₃NH₂)₄(H₂O)₂]²⁺

183
Q

Describe the colour changes when an amine is added to an aqueous solution of copper ions (such as copper(II) sulphate).

A
  • Starts are blue solution of [Cu(H₂O)
  • When some amine is added, pale blue precipitate of [Cu(OH)₂(H₂O)₄] forms
  • When excess amine is added, this dissolves and a deep blue solution forms when some of the ligands are replaced by the amine (e.g. [Cu(CH₃(CH₂)₃NH₂)₄(H₂O)₂]²⁺)
184
Q

Remember to practise drawing out the displayed equations for the addition of butylamine to copper(II) sulfate solution.

A

Pg 213 of revision guide

185
Q

Do all amines form the same complex with copper(II) ions?

A

The colours will be the same, but the final product may change because as many of the larger amine molecules may not fit around the copper ion.

186
Q

How can other amines be formed from primary amines?

A

Reacting with a halogenoalkane.

187
Q

Describe and explain the reaction type when:
• Forming a primary amine from a halogenoalkane
• Forming another amine from a primary amine

A
  • Both are nucleophilic substitution
  • Because the N in the ammonia or the primary amine has a lone pair
  • This means it can act as a nucleophile and it attracted to the δ+ carbon in the halogenoalkane
188
Q

How can an N-substituted amide by formed from a primary amine?

A

Reacting with an acyl chloride.

189
Q

What type of reaction is a primary amine reacting with an acyl chloride?

A

Acylation

190
Q

What is an acyl group?

A

RCO

191
Q

Describe what happens when an acyl chloride reacts with a primary amine.

A

Stage 1:
• The acyl chloride and primary amine react to give the N-substituted amide and HCl
(This is what you memorise in the synthesis pathway)
Stage 2:
• The primary amine also reacts with the HCl to give a salt

(See of 214 of revision guide)

192
Q

Remember to practise drawing out the displayed reaction of ethanoyl chloride and butylamine.

A

Pg 214 of revision guide

193
Q

What amide and salt are ultimately produced when ethanoyl chloride reacts with butylamine?

A
  • N-butyl ethanamide

* Butylammonium chloride

194
Q

Primary amine + Acyl chloride ->

A

Primary amine + Acyl chloride -> Amide + Salt

195
Q

Remember to practise naming ammonium salts.

A

Pg 214 of revision guide

196
Q

How is the reaction between an amine and acyl chloride carried out in a lab?

A

The acyl chloride is added to a concentrated aqueous solution of the amine.

197
Q

What change is seen when ethanoyl chloride reacts with butylamine?

A

Solid, white mixture of products.

198
Q

Write the chemical equation for the reaction between ethanoyl chloride and butylamine.

A

CH₃COCl + 2C₄H₉NH₂ -> CH₃CONHC₄H₉ + [C₄H₉NH₃]⁺Cl⁻

199
Q

What are amides derivatives of?

A

Carboxylic acids

200
Q

What is the functional group of amides?

A

-CONH₂

201
Q

Why do amides behave differently to amines?

A

The carbonyl group in amides pulls the electrons away from the rest of the -CONH₂ group.

202
Q

What are the two types of amide?

A
  • Primary amide

* N-substituted amide

203
Q

What is the difference between a primary amide and N-substituted amide?

A 
Primary amides have around the N:
• COR group
• 2 x H atom
N-substituted amides have around the N:
• COR group
• At least one alkyl group instead of an H

(i.e. In an N-substituted amide, at least one of the H’s is substituted by an R group)

204
Q

What suffix do amide have?

A

-amide

205
Q

How can you name a primary amide?

A
  • Stem of the carbon chain

* Followed by -amide

206
Q

How can you name an N-substituted amide?

A
  • Prefix to describe the R group on the N -> N-alkyl-
  • Followed by the stem of the carbon chain
  • Followed by -amide
207
Q

Remember to practise naming amides.

A

Pg 215 of revision guide.

208
Q

How can a primary amide be formed?

A

Reacting an acyl chloride with ammonia at RTP.

209
Q

How can an N-substituted amide be formed?

A

Reacting an acyl chloride with a primary amine at RTP.

210
Q

What is produced when ethanoyl chloride reacts with ammonia?

A

Ethanamide + HCl

211
Q

What is produced when ethanoyl chloride reacts with methylamine?

A

N-methylethanamide + HCl

212
Q

Break down the name: N-ethylpropanamide

A
  • N-ethyl -> There is an ethyl group attached to the N
  • propanamide -> There is a propane carbon chain attached to the N with the C=O group on the adjacent carbon

Therefore, there must also be a H attached to the N.

213
Q

What types of molecules does condensation polymerisation usually involve?

A
  • Two different types of monomer

* Each monomer has at least two functional groups

214
Q

How does condensation polymerisation work?

A
  • There are two adjacent molecules, usually each with a functional group at each end
  • The functional groups react with each other creating polymer chains
  • This releases water
215
Q

What are the different types of condensation polymer you need to know about?

A
  • Polyamides
  • Polypeptides
  • Polyesters
216
Q

What molecules join to give a polyamide?

A
  • Dicarboxylic acid

* Diamine

217
Q

What is the name for the bonds formed in a polyamide?

A

Amide links

218
Q

Describe the structure of an amide link in a polyamide.

A

-CO-NH-

219
Q

Give the general structure of a polyamide.

A

-R-CO-NH-R₁-NH-CO-

220
Q

Describe how a polyamide is formed.

A
  • A dicarboxylic acid and diamine are next to each other
  • An OH is lost from the -COOH and a H is lost from the NH₂, so overall H₂O is lost
  • This produces an amide link
  • This can happen at each end of the molecule to give a long chain
221
Q

Remember to practise drawing out how a polyamide is formed.

A

Pg 216 of revision guide

222
Q

What is the name for the bonds formed in a polypeptide?

A

Peptide links

223
Q

Describe the structure of a peptide link.

A

-CO-NH-

224
Q

Describe the general structure of a polypeptide.

A

-CH(R)-CO-NH-CH(R₁)-CO-NH-

225
Q

Describe how a polypeptide is formed.

A
  • Two amino acids are next to each other
  • An OH is lost from the -COOH and a H is lost from the NH₂, so overall H₂O is lost
  • This produces a peptide link
  • This can happen at each end of the molecule to give a long chain
226
Q

How can you break down a protein into amino acids?

A
  • Add hot aqueous 6mol/dm³ HCl
  • Reflux for 24hrs -> This produces ammonium salts
  • Neutralise using a base
227
Q

How can you determine the amino acids a protein is made out of?

A
  • Hydrolyse it (6mol/dm³ HCl, reflux for 24hrs)

* Use chromatography to identify the amino acids

228
Q

Remember to practise drawing out how a polypeptide is formed.

A

Pg 216 of revision guide

229
Q

What is the difference between polypeptides and polyamides?

A

Polypeptides are really polyamides.

230
Q

What molecules join to give a polyester?

A
  • Dicarboxylic acid

* Diol

231
Q

What is the name for the bonds formed in a polyester?

A

Ester link

232
Q

Describe the structure of an ester link.

A

-CO-O-

233
Q

Describe the general structure of a polyester.

A

-R-CO-O-R₁-O-CO-

234
Q

Describe how a polyester is formed.

A
  • Dicarboxylic acid and diol are next to each other
  • An OH is lost from the -COOH and a H is lost from the OH, so overall H₂O is lost
  • This produces an ester link
  • This can happen at each end of the molecule to give a long chain
235
Q

Remember to practise drawing out how a polyester is formed.

A

Pg 216 of revision guide

236
Q

How can you find the monomers a polymer is made of?

A

1) Find the amide or ester link. Break it down the middle.

2) Add a H or OH to both ends of both molecules to find the monomers.

237
Q

Remember to practise joining monomers to give a condensation polymer.

A

Pg 217 of revision guide

238
Q

Describe the structure of an amino acid.

A 
Central carbon surrounded by:
• Carboxyl group (COOH)
• Amino group (NH₂)
• R group
• H atom
239
Q

What is the term for a molecule having both acidic and basic properties?

A

Amphoteric

240
Q

What makes amino acids amphoteric?

A

They have:
• Carboxyl group -> Acidic
• Amino group -> Basic

241
Q

What type of amino acids are found in nature?

A
  • 2-amino acids

* Where the amino group is on carbon-2 (the carboxyl group is always carbon-1)

242
Q

What can amino acids exist as?

A

Zwitterions

243
Q

What is a zwitterion?

A

An overall neutral molecule that has both a positive and negative charge in different parts of the molecule.

244
Q

When can an amino acid exist as a zwitterion?

A
  • Near its isoelectric point

* This is the pH where the overall charge on the amino acid is 0

245
Q

What is the isoelectric point of an amino acid?

A

The pH where the overall charge on the amino acid is 0.

246
Q

Is the isoelectric point the same for all amino acids?

A

No, it depends on their R group.

247
Q

What happens to amino acids in conditions more acidic than the isoelectric point?

A
  • The -NH₂ group is protonated, making it -NH₃⁺.
  • The -COOH group is unchanged.

(See diagram of 218 of revision guide)

248
Q

What happens to amino acids at the isoelectric point?

A
  • The -NH₂ group is protonated, making it -NH₃⁺.
  • The -COOH group loses its proton, making it -COO⁻.
  • This is now a zwitterion.

(See diagram of 218 of revision guide)

249
Q

What happens to amino acids in conditions more alkaline than the isoelectric point?

A
  • The -NH₂ group is unchanged.
  • The -COOH group loses its proton, making it -COO⁻.

(See diagram of 218 of revision guide)

250
Q

Which amino acids will exist as zwitterions when dissolved in solution and will have a pH around 7?

A

Those with the same number of carboxyl groups as amino groups.

251
Q

What can be said about the optical activity of amino acids?

A

They are often chiral and therefore have two optical isomers.

252
Q

What is the 3D structure of an amino acid?

A

Tetrahedral around the carbon.

253
Q

What is the exception to the optical activity of amino acids?

A
  • Glycine
  • The R group is just a hydrogen atom.
  • So the molecule is not chiral, since there are two H’s around the central carbon.
  • This means it won’t rotate plane-polarised light.
254
Q

What molecules join to give a polypeptide?

A

Amino aicds

255
Q

What two types of chromatography can be used to identify amino acids?

A
  • Paper chromatography

* Thin-layer chromatography

256
Q

Remember to revise how to do paper chromatography for amino acids.

A

Pg 219 of revision guide

257
Q

What is the formula for the Rf value in chromatography?

A

Rf = A / B = Distance travelled by spot / Distance travelled by solvent

258
Q

How are spots of amino acid in paper chromatography made visible?

A
  • Spraying ninhydrin to turn them purple

* Dipping the paper into a jar containing a few crystals of iodine, which sublimes

259
Q

What is thin-layer chromatography?

A

The same as paper chromatography, except you use a plate covered in a thin layer of silica (SiO₂) or alumina (Al₂O₃) as the stationary phase.

260
Q

What is the general formula for Grignard reagents?

A

RMgX

Where:
• R = Alkyl group
• X = Halogen

261
Q

How is a Grignard reagent prepared?

A

Refluxing a halogenoalkane with magnesium in dry ether.

262
Q

Give the general equation for the formation of a Grignard reagent.

A

R-X + Mg -> RMgX

Note: This is done in dry ether!

263
Q

Give the equation for the formation of a Grignard reagent from bromoethane and magnesium.

A

CH₃CH₂Br + Mg -> CH₃CH₂MgBr

Note: This is done in dry ether!

264
Q

What are the two things you can make from a Grignard reagent?

A
  • Carboxylic acid

* Alcohol

265
Q

Describe how a carboxylic acid can be made from a halogenoalkane.

A

1) Grignard reagent is prepared by reacting the appropriate halogenoalkane with Mg in dry ether
2) CO₂ is bubbled through this
3) A dilute acid is added

266
Q

When using Grignard reagents to product organic compounds, what is the byproduct?

A

MgX¹X²

Where:
• X¹ = Halogen from the halogenoalkane
• X² = Anion from the acid

267
Q

Give the equation for a Grignard reagent (made from a bromoalkane) reacting to give a carboxylic acid.

A

R-MgBr + CO₂ -> R-COOH + MgBrCl

268
Q

Describe how the reaction of a Grignard reagent with CO₂ to give a carboxylic acid works.

A
  • A new C-C bond forms between the carbon atom in CO₂ and the C-Mg carbon in the Grignard reagent
  • One of the C=O bonds in CO₂ breaks to form a -COO⁻ group
  • This is then protonated by the dilute acid to form -COOH
269
Q

Butanoic acid can be synthesised from bromopropane in three steps. Give the reagents and conditions needed for each step, and the product formed at each stage of the synthesis.

A

CH₃CH₂CH₂Br ——Mg, Dry ether)——> CH₃CH₂CH₂MgBr ——(1)CO₂, dry ether (2)Dilute HCl——> CH₃CH₂CH₂COOH + MgBrCl

270
Q

Describe how a alcohol can be made from a halogenoalkane (not using aqueous KOH).

A

1) Grignard reagent is prepared by reacting the appropriate halogenoalkane with Mg in dry ether
2) Aldehyde or ketone (carbonyl compounds) are added
3) A dilute acid is added

271
Q

Give the equation for a Grignard reagent (made from a bromoalkane) reacting to give an alcohol.

A

R-MgBr + R¹COR² -> CRR¹R²OH + MgBrCl

272
Q

Describe how the reaction of a Grignard reagent with an aldehyde/ketone (carbonyl) to give an alcohol works.

A
  • A new C-C bond forms between the carbon atom in C=O of the carbonyl and the C-Mg carbon in the Grignard reagent
  • This causes the C=O bond to break
  • The oxygen is then protonated by the dilute acid to form an -OH group
273
Q

When a Gringard reagent reacts with an aldehyde, what is produced?

A

Secondary alcohol (except methanal)

274
Q

When a Gringard reagent reacts with a ketone, what is produced?

A

Tertiary alcohol

275
Q

Remember to practise drawing out Gringard reagent formation and reactions.

A

Pg 220 of revision guide

276
Q

Give the full list of homologous series you have studied.

A
  • Alkane
  • Alkene
  • Aromatic compounds
  • Alcohol
  • Halogenoalkane
  • Amine
  • Amide
  • Nitrile
  • Aldehyde/Ketone
  • Carboxylic acid
  • Ester
  • Acyl chloride
277
Q

For an alkane, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> C-C
  • Properties -> Non-polar, Unreactive
  • Typical reactions -> Radical substitution
278
Q

For an alkene, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> C=C
  • Properties -> Non-polar, Electron-rich double bond
  • Typical reactions -> Electrophilic addition
279
Q

For an aromatic compound, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> C₆H₅-
  • Properties -> Stable, Delocalised ring of electrons
  • Typical reactions -> Electrophilic substitution
280
Q

For an alcohol, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> C-OH
  • Properties -> Polar C-OH bond, Lone pair on O can act as nucleophile
  • Typical reactions -> Nucleophilic substitution, Dehydration/Elimination, Esterification, Nucleophilic substitution (acting as the nucleophile)
281
Q

For a halogenoalkane, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> C-X
  • Properties -> Polar C-X
  • Typical reactions -> Nucleophilic substitution, Elimination
282
Q

For an amine, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> C-NH₂ / C-NR₂
  • Properties -> Lone pair on N is basic and can act as a nucleophile
  • Typical reactions -> Neutralisation, Nucleophilic substitution (acting as the nucleophile)
283
Q

For an amide, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> -CONH₂ / -CONR₂
  • Properties -> N/A
  • Typical reactions -> N/A
284
Q

For a nitrile, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> C-CN
  • Properties -> Electron deficient carbon centre
  • Typical reactions -> Reduction, Hydrolysis
285
Q

For an aldehyde/ketone, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> C=O
  • Properties -> Polar C=O bond
  • Typical reactions -> Nucleophilic addition, Reduction, Oxidation (only aldehydes)
286
Q

For a carboxylic acid, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> -COOH
  • Properties -> Electron deficient carbon centre
  • Typical reactions -> Neutralisation, Esterification, Reduction
287
Q

For an ester, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> RCOOR¹
  • Properties -> Electron deficient carbon centre
  • Typical reactions -> Hydrolysis
288
Q

For an acyl chloride, give the:
• Functional group
• Properties
• Typical reactions

A
  • Functional group -> -COCl
  • Properties -> Electron deficient carbon centre
  • Typical reactions -> Nucleophilic addition-elimination, Condensation (lose HCl), Friedel-Crafts acylation
289
Q

What is a synthetic route?

A

A series of reactions used to get from one compound to another.

290
Q

If you’re asked how to make one compound from another in the exam (in multiple steps), what must you remember to include?

A

1) Any special procedures (e.g. refluxing)
2) Conditions (e.g. high temperature or pressure, or the presence of a catalyst)
3) Safety precautions (e.g. do it in a fume cupboard)

291
Q

When chemists are planning a synthetic route, what is it important to keep in mind?

A

1) Stereoisomers

2) Safety

292
Q

When chemists are planning a synthetic route, why is it important to keep stereoisomers in mind?

A

Different stereoisomers have different properties, which is important in the pharmaceutical industry.

293
Q

When chemists are planning a synthetic route, why is it important to keep safety in mind?

A

To reduce the risk to the person doing the synthesis.

294
Q

What are some examples of safety measures that can be taken in planning a synthetic route?

A
  • Fume hoods to remove toxic gases

* Water baths to heat solutions so there are no naked solutions near flammable reagents

295
Q

Remember to practise drawing out the flowchart for the synthesis of aliphatic compounds.

A

Pg 222 of revision guide

296
Q

Remember to practise drawing out the flowcharts for the synthesis of aromatic compounds.

A

Pg 223 of revision guide

297
Q

What is combustion analysis used for?

A

Looking at information from burning an organic product in order to work out its empirical formula.

298
Q

How can you work out the empirical formula of an organic compound (containing C, H and O) when you are told the mass of water and CO₂ produced when a given mass of it is burned?

A

1) Work out the moles of water and CO₂
2) Work out the moles of C and H in these
3) Work out the masses of C and H in the organic product, and therefore the remainder must be O
4) Work out the moles of O
5) Find the ratio of the moles of C, H and O
6) This tells you the empirical formula

300
Q

How can you work out the molecular formula of an organic compound (containing C, H and O) when a given mass of it is burned with a known volume of oxygen and produced a known volume of water/CO₂?

A
  • All gases at the same temperature and pressure have the same molar volume
  • This means you can use the ratios of the gases as molar ratios
  • This is used to work out the molecular formula of the organic compound since all of the C, H and O atoms in the products must be accounted for by this or the oxygen
301
Q

30cm³ if hydrocarbon X combusts completely with 180cm³ oxygen. 120cm³ carbon dioxide is produced. What is the molecular formula of hydrocarbon X?

A

• The reaction can be written as:
30X + 180O₂ -> 120CO₂ + ?H₂O
• This simplifies to:
X + 6O₂ -> 4CO₂ + nH₂O
• 6 moles of O₂ = 12 O atoms. 8 of these are accounted for by the CO₂, so the remainder must be in the H₂O. Therefore, n = 4.
• So the equation is: X + 6O₂ -> 4CO₂ + 4H₂O.
• All of the carbons and hydrogen must come from the hydrocarbon, so the molecular formula must be C₄H₈.

302
Q

When 7.2g of a carbonyl compound is burnt in excess oxygen, it produces 17.6g of carbon dioxide and 7.2g of water. Calculate the empirical formula for the carbonyl compound.

A

Stage 1
• No. of moles of CO₂ = 17.6 / 44.0 = 0.40 moles
• This means there are 0.40 moles of C
• No. of moles of H₂O = 7.2 / 18.0 = 0.40 moles
• This means there are 0.80 moles of H
Stage 2
• Mass of C = 0.40 x 12.0 = 4.8g
• Mass of H = 0.80 x 2.0 = 1.6g
• Mass of O = 7.2 - (4.8 + 1.6) = 0.10 moles
Stage 3
Molar ratio = C : H : O = 0.40 : 0.80 : 0.10 = 4 : 8 : 1

Empirical formula = C₄H₈O

A LEVEL - Chemistry (Both) (21 decks)

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