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Chemistry Y12 > Reaction Mechanisms > Flashcards

Reaction Mechanisms Flashcards

1
Q

What does a reaction mechanism describe?

A
  • how a reaction occurs and the steps involved
  • the series of bond breaking and bond making steps that occur in a reaction
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2
Q

what is the first step in any reaction mechanism?
how can bonds be broken?

A
  • breaking bonds
  • via homolytic or heterolytic fission
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3
Q

what happens in homolytic fission?
what does homolytic fission form?

A
  • the bond breaks equally and the electrons from the covalent bond are shared between the two atoms involved
  • free radicals
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4
Q

what are free radicals?

A
  • are atoms with an unpaired electron
  • they are very reactive particles
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5
Q

what are intermediates?

A
  • intermediates are a species that are formed in one step of a reaction mechanism and used in another step
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6
Q

what is heterolytic fission
what does it form?

A
  • the electrons from the bond are transferred to the more electronegative atom
  • an electron-deficient atoms (positive) also known as electrophiles
  • and an electron pair donor called a nucleophile
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7
Q

alkanes react with…?
halogenalkanes react with…?

A
  • electrophiles
  • nucleophiles
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8
Q

what else can be formed in heterolytic fission?

A

a positive carbon atom called a carbocation this only forms 3 bonds

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9
Q

Alkanes react with chlorine via what reaction mechanism?

A

Free radical substitution mechanism

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10
Q

What does the chlorination of alkanes produce?

A

A mixture of chlorinated alkanes and hydrogen chloride

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11
Q

what conditions are required for chlorine to react with methane?

A

in the presence of ultraviolet light - will react explosively even at room temp or heat

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12
Q

what is the equation for the chlorination of methane?

A

CH4 + CL2 ——–> CH3CL + HCL
methane + clorine —> chloromethane + hydrochloric acid

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13
Q

what type of reaction is this?

A

substitution reaction - 1 of the chlorine atoms replaces one hydrogen

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14
Q

this process occurs in three stages?

A

1.initiation
2. propagation
3. termination

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15
Q

initiation step 1

A

the UV light provides the energy to break the covalent bonds in some of the chlorine molecules to produce free radicals by homolytic fission
CL2 —UV—–> CL*

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16
Q

why is cl-cl bonds broken first then c-h?

A

cl-cl bonds are weaker

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17
Q

propagation step 1

A

a highly reactive chlorine radical reacts with a methane molecule
cl* + CH4 ———> CH3* + HCL

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18
Q

Propagation step 2

A

the methyl radical can then attack another chlorine molecule to produce another chlorine free radical
CH3* + CL2 ——-> CH3CL + CL*

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19
Q

why is this a chain reaction?

A

in each propagation step, a radical is used up and a new radical is formed. hence chain reaction. the chlorine radicals are made and used until there is no more chlorine to react

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20
Q

what is the overall reaction?

A

is the sum of the 2 propagation steps
CH4 + Cl2 ——> CH3Cl + HCl

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21
Q

when does the reaction stop and ecome stable?

A

when 2 radicals combine, the product will be a stable molecule and the sequence of the reaction stops.
-once all the intermediates have been used up the reaction is finished

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22
Q

termination step 1

A

the unpaired electrons in the radicals pair up to form a covalent bond e.g
CH3* + CL* ——> CH3CL
CH3* + CH3* ——-> CH3CH3

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23
Q

What do the termination steps do?

A

‘mop’ up free radicals preventing further reactions

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24
Q

why can chloromethane be further substituted?

A
  • contains 3 hydrogen atoms that can be substituted via further pairs of propagation steps
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25
Q

further substitutions can lead to the formation of:

A
  • dichloromethane (CH2Cl)
  • trichloromethane (CHCl3)
  • tetrachloromethane (CCL4)
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26
Q

How can further substitution be reduced?

A

By using an excess of methane

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27
Q

however industrial chemists would not use this method to prepare a sample of chloromethane why?

A
  • the reaction is too messy.
  • too many by-products are formed by further substitution and removal would be costly
    better to prepare halogenalkanes from alkenes
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28
Q

trends in rate og halogenalkanes?

A
  • methane reacts less readily with halogens as you go down group 7.
  • ## as halogens are less reactive as you go down
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29
Q

why is chlorination of alkanes harmful to the ozone layer?

A
  • some organic chloro-compounds escape into the atmosphere and eventually decompose to form chlorine free radicals
  • these radicals react with ozone and contribute to the depletion of the ozone layer
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30
Q

What may protect the body from free radicals?

A

antioxidants are thought to protect the body from free radicals which may cause cell damage

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31
Q

uses of halogenalkanes?

A

solvents, refrigerants, aerosol propellants , insecticides and anaesthetics

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32
Q

why has their uses become controversial?

A

recent evidence has shown some of these halogen alkanes are toxic and have harmful effects on the atmosphere

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33
Q

what halogenoalkanes are used commonly but now banned?

A

1,1,1 - tricholorethane - used in tippex and as a dry-cleaning solvent
tetrachloromethane- used as a dry-cleaning solvent

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34
Q

why are CFC’s very unreactive?

A

because c-f and c-cl bonds are very strong

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35
Q

properties of CFC’s?

A

stable compounds, have not taste, no smell, usually gases at room temp - useful

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36
Q

what could they be used as?

A

refrigerants, aerosol propellants and in packaging materials such as expanded polystyrene

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37
Q

Since CFC’s were considered very stable and unreactive they were….?

A

vastly released into the atmosphere and made

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38
Q

what is ozone? any distinct properties?

A

ozone is an allotrope of oxygen (03)
blue pale gas with a sharp smell

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39
Q

the ozone in the upper atmosphere is continually being formed and broken down by?

A

intense ultraviolet radiation

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40
Q

how is ozone formed and what does it break down into? (step 1)

A

ozone is formed when ultra-violet radiation from the sun breaks down oxygen molecules into 2 oxygen radicals
O2 —uv—-> 2 O*

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41
Q

2nd step into making ozone?

A

these oxygen radicals then react with more oxygen to form ozone
O* + O2 ——-> O3

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42
Q

then with ultraviolet radiation ozone decomposes into…?

A

an oxygen molecule and an oxygen radical
O3 —-uv——> O2 + O*

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43
Q

then these oxygen radicals can react with more …? to reform..?

A

ozone to reform O2 molecules
O* + O3 ——> 2O2

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44
Q

what is the overall result of these actions?

A

the presence of ozone in the upper atmosphere reduces the amount of harmful ultraviolet radiation from the sun that can reach the Earth’s surface
so ozone formed naturally in the upper atmosphere is beneficial

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45
Q

what was a problem with the ozone layer?

A

it was becoming thinner - 1970s found that ozone layer in antartica was thinner than expected

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46
Q

what was the reason for the destruction of the ozone layer?

A

reaction of ozone with chlorine radicals

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47
Q

why are CFC’s able to diffuse into the upper atmosphere?

A

they are extremely unreactive so do not get broken down by reacting with oxygen or water so they’re are able to diffuse into the upper atmosphere

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48
Q

what happens in the upper atmosphere?

A

in the upper atmosphere there are large amounts of uv radiation, so the large amounts of uv radiation breaks down the C-CL bonds and chlorine radicals are produced
CF2CL2 —-uv——> CL* + *CF2CL

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49
Q

Why can the uv radiation not break down the C-F bonds?

A

uv radiaiton does not have sufficient energy to breaks down c-f bonds

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50
Q

then propagation steps occur?

A

CL* + O3 —-> ClO* + O2
CIO* + O3 ——> CL* + 2O2

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51
Q

Why are chlorine radicals not used up in these steps?

A

chlorine radicals are used up in the 1st step but are reformed in the 2nd step ∴ not used up

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52
Q

this means that…?

A

the chlorine radicals CATALYSE the decomposition of ozone and contribute to the formation of a hole in the ozone layer.
because the chlorine radicals are not used up it means that small amounts of chlorine radicals can continue to destroy ozone for many years

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53
Q

what was the Montreal protocol (1987)?

A

chemists supported the international agreement to ban the use of CFC’s

54
Q

what is an alternative to CFC’s?

A

chlorine free CFC’s called HFC’S

55
Q

what are HFC’s?

A

contain hydrogen flourine and carbon

56
Q

why is this better?

A

because C- F bonds are stronger than c-cl bonds ∴ less likely to be broken down by ultraviolet radiaiton

57
Q

what is the general formula for halogenoalkanes?

A

CnH2n+1 X
X = halogen

58
Q

uses of halogenoalkanes?

A

refrigerants, solvents, pharmaceuticals

59
Q

how are halogenoalkanes classified?

A

with primary, secondary, or tertiary depending on the number of carbon atoms directly attached to the carbon with the halogen atom

60
Q

the halogen in the halogenoalkane is more electronegative so the bonds are…?

A

polar

61
Q

explain how halogenoalkanes are polar?

A

the halogen is more electronegative, electrons are attracted towards the halogen atom.
the halogen becomes slightly negative , the carbon atom becomes slightly positive (electron deficient)

62
Q

polar bonding means..?

A

halogenoalkanes have permanent dipole dipole forces between neighbouring molecules

63
Q

what is the polar c-halogen bond responsible for?

A

chemical reactions of halogenoalkanes
hence makes them more reactive than alkanes

64
Q

what are nucleophiles?

A

nucleophiles are a species that donates a lone pair to form a covalent bond

65
Q

what happens in nucleophilic substitution?

A
  • the lone pair on the nucleophile forms a covalent bond with the δ+ carbon atom
  • the carbon halogen breaks via heterolytic fission
  • ## the pair of electrons from the carbon-halogen bond pass to the carbon atom - forming a halide ion
66
Q

What is nucleophilic substitution also referred to as?

A

hydrolysis

67
Q

why do halogenoalkanes get more reactive as you go down group 7?

A
  • bonds get weaker and are more easily broken , so more reactive the halogen
68
Q

why does rate of reaction increase as you go down the group?

A
  • the carbon-halogen bond energy decreases down the group
  • therefore rate of reaction with nucleophiles increases down group 7
69
Q

since C-F is very polar so why is it unreactive?

A

because the c-f bonds is so strong that it becomes very unreactive and do not undergo nucleophilic substitution

70
Q

why are halogenoalkanes used in the synthesis?

A
  • they can be prepared from different compounds
  • they can be changed into many different homologous series
71
Q

what type of ions does halogenoalkanes undergo nucleophilic substitution with?

A
  • hydroxide ions (OH-)
  • cyanide ions (CN-)
  • ammonia (NH3)
72
Q

explain what happens when halogenoalkanes react with warm aqueous sodium hydroxide/potassium hydroxide?

A

when halogenoalkanes react with warm aqueous sodium hydroxide or potassium hydroxide ALCOHOLS are formed

73
Q

draw the reaction between 1-bromopropane and koh? [WHITEBOARD]

A

CH3CH2CH2Br + OH- ———–> CH3CH2CH2OH (PROPAN-1-OL)+ Br-

74
Q

explain the product formed when halogenoalkanes are reacted with an aqueous alcoholic solution of potassium/ sodium cyanide

A

nitriles are formed

75
Q

draw the reaction mechanism of iodopropane with KCN

A

CH3CH2CH2I + CN- ————–> CH3CH2CH2CN + I-

76
Q

hydrolysis of nitriles?

A

nitriles are readily hydrolysed in acid solution to form carboxylic acids

77
Q

propanetrile hydrolysis equation?

A

CHH3CH2CN + 2H20 + HCL ————->
CH3CH2COOH + NH4CL

78
Q

draw the Nitrile and carboxylic acid structure and ammonia structure

A
79
Q

what products are formed when halogeneoalkanes are reacted with excess ammonia (warmed in a sealed container)?

A

primary amines are formed

80
Q

draw the reaction mechanism with 1-chloropropane and 2NH3

A

[step 1]CH3CH2CH2CL + 2NH3 ——–>
CH3CH2CH2NH2 (propylamine or 1-aminopropane)+ NH4CL

[step2] - a molecule of ammonia acts as a base so accepts h+ ions

81
Q

structure of amines?

A

whiteboard drawing and answer in folder

82
Q

what can amines do?

A

amines have a lone pair of electrons on the nitrogen atom so can act as nucleophiles themselves

83
Q

why do we use an excess of ammonia

A

to prevent successive substitution. the excess ammonia reduces the chance of further reaction of the primary amine with the halogenoalkane to form secondary or tertiary amines or quaternary ammonium salts

84
Q

nucleophillic substitution and elimination

A
85
Q

when a halogenoalkane reacts with a hydroxide ions two types of reactions can occur?

A
  • nucleophillic substitution
  • elimination
86
Q

the elimination reaction is favoured by…?

A

using ethanol as a solvent, using a higher temperature and by increased branching of the halogenoalkane ( so tertiary and also secondary)

87
Q

the substitution reaction is favoured by…?

A

using warm aqueous alcoholic as the solvent

88
Q

however, secondary alkanes undergo…?
primary undergo…?

A
  • both substitution and elimination concurrently
  • mainly substitution and very little elimination
89
Q

outline the mechanism for the reaction between 2-bromobutane and potassium hydroxide (substitution and elimination)

A

Substitution forms - butan-2-ol
elimination forms - But-2-ene and But-1-ene

90
Q

What happens when 2 - bromobutane is reacted with warm aqueous sodium hydroxide?

A

a mixture of organic products are formed - butan-2-ol, But-2-ene and But-1-ene

91
Q

what is the general formula for alkenes

A

CnH2n

92
Q

the carbon - carbon double bond is…?

A

planar and has an angle of 120 degrees

93
Q

why are alkenes very reactive?

A

because the carbon-carbon double bond is electron rich. the double bond can therefore attack electron deficient species

94
Q

in the carbon-carbon double bond there are 2 types of covalent bonds?

A
  • (sigma) σ bond
  • (pi) π bond
95
Q

explain how they differ?

A

the differ in the way that the orbitals overlap so the electrons are shared differently in the two different bonds

96
Q

what is the sigma bond?

A

it can be thought of as a conventional covalent bond

97
Q

what is the π bond?

A

is a cloud of electron density above and below the molecule. the π bond is formed by the overlap of a spare unbonded, singly filled p-orbital present on each carbon atom in the bond

98
Q

why is the carbon-carbon double bond planar?

A

due to the p-orbitals overlap leading to the π bond formation

99
Q

E-Z stereoisomerism

A

(a.k.a geometrical isomerism or cis-trans isomerism)

100
Q

what are stereoisomers?

A

stereoisomers are molecules which have the same structural formula but their bonds are arranged differently in space

101
Q

what leads to the formation of e-z stereoisomers?

A

the double bonds can not rotate due to the π electron clouds above and below the plane of the bond. a large amount of energy is needed to rotate a double bond (not enough energy at room temperature) so we say the c=c has a restricted rotation leading to the formation of e-z stereoisomers

102
Q

how do we differentiate between e-z stereoisomers?

A

Z form - both groups are on the SAME side of the double bond
E form - both groups are on the OPPOSITE side of the double bond

103
Q

how do we differentiate e-z stereoisomers from more complex alkenes?

A

using Cahn-Ingold Priority rules
1. label the two carbon atoms of the c=c double bond
2. look at the two other atoms that are directly bonded to c1 and assign them a priority (higher or lower).
(the atom with the highest atomic number has the highest priority )
3. look at the other 2 atoms that are directly bonded to c2 and assigns them a priority
4. to work out their e-z isomer look at the positions of the higher priority groups

z form - higher priority groups are on the same side
e form - the higher priority groups are on the opposite side

104
Q

explain the different properties of e-z isomers?

A
  • z- isomers have higher boiling points because they are more polar than e-isomers
  • e-isomers have a higher melting point because they pack together more closely than z-isomers
105
Q

why can alkenes undergo combustion?

A

because they are hydrocarbons

106
Q

what is the difference between complete and incomplete combustion?

A
  • complete combustion occurs in a plentiful supply of oxygen whereas incomplete combustion occurs in a insufficient supply of oxygen.
  • the products of complete combustion is c02 and water
  • the products of incomplete combustion is carbon monoxide and water
107
Q

Electrophilic Addition

A
108
Q

why can alkenes undergo electrophilic addition?

A
  • the alkene c=c double bond is electron rich
  • the double bond can therefore attack electron deficient species (electrophiles)
  • electrophiles have a partial positve charge and accepts a pair of electrons .
  • alkenes undergo an addition reaction with an electrophile to make a saturated product
109
Q

outline the reaction mechanism for the reaction of ethene with hydrogen bromide

A
  • the hBr molecule ha a permanent dipole with a δ+ charge on the hydrogen atom
  • a pair of electrons from the c=c double ond forms a bond with the δ+ hydrogen atom.
  • the hydrogen-bromine bond breaks , leaving a bromide ion
  • he addition of hydrogen to the alkene forms a carbocation intermediate
  • carbocations are very reactive and are readily attacked by nucleophiles
  • the carbocation is attacked by the bromide ion that acts as a nucleophile
110
Q

What are the reactions conditions for reacting and alkene with H-Br?

A
  • aqueous solutions / gas phase
111
Q

alkenes react with cold concentrated sulphuric acid to form…?

A

alkyl hydrogensulphates

112
Q

draw and outline the reaction mechanism of sulphuric acid to ethene

A
  • electrophilic addition
  • (forms ethyl hydrogensulphate)
113
Q

hydrolysis of alkyl hydrogen sulphates with warmed water produces…?

A

an alcohol

114
Q

what is the overall reaction of an alkene with concentrated sulphuric acid followed by hydrolysis represented by a single equation?

A

CH2=CH2 + H20 ——–> CH3CH2OH

115
Q

Why is sulphuric acid not included in the equation above?

A

because in this reaction sulphuric acid is considered a catalyst

116
Q

how is ethanol produced

A

ethanol is not produced via this reaction but produced by fermentation of sugars and direct hydration of ethene

117
Q

when alkenes react with bromine they form..?

A

dibromoalkanes

118
Q

draw and ouline the addition reaction mechanism of bromine to ethene

A
  • the bromine molecule is non-polar
  • the high electron density of the c=c double bond induces a temporary dipole on the bromine molecule
  • the electron deficient δ+ bromine atom is the electrophile
  • a pair of electrons from the c=c double bond forms a bond with the δ+ bromine atom
  • this forms a carbocation intermediate and a bromide ion
  • the bromide ion then attacks the carbocation formed acting as a nucleophile
119
Q

describe the test for alkenes

A

bromine water (orange-brown) to colourless

120
Q

why do we only use a dilute solution of bromine water?

A

because bromine is a hazardous chemical

121
Q

when alkenes react in water an alternative product is also formed…?

A

so when ethene reacts with bromine water
CH2BrCH2OH is formed

122
Q

electrophilic addition to unsymmetrical alkenes

A
123
Q

when we add an electrophile to an unsymmetrical alkene it forms two products..?

A

a major and a minor product

124
Q

when is the major product formed?

A

via the most stable carbocation

125
Q

how can a carbocation be labelled?

A

primary - have 1 carbon atom directly bonded to positve carbon atom
secondary - have two atoms directly bonded to the positve carbon atom
tertiary - have 3 carbon atoms directly bonded tot the positve carbon atom

126
Q

what determines the stability of a carbocation?

A

by the presense of alkyl groups. the more alkyl groups directly attached to the positve carbon atom, the more stable the carbocation

127
Q

state what is the order of stablity?

A

tertiary > secondary > primary

128
Q

draw and outline the reaction mechanism of propene and hydrogen bromide, label the major and minor products formed

A
  • hydrogen from the hBr molecule can bond to either of the carbon atoms in the c=c double bond
  • this results in two different carbocations a primary and secondary
  • the secondary carbocation is the most stable than the primary carbocation so the secondary carbocatin is mostly formed
  • the bromide ion attacks the secondary carbocation and 2-bromopropane is formed as the major product
  • and a small amount of 1-bromopropane is formed as the minor product
129
Q

draw and outline the mechanism to show the reaction of propene with sulphuric acid, state the major and minor products formed

A
  • hydrogen from the sulphuric acid adds adds on to the alkene forming a secondary and primary carbocation
  • the secondary carbocation is more stable than the primary one
    so the sulphuric acid ion attacks the secondary carbocation to form propan-2-ol the major product
  • a small amount of propan-1-ol is also formed after hydrolysis as the minor product
130
Q

FINISH

A

Chemistry Y12 (11 decks)

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