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A2 Chemistry > Unit 4 > Flashcards

Unit 4 Flashcards

1
Q

Initial rate

A

Rate at start of reaction, where it is fastest.

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2
Q

Calculating order of particular reactant:

A

Compare 2 experiments where only that reactant is being changed.
If conc. Is doubled and rate stays same:0 order
If conc. Is doubled and rate doubles: 1st order
If conc. Is doubled and rate quadruples: 2nd order.

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3
Q

Working out orders when 2 reactant conc. Are changed at same time.

A

Where r= k[A][B]^2:
If [A] is x2, rate is x2
If [B] is x3, rate is x3^2 = x9
If this happens at the same time, rate is x2x9=18

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4
Q

Rate determining step

A

Slowest step in a mechanism.
No. of molecules is same as the order for each substance.
If rate determining step has intermediate, use what made up that intermediate from previous step.

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5
Q

Sn1 and Sn2

A 
Sn1= substitution, nucleophilic, 1 molecule in rate determining step. 
Sn2= substitution, nucleophilic, 2 molecules in RDS.
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6
Q

Kc

A

Equilibrium constant

Units depends on equation.

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7
Q

Working out units for Kc

A
  1. Put units of conc. Into equation for Kc.

2. Cancel out, including powers.

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8
Q

Calculating Kc

A

Mol. of reactant at equilibrium= initial mol. - mol. Reacted
Mol. of product at equilibrium= initial mol. + mol. formed.
1. Draw table of mol. and equilibrium mol.
2. Initial mol. Of product is always 0.
3. Work out equilibrium conc. by dividing equilibrium mol. by vol.
4. Put conc. into Kc equation.

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9
Q

When Kc has no units

A

There are equal no. Of reactants and products, ∴don’t need to divide by vol. to find conc., can just put equilibrium mol. straight into Kc expression.

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10
Q

Calculating units for k(rate constant)

A
  1. Rearrange rate equation to make k the subject.
  2. Insert units and cancel.
    Units of k=s^-1
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11
Q

Effect of temp on position of equilibrium and Kc

A

If position of equilibrium moves to left, value of Kc gets smaller as less products.

Catalysts have no effect on Kc or equilibrium.

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12
Q

Effect of pressure on position of equilibrium and Kc

A

With increased pressure, position of equilibrium will shift to side with less mol. of gas.
Kc will stay constant, as only changes with temp change.

Catalysts have no effect on Kc or equilibrium.

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13
Q

Bronsted- Lowry Acid

A

Substance that can donate a proton. (Proton donator).

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14
Q

Brondsted-Lowry Base

A

Substance that can accept a proton. (Proton acceptor).

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15
Q

Calculating pH

A

pH= -log [H+]

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16
Q

Calculating pH of strong acid

A

Strong acids completely dissociate.
Conc. of H+ ions in a monoprotic strong acid will be same as conc. of acid.
Give pH to 2dp.

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17
Q

Calculating [H+] from pH

A

[H+] = 1x10^-pH (inverse log on calculator)

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18
Q

Ionic product of water, equation

A

Kw= [H+][OH-]

At 25C, Kw=1x10^-14 mol^2dm^-6

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19
Q

pH of pure water

A

[H+]=[OH-], ∴Kw= [H+]^2, ∴[H+]=_/Kw (square root of Kw)

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20
Q

pH of strong base

A
  1. Find [H+] using Kw.

2. Use pH= -log[H+] to find pH.

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21
Q

Weak acids

A 
Only partially dissociate when dissolved in water, giving equilibrium. 
HAH^+ + A^-
Weak acid dissociation equation: 
Ka=[H+][A-]/[HA]
Larger the Ka, stronger the acid. 
pKa= -logKa, ∴Ka=10^-pKa.
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22
Q

Calculating pH of weak acid

A 
[H+]= [A-],∴Ka=[H+]^2/[HA]initial
[HA]equil = [HA]initial.
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23
Q

Strong acid and strong base neutralisation

A
  1. Work out mol. of original acid.
  2. Work out mol. of base added.
  3. Work out which one is in excess.
  4. If excess acid: work out new conc. of excess H+ ions, [H+]= mol. excess H+ /total vol.(vol. of acid + base)
  5. pH= -log[H+]
  6. If excess alkali: work out new conc. of excess OH-ions,
    [OH-]= mol. of excess OH- /total vol.
  7. [H+]= Kw/ [OH-]
  8. pH= -log[H+]
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24
Q

Weak acid and strong base neutralisation

A
  1. Work out mol. of original acid.
  2. Work out mol. of base added.
  3. Work out which is in excess.
  4. If excess acid: work out new conc. of excess HA, [HA]= initial mol. HA - mol. OH- / total vol.
  5. Work out conc. of sat formed, [A-]=mol. OH- added/ total vol.
  6. Rearrange equation for Ka to get [H+].
  7. pH= -log[H+]
  8. If excess alkali: same method as with strong acid and strong base.
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25
Q

Working out pH of weak acid at half equivalence/ neutralisation

A

When weak acid has been reacted with exactly half the neutralisation vol. of alkali, assume [HA]=[A-], ∴[H+]=Ka and pH=pKa.

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26
Q

pH of diluted strong acid

A
  1. [H+]= [H+]old x old vol./new vol.

2. pH= -log[H+]

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27
Q

pH of diluted base

A
  1. [OH-]= [OH-]old x old vol./new vol.
  2. [H+]= Kw/ [OH-]
  3. pH= -log[H+]
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28
Q

Buffer solution

A

Solution where pH does not change significantly if small amounts of acid or alkali are added to it.

There is a higher conc. of salt in buffer solution than in pure acid.
If small amounts of acid are added to buffer solution, equilibrium will shift to remove H+ ions.
As large conc. of salt ion in buffer, ratio stays constant, so does pH.

If small amounts of alkali are added, OH- ions will react with H+ ions to form H2O, equilibrium shift to produce more H+ ions.
Some molecules are changed to -ve ions, but as large conc. of salt ion, ratio and pH stays constant.

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29
Q

Basic buffer solution

A

Made from weak base and a salt of that base, eg. NH3, NH4^+Cl^-

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30
Q

Acidic buffer solution

A

Made from weak acid and a salt if that acid, eg. Ethanoic acid, sodium ethanoate.

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31
Q

Calculating pH of buffer solution

A

Salt content can be added by solution form or solid form.
Buffer can be made by partially neutralising a weak acid with an alkali as well.
Assume initial conc. of acid remains constant, as small amount has reacted.

  1. Rearrange Ka to find [H+].
  2. Calculate mol. of both solutions.
  3. Put into equation when they both have same final vol., if not calculate the one in excess, then find conc. of both solutions and put into equation.
  4. Use pH= -log[H+].
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32
Q

Calculating change in pH of buffer on addition of alkali

A

If small amount of alkali is added, mol. of acid would reduce by mol. of alkali added.
Mol. of salt would increase by same amount, so new calculation of pH can be done with new values

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33
Q

4 main types of titration curves

A
  1. Strong acid and strong base.
  2. Weak acid and strong base.
  3. Strong acid and weak base.
  4. Weak acid and weak base.
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34
Q

Equivalence point

A

Midpoint of vertical part of curve.

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35
Q

Choosing an indicator

A

They can be considered weak acids, has to be different colour to conjugate base.
Changes colour from HIn to In- over a small range.
Different indicators change colour over a different range.
End point of a titration is reached when [HIn]=[In-]
Need to pick an indicator whose end point coincides with equivalence point for the titration.
HIn In- + H+
In acidic sol., H+ ions cause equilibrium to shift to reactants, HIn is ∴an acidic colour.
In alkali sol., OH- removes H+, so equilibrium shifts to products, In- is alkaline colour.

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36
Q

Examples of indicators

A

Indicators will work if it’s pH range is on the vertical part of a titration curve, indicator will change colour rapidly and it will respond to the neutralisation point.
Eg.1. Phenolphthalein- use with strong base.
Colourless acid –> pink alkali
Eg.2. Methyl orange.
Red acid –> yellow alkali (Orange end point).

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37
Q

Optical isomerism

A

Occurs in carbon compounds with 4 different groups of atoms attached to a carbon atom (chiral carbon atom).
Have similar physical and chemical properties, but rotate plane-polarised light in different directions, by same amount.

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38
Q

Chiral (asymmetrical) carbon atom

A

A carbon atom that has 4 different groups attached.

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39
Q

Enantiomers

A

2 compounds that are optical isomers of each other.

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40
Q

Racemate/racemic mixture

A

Mixture containing 50/50 mixture of the 2 isomers (enantiomers).
Racemic mixture will not rotate plane-polarised light.

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41
Q

Formation of a Racemate

A

Formed when a trigonal planar reactant or intermediate is approached from both sides by an attacking species.
Equal chances of either enantiomer forming, so a racemate forms.

Eg. Nucleophilic addition of HCN to aldehydes and ketones when trigonal planar carbonyl is approached from both sides by HCN attacking species.
Racemate can also be formed by electrophilic addition of HBr to an unsymmetrical alkene.

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42
Q

Carbonyls

A

Compounds with C=O bond. (Aldehydes or ketones).
If C=O is on end of chain with H attached, it is aldehyde.
If C=O is in middle of chain, it is ketone. Smaller carbonyls are soluble in water as they can form H-bonds.
Pure carbonyls can’t from H-bonds, but form per many dipole bonds.

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43
Q

Reactions of carbonyls

A

C=O bond is polarised as O is more electronegative than C.
The C attracts Nucleophiles.
C=O is stronger than C=C in alkenes.

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44
Q

Oxidation reactions

A

Potassium dichromate is oxidising reagent for alcohols and aldehydes to oxidise.
Primary alcohols–> aldehydes–> carboxylic acid
Secondary alcohols–> ketones
Ketones can’t be oxidised.
Can be oxidised using Fehling’s solution or Tollen’s reagent, and used to test for presence of aldehyde groups.

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45
Q

Reduction of carbonyls

A

Reducing agents: NaBH4 or LiAlH4
Conditions: room temp and pressure.
Nucleophilic addition mechanism.
NaBH4 contains a source of nucleophilic hydride ions (H-).

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46
Q

Catalytic hydrogenation

A

Carbonyls can also be reduced using catalytic hydrogenation
reagents: hydrogen and nickel catalyst
Conditions: high pressure.

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47
Q

Addition of cyanide ions to carbonyls to form hydroxynitriles.

A 
Reaction: carbonyl--> hydroxynitrile
Reagent: NaCN and dilute H2SO4
Mechanism: Nucleophilic addition.
NaCN supplies CN- ions. 
H2SO4 supplies H+ ions.
Nucleophilic addition of HCN to aldehydes and ketones results in a Racemate, as it is approached from both sides and it is trigonal planar shape.
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48
Q

Acidity of Carboxylic acids

A

Weak acids in water, and only slightly dissociate.

They are strong enough to displace CO2 from carbonates.

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49
Q

Solubility in water of carboxylic acids

A

Smaller ones dissolve, soon after solubility rapidly reduces.
They dissolve, as they can form H-bonds to H2O molecules.

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50
Q

How carboxylic acid salts are stabilised.

A

Carboxylic acid salts are stabilised by delocalisation, so dissociation is more likely.
Delocalised ion has equal C-O bond lengths.
If delocalisation didn’t occur, C=O bond would be shorter.
Pi charge cloud has delocalised and spread out.
Delocalisation makes the ion more stable and ∴ more likely to form.
Increasing chain length pushes electron density on to the COO- ion making it more -ve and less stable, making the acid less strong.
Electronegative Chlorine atoms withdraw electron density from the COO- ion, making it less -ve and more stable, making the acid more strong.

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51
Q

Forming salts from carboxylic acid

A

With metal, alkali and carbonate.

Effervescence caused by reaction with carbonate/ production of CO2 can be used to test for carboxylic acid.

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52
Q

Esterification

A

Carboxylic acid reacts with alcohol, in presence of a strong H2SO4 catalyst, to form ester and H2O.
Reaction is slow and needs heating under reflux.
Low yields (50%)

53
Q

Uses of esters

A

Used in perfumes and flavourings. As solvents for polar organic substances, have low bp and are insoluble in water.
Used as plasticisers for polymers.

54
Q

Hydrolysis of esters

A

Can be hydrolysed by heating with acid or sodium hydroxide.

55
Q

Hydrolysis of esters by heating with acid

A

Reagents: dilute HCl
Conditions: Heat under reflux.
Reaction is reverse reaction of ester formation.
Carboxylic acid and alcohol are formed.
Reaction is reversible and doesn’t give good yield of product.

56
Q

Hydrolysis of esters by heating with sodium hydroxide

A

Reagents: dilute sodium hydroxide
Conditions: Heat under reflux
Carboxylic acid salt is the anion of carboxylic acid.
Anion is resistant to attack from weak Nucleophiles such as alcohols, so reaction is not reversible.
Reaction goes to completion.

57
Q

Fats and soaps

A

Fats and oils are esters of glycerol and long chain carboxylic acids (fatty acids).
Vegetable oils and animal fats can be hydrolysed to give soap, glycerol and long chain carboxylic acids.

58
Q

glycerol

A

Forms H-bonds very easily and is readily soluble in water.

It is used in cosmetics, food and in glues.

59
Q

Soaps

A

Fatty acid act as soaps.
Polar CO2^- end is hydrophilic and mixes with water.
Long non-polar hydrocarbon chain is hydrophobic and mixes with grease.
This allows grease and water to mix and be washed away.

60
Q

Biodiesel

A

Mixture of methyl esters of long chain carboxylic acids.
Vegetable oils can be converted into biodiesel by reaction with methanol in presence of a strong alkali catalyst.
This method is said to be carbon-neutral.
Does not take into account energy needed to extract oil, heat reaction or process fuel.
Also effects land available for food production.

61
Q

Acyl chlorides

A

Acyl chlorides are more reactive than carboxylic acids.
Cl and -OCOCH3 groups are classed as good leaving groups.
This makes acyl chlorides and acid anhydrides more reactive than carboxylic acids and esters.

62
Q

Acid anhydrides

A

They have a similar reactivity to acyl chlorides, so cause changes in functional groups.
Acyl chlorides give off HCl, whereas acid anhydrides give off RCOOH.

63
Q

Acyl chloride reaction with water

A

Acyl chloride–> carboxylic acid
Reagent: water
Conditions: room temp
Observations: steamy white fumes of HCl.
Nucleophilic addition-elimination mechanism.

64
Q

Acid anhydride reaction with water

A

Acid anhydride–> carboxylic acid
Reagent: water
Conditions: room temp
Nucleophilic addition-elimination mechanism.

65
Q

Acyl chloride reaction with alcohol

A

Acyl chloride–> ester
Reagent: alcohol
Conditions: room temp
Observations: steamy white fumes of HCl
Nucleophilic addition-elimination mechanism
This reaction for making esters is better than using carboxylic acids as the reaction is quicker and isn’t reversible.

66
Q

Acid anhydride reaction with alcohol

A

Acid anhydride–> ester
Reagent: alcohol
Conditions: room temp
Nucleophilic addition-elimination mechanism
This reaction for making esters is better than using carboxylic acids as the reaction is quicker and isn’t reversible.

67
Q

Acyl chloride reaction with ammonia

A 
Acyl chloride--> primary amide
Reagent: Ammonia 
Conditions: Room temp 
Observations: white smoke of NH4Cl
Nucleophilic addition-elimination mechanism
68
Q

Acid anhydride reaction with ammonia

A

Acid anhydride–> primary amide
Reagent: ammonia
Conditions: room temp
Nucleophilic addition-elimination mechanism

69
Q

Acyl chloride reaction with primary amides

A

Acyl chloride–> secondary amide
Reagents: primary amide
Conditions: room temp
Nucleophilic addition-elimination mechanism

70
Q

Acid anhydride reaction with primary amides

A

Acid anhydride–> secondary amide
Reagents: primary amide
Conditions: room temp
Nucleophilic addition-elimination mechanism

71
Q

Making aspirin

A

Made from 2-hydroxybenzoic acid, which has a phenol group.
Phenol group is turned into an ester by reacting with ethanoic anhydride.
Ethanoic anhydride is used instead of acid chloride because it is cheaper, less corrosive, less vulnerable to hydrolysis and less dangerous.

72
Q

2 classes of organic chemicals

A

Aliphatic- straight or branched chain organic substances.

Aromatic/ arene- 1 or more ring of 6 carbon atoms with delocalised bonding.

73
Q

Benzene structure

A

It is an arene
Each C atom is bonded to 2 C atoms and 1 H atom by single covalent σ-bonds.
1 unused e^- is left on each C atom in a p orbital, perpendicular to the plane of the ring.
The 6 p electrons are delocalised in a ring structure above and below the plane of C atoms.
It is a planar molecule, as all C-C bonds are the same.
H-C-C bond angle is 120*

74
Q

Delocalisation energy

A

Increase in stability connected to delocalisation.

Delocalised benzene is more thermodynamically stable.

75
Q

Phenyl group

A

C6H5 group attached to a compound.

76
Q

Reactions of benzene

A

Doesn’t undergo addition reactions, as it would break up the delocalised electrons.
Most of its reactions involve substituting a H for another atom or group.
Has high electron density, so attracts electrophiles, so it’s reactions are usually electrophilic substitution.

77
Q

Toxicity of benzene

A

It is a carcinogen
Methylbenzene is less toxic and reacts more readily than benzene, as methyl side group releases electrons into delocalised system, making it more attractive to electrophiles.

78
Q

Nitration of benzene

A 
Benzene--> nitrobenzene 
Reagents: conc. nitric acid in presence of conc. H2SO4 catalyst.
Electrophile: NO2^+ 
Conditions: 60*C +
Electrophilic substitution mechanism.
79
Q

Friedel crafts acylation

A

Benzene–> phenyl ketone
Reagents: acyl chloride in presence of anhydrous aluminium chloride catalyst.
Conditions: Heat under reflux
Electrophilic substitution mechanism.

80
Q

Reducing a nitrobenzene to aromatic amines

A

Reagents: Sn and HCl or Fe and HCl
Conditions: heating
Reduction mechanism
This can be done by catalytic hydrogenation (H2 with Ni catalyst).

81
Q

Effect of delocalisation on side groups with lone pairs

A

If a -OH group, Cl or a NH2 group is directly attached to benzene ring, the delocalisation in benzene ring will extend to include lone pairs on N,O and Cl.
This changes the properties and reactions of the side group.

82
Q

Chlorobenzene

A

C-Cl bond is made stronger.
Typically haloalkane substitution and elimination reactions don’t occur.
Electron-rich benzene ring will repel Nucleophiles.

83
Q

Phenol

A

Delocalisation makes C-O bond stronger and O-H bond weaker.

Doesn’t act like an alcohol- it is more acidic and doesn’t oxidise.

84
Q

Phenylamine

A

Less basic than aliphatic amines as lone pair is delocalised and less available for accepting a proton.

85
Q

Primary aliphatic amines

A

Act as Brønsted-Lowry bases, as lone pair on N is readily available for forming a dative covalent bond with a H+ and accepting a proton.
They are stronger bases than ammonia as alkyl groups are electron releasing and push electrons towards nitrogen atom, so make it a stronger base.

86
Q

Secondary aliphatic amines

A

Stronger bases than primary, as have more alkyl groups that are substituted on the N atom in place of the H atom.
So more electron density is pushed on to the N atom (as inductive effect of alkyl groups is greater than H atoms).

87
Q

Tertiary aliphatic amines

A

They are less strong bases than secondary amines.

88
Q

Base strength of aromatic amines

A

Primary aromatic amines, don’t form basic solutions, as lone pair on N delocalises with ring of e^- in benzene ring.
∴ N is less able to accept protons.

secondary amines > tertiary amines > primary amines > ammonia > aromatic amines

89
Q

Reactions of amines with acids

A

All form ammonium salts
They act as bases.
Addition of NaOH to ammonium salt will convert it back to amine.
Ionic salts will be solid crystals if water is evaporated, because of strong ionic interactions.
As ionic salts are formed, the compounds are soluble in the acid.

90
Q

Making a buffer from an amine

A

Basic buffers can be made from combining a weak base with a salt of that base.
Eg. Ammonia+ ammonium chloride, methylamine+ methylammonium chloride.

91
Q

Nucleophilic properties of amines

A

Primary amines can be formed by nucleophilic substitution reaction between haloalkanes and ammonia.
As lone pair is still available on N in amine formed, primary amine can react in same way, in successive series forming secondary, tertiary amines and quaternary ammonium salts.
Not good from making primary amines because of further reactions.
Desired product would have to be separated from other products.

92
Q

Reaction 1: forming primary amines

A

Ammonia dissolved in ethanol is initial Nucleophile.
1st step of mechanism: Nucleophile attacks haloalkane to form an intermediate.
2nd step: second ammonia removes a proton from intermediate to form amine.

93
Q

Reaction 2: forming secondary amine

A

Amine of 1st reaction has a lone pair on N, so reacts further with haloalkane.
Using excess ammonia can limit further reaction and produce more primary amines.
2nd step of mechanism: ammonia or amine can remove a proton from intermediate to form amine.

94
Q

Reaction 3: forming a tertiary amine

A

Same mechanism, with secondary amine reacting to form tertiary amine.

95
Q

Reaction 4: forming a quaternary ammonium salt

A

Excess of haloalkane will promote formation of quaternary salt.
Only 1st step of mechanism occurs.
Quaternary ammonium salts are not amines.

96
Q

Quaternary ammonium salts

A

Can be used as a cationic surfactant.
Surfactants reduce surface tension of liquids.
+ve N is attracted to -ve charges surfaces, eg. Glass, hair, plastics, etc.
This helps in use as fabric softener, hair conditioner, etc.

97
Q

Preparing amines from nitriles

A

Step 1: convert haloalkane to nitrile by using KCN in ethanol (Heat under reflux).
Step 2: reduce nitrile to amine using LiAlH4 in ether or by reducing with H2 using Ni catalyst.
A disadvantage is it is a 2 step reaction, as might have a low yield, and KCN is toxic.

98
Q

Optical activity of amino acids

A

All amino acids, except glycine, are chiral because of 4 different groups around C.
They rotate plane-polarised light.
Some amino acids have an extra carboxylic acid or amino group on R group.
These are classed as acidic or basic amino acids.

99
Q

Zwitterions

A

Neutral compounds that have both a permanent +ve and -ve charge.
Amino acids only exist as zwitterions.
Amino acids are often solids.
As it is a zwitterion, it has a relatively high melting point, due to ionic interactions.

100
Q

Acidity and basicity of amino acids

A

Amine group is basic and carboxylic acid group is acidic.

Amino acids act as weak buffers; will only gradually change pH if small amounts of acid or alkali is added.

101
Q

Proteins

A

Proteins are polymers made from combinations of amino acids.
Amino acids are linked by peptide bond, which are the amide functional group.
For any 2 different amino acids, there are 2 possible combinations in dipeptide.

102
Q

Chromatography of amino acids

A

Mixture of amino acids can be separated by chromatography and identified from the amount they have moved.
Rf value= dist. moved by amino acid/ dist. moved by solvent.
Each amino acid has its own Rf value.
Amino acids can undergo usual reactions of its functional groups, carboxylic acid and amine, eg. Esterification.

103
Q

2 types of Polymerisation

A

Addition

condensation

104
Q

Addition polymerisation

A

When saturated monomers react to form a polymer (monomers have C=C bonds).
Poly (alkenes) are inert, due to strong C-C and C-H bonds and non-polar nature of the bonds, so are non-biodegradable.
Chains form when same basic unit is repeated.

105
Q

Condensation polymerisation

A

2 different monomers add together and small molecule is given off as a side product, eg. H2O or HCl.
2 most common types of condensation polymers are polyesters and polyamides, involving formation of an ester linkage or an amide linkage.
Monomers usually have same functional group on both ends of molecule, eg. Di-amine, di-carboxylic acid, diol, etc.
Using carboxylic acid to make the ester or amide, need an acid catalyst and would only give equilibrium mixture.
The more reactive Acyl chloride goes to completion and doesn’t need a catalyst, but produces hazardous HCl fumes.

106
Q

Chemical reactivity of condensation polymers

A

Polyesters and polyamides can be broken down by hydrolysis, so are biodegradable.
Reactivity is due to polar bonds, which can attract attacking species, (Nucleophiles and acids).

107
Q

Polyesters can be hydrolysed by acid and alkali

A

With HCl: polyester split into original dicarboxylic acid and diol.
With NaOH: polyester split into diol and dicarboxylic acid salt.

108
Q

Polyamides can be hydrolysed by aqueous acids or alkalis

A

With HCl: polyamide split into original dicarboxylic acid and diamine salt.
With NaOH: polyamide split into diamine and dicarboxylic acid salt.

109
Q

Intermolecular bonding between condensation polymer chains

A

Polyesters have permanent dipole bonding between C^δ+=O^δ- groups in different chains in addition to VDWs forces between chains.
Polyamides have H-bonding between O in C^δ+=O^δ- groups and H in N^δ- - H^δ+ groups in different chains in addition to VDWs forces.
Polyamides have higher melting points than polyesters.

110
Q

3 ways to dispose of polymers

A

Landfill
Incineration
Recycling

111
Q

Purifying an organic liquid

A
  1. Put distillate of impure product into separating funnel.
  2. Wash product by adding either:
    - sodium hydrogencarbonate sol., shaking+releasing pressure of CO2.
    - saturated sodium chloride sol.
  3. Allow layers to separate in funnel, then run and discard the aqueous layer.
  4. Run organic layer into clean, dry conical flask and add 3 spatulas of drying agent (anhydrous sodium sulfate) to dry organic liquid.
  5. Carefully pour liquid into distillation flask.
  6. Distill to collect pure product.
112
Q

Drying agent in purifying an organic liquid should:

A

Be insoluble in the organic liquid.

Not react with organic liquid.

113
Q

Purifying organic solid: recrystallisation

A
  1. Dissolve impure compound in a minimum vol. of hot solvent. -Solvent has to dissolve both compound and impurities when hot and not the compound itself when cold. Min. vol. is used to enable crystallisation on cooling.
  2. Put hot filter sol. through filter paper quickly.- this will not remove any insoluble impurities and heat will prevent crystals reforming during filtration.
  3. Cool filtered sol. by inserting beaker in ice.- crystals will reform, but soluble impurities will remain in sol. form, as they are present in small amounts so sol. is not saturated. Ice will increase yield of crystals.
  4. Use Suction filtrate with a Büchner flask to separate out crystals.- water pump connected reduces pressure and speeds up filtration.
  5. Wash crystals with distilled water.- to remove soluble impurities.
  6. Dry crystals between absorbent paper.
114
Q

Loss of yield in the process of purifying organic solid (recrystallisation)

A

Crystals lost when filtering or washing.
Some product stays in sol. after recrystallisation.
Other side reactions occurring.

115
Q

Measuring melting point (determining purity)

A

If the sample is very pure, melting point will be same as in data books.
If impurities are present, melting point will be lowered and sample will melt over a range of several degrees.
Can be measured in an electronic melting point machine or by using a practical set up, where capillary tube is strapped to a thermometer immersed in heating oil.
In both cases small amount of salt is put into capillary tube.
Tube is heated slowly near melting point.
Comparing it with value in data book will determine purity level.

116
Q

Measuring boiling point (determining purity)

A

Purity of liquid can be determined by measuring boiling point.
This is done by distillation or by simply boiling a tube of sample in a heating oil bath.
Pressure should be noted, as changing pressure can change the boiling point of a liquid.

117
Q

Mass spectroscopy

A

High res. mass spectroscopy can be used to find molecular formula of compound from the accurate mass of molecular ion.
Can measure mass to 5dp, helps differentiate between compounds that appear to have similar Mr.

118
Q

Fragmentation

A

When organic molecules go through mass spectrometer, whole molecule and fragments are detected.
Forming a molecular ion:
M–>[M]^+ + e-
Molecular ion is both an ion and a free radical.
Several peaks are due to fragmentation.
Molecular ion fragments covalent bonds breaking: [M]^+ –>X^+ + Y•
The ion is responsible for the peak.
Relatively stable ions, R+ and acylium ions are common.
The more stable the ion, the greater the peak intensity.
Peak with highest m/z ratio is the molecular ion.
As charge of ion is +1, m/z ratio is equal to Mr.

119
Q

Infrared spectroscopy

A

Certain groups in a molecule absorb infrared at characteristic frequencies.
Complicated spectra can be obtained that provide info about types of bonds present in a molecule.
Above 1500= functional group identification.
Below 1500= fingerprinting
Rogue absorption a can occur and are indicators of impurities.

120
Q

NMR spectroscopy

A

2 main types:
C13 NMR
H (proton) NMR
C13 spectra is simpler than H- NMR spectra
C13 NMR spectrum, 1 peak for each set of equivalent C atoms (carbon environment).
H-NMR spectrum, 1 peak for each set of equivalent H atoms (hydrogen environment). The intensity (integration value) of each signal is proportional to no. of equivalent H atoms it represents.

121
Q

NMR: Solvents

A

Samples are dissolved in solvents without any H atoms, so it doesn’t give any peaks in H NMR.
Same solvent is used in C13 NMR, and will be 1 peak, due to solvent that will appear in spectrum.
Solvent: CDCl3 or CCl4

122
Q

NMR: calibration and shift

A

Small amounts of TMS is added to sample to calibrate the spectrum.
Same calibration compound is used for H and C13 NMR.
TMS is used because:
- its signal is always from others.
- only gives 1 signal.
- It is non-toxic.
- it is inert.
- it has low bp so can be removed from sample easily.
Spectra are recorded on a scale called chemical shift, which is how much the field has shifted away from the field for TMS.
H-NMR: more electronegative groups give greater shifts.

123
Q

Spin-spin coupling in H-NMR

A

In high res. H-NMR, each signal can be split into further lines due to inequivalent H’s on neighbouring/adjacent C atoms.
Splitting of peaks= no. of H’s on neighbouring C atoms +1.
Nuclei in identical chemical environments do not show coupling between themselves.

124
Q

Chromatography

A

An analytical technique that separates components in a mixture between a mobile phase and a stationary phase.
Separation by column chromatography depends on balance between solubility in moving phase and retention in stationary phase.
Mobile phase may be a liquid or gas.
Stationary phase may be a solid or a liquid or a solid on solid support.
Solid stationary phase separated by absorption.
Liquid stationary phase separates by relative solubility.
If SP was polar and MP was non-polar, compounds would pass through column more quickly than polar compounds as they have greater solubility in non-polar MP.

125
Q

HPLC-High Performance Liquid Chromatography

A

Stationary phase is a solid silica.

Mobile phase is a liquid.

126
Q

Gas-Liquid Chromatography

A

Used to separate mixtures of volatile liquids.
MP is an inert gas such as He, Ar,N, etc.
SP is a high BP liquid absorbed into a solid.
Tells us how many components there are in the mixture by no. of peaks, and abundance of each substance.
Area under each peak will be proportional to abundance of that component.
GC mixtures can be connected to Mass spectrometers, IR or NMR machines.
Mass spectrometer is commonly combined, spectra can be analysed or compared with a spectral database by computer for +ve identification of each component in mixture.

127
Q

Retention time

A

Time taken for a particular compound to travel from injection of sample to where it leaves the column to detector; used to identify a substance.
Some compounds have similar retention times so will not be distinguished.

128
Q

How to identify a compound.

A
  1. Work out empirical formula.
  2. Using molecular ion peak m/z value from mass spectrum, calculate molecular formula.
  3. Use IR spectra to identify main bonds/ functional groups.
  4. Use NMR spectra to give details of carbon chain.
  5. Put all together to find final structure.

A2 Chemistry (2 decks)

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