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Chemistry > Paper 2 > Flashcards

Paper 2 Flashcards

1
Q

What is a hydrocarbon

A

A compound which contains hydrogen or carbon only

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2
Q

Homologous series

A

A series of compounds that contain the same functional group and have the same general formula which differs by CH2

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3
Q

Functional group

A

Small group of atoms or a single halogen atom that give the compounds in the series particular chemical properties

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4
Q

Hydrolysis

A

H+ or OH- ions break a bond in a molecule splitting into two parts

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5
Q

How do you obtain alkane fuels

A

Fractional distillation of alkanes produces large amounts of heavier fuel. These are then cracked over suitable catalysts into smaller alkanes. these straight chained hydrocarbons are reformed into branched chain and cyclic hydrocarbons for efficient combustion by passing them over a suitable catalyst. These have a higher octane and therefore are more suitable for petrol driven cars

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6
Q

Problems which arise from the pollutants of combustion of fuels

A

Carbon monoxide
Sulphates of sulphur and nitrogen
Carbon particulates
Unburnt hydrocarbons are formed

CO is toxic
Oxides of nitrogen and sulphur are acidic

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7
Q

Alternative fuels

A

Biodiesel and alcohols
Renewable
Offset carbon released when grown
Some can be made from waste products eg coffee beans

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8
Q

What is a radical

A

Unpaired electron represented by a single dot

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9
Q

How are free radicals formed

A

Homolytic fission of a covalent bond

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10
Q

Alkanes plus oxygen in air

A

Burns balance with oxygen co2 and water

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11
Q

Alkane plus halogens

A

UV light or bright white light - chlorine and bromine

Free radical substitution

Initiation( homolytic fission of halogen)
Propagation - numerous number of products one radial reacts with one non radical to produce a radical and a non radical. The radical can be organic or a halogen)
Termination - two free radicals come together to produce a non free radical

Unhelpful in organic synthesis as there is a variety of products and further reactions would occur

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12
Q

Electrophile

A

Electron pair acceptor

Electron is attracted to the e- rich site

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13
Q

Alkenes plus hydrogen

Application

A

150degrees
Nickel catalyst

Forms alkane

Vegetable oil becomes margarine

Free radical addition

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14
Q

Alkene plus halogen

A

Room temperature, mix

In organic solvent
Halogen added across double bond

In aqueous solution
One halogen added one OH added

Chlorine and bromine

However iodine is not a strong enough electrophile unless the C=C bond is activated by an oxygen atom

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15
Q

Alkane plus hydrogen halide

A

Forms halogenoalkane
One halogen and one H added across double bond m
Mix gases at room temperature

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16
Q

Alkene plus steam

A

H3PO4 acid catalyst
Alkene vaporised

Forms alcohol

Reversible reaction so alcohol is removed by cooling and remaining gases repossessed over catalyst

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17
Q

Alkene plus potassium manganate (VII)

A

Shake at room temp
alkene + [O] + H20 -> diol

Purple manganate (VII) ions are reduced to ppt of brown manganese ( IV) oxide in neutral or colourless Mn^2+ ions in acid

NOT DICHROMATE

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18
Q

Formation of ions from covalent molecule

A

Heterolytic fission of the covalent bond

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19
Q

Addition of binary compounds to an alkene

A

Arrow from double bond to molecule (draw dipole on molecule if polar)

Covalent bond opens, one carbon has the halogen, the other carbon is positive

The negative ion (lone pair) arrow to carbon

Then compound

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20
Q

Markovnikov’s rule

A

Addition to an A-symmetric alkene, the hydrogen goes to the carbon which already has more hydrogen atoms directly attached

This is because he secondary carbocation is stabilised by the electron pushing affect compared to the primary

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21
Q

Stability of primary secondary and tertiary carbonation intermediated

A

tertiary is more stable than secondary which is more stable than primary due to the electron pushing effect

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22
Q

Test for C=C bond

A

Bromine water decolourises from orange

Mix at room temp

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23
Q

Waste polymers

A

Can be separated into specific types of polymer for
•recycling,
•incineration - producing energy (can produce toxic compounds eg CO)
•feedstock for cracking - producing a mixture of short chain alkene

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24
Q

How do chemists contribute to better use of polymers

A
  • more sustainable use of materials - less energy used making them or not using limited resources
  • developing biodegradable polymers
  • removing toxic waste gas from the incineration of plastics
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25
Q

What is a nucleophile

A

Lone pair donor, forms covalent bond with δ+ atom in another molecule

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26
Q

Halogenoalkane plus potassium hydroxide

A

Hydroxide ion acts as nucleophile

Heat under reflux and aqueous

R-BX + KOH -> R-OH + KX

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27
Q

Halogenoalkane plus aqueous silver nitrate in ethanol

A

Water acts as nucleophile

R-X + H20 -> R-OH + H+ + X-

Ag+ X -> AgX

ethanol !!
60 degrees

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28
Q

Halogenoalkane plus potassium cyanide

A

Reflux, aq and ethanol

R-X + KCN -> R-CN +KX

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29
Q

Halogenoalkane plus ammonia

A

R-X + 2NH3 -> R-NH2 + NH4X

Conc ammonia at room temp
Or heat in sealed tube

Heat in sealed tube so that it reacts fast but ammonia liberated if it was under reflux, ammonia gas would not be condensed by the reflux condenser

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30
Q

Halogenoalkane plus Ethanolic potassium hydroxide

A

R-Br + KOH -> alkene Of chain length R + H20 + KX

Hydroxide ion acts as a base

Heat under reflux with ethanol conc KOH

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31
Q

Experiment to determine relative rates of reaction of primary secondary and tertiary halogenoalkanes and chloro bromo and iodo halogenoalkanes

A

Equal amount of the halogenoalkanes are placed in a test tube water bath 60 degrees
ethanol and silver nitrate and time how long it takes the ppt to appear

Tertiary fastest secondary then primary SN1 faster than SN2

Iodo after than bromo faster than chloro because C-I bond enthalpy is the weakest, then bromo then chloro

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32
Q

Mechanism primary halogenoalkane and KOH

A

Lone pair on OH arrow from that to δ+ carbon, δ- on Br - attacks from opposite side of bromine arrow from C-Br bond to Br
-> Forms intermediate with five bonds, square bracket and Br and OH bonds dotted lines, remember charge on intermediate, -> final product and Br-

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33
Q

Primary halogenoalknes and ammonia mechanism

A

Lone pair on ammonia, dipole on c-Cl

Lone pair on ammonia arrow to carbon, then arrow from C-Cl bond to Cl

Intermediate with five bonds around carbon and four around ammonia

Then forms 3HN-C-RHH + Cl- ammonia has a positive charge

Another ammonia lone pair arrow attacks hydrogen, arrow from H-N bond to N

Forms NH4Cl + 2HN-CRHH

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34
Q

Alcohol plus oxygen in the air

A

Burning !

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35
Q

Alcohol plus PCl5

A

Dry alcohol solid PCl5
R-OH + PCl5 -> R-Cl + POCl3 + HCl

This is the test for alcohol as steamy fumes of HCl is given off

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36
Q

Alcohol plus KBr H2SO4

A

50% conc H2SO4 to prevent HBr being oxidised, heat under reflux

KBr + H2SO4 -> KHSO4 + HBr
R-OH + HBr -> R-Br + H20

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37
Q

Alcohol plus red phosphorus and I2

A

Warm, moist red phosphorous
2P + 3I2 -> 2PI3

3R-OH + PI3 -> 3R-I + H3PO3

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38
Q

Alcohol plus potassium dichromate (VI)

A

Dilute sulphuric acid
R-OH + [O] -> R=O + H20 distill as produced (hot ethanol 60 degrees)
R-OH + 2[O] -> R-OOH + H20 heat under reflux

Secondary -> just to ketone - can be refluxed

HUF, electric heater, then use test for aldehyde to determine if it was primary or secondary secondary

Dichromate orange to green

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39
Q

Phosphoric acid + alcohol

A

Warm, conc H3PO4

Alcohol -> alkene + H2O

(Or heat over aluminium oxide - separate reaction)

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40
Q

Heating under reflux

A

Volatile/ flammable / toxic substancesthe
•Place in round bottomed flask with vertical reflux condenser
•Water in at bottom and out at top
•top of reflux condenser being open
•use electric heater

No I reacted reagent escapes
Add antibumping granules
Distill off wanted product at bp

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41
Q

Solvent extraction

A 
Sparingly soluable organic product
•shake reaction mixture in separating funnel with solvent eg ethanol, cyclohexane, Dry ether. Depends on circumstance
•organic layer collected only 
•wash and dry organic layer 
•still off solvent

Suitable solvent dissolves the organic and is low boiling point so can be easily removed

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42
Q

Boiling point determination

A
  • Place a small about of a test liquid in an ignition tube and attach to thermometer with rubber band.
  • Place in beaker with water. Clamp thermometer
  • Place empty capillary tube in liquid, open end below surface
  • Heat water and stir, slowly heat until rapid stream of bubbles comes out. Note temp and stop heating
  • allow to cool, stir until bubbles stop and liquid sucked into capillary tube. Note temp
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43
Q

Drying

A

Use an anhydrous salt such as calcium chloride to remove any water from a crystal product.

Anhydrous sodium sulphate or calcium chloride

Initial mixture is cloudy, clear when dry. Then filter organic liquid

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44
Q

Chirality

A

Optical isomerism is a result of chirality in molecules with a single chiral centre. This results in optical isomerism. Where the optical isomers are object and non superimposable mirror images

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45
Q

What is optical activity

A

The ability of a single optical isomer to rotate the plane of polarisation of the plane-polarised monochromatic light in molecules containing a single chiral centre

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46
Q

Racemic mixture

A

A solution containing equimolar amounts of the two enantiomers does not rotate the plane of plane polarised light

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47
Q

Optical activity as evidence for Sn1 and Sn2 reactions

A

SN1 - chance of attack is identical from top or bottom - racemic mix. The carbocation has three pairs of bonding electrons and no lone pairs, so it’s shape is triangular planar around the positive carbon atom

SN2 - nucleophile attacks from the opposite side to the halogen therefore single optical isomer

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48
Q

Boiling and melting point of aldehydes and ketones

A

Lower than alcohols because does not hydrogen bond intermolecular

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49
Q

Solubility of aldehydes and ketones

A

Solvable as can hydrogen bond with water

They cannot hydrogen bond with themselves

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50
Q

Carbonyl compounds with Fehling/Benedicts

A

Copper sulphate and potassium tartrate
Deep blue solution

Aldehydes only -> deep blue solution forms red ppt of Cu2O copper (I) oxide

Aldehyde oxidised

R=O + [O] + OH- -> ROO- +H20

Salt of carboxlyic acid

Test for aldehyde

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51
Q

Carbonyl compounds and Tollen’s

A

Sodium hydroxide and silver nitrate dissolved in dilute ammonia, warm

Only aldehyde
Silver mirror produced
R=O + [O] + OH- -> ROO- +H20

Ag(NH3)2^+ -> Ag + 2NH3

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52
Q

Carbonyl compounds and acidified dichromate (VI) ions

A

R=O + [O] + OH- -> ROO- +H20
Aldehyde only

Orange reduced to green

(If it was manganate (VII) purple to colourless in acidic or brown ppt in alkaline)

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53
Q

Carbonyl compounds lithium aluminium hydride

A

In dry ether
Aldehyde -> primary alcohol
Ketone -> secondary

R=O + 2[H] -> R-OH

R1-CO-R2 + 2[H] -> R-CHOH-R2

Also works with hydrogen with a platinum catalyst

Does not reduce double bonds (hydrogen with platinum catalyst does)

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54
Q

Carbonyl compounds plus HCN

A

In the presence of HCN pH 8

R=O + HCN -> ROHCN

Excess potassium cyanide and some dilute sulfuric acid or hydrogen cyanide and some sodium hydroxide

pH allows enough of CN- and HCN

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55
Q

Carbonyl + HCN mechanism

A

Dipole on C=O lone pair on CN- from the KCN but as ion

Arrow from CN lone pair to carbon, arrow from double bond to oxygen

CRHCNO-

Lone pair on negative oxygen attacks hydrogen on HCN molecule, arrow from H-CN bond to the carbon

Forms CRHCNOH and CN-

If pH is too low not enough CN- if too high not enough HCN

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56
Q

Optical activity of product of HCN KCN and carbonyl compound

A

Racemic minxture, planar around C=O group, equal chance of attack from left right of planar centre

(Not the intermediate which is tetrahedral). The initial aldehyde is the planar one

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57
Q

Carbonyl plus (2,4-DNPH)

A

Test for carbonyl group

Brady’s reagent

Orange ppt

Filter off and recrystallise and do melting point determination to indentify compound

Does not do carboxylic acids

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58
Q

Iodine in the presence of alkali plus carbonyl compound

A

Does ethanal, methyl ketones, ethanol also(gets oxidised first) and secondary methyl alcohols, gently warm with iodine and sodium hydroxide

CH3COR + 3I2 + 4NaOH -> CHI3 + RCOO-Na+ + 3NaI + 3H2O

Stepwise
I2 + 2OH- -> IO- + I- + H2O
CH3COR + IO- CI3COR IO withdrawing weakens σ bond with carbon -> CHI3 + RCOO-

Pale yellow ppt of iodoform is produced

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59
Q

Boiling point of carboxylic acid

A

Higher as it can hydrogen bond

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60
Q

Solubility of carboxylic acids

A

Soluble ad can hydrogen bond

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61
Q

Preparation of carboxylic acids

A

•R-OH + 2[O] -> RCOOH + H2O
•R=O + [O] -> RCOOH
heat under reflux with acidified potassium dichromate primary alcohol

RCN + H+ + 2H2O -> RCOOH + NH4+
Reflux with dilute acid

Or hydrolysis of an Ester

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62
Q

Carboxylic acid plus lithium aluminium hydride

A

RCOOH + 4[H] -> ROH + H20

Dry ether

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63
Q

Carboxylic acid plus base

A

RCOOH + NaOH -> H2O + RCOO-Na+

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64
Q

Carboxylic acid plus PCl5

A

Solid, dry acid

RCOOH + PCl5 -> ROCl + POCl3 + HCl

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65
Q

carboxylic acid plus alcohol

A

Acid catalyst conc H2SO4

RCOOH + R’OH reversible RCOOR’

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66
Q

Acyl chloride plus water

A

RCOCl + H20 -> RCOOH + HCl

Vigorous

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67
Q

Acyl chloride plus alcohol

A

RCOCl + R’OH -> RCOOR’ + HCl

Not reversible and Cl is a better leaving group therefore better than wth carboxylic acid

Dry

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68
Q

Acyl chloride with ammonia

A

Conc ammonia, dry

RCOCl + NH3 -> HCL + RONH2
HCl + NH3 -> NH4Cl

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69
Q

Acyl chloride plus amine

A

RCOCl + R’NH2 -> RCONHR’ + HCl

Dry

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70
Q

Hydrolysis of esters

A

Acid

RCOOR’ + H20 ⇌ RCOOH + R’OH acid cat, low yield

Alkali

RCOOR’ + NaOH -> RCOO-Na+ + R’OH
Goes to completion

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71
Q

Polyester formation

A

Condensation - water is lost
Acyl chloride then HCl lost

Monomers join together with the elimination of a simple molecule such as water or hydrogen

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72
Q

Bonding in benzene

A

Kekule model and delocalised model

Overlap of p-orbitals to form π-bonds

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73
Q

Evidence for the delocalised model of the bonding in benzene

A

•Enthalpy of hydrogenation :
The predicted value is lower, more exothermic than the true value because benzene is stabilised because of the delocalisation of the π-electrons
Predicted -357 real -207 KJmol^-1
•c-c bond lengths, C=C is slightly shorter than C-C in aliphatic compounds, x-ray diffraction shows that the bond lengths between the C atoms are the same
•does not undergo the typical electrophilic addition reactions of unsaturated compounds

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74
Q

Why is benzene resistance to bromination

A

The π-bonds are delocalised in benzene and localised in π-bond of alkene
The compound is more stable

Substitution rather than addition

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75
Q

Benzene plus oxygen in air

A

Burns

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76
Q

Bromine plus benzene

A

Iron catalyst
2Fe + 3Br2 -> 2FeBr3
FeBr3 + Br2 -> FeBr4- + Br+

Bromine liquid
Dry conditions
Heat under reflux

Br substitutes onto ring for hydrogen

Benzene + Br2 -> bromobenzene + HBr

Or UV light - heat under reflux, addition occurs from free radicals

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77
Q

Benzene and nitric acid

A

Sulphuric acid conc Of H2SO4 and HNO3 catalyst warm to 50degrees in flask with reflux condenser

HNO3 + H2SO4 -> H2NO3^+ + HSO4-
H2NO3^ -> NO2+ + H2O

NO2 substitutes on ring
H lost replaces in HSO4-

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78
Q

Fridel-Crafts reaction

A

Alkylation
Dry
R-Cl + AlCl3 -> R+ + AlCl4-

R+ substitutes on ring, H takes a Cl forming HCl and reforming the AlCl3

Acylation
Dry
Acyl chloride, anhydrous aluminium chloride
R-OCl + AlCl3 -> R+=O + AlCl4-

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79
Q

Mechanism of electrophilic substitutions

A

Generate electrophile

1) Kekule
•arrow from a double bond to the electrophile
•intermediate has positive charge on the carbon next to the one the electrophile has attached too, arrow from the hydrogen carbon bond next to electrophile back to where double bond is reformed
•double bond reformed
•show catalyst regenerated by hydrogen

2) delocalised -> instead of coming from double bond, comes from Armstrong inner circle
•gets circle turns into horseshoe open opposite where the electrophile has attached, positive charge inside horseshoe,
•arrow from hydrogen bond carbon bond to positive charge in horse shoe
•show solvent getting reformed

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80
Q

Phenol and bromine water

A

Lone pair in OH is delocalised in the π-system electron density is increased. More susceptible to electrophilic substitution
Ortho and para directing

Bromine water water works, no need for FeBr3 ROOM TEMP
3 substitutes instead of one

(Same with nitration)

C6H6OHBr3 + 3HBr
White ppt

81
Q

Amines

82
Q

Amides

83
Q

Amino acids

A

H2N-CHR-COOH

84
Q

Amine plus water

A

R-NH2 + H2O -> RNH3^+ + OH-

85
Q

Amine plus acid

A

R-NH2 + HCl -> R-NH3^+Cl-

Strong acid

86
Q

Amines plus ethanoyl chloride

A

R-NH2 + CH3COCl -> R-NH-COCH3 + HCl

87
Q

Amines plus halogenoalkanes

A

Dropwise
R-NH2 + R’Cl -> R-NH2-R’ pos charge on nitrogen and Cl-

Excess

R-NHR’R’ + HCl

88
Q

Order of strength of base ammonia aromatic amines and aliphatic amines

A

Aliphatic - electron density pushes into nitro group - more available lone pair. More basic

Ammonia

Aromatic lone pair delocalised in the ring, less available lone pair , less basic

89
Q

Preparation of primary aliphatic amines

A

•halogenoalkane conc ammonia and aqueous ethanolic solution, long time in a sealed tube
R-Cl + 2NH3 -> R-NH2 + NH4Cl

•reduction of nitriles warm with LiAlH4 in dry ether, hydrolyse with dil acid
R-CN + 4[H] -> R-NH2

90
Q

Forming aromatic amides

A

Benzene plus nitric acid plus sulphuric acid forms benzene-no2

Then heat under reflux with conc HCl and Tin to reduce to benzene-NH2

Add excess NaOH
Steam distill
Then place in separating funnel and sodium chloride added to reduce solubility, the phenyl-amine later is run off and some solid anhydrous potassium carbonate is added to dry

91
Q

Preparation of amides

A

•ethanoyl chloride + ammonia

CH3COCl + NH3 -> CH3CONH2 + HCl

92
Q

Formation of polyamides

A

Condensation reaction

Acyl chloride and amine HCl lost

Amino acids - water lost

93
Q

Acidity and basisity of 2-amino acid

A

Zwitterions are formed (solid at room temp)

The ammonia group is protonated, the carboxylic acid is deprotonated

Therefore can act as an acid or base

(Also strongly attracted)

94
Q

Are 2-amino acids optically active or racemic mixture

A

Only one isomer is naturally occurring therefore it rotates the plane of plane polarised monochromatic light

95
Q

Proteins

A

•contain a peptide bond when amino acids combine, by condensation polymerisation. One hydrogen is lost on the NH2 group and the nitrogen bonds with a carbon from the carboxylic acid group
Water is lost the same no of waters as no of monomers

•can be hydrolysed to form the constituent amino acids, which can be separated by chromatography

96
Q

Gringard reagents

A

•halogenoalkane + magnesium in dry ether with trace iodine
R-I + Mg -> R-MgI

Then with carbonyl compounds

R-MgI + R’R’’CO -> R-CR’R”-OMgI (R is opposite O)
Dil acid O-H of MgI

Carbon dioxide

R-MgI + CO2 -> RCOOMgI
Dilute acid - RCOOH

All in dry ether until it is hydrolysed with dilute acid (and water -hydrolysed)

97
Q

Purification by washing

A
  • wash with sodium hydrogencarbonate in separating funnel, removes acidic impurities, release pressure CO2. Stop when no further gas produced
  • discard aq layer. Wash organic layer with water to remove unreacted sodium salts and soluable organic substances e.g ethanol
  • discard aqueous layer. Dry with anhydrous salt eg calcium chloride complete when goes clear from cloudy
  • distill and collect in range of 2 degrees either side of boiling point
98
Q

Recrystallisation

A
  • filter
  • dissolve ppt in minimum about of suitable solvent when solubility is high when hot and low when cold
  • filter hot solution again through fluted filter paper using warmed stemless funnel into conical flask-> removes insoluable impurities
  • cool solution
  • filter solvent + pure solid under reduced pressure in Buchner funnel
  • collect solid on filter paper, discard liquid- removes soluable impurities
  • wash solid, ice cold solvent and leave to drypp
99
Q

Melting point determination

A
  • Insert solid into capillary tube and attach tube, open end up to thermometer with a rubber band
  • place thermometer into bath of liquid with higher bp than mp of solid
  • heat liquid bath, stir and note temp where solid melts
  • the smaller the range the purer the solid
100
Q

Steam distillation

A
  • used to extract a volatile liquid that is immiscible with water from a Complex mixture, partially for a substance that would decompose at its boiling point or just for convenience of it boiling bellow it’s boiling point
  • heat water to form steam which enters a round bottomed flask with the substance and water. Then distill with a condenser
  • have a safety vent
  • electric heater
101
Q

Heights of peaks in proton NMR

A

Relative heights relate to the relative numbers of hydrogens in each environment

102
Q

Splitting pattern proton NMR

A

If the proton of a hydrogen atom has n hydrogen atoms on neighbouring carbon atom, its peak will be split into (n+1) subpeaks

103
Q

What is chromatography

A

Separates components of a mixture between a mobile phase and a stationary phase

The sample dissolved in a solvent and washed through a stationary phase by a mobile phase called an eluent
The competition between the sample molecules adsorbed by the stationary phase and dissolved by the eluent results in separation depending on how soluable they are in eluent and less adsorbed by stationary phase as these move faster through apparatus

104
Q

Calculating Rf values

A

For a one way chromograph

Rf=distance moved by substance/ distance moved by eluent

105
Q

Types of column chromatography

A

TLC - thin layer chromatography

HPLC - high performance liquid chromatographs

GC - gas chromatography

Different substances have different retention times due to their strengths attractions with the mobile and stationary phase

These can be used in conjunction with mass spectrometry (in applications such as forensics and sport)

106
Q

Affect of concentration on rate of reaction

A

Increase in concentration causes an increase in reaction rate. Collisions occur more frequently as more in the same volume
Successful collisions also occur more frequently

107
Q

Affect of temperature on rate of reaction

A

Molecules have higher KE large fraction possess activation energy greater the activation energy, larger proportion of collisions result in reaction. Also a increase in collision frequency

108
Q

Affect of pressure on rate of reaction

A

Increases no of moles per cm^3 -> frequency of collisions increased KE sameX frequency of successful collisions increases. Therefore rate increases

However if the reaction is gaseous catalysed by a solid catalyst, the rate is determined by the number of active sites on the catalyst surface, so increasing pressure does not alter the rate

109
Q

Affect of surface area on the rate of reaction

A

Only surface area can react, larger SA increased frequency of collisions, increased frequency of collisions, increased frequency of successful collision

110
Q

Activation energy

A

Minimum KE the colliding molecules must posses for the collisions to be successful and result in the formation of product molecules

111
Q

How to calculate rate of reaction

A

DProduct/Dt

Or -Dreactant/

Gradient of graph, either initial rate or at time t

112
Q

In terms of maxwell Boltzmann

A

Ecat is lower than initial Ea, larger fraction of molecules above ecat line

T hot line has peak shifted to the right but lower peak

113
Q

Reaction profiles for catalysed and uncatalysed reactions

A

Reactants products AH and Ea

For catalysed add a middle transition state

114
Q

Rate of reaction

A

Rate of change of concentration of product or reactant with time

115
Q

Half life chem

A

Time taken for reactants to half

116
Q

Rate determining step

A

Slowest step in a multi step mechanism

117
Q

Obtaining rate equation by titration

A
  • Measure out samples of reactants with known conc,
  • Mix and start clock
  • stir throughly
  • at regular time intervals, withdraw samples using a pipettes and quench stop reaction either with ice or cold water or solution to react with one of the reactants. Note time where half the contents of the pipette have been added to the quenching solution
  • titrate solutions to find conc of reactant
118
Q

Obtaining rate equation by colorimetry

A

Spectrophotometer can be used to measure conc of coloured species
amount of light of frequency absorbed is measured at set time intervals this depends on conc

119
Q

Measuring rate equation by mass change

A

Average rate = mass lost/time

Do the reaction in a conical flaks tarred when t=O, record mass loss over time in gradual intervals

(Some liquid may be lost as spray, plug it with cotton wool, mass change is v small)

120
Q

Measuring rate equation by volume of gas involved

A

The moles of gas measured of product when less than ten percent of acid used up
Rate inversely proportional to time for certain volume of gas to be produced

121
Q

Initial rates method

Give example

A

Initial rate = gradient of a conc time graph at t=0

Conc recatants falls by 10% or less to make initial rates valid, measure change in conc over time

Example: 
Iodine clock reaction 
H2O2 + 2I- -> I2 + 2H2O 
With starch and thiosulphate
The thiosulphate reacts with the produced iodine. Thererefore the blue black colour is perceived only after the set amount of thiosulphate has been used up
Blue black end point

Vary concentrations but keep total volume constant

Sulphur clock
S2O3^2- + 2H+ -> S + SO2 + H20
With nitric acid

Time taken for a large X on a white tile to become Invisible

122
Q

Conc time graph is flat

A

No reaction

123
Q

Conc time graph is linear

A

Zero Order

124
Q

Conc time graph has a constant half life

A

First order

125
Q

Conc time graph where half lives increase in a regular geometric manner

A

Second order

126
Q

Rate conc graph horizontal

A

Zero order

127
Q

Rate conc graph straight

A

First order

128
Q

Rate conc graph where rate - reactant squares is straight

A

Second order

129
Q

How to determine rate determining step from a rate equation or vice-versa

A

Anything involved in the rate equation is involved in or before the rate determining step

130
Q

How to gather data on the iodonation of propanone

A

Colorimeter
Acid catalyst
- dark brown I2 -> colourless

Vary concentrations and keep total volume constant

Rate = k [propanone] [H+]

131
Q

How to determine mechanism for the iodonation of propanone

A

Iodine not involved in or before the rate determining step

•Lone pair on oxygen attacks H+ pos charge moves onto carbon (CH3)2C+OH), arrow from one of the hydrogens on CH3 to form double bond - this is slow step as C-H bond is broken
CH2=C(CH3)OH
•arrow from OH bond to carbon, then from double bond to I , from the I-I bond to the other I,

Forms H3C-CO-CH2I + I-

132
Q

Rate equations for SN1 and SN2

A

SN1 R=k[halogenoalkane] not involved in rate determining step - tertiary

SN2 R=k[halogenoalkene][nucleophile] - primary

Evidence for the mechanisms involved

133
Q

How can a reaction be zero order with respect to all gaseous reactants

A

If it is catalysed by something in the solid state, the limiting factor is the availability of active sites. Unless the pressure is extremely low

Adsorption fast
Slow step is adsorbed reactants to adsorbed products
Adsorbed products to gaseous products is fast

134
Q

How can a regent which appears in the slow step not appear in the rate equation for

A

If the conc of the reagent is at least ten times that of the other reagent, the change in conc during reaction will be negligible. This means that within experimental error it appears to be constant and rate equation. Is
r = k * what appears to be constant [B]^q * [A]

135
Q

Structural isomer

A

Compounds with the same molecular formula but different structural formula

136
Q

Stereoisomers

A

Compounds with the same structural formula but which have the atoms arranged differently in space

Split into geometric and optical isomerism

137
Q

Geometric isomerism

A

Is a type of stereoisomerism caused by the presence of a functional group that restricts rotation

Eg double bonds or cyclic compounds

138
Q

Optical isomer

A

Do not have a plane of symmetry. They are chiral. They are two isomers which are non superimposable mirror images, two enantiomers.

139
Q

Factors which effect the angle through which the plane of plane polarised light is rotated through

A
  • The nature of the enantiomer
  • The concentration of the enantiomer in the solution
  • The length of reaction tube
140
Q

How to distinguish between primary and secondary and tertiary alcohols

A
  • primary and secondary alcohols turn orange acidified potassium dichromate (VI) solution to a green Cr3+ solution. Tertiary are not oxidised, remains orange
  • by distilling off the product as the former, and then doing a test with tollens, only aldehyde would reduce silver ion complex to give silver mirror, this would come from the secondary alcohol
141
Q

Test for carboxylic acid

A

Addition of sodium hydrogencarbonate or sodium carbonate.

Has evolved

RCOOH + NaHCO3 -> RCOONa + H20 + CO2

Gas turns limewater cloudy

CO2 + Ca(OH)2 -> CaCO3 + H20

142
Q

Fingerprint region

A

Region below 1500cm^-1, it shows a series of peaks that dependant on the exact compound being analysed.

143
Q

How can an infrared spectrum test purity

A

The purer the compound the closer it matches database.

Any stray peaks will be due to impurities

144
Q

Why is there a need for standard in NMR

A

The extend of the splitting depends on the strength of the magnetic field. Therefore there must be some comparison.
TMS is used - dissolved in solvent which does not contain any protons

145
Q

Chemical shift in NMR

A

Caused by the extent to which the electrons in orbit around the hydrogen nucleus are pulled away from the hydrogen (deshielding) the chemical shift is less for Ch3 than CH2 because the carbon is pulling electrons away from theee groups

146
Q

Why does splitting occur

A

The magnetic environment of a proton in one group is affected by the magnetic field of protons on the neighbouring carbon atoms
•if a neighbouring group have opposite spin from another compound it results in two different fields affecting the CH2 so their peaks are split in two

147
Q

Why is there no splitting in carbon 13 NMR

What do peak heights represent

A

The change of two C-13 atoms being next to each other in the carbon chain is about 1 in 1000

This is also why peak heights are not prioritising to the number of carbon in each environment

148
Q

High performance liquid chromatography

A
  • column packed with a solid of inform particle size- stationary phase
  • sample is dissolved in suitable solvent and added to top of column, liquid eluent is forced through column under high pressure
  • it has a high resolution because the high pressure increases the speed at which the eluent passes through the column and so reduces the extent to which the band of a component spreads out due to diffusion
  • non polar stationary phase, polar eluent
  • connect column to infrared spectrometry
149
Q

Retention time

A

The time taken for a component in the sample mixture to pass through the column

150
Q

Gas chromatography

A
  • inject a sample into a chromatography column in a thermostatically controlled oven
  • sample evaporated and is forced through column by flow of inert, gaseous mobile phase, called carrier gas (hydrogen, argon, oxygen, nitrogen or air). Column contains a liquid stationary phase that is adsorbed onto surface of inert solid
  • useful for separating mixtures of gases or volatile liquids.
  • remove the samples, measure mass and use NMR/IR compare with database

Useful for forensic work and detention of drugs in urine sample

151
Q

Which have higher boiling points

Alkene or alkane or halogenoalkane

A

Alkanes have higher bp because the rigidity of the double bond does not allow the molecules to pack together as efficiently

Halogenoalkanes are the highest because they contain more electrons so have more London forces

152
Q

Solubility of halogenoalkanes

A

Insoluble because they cannot hydrogen bond

153
Q

Distillation

A

Large enough difference in the boiling temps of organic substances then distillation can be used to separate them
•round bottomed or pear-shaped flask
•a still-headed fitted with a thermometer the bulb must be positioned level with outlet of still head
•condenser with water going in at bottom and leaving at the top
•open receiving vessel
•electric heater

Collect 2 degrees either side

154
Q

Nitriles in synthesis

A

Hydrolysed to salt of carboxylic acid heat under reflux with NaOH
RCN + H20 + NaOH -> RCOO-Na+ + NH3
(To then get carboxylic acid use a strong acid after)

Acidic hydrolysis dilute HCl
RCN + 2H2O + HCl -> RCOOH + NH4Cl

under reflux with lithium aluminium hydride LiAlH4 in dry ether and then hydrolyse with H2SO4 aq to form amine

155
Q

Acyl chlorides plus LiAlH4

A

rR-OCl + 4[H] -> rRHOH + HCl

dry ether

156
Q

Are esters soluable

A

No as they cannot hydrogen bond

157
Q

Presence of a alkyl group on a benzene

A

Ortho
Para
Directing

158
Q

Melting temp of phenol

A

Can hydrogen bond so melts higher than benzene which only has london forces

Also soluable unlike Benzene as can hydrogen bond

159
Q

Phenol plus acyl chlorides

A

Not Friedel crafts

Oxygen atom acts as nucleophile and attacks carbon atom forming an Easter

160
Q

Boiling temps of amines

A

Hydrogen bonding therefore higher than comparison alkane (also soluable)

161
Q

TLC

A

•thin layer chromatography
Stationary phase is either silica gel aluminium oxide immobilised on a flat inert sheet made from class or plastic
•unknown amino acid mixture is dissolved in suitable solvent and a spot of dissolved known amino acid is placed separately on the same plate at the same level
•the plate is then dipped in suitable eluent with spots above the level of liquid eluent
•place in sealed container
•capillary action draws up eluent
•plate left until eluent has risen up to the top of the plate
•remove plate and spray with ninhydrin and heat
•ninhydrin reacts with amino acids producing a blue purple colour

Compare with the heights reacted by known amino acids

162
Q

Which is more acidic

Phenol
Alchohol
Carboxylic acid

A

Carboxylic acid is strongest -> two oxygen groups

Phenol is second as the lone pair can be delocalised into pi system -> stabilises conjugate base

Alcohol weakest

163
Q

Drawing enantiomers

A

Use two lines, one solid triangle and one dashed line

The entanioner has the same type of line drawn to each group but the groups are reversed like a mirror image

164
Q

Why would you wash with ice cold water

A

To remove soluble impurities but to ensure as little solid as possible dissolves

165
Q

Benefit of Grignard

A

Source of nucleophile can carbon

166
Q

How do anti bumping granules prevent bumping

A

They provide a surface for bubbles to form

They distribute the heat more evenly

167
Q

Cracking

A

Collection over water/ gas syringe
Ceramic fibre/ wool soaked in substance to be cracked
Aluminium oxide
Heat under catalyst

168
Q

Why should hydrogen not be above lithium

A

The rest of group one are metals
Hydrogen does not react in the same way, has different chemical properties
Forms H- ion

169
Q

Which halogens react with alkenes

A

Chlorine and bromine

Iodine only if the C=C bond is activated by an oxygen atom

170
Q

Rates of addition of hydrogen halides to alkenes

A

HI > HBr > HCl because the HCl bond the weakest

Gaseous

171
Q

Polymerisation of an alkene

A

Heat under 1000 atm pressure in the presence of trace of oxygen
Causes radicals to be formed that initiate the polymerisation reaction

Or mix with a solution containing alkylaluminium and titanium chloride

172
Q

What is the difference between hazard and risk

A

Hazard is the potential to do harm

Risk is the probability of harm occurring

173
Q

State of halogenoalkanes

A

Chloro bromo methane and chloroethane are gases at RTP

Iodomethane and higher members are liquids

174
Q

Alcohol and HCl

A

Only works for tertiary
Conc HCL heat under reflux
Chloroalkane plus H20

175
Q

Obtaining rate by production of a solid

A

The time taken to produce enough solid to hide a cross on a piece of paper

176
Q

Rate of reaction using change in pH or conductivity

A

Measure pH at time intervals
Remember pH is logorithmic

Or when no of ions change measures electrical conductivity

177
Q

How to make an aldehyde

A

Heat at 60 degrees using an electric heater
Add potassium dichromate and did sulphuric acid
Collect ethanol in flask surrounded by iced water

Distill off directly to prevent further oxidation

Electric heater as organic stuff is flammable

178
Q

Preparation of ketones

A

Potassium dichromate sulphuric acid
Round bottomed flask
Reflux condenser
Heat

Aromatic can be done using Fredel-crafts

179
Q

Reduction of alkene group

180
Q

Steps of iodoform

A

Sodium hydroxide reacts withs with iodine to form IO- and I- and H20
The IO- ions substitute into the CH3 group next to the C=O group forming CI3COR the electron withdrawing effect of the three halogen atoms and the oxygen atom weakens the σ bond which then breaks, and then OH- adds forming iodoform

181
Q

How to prepare a carboxylic acid

A
  • oxidation of primary alcohol
  • oxidation of aldehyde
  • hydrolysis of an Easter
  • hydrolysis of a nitrile
  • iodoform then add excess strong acid
182
Q

Test for cabrolyxic acid

A

Carboxylic acid + sodium carbonate -> salt of acid plus co2 plus h20

Turns limewater cloudy

183
Q

Preparation of acyl chlorides

A

Carboxylic acid plus PCl5 -> POCl3 plus acyl chloride plus HCl

184
Q

Advantages of forming esters using acyl chloride

A

HCl gas is lost hence irreversible
Lower activation energy
Cl is a better leaving group

185
Q

Benzene plus hydrogen

A

Nickel catalyst and heat

186
Q

Why plus temp of nitration be 50

A

Bellow 50 too slow

Above 60 Second NO2 group substitutes

187
Q

Melting temp of phenol compared to bromine

A

Can hydrogen bond hence higher melting temp

Also soluble

188
Q

Nitration of phenol

A

Reacts with dilute nitric acid

1,4

189
Q

Relative acidity of alcohols phenols and carboxylic acid

A

Alcohol not very acidic - no litmus no sodium hydroxide no sodium carbonate

Phenol - litmus goes red, salt formed with NaOH, no reaction with sodium carbonate

Carboxylic acid does all three

190
Q

Hydrolysis of amides

A

Heat under reflux with aq acid or alkali

Acid
Amide + H+ + H20 -> CH3COOH + NH4+

Alkali
Amide + OH- ->salt of carboxylic acid plus NH3

191
Q

Chromatography of amino acids

A

TLC
Stationary phase is either silica gel or aluminium oxide immobilised on a flat inert sheet that is usually made from glass or plastic
•acid or mixture dissolved in a suitable solvent and a spot of the test solution is placed 2cm from bottom
•known amino acid placed at the same level
•plate dipped in mobile phase with the spots above the level of the liquid eluent
•plate placed in sealed container
•the eluent is drawn up by capillary action
•plate left until eluent has risen to top
•plate removed and sprayed with a solution of ninhydrin and then heated
•ninhydrin reacts with amino acids producing a blue purple colour

192
Q

Describe the bonding in benzene

A

Head on overlap between orbitals from neighbouring carbon atoms to form a sigma bond

The remaining orbitals overlap sideways and so electrons delocalise around the ring

193
Q

How can chemists contribute to a more sustainable use of polymers

A
  • Reprocessing polymers into simpler compounds for use in feedstock and chemical industry
  • Capture and use of energy from incineration
  • sorting and recycling of polymers
  • removal of toxic products formed during incineration
194
Q

Why does stereoisomerism occur in alkenes

A

Restricted rotation around the C=C bond because the π bond prevents it

195
Q

Describe the bonding in benzene

A

Head on overlap between orbitals from neighbouring carbon atoms to form a sigma bond

The remaining orbitals overlap sideways and so electrons delocalise around the ring

196
Q

How can chemists contribute to a more sustainable use of polymers

A
  • Reprocessing polymers into simpler compounds for use in feedstock and chemical industry
  • Capture and use of energy from incineration
  • sorting and recycling of polymers
  • removal of toxic products formed during incineration
197
Q

Why does stereoisomerism occur in alkenes

A

Restricted rotation around the C=C bond because the π bond prevents it

198
Q

Equation for hydrolysis of nitrile

A

Nitrile + 2H20 + H+ -> Carboxylic acid plus NH4+

Strong acid heat under reflux

199
Q

Addition across cyclohexene

A

Transisomer is formed

Providing that the second chlorine for example adds on opposite side to the first chlorine

Chemistry (5 decks)

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