Reactions and Conditions Flashcards

1
Q

Half-Life Equation

A

t1/2=Ln2/k

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2
Q

Iodine Peroxodisulfate reaction and how the catalyst works

A

Overall
S2O82- (aq) + 2I- (aq) = 2SO42- (aq) + I2(aq)

2Fe3+(aq) + 2I-(aq) = I2(aq) + 2Fe2+(aq)
2Fe2+(aq) + S2O82-(aq) = 2Fe3+(aq) + 2SO42-(aq)

The Fe3+ can react with either I- or S2O82- first

Fe3+ ion catalyst

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3
Q

Overall Order

A

Sum of individual orders

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4
Q

Processes in Haber Process

A
Diffusion to surface
Adsorption to surface and weakens covalent bonds and enables the reaction to take place
Reaction
Desorption from surface
Diffusion away
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5
Q

Nitration of Benzene

A

Reflux cHNO3 cH2SO4

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6
Q

Alkylation and acylation Benzene + chloroalkane/ acyl chloride

A

AlCl3 anhydrous

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7
Q

Oxidation of side chain in benzene

A

Reflux Alkaline KMnO4 then H2SO4

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8
Q

Electrophilic substitution with Phenol

A

Phenol + Br2 = Tribromophenol + 3HBr. White ppt

Occurs with phenylamine

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9
Q

Nitration of Phenol

A

dilute HNO3

concentrated HNO3 produced trinitrohenol

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10
Q

Formation of Amines with NH3

A

Haloalkane + Hot ethanolic NH3 reflux

Hot excess ammonia to avoid secondary and tertiary amines

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11
Q

Phenylamine

A

Nitrobenzene reduced to phenylamine cHCl reflux Sn catalyst

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12
Q

Diazotization

A

Phenylamine + Nitrous acid + HCl = diazonium chloride + 2H2O

HNO2 Nitrous acid
Less than 10 degrees C

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13
Q

Coupling diazonium ion with phenol

A

An alkaline solution of Phenol

Alkaline can react with OH on phenol to form O- Na+ section

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14
Q

Formation of Amides

A

Acyl chlorides +NH3/amines = Amide + HCl
Concentrated NH3 with acyl chlorides for primary amides

Amines can be used to produce a secondary amide

YOU CANNOT SIMPLY USE A CARBOXYLIC ACID AND AN AMINE. ONLY POLYMERS AND PROTEINS CAN DO THAT. CARBOXYLIC ACIDS ARE NOT REACTIVE ENOUGH

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15
Q

Hydrolysis of Amides

A

Heat under reflux HCl/NaOH

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16
Q

Hydrolysis of peptide bonds

A

With acid 6 moldm-3 HCl, 24 hours, 110 celsius

With alkaline NaOH and reflux

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17
Q

SimplifiedNernst equation

A

E= Eo + 0.059/z log [oxd]/[red]

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18
Q

Full Nernst Equation

A

E = Eo + RT/zF log Ox^n/Red^n

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19
Q

Standard Electrode Potential

A

The voltage of a half cell measured under standard conditions with a standard hydrogen electrode as the other half-cell

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20
Q

Enthalpy change of hydration

A

The enthalpy change when 1 mole of a specified gaseous ion dissolves in sufficient water to form a very dilute solution. STP

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21
Q

Enthalpy change of solution

A

The energy absorbed or released when 1 mole of an ionic solid dissolves in sufficient water to form a very dilute solution. STP

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22
Q

Lattice Energy

A

Lattice energy is the enthalpy change when 1 mole of an ionic compound is formed from its gaseous ions under standard conditions.

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23
Q

Differences between the predicted structure of benzene

A

Does not undergo addition reactions like ethene
All the bond lengths are the same
The enthalpy of hydrogenation is lower than the expected

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24
Q

Reactivity of Phenol in regard to dilute vs conc HNO3 and Br2

A

Reacts readily with bromine water. Reacts with dilute nitric acid to form 2-nitrophenol or 4-nitrophenol. With concentrated HNO3 trinitrophenol is formed.

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25
Q

Why is phenol more reactive than benzene

A

The lone pairs on the oxygen atom overlap with the pi system and delocalise in the benzene ring. Electron density increases and thus more susceptible to attack from electrophiles.

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26
Q

Species that can be reduced multiple times

A

Iron, Vanadium

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27
Q

Volume and moles

A

n=V/24

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28
Q

Purpose of Salt Bridge

A

To complete the electrical circuit and so that ionic balance is maintained

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29
Q

Full Nernst Equation

A

E=Eθ + RT/zF ln[oxidised]/[reduced]

For when the temperature is not 25 degrees

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30
Q

Simplified Nernst Equation

A

E=Eθ + 0.059/z log10[oxidised]

For metal/metal ion electrodes

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31
Q

Copper Ions with hydroxide ions colour and what type of reaction is it?

What can the hydroxide be replaced for?

A

[Cu(H20)6]2+ + 2OH- = Cu(OH2)2(H2O)4 (s) + 2H2O
Precipitation reaction
No ligand is formed
Pale blue precipitate

The exact same reaction can occur by using ammonia. When excess ammonia has been added it forms the deep blue solution.

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32
Q

Cu(OH2)2(H2O)4 precipitate with concentrated ammonia solution

A

Cu(OH2)2(H2O)4 + 4NH3 = [Cu(H2O)2(NH3)4]2+ + 2H2O + 2OH-

Pale blue precipitate disappears and a deep blue solution is formed

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33
Q

Copper ions with concentrated hydrochloric acid

A

[Cu(H20)6]2+ + 4Cl- = [CuCl4]2- + 6H2O

Yellow solution

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34
Q

Cobalt ions color

A

Pink

35
Q

Cobalt ions with hydroxide ions and what happens when excess is added

A

[Co(H20)6]2+ + 2OH- = Co(H2O)4(OH)2 (s) + 2H2O
Blue precipitate
If there is excess alkali and it is heated it turns red

36
Q

Reactions of amino acids with acids and bases

A

+H3NCH2COO- + HCl = +H3NCH2COOH + Cl-
+H3NCH2COO- + NaOH = H2NCH2COO-Na+ + H2O

The acid and base can also react with the R group if in excess

37
Q

Formation of amines through nitriles

A

Haloalkane + cyanide ion = nitrile
KCN in ethanol, reflux
Reduction of nitrile with H2 and a nickel catalyst or LiAlH4 in dry ether

38
Q

Formation of amines with amides

A

Amide reduction with LiAlH4 in dry ether

Ethanamide + 4[H] = ethanamine + H2O

39
Q

Fragmentation reaction with propanone

A
CH3COCH3 = CH3+ + CH3CO. + e-
CH3COCH3 = CH3. + CH3CO+ + e-

Fragmentation causes the loss of an electron as a result of electron bombardment.

C-O C-N C-H bonds are usually the bond that breaks

40
Q

With reference to the orientation of d orbitals in an octahedral transition metal complex explain why such complexes are usually coloured

A

dx2-y2 orbitals and dz2 orbitals are on the x y z axes of the atom and the other 3 orbitals are in between the axes. When ligands are in line with the orbitals it causes the energy levels to split. The movement of electrons between the different energy levels results in the absorption of some light. The colour produces is complementary to the frequency of light absorbed by the atom.

41
Q

How does cisplatin work

A

It is an anti-cancer drug that binds to sections of DNA in cancer cells preventing cell division.

42
Q

How does splitting of energy levels occur in octahedral complexes

A

There are 2 orbitals of dx2-y2 and dz2 and 3 orbitals at the lower energy level

43
Q

How does splitting of energy levels occur in tetrahedral complexes

A

5 degenerate orbitals. 3 above 2 below

44
Q

Meaning of degenerate

A

Degenerate means that all the orbitals are all at the same energy level

45
Q

Meaning of non-degenerate

A

The orbitals are at slightly different energy levels

46
Q

The shape of [Ni(CN)4]2-

A

Square planar

47
Q

The shape of [Co(Cl)4]2-

A

Tetrahedral

48
Q

Enthalpy of solution

A

Enthalpy of hydration-Lattice Enthalpy

49
Q

Cu2+ ions with excess ammonia

A

Deep Blue solution

[Cu(H2O)2(NH3)4]2+

50
Q

Hydrolysis of an ester using dilute acid

A

Catalysed by dilute HCl reflux. Non-reversible

51
Q

Hydrolysis of an ester using alkali

A

Using NaOH solution reflux sodium ethoxide salt is produced. Non-reversible.

52
Q

Oxidation of methyl ketones with iodoform

A

Methyl ketones and ethanol and ethanal react with NaOH and I2. The methyl group gets chopped off and the rest is oxidised normally.

53
Q

What mechanism occurs with acyl chlorides and a nucleophile attached to a hydrogen

A

Addition elimination reaction

54
Q

Dehydration of alcohol

A

cH3PO4/Al2O3

55
Q

Catalyst for Haber process

A

Iron

56
Q

Test for methandioic acid and ethandioic acid

A

Fehling’s reagent, Tollen’s reagent or acidified potassium dichromate or manganate

57
Q

Transition metal complex

A

A molecule or ion formed by a central metal ion surrounded by one or more ligands

58
Q

Why are acyl chlorides so reactive

A

The carbonyl carbon has its electrons drawn away from it by the highly electronegative chlorine and oxygen atoms giving it a large particle positive charge and more open to nucleophiles.

59
Q

What is the difference between Friedel-Crafts acylation and alkylation

A

Acylation is when an acyl chloride is added to the benzene ring and alkylation is when an alkyl group is added

60
Q

Buffer Solution Equation

A

pH=pKa + log[conjugate base/[acid]

61
Q

Relationship between Ka and pKa

A

Ka=10^-pKa

62
Q

Explain the strength of carboxylic acids with alcohols

A

The carbonyl group weakens the O-H bond. The electrons in C-O are drawn to C=O. This spreads out the negative charge (reducing its charge density) on the carboxylate ion making it less likely to bond with an H+

63
Q

Why is chloroethanoic acid more acid than ethanoic acid

A

Because the electron-withdrawing chlorine group weakens the O-H bond. The chlorine extends the delocalisation of the negative charge making it less likely to bond with an H+.

64
Q

Why is methanoic acid more acidic than ethanoic acid

A

Because the alkyl group has an electron-donating inductive effect so it increases the electron density in the COO- group making it more attractive to H+ ions

65
Q

Buffer solution in the blood and how do changes in pH affect it

A

CO2 + H2O = HCO3- + H+

When H+ changes the equilibrium move to the left or right accordingly

66
Q

Function of HCO3- ion in the blood

A

H+ + HCO3- = H2CO3

OH- + HCO3- = H2O + CO3 2-

67
Q

Conditions are nitration of phenol

A

Dilute HNO3 room temperature

Nitration of benzene needs cHNO3 and cH2SO4 and reflux

68
Q

Concentrated nitric acid and phenol

A

Trinitrophenol

If it is just diluted nitric acid at room temperature it forms 2-nitrophenol or 4-nitrophenol

69
Q

How does the solubility of carbonates vary down group 2

A

The solubility decreases down the group like with sulfates

70
Q

Explain how the solubility of sulfates varies down group 2

A

The solubility decreases. The ΔHhyd and ΔHlatt both become less exothermic because the ionic radius of the metal ions gets bigger so the charge density decrease. The ΔHhyd becomes less exothermic quicker than the ΔHlatt becomes less exothermic. So overall the ΔHsoln becomes more endothermic.

71
Q

Compare the ease of hydrolysis of acyl chlorides with chloroalkanes and aryl chlorides

A

Carbon bonded to the chlorine atom in chloroalkanes have a weaker positive partial charge than acyl chloride. Acyl chlorides have oxygen and chlorine bonded to the carbon. Therefore attack from a nucleophile is much more rapid. Aryl chlorides do not undergo hydrolysis. Lone pairs in p orbitals are delocalised into the benzene ring. This makes the C-Cl bond much stronger.

72
Q

Explain how the solubility of hydroxides varies down group 2

A

The solubility increase. The ΔHhyd and ΔHlatt both become less exothermic because the ionic radius of the metal ions gets bigger so the charge density decrease. The ΔHlatt becomes less exothermic quicker than the ΔHhyd becomes less exothermic. So overall the ΔHsoln becomes more exothermic.

73
Q

Bidentate definition

A

An ion or molecule that forms two coordinate bonds with a central metal ion and donates two lone pairs to the central metal ion

74
Q

Colour of CuCl42-

A

Yellow

75
Q

Colour of CoCl42-

A

Blue

76
Q

Reagents and conditions for nucleophilic addition of CN to aldehyde or ketones

A

NaCN with HCN catalyst

77
Q

What is a buffer solution

A

A solution that minimises changes in pH when small amounts of acid or base are added. They consist a weak acid and its conjugate base (or a weak base and its conjugate base)

78
Q

Ethanol to NaOH and I2

A

Ethanal is formed

79
Q

What is meant by stability constant

A

The equilibrium constant for the formation of a complex ion in solution

80
Q

Colour changed with MnO42- and Cr2O72-

A

Purple/pale pink to colourless

Orange to green

81
Q

Tollen reagent

A

Ammoniacal silver nitrate solution
Silver mirror formed
Oxidises aldehydes

82
Q

Fehlings solution

A

Alkaline solution of Cu2+ ions
Blue to red/orange
Oxidises aldehydes

83
Q

Addition of water to ethene

A

Conc H3PO4 300 Celcius forms alcohol

84
Q

What type of reaction is the coupling reaction between diazonium ions and phenol

A

Electrophilic substitution