Reactions and Conditions Flashcards

(84 cards)

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
Why is phenol more reactive than benzene
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.
26
Species that can be reduced multiple times
Iron, Vanadium
27
Volume and moles
n=V/24
28
Purpose of Salt Bridge
To complete the electrical circuit and so that ionic balance is maintained
29
Full Nernst Equation
E=Eθ + RT/zF ln[oxidised]/[reduced] | For when the temperature is not 25 degrees
30
Simplified Nernst Equation
E=Eθ + 0.059/z log10[oxidised] | For metal/metal ion electrodes
31
Copper Ions with hydroxide ions colour and what type of reaction is it? What can the hydroxide be replaced for?
[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.
32
Cu(OH2)2(H2O)4 precipitate with concentrated ammonia solution
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
33
Copper ions with concentrated hydrochloric acid
[Cu(H20)6]2+ + 4Cl- = [CuCl4]2- + 6H2O | Yellow solution
34
Cobalt ions color
Pink
35
Cobalt ions with hydroxide ions and what happens when excess is added
[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
Reactions of amino acids with acids and bases
+H3NCH2COO- + HCl = +H3NCH2COOH + Cl- +H3NCH2COO- + NaOH = H2NCH2COO-Na+ + H2O The acid and base can also react with the R group if in excess
37
Formation of amines through nitriles
Haloalkane + cyanide ion = nitrile KCN in ethanol, reflux Reduction of nitrile with H2 and a nickel catalyst or LiAlH4 in dry ether
38
Formation of amines with amides
Amide reduction with LiAlH4 in dry ether Ethanamide + 4[H] = ethanamine + H2O
39
Fragmentation reaction with propanone
``` 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
With reference to the orientation of d orbitals in an octahedral transition metal complex explain why such complexes are usually coloured
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
How does cisplatin work
It is an anti-cancer drug that binds to sections of DNA in cancer cells preventing cell division.
42
How does splitting of energy levels occur in octahedral complexes
There are 2 orbitals of dx2-y2 and dz2 and 3 orbitals at the lower energy level
43
How does splitting of energy levels occur in tetrahedral complexes
5 degenerate orbitals. 3 above 2 below
44
Meaning of degenerate
Degenerate means that all the orbitals are all at the same energy level
45
Meaning of non-degenerate
The orbitals are at slightly different energy levels
46
The shape of [Ni(CN)4]2-
Square planar
47
The shape of [Co(Cl)4]2-
Tetrahedral
48
Enthalpy of solution
Enthalpy of hydration-Lattice Enthalpy
49
Cu2+ ions with excess ammonia
Deep Blue solution | [Cu(H2O)2(NH3)4]2+
50
Hydrolysis of an ester using dilute acid
Catalysed by dilute HCl reflux. Non-reversible
51
Hydrolysis of an ester using alkali
Using NaOH solution reflux sodium ethoxide salt is produced. Non-reversible.
52
Oxidation of methyl ketones with iodoform
Methyl ketones and ethanol and ethanal react with NaOH and I2. The methyl group gets chopped off and the rest is oxidised normally.
53
What mechanism occurs with acyl chlorides and a nucleophile attached to a hydrogen
Addition elimination reaction
54
Dehydration of alcohol
cH3PO4/Al2O3
55
Catalyst for Haber process
Iron
56
Test for methandioic acid and ethandioic acid
Fehling's reagent, Tollen's reagent or acidified potassium dichromate or manganate
57
Transition metal complex
A molecule or ion formed by a central metal ion surrounded by one or more ligands
58
Why are acyl chlorides so reactive
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
What is the difference between Friedel-Crafts acylation and alkylation
Acylation is when an acyl chloride is added to the benzene ring and alkylation is when an alkyl group is added
60
Buffer Solution Equation
pH=pKa + log[conjugate base/[acid]
61
Relationship between Ka and pKa
Ka=10^-pKa
62
Explain the strength of carboxylic acids with alcohols
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
Why is chloroethanoic acid more acid than ethanoic acid
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
Why is methanoic acid more acidic than ethanoic acid
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
Buffer solution in the blood and how do changes in pH affect it
CO2 + H2O = HCO3- + H+ | When H+ changes the equilibrium move to the left or right accordingly
66
Function of HCO3- ion in the blood
H+ + HCO3- = H2CO3 | OH- + HCO3- = H2O + CO3 2-
67
Conditions are nitration of phenol
Dilute HNO3 room temperature Nitration of benzene needs cHNO3 and cH2SO4 and reflux
68
Concentrated nitric acid and phenol
Trinitrophenol | If it is just diluted nitric acid at room temperature it forms 2-nitrophenol or 4-nitrophenol
69
How does the solubility of carbonates vary down group 2
The solubility decreases down the group like with sulfates
70
Explain how the solubility of sulfates varies down group 2
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
Compare the ease of hydrolysis of acyl chlorides with chloroalkanes and aryl chlorides
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
Explain how the solubility of hydroxides varies down group 2
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
Bidentate definition
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
Colour of CuCl42-
Yellow
75
Colour of CoCl42-
Blue
76
Reagents and conditions for nucleophilic addition of CN to aldehyde or ketones
NaCN with HCN catalyst
77
What is a buffer solution
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
Ethanol to NaOH and I2
Ethanal is formed
79
What is meant by stability constant
The equilibrium constant for the formation of a complex ion in solution
80
Colour changed with MnO42- and Cr2O72-
Purple/pale pink to colourless | Orange to green
81
Tollen reagent
Ammoniacal silver nitrate solution Silver mirror formed Oxidises aldehydes
82
Fehlings solution
Alkaline solution of Cu2+ ions Blue to red/orange Oxidises aldehydes
83
Addition of water to ethene
Conc H3PO4 300 Celcius forms alcohol
84
What type of reaction is the coupling reaction between diazonium ions and phenol
Electrophilic substitution