F335 Key Points Flashcards
1
Q
Nitrous oxide gas, N2O, is formed in the soil by denitrifying bacteria.
Give the systematic name for nitrous oxide.
A
N2O
Dinitrogen oxide
Nitrogen(I) oxide
Dinitrogen monoxide
2
Q
A N/O dative bond model
N2O
A
N =(triple)N –>O
Linear
3
Q
Suggest the reason why denitrification is a problem for crop production
A
Nitrogen in compounds or nitrates is less available to plants/crops
Is needed by plants crops
Makes plants/crops grow.
Is a fertiliser.
4
Q
Why is nitrous oxide soluble in fat(propane-1,2,3-triol) in terms of imb?
A
Instantaneous dipole-induced dipole for compounds, between compounds
Pd-Pd for compounds, between compounds
Intermolecular bonds are similar
IMB formed are similar to those broken
5
Q
Atom economy
A
Atom economy = mr useful products/mr reactants
6
Q
Tertiary alcohol
A
Carbon with OH is attached to three other carbons
No hydrogens
7
Q
Primary and secondary alcohols are readily acidified by potassium dichromate
But an-2-ol with acidified potassium dichromate (VI)
A
From orange to green
Name of product: butone
8
Q
(CH3)3COH
A
2 methyl Propan-2-ol
9
Q
2 methyl Propan-2-ol has a lower boiling point than Butan-1-ol
A
Idid
Imb are weaker
Less energy required to break imb
Molecules, chains. Can’t get as close together, less areas of contact.
10
Q
Suggest why it is important that the energy for this reaction comes from the Sun rather than burning fossil fuels.
A
The energy source must not produce Carbon dioxide, CO2.
Fossil fuels produce carbon dioxide
11
Q
NMR splitting
A
Number of peaks is one Maori than the number of protons on the adjacent Carbon
12
Q
NMR number of peaks (excluding splitting)
A
The number of printing environments
13
Q
CH3(-C) NMR peak ( next to quartet)
A
0.7-1.6
14
Q
O-CH2
NMR peak next to triplet
A
3.3-4.8
15
Q
Triplet
A
For hydrogens on C next to CH2
16
Q
Quartet
A
For hydrogens on C next to CH3
17
Q
Kc
2CO2 –> 2CO + O2
A
Kc=[products]^mol / [reactants]^mol
Kc = [CO]^2 [O2]/[CO2]^2
18
Q
Disadvantage of using this method of carbon capture
CO2+Ca(OH)2–>CaCO3 + H2O
A
Uses a lot of Ca(OH)2
CO2 is emitted in manufacture of Ca(OH)2
19
Q
CO2+Ca(OH)2–>CaCO3 + H2O
A
Acid base reaction
Not precipitation
20
Q
Mr of benzene from mass spec
A
It will be the value of the highest mass peak
The molecular ion peak
Peak furthest to the right
21
Q
CH2=CH-C=(triple) -CH=CH2
Benzene does not react with HBr
Casts doubt on model
A
Alkene groups react with HBr
They undergo addition reaction
22
Q
Kekulé accounts for some but not all electron diffraction data
A
Accounts for
Bond angle being the same/120
Three bonds
Does not account for
Bond lengths equal
But under kekulé model, Single bonds are longer than double bonds, so unequal
23
Q
Explain the meaning of the circle in benzene
A
Delocalised electrons
One electron from each Carbon
Two rings above and below Carbon ring
24
Q
Interaction formed in hydration of ions in a solution
A
Ion-dipole
25
Proton NMR spectrum of benzene
Chemical shift range
Number of peaks
6.4 -8.2ppm
| One peak
26
Systematic name C6H5Br
Bromobenzene
27
Compounds contains several benzene rings are used as dyes
Electron is excited as a result of light energy/photons
Frequency absorbed depends on energy change
/\E= hv
Dyes/ coloured compounds absorb light in the visible spectrum
Energy levels are closer
More delocalisation
More conjugation
More than one benzene ting
Larger chromosphere
Frequency of UV light is greater than visible
Dues transmit the complementary colour
28
C6H6Br2 -->C6H5Br + HBr
The reaction is electrophilic substition
Define electrophilic and describe behaviour
Electrophile is a partially positively charged electron deficient species
Electrophilic accepts a pair of elections and forms a covalent bond.
Bromine molecules is polarised and forms dela positive Bromine
Positive end attacks - is the electrophile
29
Suggest how trimetrexate inhibits the enz but not itself easily hydrogenated
Inhibits enzyme
It fits into the active site
Blocks the site - prevents substrate from binding
Trimetrexate has an arene ring - is aromatic
It dies not have C=N, double bond, alkene
30
Trimetrexate synthesis
Place a methyl group on an aromatic ring
Benzene to methylbenzene
Reagents and conditions
CH3Cl - chloromethane
AlCl3 - aluminium chloride
Reflux or anhydrous conditions or ionic solvent
31
Ionisation of weak acid, HA, in water
HA H+ + A-
HA is conjugate acid
A- is conjugate base
32
Oxonian version of weak acid, HA, ionisation
HA + H2O H3O+ + A-
33
Acidity constant, Ka for HAH+ + A-
Ka = [H+] [A-] / [HA]
34
Ka to pKa
Ka x - log
35
pH
[H+] x -log
36
Weak acid calculation inaccuracies
| Approximations made
Concentration of acid at equilibrium = concentration of acid initially
[H+] compared with [HA] is not negligible
0. 1 M HA
2. 24 x 10^-2 M H+
37
Weak acid in blood stream
| [A-]/[HA]
= Ka/[H+]
38
HCO3-
Hydrogencarbonate(IV)
39
pH of blood is buffered when a small amount of acid is added
[H+] increases so equilibrium position moves to left to form more CO2
Excess CO2 and HCO3- formed pH virtually unchanged.
pH is virtually unchanged
40
Sum of Enthalpy changes of hydration
Lattice Enthalpy -/+ Enthalpy Change Solution
41
Pharmacophore
The part of the molecule either binds to a receptor/active site
is responsible for the medicinal, pharmacological action - acts as the drug.
42
Statins inhibit a reaction in the body's synthesis of cholesterol. This reaction is enzyme catalyses. Suggest how the reaction is inhibited.
Statin bonds to active site of enzyme
Blocking active site, enzymes, receptor
Substrate cannot bind
Statin binds more strongly
Fewer active sites
Reduces enzymes activity
43
Discovered from work on fungi.
Suggest the sequence of events that led from one such substance of the production of a synthetic Statin for medicinal use
Show that a compound prevented cholesterol synthesis and therefore has medicinal properties
Carry out clinical trials
Analyse structure and identify pharmacophore
Synthesise similar molecules
44
Ethanoate ion
CH3COO-
45
Cu (AsO2)2
Arsenic oxidation state
Copper here is 2+
+3
AsO2 = -1
O2=-4
As=+3
46
Cu(AsO2)2
Copper (II) arsenate (III)
47
```
Why copper (II) compounds are coloured
Why different ligands cause different colours
```
Electrons promoted, excited
Light absorbed
/\E = hv : freq, wavelength depends on energy gap
Ligands affect size of gap, splitting, excitation energy
Complementary colour transmitted, reflected
48
What feature of this molecule gives rise to to being coloured?
Delocalised electrons
Conjugated electrons, system
Pi system
Possesses chromosphere
49
Type of electrophilic substitution which adds a conjugated system
Coupling electrophilic substitution
50
Name the -SO3H group
Sulfonic acid
51
Ionic substances are soluble in water
Electrostatic, ionic bonds broken
Ion-dipole bonds formed
H bonds in water broken
Similar strength between bonds broken and made
52
Shape of amine group
| -NH2
-NH2
109'
Four areas of electron density, including the lone pair on N
53
NH4+ bonding
Three covalent bonds, one dative covalent bond
54
Cyanate ion
[N=tripleC-O]-
55
Why are modern chemists not surprised that organic compounds can be made from inorganic compounds?
Both organic and inorganic contain the same elements
56
Entropy change of a system is products -reactants, and times according to
Moles
57
Negative entropy
| N2 + 3H2 ---> 2NH2
Fewer moles in products
| Fewer ways of arranging as fewer molecules
58
/\S tot
| Effect of raising temperature
```
/\S tot = /\S system + -/\H/T
T is greater
/\S tot is less positive, lower
Therefore the forward reaction is less likely to occur
Equilibrium lies further to reactants
Decreased yield
Fewer products
```
59
NH3 + HNO3 --> NH4NO3
Atom economy
Hazard
100%
Ammonia is toxic, alkaline
Nitric acid, HNO3, is powerfully oxidising, toxic, acidic
Ammonium nitrate, NH4NO3, is explosive
60
Ethanoic acid
CH3COOH
61
-S-
| Bond angle
Two lone pairs, two bonds
| Four areas electron density which repeal each other and best as far apart as possible to miniseries repulsion.
62
Describe and explain the effect that carbonic anhydride will have on the position of the eq
CO2 + H2O HCO3- + H+
None as catalysts do not affect equilibrium positions
It speeds up forward and backward reactions equally( and affects reaction rate only)
Speeds up achievement of equilibrium
63
Describe and explain how a lower amount of dissolved CO2 affects the pH of the blood
CO2 + H2O HCO3- + H+
Lower [CO2] moves equilibrium position to reactants/ left
| pH rises.
64
Inhibition of carbonic anhydrase by acetazolamide affects the change caused by decrease in [CO2] (and equilibrium shift to left/ reactants, decreasing [H+], increasing pH) and so combats altitude sickness
Equilibrium change happens much more slowly, does not happen without catalyst, pH does not change much, changes slowly
65
pH to [H+]
10^-pH
66
Explain what is meant by the term a buffers solution
CO2 + H2O
A buffer solution resists pH change when acid/alkali added in small quantities
Equilibrium moves to left when acid in added removing acid/H+
Large concentration of HCO3- reacts with
H+
67
There is considerable concern over rising levels of carbon dioxide in the atmosphere
Methods for removing CO2 from the atmosphere
Increasing photosynthesis by planting more trees
| Capture and storage in the oceans, in former oil, gas wells, porous rock
68
2-methylpropanyl methanoate
| Acid hydrolysis
Methanoic acid
| 2-methylpropan-1-ol
69
Two esters
Mass spectrometry
IR spectroscopy
Proton NMR spectroscopy
Mass Spectrometry - Mr
Different fragments, e.g ester B: CH3CH2CH2CO+
Infrared spectroscopy - functional groups
Fingerprint compared with known values to distinguish between esters. The esters both have the same Mr and bonds.
Proton NMR spectroscopy - environments of protons.
Ratio of areas 6:1:2:1 and 3:2:2:3
Six protons in one environment in A only
Different shifts - B has 2.0-2.7, A has 9.4-10.0
Splitting
70
Weak acids
| Ionisation ethanoic acid
CH3COOH CH3COO- + H+
71
Ethanoic acid, weak acid, in pure water
Moles Ethanoate in concentration
pH 3.2
CH3COOH CH3COO- + H+
[H+] = [CH3COO-]=10^-3.2= 6.3 x 10^-4
| 2sf
72
Suggest how the initial and equilibrium amounts of acid could be measured
Titration with alkali/base of standard/known concentration
73
Classify coupling reaction
Electrophilic substitution
74
Condition for aminosation
Below 5'C
75
Coupling of brown and yellow dye produces organise dye
Orange is between brown and yellow
One has one more NH2 and the other one less
Having two amine groups is between 1 and 3
76
Explain why these three dues are coloured and why the colour varies between them
Electrons are excited
Move go higher energy level
Absorption of light in the visible spectrum
Freq/ wavelength depends on energy gap, /\E = hv
Different number amine groups
Different amounts of delocalisation, and different chromosphere
Affect
/\E - energy to excite electron, energy absorbed
NOT when electron moves down - colour, not emission spectrum.
77
Benzene undergoes mainly substitution
| Rather than addition
Delocalisation
Stability retained in substitution as delocalisation maintained
Addition product loses delocalisation and is less stable (has higher energy)
Energy is required to break delocalisation in addition
78
SF6
| Shape and angle
Octahedral
| 90'
79
Bonding
Lithium
Sulphur hexafluoride
Lithium fluoride
Lithium - metal
Sulphur hexafluoride - covalent
Lithium fluoride - ionic
80
Li2S
Lithium sulfide
Lithium (I) sulfide
81
SF6 and hexanedioic acid
SF6 has I'd-I'd bonds
Hexadioic acid has hydrogen bonding
Imb in acid stronger than imb in SF6
More energy required to separate molecules, break or overcome imb. to boil, melt, acid
82
Formula of posassium chlorate (VII)
KClO4
83
Need for a fuse
| Large positive entropy change total
Shows reaction is spontaneous, likely to occur, feasible
Fuse implies high activation Enthalpy
Slow reaction
84
Fragments called synthons from imagined breaking of bonds
| Explain why synthons are not often used to make the required compound
They do not exist
They are not very stable
They cannot be obtained
They are theoretical
85
Predict and explain the effect of increasing the temp on the position of the equilibrium.
How will this affect the equilibrium constant?
Equilibrium position moves to the reactants (left)
It moves in endothermic direction
Equilibrium constant falls
Kc(decreases)= [products]/[reactants - gets larger]
86
Why is a compound converted into analysts be used as the herbicides
The salt will be much more soluble
| Salt is also less acidic
87
C6H3Cl2OCH2COO- (CH3)/ NH2+
..
88
Herbicide toxic to animals and humans
| Suggest two arguments that farmers might give to be allowed to continue to use.
Its usefulness. No alternatives.
It can be used with care: used in low concentrations, small amounts
Greater yield of crop, greater profit, and cheaper food
89
C-Halogen
| Polarity
C and Halogen differ in electronegativity. Halogen is more electronegative.
90
CBr3CBr3
| Has no overall dipole
All dipoles cancel
91
CH2BrCH2Cl
The dipoles do not cancel
| C-Cl is more polar than C-Br
92
Francis fractionally distilled the oil and identified its components using their densities and poling points. Today, the oil might be vaporised and analysed using a chromatographic technique.
Gas-liquid chromatography
93
CH2BrCH2Br and CH2BrCH2Cl
| Dissolve in each other, but do not dissolve in water
Hydrogen bonds in water
Dipole-dipole bonds between halogens (pd-pd + id-id)
Similar imb are formed in the mixture as they are found in one separate substances imb formed have similar energy, strength to those broken
H bond stronger than imb between water-halogenalkanes. H bonds are not formed between halogen alkane and water. Energy required to break H bonds is not provided by imb between halogenoalkane and water
94
Define electrophile
Electron deficient reagent
Attracted to area of high electron density
Receives electrons - lone pairs and forms covalent bond
95
Ethene plus bromine
NaCl solution
CH2ClCH2Br vs. CH2ClCH2Cl
Cl- is not electrophile
Cl- cannot attack
Cl- is not attracted
Cl- attacks intermediate carbocation once Br is there
96
CH2BrCH2Br --NaOH--> CH2(OH)CH2(OH)
| --Acid dichromate, Reflux-> HOOCCOOH
NaOH
Acid dichromate
Reflux
97
CH3CHBr2
| NMR
Two peaks hence two hydrogen environments
Hydrogens in ratio 3:1
CH3/3H split into two by CH (one hydrogen adj C)
CH/1H split into four by the CH3 ( three H on adj C)
98
HPO42-
| Systematic name
Hydrogen phosphate (V)
99
Formula of conjugate acid of H2PO4-
| H2PO4- H+ + HPO42-
H2PO4-
100
Explain buffer solution and explain why buffer solutions are important in human cells
Resits change in pH when acid or alkali added in small quantities
Cell enzymes work are specific optimum pH change in pH can stop enzymes working - they are denatured.
101
Ka=[H+][HPO4-]/[H2PO4-]
6.2x10^-8M = [H+]^2 / 0.1M
102
H2PO4- H+ + HPO42-
Increase [H+] moves equilibrium position to left (reactants).
This removes H+ and restores [H+]/pH
[HPO42-] is large
103
Ka=[H+][HPO4-]/[H2PO4-]
| pH of a buffer solution where [HPO4-] and [H2PO4-] are equal
[H+] = Ka
pH=pKa
= -log [6.2x10-8]
104
Nitrification (+NO2, nitrous oxide)
| Reagent
HNO3
105
[H+]/Ka = [Ka]/[A-]
[Ka]/[A-]
