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F335 OCR B Chemistry > F335 Key Points > Flashcards

F335 Key Points Flashcards

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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

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2
Q

A N/O dative bond model

N2O

A

N =(triple)N –>O

Linear

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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.

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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

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5
Q

Atom economy

A

Atom economy = mr useful products/mr reactants

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6
Q

Tertiary alcohol

A

Carbon with OH is attached to three other carbons

No hydrogens

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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

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8
Q

(CH3)3COH

A

2 methyl Propan-2-ol

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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.

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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

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11
Q

NMR splitting

A

Number of peaks is one Maori than the number of protons on the adjacent Carbon

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12
Q

NMR number of peaks (excluding splitting)

A

The number of printing environments

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13
Q 
CH3(-C)
NMR peak ( next to quartet)
A

0.7-1.6

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14
Q

O-CH2

NMR peak next to triplet

A

3.3-4.8

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15
Q

Triplet

A

For hydrogens on C next to CH2

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16
Q

Quartet

A

For hydrogens on C next to CH3

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17
Q

Kc

2CO2 –> 2CO + O2

A

Kc=[products]^mol / [reactants]^mol

Kc = [CO]^2 [O2]/[CO2]^2

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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

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19
Q

CO2+Ca(OH)2–>CaCO3 + H2O

A

Acid base reaction

Not precipitation

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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

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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

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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

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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

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24
Q

Interaction formed in hydration of ions in a solution

A

Ion-dipole

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25
Q

Proton NMR spectrum of benzene
Chemical shift range
Number of peaks

A

6.4 -8.2ppm

One peak

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26
Q

Systematic name C6H5Br

A

Bromobenzene

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27
Q

Compounds contains several benzene rings are used as dyes

A

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

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28
Q

C6H6Br2 –>C6H5Br + HBr
The reaction is electrophilic substition
Define electrophilic and describe behaviour

A

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

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29
Q

Suggest how trimetrexate inhibits the enz but not itself easily hydrogenated
Inhibits enzyme

A

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

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30
Q

Trimetrexate synthesis
Place a methyl group on an aromatic ring
Benzene to methylbenzene
Reagents and conditions

A

CH3Cl - chloromethane
AlCl3 - aluminium chloride
Reflux or anhydrous conditions or ionic solvent

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31
Q

Ionisation of weak acid, HA, in water

A

HA H+ + A-

HA is conjugate acid
A- is conjugate base

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32
Q

Oxonian version of weak acid, HA, ionisation

A

HA + H2O H3O+ + A-

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33
Q

Acidity constant, Ka for HAH+ + A-

A

Ka = [H+] [A-] / [HA]

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34
Q

Ka to pKa

A

Ka x - log

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35
Q

pH

A

[H+] x -log

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36
Q

Weak acid calculation inaccuracies

Approximations made

A

Concentration of acid at equilibrium = concentration of acid initially
[H+] compared with [HA] is not negligible

  1. 1 M HA
  2. 24 x 10^-2 M H+
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37
Q

Weak acid in blood stream

[A-]/[HA]

A

= Ka/[H+]

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38
Q

HCO3-

A

Hydrogencarbonate(IV)

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39
Q

pH of blood is buffered when a small amount of acid is added

A

[H+] increases so equilibrium position moves to left to form more CO2
Excess CO2 and HCO3- formed pH virtually unchanged.
pH is virtually unchanged

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40
Q

Sum of Enthalpy changes of hydration

A

Lattice Enthalpy -/+ Enthalpy Change Solution

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41
Q

Pharmacophore

A

The part of the molecule either binds to a receptor/active site

is responsible for the medicinal, pharmacological action - acts as the drug.

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42
Q

Statins inhibit a reaction in the body’s synthesis of cholesterol. This reaction is enzyme catalyses. Suggest how the reaction is inhibited.

A

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

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43
Q

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

A

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
Q

Ethanoate ion

A

CH3COO-

45
Q

Cu (AsO2)2
Arsenic oxidation state
Copper here is 2+

A

+3

AsO2 = -1
O2=-4
As=+3

46
Q

Cu(AsO2)2

A

Copper (II) arsenate (III)

47
Q 
Why copper (II) compounds are coloured
Why different ligands cause different colours
A

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
Q

What feature of this molecule gives rise to to being coloured?

A

Delocalised electrons
Conjugated electrons, system
Pi system
Possesses chromosphere

49
Q

Type of electrophilic substitution which adds a conjugated system

A

Coupling electrophilic substitution

50
Q

Name the -SO3H group

A

Sulfonic acid

51
Q

Ionic substances are soluble in water

A

Electrostatic, ionic bonds broken
Ion-dipole bonds formed
H bonds in water broken
Similar strength between bonds broken and made

52
Q

Shape of amine group

-NH2

A

-NH2
109’
Four areas of electron density, including the lone pair on N

53
Q

NH4+ bonding

A

Three covalent bonds, one dative covalent bond

54
Q

Cyanate ion

A

[N=tripleC-O]-

55
Q

Why are modern chemists not surprised that organic compounds can be made from inorganic compounds?

A

Both organic and inorganic contain the same elements

56
Q

Entropy change of a system is products -reactants, and times according to

A

Moles

57
Q

Negative entropy

N2 + 3H2 —> 2NH2

A

Fewer moles in products

Fewer ways of arranging as fewer molecules

58
Q

/\S tot

Effect of raising temperature

A 
/\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
Q

NH3 + HNO3 –> NH4NO3
Atom economy
Hazard

A

100%

Ammonia is toxic, alkaline

Nitric acid, HNO3, is powerfully oxidising, toxic, acidic

Ammonium nitrate, NH4NO3, is explosive

60
Q

Ethanoic acid

A

CH3COOH

61
Q

-S-

Bond angle

A

Two lone pairs, two bonds

Four areas electron density which repeal each other and best as far apart as possible to miniseries repulsion.

62
Q

Describe and explain the effect that carbonic anhydride will have on the position of the eq

CO2 + H2O HCO3- + H+

A

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
Q

Describe and explain how a lower amount of dissolved CO2 affects the pH of the blood
CO2 + H2O HCO3- + H+

A

Lower [CO2] moves equilibrium position to reactants/ left

pH rises.

64
Q

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

A

Equilibrium change happens much more slowly, does not happen without catalyst, pH does not change much, changes slowly

65
Q

pH to [H+]

A

10^-pH

66
Q

Explain what is meant by the term a buffers solution

CO2 + H2O

A

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
Q

There is considerable concern over rising levels of carbon dioxide in the atmosphere
Methods for removing CO2 from the atmosphere

A

Increasing photosynthesis by planting more trees

Capture and storage in the oceans, in former oil, gas wells, porous rock

68
Q

2-methylpropanyl methanoate

Acid hydrolysis

A

Methanoic acid

2-methylpropan-1-ol

69
Q

Two esters
Mass spectrometry
IR spectroscopy
Proton NMR spectroscopy

A

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
Q

Weak acids

Ionisation ethanoic acid

A

CH3COOH CH3COO- + H+

71
Q

Ethanoic acid, weak acid, in pure water
Moles Ethanoate in concentration
pH 3.2
CH3COOH CH3COO- + H+

A

[H+] = [CH3COO-]=10^-3.2= 6.3 x 10^-4

2sf

72
Q

Suggest how the initial and equilibrium amounts of acid could be measured

A

Titration with alkali/base of standard/known concentration

73
Q

Classify coupling reaction

A

Electrophilic substitution

74
Q

Condition for aminosation

A

Below 5’C

75
Q

Coupling of brown and yellow dye produces organise dye

A

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
Q

Explain why these three dues are coloured and why the colour varies between them

A

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
Q

Benzene undergoes mainly substitution

Rather than addition

A

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
Q

SF6

Shape and angle

A

Octahedral

90’

79
Q

Bonding
Lithium
Sulphur hexafluoride
Lithium fluoride

A

Lithium - metal
Sulphur hexafluoride - covalent
Lithium fluoride - ionic

80
Q

Li2S

A

Lithium sulfide

Lithium (I) sulfide

81
Q

SF6 and hexanedioic acid

A

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
Q

Formula of posassium chlorate (VII)

A

KClO4

83
Q

Need for a fuse

Large positive entropy change total

A

Shows reaction is spontaneous, likely to occur, feasible
Fuse implies high activation Enthalpy
Slow reaction

84
Q

Fragments called synthons from imagined breaking of bonds

Explain why synthons are not often used to make the required compound

A

They do not exist
They are not very stable
They cannot be obtained
They are theoretical

85
Q

Predict and explain the effect of increasing the temp on the position of the equilibrium.

How will this affect the equilibrium constant?

A

Equilibrium position moves to the reactants (left)
It moves in endothermic direction
Equilibrium constant falls

Kc(decreases)= [products]/[reactants - gets larger]

86
Q

Why is a compound converted into analysts be used as the herbicides

A

The salt will be much more soluble

Salt is also less acidic

87
Q

C6H3Cl2OCH2COO- (CH3)/ NH2+

A

..

88
Q

Herbicide toxic to animals and humans

Suggest two arguments that farmers might give to be allowed to continue to use.

A

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
Q

C-Halogen

Polarity

A

C and Halogen differ in electronegativity. Halogen is more electronegative.

90
Q

CBr3CBr3

Has no overall dipole

A

All dipoles cancel

91
Q

CH2BrCH2Cl

A

The dipoles do not cancel

C-Cl is more polar than C-Br

92
Q

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.

A

Gas-liquid chromatography

93
Q

CH2BrCH2Br and CH2BrCH2Cl

Dissolve in each other, but do not dissolve in water

A

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
Q

Define electrophile

A

Electron deficient reagent
Attracted to area of high electron density
Receives electrons - lone pairs and forms covalent bond

95
Q

Ethene plus bromine
NaCl solution
CH2ClCH2Br vs. CH2ClCH2Cl

A

Cl- is not electrophile
Cl- cannot attack
Cl- is not attracted
Cl- attacks intermediate carbocation once Br is there

96
Q

CH2BrCH2Br –NaOH–> CH2(OH)CH2(OH)

–Acid dichromate, Reflux-> HOOCCOOH

A

NaOH
Acid dichromate
Reflux

97
Q

CH3CHBr2

NMR

A

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
Q

HPO42-

Systematic name

A

Hydrogen phosphate (V)

99
Q

Formula of conjugate acid of H2PO4-

H2PO4- H+ + HPO42-

A

H2PO4-

100
Q

Explain buffer solution and explain why buffer solutions are important in human cells

A

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
Q

Ka=[H+][HPO4-]/[H2PO4-]

A

6.2x10^-8M = [H+]^2 / 0.1M

102
Q

H2PO4- H+ + HPO42-

A

Increase [H+] moves equilibrium position to left (reactants).
This removes H+ and restores [H+]/pH

[HPO42-] is large

103
Q

Ka=[H+][HPO4-]/[H2PO4-]

pH of a buffer solution where [HPO4-] and [H2PO4-] are equal

A

[H+] = Ka
pH=pKa
= -log [6.2x10-8]

104
Q

Nitrification (+NO2, nitrous oxide)

Reagent

A

HNO3

105
Q

[H+]/Ka = [Ka]/[A-]

A

[Ka]/[A-]

F335 OCR B Chemistry (1 decks)

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