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Aqa a level chemistry > Organic chem - analysis > Flashcards

Organic chem - analysis Flashcards

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1
Q

the molecular (parent) ion in the mass spectrum of a hydrocarbon containing 13C and H only

A

is the peak with highest mass to charge ratio

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2
Q

in a mass spectrum of butane C4H10 where would a peak be seen for the molecular ion if it has a charge of 2+

A

29

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3
Q

suggest what issue would arise when looking at the mass spec of propanal and propanone

A

-the molecular ion peak would be the same as they have the same mr (molecular mass)

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4
Q

deduce how many m/z values will be produced in the mass spectrum of butanone

A

CH3+
CH3CH2+
CH3CH2C=O+
CH3CH2C=OCH3+

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5
Q

state the meaning of the term molecular ion

A

an ion with the highest mz value

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6
Q

state why the precise RAM for the 12C isotope is exactly 12.00000

A

by definition

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7
Q

explain why there are two molecule ion peaks

A

chlorine has 2 isotopes

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8
Q

purpose of mass spec

A

-Purpose of mass spec is to identify the relative abundance of different isotopes -> the relative atomic mass of a compound and the molecular structure of an organic compound

-m/z = mass to charge ratio

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9
Q

Br isotope

A

79-81 Bromine is the most abundant isotope for Br2

3 molecule ion peak for bromine

All species in mass spec are positively charged bc they are ions

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10
Q

what would happen if there was a 2+ charge on a peak

A

+2 charge on mass spec graph would cause the m/z ratio to halve e.g 160/2 = 80

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11
Q

molecular ion

A

-the largest mz value represents the molecular ion peak –> the peak that occurs when the whole molecule is ionised (M (g) –> M+ + e-

-calculate mr = largest significant ion peak with no C13 peaks

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12
Q

fragmentation

A

-the issue is overcome as different molecules have fragmentation patterns

-high energy electron (molecule in sample) –> cation with an unpaired electron (radical cation)

-Electrons are fired at the molecules to ionise them

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13
Q

fragmentation pattern

A

-bombarding molecules with high energy electrons not only ionises them but it also can split them into fragments

-as a result, the mass spectrum consists of a fragmentation pattern (molecule shown as 2 parts –> positive ion on molecule –> fragmentation –> positively charged fragment and uncharged fragment

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14
Q

readily bonds breaking

A

-molecules break up more readily at weak bonds which give rise to more stable fragments

-The mass to charge ratio indicates the mass of the fragment so we can deduce the structure of the fragment

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15
Q

M/Z and fragment

A

15

CH3+

17

OH+

29

CH3Ch2+ or CHO+

43

CH3CH2CH2+ or CH3CO+

31

CH2OH+

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16
Q

which compound forms a molecular ion with a different precise molecular mass from the other three

A

dimethylpropane

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17
Q

what is the angle of C-C-C and C-O-H bond

A

C-C-C = 109.5

C-O-H = 104.5

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18
Q

suggest what you would expect to see when cyclohexanol reacts with sodium

A

fizzing
white solid forms

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19
Q

give a possible reason why there is no molecular ion peak in the mass spectrum of 2-methylpropan-2-ol

A

molecular ion is unstable

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20
Q

amine formula

A

N2H

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21
Q

How can we distinguish between propan-1-ol and propan-2-ol

A

different fingerprint regions -> compare absorption space below 1500

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22
Q

suggest two possible identities of C. Justify your answer and explain how you could further confirm the exact identity of C

A

propanal or propanone
peak at 1700 due to C=O
use fingerprint region to compare literature

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23
Q

IR spec

A

-IR spec allows us to identify what functional groups are present in organic molecules. This is possible because most compounds will absorb infrared radiation

-Certain bonds (bonds with a functional group) will absorb IR at specific frequencies which causes bonds in the molecules to bend and stretch

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24
Q

detecting the absorption

A

-the absorptions are detected and analysed on an absorption spectrum

-this spectrum is a plot of transmittance against wavenumber

-it is possible to correlate an absorption in the region 4000 to 1500cm-1 on the spectrum to particular bonds

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25
Q

How to read the spectrum

A

1) State the type and frequency range of the peak in the spectra

2) Identify the bond responsible

3) State the molecule and if given the molecular formulae you must nam e the compound

In order to interpert IR you need to determine what bonds are present e.g ethan-1,2-diol has OH, CO, CH and CC

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26
Q

fingerprint region

A

-the region below 1500 is usually called the fingerprint region as it has many peaks which are difficult to assign

-the pattern of these peaks is unique to a particular compound

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27
Q

broad vs narrow peak

A

Broad peak = greater absorption due to longer bond e.g O-H bond at 3500-3000

Sharp peak = smaller absorption e.g ketone (C=O)

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28
Q

peak for carboxylic acids

A

Carboxylic acids -> broad peak at 3000-3500 due to OH and narrow peak at 1700 due to C=O

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29
Q

peak for alcohols

A

Alcohols –> broad peak at 3500 due to OH and 1100 due to C-O

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30
Q

peak for aldehydes and ketones

A

Aldehydes and Ketones –> peak at 1700 due to C=O

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31
Q

peak for alkenes and alkanes

A

Alkenes –> peak at 1650 due to C=C

Alkanes –> C-H peak at 2850 – 3000

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32
Q

same molecular formula =

A

same mr

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33
Q

suggest why these granules prevent bumping

A

form smaller bubbles

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34
Q

no ozone depletion =

A

doesnt contain Cl

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35
Q

tests

A

COOH -> sodium carbonate –> effervesence/milky

OH and CHO –> acidified K2Cr2O7

oxidise alkenes in conc sulfuric acid –> potassium manganate (purple - colourless)

Br or any halogen –> silver nitrate = cream precipitate

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36
Q

explain why determinging the precise relative molecular mass could not be used to distinguish between the two compounds

A

-same molecular formula so their mr value would be the same

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37
Q

explain why 1,1,1,-triflouroethane does not lead to the depletion of the ozone

A

-doesnt contain C-Cl bonds so cant form chlorine radicals

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38
Q

Z isomers

A

according to priorty of mr first!!!

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39
Q

explain why the water should enter the condeser at the bottom and not the top

A

-to ensure full condensation as more of the condenser fills with water

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40
Q

why does rate of decomposition of gas increase at higher temperatures

A

-more Ea
-more frequenct and successful collisions

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41
Q

test for aldehyde + tollens reagent

A

aldehyde = fehlings = blue solution to red precipitate

aldehyde = tollens = silver mirror

42
Q

carboxylic acid + sodium carbonate

A

effercvescence of CO2

43
Q

CH3CHO

A

ethanal

44
Q

propanone

A

(CH3)2CO (smallest possible ketone)

45
Q

propanal

A

CH3CH3CHO

46
Q

reduction of propanal equation

A

Reduction of propanal:

-CH3CH2CHO + 2[H] –> CH3CH2CH2OH

47
Q

nucelophilic addition

A

-Aldehydes can be reduced to primary alcohols, and ketones to secondary alcohols, using NaBH4 in aqueous solution. These reduction reactions are examples of nucleophilic addition.

-nucelophilic addition of carbonyl compounds with KCN (potassium cyanide) followed by a dilute acid to produce hydroxynitriles

-Aldehydes and unsymmetrical ketones form mixtures of enantiomers when they react with KCN followed by dilute acid.

48
Q

carbonyl compounds

A

-carbonyl = compounds with C=O bond (aldehydes or ketones)

-If the C=O is on the end of the chain with an H attached it is an aldehyde.

-If the C=O is in the middle of the chain it is a ketone

49
Q

bonding in carbonyl compounds

A

-intermolecular forces in carbonyl atoms = van der waals and permanent dipole-permanent dipole

-smaller carbonyls are soluble in water bc they can form hydrogen bonds

-In comparison to the C=C bond in alkenes, the C=O is stronger and does not undergo addition reactions easily.

50
Q

reaction of carbonyls

A

-The C=O bond is polarised because O is more electronegative than carbon. The positive carbon atom attracts nucleophiles.

Aldehyde + [O] –> carboxylic acid

51
Q

tollens reagent

A

-mix aqueous ammonia and silver nitrate

-heat gently

-aldehydes oxidised into carboxylic acid

-silver mirror

-no change for ketones

-CH3CHO + 2Ag+ + H2O –> CH3COOH + 2Ag + 2H+

52
Q

fehlings solution

A

-Blue Cu2+ ions

-heat gently

-copper ions reduced to copper oxides

-form carboxylic acid

-CH3CHO + 2Cu2+ + 2H2O –> CH3COOH + Cu2O + 4H+

53
Q

reaction of carbonyls into alcohol

A

-reagent = NaBH4 in aqueous ethanol

-conditions = room temperature and pressure

-aldehydes reduced to primary alcohol

-ketones reduced to secondary alcohol

54
Q

catalytic hydrogenation

A

-hydride ions in NaBH4 are attracted to the positive carbon in the C=O bond

Catalytic hydrogenation:

-Reagent = hydrogen and nickel catalyst

-Conditions = high pressure

-CH3CHO + H2 –> CH3CH2OH

55
Q

addition of hydrogen cyanide to carbonyls to form hydronitriles

A

-carbonyl –> hydroxynitrile

-reagent = potassium cyanide and dilute sulfuric acid

-room temperature and pressure

-nucelophilic addition

-CH3COCH3+ HCN –> CH3C(OH)(CN)CH3 (ethanone + potassium cyanide –> 2-hydroxy, 2-methylpropanitrile

56
Q

hazards + nucelophilic addition

A

-Nucleophilic addition of HCN to aldehydes and ketones (unsymmetrical) when the planar carbonyl group is approached equally from both sides by the HCN attacking species: results in the formation of a racemate.

-advantage of using KCN or NaCN is that there will be a higher concentration of the CNion as these compounds will completely ionise. HCN is a weak acid an will only partially ionise

-We could use HCN for this reaction but it is a toxic gas that is difficult to contain. KCN/NaCN are still, however, toxic, because of the cyanide ion.

-Bond angle around carbonyl group 120degrees = trigonal planar)

-carbonyl group = polar due to difference in electronegativity between oxygen and carbon

-Aldehydes and Ketones can be reduced using LiAlH4 in a dry ether solvent since it reacts violently with water

57
Q

carboxylic acids and derivatives

A

-carboxylic acids are weak acids. Both carboxylic acids and alcohol react in the presence of a catalyst to give esters

-common uses of esters = plasticiers, solvents, perfumes etc

-Vegetable oils and animal fats are esters of propane-1,2,3-triol (glycerol)

58
Q

esters

A

-esters can be hydrolysed into acid or alkaline conditions to form alcohols and carboxylic acids

-Vegetable oils and animal fats can be hydrolysed in alkaline conditions to give soap (salts of long-chain carboxylic acids) and glycerol.

-Biodiesel is a mixture of methyl esters of long-chain carboxylic acids. Biodiesel is produced by reacting vegetable oils with methanol in the presence of a catalyst

59
Q

benzene ring

A

-Benzene ring –> Delocalisation of p electrons makes benzene more stable than the theoretical molecule cyclohexa-1,3,5-triene

60
Q

electrophilic subsitution

A

-Electrophilic attack on benzene rings results in substitution, limited to monosubstitutions. Nitration is an important step in synthesis, including the manufacture of explosives and formation of amines. Friedel–Crafts acylation reactions are also important steps in synthesis.

61
Q

condensation polymers

A

-Condensation polymers are formed by reactions between:

  • dicarboxylic acids and diols
  • dicarboxylic acids and diamines
  • amino acids.
  • The repeating units in polyesters (eg Terylene) and polyamides (eg nylon 6,6 and Kevlar) and the linkages between these repeating units.

-polyalkenes are chemically inert and non-biodegradable

-polyesters and polyamides can be broken down by hydrolysis and are biodegradable

62
Q

ethanedioic acid

A

carboxyl group on both carbons

63
Q

acidity of carboxylic acid

A

The carboxylic acid are only weak acids
in water and only slightly dissociate, but
they are strong enough to displace
carbon dioxide from carbonates.

small carboxylic acids can dissolve in water (form hydrogen bonds)

64
Q

delocalisation

A

The carboxylic acid salts are stabilised by delocalisation,
which makes the dissociation more likely.

The delocalised ion has equal C-O bond lengths. If
delocalisation did not occur, the C=O bond would
be shorter than the C-O bond.

The pi charge cloud has
delocalised and spread out. The
delocalisation makes the ion
more stable and therefore more
likely to form.

65
Q

strength of carboxylic acids

A

Increasing chain length pushes
electron density on to the COO-ion, making it more negative and
less stable. This make the acid
less strong.

Propanoic acid is less acidic than ethanoic acid

Electronegative chlorine atoms
withdraw electron density from
the COO- ion, making it less
negative and more stable. This
make the acid more strong. (chloroethanoic acid is more acidic than ethanoic acid)

66
Q

test for carboxylic acids

A

solid Na2CO3 = production of CO2 (effervescene)

67
Q

methanoic acid can be oxidised into

A

carbonic acid (H2CO3) which decomposes to give CO2

68
Q

esterification

carboxylic acid + alcohol –> ester + water

A

Carboxylic acids react with alcohols, in the
presence of a strong acid catalyst, to form
esters and water.

Esters have two parts
to their names, eg
methyl propanoate.

69
Q

ethanoic acid + ethanol –> ester + water

A

The reaction is reversible. The
reaction is quite slow and needs
heating under reflux, (often for
several hours or days). Low yields
(50% ish) are achieved. An acid
catalyst (H2SO4
) is needed

70
Q

polymers + esters

A

Although polar, they do not form hydrogen bonds (reason: there is no
hydrogen bonded to a highly electronegative atom).
They have a lower b.p. than the hydrogen-bonded carboxylic acids
they came from. They are also almost insoluble in water.
Esters can be used as
plasticisers for polymers.
Often pure polymers have limited flexibility because the polymer chains
cannot move over each other.
Incorporating some plasticiser into the polymer allows the chains to move
more easily and the polymer can become more flexible.

71
Q

how can esters be hydrolysed

A

-heat with acid or NaOH

acid = dilute HCl under reflux –> form carboxylic acid and alcohol

NaOH = dilute NaOH and reflux,
methyl propanoate –> sodium propanoate + methanol

The liquid ethyl benzoate can be hydrolysed by sodium hydroxide by heating under reflux
for 30 minutes.

72
Q

solubility

A

Sodium benzoate is soluble in water because it is ionic. Benzoic acid, however, is insoluble. This is
because even though the polar COOH group can form hydrogen bonds, the benzene ring is non-polar.
In organic compounds there are often polar parts and non-polar parts. The solubility in water of a
compound will controlled by whether the polar or non polar part is of greater importance.

73
Q

functional group for ester

A

-ester = O-C=O (synthetic aromatics)

74
Q

production of an ester

A

Alcohol + carboxylic acid –> ester + alcohol

e.g ethanol + propanoic acid –> ester + H2O

75
Q

naming esters

A

-first part of the name = alcohol = yl

-last part of the name = oate = COOH

Ethylethanoate

Propanol + butanoic acid = propylbutanoate + water

76
Q

polarity

A

More polar = more soluble

Longer chain = less soluble

77
Q

tollens and Fehlings

A

Tollens reagent and Fehlings reagent are oxidising agents to distinguish between a ketone and aldehyde

Tollens:

Aldehyde = silver mirror

Ketone = no reaction

Fehlings:

Aldehyde = brick-red copper oxide ppt

Ketone = no reaction

78
Q

reduce aldehydes and ketones back into an alcohol

A

Reduce aldehydes and ketones back into an alcohol using reducing agent (NaBH4)

e.g propan-2-one + 2[H] –> propan-2-ol

79
Q

reducing agents =

A

hydride
H-

80
Q

reduction of carbonyls

A

Reagents: NaBH4 In aqueous ethanol
Conditions: Room temperature and pressure

81
Q

when can nucelophilic subsitution occur

A

-Why can nucelophilic subsitution occur = halogen is more electronegative than carbon. C-X bond is polar so partial positive carbon is attracted to lone pair of electrons

82
Q

nucelophilic addition

A

Nucelophilic addiiton of ethanal using NaBH4 where H- is the nucelophile

NaBH4 contain a source of nucleophilic hydride ions (:H-) which are attracted to the positive carbon in the C=O bond.

Ketone or aldehyde –> alcohol

83
Q

addition of hydrogen cynaide to carbonyls to form hydroxynitrtiles

A

-e.g propanone + HCN –> 2-hydroxy-2-methylpropanenitrile

When naming hydroxy nitriles the CN becomes part of the main chain and carbon no1 = 2-hydroxy-2-methylpropanenitrile 1

84
Q

aldehyde =

A

end of carbon chain

85
Q

true or false = nucelophile is repelled by C=C

A

true

86
Q

naming nitriles

A

-naming nitriles in nucelophilic addition –> ethanone –> 2-hydroxylpropenitrile

-propanone –> 2-hydroxyl-2-methylpropenitrile

CN group counts as the first carbon of
the chain. Note the stem of the name is
different : butanenitrile and not
butannitrile.

87
Q

racemic mixture

A

-planar carbonyl group can be attacked from each side with an equal probability

-four different groups attached to carbon atom (chiral)

-equal amounts of the 2 enantiomers

88
Q

enatinomers

A

Enantiomers = Enantiomers are a pair of molecules that exist in two forms that are mirror images of one another but cannot be superimposed one upon the other. Enantiomers are in every other respect chemically identical

89
Q

chiral vs achiral

A

Chiral carbon = carbon atoms that are attached to 4 different substituents (asymmetrical)

Achiral compound = optically inactive

90
Q

optically active

A

-optically active = consists of more than 1 carbon atom and least 1 chiral carbon. Ability of a substance to rotate the plane of polarization of a beam

91
Q

by considering the mechanism formed explain why the product formed is optically inactive

A

-achiral compound
-polar C=O group
-attacked equall from either side
-racemic mixture formed

92
Q

state how to distinguish between 2 enantiomers

A

-plane polarized light
-rotate in opposite directions

93
Q

why does the structure not show stereoisomerism

A

doesnt have a chiral centre
symmetrical

94
Q

aldehyde pre-fix

A

formyl

95
Q

ketone pre-fix

A

oxo

96
Q

nitrile pre-fix

A

cyano

97
Q

optical isomerism

A

Optical isomerism occurs in carbon compounds with 4
different groups of atoms attached to a carbon (called
an asymmetric carbon).

98
Q

racemate

A

A mixture containing a 50/50 mixture of the two isomers (enantiomers) is described as being a racemate or racemic mixture.

A racemic mixture (a mixture of equal amounts of the two
optical isomers) will not rotate plane-polarised light.

Nucleophilic addition of HCN to
aldehydes and ketones (unsymmetrical)
when the trigonal planar carbonyl group
is approached equally from both sides by
the CN- attacking species: results in the
formation of a racemate (equal amounts
of both enantiomers)

99
Q

carboxylic acid solubility and acidity

A

The carboxylic acid are only weak acids
in water and only slightly dissociate, but
they are strong enough to displace
carbon dioxide from carbonates.

The smaller carboxylic (up to C4)
acids dissolve in water in all
proportions but after this the solubility
rapidly reduces. They dissolve
because they can hydrogen bond to
the water molecules.

100
Q

explain why NaBH4 reduces 2-methylbutanal has no reaction with 2-methylbut-1-ene

A

-hydride ion attackes C=O bond as it is attracted to the positive carbon
-double bond = electron dense so repels H- ion

101
Q

explain how the structure of ethananl leads to the formation of the two isomers

A

-ethananl = planar carbonyl group
-equal chance of attack from equal sides

102
Q

reflux (3 marK)

A
  • heating under reflux allows increase of temperature (boil)
  • volatile solvents, products and reactants are not lost
  • vapour returned converted back into liquid to return to mixture
  • any ethanol and ethanal having evaporated stay in mixture to be furtehr oxidised into ethanoic acid

Aqa a level chemistry (18 decks)

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