Organic chem - analysis Flashcards
the molecular (parent) ion in the mass spectrum of a hydrocarbon containing 13C and H only
is the peak with highest mass to charge ratio
in a mass spectrum of butane C4H10 where would a peak be seen for the molecular ion if it has a charge of 2+
29
suggest what issue would arise when looking at the mass spec of propanal and propanone
-the molecular ion peak would be the same as they have the same mr (molecular mass)
deduce how many m/z values will be produced in the mass spectrum of butanone
CH3+
CH3CH2+
CH3CH2C=O+
CH3CH2C=OCH3+
state the meaning of the term molecular ion
an ion with the highest mz value
state why the precise RAM for the 12C isotope is exactly 12.00000
by definition
explain why there are two molecule ion peaks
chlorine has 2 isotopes
purpose of mass spec
-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
Br isotope
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
what would happen if there was a 2+ charge on a peak
+2 charge on mass spec graph would cause the m/z ratio to halve e.g 160/2 = 80
molecular ion
-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
fragmentation
-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
fragmentation pattern
-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
readily bonds breaking
-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
M/Z and fragment
15
CH3+
17
OH+
29
CH3Ch2+ or CHO+
43
CH3CH2CH2+ or CH3CO+
31
CH2OH+
which compound forms a molecular ion with a different precise molecular mass from the other three
dimethylpropane
what is the angle of C-C-C and C-O-H bond
C-C-C = 109.5
C-O-H = 104.5
suggest what you would expect to see when cyclohexanol reacts with sodium
fizzing
white solid forms
give a possible reason why there is no molecular ion peak in the mass spectrum of 2-methylpropan-2-ol
molecular ion is unstable
amine formula
N2H
How can we distinguish between propan-1-ol and propan-2-ol
different fingerprint regions -> compare absorption space below 1500
suggest two possible identities of C. Justify your answer and explain how you could further confirm the exact identity of C
propanal or propanone
peak at 1700 due to C=O
use fingerprint region to compare literature
IR spec
-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
detecting the absorption
-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
How to read the spectrum
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
fingerprint region
-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
broad vs narrow peak
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)
peak for carboxylic acids
Carboxylic acids -> broad peak at 3000-3500 due to OH and narrow peak at 1700 due to C=O
peak for alcohols
Alcohols –> broad peak at 3500 due to OH and 1100 due to C-O
peak for aldehydes and ketones
Aldehydes and Ketones –> peak at 1700 due to C=O
peak for alkenes and alkanes
Alkenes –> peak at 1650 due to C=C
Alkanes –> C-H peak at 2850 – 3000
same molecular formula =
same mr
suggest why these granules prevent bumping
form smaller bubbles
no ozone depletion =
doesnt contain Cl
tests
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
explain why determinging the precise relative molecular mass could not be used to distinguish between the two compounds
-same molecular formula so their mr value would be the same
explain why 1,1,1,-triflouroethane does not lead to the depletion of the ozone
-doesnt contain C-Cl bonds so cant form chlorine radicals
Z isomers
according to priorty of mr first!!!
explain why the water should enter the condeser at the bottom and not the top
-to ensure full condensation as more of the condenser fills with water
why does rate of decomposition of gas increase at higher temperatures
-more Ea
-more frequenct and successful collisions
test for aldehyde + tollens reagent
aldehyde = fehlings = blue solution to red precipitate
aldehyde = tollens = silver mirror
carboxylic acid + sodium carbonate
effercvescence of CO2
CH3CHO
ethanal
propanone
(CH3)2CO (smallest possible ketone)
propanal
CH3CH3CHO
reduction of propanal equation
Reduction of propanal:
-CH3CH2CHO + 2[H] –> CH3CH2CH2OH
nucelophilic addition
-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.
carbonyl compounds
-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
bonding in carbonyl compounds
-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.
reaction of carbonyls
-The C=O bond is polarised because O is more electronegative than carbon. The positive carbon atom attracts nucleophiles.
Aldehyde + [O] –> carboxylic acid
tollens reagent
-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+
fehlings solution
-Blue Cu2+ ions
-heat gently
-copper ions reduced to copper oxides
-form carboxylic acid
-CH3CHO + 2Cu2+ + 2H2O –> CH3COOH + Cu2O + 4H+
reduction of carbonyls into alcohol
-reagent = NaBH4 in aqueous ethanol
-conditions = room temperature and pressure
-aldehydes reduced to primary alcohol
-ketones reduced to secondary alcohol
catalytic hydrogenation
-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
addition of hydrogen cyanide to carbonyls to form hydronitriles
-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
hazards + nucelophilic addition
-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
carboxylic acids and derivatives
-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)
esters
-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
benzene ring
-Benzene ring –> Delocalisation of p electrons makes benzene more stable than the theoretical molecule cyclohexa-1,3,5-triene
electrophilic subsitution
-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.
condensation polymers
-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
ethanedioic acid
carboxyl group on both carbons
acidity of carboxylic acid
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)
delocalisation
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.
strength of carboxylic acids
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)
test for carboxylic acids
solid Na2CO3 = production of CO2 (effervescene)
methanoic acid can be oxidised into
carbonic acid (H2CO3) which decomposes to give CO2
esterification
carboxylic acid + alcohol –> ester + water
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.
ethanoic acid + ethanol –> ester + water
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
polymers + esters
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.
how can esters be hydrolysed
-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.
solubility
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.
functional group for ester
-ester = O-C=O (synthetic aromatics)
production of an ester
Alcohol + carboxylic acid –> ester + alcohol
e.g ethanol + propanoic acid –> ester + H2O
naming esters
-first part of the name = alcohol = yl
-last part of the name = oate = COOH
Ethylethanoate
Propanol + butanoic acid = propylbutanoate + water
polarity
More polar = more soluble
Longer chain = less soluble
tollens and Fehlings
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
reduce aldehydes and ketones back into an alcohol
Reduce aldehydes and ketones back into an alcohol using reducing agent (NaBH4)
e.g propan-2-one + 2[H] –> propan-2-ol
reducing agents =
hydride
H-
reduction of carbonyls
Reagents: NaBH4 In aqueous ethanol
Conditions: Room temperature and pressure
when can nucelophilic subsitution occur
-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
nucelophilic addition
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
addition of hydrogen cynaide to carbonyls to form hydroxynitrtiles
-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
aldehyde =
end of carbon chain
true or false = nucelophile is repelled by C=C
true
naming nitriles
-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.
racemic mixture
-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
enatinomers
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
chiral vs achiral
Chiral carbon = carbon atoms that are attached to 4 different substituents (asymmetrical)
Achiral compound = optically inactive
optically active
-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
by considering the mechanism formed explain why the product formed is optically inactive
-achiral compound
-polar C=O group
-attacked equall from either side
-racemic mixture formed
state how to distinguish between 2 enantiomers
-plane polarized light
-rotate in opposite directions
why does the structure not show stereoisomerism
doesnt have a chiral centre
symmetrical
aldehyde pre-fix
formyl
ketone pre-fix
oxo
nitrile pre-fix
cyano
optical isomerism
Optical isomerism occurs in carbon compounds with 4
different groups of atoms attached to a carbon (called
an asymmetric carbon).
racemate
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)
carboxylic acid solubility and acidity
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.
explain why NaBH4 reduces 2-methylbutanal has no reaction with 2-methylbut-1-ene
-hydride ion attackes C=O bond as it is attracted to the positive carbon
-double bond = electron dense so repels H- ion
explain how the structure of ethananl leads to the formation of the two isomers
-ethananl = planar carbonyl group
-equal chance of attack from equal sides
reflux (3 marK)
- 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
NaOH produces blue ppt =
copper present
A student is testing a salt labelled ‘compound Z’.
Testing for cation: Flame tests of Z give a brick red colour.
Testing for anion: When dilute hydrochloric acid is added to Z, bubbles of gas are evolved. When this gas is bubbled through limewater, the limewater turns cloudy.
CaCO3
A student is testing a salt labelled ‘compound A’.
Testing for cation: Flame tests of A give a brick red colour.
Testing for anion: When dilute hydrochloric acid followed by barium chloride solution is added to A, a white precipitate forms.
CaSO4
flame test gives green colour =
barium chloride
The student adds excess dilute sodium hydroxide, and the white precipitate dissolves
aluminium (clourless solution)
amine base strength =
depends on nitrogens lone pair avaliability + mention positive inductive effect
recrystallisation
-dissolve impure lidocane in hot hexane
-filter mixture while hot
-remove any insoluble impurities
-all solution to cool to crysallise
-filter under reduced pressure
-remove soluble purities
-wash
nitric acid =
prevents precipitation of other silver compounds
why can electrospray ionisation not be used to distinguish between the 2 isomers
all have the same mr
