unit 4 Flashcards
(117 cards)
1
Q
complete the equation for reaction of sodium 2-hydroxybenzoate with soda lime (NaOH), which, on strong heating, can give phenol as one of the products
phenol with COO- Na+ + NaOH —>
[1]
A
phenol + Na2CO3
(accept C6H5ONa + NaHCO3)
2
Q
give the systematic name of a compound of formula C5H10O2 that gives propan-1-ol as one of the products when it is warmed with aqueous sodium hydroxide [1]
A
1-propyl ethanoate (accept propyl ethanoate)
3
Q
state the reagents necessary to convert hexane-1,6-diamine to hexane-1,6-dioic acid in two stages:
H2N(CH2)6NH2 —A—> HO(CH2)6OH —B—> HOOC(CH2)4COOH
A=
B=
[2]
A
A= nitric (III) acid / HNO2 / NaNO2 and HCl
B= acidified dichromate / Cr2O7 2- and H+
4
Q
- N,N-Diethyl-3-methylbenzamide, commonly known as DEET, has proved to be one of the most effective insect repellents
- DEET is a colourless liquid in white light
explain in terms of the electromagnetic spectrum, why DEET is colourless [1]
A
- it doesnt absorb in the visible region
5
Q
- an alternative insect repellant is dimethyl benzene-1,2-dicarboxylate, DMP
- this can be made by the oxidation of 1,2-dimethylbenzene, followed by acidification of the resulting product and then esterification
i) state the oxidising agent used [1]
ii) explain why it is necessary to acidify the product of this oxidation [1]
A
i) alkaline potassium manganate (VII) solution / MnO4 - + OH-
ii) as the reaction is carried out in alkaline solution the salt of the acid rather than the acid itself is produced
6
Q
- another insect repellent is the 1-butyl ester of 2-aminobenzoic acid, known as BA
- this can be produced from 2-aminobenzoic acid by esterification with butan-1-ol (and HCl catalyst)
- compound T (+NH3 Cl- coming off benzene ring) forms as an intermediate, before the addition of Na2CO3 is added to form 1-butyl 2-aminobenzoate
explain why compound T is initially produced in this reaction [1]
A
- the ester contains a basic -NH2 group which can accept a proton
7
Q
- the esterification of 2-aminobenzoic acid is a reversible reaction and eventually the mixture will reach equilibrium
- use the information below to suggest why that addition of hexane to the stirred aqueous mixture of reactants increases the yield of the product of BA
• hexane and the aqueous reaction mixture are immiscible
• 2-aminobenzoic acid is more soluble in water than in hexane
• BA is more soluble in hexane than in water
[3]
A
- as BA is produced it dissolves preferentially in the hexane
- this removes BA from the reaction and moves the position of equilibrium to the right
- 2-aminobenzoic acid remains largely in the aqueous alcohol mixture
8
Q
- the table shows the colour seen and the colour absorbed for a solution of Mordant Orange 1 (an azo dye) as the pH increases
- colour seen: yellow - orange red
- colour absorbed: violet - greenish-blue
- pH increasing ——————>
state how the wavelength of the light ABSORBED changes with pH, giving your reasoning [1]
A
- the wavelength of the light absorbed increases as the pH increases;
- in the visible spectrum, violet has the shortest wavelength and this increases as the colour moves towards red
9
Q
1,4-diaminobenzene can be made from 4-nitrophenylamine
state the reagent(s) necessary for this reaction [1]
A
tin (or iron) and concentrated hydrochloric acid
10
Q
state one difference between condensation polymerisation and addition polymerisation [1]
A
- in condensation polymerisation a small molecule e.g water/HCl is lost; this doesnt occur with addition polymerisation
- accept alternatives e.g different number of monomer types, functional groups, atom economy
11
Q
what is the molecular formula of oct-1-en-3-ol? [1]
A
C8H16O
12
Q
- 2,4-DNPH can be used as a reagent to identify aldehydes and ketones
- explain why 2,4-DNPH is an appropriate reagent to use in their identification [1]
A
- the derivatives formed using 2,4-DNPH have suitable melting temperatures for identification / precise melting temperatures
13
Q
- for a number of years CFC‘s were used in refrigeration and air-conditioning
- the use of these compound is in decline as they damage the ozone layer.
- a typical CFC is chlorotrifluoromethane
give an equation that shows the formation of a chlorine radical from chlorotrifluoromethane [1]
A
F3CCl —> F3C• + Cl•
14
Q
- refluxing a sample of a compound with aqueous sodium hydroxide did not show the presence of chloride ions in the resulting mixture
= no chloride ions on reflux with NaOH therefore chlorine atoms are bonded directly to benzene ring, not in alkyl side-chains
A
15
Q
suggest a simple laboratory method that would help to confirm that a compound is 2,4-D
A
- take its melting temperature and compare to a book value
- mix a sample with some actual 2,4-D, and take its melting temperature - see if the value is unchanged
16
Q
- a diagram shows a gas chromatography of a typical mixture
- the peak areas shown give the molar % and in this case also the % by mass of each component present in the mixture
-explain why the molar % of each compound CANNOT be used to find the % by mass of each component if the relative molecular mass of each component differs [1]
A
- if the relative molecular masses are different, the number of moles present in the chromatogram is not directly related tot he mass of each component
17
Q
A = 2-methylbutyl ethanoate - b.p = 134°C
C= pentyl ethanoate - b.p = 147°C
- suggest a reason for the difference in boiling temperatures between compounds A and C [1]
A
- A has weaker intermolecular forces therefore (less energy is needed to overcome these forces and) it has a lower boiling temperature
18
Q
- butylamine can be made from propan-1-ol in a three stage process
- give the reagent used and the structural formula of the product for each stage [3]
A
STAGE 1:
- KBr/H2SO4 - CH3CH2CH2Br
- KCN - CH3CH2CH2CN
- LiAlH4 - CH3CH2CH2NH2
19
Q
- chloroethanoic acid is produced from ethanoic acid by passing chlorine gas into it under suitable conditions
CH3COOH + Cl2 —> ClCH2COOH + HCl - this reaction involves radicles
- in the first stage chlorine radicals are formed
- they then attack the ethanoic acid molecule producing a molecule of hydrogen chloride and a new radical
- suggest the formula of the new radical formed from the acid during this reaction [1]
A
•CH2COOH
20
Q
- aminoethanoic acid is isolated from a solution of products by adding methanol to the liquid
- the acid is precipitated as a white solid
- suggest why aminoethanoic acid is insoluble in methanol [2]
A
- it exists as zwitterions / ionic compound
- ionic compounds are not (generally) soluble in covalent solvents
21
Q
explain why an aqueous solution of aminoethanoic acid doesnt affect the plane of plane polarised light [1]
A
- it doesn’t contain a chiral centre / asymmetric carbon atom
22
Q
- hydroxyethanoic acid, CH2(OH)COOH, can also be prepared by reacting chloroethanoic acid with aqueous sodium hydroxide
- explain why the reaction mixture needs to be acidified to produce hydroethanoic acid [1]
A
- the (sodium) salt of the acid is formed
23
Q
- alkenes react with ozone to give an intermediate compound, which can then react further to give aldehydes or ketones
- 1 mol of an alkene T reacts with ozone to give 2 mol of the same carbonyl compound, U
- explain what can be deduced about the structural formula of alkene T [1]
A
- either:
- the groups bonded to each carbon atom of the C=C must be the same
- it must by a symmetrical alkene
24
Q
NOT TO ANSWER JUST TO READ - INFO
- the melting temperatures of the 2,4-dinitrophenylhydrazine derivtives can be used to identify the original carbonyl compound
- the melting temperatures of some of these derivatives are shown in a table
- the 2,4-dinitrophenylhydrazine derivative of compound U was slightly impure and melted at a temperature of a 110-113°C
- use the information given to deduce the formula of compound U
A
- melting temperature cannot be higher than the literature value
- so compound U is the compound with melting temperature 106°C
25
- if asked to draw an alkene, draw it as an E/Z isomer
26
what are two factors to consider when choosing what method to produce the same product [2]
- cost
- temperature conditions
- pressure used
- yield
- rate
- separation of products
27
- the densities of biodiesel and propane-1,2,3-triol are shown in the table
- these two liquids are immiscible (insoluble in each other)
- biodiesel - density = 0.85 gcm^-1
- propane-1,2,3-triol - density = 1.26gcm^-1
- sketch and name a piece of apparatus that could be used to separate a mixture containing 25cm^3 of each liquid
- the two liquids should also be labelled [2]
- name = separating funnel
- biodiesel as top layer, propane-1,2,3-triol as bottom layer
28
- the visible spectrum of an azo dye at a certain pH shows an absorption maximum at 410nm
- calculate the frequency that corresponds to this wavelength [2]
- f = c / λ
- 3.00 x10^8 / 410 x10^-9
= 7.32 x10^14 Hz
29
suggest why an amide can act as a base [1]
- it contains a NITROGEN atom that has a lone pair of electrons / is a proton acceptor
30
- when ethanamide is heated with a dehydrating agent the organic product is ethanenitrile
- use the data booklet to state and explain how characteristic infrared absorption values change during the reaction
- in your answer you should relate the relevant bonds to their corresponding absorption frequencies [2]
- the C=O absorption at 1650-1750cm^-1 decreases / the N-H absorption at 3300-3500cm^-1 decreases
- the C≡N aborption at 2100-2250cm^-1 increases
31
- some amines can be made by the reaction of a chlorine-containing compound and ammonia
- for example (chloromethyl)benzene reacts with ammonia to give phenylmethylamine (+NH4Cl)
- suggest why phenylamine cannot easily be made by this method from chlorobenzene and ammonia [1]
- the benzene ring / negative electron cloud is not (easily) susceptible to attack by nucleophiles
- the benzene ring / negative electron cloud is usually attacked by electrophiles
- the C-Cl bond in chlorobenzene is stronger than the C-Cl aliphatic bond
32
- 2-hydroxy-5-nitrobenzenecarboxylic acid is made by reacting 2-hydroxybenzenecarboxylic acid and aqueous nitric acid
- a basic outline of the method is given below:
- add 4.00g of 2-hydroxybenzenecarboxylic acid to 100cm^3 of nitric acid of concentration 2moldm^-3
- heat the mixture to 60°C for 10minutes
- as the reaction proceeds brown fumes containing toxic nitrogen dioxide are produced
- add the products to 300cm^3 of cold water
- a yellow solution and white crystals of 2-hydroxy-5-nitrobenzenecarboxylic acid are obtained
.
- use this outline method to produce a more detailed method suitable for others to use to be able to produce dry white crystals of the acid
- as part of your answer you should give the type and size of apparatus used and mention any necessary health and safety factors
- details of subsequent recrystallisation are NOT required [6QER]
- flasks or beakers - suitable sizes
- use of a fume cupboard
- use of thermometer and method of maintaining 60°C temperature
- stirring
- filtration
- filtered crystals washed with water
- until filtrate is colourless
- reaction flask washed with water and contents added to filtering apparatus
- crystals removed and dried in a warm oven / at room temperature/ below the melting temperature of the product
33
- a student was given a sample of a polyester P
- he was told that the repeating unit was as follows -O-C=O-⏣-C=O-O-CH2-⏣-CH2-
- polyesters such as compound P can be hydrolysed by heating with aqueous sodium hydroxide
- complete the balanced equation for this hydrolysis:
-O-C=O-⏣-C=O-O-CH2-⏣-CH2- + __NaOH
+ 2NaOH —>
Na+ O- -C=O-⏣-C=O- O- Na+ + HOCH2-⏣-CH2OH
34
DONT ANSWER JUST INFO:
- gas chromatography can be used to separate the dimethylbenzene isomers
- one of the dimethylbenzene isomers has the 13C NMR spectrum below
- explain how this spectrum shows that the isomer is 1,3-dimethylbenzene [3]
- 5 environments correctly identified from the 13C NMR spectrum and indicated on drawn structure of 1,3-dimethylbenzene
- signals at 126-136 δ identified as aromatic
- signal at 20 δ identified as aliphatic / CH3
35
- the isomer of dimethylbenzene used to produce polyester P is 1,4-dimethylbenzene
- in industry this isomer is oxidised by air to produce benzene-1,4-dicarboxylic acid
HOOC-C6H4-COOH
- during this reaction a number of other aromatic oxidation products are made
- one of these has the empirical formula C4H3O
- suggest a structure for this product giving your reasoning
- state how it could be produced in the reaction between 1,4-dimethylbenzene and atmospheric oxygen [5]
- must be an aromatic compound so at least 6 carbon atoms
- molecular formula therefore likely to be C8H6O2
- there are 2 additional carbon atoms therefore likely to be a dialdehyde OR cannot be acid-aldehyde or alcohol-aldehyde or acid-alcohol as molecular formula doesnt fit
- made by partial oxidation (of each methyl group)
- formula is CHO - ⏣ - CHO
36
- after use, a bottle containing decanedioyl dichloride can become contaminated with decanedioic acid
- suggest how decanedioyl dichloride had become contaminated with dicanedioic acid [1]
- moisture / water gained entry to the bottle and hydrolysed the decanedioyl dichloride when it was previously opened
37
- ethane reacts with chlorine by free radical substitution
- explain why ultraviolet light is needed for this reaction and give an equation for the formation of chloroethane [2]
- ultraviolet light is needed to break the Cl-Cl bond / produce chlorine free radicals
C2H6 + Cl2 —> C2H5Cl + HCl
38
- the amide urea, CO(NH2)2 is made in industry by reacting ammonia and carbon dioxide
- suggest a reagent that could be used to produce ammonia from the amide urea, when the mixture is heated [1]
sodium hydroxide /NaOH / OH-
39
1,4-dimethylbenzene was oxidised to benzene-1,4-dicarboxylic acid
- state a suitable oxidising agent, used in alkaline solution, for this reaction [1]
potassium manganate (VII) / manganate (VII) / KMnO4 / MnO4 -
40
- a sample of 1-butyl butanoate (boiling temperature 164°C) is contaminated with around 10% of butanoic acid (boiling temperature 162°C)
- describe how you would obtain a pure sample of 1-butyl butanoate from this mixture [3]
- add (excess) aqueous sodium carbonate / sodium hydrogencarbonate (to neutralise the acid)
- separate off the aqueous layer (using a separating funnel)
- distill the organic layer (collecting the ester at 164°C)
41
- the reaction mixture in an experiment was gently warmed to start the reaction and the rate of exothermic reaction was controlled by appropriate cooling
- suggest a simple method of heating and subsequent cooling of the reaction mixture [1]
hot AND cold water baths
42
- a by-product of a reaction was oxopropanedioic acid
COOH
I
C = O
I
COOH
- state and explain how the C13 NMR spectrum of this compound would differ from that of ethanedioic acid [2]
- ethanedioic acid - both carbon atoms are in equivalent environments - therefore one signal
- oxopropanoic acid - 2 carbon atoms are in the same environment and the other carbon atom is in a different environment - therefore two signals
43
state the reagent used to produce N-phenylethanamide from phenylamine [1]
ethanoyl chloride / CH3COCl
44
- N-phenylethanamide is nitrated by using using a mixture of concentrated nitric and sulfuric acid
- by analogy with the nitration of benzene suggest how you could modify the nitration reaction to ensure that the dinitro-product is formed [1]
- run the reaction at a higher temperature / above 50°C
- use the excess of the nitrating agent
- add the N-phenylethanamide to the nitrating mixture rather than the reverse addition
45
- 2,4-dinitrophenylamine has the melting temperature of 180°C and is recrystallised from a 1:1 mixture of ethanol and water
- outline the essential stages of this recrystallisation process so that pure dry crystals of 2,4-dinitrophenylamine are obtained
- you should assume that a hot water bath is available for use [4]
- dissolve in MINIMUM volume of the HOT solvent
- filter hot if necessary
- allow to cool
- filter, (wash) and dry at <180°C / room temperature / window sill / warm oven
46
- another use for the 2,4-dinitrophenylamine is the preparation of the azo dye ‘permanent red 2G’
- this is prepared from the diazotisation of 2,4-dinitrophenylamine and then coupling the product with naphthalen-2-ol
- state the reagents and conditions used to produce an aqueous solution of 2,4-dinitrobenzenediazonium chloride from 2,4-dinitrophenylamine [2]
any of the following:
- nitrous acid
- HNO2 / HONO
- sodium nitrite / sodium nitrate (III) + hydrochloric acid
- NaNO2 + HCl
.
- temperature of 10°C or less
47
- permanent red 2G appears red in white light
- state and explain the colour observed, if any, when Permanent Red 2G is viewed in:
a) blue light
b) red light
a) black/ no colour - as there is no red light to be transmitted / all the light is absorbed
b) red as red light is not absorbed
48
- there are environmental concerns about the use of herbicides based on 2,4-dinitrophenylamine
- naturally occurring herbicides, such as limonene, are considered to be more acceptable
- limonene is an unsaturated compound that contains two carbon carbon double bonds in each molecule
.
a) - in an experiment 4.08 g of linonene just removed the colour from 9.58 g of bromine
- use this information to calculate the relative molecular mass of limonene [3]
b) when limonene reacts completely with hydrogen under suitable conditions, menthane is produced
- use the information from the question and answer to part a) to find the relative molecular mass of menthane [2]
a) - limonene contains two C=C double bonds therefore 2 mol of Br2 react with 1 mol of limonene
- therefore 2 x 159.8g / 319.6g bromine react with 1 mol of limonene
- moles of bromine used = 9.58/159.8 = 0.060
- moles of limonene = 0.030
- Mr of limonene = 4.08 / 0.030 = 136
b) - there are two C=C bonds present therefore two moles of H2 are needed for full hydrogenation
- the Mr therefore increases by 4, hence 140
49
- secondary protein structure is concerned with hydrogen bonding between certain groups
- state the groups present that take part in this hydrogen bonding and explain why it occurs [2]
- groups that take part in hydrogen bonding are N-H and C=O
- this occurs because of polarisation in the bonds (δ- N-H δ+)
50
- there are two common methods of producing 1-butyl ethanoate
- butan-1-ol can be reacted with either ethanoyl chloride or ethanoic acid
- both of these reactions can be considered as equilibrium reactions:
-CH3CH2CH2CH2OH + CH3COCl ⇌ CH3COOCH2CH2CH2CH2CH3 + HCl
- CH3CH2CH2CH2OH + CH3COOH ⇌ CH3COOCH2CH2CH2CH2CH3 + H2O
a) use the equations to suggest why the method using ethanoyl chloride gives a higher yield of 1-butyl ethanoate than the method using ethanoic acid [3]
b) suggest ONE practical disadvantage of the method that uses ethanoyl chloride [1]
a) hydrogen chloride is produced as a gaseous co-product, this is lost from the reaction
- moves the position of equilibrium to the right
- the method using ethanoic acid produces water as a co-product and the reaction reaches a position of equilibrium
b)
- hydrogen chloride (gas) is irritant
- the reaction needs to be done in fume cupboard
- the reaction needs anhydrous conditions
51
- when ethene is passed into aqueous chlorine one of the products is 2-chloroethanol
- the presence of the C-Cl bond in this compound can be shown by a simple test tube reaction
- outline the practical steps that are used in this method
- you can assume that 2-chloroethanol is soluble in water [4]
- warm 2-chloroethanol with aqueous sodium hydroxide
- acidify the mixture with (aqueous) nitric acid
- add aqueous silver nitrate
- white precipitate (of silver chloride) confirms the presence of a C-Cl bond
52
- over the last 30 years the production and use of CFCs has declined, mainly because of their adverse environmental effects
- using fluorotrichloromethane as your example, outline how the compound causes a reduction in the amount of ozone present in the upper atmosphere and state TWO problems that can be causes by this reaction [6QER]
- in the upper atmosphere a C-Cl bond is broken by UV radiation
- giving a chlorine atom / radical
- CFCl3 —> •CFCl2 + Cl•
- chlorine radical attacks ozone
- Cl• + O3 —> ClO• + O2
- the chlorine radical is regenerated, therefore reaction is catalytic
- ClO• + O3 —> Cl• + 2O2
- environmental effects, for example
- skin cancer
- plant death
- global warming
53
state what is meant by a chiral centre [1]
- an atom that has four different groups or atoms bonded to it
54
- butan-2-ol (Mr 74) can be dehydrated to but-1-ene and but-2-ene
- state a dehydrating agent for this reaction [1]
- (concentrated) sulfuric acid / aluminium oxide
55
- benzene only undergoes electrophilic substitution
- ammonia (a nucleophile) is going to be repeller by benzene because both electron negative
56
why is benzene more stable?
- delocalisation between three C=C in benzene
- intraction between these C=C
- interactions stabilises the molecule
57
why is 2,6-dichlorophenol a stronger acid than ethanol?
- the ion formed from phenol is able to delocalise -ve charge throughout the ring
- enthanol has no opportunity for delocalisation so charge localised on 1 atom
(ethanol isnt acidic because it cant release a proton/H+)
58
if something forms an ion, the ion must be more stable
59
out of water, carboxlic acids, phenols and alcohols, which is most acidic?
c/a > phenols > water > alcohols
60
why is a tertiary carbocation more stable?
- because carbons push electrons away towards e.g N
61
something is more likely to be stable if the -ve charge can spread around the compound / if can delocalise
62
a nucleophile e.g OH- cant get close to benzene ring to attack it because
63
what do both phenols and carboxylic acids react with strong bases e.g NaOH to make?
salts
64
why is phenol more reactive than benzene? e.g why does Br2 react with phenol to form a white ppt and decolourises orange bromine water, but has no reaction with benzene
- the lone pair of electrons on O in phenol is shared around ring
- which increases the electron density in the ring
- so increases the chance to attack bromine which is an electrophile
.
- the increased electron density activates the ring towards electrophilic attack
65
explain why 3-aminopentane is a stronger base than ammonia
NH3 vs CH3CH2CNH2CH2CH3
- a base accepts protons / if an increase in lone pairs or lone pairs more electron dense, its a better base
.
- 3-aminopentane is a stronger base because carbons push electrons away
- so push towards N
- the lone pair is more available in 3-aminopentane because the carbon group is electron releasing
66
what acid is usually used as a reagent to protonate something?
sulfuric acid / H2SO4
67
what are the temperatures needed in the carbonyl tests?
- tollens = needs to warm it
- fehlings = needs to warm it
(iodoform is done in alkaline conditions)
68
double angles are planar
69
- why does a racemic mixture form?
- because double bonds are planar
- CN- can attack below the plane or above the plane
- there’s an equal chance of attack from CN- above and below the plane, resulting in an equal chance of both optical isomers forming
70
draw a structure for a compound of formula C3H8O that shows an infrared absorption peak between 1000 and 1300cm^-1 but no peak between 2500 and 3550 cm^-1, other than that at 2800-3100cm^-1 corresponding to C-H bonds
- give your reasoning [2]
CH3CH2OCH3
- C - C - O - C - H -
- it had a C-O bond (at 1000-1300cm^-1) but no O-H bond (at 2500-3550cm^-1) so it cant be an alcohol
71
- benzene-1,2-dicarboxylic acid (or its sodium salt) is heated with sodalime
- state the type of reaction that occurs and give the name of the product [2]
- decarboxylation
- benzene
72
draw the structure of a compound that had an empirical formula CH4N [1]
either:
H2N-CH2-CH2-NH2
H3C - N(H) - N(H) - CH3
(CH3)2N - NH2
73
- briefly outline how a mixture could be ‘heated under reflux at 100°C’ [2]
- steam / boiling water bath / heating mantle / should be used to heat the mixture
- condenser should be attached vertically to the flask so that the cold water jacket condenses the vapour and returns them to the flask
74
state how a method implies that glucose penthaethanoate is insoluble in water [1]
the compound is PRECIPITATED when the mixture is poured into a large excess of water
75
- during recrystallisation glucose pentaethanoate is dissolved in the ‘minimum quantity of hot ethanol’
- why was the minimum quantity used? [1]
so that the maximum amount of glucose penthaethanoate can crystallise out on cooking
76
- ethanol can be made from glucose by fermentation
C6H12O6 —> 2C2H5OH + 2CO2
- in an experiment 0.2 mol of glucose was dissolved in water and fermented using yeast
- at the end of the reaction all the glucose had been converted into ethanol and carbon dioxide
- if the volume of the resulting solution was 2.03 dm^3 calculate the concentration of ethanol in gdm^-3 [3]
- 0.200 mol of glucose gives 0.400 mol of ethanol
- 0.400mol ethanol in 2.03dm^3
- 18.4 g in 2.03dm^3
- therefore concentration of glucose is 9.06g dm^-3
77
- the characteristic smell of some varieties of blue cheese is you mainly to heptan-2-one
CH3CH2CH2CH2CH2-C(=O)-CH3
CH3CH2CH2CH2-C(=O) - CH2CH3
- a teacher was asked whether hepan-2-one could be clearly distinguished from heptan-3-one by some suggested methods
- his responses are shown in the table:
- method vs clearly identified?
• mass spectroscopy = yes
• gas liquid chromatography = no
• boiling temperature = yes
• chemical analysis for C, H and O = no
• reaction with alkaline iodine = yes
• reaction with tollens’ reagent = no
- explain why each response is correct [6QER]
MASS SPECTROSCOPY
- each compound has its own mass spectrum
- although the molecular ion will be the same (at 114) the fragmentation pattern of each compound will be different
GAS LIQUID CHROMATOGRAPHY
- the retention times for both compounds are the same
BOILING TEMPERATURES
- the boiling temperatures of the two compounds must be different
CHEMICAL ANALYSIS
- since both compounds have the same molecular formula, C7H14O, the elemental analysis for each compound will be the same
REACTION WITH ALKALINE IODINE
- only methyl ketons will give a positive test, hence only hepan-2-one will react in this way
WARMED WITH TOLLENS REAGENT
- neither compound is an aldehyde, so no silver mirror is see
78
- a reaction sequence for obtaining pentane from pentan-2-one is shown below:
pentan2-one —LIALH4–> compound F —H2SO4–> CH3CH2CH2CH2=CH2 OR CH3CH2CH2=CHCH3
- state the role of sulphuric acid when it reacts with compound F and explain why two products are formed [2]
- the acid acts as a dehydrating agent
- water can be eliminated from either side of the -CH(OH)- group
79
- ethanoic acid is more acidic than phenol
- state a chemical test that you could use in the lab to show this difference in acidity
- you are given an aqueous solution of each compound [2]
- add NaHCO3 / Na2CO3
- ethanoic acid produces effervescence / gives of carbon dioxide, phenol doesnt
80
suggest why 4-nitrophenol is much more acidic than phenol [1]
the anion formed is more stable than the phenoxide ion / the O-H bond is weakened by the presence of the NO2 electron withdrawing group
81
- describe the structure and bonding in benzene
- explain why benzene is resistant to addition reactions and why its usual mode of reaction is by substitution [4]
- benzene exists as a six membered (planar) ring
- it has a molecular formula C6H6
- it has a delocalised electron structure/pi cloud
- stability is lost if addition occurs, hence substitution is the usual reaction
82
- the chlorination of methylpropane produces a number of polychlorinated alkanes
- state how these polychlorinated products are formed [1]
attack by a chlorine radical / atom on a chlorinated alkane product
83
explain why a compound can only show a singlet on a high resolution NMR spectrum
- all the protons are equivalent/in equivalent environments
84
give the equation for the complete combustion of methylpropane [1]
(CH3)3CH + 13/2 O2 —> 4CO2 + 5H2O
85
stage 1 is an example of a HYDROLYSIS reaction
- state what is meant by this term [1]
a reaction that involves DECOMPOSITION by WATER
86
- a mixture of ammonium sulfate in solution, together with a precipitate of an organic acid is formed by a reaction
- state how you would obtain pure crystals of the acid from this mixture [1]
filter, wash (with water) and dry
87
- the melting temperature of a compound is 209°C
- state how this figure would change if the sample of the compound was impure [1]
- melts at a lower temperature / melts over a range of temperatures
88
for tollens, when aldehyde present, the aldehyde reduces Ag+ ions to Ag
89
explain why 2,6-dichlorophenol does NOT react with aqueous sodium hydroxide to give compounds such as benzene-1,2,3-triol [1]
- reaction would involve nucleophilic attack on the phenol by an OH-
- the ring is electron rich and not susceptible to attack by nucleophiles
90
- hexane-1,6-diol reacts with hexane-1,6-dioic acid to give a polyester
- state why this is a condensation reaction [1]
small molecule / water is eliminated
91
differences in boiling temperatures could be due to:
- longer carbon chains = more VdW forces
- hydrogen bonding vs VdW
- the more branches the chain, the lower the boiling temperature
92
explain why a dye is yellow in white light [1]
- because the blue end / the rest of the visible spectrum is absorbed
93
give the formula of the triphenylmethyl radical [1]
(C6H5)3C•
94
if an atom economy question says to an appropriate number of significant figures, go to 3 s.f
95
- nitrobenzene can be made by the nitration of benzene
- state the formula of the nitrogen-containing electrophile that takes place in this reaction [1]
NO2 +
96
- compound M reacts with an excess of alcoholic potassium hydroxide solution to give compound N, which has the formula R-C≡C-R’
- suggest the role of the alcoholic potassium hydroxide in this elimination reaction [1]
it is acting as a base / it removes protons
97
green chemistry:
- percentage yield
- reaction time / time needed
- need of catalyst
- energy cost
- temperature
98
explain why this polymer is described as an addition polymer and not as a polyester [2]
- the polymer is made from an alkene / the monomer has a C=C / no additional compound is formed (when polymerisation occurs) (1)
- a polyester contains a C(=O)O group in the chain / there is no ester LINKAGE / a polyester is made from an alcohol and a carboxylic acid / acid chloride
99
- when NaOH react with 4-aminobenzamide, only the amide group reacts
- state the nature of the attacking reagent and explain why the NH2 group bonded to the benzene ring is not removed [2]
- the attacking reagent /OH- is a nucleophile
- lone pair on N becomes part of delocalised system / C-N bond is stronger when directly attached to ring
100
- write the formula of the zwitterion of tyrosing, then:
tyrosine is described as a hydrophobic amino acid, as its solubility in water is very low. use the formula of tyrosine to give a reason of thsi low solubility [1]
- the polar structure / OH group is a small part of the overall molecule so hydrogen bonding is at a ‘minimum’
101
- an experiment was repeated with a different form of tyrosine. this batch gave off less nitrogen gas under the same conditions.
- suggest one reason for this different result [1]
- not all the nitrogen was collected
- impure sample of tyrosine
- incomplete reaction
102
when compound A is made by the method in part ii) above, an equimolar mixture of both enantiomers is produced
- state how a solution of this mixture affects the plane of plane polarised light [1]
it doesnt rotate the plane of plane polarised light
103
- 1,2-diaminoethane reacts as a base
- explain how this compound acts as a base [1]
- contains a NITROGEN atom that has a LONE PAIR OF ELECTRONS (which can accept a proton)
104
benzene-1,4-dioic acid + NaOH —>
benzene, with COO- Na+ either end
105
the solubility of unsaturated carboxylic acids in water decreases as the chain length increases
- suggest why [2]
- polar OH / COOH groups dissolve by hydrogen bonding
- but as the hydrocarbon chain length increases the polar groups form an increasingly smaller part of a largely non-polar molecule
106
an acid chloride + H2O —> carboxylic acid + HCl
- state the type of reaction occurring [1]
hydrolysis
107
- benzene reacts with bromine in the presence of a catalyst
- give the mechanism for the reaction and explain how the Br-Br bond becomes polarised during the reaction
- suggest why, in the absence of this catalyst, there is very little reaction between benzene and bromine under normal conditions [6QER]
- the Br-Be bond becomes polarised (δ+ and δ-) by
- electron movement / attraction from the pi electron system
- polarisation from the ‘electron deficient’ iron atom in FeBr3
- mechanism shows the catalyst FeBr3 regeneration
- mechanism shows the production of the co-product, hydrogen bromide
- in the absence of a catalyst, there is reduced / little induces polarisation of the bromine molecule by the pi electron system
108
- explain how you could use a chemical test to distinguish between the following two compounds
- chloromethylbenzene vs chlorobenzene
- chlorobenzene has delocalisation, so C-Cl bond strengthened, because contains partial bond character
- if do silver nitrate test:
- add NaOH to liberate Cl- (in A not B)
- because the C-Cl in A is weaker and because the OH- is a nucleophile, so in B would be repelled by benzene ring and cant attack
- add nitric acid
- add AgNO3
- white ppt for A not for B
109
if it asks about purity, either recrystallisation or gas chromatography
but specifically:
- if separation = gas chromatography
- if purify = recrystallisation
110
gas chromatography separates mixture of substances
111
whats a problem with gas chromatography?
- if have a mixture of two things with similar polarity, they will come out of gas chromatogram at same time
112
- a student decided to take a crude product and make it pure
- describe the stages she would take, and how she could test the purity of the substance to ensure it was pure
- min volume of hot solvent
- filter it hot —> remove any insoluble impurities
- allow to cool and crystallise
- wash crystals with ice cold solvent
- dry in a warm oven below melting point until constant mass
- to test purity:
if melting temperature lower or over a range, is impure
113
- explain why the following reaction results in the formation of a racemic mixture
- your answer should include references to the mechanism that is taking place:
CH3-C(=O)H —> CH3-C-OH-H - C ≡N
- C=O is planar
- so CN- can attack above or below
- so 50:50 chance of attacking above or below
- so 50% 50% each enantiomer
(if got bulky group its harder to attack from that side / ground that surround the C=O)
114
if Cl/Br on benzene ring, doesn’t get substituted by e.g a nucleophilic substitution
115
- azo dye = diazonium ion + some phenol
- something is an azo dye if: N=N
116
if area of 2 on peak on H NMR, usually CH2
117
what are the conditions for coupling reactions:
- below 5°C
- alkaline (NaOH)
