Q1.          (a)     Name compound Y, HOCH2CH2COOH
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(1)
(b)     Under suitable conditions, molecules of Y can react with each other to form a polymer.
(i)      Draw a section of the polymer showing two repeating units.

 

(1)
(ii)     Name the type of polymerisation involved.
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(1)
(c)     When Y is heated, an elimination reaction occurs in which one molecule of Y loses one molecule of water. The organic product formed by this reaction has an absorption at 1637 cm–1 in its infrared spectrum.
(i)      Identify the bond that causes the absorption at 1637 cm–1 in its infrared spectrum.
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(1)
(ii)     Write the displayed formula for the organic product of this elimination reaction.

 
 
(1)

(iii)     The organic product from part (ii) can also be polymerised.
Draw the repeating unit of the polymer formed from this organic product.

(1)
(d)     At room temperature, 2-aminobutanoic acid exists as a solid.
Draw the structure of the species present in the solid form.

 
 
(1)
(e)     The amino acid, glutamic acid, is shown below.

Draw the structure of the organic species formed when glutamic acid reacts with each of the following.
(i)      an excess of sodium hydroxide

 
(1)
(ii)     an excess of methanol in the presence of concentrated sulfuric acid

 
(1)
(iii)     ethanoyl chloride

 
(1)

(f)      A tripeptide was heated with hydrochloric acid and a mixture of amino acids was formed. This mixture was separated by column chromatography.
Outline briefly why chromatography is able to separate a mixture of compounds.
Practical details are not required.
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(3)
(Total 13 marks)
Q2.A peptide is hydrolysed to form a solution containing a mixture of amino acids. This mixture is then analysed by silica gel thin-layer chromatography (TLC) using a toxic solvent. The individual amino acids are identified from their Rf values.
Part of the practical procedure is given below.
1.       Wearing plastic gloves to hold a TLC plate, draw a pencil line 1.5 cm from the bottom of the plate.
2.       Use a capillary tube to apply a very small drop of the solution of amino acids to the mid-point of the pencil line.
3.       Allow the spot to dry completely.
4.       In the developing tank, add the developing solvent to a depth of not more than 1 cm.
5.       Place your TLC plate in the developing tank.
6.       Allow the developing solvent to rise up the plate to the top.
7.       Remove the plate and quickly mark the position of the solvent front with a pencil.
8.       Allow the plate to dry in a fume cupboard.

(a)     Parts of the procedure are in bold text.
For each of these parts, consider whether it is essential and justify your answer.
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(4)

(b)     Outline the steps needed to locate the positions of the amino acids on the TLC plate and to determine their Rf values.
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(4)
(c)     Explain why different amino acids have different Rf values.
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(2)
(Total 10 marks)

Q3.(a)     A chemist discovered four unlabelled bottles of liquid, each of which contained a different pure organic compound. The compounds were known to be propan-1-ol, propanal, propanoic acid and 1-chloropropane.
Describe four different test-tube reactions, one for each compound, that could be used to identify the four organic compounds.
Your answer should include the name of the organic compound, the reagent(s) used and the expected observation for each test.
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(8)
(b)     A fifth bottle was discovered labelled propan-2-ol. The chemist showed, using infrared spectroscopy, that the propan-2-ol was contaminated with propanone.
The chemist separated the two compounds using column chromatography. The column contained silica gel, a polar stationary phase.
The contaminated propan-2-ol was dissolved in hexane and poured into the column.
Pure hexane was added slowly to the top of the column. Samples of the eluent (the solution leaving the bottom of the column) were collected.
•        Suggest the chemical process that would cause a sample of propan-2-ol to become contaminated with propanone.
•        State how the infrared spectrum showed the presence of propanone.
•        Suggest why propanone was present in samples of the eluent collected first (those with shorter retention times), whereas samples containing propan-2-ol were collected later.
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(4)
(Total 12 marks)

Q4.(a)     Because of the toxic nature of the copper(II) ion, a wide range of alternative anti-fungal drugs has been developed for use in agriculture. One example is Zineb.
 
(i)      The negative ion in Zineb could act as a bidentate ligand.
On the structure above, draw a ring around each of two atoms that could provide the lone pairs of electrons when this ion acts as a bidentate ligand.
(1)

(ii)     Calculate the Mr of Zineb. Give your answer to the appropriate precision.
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(1)

(iii)    Name the functional group formed at each end of the negative ion when all the sulfur atoms in the structure of Zineb are replaced by oxygen atoms.
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(1)

(b)     Zineb has been investigated for harmful effects. Generally, Zineb has been found to be safe to use in agriculture. It is only slightly soluble in water and is sprayed onto plants. A breakdown product of Zineb is ethylene thiourea (ETU), which is very soluble in water. The structure of ETU is shown below.
 
Determine the percentage, by mass, of sulfur in ETU (Mr = 102.1).
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(1)
(c)     Chromatography is a technique used to show the presence of a small amount of ETU in Zineb.
Outline how this technique is used to separate and identify ETU from a sample of Zineb powder.
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(4)
(Total 8 marks)
Q5.Lysine and alanine are two amino acids.

lysine	 
alanine	

(a)     Give the IUPAC name of lysine.
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(1)
(b)     Draw structures to show the product formed in each case when lysine reacts with
(i)      an excess of aqueous HCl

 
 
(1)
(ii)     an excess of aqueous NaOH

 
 
(1)
(iii)    methanol in the presence of a small amount of concentrated H2SO4

 
 
(1)

(c)     The mass spectrum of alanine gives a major peak at m / z = 44
Write an equation for the fragmentation of the molecular ion of alanine to give an ion that produces this peak.
In your answer, draw the displayed formula for this fragment ion.

 
 
(2)

(d)     Draw a dipeptide formed from one molecule of lysine and one molecule of alanine.

 
 
(1)
(e)     The dipeptide in part (d) is hydrolysed in acid conditions and the mixture produced is analysed by column chromatography. The column is packed with a resin which acts as a polar stationary phase.
Suggest why lysine leaves the column after alanine.
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(2)
(Total 9 marks)

M1.          (a)     3-hydroxypropanoic acid
allow 3-hydroxypropionic acid
must be correct spelling
1
(b)     (i)      must show trailing bonds

or can start at any point in the sequence, e.g.

not allow dimer
allow –O–CH2CH2COOCH2CH2CO–
or –CH2CH2COOCH2CH2COO–
ignore ( ) or n
NB answer has a total of 6 carbons and 4 oxygens
1

(ii)     condensation (polymerisation)
Allow close spelling
1
(c)     (i)      C=C or carbon-carbon double bond
1
(ii)

must show ALL bonds including O–H
1
(iii)     must show trailing bonds

allow polyalkene conseq on their c(ii)
ignore n
1

(d)

allow NH3+—
allow COO–
1
(e)     (i)

In (e), do not penalise a slip in the number of carbons in the -CH2CH2- chain, but all must be bonded correctly
NB two carboxylate groups
Allow COONa or COO– Na+ but not covalent bond to Na
allow NH2–
1
(ii)

In (e), do not penalise a slip in the number of carbons in the
-CH2CH2- chain, but all must be bonded correctly
NB two ester groups
allow NH2– or +NH3–
1

(iii)

In 4(e), do not penalise a slip in the number of carbons in the -CH2CH2- chain, but all must be bonded correctly
allow anhydride formation on either or both COOH groups (see below) with or without amide group formation

1
(f)      M1    phase or eluent or solvent (or named solvent) is moving or mobile
1
M2    stationary phase or solid or alumina/silica/resin
1
M3    separation depends on balance between solubility or affinity
(of compounds) in each phase
OR
different adsorption or retention
OR
(amino acids have) different Rf values
OR
(amino acids) travel at different speeds or take different times
1
[13]
M2.(a)     Wear plastic gloves:
Essential – to prevent contamination from the hands to the plate
1
Add developing solvent to a depth of not more than 1 cm3:
Essential – if the solvent is too deep it will dissolve the mixture from the plate
1
Allow the solvent to rise up the plate to the top:
Not essential – the Rf value can be calculated if the solvent front does not reach the top of the plate
1
Allow the plate to dry in a fume cupboard:
Essential – the solvent is toxic
Allow hazardous
1
(b)     Spray with developing agent or use UV
1
Measure distances from initial pencil line to the spots (x)
1

Measure distance from initial pencil line to solvent front line (y)
1

Rf value = x / y
1

(c)     Amino acids have different polarities
1
Therefore, have different retention on the stationary phase or different solubility in the developing solvent
1
[10]

M3.(a)     If 2 stage test for one compound, award no marks for that compound, eg no mark for ROH or RX to alkene then Br2 test. If reagent is wrong or missing, no mark for that test; if wrong but close/incomplete, lose reagent mark but can award for correct observation. In each test, penalise each example of wrong chemistry, eg AgClr2
propan-1-ol
acidifiedpotassiumdichromate
sodium
Named acid + conc H2SO4
named acyl chloride
PCl5
M1
1
(orange) turns green
effervescence
Sweet smell
Sweet smell /misty fumes
Misty fumes
M2

1

propanal
add Tollens or Fehlings / Benedicts
acidifiedpotassiumdichromate
Bradys or 2,4-dnph
if dichromate used for alcohol cannot be used for aldehyde
M3
1
Tollens: silver mirror or Fehlings/ Benedicts: red ppt
(orange) turns green
Yellow or orange ppt
M4
1
propanoic acid
Named carbonate/ hydrogencarbonate
water and UI (paper)
Named alcohol + conc H2SO4
sodium or magnesium
PCl5
if sodium used for alcohol cannot be used for acid
M5
1
effervescence
orange/red
Sweet smell
effervescence
Misty fumes
if PCl5 used for alcohol cannot be used for acid
M6
1
1-chloro propane
NaOH then acidified AgNO3
AgNO3
If acidification missed after NaOH,no mark here but allow mark for observation
M7
1
white ppt
white ppt
M8
1
(b)      oxidation (of alcohol by oxygen in air)
M1
1
absorption at 1680 -1750 (due to C=O)
Must refer to the spectrum
M2
1

comparison of polarity of molecules or correct imf statement:propanone is less polar OR propan-2-ol is more polarOR propanone has dipole-dipole forcesOR propan-2-ol has hydrogen bonding
M3
1

about attraction to stationary phase or solubility in moving phasePropan-2-ol has greater affinity for stationary phase or vice versaOR propanone is more soluble in solvent/moving phase or vice versa
M4
1
[12]
M4.(a)     (i)      Two rings only around nitrogen or sulfur
Lose this mark if more than 2 atoms are ringed.
Do not allow two atoms at the same end of the ion.
1
(ii)     275.8
Accept this answer only. Do not allow 276
1
(iii)    Carboxylate / COO–
Allow salt of carboxylic acid or just carboxylic acid.
1
(b)     (32.1 / 102.1) = 31.4%
Do not penalise precision but do not allow 1 significant figure.
1

(c)     Zineb is mixed with a solvent / water
Max=2 if M1 missed
1

Use of column / paper / TLC
Lose M1 and M2 for GLC
1
Appropriate collection of the ETU fraction
OR Appropriate method of detecting ETU
Allow ETU is an early fraction in a column or collecting a range of samples over time, lowest retention time / travels furthest on paper or TLC (allow 1 mark for having the longest retention time in GLC).
1
Method of identification of ETU (by comparison with standard using chromatography)
If method completely inappropriate, only M1 is accessible
1
[8]
M5.(a)    2,6-diaminohexanoic acid
Ignore additional , or – or spaces.
1
(b)    (i)      

NB both N must be protonated.
Allow −NH3+    allow CO2H    Allow −+H3N.
Penalise – C4H8 – here.
1
(ii)     

Allow CO2−.
Allow −H2N.
Allow –COONa but penalise O–Na bond shown.
1
(iii)    

Allow CO2CH3.
Allow −NH3+ or –H2N.
1
(c)     

1 for displayed formula of fragment ion.
1 for molecular ion of alanine AND radical.
Allow molecular ion without brackets and fragment ion in brackets with outside +.
Allow dot anywhere on radical.
Allow [C3H7NO2] for molecular ion.
2
(d)     

OR

OR

Dipeptide, not repeating unit /.
Allow CO2H Allow –H2N.
Allow –CONH–.
1

(e)     M1        In acid lysine has double positive or more positive charge
1

M2        (Lysine ion) has greater affinity / greater attraction / adheres better / sticks              better to polar / stationary phase
M2 only scores after a correct M1.
Ignore greater retention time.
1
[9]