Question 1 2 Total
1.1 1.2 1.3 1.4 1.5 1.6 2.1 2.2 2.3 2.4 2.5
Marks 2 3 6 4 6 1 2 2 4 2 2 34
Carbohydrates are the most important providers of energy for living cells. Monosaccharide glucose is a source of energy for the living cell, but for persons who suffer from diabetes glucose may be dangerous. High level of glucose may lead to cardiovascular diseases and even death. That is why people avoid consuming too much carbohydrates and glucose particularly.
1. Determination of reducing sugars in fruit juice
One of the technique for determination of reducing sugars in different samples includes the use of Fehling's reagent. A 10.00-mL aliquot of fruit juice (assuming the initial sample contained only glucose and fructose) was transferred into a titration flask and Fehling's reagent was added. This reagent was prepared by mixing 50.00 mL of 0.04000 M copper sulfate (solution A) and potassium-sodium tartrate and potassium-sodium hydroxide (solution B). Solution C thus obtained, was then heated and red precipitate was formed.
Glucose
1.1. Write the balanced ionic equation of chemical reaction occurring upon heating of the solution C. Use Cu2+ for initial copper solution.
C6H12O6 + 2 Cu2+ + 5OH-= C6H11O7
+ Cu2O+ 3H2O If C6H12O7 instead of C6H11O7
-Hereinafter if an equation is not balanced, then points/2.
2 points 1 point
After that 10 mL of 10% solution of potassium iodide and 1 M sulfuric acid were added to the flask.
The mixture was covered with watch glass and was then placed in a dark place. An excess of iodine was then titrated with 0.05078 М sodium thiosulphate solution. 11.87 mL of the titrant was required to reach the endpoint.
1.2. Write the balanced equation(s) in molecular or ionic form for all the reactions taking place in the flask.
2CuSO4 + 4KI = 2CuI + I2 + 2K2SO4
or 2Cu2+ + 4I– = 2CuI + I2
KI +I2 = KI3
or I– + I2 = I3–
C6H11O7- + H2SO4 = C6H12O8 + HSO4 -2Na2S2O3 + I2 = 2NaI + Na2S4O6
or 2S2O32– + I2 = 2I– + S4O62–
2 points
not marked not marked
1 point
1.3. Consider all fructose was transformed into glucose under the experimental conditions; calculate the total mass content of sugars (in g/L) in a fruit juice. Mw = 180.16 g/mol.
Total amount of copper(II) is 50.00 mL * 0.04000 M = 2.0000 mmol.
Obviously, there is an excess of iodine and the remaining iodine was titrated with sodium thiosulphate: 11.87 mL * 0.05078 M = 0.6028 mmol.
2.0000 – 0.6028 mmol = 1.3972 mmol of copper(II) was required to oxidize the sugars.
ν(sugars) = ν(Cu2+)/2 = 0.6986 mmol in 10.00 mL C(sugars) = 0.6986 mmol/10.00 mL = 0.06986 M mass content = 180.16 g/mol * 0.06986 M = 12.6 g/L
6 points
A new 10.00-mL aliquot of the same juice was treated with a 10.00-mL portion of acidified potassium iodate(V) solution (0.01502 М) and 10 mL of 10 % solution of potassium iodide. After the mixture turned brown, an excess of sodium hydroxide solution was added. The flask was then covered with a watch glass and put into a dark place. The obtained solution was acidified and titrated with 0.01089 M solution of sodium thiosulphate. The average titrant volume used for titration was 23.43 mL. Note that fructose is not converted into glucose under these conditions.
1.4. Write all the balanced equations for the described reactions in molecular or ionic form.
KIO3 + 5KI + 3H2SO4 = 3I2 + 3K2SO4 + 3H2O
IO3– + 5I– + 6H+ = 3I2 +3H2O 2 points
Only glucose was oxidized with iodine 2 points
2H2O I2 3NaOH
Na
2NaI
or
3 OH
-I2 2I- 2H2O
Na
H+ + OH- = H2O not marked
2Na2S2O3 + I2 = 2NaI + Na2S4O6 not marked
1.5. Calculate the mass content of each sugar (in g/L) in the juice.
Total amount ν(I2) = 3ν(IO3–
) =3*0.01502 M * 10 mL = 0.4506 mmol 1 pt ν(S2O3
2-)=23.43 mL*0.01089 M = 0.2552 mmol 1 pt ν(S2O3
2-)/2=ν(I2) = 0.1276 mmol 1 pt 0.4506 mmol – 0.1276 mmol = 0.3230 mmol of iodine was used to oxidize glucose
C(glucose) = 0.3230 mmol/10.00 mL = 0.03230 M 1 pt
mass content of glucose = 180.16 g/mol *0.03230 M = 5.82 g/L 1 pt mass content of fructose = 12.6 – 5.82 = 6.78 g/L 1 pt
6 points
1.6. One bread exchange unit (1 BEU) corresponds to the content of 12 g of digestible carbohydrates in product. How many BEU are in one glass (200 mL) of juice?
0.2 L*5.82 g/L = 1.16 g of digestible carbohydrates, it is 0.1 BEU 1 point Or 0.2 L*12.6 g/L = 2.52 g, it is 0.2 BEU 1 point
2. Diagnosis of diseases
The derivative of glucose, 2-deoxy-2-(18F)fluoro-D-glucose (FDG), is the most common radiopharmaceuticals for diagnosis of cancer using positron emission tomography. The first step of FDG preparation is to produce a radionuclide fluoro-18 by nuclear reaction in a cyclotron. The next
nucleophilic substitution. 2-deoxy-2-(18F)fluoro-D-glucose once injected into the patient actively accumulates in cells of malignant tumors; this process is accompanied by decomposition of fluorine-18. This radionuclide is a β+ emitter – nucleus emits a positron (anti-electron). Positron interacts with an electron and after that annihilation occurs, which can be detected. This allows determining precisely the tumor sizes and type.
2.1. Complete the nuclear reactions leading to various fluorine isotopes.
a) 18O + 11H → …+ 18F n 0.5 points
b) … + 12D → 18F + α 20Ne 0.5 points
c) 19F + 12D → 20F + … 0.5 points
d) 16O + … → 18F + 11H + n α or 0.5 points
2.2. The decay mode of unstable light nuclei depends on the ratio between the number of neutrons and protons in them. If this ratio is greater than that for a stable isotope then the nucleus decays in a β–-mode, if it is smaller – in a β+-mode.
Determine the type of decay for the nuclei in the table:
Nucleus 11С 20F 17F 14C
Decay mode β+ β- β+ β–
0.5 points 0.5 points 0.5 points 0.5 points
When nuclear reaction (a) is used for fluorine-18 preparation, the target material is presented as water enriched with H218
O. The presence of usual water H216
O leads to a side nuclear reaction with
16O, leading to the formation of isotope 17F.
2.3. It is known that within five minutes after completion of irradiation of the target the ratio of radioactivities of 18F and 17F is 105. Assuming that irradiation time is short, the radioactivity of each isotope is proportional to the nuclear reaction yield and the mole fraction of a component in the irradiated target, calculate the mass fraction of H218O in the target. t1/2(18F) = 109.7 minutes, t1/2(17F) = 65 seconds. The ratio between nuclear reactions yields is / = 144.7.
Radioactivity is:
A = λN, where N is the number of atoms, λ = ln 2 / t1/2 1 point The initial ratio of radioactivities:
( ) ( )
After 5 minutes the ratio changed due to radioactive decay of fluorine:
18
2.4. Calculate the yield of labeling D-glucose with 18, if initial radioactivity of a fluorine-18 sample was 600.0 MBq and radioactivity of the obtained 2-deoxy-2-(18F)fluoro-D-glucose is 528.2 МBq. Synthesis time is 3.5 minutes.
During the synthesis, the radioactivity will decrease:
3.5 0
2.5. Biological half-life (through the excretory organs) of 2-deoxy-2-(18F)fluoro-D-glucose is 120.0 minutes. How much radioactivity (in MBq) will remain in the patient ten hours after injection of FDG with the initial radioactivity of 450.0 MBq.
Radioactivity is excreted by radioactive decay and through the excretory organs (e.g. kidneys).
The excretion process may be considered as two competitive first-order reactions. Activity after one hour is:
Problem 6. Bread is the stuff of life me that this is not perfect, you are my enemy forever."
The principle bread flavour component was identified in 1969 as compound X which occurs in equilibrium with its tautomer Y in a 2:1 ratio. Unfortunately, both
forms are labile, and after some hours bread has no the same nice smell.
This tautomeric mixture of X and Y was synthesized in 1993 from piperidine by the reaction sequence given in Scheme 1. It is noteworthy that the initial ratio of X and Y was 1:4; on standing this ratio gradually changed to an equilibrium one.
Scheme 1.
Compound B which is characterized by 3-fold axis of symmetry (i.e., rotation by 120° results in a molecule indistinguishable from the original) occurs in equilibrium with its diastereomer C.
The interconversion of these two forms proceeds via intermediate A which is also intermediate in B and C formation as well as their transformation to D. Compounds A, B, and C have the same elemental composition: ωC = 72.24%, ωH = 10.91%, ωN = 16.85%.