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Problem 1: Separation and Identification of Ions
A student studied the chemical reactions between cations, A , B , C , D , E in nitrate aqueous solutions and anions X , Y , Z , Cl , OH in sodium aqueous solutions as well as an organic ligand L. Some precipitation (ppt) products and colored complexes were found as shown in Table 1: Table 1
2+ 2+ 2+ 2+ 2+ 2+

A B

X *** Yellow ppt White ppt *** ***

-

Y *** White ppt Brown ppt Red ppt Red ppt

-

Z *** *** Brown ppt *** White ppt

-

Cl *** ***

-

2+

OH White ppt *** Black ppt *** ***

-

L *** BLn Complex CL , CL2 Complexes *** ***
2+ 2+ 2+

C D E

2+

White ppt *** ***

2+

2+

*** = No reaction, 1-1 Design a flow chart for the separation of A , B , C , D , E in a nitrate aqueous solution by using various aqueous solutions containing anions X , Y , Z , Cl , OH , respectively, as testing reagents. Write down the product of the chemical reaction for each step in the flow chart. Design a flow chart for the separation of anions X , Y , Z , Cl , OH in a sodium aqueous 2+ 2+ 2+ 2+ 2+ solution by using various aqueous solutions containing cations A , B , C , D , E , respectively, as testing reagents. Write down the product of the chemical reaction for each step in the flow chart. The white ppt BY2 and brown ppt CY2 have low solubilities in water with solubility products -8 -13 o (Ksp) of 3.20 × 10 and 2.56 × 10 , respectively at 25 C. Calculate the solubility of BY2. Calculate the solubility of CY2. A series of solutions containing B and L were prepared in 50 mL volumetric flasks by -3 2+ adding a 2 mL of 8.2 × 10 M solution of B to each flask. Varying amounts of a 1.0 × -2 10 M solution of the ligand L are added to each flask. The solution in each volumetric flask was diluted with water to the mark (50 mL). The absorbance (A) of Complex BLn was measured at 540 nm for each solution in a 1.0 cm cell. The data are summarized in Table 2+ [Mole Ratio 2. (Both B and ligand L show no absorption (A = 0) at 540 nm.) Method] Calculate the value of n (Coordination number) in the complex BLn .
1
2+ 2+ 2+ 2+ 2+ 2+ 2+

1-2

1-3

1-3-1 1-3-2 1-4

1-4-1

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1-4-2

Calculate the formation constant (Kf) of complex BLn . Table 2 Added L VL (mL) 1.00 3.00 5.00 7.00 9.00 Absorbance (A) 0.14 0.40 0.55 0.64 0.66 Added L VL (mL) 2.00 4.00 6.00 8.00 10.00 Absorbance (A) 0.26 0.48 0.60 0.66 0.66
2+

2+

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1-5-1

Which cation (B or C ) precipitates first? What is the [Y ] when this happens? (Ksp= -8 -13 o 3.20 × 10 for BY2 and Ksp = 2.56 × 10 for CY2, at 25 C.) [Separation by Precipitation] What are the concentrations of Y and the remaining cation when complete precipitation of the first precipitating cation has occurred (assume that the concentration of the first -6 2+ cation in solution after complete precipitation is ≤ 10 M)? Is it possible to separate B 2+ and C by the precipitation method with Y ion as a precipitating agent?
-

2+

2+

1-5-2





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Problem 2: Preparation and Applications of Radioisotopes 中华化学竞赛网www.1000hx.com
Radioisotopes can be used in medical diagnosis and therapy as well as industrial analysis. Many radioisotopes, e.g. P-32 (Mass number = 32) and Co-60 can be generated by the irradiation of neutrons in a nuclear reactor. However, some radioisotopes in nature, e.g. C-14 and T-3 (Tritium), can be produced by the bombardment of nitrogen N-14 atoms in the atmosphere by neutrons in the cosmic ray. (Atomic numbers of T & H, C, N, P, Co and neutron

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1-5

Solid NaY (soluble) was added very slowly to an aqueous solution containing 0.10 M in B 2+ and 0.05 M in C prepared from their respective nitrate aqueous salts.

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Radioisotope C-14 can be used as a reagent in C-14 dating. The activity (A) in terms of counts per minute (cpm) of the radioisotope C-14 is proportional to the number (N) of C-14 atoms as follows : [C-14 Dating] A= N (1)

N = No e

-λt

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Where ε is the detection coefficient of a detector for C-14 and λ is the decay constant of C-14. An initial amount (No) of C-14 can be reduced to the amount (N) of C-14 by decay after a given time (t) as follows: (2)





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2-1

Write down the equations for the nuclear reactions for the production of C-14 and T-3 by the bombardment of nitrogen N-14 atoms in the atmosphere with neutrons in the cosmic ray.

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are 1, 6, 7, 15, 27 and 0, respectively.

P-32 can be denoted as

)



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The half life (t1/2) of C-14 is 5730 years which is defined as the time required for 50% of the number of the radioisotope C-14 atoms in a sample to undergo decay, that is N = 1/2 No. It is well known that activity (Ao) of C-14 in a living animal or plant is kept to be around 16.5 cpm/g of carbon. After the death of the animal or plant, the activity (cpm/g of carbon) of C-14 in the body of the living animal or plant is decreased by the time passed. 2-2-1 Give the equation showing the relation between the original activity (Ao) and final activity (A) as function of time for a living animal or plant. The activity of C-14 in an ancient wood boat is found to be 10.2 cpm/g of carbon. Calculate the age of the ancient boat.

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2-2-2

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2-3

Rp = N δ

(3)

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The radioisotope P-32 is a very important leveling reagent for biological research and can be produced by the bombardment of P-31 by a neutron in a nuclear reactor. The production rate (Rp) of P-32 can be estimated as :

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and

A = λ N* = Rp – Rd

(5)

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Where λ is the decay constant of P-32, t is the neutron irradiation time in the nuclear reactor and the half life (t1/2) of P-32 is 14.3 days. 2-3-1 A 10 mg sample of pure H3PO4 is irradiated by neutrons with a neutron flux of 1.00 x 10 -2 -1 n cm sec for one hour in a nuclear reactor. Calculate the activity of the sample in cps 10 (counts / second) and Ci. (Ci = Curie, 1 Ci ≈ 3.7 x 10 cps, atomic weight: H=1, P=31, O=16)

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Rd = N

δ(e

-λt

)

(4)

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Where N and δ are the number of atoms and neutron capture cross section (≈ 0.9 x 10 2 2 cm /atom) of P-31, respectively, and is the neutron flux (neutron / (cm sec)) of the nuclear reactor. If the detection efficient (ε) of the detector for P-32 is 1.0, the decay rate (Rd) and the activity (A) of P-32 in the nuclear reactor can be approximately estimated as a function of the number (N*) of P-32 atoms as follows:

-24



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2-3-2

Problem 3: Ion Exchangers 中华化学竞赛网www.1000hx.com
Ion exchangers can be employed to adsorb and separate cations and anions. They can be prepared from organic or inorganic materials. An organic, cationic ion exchanger can be synthesized by the polymerization of styrene / divinyl benzene followed by sulfonation with H2SO4,





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The radioisotope P-32 can be used to measure the volume of water in a pool or the blood volume of an animal. A 2.0 mL solution of 1.0 Ci/mL P-32 was injected into a pool. After mixing well, the activity of 1.0 mL of water in the pool was found to be 12.4 cps (counts / second). Calculate the volume of water (L) in the pool. (Ci = Curie, 1 Ci ≈ 3.7 10 x 10 cps)

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as shown in Scheme 1:
C
C
N a2S 2O 8

C C

C

C

C

C
H 2 SO
4

C

C

C

C
SO 3 - H +

+
C
C

C

C

C

C

C

C

C

C

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[ p o lym er ]

[ c a tio n ic e x c h a n g e r ]
R -H +

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Cationic ion exchanger (denoted as R H ) can be employed to adsorb the cations, M ,the chemical reaction and the equilibrium constant Kc as well as the distribution coefficient Kd can be expressed as follows:

-

+

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(1) (2)
+ z+

[Scheme 1]

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R H + M = RM + H , Kd = [RM] / [M ]
+ +

-

+

+

+

Kc = [RM][H ] / ([M ][RH])

+

+





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R H + MOH = R M + H2O and 3-1 3-1-1 z R H + M(OH)z = (R )zM + z HCl
+ +

-

+

-

+

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The cationic ion exchanger R H can be transformed into the ion exchanger R M or R M by the - + reaction of R H with a metal hydroxide (M(OH)z). The approximate equations are: (3) (4)

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R Na .

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3-1-2

If another ion exchanger R H is employed instead of R Na .

-

+

-

+ -

(a) Give the chemical
+





exchanger, R H or R Na , is suitable for drinking purpose and give the reason.





3-2

An organic, anionic ion exchanger (denoted as R Cl ) can also be synthesized by the polymerization of styrene / divinyl benzene followed by the reaction of the resulting polymer, poly (styrene / divinyl benzene), with the Lewis acid AlCl3 and tertiary amine NR′3, as shown in Scheme 2:

+

-



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-

+

-

+



equation for the adsorption of Ca

2+

by the ion exchanger R H and (b) tell which ion

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+

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Give the chemical equation for the adsorption of Ca

2+

by the cationic ion exchanger

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A cationic ion exchanger R Na was employed to remove CaCl2 in tap water,

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+

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exchanger R Cl with 3.0 M of NaOH by the equation: R Cl + NaOH = R OH + NaCl
+ + -

+

-

10

The anionic ion exchanger R OH can be obtained from the chemical reaction of the ion

3-2-1

Tell how the removal of H+ from a solution of HCl can be achieved with an anionic ion exchanger and give the chemical equation for the process. Tell how the amount of SO4 in tap water can be estimated by using an anionic ion + exchanger R OH . Give all of the chemical equations involved in the process.
-





中 华

3-2-2

The capacity (S) of the cationic ion exchanger R H for an adsorbed ion can be expressed in moles of the adsorbed ion per gram of the ion exchanger in 1.0 mL of aqueous solution and can be calculated by using the following equation:

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10

w.

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1 / Kd = [M ] / (S(10 )) + [H ] / (S Kc(10 ))



+

3

+

3

(7)







3-4

Ion exchangers can be employed as stationary phase materials in liquid chromatography to + adsorb and separate various ions. For example, the anionic ion exchanger R OH can be used to separate X and Y ions with the eluent NaOH. The chromatogram for separation of X and Y ions using a 30 cm of anionic ion exchange column is shown in Figure 1.
-

Where t1, t2 and to are the retention times (tR) for X , Y and the pure eluent (NaOH) to traverse the column, respectively. ω1 and ω2 are the peak-widths for X and Y . The number of theoretical plates N and the plate height H (height equivalent of the theoretical plates) of the column can be estimated as shown below: N = 16 (tR / ω) 2 and H=L/N



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(8) (9)



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3-3

Show that the relationship between Kd, S, Kc, [M ] and [H ] as shown by the equation:

+

+

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The capacity (S) of the cationic ion exchanger R H for M ions in an aqueous solution can be estimated from the equilibrium constant Kc, the distribution coefficient Kd and the concentrations + + of M and H ions in the aqueous solution.

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.c

-

+

+

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S = ([RM] + [RH]) × 10

-3

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2-



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(5) (6)

+

-

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t2 t1
YX-

t0

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Retention Time / min
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-



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R = 2 (t2 - t1) / (ω1 + ω2) and
α = (t2 - t0) / (t1 - t0)

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where L is the length of the column. The resolution (R) of the column and the separation factor (α) for X and Y also can be estimated using the following equations: (10) (11)



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Figure 1.

Liquid Chromatogram for X and Y ions

3-4-1 3-4-2 3-4-3 3-4-4 3-5

Calculate the average number of theoretical plates N of the column. Calculate the plate height H of the column.

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z+

tR

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0x ue

Calculate the resolution (R) of the column for X and Y ions.

-

-

Some examples of Zeolites are shown in Figure 2.







(M = Na , K or Ca , Mg ) are the best known examples of inorganic ion exchangers.

z+

+



Some ion exchangers are derived from inorganic matters. Zeolites [(M )(Al2O3)m / (SiO2)n]
+ 2+ 2+

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w.

10

Calculate the separation factor (α) for X and Y ions.

-

-



A Na -Zeolite (denoted as Z-Na ) with a pore size of 13 ? is an important ion exchanger for the removal of Ca
2+

+

+

or Mg

2+

ion from tap water.

Zeolites with definite pore sizes also behave as Thus, the zeolite

highly selective adsorbents for various molecules, e.g. H2O and iso-butane. can be used as a molecular sieve.

The zeolite can also be used as a catalyst by adsorption of a

petroleum component, e.g. iso-butane, in petroleum resulting in the enhancement of rate of the



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cracking of the adsorbed component.

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3-5-1

Give the chemical equation for the removal of Ca ion exchange column.

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2+

ions from tap water with Z-Na zeolite

3-5-2

Give the chemical equation for the adsorption of K with Z-Na zeolite.

+





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Problem 4: Determination of Calcium Ion by Precipitation Followed by Redox Titration

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+ +

Figure 2.

00 h

Various types of Zeolites

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10

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-8

The calcium content of an aqueous sample can be determined by the following procedure:

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(this typically takes 15 min). Step 3

CaC2O4 precipitates out.

The hot solution is filtered and the solid CaC2O4 is washed with ice-cold water to









remove excess C2O4 ions.

2-

Step 4

The insoluble CaC2O4 is dissolved in hot 0.1 M H2SO4 to give Ca point is observed.

The dissolved H2C2O4 is titrated with standardized KMnO4 solution until the purple end

Relevant reactions and equilibrium constants: CaC2O4(s) → Ca
2+ (aq)

+ C2O4

2(aq)

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Ksp = 1.30x10



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Step 2

Urea ((NH2)2CO) is added and the solution gently boil until the indicator turns yellow



2+

ions and H2C2O4.

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10

thorough mixing with Na2C2O4 solution.

00

Step 1

A few drops of methyl red are added to the acidified aqueous sample, followed by

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Ca(OH)2(s) → Ca

2+ (aq) -

+ 2OH (aq)
+ (aq)

-

Ksp = 6.50x10

-6

H2C2O4(aq) ? HC2O4 (aq) + H HC2O4 (aq) ? C2O4 H2O ? H 4-1
+ (aq) 2(aq) -

Ka1 = 5.60x10 Ka2 = 5.42x10 Kw = 1.00x10

-2

+H

+ (aq)

-5

+ OH (aq)

-14

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2). 4-2

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Write a balanced equation for the reaction that takes place upon the addition of urea (Step

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procedure and found to require 27.41 mL of a 2.50 x 10 M KMnO4 solution in the final step. Find the concentration of Ca 4-3
2+

-3

ions in the sample.

Calculate the solubility of CaC2O4 in an aqueous solution buffered at pH 4.0.





中 华

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In the above analysis, a possible source of error was neglected.

Step 1 will be incomplete if an excess of C2O4 ions is added, due to the following reactions: Ca
2+ (aq)

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2-



activity coefficients)

The precipitation of CaC2O4 in

+ C2O4

2(aq)

→ CaC2O4(aq)

Kf1 = 1.0 x 10
2(aq)

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CaC2O4(aq) + C2O4 4-4

2(aq)

→ Ca(C2O4)2

Kf2 = 10 and C2O4 ions in solution after optimal

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3

The calcium content of a 25.00 mL aqueous sample was determined using the above

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(Neglect

Calculate the equilibrium concentrations of Ca

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precipitation of CaC2O4 is reached.

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activity coefficients. Any assumptions made during calculation must be clearly stated.)







Problem 5: Nitrogen in Wastewater
In natural water and waste water, the forms of nitrogen of greatest interest are nitrate, nitrite, ammonia, and organic nitrogen. All these forms of nitrogen, as well as nitrogen gas, are biochemically interconvertible and are components of the nitrogen cycle. 5-1 The Macro-kjeldahl method, in combination with a titration method, is often used in the





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w.

4-5

Calculate the concentrations of H and Ca

+

2+

in a saturated solution of CaC2O4. (Neglect

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.c

2+

2-

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determination of organic nitrogen in wastewater. HgSO4 are added to the sample solution.

In the first step, H2SO4, K2SO4, and

After digestion, the solution is neutralized by the

addition of concentrated NaOH. The gas liberated by the treatment is then distilled into a solution of excess boric acid and the latter subsequently titrated with 0.02 N H2SO4. 5-1-1 5-1-2 5-1-3 5-1-4 5-1-5 Identify the product formed in the digestion step. Identify the gas liberated upon the addition of NaOH.

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Write a balanced equation for the final titration step.



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Which of the following indicators is most suitable to be used in the final titration step: Methyl orange (transition range pH 3.1 - 4.4), phenolphthalein (transition range pH 8.0 -



中 华

5-2

Nitrite is known to cause the illness methemoglobinemia in infants. nitrite can be determined by a colorimetric method. of a series of standard nitrite solutions.

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9.6) is chosen as the indicator. Explain your choice.



The method requires the preparation

However, nitrite is readily oxidized in the presence

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nitrite stock solution.

The purple color of the solution due to the presence of excess

10

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5-2-1 5-2-2 5-2-3







Write a mathematical equation for calculation of nitrogen concentration. A: mg/ml N in stock NaNO2 solution B: total ml standard KMnO4 used C: molarity of standard KMnO4 E: molarity of standard Na2C2O4 D: total ml standard Na2C2O4 solution F: ml stock NaNO2 solution taken for titration
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Write a balanced equation for the back titration.





Write a balanced equation for the reaction of the nitrite solution with KMnO4.

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Na2C2O4 and the mixture back titrated with standard permanganate solution.

10

permanganate was subsequently discharged by the addition of a known quantity of

00

hx

.c

adding a known excess of standard KMnO4 solution and H2SO4 solution are added into the

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achieve high accuracy in the subsequent analysis. The standardization is carried out by

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of moisture and hence standardization of the stock nitrite solution is required in order to

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Write a balanced equation for the reaction between the liberated gas and boric acid.

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In the laboratory,

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Problem 6: Use of Isotopes in Mass Spectrometry
Many elements in the periodic table have more than one isotope.
35

The atomic mass of an

element is calculated based on the relative abundance of the isotopes. As an example, the atomic mass of chlorine is about 35.5 because the abundance of Cl
37

is about three times the
81

abundance of Cl . In mass spectrometry, instead of average atomic mass, the isotope peaks are observed. (Cl
35

75.77%, Cl

37

24.23%, C 98.9%, C 1.1%, Br

12

13

79

50.7%, Br

49.3%)

Isotopes are quite useful in quantitative mass spectrometry.

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gas chromatography / mass spectrometry.

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the qualitative criteria in the analysis of 2,3,7,8, tetra chlorinated dioxin (2,3,7,8-TCDD) by Calculate the theoretical ratio of the two ions. The intensities of the isotopic species can be found by applying the following formula: (a+b) , where a is the relative abundance of the light isotope, b is the relative abundance of
n



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中 华

6-2

Molecular ion is often selected in quantitative analysis. The intensity of the molecular ion needs to be corrected if the signal is interfered by other compounds. for quantitative analysis.

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the heavy isotope, and n is the number of chlorine atoms present.

10

6-1

In addition to the retention time (migration time), the ratio of M and M+2 ions was used as

non-halogenated compound with a molecular weight of 136, the molecular ion was selected Propose a mathematical equation for calculation the corrected signal, if the analyte co-elutes (same migration time) with the compound n-butyl bromide.

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In the analysis of a

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Problem 7: Atomic Orbitals
One way to describe the shape of atomic orbitals of H-atom is in terms of the nodal surfaces, or nodes, where the electron has zero probability. nodes increases as n increases. “l” angular nodes. 7-1 Describe the electron probability distribution for the 1s, 2s and 3s orbitals. How many According to wave mechanics, the number of For given set of orbitals nlm, there are “n-l-1” spherical nodes,

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does each orbital have respectively? 7-3



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Imagine that you are traveling along the z axis, beginning your journey at a distance far



中 华

3s, 2pz and 3pz. 中华化学竞赛网www.1000hx.com

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10

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Part 1: Ion-Dipole Interactions

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The bonding of an ion, such as Na+, with a polar molecule, such as water, is an example of an



ion-dipole interaction. Shown below are a sodium ion, a water molecule, and a crown ether





O O O O O

Na

+

H2O
O

8-1-1

Draw the geometrical structure of the product resulting from the interaction between the



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sodium ion and water molecules.

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compound.





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dipole-dipole interaction are two common types of intermolecular forces.

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Intermolecular forces occur between, rather than within, molecules. Ion-dipole interaction and

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Problem 8: Intermolecular Forces

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How many nodal surfaces would you pass through for each of the following orbitals: 1s, 2s,



from the nucleus on the z axis, passing through the nucleus to a distant point on the –z axis.

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7-2

Describe the electron probability distribution for the 2pz and 3pz orbitals. How many nodes

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nodes does each orbital have respectively?

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8-1-2

Draw a diagram showing the interaction between the sodium ion and the crown ether molecule.

Part 2: Dipole-Dipole Interactions A hydrogen bond may be regarded as a particular kind of dipole-dipole interaction. electronegative atom, such as nitrogen, oxygen, or fluorine. Strong

hydrogen bonding forces are seen among molecules in which hydrogen is bound to a highly

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δ–

O–Hδ+···Nδ–

even in a vapor. 8-2-1



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In gaseous hydrogen fluoride, many of the HF molecules are associated into (HF)6.



中 华

8-2-2

Draw a diagram showing the hydrogen-bonding interactions between two acetic acid (CH3CO2H) molecules.

Part 3: Hydrogen-bonding in Living Matter

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Some chemical reactions in living matter involve complex structures such as proteins and DNA, and in these reactions certain bonds must be easily broken and reformed. Hydrogen bonding is the only type of bonding with energies of just the right magnitude to allow this.

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学 习



Draw the structure of this hexamer.

网 ww w.

of the order of 15 to 40 kJ/mol.

Hydrogen bonding is so strong that, in some cases, it survives

ww w. 10 0x ue xi .c 竞 赛 网 ww w. 10

Compared to other intermolecular forces, hydrogen bonds are relatively strong; their energies are

10

00 h

om

0x ue

The key to DNA’s functioning is its double-helical structure with complementary bases on the two



ww

strands.

The bases form hydrogen bonds to each other.

w.

10



H H N N N H N N R H H H

O N N N H

N N

H







Adenine (A)
H H H N R O N N H H H3C

Guanine (G)
O N N R O H

The organic bases found in DNA
12

圣才学习网http://www.100xuexi.com



华 化

Cytosine (C)

Thymine (T)



R

00

hx

.c

om

中华化学竞赛网http://www.1000hx.com

8-3

There are two kinds of hydrogen-bonded base pairs, T-A and G-C, in DNA. two base pairs, showing the hydrogen-bonding interactions.

Draw these

Problem 9: Crystal Packing
There are three cubic unit cells for the atomic solids, namely, simple cubic, body-centered cubic

x. c

om





中 华

9-1 9-2

How many nearest neighbor atoms are in each packing, respectively? In each packing, the packing efficiency is defined by

xi .c om

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中华化学竞赛网www.1000hx.com

学 习



ww w. 10 0x ue xi .c 竞 赛 网 ww w. 10 00 hx .c om

and face-centered cubic. They are illustrated in the following figures:

竞 赛

网 ww w.

fv =

10





determination, the emitted X rays were diffracted by a LiF crystal (d = 201 pm), and the first-order diffraction was detected at an angle of 34.68 . X-ray emitted by the metal.
o

Calculate the wavelength of the



华 化

13

圣才学习网http://www.100xuexi.com



9-4

X-ray diffraction is commonly used for the determination of crystal structures. In one such





of a silver atom is 144 pm. Calculate the density of silver.



ww

9-3

Silver crystallizes in a cubic closest packed structure, i.e. face-centered cubic. The radius

w.

10

What is the value of fv in each type of packing, respectively?

0x ue

volume occupied by spheres in the unit cell volume of the unit cell

00 h

om

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Problem 10: Applications of Transition Metals
The transition metal elements are widely distributed in the Earth’s crust. Many of these elements

find uses in everyday objects: iron pipes, copper wiring, chromium auto parts, etc.

Part 1: Properties of Chromium

00 h

网 ww w.

10-1-1

In an acidic solution, the yellow chromate ion (CrO4 ) changes into the orange
2-

dichromate ion (Cr2O7 ).

Write the equation for the reaction.

10-1-2

What is the oxidation state of each metal center in the chromate ion and the dichromate





中 华

10-1-3 10-1-4 10-1-5

Is this a redox reaction?

Explain.

What is the main factor that controls the equilibrium position? Draw the three-dimensional structures of CrO4 and Cr2O7 .
22-

xi .c om

圣 才

学 习



ion?

ww w. 10 0x ue xi .c
2-

竞 赛

10

om
(The atomic mass of Cr is

their use in pigments for artists paints and ceramic glazes.

x. c

alludes to its many colorful compounds. The bright colors of chromium (VI) compounds lead to

om

Chromium is a silvery-white, lustrous metal, whose name (from the Greek chroma, meaning color)

0x ue

10

w.

Cr2O7 solution where it serves as the cathode of an electrolytic cell.

ww





half-reactions and the overall cell reaction.



10-2-2



How many moles of oxygen gas will be evolved for every 52.0 g of chromium deposited?

10-2-3

If the current is 10.0 amperes, how long will it take to deposit 52.0 g of chromium onto the bumper?

10-2-4

What is the chemical reason that chromium is so useful for decorative plating on metals?



华 化

14

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10-2-1

Given that oxidation of H2O occurs at the anode, write equation for the two

网 ww

51.996; 1 F = 96,485 C.)



w.

10

2-

00

An antique automobile bumper is to be chrome plated.

The bumper is dipped into an acidic

hx

.c

Part 2: Uses of Chromium

om

中华化学竞赛网http://www.1000hx.com

Problem 11: Electrochemistry of Inorganic Compounds
Some inorganic compounds exhibit a variety of oxidation states, for example, many Mn compounds are known with oxidation states ranging from 0 to +7. reduction Mn
2+

The standard reduction For Mn in acidic

potential of a half reaction is measured against the hydrogen electrode. In this problem, the
2+ + 2 e → Mn, E° = -1.18V is expressed as Mn (-1.18) Mn. 3+

solution the reduction series: Mn

Mn Mn
3+

2+

Mn can be represented as follows:
2+

om

x. c

(1.5) Mn

(-1.18) Mn

X(N) / X(0) against the oxidation number, N, of the element, is used to indicate the most stable species of the compounds in different oxidation states. The Frost diagram of Mn
3+

网 ww w.

10

plot of nE° (n is the number of electron transferred in the half reaction) of the reduction couple / Mn
2+



竞 赛

shown below.



中 华

xi .c om

圣 才

学 习



ww w. 10 0x ue xi .c 竞 赛 网 ww w.
2+ (aq) -8

A redox reaction takes place spontaneously if the redox potential is positive. A Frost diagram, a

00 h

om

/ Mn is

0x ue

w.

the voltaic cell consisting of Mn(s) | Mn

(1M) || Mn



O2 (0.70) H2O2 (1.76) H2O.





11-2

For oxygen, the standard reduction potential in acidic solution can be represented as: Construct the Frost diagram for oxygen, and use the Could H2O2 undergo disproportionation information contained in the diagram to calculate the reduction potential of the half reaction for reduction of O2 to H2O. spontaneously?



ww

concentration of Mn

2+

/ MnCO3 | Mn(s), if the

in the right hand side of the cell is 1.0 10 M.

Xenon difluoride can be made by placing a vigorously dried flask containing xenon gas and fluorine gas in the sunlight. XeF2(aq) + 2H 11-3 The half-reaction for the reduction of XeF2 is shown below:



华 化

+ (aq)

+ 2e → Xe(g) + 2HF(aq)

-

E° = 2.32V

Use the VSEPR model to predict the number of electron-pairs and molecular geometry of
15

圣才学习网http://www.100xuexi.com





10

2+ (aq)

10

00

of MnCO3 is 1.8×10 .

-11

Use the Nernst equation to determine the potential at 25°C for

hx

11-1

The reduction potential depends on the concentration of the species in solution.

The Ksp

.c

om

中华化学竞赛网http://www.1000hx.com

XeF2.
o

Show that XeF2 decomposes in aqueous solution, producing O2, and calculate the Would you expect this decomposition to be favored in an acidic or a Explain.
+ -

E for the reaction. basic solution ?

2 H2O → O2 + 2H + 4e

E = -1.23V

o

Problem 12: Metal Carbonyl Compounds

x. c

om

00 h

carbonyl compounds.

网 ww w.

10

Carbon monoxide, as a two electron donor ligand, coordinates to transition metals to form metal For example, iron forms the pentacarbonyl metal complex, Fe(CO)5. Nickel tetracarbonyl, Ni(CO)4, has been used for the purification of Ni metal in the Mond process. Cobalt and manganese react with CO to form dinuclear complexes Co2(CO)8 and Mn2(CO)10,



竞 赛

Electron counts of these metal carbonyl complexes show that they obey the 18-electron rule. respectively. (Electronic configuration of Mn is [Ar](3d)5(4s)2)
-



中 华

metal centers is considered essential in order for the compounds to obey the 18 electron rule.

ferrocene (C5H5)2Fe, a classical compound, obeys the 18 electron rule. The reaction of W(CO)6 with sodium cyclopentadienide NaC5H5 yields an air sensitive compound

xi .c om

A.

Oxidation of A with FeSO4 yields compound B.

圣 才

The cyclopentadienyl anion C5H5 has also been widely used as a η5-ligand.

学 习



A metal-metal bond between the For example,

Compound A can also be prepared from
-1

ww w. 10 0x ue xi .c
-1

om 赛
The chemical shifts (δ) of the CH2 and CH

0x ue

w.

ww



composition was found for compounds C and D.



temperature compound C undergoes a transformation to yield compound D.

网 ww

propargyl bromide (HC≡CCH2Br) gives compound C containing a metal-carbon σ-bond. At room The same chemical
1







spectra are listed in the following table.
1

H NMR

HC≡CCH2Br 3.86 2.51

C 1.90 1.99 2.8



resonances and coupling constants JH-H of propargyl bromide, C and D in the respective H NMR

δ (CH2) δ (CH) JH-H (Hz)

华 化

2.7



16

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D 4.16 5.49 6.7

w.

10

synthesis of organometallic compounds containing metal–carbon bonds.

10

The reaction of A with

00

and 2010 cm .

-1

Compound A is a strong nucleophile and a good starting material for the

hx

spectrum, A shows absorption bands at 1744 and 1894 cm and B absorption bands at 1904,

.c

the reaction of B with Na/Hg, a strong reducing agent.

In the 1600-2300 cm region of the IR

om

中华化学竞赛网http://www.1000hx.com

W(CO)6

NaC5H5

A

FeSO4 Na/Hg B

HC

CCH2Br C metal migration D

x. c

om

12-1 12-2 12-3

Explain the differences in the IR spectra of A and B. Draw chemical structures for A, B, C and D.

00 h

The transformation of C to D involves a migration of the metal on the propargyl ligand. If DC≡CCH2Br is used for the synthesis of C, draw the structures of C and D.





中 华

xi .c om

圣 才

学 习



ww w. 10 0x ue xi .c 竞 赛 网 ww w. 10

竞 赛

网 ww w.

10

om 学 00 hx .c om











ww

w. 中
17

10
圣才学习网http://www.100xuexi.com

华 化

0x ue

中华化学竞赛网http://www.1000hx.com

Problem 13: Carbocations and Aromaticity 中华化学竞赛网www.1000hx.com
Carbocations are reactive intermediates having a charge of +1 on the central carbon atom, with three groups attached. The carbocation center is electron deficient and lies in the same plane of the three attached atoms. Proton NMR is one of the primary instrumental methods applied to determine the structure and properties of carbocations. In a superacid media, such as SbF5, SbF5 is a strong Lewis
-

om

stable carbocations can be generated and directly observed by NMR.

x. c

acid that complexes with a weak base such as F to form SbF6 .

-

F H3C H3 C CH3

SbF5

A





中 华

and the other, spectrum II, revealed a doublet at δ 1.30 with a coupling constant J = 20 Hz. Which spectrum was obtained from (CH3)3CF in SbF5? 13-3 Tropylium ion B is one of the most stable carbocations. How many π electrons are there

xi .c om

in the tropylium ion?

圣 才

学 习

in SbF5, respectively.

One spectrum, denoted as spectrum I, showed a singlet at δ 4.35,



13-2

Two proton NMR spectra of (CH3)3CF were obtained using neat (CH3)3CF and (CH3)3CF

ww w. 10 0x ue xi .c 竞 赛 网 ww w. 10

13-1

What is the product A in the following reaction?

竞 赛

网 ww w.

10

00 h

om 学
What does the proton

H

H + H B

0x ue





13-5

The chemical shift of benzene in the 1H NMR spectrum is δ 7.27.

NMR spectrum of B look like? (a) A singlet at δ 9.17. (b) A singlet at δ 5.37 (c) A triplet at δ 9.17. (d) A Triplet at δ 5.37.





13-4

Is tropylium ion B an aromatic structure?



ww

w.

10

H

H

Explain.



华 化

18

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00

hx

H

H

.c

om

中华化学竞赛网http://www.1000hx.com

13-6

4-Isopropyltropolone C was the first example of a non-benzenoid aromatic compound. University in 1938. Draw resonance structure(s) to illustrate the aromaticity of C.
O OH

It

was isolated from cypress trees in Taiwan by Professor T. Nozoe at the National Taiwan

om

C

x. c

What is the structure of D?



竞 赛

网 ww w.

10

with one mole of tris(2,4-pentanedionato)iron(III) (Fe(acac)3) to form a red complex D.



中 华

Problem 14: Photochemical Ring Closure and Opening
1,3,5-Hexatriene is known to undergo light-induced cyclization to give 1,3-cyclohexadiene. (A) with UV-light gives cyclohexadiene (B).

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学 习



ww w. 10 0x ue xi .c 竞 赛 网 ww w. 10

13-7

The proton of the OH group in tropolone is acidic. Three moles of tropolone C can react

00 h

photochemical reaction is reversible and stereospecific. Thus, irradiation of (2E,4Z,6E)-octatriene The choice of the wavelength of light depends on the absorption maximum of the compound to be irradiated, and the absorption maximum is

0x ue

w.

10

H3C

CH3

H3C

CH3

A

B

14-1

What is the chemical name of the starting triene (C) for the related reaction shown below?





ww



irradiation





H3 C

CH3

D

A similar reaction mechanism is involved in the synthesis of biologically active molecules. For example, in the presence of sunlight, 7-dehydrocholesterol (E) undergoes an electrocyclic ring opening reaction to give provitamin D3 (F), which can be further transformed through a [1,7]-hydrogen shift to yield vitamin D3 (G).



华 化

19

圣才学习网http://www.100xuexi.com



C

00

hx

irradiation

.c

om

related to the number of conjugated double bonds in a chain.

xi .c om

om

The

中华化学竞赛网http://www.1000hx.com

H3C H3C H3C H HO H

(CH2)3

CH3 CH3 sunlight provitamin D3 (F) [1,7] H-shift

H3C H3C

(CH2)3

CH3 CH3

H

7-dehydrocholesterol (E) HO

vitamin D3 (G)

14-2

x. c

expect to absorb light with the higher energy? 14-3 What is the chemical structure of F?

om

Of the two compounds 7-dehydrocholesterol (E) and vitamin D3 (G), which would you (E or G)

00 h

of colorless compound H with UV light gives colored compound I.

网 ww w.

This principle has been elaborated to develop photochromic materials. For example, irradiation The color change is reversed upon exposure to visible light. 14-4



Give the structure of colored compound I.
UV light

中 华

学 习

CH3 CH3 H 3C O O



H

O

colorless

Aromatic hydrocarbons are usually highly fluorescent.

圣 才

visible light

colored compound

However, a neighboring amino

xi .c om

substituent may quench the fluorescence. below.

This quenching mechanism is due to Photoinduced



I

ww w. 10 0x ue xi .c 竞 赛 网 ww w. 10

竞 赛

10

om
In the presence of a
20

Electron Transfer (PET), which can be clearly illustrated by the molecular orbital diagrams shown

0x ue

up to the lowest unoccupied molecular orbital (LUMO) (state b).

process, and retrieves the fluorescence of the aromatic chromophore (step 3).







Coordination of the amine lone-pair electrons to proton or metal ions efficiently inhibits the PET



of the excited chromophore (step 2), and thus blocks the normal fluorescent pathway (state c).

ww

neighboring amino group, one of the lone-pair electrons at the nitrogen atom moves to the HOMO

w.

10



圣才学习网http://www.100xuexi.com



华 化



00

chromophore (state a) will pump an electron from the highest occupied molecular orbital (HOMO)

hx

Upon irradiation with a light of suitable wavelength (step 1), the initial aromatic

.c

om

中华化学竞赛网http://www.1000hx.com

(a) LUMO hν (1)
amine lone-pair electrons

(b) LUMO
PET (2)

(c) LUMO

HOMO

HOMO
M+ (3)

HOMO
no fluorescence

N

x. c

om
(d) LUMO

00 h

竞 赛

HOMO
N M





中 华

Many interesting and sensitive proton or metal ions fluorescent sensors have been developed based on the manipulation of the PET process. pH-sensor.

xi .c om

圣 才

For example, compound J is used as a

N

学 习



ww w. 10 0x ue xi .c 学 竞 赛 网 ww
In this molecular The

网 ww w.

10

om
21

0x ue

10

w.

14-5

Do you expect that compound J is fluorescent in an alkaline solution (pH = 10.0)?

Problem 15: Stereochemistry 中华化学竞赛网www.1000hx.com
A simple two-dimensional representation for showing the three-dimensional arrangement of groups bonded to a carbon center is called a “Fischer projection”. representation, the intersection point of two perpendicular lines represents an sp3 center. plane relative to the observer.

The vertical lines connecting A and C to the central carbon atom

圣才学习网http://www.100xuexi.com



华 化

horizontal lines connecting B and D to the central carbon atom represent the bonds that are out of











ww

w.

10

J

00

Cl

hx

.c

om

中华化学竞赛网http://www.1000hx.com

represent the bonds that are directed away from the observer into the plane of the page.
A D C B D C A B into plane out of plane

15-1

ChiraPhos was developed by Prof. Kagan and has found useful applications in asymmetric synthesis. Using the Fischer projection shown below, assign the absolute the Cahn-Ingold-Prelog sequence rule. configuration of the chiral centers of ChiraPhos in terms of the R/S notation according to

x. c

om

ChiraPhos



竞 赛

15-2

One of the stereoisomers of ChiraPhos is a meso compound.

X and Y in the Fischer projection shown below?
X H3C Y PPh2



中 华

2 3

H

meso-ChiraPhos

It is very useful and common to use Fischer projections for the stereo representation of

0x ue









(open chain)

(cyclic structures)

The hemiacetal formation generates two stereoisomers, which are called “anomers”.
o o



D-glucose

α-anomer

β-anomer





H HO H H

CH 2 OH 3 H 4 OH 5 OH CH2OH

10

HO C H OH HO H H OH H O CH2OH

ww

网 ww

or

H C H OH HO H H OH H O CH2OH

w.



w.
The pure

α-anomer of D-glucose has a specific rotation of +112.2 , whereas the pure β-anomer has a specific rotation of +18.7 . In water, either one of the two anomers gives an equilibrium mixture with a specific rotation of +52.6 . 15-3 Calculate percentage of α-anomer in the equilibrium mixture of D-glucose in water.
o



华 化

22

圣才学习网http://www.100xuexi.com

10

00

O

H

OH

hx

structures through the hemiacetal formation between the C5-OH and the C1-aldehyde group.

.c

om

D-glucose.

It is interesting to note that the open-chain glucose can be interconverted with cyclic

xi .c om

carbohydrates.

For example, the following Fischer projection represents the structure of

圣 才

学 习

H



ww w. 10 0x ue xi .c

网 ww w.

CH3 H 2 PPh2 Ph2P 3 H CH3

10

00 h

What are the identities of

om

中华化学竞赛网http://www.1000hx.com

15-4 15-5 15-6

Which is the more stable anomer in water? (α or β) Draw the chair conformation of the β-anomer. What is the common intermediate for the interconvertion between the α- and β-anomers?

The addition reaction of HCN to an aldehyde generates a cyanohydrin, which can be further

x. c
O

om

reduced to form an α-hydroxyaldehyde.

00 h

HCN R

reduction R

10

CN

High homologs of carbohydrates, such as D-talose, can be synthesized from D-glyceraldehyde by

竞 赛

repeating the same reaction conditions three times as shown below.
CHO HO H H OH CH2OH

网 ww w.

cyanohydrin



中 华

H

2. reduction

D-glyceraldehyde

stereoisomer

圣 才

学 习

CHO OH CH2OH



1. HCN



xi .c om

15-7

How many pairs of enantiomers will appear in the final product mixture?

ww w. 10 0x ue xi .c
O

R

H

H

HO HO HO H

CHO H H H OH CH2OH

D-talose + stereoisomers

om 网 ww
R

OH

OH

0x ue

10

w.









O R yeast

O O R +



racemic mixture



华 化

23

圣才学习网http://www.100xuexi.com



Villiger reactions

O



Table 1. Yeast-mediated kinetic resolution of racemic 2-substituted cyclohexanone via Baeyer



ww

racemic 2-substituted cyclohexanone via Baeyer Villiger reactions (Table 1).

w.

synthetic methodology.

Following are the data for the yeast-mediated kinetic resolution of

10

00

organic synthesis has become one of the fastest growing areas for the development of new

hx

in living organisms. Because of their striking catalytic power and specificity, applying enzymes in

.c

Enzymes are remarkable biological catalysts that control the pattern of chemical transformations

om

中华化学竞赛网http://www.1000hx.com

O O R

O R

Entry 1 2 3

R Et n-Pr Allyl

Yield (%) 79

ee% 95 97 98

Yield (%) 69 66 58

ee% 98 92 98

x. c

om

54

59

15-8 15-9

What is the ratio of (R)/(S) isomers of 6-allylcaprolactone in entry 3?



竞 赛

MCPBA (meta-chloroperbenzoic acid) is a common oxidizing agent for Baeyer Villiger

reactions. Using MCPBA as an oxidizing agent, instead of yeast, for the above reaction,



中 华

Problem 16: Organic Synthesis 中华化学竞赛网www.1000hx.com

10

w.

ww



9 Fluorene Biphenyl







substituents have been introduced at the C9 position of fluorene.



applications in flat-panel-display technology.

In order to avoid intermolecular interactions, bulky One example of this is

compound C, an interesting and useful building block in the development of a highly efficient blue light emitting material, the synthesis of which is shown in the following scheme.
NH2 1) NaNO2, HCl 0-5 oC A 2) KI 1) Mg, Et2O HOAc, HCl B reflux
O



Many fluorene derivatives have been developed that have subsequently found promising





华 化

2) 3) H2O



24

圣才学习网http://www.100xuexi.com

网 ww

8

1

C (C25H16)

w.

7

2

10

6

5

4

3

00

hx

fluorescence than biphenyl.

0x ue

.c

For example, fluorene, a methylene-bridged biphenyl, exhibits a higher quantum yield of

om

search for highly efficient luminescent materials which can be either small molecules or polymers.

xi .c om

One of the primary demands for the development of organic light emitting diodes (OLEDs) is the

圣 才

学 习



what will be the ee% of the caprolactone product?

ww w. 10 0x ue xi .c

ee: enantiomeric excess

网 ww w.

10

00 h

om

中华化学竞赛网http://www.1000hx.com

16-1

Draw the structures of A, B, and C.

Liquid crystals have become a part of our daily life, from wristwatches, pocket calculators, to full color flat-panel displays. Liquid crystal molecules usually contain a rigid core with a flexible terminal alkyl chain as shown in the example below.
CN Flexible terminal alkyl chain

om

Rigid core

x. c

bromide or iodide with arylboronic acid, known as the Suzuki coupling reaction.



竞 赛

网 ww w.

kind of structure can be effectively synthesized through palladium catalyzed coupling of an aryl



中 华

A typical example of a Suzuki coupling reaction is shown below.

学 习



Terphenyl

Br

+

B(OH)2

Pd(0) catalyst

xi .c om

The following is a synthetic scheme for the preparation of two liquid crystal molecules, 4-cyano-4’-pentylbiphenyl and G.

圣 才

phenylboronic acid in the presence of a palladium catalyst to give biphenyl.

ww w. 10 0x ue xi .c
C5H11

Biphenyl and terphenyl are common basic skeletons for the rigid core of liquid crystals. This

10

00 h

Bromobenzene reacts with

om

0x ue

10

w.

ww

D

E



AlCl3

KOH, heat

F F









G

16-2

What are the structures of D, E, F, and G?



华 化

25

圣才学习网http://www.100xuexi.com



Pd(PPh3)4 Na2CO3 MeOCH2CH2OMe, H2O



C8H17O



网 ww

Br2

C4H9COCl

NH2NH2

F

B(OH)2

w.

CuCN DMF

10

00

NC

hx

.c

om

中华化学竞赛网http://www.1000hx.com

Problem 17: Spectroscopy and Polymer Chemistry 中华化学竞赛网www.1000hx.com
Organic polymers have had a great impact on our daily lives. polymers are produced every year. Millions of tons of different kinds of They are widely used as

Synthetic organic polymers have a variety of uses, from

textiles to computer chips, and to the life saving artificial heart valve. (PVA) is an important example of a water-soluble polymer.

plastics, adhesives, engineering materials, biodegradable plastics, and paints. Poly(vinyl alcohol) One synthetic route to PVA is

Polymer B shown above is also a major component in chewing gum.

网 ww w.

10

Monomer A

Polymer B

Poly(vinyl alcohol) (PVA)

竞 赛

suggests a ratio of C:H:O = 56:7:37. identical composition of C, H, and O.

In addition, elemental analysis of B gives an almost The following IR and 1H NMR spectra were recorded for





中 华

xi .c om

圣 才

学 习



Monomer A.

ww w. 10 0x ue xi .c 竞 赛 网 ww w. 10

polymerization

00 h

Elemental analysis of A

om
H NMR spectrum of Monomer A

Scheme 1

x. c

om

summarized in Scheme 1.

0x ue









ww

w.

10



1



华 化

26

圣才学习网http://www.100xuexi.com



00

hx

.c

om

中华化学竞赛网http://www.1000hx.com

110 100
3503 3094 1434 849 977 1295 951 876

90

网 Transmittance (%T) ww w. 10 00 hx .c om
80 70 60 50 40

1021 1372

1648

1761



中 华

wavenumber (cm )
IR Spectrum of Monomer A 17-1 17-2 17-3 17-4 17-5 17-6 What is the molecular formula of A?

xi .c om

圣 才

学 习

30 4000



3500

3000

2500

2000



-1

ww w. 10 0x ue xi .c
1217

1138

竞 赛

1500

1000

om w. 10 00 hx .c om

Which functional group would give an IR absorption band at 1761 cm-1? What is the chemical structure of A?

ww







17-7



How many pairs of enantiomers would be obtained from polymer B with a molecular

weight of 8600, assuming that the polymer is terminated by hydrogen abstraction and the mass of the terminal groups can be ignored? Compound C, an isomer of A, is also an important monomer for the synthesis of polymers. On the basis of its 1H NMR and IR spectra provided below, deduce the chemical structure of C.



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27

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Provide a method for the transformation of B to PVA.



网 ww

Draw a portion of polymer B.

w.

10

0x ue

Show at least three repeat units.



中华化学竞赛网http://www.1000hx.com

x. c

om

00 h





中 华

H NMR spectrum of Monomer C

xi .c om

100
35533445 3632

2697 2587 2062

圣 才

1

3107

2856

学 习

1945



ww w. 10 0x ue xi .c
2999 662 2955 1634

竞 赛

网 ww w.

10

om 竞 赛 网 ww
988 1069 1404 1279 1205 1731

才Transmittance (%T) 学 习 网 ww w. 10 0x ue

60

40



20

4000

3500

3000

2500

2000
-1



1439

1500

1000

wavenumber (cm )
IR Spectrum of Monomer C



华 化

28

圣才学习网http://www.100xuexi.com

w.
812

854

10

00

hx

80

.c

om

中华化学竞赛网http://www.1000hx.com

Polymer D is an acid sensitive macromolecule. On treatment of D with a protic acid, gases E and F are released, and a new polymer G is formed. Gas E turns an aqueous solution of Ca(OH)2 into a turbid solution while gas F reacts with bromine to give a colorless solution of H.
H+ E aqueous Ca(OH)2 turbid solution

n

+

F Br2

+

G

x. c

O

om
O

00 h

17-8

Draw the chemical structures of E, F, G and H?

acid-generator (PAG).

竞 赛

Polymer D has been used to formulate a photo-imaging material by mixing it with a photo After being coated onto a substrate and exposed to light, protons are





中 华

baking after exposure is necessary to accelerate the acid catalyzed reaction.

学 习



generated from the PAG, catalyzing a chemical reaction within the polymer matrix.

baking and using an aqueous basic developer to wash away the acidic materials, a patterned substrate I is obtained.
Light

xi .c om

0x ue

10

w.

ww









The dark areas represent a polymeric coating that is structurally different from the original one.



华 化

29

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(a)

(b)

(c)

网 ww

17-9

Which of the following diagrams best illustrates the patterned structure of substrate I?



(d)

w.

10

Figure 1

00

hx

Patterned mask Polymer D + PAG Substrate

圣 才

through a mask (Figure 1), a latent image of the mask is formed in the polymer matrix. After

ww w. 10 0x ue xi .c

D

网 ww w.

10

Sometimes

If light is applied

om .c om

O

H (colorless)

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Problem 18: Crown Ether and Molecular Recognition 中华化学竞赛网www.1000hx.com
The discovery of crown ethers was made by Charles Pederson in 1967. contributions to supramolecular chemistry. The following schematic diagram shows a synthetic route to the linear diol A. the minor product B. However, because He was rewarded with

a share of the 1987 Nobel Prize for chemistry with Jean-Marie Lehn and Donald Cram for their

x. c

om

of the presence of catechol in the starting material, the resulting product was a mixture of A and

00 h

10

网 ww w.

OH

OH + Cl O Cl 1) NaOH 2) H3O+ O

竞 赛

O O Contaminated with catechol shown below OH OH Catechol







中 华

The molecular weight of B was found to be 360.1 and it had an elemental composition of C:H:O = 66.5:6.7:26.6. The 1H NMR spectrum of B showed four sets of proton signals. Two of which

xi .c om

were observed at δ 7.0-7.5 and the other two at δ 3.7-4.2.

圣 才

学 习

The integration ratio of the four sets

ww w. 10 0x ue xi .c
HO O O A + B

om 中 华 化
30

of signals was 1:1:2:2. Compound B strongly binds with the potassium ion. A striking example

0x ue

w.

ww





18-2

What is the major function of H3O+ in the above reaction?





The following is the synthetic route to [2.2.2]cryptand:



(b) To neutralize NaOH. (c) To remove the tetrahydropyran group. (d) To act as a buffer to control the pH of the solution.

圣才学习网http://www.100xuexi.com





(a) To activate ClCH2CH2OCH2CH2Cl.



网 ww

w.

10

18-1

What is the chemical structure of B?

10

00

hx

is the use of B to assist the dissolution of KMnO4 in benzene to give a purple coloration.

.c

om

中华化学竞赛网http://www.1000hx.com

O O O OH HO O O SOCl2 C NH2

O H2N LiAlH4 D E

high dilution O O O N O O [2.2.2]Cryptand

C

1, B2H6 F

O N

x. c

om

2. H2O

18-4

Why are high dilution conditions required in the synthesis of D from C?

竞 赛

(a) Since the reaction between C and the diamine is highly exothermic, a high dilution condition is used to dissipate the heat released from the reaction.





中 华

(d) The solubility of the starting material is low.

0x ue

crown ethers.

10

w.

O O

ww

O







12-C-4

15-C-5

18-C-6

21-C-7



Table 1. Radii of different alkali metal cations and diameters of the cavities of crown ethers. Cation (radius, pm) Li+ (68) Na+ (98) K+ (133) Cs+ (165) 18-5 Cavity of the crown ether (diameter, pm) 12-C-4 (120-150) 15-C-5 (170-220) 18-C-6 (260-320) 21-C-7 (340-430)



On the basis of the above data, match the experimental curves I-III in Figure 1 with



appropriate cyclohexyl crown ethers G-I .
31

圣才学习网http://www.100xuexi.com

华 化







O

O

O O

O

O O

O

网 ww

O

O

O

O

O

w.

O

O

O

O

O

O

Cavity

10

00

hx

.c

Table 1 shows the radii of different alkali metal cations and the diameters of the cavities of several

om

host cavity of the crown ether and the guest cation, and the number of donor atoms of the host.

xi .c om

The affinity of a metal cation is controlled by several factors such as size matching between the

圣 才

(c) Thermal equilibrium is favored to give D in higher yield under high dilution conditions.

学 习

(b) A high dilution condition is employed in order to inhibit oligomer or polymer formation.



ww w. 10 0x ue xi .c

18-3

Draw the structures of C-F.

网 ww w.

10

00 h

om

中华化学竞赛网http://www.1000hx.com

log Kf 6.0 III 5.0 4.0 II 3.0 O O O O O O O O O G O O O O O O H

2.0 1.0

x. c
I

om

00 h

Na

K

Cs

Cation Radius

Figure 1. Complexation of Crown Ethers with Cations in Methanol





中 华

Problem 19: Enzyme Catalysis 中华化学竞赛网www.1000hx.com
In biological systems, oxidases catalyze the following reaction O2 + 4H
+

+ 4e

-

→ 2H2O

0x ue

10

10

00
-1

FeIIcyt c → FeIIIcyt c + e

-

w.

ww







19-1 19-2 19-3



How many moles of cyt c were oxidized each second? How many moles of oxygen were consumed each second? What is the turnover number for oxidase? (Turnover number: the number of product produced by one catalyst per second)



华 化

32

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solution containing 2.7 x 10 M oxidase, and sufficient cyt c and oxygen.

-9



The absorbance of cyt c at 550 nm was observed to decrease at a rate of 0.1 A/sec in a 5 mL



网 ww

respectively.



w.

The extinction coefficients of FeIIcyt c and FeIIIcyt c at 550 nm are 27.7 and 9.2 mM cm ;

hx

.c

which contains an iron center.

The half reaction is

om

The reaction is the key to respiration in living systems. The electrons come from Cytchrome c,

xi .c om

圣 才

学 习



ww w. 10 0x ue xi .c

竞 赛

网 ww w.

10

om

I

中华化学竞赛网http://www.1000hx.com

Problem 20: Work in Thermodynamics
Given 10 liters of an ideal gas at

0° C and 10 atm, calculate the final volume and work done,

under the following three sets of condition, to a final pressure of 1 atm 20-1 20-2 20-3 Isothermal reversible expansion Adiabatic reversible expansion

00 h

is the gas constant.]



竞 赛

网 ww w.

[Note that the molar heat capacity at constant volume is given by the relation: C v =



中 华

Problem 21: Kinetics — Atmospheric Chemistry

The following second-order reaction is important in air pollution.

2 NO2 → 2 NO + O2

0x ue

w.

ww







Problem 22: Kinetics and Thermodynamics 中华化学竞赛网www.1000hx.com
The concept of kinetic versus thermodynamic control of reaction products has frequently been employed in organic synthesis to direct product formation, for example, in sulfonation, Diels-Alder, isomerization and addition reactions. Here the interconversion of two different products can be It is normally represented as a achieved competitively by controlling the reaction conditions. competitively toward products B and C.

concurrent reaction scheme, as shown below for the case where the reaction of A proceeds





华 化

33

圣才学习网http://www.100xuexi.com







网 ww



w.

10

constant.

10

00

a temperature of 600 °C, the reaction is 50% complete after 3 min.

Calculate the rate

hx

21-2

It was found that when a 2 liter vessel is filled with NO2 at a pressure of 600 mm of Hg and

.c

om

originally containing pure NO2, at the time t.

xi .c om

21-1

Derive an integrated relationship between the total pressures in the reaction vessel,

圣 才

学 习



ww w. 10 0x ue xi .c

3 R , where R 2

10

om

released to 1 atm and the gas expands adiabatically at constant pressure.

x. c

Irreversible adiabatic expansion carried out as follows: Assume the pressure is suddenly

om

中华化学竞赛网http://www.1000hx.com

k1 k2 ← ?? ? ?→ B A C, ??→ ←?? k ?1 k? 2

The energy profile for the reaction is depicted in the Figure below.

x. c

om

00 h





中 华

0x ue

10

w.

ww



favored?





Problem 23: Phase Diagram 中华化学竞赛网www.1000hx.com
The phase diagram is a convenient way to indicate the phases of a substance as a function of temperature and pressure. Answer the following questions based on the phase diagram of water given below: 23-1 23-2 What phases are present at A, B, and C? Why does ice not sink in its own liquid?
34



圣才学习网http://www.100xuexi.com



华 化







网 ww

product.

When the temperature of the system increases, which reaction process is



w.

22-3

B is called the kinetic-controlled product, while C is called the thermodynamic-controlled

10

00

exceeds 4 days.

hx

22-2

Using the same rate constants, estimate the product ratio B/C when the reaction time

.c

om

ratio B/C in the first 4 min of the reaction.

xi .c om

22-1

Given the rate constants k1=1, k-1=0.01, k2=0.1, and k-2=0.0005 min-1, estimate the product

圣 才

学 习



ww w. 10 0x ue xi .c

竞 赛

网 ww w.

10

om

中华化学竞赛网http://www.1000hx.com

23-3

When water freezes, it expands. which may be expressed by

Explain this observation using the Clapeyron equation,

?H dp = . Here, ?H and ?V denote the change of molar dT T?V

enthalpy and molar volume of water, respectively. 23-4 A glass container partially filled with water is connected to a vacuum pump. changes will be observed when the pump is turned on? What

temperature is 0 C. What changes will be observed along the skating track on the

x. c
o

om

23-5

A man is skating on the surface of a sheet of ice where the air pressure is 1 atm and the surface of the ice, assuming the ice can withstand his weight without cracking?

00 h





中 华

xi .c om

圣 才

学 习



ww w. 10 0x ue xi .c 竞 赛 网 ww w. 10 00 hx .c om

竞 赛

网 ww w.

10





Problem 24: Standard Deviation in One - Dimensional Quantum Mechanics





ww

w.

10

0x ue

expressed by a standard deviation, σ. σ=



Some measurable physical quantity, if measured numerically, may lead to an uncertainty which is Such a standard deviation is defined as

< G 2 > ? < G >2 ,

where G is a measurable physical property; <G> is the average value of G; <G2> is the average value of G2. The average values, <G> and <G2>, can be obtained by integrating the corresponding physical quantity multiplied by its probability distribution over all the values of G. This definition may be applied to both classical and quantum mechanical worlds. Two examples



related to the estimate of σ, one for the kinetic property of gaseous molecules and the other for the
35

圣才学习网http://www.100xuexi.com

华 化



om

中华化学竞赛网http://www.1000hx.com

particle motion in one dimension, are given in the following. 24-1 The distribution of speeds of gaseous molecules at a fixed temperature can be described by the following probability density, called the Maxwell-Boltzmann distribution,
3/ 2

? M ? F ( v) = 4πv 2 ? ? ? 2πRT ?

? ? Mv 2 ? exp? ? 2 RT ? ? ? ?

24-2

Suppose a particle moving in the x direction has a normalized wavefunction,

竞 赛

? = [(1 / 2π ) exp(? x 2 / 2)]1 / 2 ;

?∞ ≤ x ≤ ∞ ,



Calculate the average position, <x>, and the standard deviation,



中 华

24-3

In quantum mechanics, momentum for one dimension can be expressed by an operator, i.e., p= ?

ih d , where h is the Planck’s constant. Calculate the average momentum, 2π dx

0x ue

24-4

Calculate the uncertainty product of position and momentum, σxσp, for the above quantum

10

w.

ww











0

x 2 n +1 exp(? ax 2 )dx =

n! 2a n +1

where n = 0,1,2,3…

Problem 25: A Particle in a 2-D Box Quantum Mechanics 中华化学竞赛网www.1000hx.com
The π electrons of the iron-heme of a hemoglobin molecule can be visualized as a system of free



electrons moving in a two-dimensional box.

华 化

According to this model, the energy of the electron is
36

圣才学习网http://www.100xuexi.com











0

1 ? 3 ? 5 ? ? ? (2n ? 1) ? π ? x exp(?ax )dx = ? 2 n +1 ? 2 n +1 ? ?a
2n 2

1/ 2



网 ww

Some useful integrals are given below:



w.

10

mechanical example.

00

hx

.c

om

in part 2.

xi .c om

<p>, and the standard deviation, σp, for the particle with the same wavefunction described

圣 才

学 习

distribution of the particle after a large number of measurements of x.



ww w. 10 0x ue xi .c

网 ww w.

10

K mol )

-1

-1

00 h

σ x , of the position

om

σv, of the distribution of speeds of the O2 molecules at 300 K. (O2 = 32 g/mol, R = 8.31 J

x. c

and R is the gas constant.

om

where v is the speed of molecule, M is the mass of molecule, T is the temperature in Kelvin, Calculate the average speed, <v>, and the standard deviation,

中华化学竞赛网http://www.1000hx.com

limited to the values

where h=6.63 x 10 me=9.11 x10 25-1 25-2
-31

-34

Js

is the Planck constant; nx and ny are the principal quantum numbers;

kg is the electron mass; L is the length of the box.

Given the molecule contains 26 electrons, determine the electron population of the highest

x. c

om

Construct an energy level diagram showing the relative ordering of the lowest 17 orbitals.

paramagnetic.

竞 赛

25-4

Light is absorbed only when the condition hν = ? E is satisfied.

网 ww w.

25-3

Assuming Hund's rule can be applied to this system, predict whether or not this system is

box is 1 nm, what is the longest wavelength of light that can lead to excitation?





中 华

Problem 26: Spectral Analyzer 中华化学竞赛网www.1000hx.com

0x ue

Beam splitter

10

w.

ww





Slit



Dye cell-2

Dye cell-1





DFDL

<

preamplifier
Mirror Mirror

Wavemeter

Grating

华 化

oscillator



Fig. 1. Block diagram of the distributed-feedback dye laser.
37

圣才学习网http://www.100xuexi.com







网 ww

Cylindrical lens

w.

10

<

Laser (355 nm)

00

hx

.c

om

consists of an oscillator and a preamplifier.

xi .c om

The configuration of the distributed feedback dye laser (DFDL) system is shown in Fig. 1, and

圣 才

学 习



your result in nm. [The speed of light, c = 3.00 x10 m/s]

8

ww w. 10 0x ue xi .c

occupied orbitals in the ground state.

10

00 h

If the length L for this 2D Express

om

中华化学竞赛网http://www.1000hx.com

The oscillator is made of a quartz cuvette (dye cell-1) equipped with a dye circulation pump. detailed construction of the oscillator is shown in Fig. 2.
Dye cell n

The

??

x. c

λDFDL

om

λDFDL

00 h

λP

λP





中 华

determines the wavelength of the laser emission. can be calculated using the following equations:

The wavelength of the laser emission, λDFDL,

λDFDL = 2nΛ Λ = λ p / 2 sin θ

0x ue

turn, can be determined from the angle of incidence of the pump beam.

The DFDL beams come

10

w.

ww









(a) 374 nm (b) 474 nm (c) 574 nm (d) 674 nm (e) 774 nm



华 化

38

圣才学习网http://www.100xuexi.com







26-1

What would be the wavelength of the DFDL when the angle of θ is 60.00° and the refractive index of the medium is 1.40?

网 ww

dye cell-2).



w.

power of the DFDL can also be amplifier by passing it through a preamplifier (the second dye cell;

10

from two sides of the cell.

The wavelength of the DFDL can be measured by a Wavemeter. The

00

hx

.c

normal surface. The laser wavelength can also be determined from the fringe spacing, which, in

om

where n is the refractive index of the medium; Λ, the fringe spacing; and θ, the angle from the

xi .c om

圣 才

学 习

subsequently focused onto the dye solution to form an interference pattern, the spacing of which



Two laser beams (λP = 355.00 nm) are reflected by two rotating dielectric mirrors and

竞 赛

Fig. 2. The detailed construction of a DFDL oscillator

ww w. 10 0x ue xi .c

网 ww w.

10

om

中华化学竞赛网http://www.1000hx.com

Problem 27: Time-of-Flight Mass Spectrometer 中华化学竞赛网www.1000hx.com
There are numerous ways in which we may separate and select ions in a mass spectrometer. Magnetic and electric fields, as well as radio frequency are often used to separate ions in mass spectrometers. The time-of-flight (TOF) mass spectrometer is the simplest type of common With the introduction of matrix-assisted laser mass analyser and it has a very high sensitivity.

desorption / ionization (MALDI) or electrospray ionization (ESI), which is used to introduce and range became achievable. It is now possible to desorb and analyze ions with molecular weights

x. c

om

ionize macromolecules such as proteins, DNA, and polymers, the measurement of a large mass upwards of one million atomic mass units (amu; 1 amu = 1.6605 × 10
-27

accelerate the ions from the source into the field-free “drift” zone of the instrument. different velocities depending on their m / z value.

网 ww w.

being employed.

A high voltage (HV) potential is applied across the source to extract and The

fundamental operating principle of TOF is that ions of the same kinetic energy will move with

This can be seen in the following equation, the





中 华

KE = ? mν

This relationship may be rearranged to give velocity in terms of kinetic energy and m / z. ν = (2 ? KE/m)?

0x ue

10

w.

ww





microseconds (?s) (depending on the flight distance). by a potential is given by:

The kinetic energy of an ion accelerated









KE = zeV





网 ww

energies.

At these energies, flight times will be in the range of a few tens to a few hundreds of

w.

In most modern TOF mass spectrometer, ions are generally accelerated to kilo-electron volt (keV)

10

00
-19

t = L / ν = L / (2 ? KE/m)?

Where z is the number of charges on the ion, e is the fundamental unit of charge (1.6022 × 10 and V is the applied accelerating potential in volts. 27-1 An average protonated cytochrome has a molecular weight of 12,361 amu.
+

What will be

the velocity of the (MH ) ion of the cytochrome when accelerated by a potential of 20,000 volts? (a) 18000 m/s



华 化

39

圣才学习网http://www.100xuexi.com

hx
C)

.c

equation:

om

the extraction time from the source), the time of flight (t) can be calculated by the following

xi .c om

If the distance from the point of ion formation to the detector at some fixed point is L (neglecting

圣 才

学 习

2



classical equation for kinetic energy.

ww w. 10 0x ue xi .c

sample ions are generated in a source zone of the instrument, by whatever ionization method is

竞 赛

10

00 h

om

kg). In general, the

中华化学竞赛网http://www.1000hx.com

(b) 28000 m/s (c) 38000 m/s (d) 48000 m/s (e) 58000 m/s 27-2 If the detector is positioned 1.00 m from the point of ion formation, the time of flight for the

00 h

(c) 60 ?s



竞 赛

(d) 70 ?s (e) 80 ?s

网 ww w.

10

(b) 50 ?s



中 华

xi .c om

圣 才

学 习



ww w. 10 0x ue xi .c 0x ue 中 华 化
40

om 学 竞 赛 网 ww w. 10 00 hx .c om

(a) 40 ?s










圣才学习网http://www.100xuexi.com

ww

w.

10

x. c

om

ion will be approximately:


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