Titanium complexes have been investigated for their antitumor activity. Many factors including isomerism and sizes have shown to affect the potency of the complexes. This question deals with the synthesis and characterization of some titanium complexes.
7-A1) A reaction of 2 equivalents of 2-tert-butylphenol, 2 equivalents of formaldehyde, and N,N'-dimethylethylene-1,2-diamine under acidic conditions at 75 C affords three major products with the same chemical formula of C26H40N2O2, as shown in the equation below.
Draw the structure of each product.
Ans
(4.5 points) Score distribution: +1.5 points for each product If phenolic OH is used as a nucleophile for the iminium ion, get 0.5 point.
Reasonable structures with missing Cs results in 0.25 deduction
Problem 7
6% of the total
7-A2) If 2,4-di-tert-butylphenol is used as a substrate instead of 2-tert-butylphenol using the same stoichiometry as that in 7-A1), only one product X was obtained. Draw the structure of X.
Ans
(1.5 points) 0 point for other isomers (meta-substitutions, etc.) If 2,6-di-tert-butylphenol is drawn (with correct substitution), 0.25 deduction.
A reaction between X from 7-A2) and Ti(OiPr)4 [iPr = isopropyl] in diethyl ether under an inert atmosphere resulted in the six-coordinate Ti complex Y, as a yellow crystalline solid and isopropanol at room temperature.
(equation 1) UV-Vis spectra of X, Ti(OiPr)4, and Y reveal that only the product Y has an absorption at = 370 nm. By varying the volumes of X and Ti(OiPr)4, each with the concentration of 0.50 mol dm-3, and using benzene as the solvent, the absorbance data at = 370 nm are given below:
Volume of X
7-A3) Fill in appropriate values in the table provided below.
mole of X
(2 digits after the decimal)
(0.25 points for each correct value in the left column) Plot a graph showing a relationship between mole of X
mole of X + mole of Ti(OiPr)4 and absorbance in the space provided below.
Ans
mole of X
mole of X + mole of Ti(OiPr)4 (0.25 point for each data)
The trendlines are not considered for scoring.
The value of mole of X
mole of X + mole of Ti(OiPr)4 which maximizes the amount of the product Y represents the stoichiometry of X in the chemical formula of Y. Based on the graph above, what is the molar ratio between Ti:X in the complex Y?
The molar ratio between Ti:X in the complex Y is ...1:1...or 1...
(2 points for the ratio)
1 point for the correct answer without the graph If the ratio is >1.2 or <0.8 (0 point)
Note: Based on the given data, the turning point in Job's plot occurs at mole fraction of X ~ 0.5. As a result, we conclude that the product has the ratio of Ti:X = 1:1.
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00
Absorbance
7-A4) The Ti complex Y is six-coordinated. The IR spectrum of Y does not contain a broad absorption band in the range of 3200–3600 cm-1. Y exists as three diastereomers. Ignoring stereochemistry at N atoms, draw clearly the structures of all three diastereomers.
Note that you do not need to draw the complete structure of the ligand. Only identify donor atoms that involve in coordination with titanium and the ligand framework between the donor atoms can be drawn as follows:
For example: can be drawn as:
**If you did not get a structure of X from 7-A2), use the following ligand symbol to represent X (A and Z are donor atoms):
Ans
A B C
(6 points) Score distribution: 1.5 points for each isomer 1.5 points if the proposed structures do have three possible diastereomers.
Bidentate ligands will not be considered for partial credits.
Any O-H functional groups in the structure will get 0.25 point deduction.
Without any reasonable monodentate ligands, 0.25 point deduction.
7-A5) Under certain conditions, the reaction shown in equation 1 affords only one diastereomer of Y. Given that structures of Y are "fixed" (no intramolecular movement), the
1H NMR spectrum of Y in CDCl3 shows four singlet resonances at 1.25, 1.30, 1.66, and 1.72 corresponding to the tert-butyl groups. Draw a structure of the only possible diastereomer of Y.
(You do not need to draw the complete structure of the ligand. Only identify donor atoms that involve in coordination and the ligand framework between the donor atoms can be drawn as shown in 7-A4))
or
(2 points) Only six-coordinated structures featuring tert-butyl groups in the chelate ligand structure will be considered for any credits.
Note for mentors: The 1H NMR spectra of isomers A and B contain two resonances assignable to the tert-butyl groups.
Problem 8 A
Total
A1 A2 A3 A4 A5
Total 6 5.5 3 4 1.5 20
Score Problem 8: Silica Surface
Silica exists in various forms like amorphous and crystalline. Silica can be synthesized via sol-gel process by using silicon alkoxides like tetramethoxysilane (TMOS) and tetraethoxysilane (TEOS) as the details below:
a. Hydrolysis
b. Water condensation
c. Alcohol condensation
Problem 8
5% of the total
In bulk silica, all silicon atoms are tetrahedrally bonded to four oxygen atoms giving three-dimensional solid network. The silicon environments found inside silica is presented below:
8-A1) Three silicon atom environments (similar to the example above) are commonly observed at the silica surface. The three structures of the silicon environments must be drawn in the provided boxes.
Silica can be used as an effective metal ion adsorbent in water. The proposed structure for metal-silica complex is as follows:
I II
x y z
Answer (total = 6 points)
Marking scheme Draw 1 structure => 2 points Draw 2 structures => 4 points Draw 3 structures => 6 points
8-A2) After Cu2+ is being adsorbed, the color of silica changes from white to pale blue. The visible spectrum shows a broad absorption band (with a shoulder) at max = 550 nm. If Cu2+
can bind with silica and adopt the structure similar to II, draw the splitting diagram of the d-orbitals of Cu2+ ion including the label of the d orbitals in the complex and specify the corresponding electronic transition(s) for the visible absorption.
The splitting diagram:
The corresponding electronic transition(s) (indicate the lower energy d-orbital and higher energy d-orbital)
Answer (total = 5.5 points) Cu2+: 1s2 2s2 2p6 3s2 3p6 3d9
Cu2+-complex tetragonal distortion Tetragonal elongation or tetragonal compression
Electronic Transition:
1. Tetragonal compression: dxy dz2 and dxz , dyz dz2
2. Tetragonal elongation: dxz , dyz dx2-y2 and dxy dx2-y2
Marking scheme
Draw one splitting diagram with d-orbital label; tetragonal compression or tetragonal elongation => 3.5 points
Specify the correct electronic transitions according to the drawn diagram => 2 points For partial credits:
- The label of certain orbital is missing => 0.5 point is deleted for each missing d-orbital label
- Draw a regular octahedral field splitting with d-orbital label => 1.5 points and specify the corresponding electronic transition => 0.5 point
- Only the splitting diagram without d-orbital label => 0 point - Write the third electronic transition => -0.5 point
8-A3) If the first row transition metal ions form complexes with silica analogous to Cu2+, which metal ion(s) do(es) have the analogous to electronic transition(s) to Cu2+? The metal ion(s) must be in +2 or +3 oxidation state. Please note that the silanol groups (Si-OH) and water are weak field ligands.
However, silica is randomly bonded to various types of metal ion. To increase the selectivity, modification of silica surface has been performed by grafting with various organic molecules like 3-aminopropyltrimethoxysilane and 3-mercaptopropyltrimethoxysilane.
8-A4) If Hg2+ is only bonding to sulfur binding sites in silica-SH, the symmetric complex of [Hg(silica-SH)2]2+ is formed. Draw the structure of [Hg(silica-SH)2]2+, specify the direction of the bond axes, and draw the corresponding d-orbital splitting. (You may use R-SH instead of drawing the whole structure of silica-SH.)
Answer (total = 3 points)
Cr2+, Mn3+
(1.5 point for each metal ion)
(the extra incorrect metal ion => -0.5 each)
The structure: d-orbital splitting diagram :
Answer (total = 4 points) Linear structure:
Marking scheme Draw the correct structure => 1 point Draw the correct d-orbital splitting diagram (no need to fill in the electrons)- corresponding
to the specified axes =>0.5 point for the axes
=>2.5 points for the diagram For partial credits:
- Draw the correct structure without axes but possible splitting diagram => 1+2 points - Draw the correct structure with wrong axes but possible splitting diagram => 1+2 points
8-A5) Specify true or false for the following statements:
a) d-d transition is found in [(Hg(silica-SH)x)]2+
True False
b) The [Cu(silica-NH2)x]2+ in a similar geometry, is expected to have a color similar to other copper(II) amine complexes.
True False
c) In the visible absorption spectra, max of [Cu(silica-NH2)x]2+ is greater than that of [Cu(silica-OH)x]2+.
True False
Answer (total = 1.5 points)
a) d-d transition is found in silica-SH-Hg2+.
True False
Explanation : Hg2+ is a d10- metal ion in which d-d transition is not found.
b) The [(Cu(silica-NH2)x]2+ is expected to have a color similar to other copper(II) amine complexes.
True False
Explanation : Various copper amine complexes like [Cu(en)(H2O)4]2+, [Cu(en)2(H2O)2]2+ and [Cu(en)3]2+ have deep blue color. [(Cu(silica-NH2)x]2+
containing similar ligands to these copper complexes is expected to have a similar color.
c) In the visible absorption spectra, max of [(Cu(silica-NH2)x]2+ is greater than that of [(Cu(silica-OH)x]2+.
True False
Explanation : R-NH2 is a stronger field ligand as compared to R-OH. This results in a larger energy gap or a smaller max of [(Cu(silica-NH2)x]2+ as compared to that of [(Cu(silica-OH)x]2+.
Marking scheme 0.5 point for each correct answer
Problem 9 A
Total
A1 A2 A3
Total 6 6 11 23
Score
Problem 9: Into the Unknown
9-A1) Organic compound A is chiral and contains only three elements with the molecular weight (MW) of 149 (rounded to an integer).
1H NMR spectrum of compound A shows among others, three types of aromatic protons, and its 13C NMR spectrum shows eight signals, of which four signals are in the range of 120-140 ppm.
Compound A can be prepared by treating a carbonyl compound with methylamine followed by NaBH3CN. Write all possible structural formulae of compound A. No stereochemistry is required, and do not include stereoisomers.
2 points each Grading Scheme:
No points will be given if the structure drawn does not have molecular weight = 149.
No points will be given if the structure drawn contains more than three elements.
Partial credits will be given to each incorrect structure as follows
Contains a benzene/aromatic ring = 0.25 points
Mono substituted aromatic ring = 0.25 points
Contains –NHCH3 group = 0.25 points
Contains 1 chiral carbon = 0.25 points
If two or three structures look exactly the same or they are stereoisomers, partial credits will be given to only one structure)
0.25 points will be deducted if H on O, N or C (e.g. –C=CHCH3) is missing.
Problem 9
6% of the total
9-A2) One of the position isomers of compound A (structure A1, A2 or A3) can be synthesized from compound B or C and D as shown in the diagram below. Write down the structural formulae of compounds B-F, and the position isomer of compound A.
Grading Scheme for Structures B-F: (1 point for each structure)
0 points if the structure drawn does not contain a benzene/aromatic ring.
0 points if the molecular weight of the structure drawn does not match the molecular weight given.
0.25 points will be deducted if H on O, N or C (e.g. –C=CHCH3) is missing
Partial credits (maximum 0.5 points for each structure and maximum 1 point for each reaction) will be given to other structures if
both the starting material and the product are incorrect and the starting material can be converted to the corresponding product with a given name reaction or reagent(s).
9-A3) Compound A is the (R)-form of one of structures A1-A3. It can be prepared from vicinal diols X and Y as shown in the diagram below. Both diols are structural isomers, and each structure contains one carbon less than that of compound A. Write down the structural formulae of compounds G-N, X, Y and the (R)-form of compound A. You must show stereochemistry of all compounds.
Grading Scheme for Structures G-N, X and Y: (1 point for each structure)
Same grading scheme as in question 9-A2 and
0.5 points will be deducted for incorrect stereochemistry.
For structures Y, L and M, 0.25 points will be deducted for incorrect stereochemistry at each chiral carbon.
Problem 10 A B
Total A1 B1 B2
Total 20.5 4 5.5 30 Score
Problem10: Total Synthesis of Alkaloids
Alkaloids are a class of nitrogen-containing natural products. Their structural complexity and potent biological activities has drawn attentions. Two representative examples of alkaloids –sauristolactam and pancratistatin are highlighted in following questions.
Part A
Sauristolactam possesses excellent cytotoxicity against various cancer cell lines. It could be prepared by following synthetic sequence below. (1H-NMR spectra were recorded in CDCl3 at 300 MHz.)
Problem 10
7% of the total
10-A1) Draw the structures of A-G in the sequence. Provide your answers on the following blank sheet.
The structures of A-G.
+3 points for correct structure -1.5 point if benzylation at positions other
than phenolic oxygen
-1.5 if formylation at positions other than para to OMe
+2 points for oxidation of CHO to COOH regardless of position
Otherwise, 0 point
+1.5 points for esterification of COOH regardless of position ans structure of SM
Otherwise, 0 point
+1.5 points if O-debenzylation is implied, regardless of position and structure of SM +1.5 points for O-acetylation, regardless of
position and structure of SM Otherwise, 0 point
+3 points for single bromination on aromatic ring, regardless of position and structure of
SM Otherwise, 0 point
+3 for single benzylic bromination on aromatic methyl only
+1 point for bromination on other methyl groups
Otherwise, 0 point
+5 points for correct structure +2 points for SN2 of MeNH2 on benzylic
bromide
+1.5 points each for deacetylation and lactamization
-2 if structure of product G cannot correspond correctly to final product
Part B
Pancratistatin, isolated from a Hawaiian native plant, spider lily, exhibits potent in vitro and in vivo inhibitory activity of cancer cell growth in addition to its excellent antiviral activity.
Pancratistatin could be successfully synthesized via intermediates X1 and X2. The synthesis of these intermediates are shown in the following schemes.
10-B1) Draw the structures of A and B.
Grading scheme:
Compound A: 2 points. Wittig reaction.
2 points for correct answer. 1 point for product with (Z)-siomer.
0 point for other answers.
Compound B: 2 points. Simple hydration/oxidation. 2 points for correct answer.
1 points if the answer is lactol. (no oxidation) 1 point if the answer is phenyl ketone.
0 point for other answers.
10-B2) Intermediate X1 (a single enantiomer with the stereochemistry shown) is labeled with deuterium with configuration as indicated below, propose the 3-D chair structure of compound E and the structure of compound F, with stereochemistry. Is Y a proton (1H) or a deuterium (2H)?
Grading scheme:
Compound D: Iodolactonization. (3 points)
Student needs to give the correct structure and stereochemistry with given absolute configuration of deuterated starting material – ability to analyze and present the 3D structure of the starting material and the right product.
3 points for complete answer. Any style of drawing is acceptable.
1 points for correct structure of iodolactone without stereochemistry +0.5 for correct stereochemistry of deuterium.
+1 for correct stereochemistry of lactone +0.5 for correct stereochemistry of iodine
Compound E: E2 Elimination. (2 points)
Student needs to realize the anti-stereochemistry required for E2 elimination and that the deuterium is selectively removed by base (over proton) during the elimination.
2 points for complete answer. Any style of drawing is acceptable.
1 points for correct structure but elimination of H instead of D.
0.5 points for recognizing elimination although E1 or E2 is impossible to occur given the structure of compound E
0 point for other answers.
Problem 11 A
Total A1 A2
Total 10 2 12
Score