09-107 onors Chemistry ame Exam III Please read through each question carefully, and make sure you provide all of the requested information. 1. A series of octahedral metal compounds are made from 1 mol 3+, 3 mol -, and excess C. The compound incorporates as much C as needed to fill in the vacant inner sphere binding sites. All three - ions must be incorporated into the structure to balance the +3 charge on. The data helps you decide if those - s are inner-sphere or outer-sphere ligands. The C ligands fill in the remaining octahedral sites. a) Compound I exists as two isomers. If we treat the isolated compounds with Ag 3, 1 mol of Ag is formed. What are the structures of the two isomers? What is the relationship between the two isomers? [8 points] Molecular formula: [ 2 (C) 4 ] the outer sphere reacts with Ag 3 C C C C C C C C cis trans b) Compound II also exists as two isomers. A conductivity meter shows no activity- a solution of the compound does not conduct electricity. What are the structures of the two isomers? What is the relationship between the two isomers? [8 points] Metal complex must be neutral: [ 3 (C) 3 ] C C C C C C fac mer Exam III Distributed on Wednesday, December 5, 2001 Page 1 of 7
09-107 onors Chemistry ame 2. The following compound is typical of the Collins type ligands. (We ll be discussing this more when we discuss catalysis on Tuesday.) y x a) Circle the atom(s) that bind to the metal. [2 points] The 4 nitrogens coordinate with the metal. b) Given the arrangement of the ligand on the coordinate axes, which d-orbital interacts the most with the ligand? [6 points] The d xy orbital interacts the most- the lobes of that orbital are off-axis. 3. Amphetamines were initially studied for their properties as appetite suppressants and cough medicines. Amphetamine has the following structure: 2 2 Phenylalanine In your brain, amphetamines interact with a neuroreceptor that typically binds endorphins, causing a release of dopamine and noradrenaline and resulting in euphoria. The natural endorphins are polypeptides and the neuroreceptor recognizes them because they have a particular amino acid at the end of the molecule. This amino acid is mimicked by amphetamine. a) What amino acid is amphetamine mimicking in structure? [4 points] Phenylalanine (Phe) b) Is that amino acid hydrophobic or hydrophilic? [4 points] ydrophobic Exam III Distributed on Wednesday, December 5, 2001 Page 2 of 7
09-107 onors Chemistry ame 4. I perform the following ligand replacement reaction: [Fe(en) 3 ] 4+ + 6 - [Fe() 6 ] 2- + 3 en a) Draw the d-orbital splitting diagrams for the two compounds. [8 points] Fe 4+ = [Ar]3d 4, en is a low spin ligand, - is a high spin ligand. [Fe(en) 3 ] 4+ [Fe() 6 ] 2- b) Which compound will interact more strongly with a magnet? [2 points] [Fe() 6 ] 2- c) If the starting material is a blue-green color, what is the size of o, in kj/mol. [4 points] Blue-green absorbs red-orange λ = 685 nm = 6.85 x 10-7 m E = hc ( = 6.626 x 10-34 J s)2.998 x 10 8 m s -1 λ 6.85 x 10-7 m ow convert to KJ/mol 2.90 x 10-19 J x 1 kj 1000 J x 6.022 x 1023 mol ( ) = 175 KJ/mol = 2.90 x 10-19 J d) What color could you expect the product to be? [4 points] The product has a smaller o. As E goes down, the wavelength absorbed goes up. The only color with a higher wavelength is red. If red is absorbed, we see green. Exam III Distributed on Wednesday, December 5, 2001 Page 3 of 7
09-107 onors Chemistry ame 5. Show how the ethanolamine ion ( 2 - - - - ) can act as a bidentate ligand. You may need to draw a Lewis structure for this. Draw both isomers for Co(ethanolamine) 3. [8 points] 2 M Co enantiomers Co 6. The unit cell below describes a fictitious compound containing Tungsten (W), xygen and Chlorine. xygen Chlorine Tungsten a) ow many chlorine atoms are in the unit cell? [3 points] is at a vertex shared by 8 unit cells 4 atoms x 1 8 = 1 2 a atom b) ow many total atoms are in the unit cell? [4 points] There is 1 W atom, 0.5 atoms and 0.5 atoms = 2 atoms total. c) What is the coordination number of one of the oxygen atoms? [3 points] An atom at the vertex of a body-centered cubic cell has a coordination number of 8. The nearest neighbor is the W atom in the center. d) Is this a close-packed structure? (circle one) [2 points] Yes o Exam III Distributed on Wednesday, December 5, 2001 Page 4 of 7
09-107 onors Chemistry ame 7. Use the following 6 amino acids to answer the following questions: 2 C 2 2 2 3 C 3 2 Asparagine (Asn) Lysine (Lys) Isoleucine (Ile) 2 2 2 C Phenylalanine (Phe) Tryptophan (Trp) Glutamic Acid (Glu) rder the amino acids from the most hydrophobic to the most hydrophilic. [6 points] Ile Phe Trp Asn Glu Lys onpolar ot polarizable onpolar Polarizable Slightly Polar Polarizable Polar Polar Charged at bio p. Polar Charged at bio p If protons ( + ions) are required for a reaction to proceed, which amino acid(s) can supply that proton? [4 points] Glu Which amino acid(s) would be found in an active site that recognizes benzene? [4 points] benzene Phe, Trp, (Ile) Which amino acid(s) could exist as a positively charged side chain? [4 points] Lys, (Trp) Which amino acid(s) are both hydrogen bond donors and acceptors? [4 points] Trp, Asn, (Glu), (Lys) Which amino acid(s) would you expect to find on the periphery of a protein that is bound to the surface of a membrane (extrinsic protein)? [4 points] All polar/hydrophilic AAs, Lys, Glu, Asn, (Trp) Which amino acid is the most polarizable? [4 points] Trp most pi electrons! Exam III Distributed on Wednesday, December 5, 2001 Page 5 of 7
09-107 onors Chemistry ame Extra Credit Question: [5 bonus points] The following questions are based on Dr. Armitage s lecture, and the discussion that followed. The figures below show the G-C and A-T base pairs, and the hydrogen bonds that are formed. Is it more common to have a mutation with a G-C base pair or and A-T base pair? Why? ow can we, as chemists, modify a base to rectify this problem? You may draw your suggestion on the figure. A-T pair G-C pair The A-T pair typically has the most mutations- there are only two hydrogen bonds holding that pair together, compared to 3 -bonds for the G-C pair. To remedy this, we can add an 2 group at the position to the arrow, to create a third -bond between A-T. For official use only: 1) /16 2) / 8 3) / 8 4) /18 5) / 8 6) /12 7) /30 subtotal /100 Extra Credit: Total /100 Exam III Distributed on Wednesday, December 5, 2001 Page 6 of 7
09-107 onors Chemistry ame Useful Stuff: h = 6.626 x 10-34 J s c = 2.998 x 10 8 m/s A = 6.022 x 10 23 mol -1 720 nm 400 nm V R 650 nm 470 nm B G Y 590 nm 520 nm 2 2 2 2 2 2 C 3 Alanine (Ala) C Glutamic Acid (Glu) C 2 Arginine (Arg) 2 C 2 Glutamine (Gln) 2 C 2 Asparagine (Asn) Glycine (Gly) C S Cysteine (Cys) Aspartic Acid (Asp) 2 istidine (is) 2 3 C 3 Isoleucine (Ile) 2 2 2 2 C 3 C 3 Leucine (Leu) 2 Lysine (Lys) S C 3 Methionine (Met) Phenylalanine (Phe) Prolline (Pro) 2 2 2 2 2 C 3 C 3 C 3 Serine (Ser) Threonine (Thr) Valine (Val) Tryptophan (Trp) Tyrosine (Tyr) Exam III Distributed on Wednesday, December 5, 2001 Page 7 of 7