Page 1 id#___________________ CHE538 Exam 2, Principles of Organic Chemistry Fall 2004 Warning: write/print neatly! If I can’t read it you don’t get credit. 1. (30 pts.) Think about the photolysis of diazomethane to produce methylene and N2. H2CN2 hv CH2 + N2 a. (5 pts.) Draw two low-energy resonance structures of diazomethane (reactant above). SHOW ALL ELECTRONS—put lone pairs as dots, and bonds as lines. Answer 1a: H C N N H H C N N H b. (5 pts.) In a Hückel molecular orbital explanation of why and how the bond breaks when exposed to light how many atomic orbitals will you use and how many electrons are there in the pi system of your MO diagram? NUMBER OF AOs _3___ NUMBER OF MOs _3__ NUMBER OF electrons in MOs _4__ 1b. c. (10 pts.) Qualitatively draw the π-orbitals of diazomethane and qualitatively place them on an energy vertical axis relative to one another. Show how the coefficients (relative sizes of the AOs) are perturbed by the unequal electronegativities of the C and N atoms. C N N ψ3 C N N ψ2 C N N ψ1 1c: d. (10 pts.) Short Answer. Use no more than ten words. Why does the bond break? You do not have to get the other parts of the question right to get full credit here. Refer to your MO diagram if this is convenient. 1d: The antibond is populated after being excited by the photon. Page 2 id#___________________ 2. (8 pts.) Samarium (II) iodide is a one-electron reducing agent. a. (3 pts.) Draw a Lewis (line) structure the intermediate of the first elemental step when R−CH=O reacts with SmI2. b. (5 pts.) Draw a crude depiction of the HOMO and LUMO of R-CH=O. Your picture has to include the perturbation of the ethylenic orbitals by the electronegative oxygen atom. Put the electrons in the MOs. On which atom is there the most spin density of the one-electron reduction product and why? a) R C O H or SmI R C O H +I b) R C O H R C O H Spin density is greatest on Carbon. or SmI2 R C O H Page 3 id#___________________ 3 (50 pts.) PUT YOUR ANSWERS on page 4. Consider the following reaction: 3-4 N N ? N 3-1 3-45 + O N N N + N2 3-46 O 3-3 O N 3-2 H H + N O H H 3-5 65% 3-6 22% http://userpage.chemie.fu-berlin.de/~shecht/Huisgen.pdf There are at least two different routes through which the reaction could proceed. (1) Molecule 3-4 could be an intermediate or (2) 3-45 and 3-46 could be transition states. 3. a. (5 pts.) Briefly describe what the difference is between an intermediate and a transition state (two short sentences). b. (15 pts.) How would you tell the difference between these two reaction profiles? Describe two experiments to tell the difference between the reaction going through intermediate 3-4 or through transition states 3-45 and 3-46. EXPLAIN HOW YOUR EXPERIMENTS WILL HELP YOU DECIDE! FOR EXAMPLE, WHAT WOULD YOU EXPECT TO OBSERVE IN YOUR EXPERIMENT IF THE FIRST HYPOTHESIS IS CORRECT? ETC. c. (10 pts.) What if the cyclization of 3-1 and 3-2 is the slow step (ratedetermining)? What will the effect be of using the following substrate, D D ? Include an explanation with your answer.. d. (10 pts.) Draw a line structure of intermediate 3-4. Show all electrons as either lines for bonds or dots for lone pairs. e. (10 pts.) Assume we can measure the rate of the cyclization of 3-1 with 3-2 to produce 3-3 and/ or the rate of formation of 3-5 and 3-6 from 3-3, what should we expect for the ∆S¥ parameters of these two processes (greater than or less than zero?). Discuss both processes separately. Include a short justification (explanation) of your answers. Page 4 id#___________________ 3a. __Intermediates have the potential energy minimized for all coordinates. Transition states have the potential minimized for all but one coordinated, the reaction coordinate._ 3b. (1) __Complete credit was given for answers like a) run a free E correlation by substituting the phenyl ring. Look for charge build up on N with positive ρ. b) Try to isolate the intermediate and subject it to the reaction conditions. Check the product ratio. If it is identical then the intermediate is probably on the reaction path. c) trap the intermediate with a radical clock. A group like a cyclopropyl group that would open if a cationic or radical reaction center were visited. d) Suggestions that solvent dependence in the involving zwitterionic intermediate 3-3 would be more pronounced than pathways involving the two TSs. e) Measurement of the activation parameters of the rate constants. The route involving the two TSs would have ∆S>0, whereas the route involving intermediate 3-3 would have DS<0. f) Run the experiment with an alkyl group attached directly to the sp2 carbon atom (2-alkylbicyclo[2.2.1]hept-2-ene) to block one face in TS 3-46. g) Use the molecule in part c of this question. Measure the rate of the disappearance of 3-3. The SKIE will be strongly inverted for the rate of the disappearance of 3-3 if the intermediate if formed. 3b. (2) _______________________________________________________________ Draw structures to describe your proposed experiments for 3b if you wish. Label these and refer to them in your descriptions above. 3c. ___The rate should decrease when D replaces H. There are stronger bonds in reactant than the product. _____ 3d. H C N O The diradical was accepted also. 3e. ___Activation entropy (∆S¥) < 0 for the process that combines two molecules: 3−1+3−2. From intermediate 3−4 the process giving 3−5 and 3−6 has ∆S¥ < 0 because it is a ring closure. From 3−3 to 3−5 or 3−6 via TS 3−45 or TS 3−46, the reaction has ∆S¥ > 0 due to loss of N2. Page 5 id#___________________ (12 pts.) 4. (5 pts.) Write out the secular determinant for fulvene. 6 5 1 2 4 3 Use the numbers given in the structure otherwise we will have no way of grading your answer. a. (5 pts.) Simplify the secular determinant using a symmetry element. Write two determinants one symmetric and the other dissymmetric to your symmetry element. b. (2 pts.) Write the symmetric group orbitals and the dissymmetric group orbitals. X 1 0 0 1 0 1 X 1 0 0 0 0 1 X 1 0 0 0 0 1 X 1 0 1 0 0 1 X 1 0 0 0 0 1 X 4a.) Φ d1 Φ d1 Φ d2 2X+2 Φ d2 2 χ χ 0 0 5 6 2 2X 2 0 χ 5 0 2 X 1 χ 6 0 0 1 X 4b) 4c) sym Φs1 = X2 + X3, Φs2 = X1 + X4, X5, X6 dissym Φd1 = X2 - X3, d1 Φd2 = X1 – X4, Φ d1 Φ d1 2X-2 Φ d2 2 Φ d2 2 2X