Hello, dear friend, you can consult us at any time if you have any questions, add WeChat: daixieit

CHEM 4523 2022 Take Home Exam

Instructions:  Complete all parts of all questions.  Submit by April 23^th, 12:00 noon. My preference is to get paper submissions from you, although if digital is what you can do, please just email them to me directly.

1. The synthesis of [5]radialene was achieved in 2015 for the first time (J. Am. Chem. Soc., 2015, 137, 14653), 50 years after the first synthesis of [3]radialene.  

a)  [5 pts] Draw a VB orbital diagram for [3]radialene.

b) [5 pts] Draw a “top down” view of one Hückel molecular orbital (HMO) for [5]radialene. Basically just draw in the different lobes with appropriate sizes and phases (corresponding to c-values). Draw the MO that corresponds to your birthday according to the chart below. All of the Hückel MOs have been calculated for you and appear at the end of the exam as an appendix.

Month born MO to draw

1, 11 ψ₁

2, 12 ψ₂

3 ψ₃

4 ψ₄

_{etc. etc.}

c) [5 pts] Calculate the free valence for any atom in [5] and [3]radialene.  (Only calculate one atom per molecule.)

d) [5 pts] Imagine that [3]radialene yeets one electron to make [3]radialene⁺, a radical cation. Draw a Lewis structure of this cation along with 3 resonance structures.

e) [5 pts] In practice, both radialenes polymerize at room temperature via Diels-Alder reactions.  Draw examples of radialene dimers that would be formed by Diels-Alder reactions for both [3] and [5]radialene.  

f)  [5 pts] [5]Radialene polymerizes much more rapidly than [3]radialene (which is why it took 50 years longer to make it).  Can you suggest a reason why this might be so? (In the paper cited above, the authors discuss why – you might want to read this to help inform your answer. It is discussed on page 14656 starting at the top of the right-hand column.)

2. Photolysis of diene A below in benzene leads to eventual formation of 1,2,3,4-tetrahydronaphthalene and 1,3-butadiene (A. Gilbert and R. Walsh, J. Am. Chem. Soc., 1976, 98, 1606).  The mechanism is believed to follow a cascade of five sequential pericyclic reactions.  The structures of all intermediates B to F have been drawn in for you.

a)  [10 pts] Classify the pericyclic reaction occurring in each step (i.e. type of reaction, along with [m,n] for sigmatropic, [m + n] for cycloadditions, or # of electrons moving in the transition state for electrocyclic reactions) and draw in the curly arrows. Indicate allowed stereochemistry (i.e. con/dis) if appropriate.

b) [5 pts] Redraw compound C showing the correct stereochemistry for the labelled hydrogens based on the Woodward-Hoffman selection rules.

3. [20 pts] Reaction between a sulfur ylide and an aldehyde provides a convenient route to the synthesis of epoxides. After a number of mechanistic studies, a consensus has now been reached on the reaction mechanism, and this is shown below:

A study by Canadian chemist Cathleen Crudden (Queens University) aimed to deduce which of the steps is rate determining (Org. Lett., 2007, 9, 5481-5484). To this end, she studied the reaction rates with a variety of aldehydes. When the aldehyde C-H was replaced with deuterium, a kinetic isotope effect (KIE) of 0.93 was measured for the reaction. A Hammett plot was constructed using a variety of meta and para substituted aldehydes and a ρ value of +2.50 was obtained.

Discuss the significance of the measured KIE and ρ value for this reaction – what do they tell us about the rate determining step?

4. Rationalize the following observations

(a) [5] The difference between the s_p and s ⁺ values for the methoxy group are large (-0.27 vs -0.78), whereas the difference between these values for the methyl group are much more modest (-0.14 vs -0.30).

The σ_p and σ^- values for some common electron withdrawing groups are shown:

(b) [5] For the first 3 EWG listed, the σ- values are significantly higher than the σ_p. The CF₃ group stands out since it has a σ- value very similar to its σp value.

(c) [5] The methoxy group has very different electronic effects in the para vs meta position. s_p = -0.27 and s_m = +0.12.

(d) [5] Fluorine, despite being practically the most electronegative atom, is not a particularly good EDG or EWG at the para position, with s_p = 0.05.

5. [15 points] 1,3-bis(1-cyclopentadienyl)propane and acetylene dicarboxylic acid give product B in >70% yield after sitting at room temperature for 7 days.  This product contains alkene double bonds that can be reduced by hydrogenation to give the product on the right.  Formation of B takes place via the presumed intermediate A.

Draw structures for A and B and show the mechanism of all pericyclic transformations using curly arrows.

BONUS [5 points]

Preamble: OK besties, I’ve been feeling under a lot of pressure to come up with a bonus question as tight as my last one, no cap. I want one that slaps – but I know you would all get salty if I delayed releasing the exam to craft something epic. So while not especially sick, hopefully this bonus at least doesn’t make you down bad. It would be cash money of you to give the question some serious thought and come up with a boss response. Stay Woke!

Let me know something cool, weird, or interesting you learned in the course which has stuck with you. It could be something that was core to the course, some side story or fact, or something you came across in your real life which you linked in your brain to something we did in the course. (extra karma if you can squeeze in cringey slang into your answer)

Postamble: In all seriousness, I think you are all amazing and it has been an honour having you all in my class. Despite all the curve balls this semester has thrown at us (strike, COVID, hybrid, etc.) you all handled it like champs. You deserve a huge congratulations, and also a break! If you are graduating, please stay in touch! If you are back next year, I hope to see you around Elliott!

Supplementary Data

Below are the results of Huckel MO calculations for [3]radialene and [5]radialene. You can see first a screenshot of the calculator, with the structure drawn. Most importantly in this image, you can see the labels for all atoms in the structure. On the right of the image, you can see the MO diagram (in terms of a and b of course). Below the screenshot are the results for all of the MOs. It lists all the MOs from lowest energy to highest. You can see the c-value for each atom given by f_n. The value of n corresponds to the atom label in the diagram. In other words, f₃ will be the c-value for atom 3 in that particular MO. You will need these values for Q1.

If it says after the MO ‘degeneracy = 2’, it means there are two MOs with the same energy, and it will go ahead and list both. All MO energies are in terms of the absolute value of b. All MOs were calculated using the online HMO calculator available here: https://www.ebert.cup.uni-muenchen.de/html/elch/spec/spec_shmo_en.html  Please feel free to go here and play around!

[3]Radialene

E₁ = α - 2.414|β|; degeneracy = 1
ψ₁ = 0.53ϕ₁ + 0.53ϕ₂ + 0.53ϕ₃ + 0.22ϕ₄ + 0.22ϕ₅ + 0.22ϕ₆
E_2,3 = α - 0.618|β|; degeneracy = 2
ψ₂ = 0.38ϕ₁ – 0.36ϕ₂ + 0.62ϕ₄ – 0.04ϕ₅ – 0.58ϕ₆
ψ₃ = –0.19ϕ₁ – 0.24ϕ₂ + 0.43ϕ₃ – 0.31ϕ₄ + 0.69ϕ₅ – 0.38ϕ₆
E₄ = α + 0.414|β|; degeneracy = 1
ψ₄ = –0.22ϕ₁ – 0.22ϕ₂ – 0.22ϕ₃ + 0.53ϕ₄ + 0.53ϕ₅ + 0.53ϕ₆
E_5,6 = α + 1.618|β|; degeneracy = 2
ψ₅ = –0.34ϕ₁ – 0.36ϕ₂ + 0.69ϕ₃ + 0.21ϕ₄ – 0.43ϕ₅ + 0.22ϕ₆
ψ₆ = 0.61ϕ₁ – 0.60ϕ₂ – 0.37ϕ₄ + 0.37ϕ₆

[5]Radialene

E₁ = α - 2.414|β|; degeneracy = 1
ψ₁ = 0.41ϕ₁ + 0.41ϕ₂ + 0.41ϕ₃ + 0.41ϕ₄ + 0.41ϕ₅ + 0.17ϕ₆ + 0.17ϕ₇ + 0.17ϕ₈ + 0.17ϕ₉ + 0.17ϕ₁₀
E_2,3 = α - 1.356|β|; degeneracy = 2
ψ₂ = 0.33ϕ₁ + 0.47ϕ₂ – 0.04ϕ₃ – 0.49ϕ₄ – 0.27ϕ₅ – 0.36ϕ₆ + 0.35ϕ₈ + 0.24ϕ₉ – 0.20ϕ₁₀
ψ₃ = 0.39ϕ₁ – 0.19ϕ₂ – 0.51ϕ₃ – 0.12ϕ₄ + 0.43ϕ₅ – 0.09ϕ₆ – 0.37ϕ₇ – 0.14ϕ₈ + 0.29ϕ₉ + 0.32ϕ₁₀
E_4,5 = α - 0.477|β|; degeneracy = 2
ψ₄ = 0.27ϕ₁ – 0.18ϕ₂ + 0.14ϕ₄ – 0.25ϕ₅ + 0.30ϕ₆ + 0.05ϕ₇ – 0.37ϕ₈ + 0.56ϕ₉ – 0.53ϕ₁₀
ψ₅ = 0.06ϕ₁ – 0.21ϕ₂ + 0.27ϕ₃ – 0.23ϕ₄ + 0.11ϕ₅ – 0.49ϕ₆ + 0.57ϕ₇ – 0.43ϕ₈ + 0.13ϕ₉ + 0.22ϕ₁₀
E₆ = α + 0.414|β|; degeneracy = 1
ψ₆ = –0.17ϕ₁ – 0.17ϕ₂ – 0.17ϕ₃ – 0.17ϕ₄ – 0.17ϕ₅ + 0.41ϕ₆ + 0.41ϕ₇ + 0.41ϕ₈ + 0.41ϕ₉ + 0.41ϕ₁₀
E_7,8 = α + 0.738|β|; degeneracy = 2
ψ₇ = 0.32ϕ₁ + 0.29ϕ₂ – 0.14ϕ₃ – 0.37ϕ₄ – 0.10ϕ₅ + 0.51ϕ₆ + 0.18ϕ₇ – 0.39ϕ₈ – 0.43ϕ₉ + 0.13ϕ₁₀
ψ₈ = 0.20ϕ₁ – 0.24ϕ₂ – 0.35ϕ₃ + 0.36ϕ₅ + 0.47ϕ₇ + 0.32ϕ₈ – 0.28ϕ₉ – 0.49ϕ₁₀
E_9,10 = α + 2.095|β|; degeneracy = 2
ψ₉ = 0.04ϕ₁ – 0.37ϕ₂ + 0.55ϕ₃ – 0.53ϕ₄ + 0.30ϕ₅ + 0.25ϕ₆ – 0.26ϕ₇ + 0.18ϕ₈ – 0.14ϕ₁₀
ψ₁₀ = 0.57ϕ₁ – 0.43ϕ₂ + 0.13ϕ₃ + 0.22ϕ₄ – 0.49ϕ₅ – 0.10ϕ₆ – 0.06ϕ₇ + 0.21ϕ₈ – 0.27ϕ₉ + 0.23ϕ₁₀