Flashcards in Exam 3 Deck (92)
How would stereochemistry of products of Diels-Alder change if the reaction was stepwise?
Would see more configurations
2^n configurations at each stereocenter
*We don't see this because Diels-Alder is not stepwise
Where is a dashed constituent on the six-membered ring located on the diene?
(when attacks from the top)
Where is a wedge/up constituent on the six-membered ring located on the diene?
(when attacks from the top)
How can you tell if a Diels-Alder reaction has occurred?
Look at the number of configurational isomers
If see less than 2^n, then we know at some point the steps were concerted which indicate a Diels-Alder
Are endo constituents wedges or dashes?
Are exo constituents wedges or dashes?
Is cyclohexane more stable than benzene? Why or why not?
Benzene's pi-bonds make it less stable than cyclohexane which is made entirely of unreactive sigma bonds
How much more stable is cyclohexane compared to benzene?
Why reaction is rearomatization of electrophilic aromatic substitution similar to?
Form carbocation, than kick off proton/leaving group
How much aromatic stabilization does benzene have? Compared to what?
Benzene is 36 kcal/mol more stable than the hypothetical 1,3,5 cyclohexatriene (bonds not delocalized)
How many nodes does psi-6 of benzene have?
How many nodes dow psi-3 of benzene have?
Why is benzene not capable of a Diel's Alder reaction with its self?
Both molecules would lose their aromaticity
This is highly unfavored
Why does Friedel-Craft alkylation often result in over-alkylation?
You are adding an alkyl group to the aromatic ring
This alkyl group then activates by making a better nucleophile
How can one experimentally stop over-alkylation of Frield-Craft?
Add a large excess of aromatic rings, so it is unlikely for the electrophile to interact with the aromatic ring
An activating group is often an electron donating group that makes a reaction more likely to occur
For example: EDG makes electrophilic alkylation more likely to occur since making aromatic ring a better nucleophile
Why are halogens O/P-directing yet deactivating?
Halogens have a lone pair that can act as an EDG and be in conjugation with aromatic system, BUT due to electrogativity and strong inductive effect they have some EWG properties
This makes halogens have an overall slower rate of electrophilic addition compared to unsubstituted benzene
This makes them an O/P-director and also a deactivator
What is a good approach to explaining why something is a good O/P or meta director?
Draw the opposite and explain why this is so bad (normally due to close positive charges)
Ex: if something is o/p, draw meta intermediate and explain why this sucks
Rules for drawing poly-ene conjugated MOs:
1) Orbitals alternate in symmetry with respect to the pseudo mirror plane passing through the midpoint of the chain
2) Lowest energy MO has 0 nodes; the number of nodes increases by one going to the next highest level (rings are different-can have degenerate orbitals)
3) Highest energy MO has a node between every adjacent orbital
4) In chains with an odd number of atoms, where the central atom lies in the mirror plane, antisymmetric MOs must have 0 contribution from the central atom. There is a nonbonding level to which alternate p-orbitals make 0 contribution
What are the conditions for kinetic control?
shorter reaction times
What are the conditions for thermodynamic control?
longer reaction times
Under kinetic control is the reaction at equilibrium?
No since the reaction is not allowed to reverse
Under thermodynamic control is the reaction at equilibrium?
Why does reversibility matter for kinetic versus thermodyanmic control?
Whatever product forms the fastest will be the product that is "stuck" since the reaction cannot reverse to make the slower, but more stable, product
Why do high temperatures matter for thermodynamic control?
High temperatures allow the products to have enough energy to overcome higher barriers for the reverse reaction to occur
Why will the thermodyanmic product start to build up at equilibrium?
the energy barrier for the reverse reaction to occur away from the thermodyanmic product is higher (since product is more stable) and makes less thermodyanmic product reverse compared to the kinetic product
Why would we not originally expect endo product to be favored in Diels-Alder? What makes it favored?
Would not expect it to be favored due to steric crowding
Endo product is favored due to secondary orbital stabilization
3 types of polycyclic molecules:
Spiro polycyclic molecule
connected at a single carbon
looks like an "X" between the two rings