MCAT Questions

[Chouster]

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Hi everyone, please help me with the following questions. These are things I'm having slight confusion on and need clarification:

1) Aromaticity - I realize this may not appear directly on the test (without the use of a passage but still nice to know). Im still confused as to when a heterocyclic ring can have an atom, such as nitrogen, donate its lone pair or not to the delocalized pi system. For example, pyrrole and pyridine. Pyridine does not donate its lone pair because its perpendicular to the p-orbitals around the ring whereas for pyrrole its not. While this may make sense to some of you, it is Greek to me. How do you know that the nitrogen in pyridine is perpendicular to begin with? This has always been a source of confusion.

2) Common transition metal ions - I think we're supposed to know that iron forms Iron(II) and Iron(III) and that Cu forms Cu(II) and Cu(III) and that Zn forms Zn(II), Ag forms Ag+, . Also, I just discovered that Ni forms Ni(II). What other transition state ions are we supposed to have memorized? Is there some rule I can use to figure out which ions they will readily form (e.g. in solution) or is this pure memorization now.

3) Equilibrium constants - Say we have A + B -> C + D. Keq then = [D][C]/[A]. If we add some more A to the system what happens to the concentrations? By Le Chatlier's principle, more C and D are going to be formed which increases the numerator, but then that means the denominator will be decreased thus increasing Keq?? This contradicts what we've been taught (assuming this is all at constant temperature). What is going on with the individual concentrations of A, B, C, and D while maintaining a constant Keq?

That's it for now. Thanks!

 

[mdsquared]

Since I have not taken organic yet I will only try to answer question 3. To see which way the reaction will shift you need to know K, and the concentration of A, B, C, & D. Then you do Q=[C][D]/[A] (plug in the concentrations), if Q>K then the reaction will shift to the left to maintain equilibrium, if Q<K then the reaction will shift to the right.

Keq then = [D][C]/[A]. If we add some more A to the system what happens to the concentrations? By Le Chatlier's principle, more C and D are going to be formed which increases the numerator, but then that means the denominator will be decreased thus increasing Keq??

If you add more A to the system the denominator gets bigger and K will be less. If K decreases then Q>K and the reaction will shift away from the products side to the reactants.

I think this is correct, but the best way to get answers is to go to your professors and ask them.

 

[ExplodingBoy]

The whole concept of the orbital being parallel to the other p orbitals, or perpendicular to them can get confusing at first, so here is the easy way to think about it. (How I initially remembered it.)

Pyridine CANT donate it's lone pair, because that lone pair has gotta be in an sp2 hybridized orbital, which as you know, doesnt have anything to do with aromaticity. The only orbital that you are interested in is the p orbital. Since the N in pyridine is already has a pi bond, that means that its p orbital is used up, and all it has left is 3 sp2 orbitals.

The N in Pyrrole however, can donate. If the N is hybridized to sp2 in pyrrole, then it is going to have a spare p orbital, which those lone pair of electrons will be in. We know that all 3 of the sp2 orbitals are used up because the N has 3 sigma bonds around it. Since the lone pair is in that p orbital, they are free to run around the ring.

Just remember to imagine a flat ring, and at each atom that has a p orbital, the p orbital is perpendicular to the ring. Its stick straight up and down through it. The p orbitals of the ring then all kinda blend together above and below the ring, allowing any electrons in those p orbitals to zoom through the p orbital track.

fastest way to check is to see if the N already has a pi bond. if it does, its not going to be donating its lone pair.

hope that cleared things up...