MendelSet works best with JavaScript enabled. Please enable JavaScript.
If you are using NoScript or another JavaScript blocker, please add MendelSet.com to your whitelist.
Keywords: acid-base, alpha carbons, relative acidity, resonance
Description: Please Read Chapter 3 of "2nd Language" and do the associated problems. In addition, complete the following problem set.
Total Problems: 10
Keywords: elementary acid-base, formal charge, introduction, resonance, review
Description: Second homework set for Chem 420
Total Problems: 13
Textbook and Chapter: Carey and Giuliano 8th Ed. (2010), Chapter 1
Keywords: formal charge, Lewis structures, resonance
Description: This mendel set goes through the basics of structure:
Total Problems: 7
The nitrosyl cation is shown below. Also shown are several proposed resonance arrows, only one of which is correct.
Draw the resonance forms that would follow from each set of arrows, and include formal charges. Which one is the correct resonance form? Explain your reasoning.
When evaluating possible Lewis structures the most important rule to follow is the octet rule: every atom must be surrounded by 8 and only 8 electrons (atoms in period 3 or below on the periodic table have d orbitals and so can contain more than 8 electrons, but that exception does not apply here).
Do any of the resonance forms drawn have atoms that break the octet rule? Yes- they all do! Except for resonance arrows in D. So D is the correct resonance form.
Pyrrole is an example of a heteroaromatic compound: it contains a heteroatom (atom that is not carbon or hydrogen, such as N, O, S, etc.), and is aromatic.
Because pyrrole is aromatic, we should be able to draw many resonance forms- usually as many resonance forms as sides (in this case, five sides, so five resonane forms).
Draw all resonance forms for pyrrole. (I've started you off.)
One of the rules for aromaticity is that all atoms shoudl be sp2 hybridized. But the nitrogen in pyrrole is sp3 hybridized, so how is it still aromatic? Because in 4/5 of its resonance forms the nitrogen is sp2 hybridized; the real picture of pyrrole looks more like the structure on the left (dashed circle) than any individual resonance form.
Draw all resonance forms for each species.
(This is similar to problem 569, which involved the allylic position.)
Notice that arrows always go from high electron density to low electron density. So for negatively charged species, the arrow starts at the lone pair. For positively charged species, the arrow starts at the double bond and goes towards the positive charge.
(Radical resonance uses hooks instead of arrows and so is a little different.)
Rank the following electrophiles in order of decreasing reactivity with NaN3 in DMF. (1 = most reactive)
The only differences among these three molecules is their leaving groups, so whichever has the best leaving group will react fastest with the nucleophile (azide, N3-). Stable molecules make good leaving groups, so the best leaving group of the three is SO3R-, which has has three resonance forms (see problem 535). So the third compound will react the fastest with NaN3.
Rank the following anions in order of decreasing stability (1 = most stable)
Charged molecules are generally less table than neutral ones, and each of the molecules below has a negatively charged oxygen.
But not all negative charges are equal; some oxygens are "closer to neutral" than others. How? Because resonance stabilizes charges by sharing electron density over multiple atoms (this is called electron delocalization).
Hydroxide (HO-) doesn't have resonance, so the oxygen has 100% of the negative charge to itself. On the other hand, the sulfonate ester (SO3R-) has three resonance forms, so each oxygen only has ~33% of a negative charge. So the sulfonate ester is the most stable anion.
In general, the more resonance forms a molecule has, the more stable it is.
Draw all possible resonance forms for each structure below. Use curved arrows.
Note that some structures only show charge, and not implied protons or lone pairs!
Notice that when drawing resonance forms with positive charges, the arrows never come from the positive charge. Arrows only come from π (pi) electrons- lone pairs or double/triple bonds.
