Can you explain in more detail the separation of enantiomers into diastereomers? (A mixture of enantiomers is treated with a chiral reagent to convert them to two new compounds, which have opposite centers from enantiomers and same centers from the chiral reagent, so they are diastereomers. Next, the diastereomers are separated and cleaved back into original enantiomers and the chiral reagent back. - GS).

What should we know regarding fluorescence? I recall discussing the topic in the past but am unable to find my notes regarding. (When a molecule absorbs the energy of light, it can cause a chemical reaction (photochemistry), or re-emit back as a light of different energy (fluorescence), or “wiggle out” by molecular vibrations to heat. Rigidity of a molecule reduces “wiggling out” the energy, which increases the efficiency of fluorescence. Therefore, we can control fluorescence by affecting the molecule by pH, presence of metals, DNA, etc. That is how many chemical sensors work. - GS).

E1 reactions are not stereoselective, so would the products be present in a racemic mixture like Sn1? (No. Unlike SN1, E1 does not produce a chiral center because the C=C bond is planar. - GS).

So would the addition of acetic anhydride to an amine be acylation? (Yes. It is actually acyl nucleophilic substitution, because half of the anhydride molecule leaves. - GS). Also are there the same issues with acylation as there are with alkylation when trying to make a primary amine or not because acylation is substitution? (No. The C=O group in an acyl blocks the lone electron pair on nitrogen from the second acylation. That is why acylation can be used as a get-around way to avoid the issue with multiple alkylations. Example - Gabriel synthesis. - GS).

Why would the hydrogens attached to the carbons in an ether and amine not be able to be hydrogen bond donors? (The small difference in electronegativity between C and H leads to the low polarity of the C-H bond, which leads to a very small positive charge on H, so it does not interact with the lone electron pair of the H-bond acceptor efficiently. - GS).

For ortho meta para directors, it is the group that is already attached to the ring that “tells” the group that is adding where to go or is it the group that is adding that knows where to add “on its own”? (The group, which is already in the ring, because it participates in the stabilization of the intermediate sigma-complex. The incoming group is separated from the positively charged conjugated system of the sigma-complex by an sp3-carbon, so its influence is minimal.

When looking at strong bases resulting in E rxns why does the major product favor Hoffman if the LG is poor? Such as F. (Reluctance of the leaving group to leave deviates the mechanism from the synchronous E2: Abstraction of a proton by a base leads to accumulation of negative charge on the carbon (in a classical E2, the developing negative charge is immediately carried away by the leaving group). As a consequence, the reaction pathway is determined by the stability of a carboanion. The trend of stability of carbanions is the opposite to that of carbocations, that is the carbanion is more stable at the less substituted carbon. Therefore, the less substituted C=C is produced, which is against Zaitsev rule. - GS).

N-14 NMR

How can 14^N be NMR active if it has 7 N and 7 Protons? I understood the text as both must be even numbers for the nucleus to be active (TA). (Not quite. If the number of neutrons and protons is odd, the nucleus is always magnetic, because the magnetic moments of individual nuclei can not compensate for each other. If this number is even, they usually do, and the nucleus is not magnetic. Sometimes, the magnetic moments of nuclei are aligned (two nuclei have the same magnetic moments, instead of the opposite moments), and then the nucleus is magnetic. 14N is an example (so-called quadrupole nucleus). Another simpler example is deuterium (2H) - GS.