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Dear Professor Katz/Professor Hudson,
Problem Set 4, Question 3, Part d.
While I was completing the problem, I had noted that there were two stereocenters.
1. Carbon attached to F, Br, COH, and the Ethyl group. (Central Carbon)
2. Carbon attached to O, CH3, H, and CFBRH. (Carbon to the "right")
From this, I had thought that there were four enantiomers, because there were two stereocenters. In the answer key, only one stereocenter is noted.
I understand that the molecule is chiral, but why is the second carbon attached to four different groups not a stereocenter?
That's right, the compound in part d has 2 stereocenters. The enantiomer has both stereocenters inverted to their other configurations (the notation may be a bit confusing - the hydroxy group going from "up" on the left structure to "down" on the right structure implies that the hydrogen atom at that position points "down" on the left structure and "up" on the right structure). Keep in mind that there may be 4 possible stereoisomers for that compound, but only one mirror image of the structure as drawn (enantiomer).
Hello! Problem set 2, question 6a:
What makes the 1st and 4th carbons have different environments? Arent they both carbons bonded to one other carbon and three hydrogens?
They are both methyl groups, but they are attached to carbon atoms that are not identical. Along the same line of reasoning, one methyl is farther away from the bromine atom than the other, making them in different environments. Remember, they have to be truly identical and usually that also means symmetry-related.
I am wondering whether I would use Diethyl ether as the solvent that I place solid magnesium metal in when I'm doing a Gringard reaction.
I am wondering whether I would use diethyl ether as a solvent that I place solid lithium metal when I'm doing a Gringard reaction.
I am wondering why I use Diethyl ether as a solvent in the Gringard reaction.
Grignard and organolithium compounds are very basic, so you have to use solvents that they can't deprotonate. Alkanes (such as hexane) and ethers (such as diethyl ether or THF) are commonly used. Ethers are especially good at dissolving them too.
I am wondering why both methyl alcohol and EtOH are stronger acids than isopropyl alcohol.
The stability of the conjugate bases (alkoxides) are very dependent on solvation. So in solution, the more steric bulk around the anion, the less stable it becomes (because the steric bulk reduces the number of stabilizing interactions with the solvent). Following the acidity trend above, MeOH > EtOH > iPrOH > tBuOH.
I am wondering why a carbocation bonded to three other carbons is more stable than a carbocation bonded to 2 other carbons is more stable than a carbocation bonded to one other carbon is more stable than a carbocation bonded to no other carbons similar to homework question 7b1.
Hyperconjugation allows the nearby bonds to add electron density and help stabilize the carbocation. More carbon substituents = more hyperconjugation. See page 311 for one picture of this (but not use of the term). Another picture is on page 454.
I had a question on the practice exam, 5b, why are the molecules identical rather than conformational isomers?
The two structures in 5b can be superimposed by reorientation of the entire molecule in space (without the need for bond rotations). This makes them identical, while requiring allowed single bond rotations would mean they were conformational isomers.
What functional groups will we need to know for the exam?
You should be very comfortable with alcohols, alkenes, alkynes, (cyclo)alkanes, and alkyl halides. Given an IUPAC name for a compound containing any of these, you should be able to draw the structure. Likewise, you should be able to provide an IUPAC name for compounds containing any of these groups if given a structure. We didn't go into the specific IUPAC nomenclature for aldehydes, ketones, carboxylic acids, esters, ethers, epoxides, and amines, but you should still be able to recognize these.
I am wondering what pKa trends are important to know or memorize for the class in general. I think that knowing the pKa trend that alkynes are more acidic than ammonia which is more acidic than alkanes is useful.
There are several helpful trends to know, but which to use (and how) will depend on the particular situation that you are analyzing.
For second row atoms, electronegativity is a big deal. The more electronegative, the better stabilized the conjugate base (all other things being equal).
Anions in larger orbitals (less change density) are generally more stable than anions in smaller orbitals.
Resonance of course affects (increases) anion stability.
For the same atoms type, hybridization changes the anion (conjugate base) stability a lot. The more s-character, the more stable (all other things being equal).
Hi, I'm wondering what functional group values we need to know for analyzing IR.
The best source for IR absorptions to know (and functional groups) is the IR absorptions handout on the Lectures page. There are also lists in the book, but I don't find those as helpful!
I was wondering if we can get an extra day(s) to get to do the learnbacon for alkenes and alkynes since we just learned some new information regarding Alkene/Akyne reductions today?
Thank you for your help!
Done! (now due at Final Exam time)
Hello! I was wondering if you can use H+ and H2O for reaction 13d on the practice final exam?
In that case, no you can't use H+/H2O to get that product. If you protonate and make the carbocation, it will shift to the tertiary position before the water attacks.
Why is it that adding water to a dry alcohol "decouples the OH hydrogen from all adjacent hydrogens"?
OH protons are hydrogen bonding and in exchange with the solvent (in most cases). Since coupling is through bonds, the exchange disrupts the coupling. It's not water specifically, just a variety of hydrogen bonding environments and exchange rates.
I am wondering what the mechanism for the reaction on exam 3 part 4 that forms C7H14O.
The acid first protonates the double bond, forming a tertiary carbocation. The alcohol then attacks the carbocation to form a 6-membered ring (and then gets deprotonated by the solvent to yield the product).
For 15 on the practice exam, why is it 70% one enantiomer? %ee is 40%, and I have in my notes that that means 40% is not racemic, but how do I determine what percentage of that is one enantiomer? (If you can put this algebraically, it will make far more sense to me than just explaining it in logic.)
%ee is equal to the difference between the major and minor enantiomer percentages. So you would have 70%(major)-30%(minor) = 40%ee
If we're speaking in terms of percentages, the percent of the '+' enantiomer (lets call it X) and the percent '-' enantiomer (lets call it Y) have to add up to 100%. X + Y = 100. We know that we have an excess of 40% of X relative to Y. So, X - 40 = Y. So, substitute X - 40 in to your first equation for Y. X + X - 40 = 100. 2X = 140, or X = 70.
I am wondering whether grades have been submitted for CH241.
Yup, they have been submitted.