Chemistry 14C Discussion Board

DBE

chilicheese21 — Sat, 07 Nov 2009 07:45:10 GMT

Can someone explain to me how DBE can be calculated and how it really affects the calculation of a molecular structure?

Problem #15

kamepan — Sat, 07 Nov 2009 07:42:59 GMT

For molecule L, I was confused as to how the answer is stereoisomer and diastereomer. First of all, I thought that the molecule had only one stereocenter to begin with, hence it can only have an enantiomer. Wouldn't a diastereomer require 2 stereocenters, one of the stereocenters switched? I knew the answer wasn't enantiomer like molecule K was, but maybe someone could help clarify to me as to why Molecule L was a diastereomer when there was only one stereocenter?

Forum: Midterm Exam 1

Problem #22

kamepan — Sat, 07 Nov 2009 07:37:05 GMT

Can someone explain to me as to why the second structure isn't aromatic? So even if the other ring isn't in the same plane, you have to take its bonds into account as well? Even so, wouldn't that be 10? (4n+ 2=10 n=2?) And if you did only the larger ring, wouldn't that still fit the rule? How do you determine if something is aromatic or not if both of the rings aren't in the same plane?

Forum: Midterm Exam 1

pg 128 in the course reader

kamepan — Sat, 07 Nov 2009 07:13:52 GMT

I was looking at the structure on the bottom left of the first slide (I believe it's a benzene ring with a methyl attached it it) and I couldn't understand how the hydrogens attached to the ring near the methyl groups were equivalent. I understand that the mirror "line" is drawn horizontally to help show symmetry...but isn't one hydrogen attached the carbon (that is attached to the methyl) a single bond while the other one is attached to the carbon with a double bond? If so, how are they equivalent hydrogens? I thought everything about them was supposed to be exactly the same?

Forum: Old Exam and Miscellaneous Questions

Problem #6

kamepan — Sat, 07 Nov 2009 06:48:59 GMT

I just have a question about the numbers for the stretching frequency that were given for this particular problem- I might be missing something but where did those specific numbers come from? Are those just general estimates based off of what we are supposed to memorize for the stretching frequencies of the different functional groups?

Forum: Practice Problems

q4 on pg 108

amschneider — Sat, 07 Nov 2009 06:23:31 GMT

On question 4, where it is asking how to determine if a peak in the MS is the molecular ion. If M, M+1, M+2 correspond to the highest peaks, unless there is significant fragmentation, how is it that you determine if a molecular ion is prone to easily fragment? What aspect of the molecule is that based on?

Forum: Practice Problems

H-NMR (Signal Splitting)

kturner5 — Sat, 07 Nov 2009 05:46:49 GMT

How do I identify when signal splitting is occuring? How does it occur?

Forum: Old Exam and Miscellaneous Questions

chemical shifts

lcsiu — Sat, 07 Nov 2009 04:47:27 GMT

I know that electronegativity influences chemical shift, but what other factors influence the chemical shift?

Forum: Old Exam and Miscellaneous Questions

General Questions about Solving Spec Problems

if90 — Sat, 07 Nov 2009 01:18:31 GMT

When figuring out the DBE (Degree of Bond Equivalency) in a Spec problem, is it right to assume that if the value of DBE is 4 or 5 the final structure will consists of a benzene ring? Also, if the DBE is 1, I understand that there is no pi bonds/ring, so is it right to assume that the NMR would be CH₂ or CH_3. Is this true for all problems or can there be exceptions?

Forum: Old Exam and Miscellaneous Questions

Analysis of Infrared Spectroscopy

if90 — Sat, 07 Nov 2009 00:57:28 GMT

When analyzing infrared spectroscopy, if there isn't any peaks in a particular zone, on an exam do we just write nothing or would it be best if we state what we could possible see in that zone? Also, for zone 2 if there is no OH, it is obvious that we know this because there is no broad peak, however would it be best to state "no OH because no broad peak"?

Forum: Old Exam and Miscellaneous Questions

Problem # 12(b)

monhernandez — Sat, 07 Nov 2009 00:51:25 GMT

For this problem eleven atoms were said to lie in the same plane due to the pyridine ring. The exam key said that these were 6 carbons, 4 hydrogens, and 1 nitrogen. I understand where the 4 hydrogens and 1 nitrogen came from, but where did the 6th carbon come from since the ring only has five? Is the 6th carbon the one that is connected to, but not directly on the pyridine ring? If so why does it get included?

Forum: Midterm Exam 1

Hydrogens in benzene rings

Eliarias04 — Sat, 07 Nov 2009 00:40:44 GMT

Hydrogens in benzene rings will often appeas as multiplets. Why?

Forum: Old Exam and Miscellaneous Questions

HELP with studying!

soccergrl5 — Sat, 07 Nov 2009 00:08:52 GMT

I tend to get scatter brain when it comes to studying for Chem....is there any advice on what to do first such as problems or looking at the class webpage, etc. Any suggestions is appreciated =)

Forum: What Works For You?

Solving For Implications

bcafelover — Fri, 06 Nov 2009 23:49:58 GMT

Does anyone know of stematic for solving for implications?

I keep missing most of the implication for one given molecule. Please help!

Forum: Practice Problems

Problem 22

davidpollack — Fri, 06 Nov 2009 20:48:08 GMT

I am unclear why the last molecule in the row is not aromatic. According to the answer key, it is because it break's Huckel's rule (4n + 2 pi electrons). However, one of the closed rings is a benzene ring, so it has 6 pi electrons. Why does that benzene ring not count towards aromaticity?

Forum: Midterm Exam 1