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Test 1

Organic II

QuestionAnswer
As bond increases, wavenumber (IR) ________. Increases
As mass increases, wavenumber (IR) ________. Decreases
IR peak above 3000 Sp2 C-H bond (unsaturated)
IR peak below 3000 Sp3 C-H bond
Sharp IR absorption around 3300 with a smaller one about 2200 Alkyne (C-H and C triple C)
C-H IR _______ as bond goes from alkane -> alkene -> alkyne Increases
Steps for IR (with molecular equation) 1. Unsaturation? 2. Possible structures/ look at key absorptions 3. Compare to IR/ build structure
Put in order of high -> low wavenumber: N-H, C-H, O-H O-H > N-H > C-H higher bond strength (electronegativity), greater wavenumber
Equation for saturation for molecule with N 2n+1=number of H needed to be saturated where n=C+N
Amine IR Low absorption (neither sharp or broad) around 3300-3500. Number of peaks in it are number of N-H bonds
Equation for saturation with molecule with O 2n+2 Ignore the O (n= C)
Sharp, large IR absorption around 1700-1800 Carbonyl group C=O
Carboxylic acid IR C=O absorption around 1700 and an OH frequency that often overlaps with C-H (around 2500-3300)
Ester IR C=O absorption around 1735-1800 with no OH absorption. Will be about 10 higher than ketones
Aldehyde IR "Fangs". Two sharp absorptions around 100cm-1 apart that are same intensity with one being around 1750cm-1. Second peak can be covered with sp3 C=H
What to do if you don't know if 1H NMR signals are same? Substitution test, see if molecule is homotopic/enantiotropic. If het/diast they are different.
NMR splitting tells you About the number of H's neighboring it.
NMR signaling tells you Number of equivalent H's on C
Splitting names Singlet, doublet, triplet, quartet, pentet, sextet, septet
Splitting equation Splitting=number of neighboring H + 1
NMR steps: 1. Saturation if given molecule 2. Draw parts 3. Put together using NMR (greater chemical shift, closer to electronegative atoms/more sub C/double or triple bonds)
What happens to chemical shift where a CH2 is surrounded by a Br on one side and a carbonyl group on the other? The chemical shifts relating to each molecule (in response to Br and in response to carbonyl) are added minus a little
NMR chemical shift around 7 Aromatic ring
Other giveaways for aromatic ring LOTS of unsaturation (like C6H8), multiplet at high saturation
If Ca - Cb J value = Cb - Cc J value in CaCbCc molecule, there is an NMR of: An apparent sextet of a multiplet
NH and OH frequently show up on NMR as: 1H singlets
Topicity if substitution test makes constitutional isomers: Heterotopic, Hs are non-equivalent
Topicity if substitution test makes conformational isomers: Homotopic
Topicity if substitution test makes enantiomers: Enantiotopic
Topicity if substitution test makes a chiral center: Enantiotopic
Topicity if substitution test doesn't make a chiral center: Homotopic
Topicity if substitution test makes diastereomers: Diastereotopic, are not equivalent in NMR
Number of signals in 13C NMR means: Number of equivalent C's
Common indicators of diastereotopic Hs: 1. Alkene 2. Existing chiral center 3. Ring
13C NMR shifts 100-160ppm Aromatics and alkenes
13C NMR shifts >180ppm Aldehydes and ketones
13C NMR shift 160-180ppm Carboxylic acids
13C NMR shift <80ppm sp3 C
In m/z, the peak showing mass of original molecule is called: Parent (molecular ion) peak. M+
M/Z tells you: Mass of whole molecule, masses of fragments (good at telling constitutional isomers apart)
Fragmentation in m/z detects the fragment of radical cation with the: Charge
Alpha splitting: Occurs with alcohols and carbonyls, C bonds will split on each side of the OH/C=O
McLafferty's principle: Certain type of m/z splitting that generated an EVEN m/z fragment due to an added H, needs a propyl chain to do, and produces an enol (alkene + alcohol) from a carbonyl group
Created by: RunningMads
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