Spectroscopy

Beer Ambert Law

 

Spectroscopy || Beer- Lambert’s Law

 

SPECTROSCOPY 
Beer-Lambert Law

 

 

 

UV Spectroscopy

 

https://www.youtube.com/watch?v=NhmSGCqVR7k&t=479s&ab_channel=Knowbee

 

-basically confirms the presence of a conjugated system if there is any absorbance at all in the UV range

 

e.g. β-Carotene, porphyrin, tryptophan

 

beta-Carotene, Antioxidant (CAS 7235-40-7) (ab142849) | Abcam

 

Porphyrins : the Colors of Life – PhysicsOpenLab

 

L-Tryptophan 73-22-3 | TCI AMERICA

 

 

Molecular Energy Levels - an overview | ScienceDirect Topics

 

 

IR Spectroscopy

 

Introduction to infrared spectroscopy | Spectroscopy | Organic chemistry | Khan Academy

 

 

Bonds as springs | Spectroscopy | Organic chemistry | Khan Academy

 

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c 
11 
7.1

where ~v is wavenumber (1/wavelength)

 

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Properties of Molecules

 

Untitled Document

 

 

 

-only vibrations that produce a change in dipole moment are detected by IR spectroscopy

-dipole moment = charge x distance

Electric Dipole

large dipole moments produce larger peaks (more absorbance)

-smaller dipole moments produce smaller peaks

 

 

How can I distinguish functional group region and fingerprint region in a infrared  spectrum? | Socratic

 

INFRARED SPECTROSCOPY (IR)

 

Interpreting IR Specta: A Quick Guide – Master Organic Chemistry

 

For the MCAT:

Single 
H: 3300-3500 
-H: 3300-3500 
H: 2800-3300 
Double 
c=o: 1700-1750 
c=c: 1600 
aromatic: 1500-1600

 

Signal characteristics – wavenumber | Spectroscopy | Organic chemistry | Khan Academy

 

 

 

 

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IR spectra for hydrocarbons | Spectroscopy | Organic chemistry | Khan Academy

 

 

-ketones produce large signals due to large dipole moment

-alkenes produce smaller signals due to smaller dipole moment

completely symmetrical alkenes produce no signal due to zero dipole moment

Signal characteristics – intensity | Spectroscopy | Organic chemistry | Khan Academy

 

 

-hydrogen bonding in solution produces broad peaks (-OH, -NH)

Signal characteristics – shape | Spectroscopy | Organic chemistry | Khan Academy

 

 

 

-for a primary amine (H2NR), the two hydrogens conncected to the nitrogen give two different signals next to eachother due to (A) symmetric streching and (B) asymmetric stretching

-also happens to anhydrides for the same reason

Symmetric and asymmetric stretching | Spectroscopy | Organic chemistry | Khan Academy

 

 

Factors effecting IR frequency absorption:

  1. Inductive Effect: increases energy (bc/ brings more lone pair electons into bond, thus increasing k)
  2. Resonance: decreases energy (bc/ kicks electrons out of bound, thus decreasing k)

-groups that allow a carbonyl bond to kick its pi electrons to the oxygen will decrease the bond energy

  1. Hydrogen bonding (i.e. O-H, N-H peaks): makes broad peak
  2. larger dipole moments produces larger peak

-completely symmetrical bonds produce no peak

 

1850 
o 
wavenumber (Ilcm) 
o 
A-łło 
1650 
o 
R NH2 
16 q O

 

IR signals for carbonyl compounds | Spectroscopy | Organic chemistry | Khan Academy

 

 

IR spectra practice | Spectroscopy | Organic chemistry | Khan Academy

 

 

 

 

UV/Vis Spectroscopy

 

UV/Vis spectroscopy | Spectroscopy | Organic chemistry | Khan Academy

 

A버•3

 

Absorption in the visible region | Spectroscopy | Organic chemistry | Khan Academy

 

 

Conjugation and color | Spectroscopy | Organic chemistry | Khan Academy

 

 

 

 

 

 

Nuclear Magnetic Resonance

 

NMR spectroscopy visualized

 

ScienceSke c 
NMR Spect c y 
Visualized

 

magnetic atoms: odd number of proton and/or odd number of neutrons

 

Introduction to NMR Spectroscopy Part 1

 

ooueUOSO.J 
ينفالف

 

Introduction to NMR Spectroscopy Part 2

 

effective magnetic field 
Beffective = - 
d eshielded 
higher E

 

What is the function of the magnet in NMR? A) It limits the nuclear spin to  certain allowed spin - Brainly.com

 

Orgo Chapter 13 Flashcards | Quizlet

 

-during NMR, a RF pulse brings the proton from the lower energy alpha spin state to the higher energy beta spin state

 

diamagnetic 
shielding

 

= hv=h B 
ΔΕ 
2π

 

effective magnetic field 
Beffective — 
Bapplied—

 

magnet 
NMR 
effective magnetic field 
= Bapplied — 
applied

 

shielded 
Beffective 
upfield 
shielded (deshielded) 
Beffective 
downfield

 

Number of Signals in NMR 1st Aspect of NMR

 

 

Shifting of Signals in NMR 2nd Aspect of NMR Part 1

 

s,mple e,oblem 
снз 
нзс -оснз 
HNHR vhe

 

Sample Problem 3) Provide a rough sketch of the HNMR spectrum of the following 
molecule: 
CH 
CCH2CH 
CH 
shifting: 
30 > 10

 

Sample Problem 4) Provide a rough sketch of the HNMR spectrum of the following 
molecule: 
CH3'l

 

distance downfield from TMS (Hz) 
operating frequency of the NMR (MHz) 
ppm

 

Shifting of Signals in NMR 2nd Aspect of NMR Part 2 Anisotropy

 

 

Aromatic ring:

Bapplied - 
Beffective = 
Beffective 
downfield

 

Sample Problem 5) Provide a rough sketch of the HNMR spectrum of the following 
CH 
olecule: 
8-6.5 ppm 
2.3 ppm

 

Aromatic protons 
5=7-8 m 
circulation 
or electrons 
(ring current) 
induced 
induced magnetic field 
Vinyl (Olefinic) protons, 
ö = 5-6 ppm 
Bindu ced 
induced field 
reinforces the 
external field 
(deshielding) 
c 
c 
H 
Binduced 
induced field 
reinforces the 
external field 
(deshielding) 
Acetylene protons 
2.5 ppm 
shields 
ind u ced 
the proton 
c 
induced 
induced

 

-aromatic rings and double bonds deshield protons (shift downfield)

-triple bonds shield protons (shift upfield)

 

Shifting of Signals in NMR 2nd Aspect of NMR Part 3

 

ё-оРГ 
12 9.0 
8.0 
6.5 
4.5 
2.5 
1.5

 

 

юн 
АН 
12 
10 
—снз 
-снг- 
н с=с-н 
x=F, Cl, Вг 
н-с-х 
нс-о 
н-СЛ 
н-с-с=о 
н-сс=с 
(i.e. electronegative atom)

 

Integration of Signals in NMR 3rd Aspect of NMR

 

NMR 
CICH2CH3

 

Splitting of Signals in NMR   4th Aspect of NMR Part 1

 

СНЗСНСНЗ 
N+l=3 N+l=7

 

Splitting of Signals in NMR   4th Aspect of NMR Part 2

 

 

Splitting of Signals in NMR  4th Aspect of NMR Part 3

 

quartet

 

Spin Spin Splitting – N+1 Rule – Multiplicity – Proton NMR Spectroscopy

 

Spin-Spin Splitting 
Singlet 
Doublet 
Triplet 
Quartet 
—IIIL

 

 

 

 

 

Nuclear Magnetic Resonance (Advanced)

 

NMR Coupling Constants in Organic Chemistry

 

 

can use J values for two things:

  1. to see which set of hydrogens are next to eachother
  2. to see whether the hydrogens around a double bond are cis or trans

 

 

-if J values are the same, then those two sets of hydrogen are next to eachother

 

doublet 
trans 
doublet 
ab 
doublet 
-15 Hz 
doublet 
cis 
Hz 
ba

 

approximate Jab (Н 
—с—с— 
—с—с—с—

 

Splitting Diagrams NMR Organic Chemistry

 

асњсњс

 

Complex Splitting:

-occurs when the neighboring hydrogens are in different electronic environments

 

_oq 
_(q 
___eq. 
'1• 
gÅozÅDßDZD

 

-quartet of triplets

bc---• bc 
bc--

 

Enantiotopic and Diastereotopic Hydrogens in Organic Chemistry

 

сн 
он 
Need this exphined in тоге 
Select the lectur•e entitled 
Diaster•eome 
снз 
он

 

Homotopic, Enantiotopic, Diastereotopic, and Heterotopic Protons

 

SPECTROSCOPY

 

-Hydrogens that are diastereotopic are in different chemical environments and so have different chemical shifts and thus are represented by different chemical peaks

 

Че4ет+ог;с 
воз 
сч

 

OH and NH Signals in HNMR in Organic Chemistry

 

Оосњснз 
НОСН2СНз

 

Deuterium in HNMR in Organic Chemistry

 

носњснз 
- осњснз 
proton 
exchange