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Spectroscopy
Is the branch of science that studies the interaction between light and matter. It is mainly used for structure determination
Principle of Spectroscopy
Measuring the response of molecules after absorbing a certain amount of energy.
Electromagnetic Radiation
Electromagnetic Radiation has wave-like & particle-Like properties
Infrared Spectroscopy “IR”
The functional groups present or absent
Ultraviolet-Visible Spectrometry “UV-VIS”
Conjugated Systems
Identification of Functional Groups by IR (How does it work???)
• Stretching of bonds is a change of state “length”. \n • Each change of state requires a specific amount of energy. \n • This amount of energy is equal to the difference in energy between the two states “energy gap” \n • Each bond has its own specific energy gap. \n • We irradiate the molecule with light having all frequencies (energy) and then we determine which one \n “frequency” is missing. \n • The missing frequency was absorbed by the molecule.
Jones Oxidation
Oxidizes primary & secondary alcohols and aldehydes (turns Blue Green)
Lucas Test
It is used to differentiate between primary, secondary & tertiary alcohols
The reagents used are HCl with ZnCl2
Primary- NR
Secondary-turbid solution in a couple of minutes
Tertiary-turbid solution quickly
Ferric Chloride Test
• It is used to test for phenols.
• Phenol forms red-blue-violet complex with Fe(III)
Derivative Test
• Convert aldehydes and ketones into other derivatives.
• 2,4-dinitrophenylhydrazine & semicarbazone
• They convert liquid aldehydes and ketones into solids.
Tollen’s Test
• It is used to test for aldehydes
• The reagents used are AgNO3 and HO-
• The aldehyde is oxidized to carboxylic acid.
• Ag(I) is reduced to Ag (silver mirror)
Iodoform Test
• Test for presence of methyl ketones
• Reagents used are iodine (I2) and (HO-)
• It gives a carboxylate and Iodoform (ICl3) (yellow precipitate)
Bromine Test
• Test for presence of double or triple bonds
• Reagents used are iodine (Br2) (red color)
• Bromine adds to the double bond or triple bonds and since it is consumed the red color disappears
Permanganate Test
• Test for presence of double or triple bonds
• Reagents used are KMnO4 & HO- (purple color)
• It adds two hydroxyl groups to the double bond
• Since the permanganate is consumed the color disappears
Atomic size
As atomic size decreases, the bond length decreases and the wavenumber increases.
AS bond length decreases,
wavenumber INCREASES
Bond Strength
As bond strength increases, the wavenumber increases
Effect of Hybridization
Effect of Resonance
Resonance decreases the wavenumber
Intensity
As bond polarity increases, the signal intensity increases
Effect of Hydrogen Bonding “Alcohols”
•Concentrated alcohols (too many H-Bonds) will have a broad signal
• Diluted alcohols give narrower signals
IR Single Bonds
appear below 1500 cm-1
IR Double bonds
appear between 1600-1850 cm-1
IR Triple Bonds
appear between 2100-2300 cm-1
X-H (O-H, N-H, C-H)bonds
appear between 2700-4000 cm-1
Carboxylic acid IR
Alkene
note the peaks between 1600-1850
to conform a C=C-H bond, there should be a small peak at 3100
Alkyne
note peaks around 2100-2300
C(triple)C-H will have peak at 3300
Alcohol
Board band around 3200-3600
Primary Amine
two short peaks after 3000
Secondary Amine
one peak after 3000
Conjugation ==
lower wavenumber 1750 to 1680
IR tips
Hint: Tetrasubstituted alkenes and internal alkynes have no signals in the designated regions.
As conjugation increases
gap decreases, λmax increases
Woodward Fieser rules: base line
217 nm
Woodward Fieser rules: each extra double bond
add 30 for each
Woodward Fieser rules: auxochromic alkyl group
add 5 for each
Woodward Fieser rules: each exocyclic double bond
add 5 for each
one carbon of the double is in a defined ring and the other carbon of the double bond is not
Woodward Fieser rules: homoannular diene
add 39
Which of the following is the least soluble in water?
A. Alcohols
B. Secondary Amines
C. Primary Amines
D. Carboxylic acids
B. Secondary Amines
Dehydration of cyclohexanol follows:
A. Markovnikov’s Rule
B. Zaitsev’s Rule
C. Evelyn Effect
D. Hallalov Rule
B. Zaitsev’s Rule
In the dehydration mechanism, the first step is:
A. Formation of the carbocation
B. Protonation of the “OH” group
C. Loss of a proton
D. Rearrangement
Protonation of the “OH” group
Dehydration reaction is done in a distillation apparatus:
A. The alkene is toxic
B. To push the equilibrium to the right
C. To push the equilibrium to the left
D. To condense the reactants back
B. To push the equilibrium to the right
Draw the product of the dehydration reaction of
SN1 reactions
-Unimolecular
-polar protic solvents
-carbocation
-tertiary and secondary halides
-AgNO3 in ethanol
-SN1 depends on ALKYL HALIDE
SN2 reactions
-Bimolecular
- polar aprotic solvents
-primary and secondary halides
-NaI in acetone
-SN2 depends on BOTH ALKYL HALIDE AND NUCLEOPHILE
Zaitzev’s rule
an elimination reaction the major product is the more stable alkene with the more highly substituted double bond
Markovnikov’s rule
When HZ is added to unsymmetrical alkene, the hydrogen will add to the carbon with the most number of hydrogens
E1 mechanism
E2 mechanism
Evelyn effect
the product composition changes during the course of the reaction
Its hypothesized that the reason is because of the starting mixture of cis- and trans- isomers
What equipment is used in fractional distillation?
Vigreux column
Mass Spectrometry
It is used to detect the molecular weight and molecular formula
Ionization
Bombard molecule with a high energy electron beam to eject an electron and form a radical cation
Radical Cation
Also called, molecular ion or the parent ion and it has a mass equal to that of the original \n compound. Radical cations are unstable and rapidly fragment into smaller ions and radicals.
base peak
highest peak
molecular ion
its weight is equal to the molecular weight of the compound
base peak does not always equal
the molecular ion peak
Nitrogen Rule
• Odd Molecular weight = Odd number of nitrogen \n atoms \n • Even Molecular weight = No nitrogen or even number \n of nitrogen atoms.
Bromine in mass spec
Will have a 1:1 ratio
Chlorine in mass spec
Will have a 3:1 ratio
Analysis (M+1)+. Peak
Intensity (M+1) = (# of carbons x (% abundance) x Intensity [M]) / 100
In our reduction experiment, hydrogen has two roles
making the catalyst and reducing the trans cinnamic acid
A Grignard reagent is
a carbon nucleophile
made from alkyl halide and Mg
cannot react with water if we want a successful Grignard reaction
Methyl benzoate