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Give the formula and structure of the compound with this IR and a molecular ion peak at 116.
How many peaks in the 13C NMR spectra?
OHOH
F
OH
F
F
2 3 4 3
Cl
FHO
Cl
FCl
Cl
ClCl
6 4 2
Which compound matches this 13C NMR spectrum?
Br
BrBr
Br
Br
Br
Br
Br Br
Br
Br
Br
O
OH
AB
C
OA
BC
Br
B
CA
Rank the labeled C’s of each compound in order of increasing chemical shift?
B ~ 33ppm
C ~ 35ppm
A ~ 115ppm
A ~ 8ppm
B ~ 37ppm
C ~ 210ppm
C ~ 9ppm
B ~ 30ppm
A ~ 177ppm
Information gained from a 1H NMR Spectrum
• Number of Signals
• Position of Signals
• Intensity of Signals
• Spin-Spin Splitting of Signals
How many different types of protons are in each compound?
H3C
O
CH3
HAHA
1 type
H3C
H2C
ClHA
HB
2 types
H3C
H2C
O
CH3
HA
HB
HC
3 types
H3C
CH
H3C
CH2
H2C CH3HA
HA
HB
HC
HD
HE
5 types
Determining the number of different protons in compounds with bonds.
Cl
Cl
H
H
cis to Cl
cis to Cl
1 type
Cl
Br
H
H
cis to Cl
trans to Cl
2 types
Cl
H
H
H
cis to Cl
cis to H
3 types
Determining the number of different protons in compounds with rings..
H
H
H
H
H
H
All H’s are equivalent, 1 type
H
H
H
H
Cl
H
cis to Cl
cis to H3 types
H
H
H
H
CH3
CH3
Protons on methyls are equivalent
Each of these is cis to a methyl 2 types
O
OO
OH
O
O HO
OH
Spin-Spin Coupling
When determining the spin-spin coupling, look at the number of protons on the adjacent carbon. For the methyl group, look at the methylene group. There are 2 protons, so using the N+1 rule tells us that the peak should be a triplet in a 1:2:1 ratio.
For the methylene group, look at the methyl group. There are 3 protons, so using the N+1 rule tells us that the peak should be a quartet in a 1:3:3:1 ratio.
Protons attached via a double bond show a unique splitting pattern., a doublet of doublets.
O
HO
Hc
Ha
Hb
OOH
O
OH
An unknown molecule A has 4 signals in the 1H NMR spectrum. Which of the following corresponds to molecule A
How many nonequivalent protons does the following structure have?
4
Reading from left to right, what multiplicity would be found for the three nonequivalent sets of protons in the 1H NMR spectrum of the following compound?
d, d, s
Introduction
Homolytic bond cleavage leads to the formation of radicals(also called free radicals)
Radicals are highly reactive,, short lived species
Single headed arrows are used to show the movement fo single electrons.
Production of Radicals
Homolysis of relatively weak bonds such as O-O and X-X bonds can occur with the addition of energy in the form of heat or light.
RCH2 R2CH R3CH
Carbon radicals are categorized as primary (1°), secondary (2°) and tertiary (3°) based on the number of attached R groups.
1° 2° 3°A carbon radical is sp2 hybridized with a trigonal planar geometry with the unpaired electron in the unhybridized p orbital.
Bond dissociation energy is used as a measure of radical stability.
Two different radicals can be formed with the cleavage of a C-H bond.
Basically, the more alkyl groups attached to the radical carbon the more stable it is. Also the more stable the radical, the less energy it takes to break the C-H bond.
What type of radical are each of the following?
H3C
H3C
CH2
CH3
•
H3C
CH
H3C
CH
CH3
•H3C
CH
H2C
CH2
CH3•
1°2°3°
Of these three radicals, which is the most stable?
H3C
H3C
CH2
CH3
•
•H + X + H X•
Radical Reactions of Alkanes
Abstraction of a H from a C-H bond in which one electron is sued to form H-X while the other is left on the new alkyl radical.
X
X
•
•
A radical can also add to a alkene by adding onto a double bond and leaving the other carbon that was part fo the double bond as a radical.
X + X X X
Radicals are highly reactive and unstable and usually react quickly with a sigma or pi bond. However sometimes they can react with another radical.
O O + X O O X
When oxygen, a diradical, is present it acts as a radical inhibitor or scavenger. Meaning it prevents the radical from attacking any alkanes or alkenes.
• •
• • • •
In the presence of heat and light, alkanes and halogens will react to form alkyl halides.
H3C H + Cl2 H3C Cl + HCl
H+ Br2
Br
+ HBr
H3CH2C CH3 + 2Cl2 H3C
H2C CH2Cl
+
H3CHC CH3
Cl
Predict the products from the monobromination of the foloowing compound?
CH CH
H3C
H3C
CH3
CH3Br2
hv/ heat CH CH
H3C
H3C
CH2Cl
CH3CH C
Cl
H3C
H3C
CH3
CH3
CH CH
H3C
H3C
CH3
CH2Cl CH CH
ClH2C
H3C
CH3
CH3
Step 1 - Initiation
Cl Cl hv/heat2Cl •
Step 2 – Propagation
H3CH2C H + Cl H3C CH2 + HCl
H3C CH2 +Cl Cl H3CH2C Cl + Cl
• •
• •
Step 3 - Termination
Cl Cl+ Cl Cl
H3C CH2 CH3H2C+ H3CH2C
H2C CH3
H3C CH2+ Cl H3C CH2Cl
In each step of the propagation a bond is broken and formed. And because the overall step has a -H it is exothermic. Step 1 is called the rate determini g step because it is higher in energy.
Transition States
Cl----H----CH2CH3 Cl---Cl---- CH2CH3
H3CH2C CH3 +Cl Cl H3C
H2C CH2Cl + H3C
HClC CH3
There are 6 Methyl H’s and 2 Methylene H s. Based on this, the ratio fo the two products should be 3:1(primary to secondary).
However, the ratio is 1:1.
The more stable the radical being formed is, the easier it is to cleave the C-H bond.
Which C-H bond in each compound is most reactive?
H H
H
H
Chlorination Vs. Bromination
1:1
H3CH2C CH3 + Cl2 H3C
H2C CH2Cl + H3C
HClC CH3
H3CH2C CH3 + Br2 H3C
H2C CH2Br + H3C
HBrC CH3
99%
Chlorination is faster and nonselective. This is due to it’s rate determining step being exothermic.
Bromination is slower and chooses the most stablew radical. This due to it’s rate determining step being endothermic.
H3C
H2C CH
CH3
CH3
+ BR2
H3C
H2C CCl
CH3
CH3
+ Br2Br
Halogenation is useful in the formation of alkenes.
+ Cl2
Cl
+ K+-OH
An elimination in the presence of a strong base is responsible for the formation of the alkene.
Cl -OH
H
+ H2O + KBr
H+
+
+ -HSO4
HOHO
H
H
-HSO4
OH
aq. H2SO4
OH+
Conversion of the alkene to an alcohol via nucleophillic substitution is an extension of the utility of radical halogenation.
+HgOAc
+HgOAc
MeOH
H
CH3
HgOAc
OCH3
HgOAc
NaBH4
OCH3
Hg(OAc)2
NaBH4
OCH3
Oxymercuration-demercuration of an alkene results in the formation of an ether.
Radical halogenations give a racemic mixture of pRodcuts when possible. This halogenation of an achiral compound results in 3 products. A primary and secondary alkyl halide. The secondary halide exist as a pair of enantiomers due to the creation of a stereogenic center upon halogenation.
H3CH2C
H2C
Cl2H3C
H2C
H2C H3C
HClC
H2C+CH3 CH2Cl CH3
H3CCH2CH3
BrH
Cl
H3C
Br
CH2CH3
Cl2
Cl2
H3C CH2CH3
Br Cl
H3C CH2CH3
Cl Br
H3C
CH2
CH3
Br H
Cl
H3C CH
CH3
Br H
Cl2
Cl2
H3CCH3
Br
Cl
H
H
H3CCH3
Br
H
H
Cl
enantiomers
diastereomers
H3C CH2
CH3
Br HCl2
ClH2C CH2
CH3
Br H
+
H3C CH2
CH2Cl
Br H
Only achieve enantiomers or diastereomers if the halogenation takes place at a stereogenic center.