Short Answer:
1) What type of electromagnetic radiation is used in nuclear
magnetic resonance?
radio
2) What is the most abundant peak in a mass spectrum called?
Base peak
3) What would the proton NMR peak look like for the
indicated hydrogen?
CH3
H3C
CH
O
CH3
Because the two sets of adjacent
protons are equivalent this peak
would follow the n+1 rule and be a
septet.
Nuclear Magnetic Resonance Spectroscopy
1H
NMR—Spin-Spin Splitting
When two sets of adjacent protons are different from each other
(n protons on one adjacent carbon and m protons on the other),
the number of peaks in an NMR signal = (n + 1)(m + 1).
Figure 14.8
A splitting diagram for the Hb
protons in 1-bromopropane
4) To which end of an alkene does the hydrogen add in
hydrohalogenation without a radical initiator?
The hydrogen adds to the least substituted end of the double bond
so that the most stable carbocation is formed.
5) What is another name for the 1,4-addition product?
Conjugate addition product
Predict the products.
1)
+
H
Br
Br
Br+
This cation is 2° and 3°. Other cation is 2° and 2°.
Thermodynamic product is most substituted alkene.
2)
Br
Br
H
Br
Br
+ Br
3)
Br
H
Br
Br
4)
O
O
H
H3CO
H3CO
H
5)
CH3
H3C
Cl2
CH3
hv
CH3
H2
C
Cl
H
CH2
Cl
CH2Cl
6)
NBS
hv
Br
H
Br
Br
Br
+
Br
Mechanism: Draw out the Mechanism for the following reaction.
H3C
1)
CH
H3C
Br2
CH3
C
H2
ROOR
H3C
H3C
RO
Br
Br
C
C
H2
Step 1 - Initiation
Br
Br
CH3
Step 2- Propagation
Br
H
Br
+ HBr
Br
Br
Step 3 - Termination
Br
Br
Br2
Br
bR
+
Br
2)
+
heat
H3CO2C
CO2CH3
CO2CH3
CO2CH3
CO2CH3
CO2CH3
H3CO2C
CO2CH3
+
CO2CH3
CO2CH3
Spectroscopy
1)
Using the MS and IR spectra attached (1A and 1B) propose
the formula and structure of this compound.
MS shows a molecular ion peak at 106 and a M+2 peak at 108..
So 106-35=71 so 71/12=5 carbons so 71-60=11 hydrogens so
C5H11Cl 2(5)+2-11-1=0
However, there is a carbonyl peak in the IR
So need to add an oxygen.
-CH4 gives C4H7ClO so 2(4)+2-7-1=2/2=1
So this is taken by the C=O bond.
One more thing, there is no peak at 2750 so no aldehyde, our
carbonyl is a ketone
O
O
O
Cl
Cl
Cl
The first one can be eliminated because of the base peak at 43
in the MS, a loss of 63 accounts for the loss of a –C2H4Cl
group.
2) Using the MS, IR and proton NMR (2A, 2B, 2C) propose a
possible formula and structure.
So the molecular ion peak is 165. Odd number means a nitrogen.
165-14=151/12=12 carbons so 151-144=7 hydrogens
So C12H7N 2(12)+2-7+1=20/2=10 way to high
So lets take off a C and add 12 Hs C11H19N 2(11)+2-19+1=6/3
Now lets look at the IR. There is a small peak at 3243 which tell
sus thee is some sort of amine present. Also there is a peak at 1649
and with a nitrogen present this tells us there is an amide.Also there
is a C=C double bond at 1622.
So we need to add an oxygen
-CH4 and Get C10H15NO so 2(10)+2-15+1=8/2=4
Alright now looking at the proton NMR we see a doublet of
doublets in the aromatic region (6 to 7ppm) which tells that we
have a para substituted benzene ring.
So the last structure we came up with, , had a DOUS of 4. This
only covers the benzene ring so we need to ad one more degree for
the C=O in the amide.
We know adding an oxygen well take care of this and give us
C9H11NO2 so 2(9)+2-11+1=10/2=5
O
NH2
So there are 5 peaks in the proton NMR, two of those are the doublet
of doublets in the aromatic region and another one is the small peak
the furthest downfield which is the amine proton.
So that leaves 2 carbons and 1 oxygen. And we know that
there is only the para position on the ring to add anything to.
So lets look at the remaining two peaks in the NMR they are
both singlets which tells us there are no adjacent protons that
have an effect.
This means that they are either attached to an oxygen or
O
nitrogen.
O
NH2
So that only leaves one place to add the other carbon.
O
O
HN