1) Draw a structure consistent with the following data:
•The MS shows a molecular ion at 59 amu.
•The IR spectrum shows a double-humped strong absorbance at
around 3300 cm–1 (the only absorbance in the functional group
region).
Odd molecular ion peak tells you there is a nitrogen.
59-14=45 45/12=3 carbons 45-36=9 hydrogens
C3H9N. 2(3)+2-9+1=0
The peak at 3300 cm–1 tells us that the N is part of an amine
NH2
2) Either 2-butanone, 2-methyl-2-nitropropane, 3-pentanone, 1nitropropane, nitroethane, or 2-bromopropane is responsible for the 1H
NMR spectrum shown. Draw the structure of the responsible
compound. (doublet and heptet)
There are two signals in the spectra so we can eliminate 2-butanone,
2-methyl-2-nitropropane and 1-nitropropane because they have
either more or less than 2 types of protons.
Next you can eliminate 3-pentanone because it would have a
triplet and a quartet which is not seen in the spectra.
Also nitroethane can be eliminated because it would have a
doublet and a quartet.
While the answer is 2-bromopropane which has a doublet
and a heptet.
Br
3) Either 2-butanone, 2-methyl-2-nitropropane, 3-pentanone, 1nitropropane, nitroethane, or 2-bromopropane is responsible for
the 1H NMR spectrum shown. Draw the structure of the
responsible compound. (quartet, singlet and a triplet)
Because there are 3 signals it is either 2-butanone or 1nitropropane.
1-nitropropane should have a doublet, a quartet and a triplet
While our answer, 2-butanone should have a singlet, a quartet and
a triplet..
O
O2N
4) The molecular formula of a compound is C6H12O. Determine the
structure of the compound based on its molecular formula and its
13C NMR spectrum. (4 PEAKS)
First 2(6)+2-12=2/2=1 so either a ring or double bond.
No peak shows up in the double bond region, C=C or C=O.
So that leaves a ring.
Four peaks and with this structure we have 4 different
types of carbon.
OH
5) Identify the compound with molecular formula C3H5Cl3 that
gives the following 13C NMR spectrum. (The resonance at 0 ppm is
due to the TMS standard, not the unknown.)
First, 2(3)+2-5-3=0 so no double bonds or rings.
Secondly there are 3 peaks, so 3 different kinds fo carbon.
So that leaves two choices that are correct.
Cl
Cl
Cl
Cl
Cl
Cl
6) 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.
7) What type of electromagnetic radiation is used in
nuclear magnetic resonance?
Radio waves
8) What is the most abundant peak in a mass spectrum
called?
Base peak
9)
Using the MS and IR spectra attached (1A and 1B) propose
the formula and structure of this compound. (106 and 108)
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.
10) What is the major product of the following reaction?
HBr
peroxide
H
Br
Br
Br
Br
11) What is the major product of the following reaction?
Br2
CCl4
Br
Br
Br
Br
Br
Br
12) What is the major product of the following reaction?
HBr
H
H
Br
Br
Br
8) What alkene should be used to synthesize the following alkyl
bromide?
Br
H
+
Br
Br
9) What alkene, when allowed to react with HBr, would produce
the following alkyl bromide? (There is more than one correct
answer.)
H
Br
Br
Br
10) What five-carbon alkene will give the same, single product
whether it reacts with HBr in the presence or the absence of a
peroxide?
Br
H
Br
Br
Br
H
Br
Br
Br
11) Draw the major product of the following reaction, including
its stereochemistry. (Use squiggly bonds to indicate a reaction that
is not stereoselective.) Assume only one equivalent of the reagent
is available to react with the substrate.
HBr
H
Br
Br
Br
12) Draw the major product of the following reaction, including its
stereochemistry. (Use squiggly bonds to indicate a reaction that is
not stereoselective.) Assume only one equivalent of the reagent is
available to react with the substrate.
HBr
H
Br
Br
Br
13)
Br
Br
H
Br
Br
+ Br
14)
Br
H
Br
Br
15)
NBS
hv
Br
H
Br
Br
Br
+
Br