1 Draw a structure consistent with the following data:

1

•  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 cm1 (the only
  absorbance in the functional group region).

Odd molecular ion peak tells you there is a
nitrogen. 59-1445 45/123 carbons 45-369
hydrogens C3H9N. 2(3)2-910
The peak at 3300 cm1 tells us that the N is part
of an amine
2
2)   Either 2-butanone, 2-methyl-2-nitropropane,
3-pentanone, 1-nitropropane, nitroethane, or
2-bromopropane is responsible for the 1H NMR
spectrum shown. Draw the structure of the
responsible compound. (doublet and heptet)
3
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.
4
3)  Either 2-butanone, 2-methyl-2-nitropropane,
3-pentanone, 1-nitropropane, nitroethane, or
2-bromopropane is responsible for the 1H NMR
spectrum shown. Draw the structure of the
responsible compound. (quartet, singlet and a
triplet)
5
Because there are 3 signals it is either
2-butanone or 1-nitropropane.
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..
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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)
7
First 2(6)2-122/21 so either a ring or double
bond.
No peak shows up in the double bond region, CC
or CO.
So that leaves a ring.
Four peaks and with this structure we have 4
different types of carbon.
8
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.)
9
First, 2(3)2-5-30 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.
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6) What would the proton NMR peak look like for
the indicated hydrogen?
Because the two sets of adjacent protons are
equivalent this peak would follow the n1 rule
and be a septet.
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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
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9)      Using the MS and IR spectra attached (1A
and 1B) propose the formula and structure of this
compound. (106 and 108)  
13
MS shows a molecular ion peak at 106 and a M2
peak at 108.. So 106-3571 so 71/125 carbons so
71-6011 hydrogens so C5H11Cl 2(5)2-11-10
However, there is a carbonyl peak in the IR
14
So need to add an oxygen. -CH4 gives C4H7ClO so
2(4)2-7-12/21 So this is taken by the CO
bond.
One more thing, there is no peak at 2750 so no
aldehyde, our carbonyl is a ketone
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.
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10)  What is the major product of the following
reaction?
16
11)  What is the major product of the following
reaction?
Br2
CCl4
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12)  What is the major product of the following
reaction?
HBr
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 8) What alkene should be used to synthesize the
following alkyl bromide?
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9)  What alkene, when allowed to react with HBr,
would produce the following alkyl bromide? (There
is more than one correct answer.)
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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?
21
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.
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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.
23
13)
24
14)
25
15)