1 Molecular ion peak of 122 and 124 3:1

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1) Molecular ion peak of 122 and 124 (31)
122-35 87/12 7 carbons
87-84 3 hydrogens C7H3Cl
(2(7) 2 3 1)/2 6 DOUS
C6H15Cl 2(6) 2 15 1 -2/2 -1
C5H11OCl (2(5) 2 11 1)/2 0
C4H7O2Cl (2(4) 2 7 1)/2 1
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You have a carbonyl peak, so its the last
formula, C4H7O2Cl And the peaks at 1000 and 1200
show a C-O so ester
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The singlet shows that you have 1 carbon attached
to the ester and the 2H shows this is where the
Cl is
1.310 triplet 3H 4.252 quartet 2H 4.062
singlet 2H
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And the triplet and the quartet represent the
ethyl group attached to the carbonyl.
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The carbon between 160-180 is your carbonyl, so 3
other types of carbon.
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2) Molecular ion peak at 150 and 152 (11)
150 79 71/12 5 carbons
71 60 11 hydrogens C5H11Br
2(5) 2 11 1 0
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Nothing but C-H stretches so C5H11Br
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3.417 triplet 2H 0.92 doublet
6H 1.86 sextet 2H 1.64 multiplet
1H
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3.417 triplet 2H 0.92 doublet
6H 1.86 sextet 2H 1.64 multiplet
1H
With 0 degrees of unsaturation we know this is a
chain, so now what does it look like?
We have a multiplet with an integration of 1H and
a doublet with an integration of 6 so that tells
us its an isopropyl group.
So that is 3 carbons and 7 hydrogens, leaving 2
carbons, 4 hydrogens and a bromine.
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3.417 triplet 2H 0.92 doublet
6H 1.86 sextet 2H 1.64 multiplet
1H
So the sextet must be the carbon adjacent to the
isopropyl group, because (11)(21) 6
So that leaves the triplet that integrates to 2
hydrogens, the triplet tells you it is next to a
CH2 group and the 2 H integration tells you the
bromine is attached b/c it should have 3 H
otherwise.
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The carbon matches this structure b/c it tell
syou there are 4 types of carbon.
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3) Molecular ion peak of 73.
73 14 59/12 4 carbons
59 48 11 hydrogens so C4H11N
2(4) 2 11 1 0
C3H7NO 2(3) 2 7 1 2/2 1
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Carbonyl peak tells you tha tit is C3H7NO also
there are no N-H stretch so its a tertiary amide.
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8.019 singlet 1H 2.970 singlet 3H 2.883
singlet 3H
Singlet at 8.019 tells you it is attached to the
carbonyl carbon.
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So we have the following structure thus far
8.019 singlet 1H 2.970 singlet 3H 2.883
singlet 3H
So two more singlets each representing 3 H so
methyl groups.
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4) Name the following compound
2,2-dimethylpropanoyl chloride
5) Name the following compound
N,N-dibenzylformamide
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6) Name the following compound
Isopropyl formate Isopropyl methanoate
7) Name the following compound
Benzoic propanoic anhydride
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8) Draw 3-methylhexanoyl chlordie.
9) Draw isopropyl 3-ethyl-4-methylpentanoate
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10) Draw N-tert-butyl-N-ethyl-2-isopropy-3,4-dime
thylpentanamide
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