1) Molecular ion peak of 122 and 124 (3:1)
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
You have a carbonyl peak, so it’s the last formula, C4H7O2Cl
And the peaks at 1000 and 1200 show a C-O so ester
1.310 triplet 3H
4.252 quartet
2H
4.062 singlet 2H
The singlet shows that you have 1
carbon attached to the ester and
the 2H shows this is where the Cl
is
And the triplet and the quartet represent the ethyl
group attached to the carbonyl.
The carbon between 160-180 is your carbonyl, so 3 other types of
carbon.
O
Cl
O
2) Molecular ion peak at 150 and 152 (1:1)
150 – 79 = 71/12 = 5 carbons
71 – 60 = 11 hydrogens C5H11Br
2(5) + 2 – 11 – 1 = 0
Nothing but C-H stretches so C5H11Br
3.417 triplet 2H
0.92
doublet 6H
1.86 sextet 2H
1.64
multiplet 1H
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.
R
So that is 3 carbons and 7 hydrogens, leaving 2
carbons, 4 hydrogens and a bromine.
3.417 triplet 2H
0.92
doublet 6H
1.86 sextet 2H
1.64
multiplet 1H
R
So the sextet must be the carbon adjacent to the
isopropyl group, because (1+1)(2+1)= 6
R
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.
Br
Br
The carbon matches this structure b/c it tell syou there
are 4 types of carbon.
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
Carbonyl peak tells you tha tit is C3H7NO also there
are no N-H stretch so it’s a tertiary amide.
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.
O
So we have the following structure thus
far:
R
H
8.019 singlet 1H
R
2.970 singlet 3H
2.883 singlet 3H
So two more singlets each representing 3 H so
methyl groups.
O
H
N
N
4) Name the following compound:
O
Cl
2,2’-dimethylpropanoyl chloride
5) Name the following compound:
N
O
N,N-dibenzylformamide
H
6) Name the following compound:
O
O
Isopropyl formate
H
Isopropyl methanoate
7) Name the following compound:
O
Benzoic propanoic anhydride
O
O
8) Draw 3-methylhexanoyl chlordie.
O
Cl
9) Draw isopropyl 3-ethyl-4-methylpentanoate
O
O
10) Draw N-tert-butyl-N-ethyl-2-isopropy-3,4dimethylpentanamide
O
N
O
O
Cl
Cl2
H2O, HCl
H2NC(CH3)3
I2 excess
O
NaOH
NHC(CH3)3
O
O
O
CHI3
H3 O+
OH
O
O
LDA, THF
CH3CH2CH2Br
LiAlH4
Br2
O
H2O
CH3COOH
Br
LiBr
DMF
Li2CO3
O
OH
O
O
O
Br
NaOCH3
Br2
CH3COOH
LDA
THF
CH2CHCH2BR
O
O
LDA
THF
CH3Br
OH
LiAl4
H2O
OCH3
O
CH3OH
O
O
H+
HO
DIBAL-H
H2O
CH3Li
OH
PCC
O
H2 O
O
CH3CH2Br
O
O
Li
CuI
O
Br
Li2CO3
LiBr
DMF
(CH3CH2)2CuLi
H2 O
O
O
LDA, THF
Br
BrCH2CH2CH2CH2Br
NaOCH2CH3
O
O
O
O
LDA, THF
NaOCH2CH3
CH3CH2Br
CH3CH2Br
Br2
CH3COOH
O
O
Li2CO3
LiBr
DMF
Br
Br
Br
Br2
CH3COCl
O
FeBr3
AlCl3
OH
HO
O
O
Mg
O
TsOH
MgBr
O
Br
O
O
O
MgBr
O
O
H 2O
H2 O
O
OH
O
PCC
O
O
H3O+
O