O
O
O
O
NaOCH2CH3
H3CH2CO
OCH2CH3
H3CH2CO
OCH2CH3
CH3CH2CH2CH2CH2Br
O
HO
O
O
OH
H3CH2CO
H3O+
O
OCH2CH3
O
HO
O
O
HO
OH
heat
O
O
NaOCH2CH3
O
O
OCH2CH3
OCH2CH3
(CH3)2CHCH2CH2CH2Br
O
O
O
O
OH
H3O+
OCH2CH3
O
O
O
OH
heat
O
O
NaOCH2CH3
Ph
Ph
Ph
Ph
O
Ph
Ph
CH3CH2OH
Ph
Ph
O
O
O
O
O
O
O
NaOH
O
O
O
O
O
H2O
O
O
O
O
O
O
+ OH
O
O
O
O
O
O
H2O
O
O
O
O
OH
+ H2O
+ OH
1) Which of the following alkanes will give more than one
monochlorination product when treated with chlorine and light?
a) 2,2’-dimethylpropane, b) cyclopropnae, c) ethane, d) 2,3dimethylbutane
Cl
Cl2
hv
Cl2
hv
Cl
H3C
CH3
Cl2
H2
C
H3C
hv
Cl
Cl
Cl2
hv
Cl
2) Which type of halogenation is most selective?
Bromination is most selective, always occurring
at the site of the most stable radical.
3) Which of the following absorbs at the highest frequency?
a) 1,3,5-hexatriene, b) 1,3,5,7-octatetraene, c) 1,7-diphenyl-1,3,5heptatiene, d) 1,6-diphenyl-1,3,5-heptatriene
4) Draw 3,5-difluoroanisole.
F
O
F
1 5) Circle each of the following compounds that is aromatic.
4n + 2 = 10
pi electrons
4n + 2 = 2
pi electrons
1) 6) Which of the following compounds will undergo FredalCrafts alkylation?
a)
a) benzoic acid, b) nitrobenzene, c) aniline, d) toluene
CH3
7) Which of the following compounds is most acidic?
O
O
OH
OH
O2N
H2N
O
O
OH
OH
O
OHO
O2N
8) Which of the following gives a secondary alcohol when treated
with methyl grignard?
a) butyl formate, b) 3-pentanone, c)pentanal, d) methyl butanoate
O
O
CH3MgBr
O
H
H2O
OH
O
CH3MgBr
H2O
OH
O
CH3MgBr
H
OH
H
H2O
O
O
CH3MgBr
O
H2O
OH
9) Draw N-ethyl-N-propylformamide.
O
N
H
10) What does a positive iodoform test tell you is
present
It indicates the presence of a methyl ketone.
11)
O
O
O
Cl
O
O
NaOH
O
O
H2 O
O
O
O
12)
O
O
O
H3O+
O
H2O
O
O
HO
OH
13)
O
O
NaBH4, MeOH
O
OH H+, H O, heat
2
O
14)
O
O
I2
NaOH
+ CHI3
O
15)
O
O
Br
Br2
aq. NaOH
16)
Br
Br
O
O
HCl
H2O
17)
NaOH
heat
O
O
O
18)
O
O
O
NaOCH2CH3
+
O
O
H+, H2O, heat
19)
O
O
O
NaOCH2CH3
OCH3
H3O+
O
20)
O
O
O
O
OCH3
NaOCH2CH3
+
H3CO
OCH3
CH3CH2OH
O
H3CO
O
O
O
O
H3O+
heat
O
NaOCH2CH3
OCH2CH3
O
OCH2CH3
CHCCH2Br
HOCH2CH2OH
H2
Pd-C
O
O
H3O+
O
O
O
O
NaOH
CH3I
O
TsOH
O
O
O
Br
Br2
Li2CO3
CH3COOH
LiBr
DMF
NaOCH2CH3
O
H2O
O
OCH2CH3
O
O
O
O
H3O+
heat
H3CH2CO
O
O
O
O
O
1) NaOEt, EtOH
+
O
OCH2CH3
2) H+, H2O, heat
O
O
O
O
OCH2CH3
H
OCH2CH3
OH
O
O
O
O
O
OCH2CH3
OCH2CH3
O
O
H
H
O
H
OH2
O HO
OH
O
O
OCH2CH3
OCH2CH3
O
O
H
O HO
O HO
H2O
OH
H2O
OCH2CH3
OCH2CH3
O
O
O HO
+
H3O
OH
OCH2CH3
O HO
OH
H
O
CH2CH3
O
O
O
O HO
O
OH
O
H2O
OH
O
O HO
O
O
O
O
Molecular ion peak is a single peak at 164, so no halogens or
nitrogen.
164/12 = 13 x 12 =156
164 – 156 = 8 hydrogens
C13H8
C12H20
C11H16O
C10H12O2
2(13) + 2 – 8 = 20/2 = 10
2(12) + 2 – 20 = 6/2 = 3
2(11) + 2 – 16 = 8/2 = 4
2(10) + 2 – 12 = 10/2 = 5
Looking at this you have sp2 and sp3 hydrogens as well at a
carbonyl.
You have 5 aromatic protons, so most likely a benzene
ring is present, so that takes 4 of your five degrees of
unsaturation. And the final degree is taken by the
carbonyl.
8 different types of carbon, 4 of those are taken up by a
monosubstituted benzene ring. You know this because of the
4 peaks in the 120 to 140 range. And the fact that there are 5
aromati protons.
So what is the substituent . We have 4 carbons
and 2 oxygens left, plus 7 hydrogens.
You know there is a carbonyl so that is one carbon plus one
oxygen. What is the other oxygen?
So it could either be an ether, an alcohol, or part of an ester.
We know it isn’t an alcohol b/c there is no large peak around
3400 in the IR.
So ester or ether?
This is not an ether b/c at least one of the remaining 3 carbons
would be a singlet somewhere up field and that isn’t present.
So it must be an ester.
O
O
O
O
O
O
O
O
O
O
O
O
So here are all the possible esters, which one is it?
For the nonaromatic protons we see a triplet a
multiplet and a quartet, The multiplet tells you that
the three carbons are in a group together.
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O