O

O

OCH

2

CH

3

H

3

CH

2

CO

NaOCH

2

CH

3

O

O

OCH

2

CH

3

H

3

CH

2

CO

CH

3

CH

2

CH

2

CH

2

CH

2

Br

O

O

OCH

2

CH

3

H

3

CH

2

CO

H

3

O

+

O

O

OH

HO

O

O

OH

HO

heat

O

HO

O

O

OCH

2

CH

3

NaOCH

2

CH

3

O

O

OCH

2

CH

3

(CH

3

)

2

CHCH

2

CH

2

CH

2

Br

O

O

OCH

2

CH

3

H

3

O

+

O

O

OH

O

O

OH

heat

O

Ph

O

Ph

NaOCH

2

CH

3

Ph

O

Ph

O

Ph

O

Ph

O

CH

3

CH

2

OH

Ph

O

Ph

O

O

O

NaOH

O

O

O

O

O

O

H

2

O

O

O

O

+ OH

O

O

O

O

O

O

O

O

O

H

2

O

O

O

OH

+ OH

O

O

+ H

2

O

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) 2,3-dimethylbutane

 

 

Cl

2

hv

Cl

2

hv

CH

3

H

3

C

Cl

2

hv

Cl

2

hv

Cl

Cl

H

2

C

H

3

C

Cl

Cl

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-1,3,5-heptatiene, d) 1,6-diphenyl-1,3,5-
heptatriene

4) Draw 3,5-difluoroanisole.

F

F

O

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 Fredal-Crafts alkylation?

a)     a) benzoic acid, b) nitrobenzene, c) aniline, d)
toluene

CH

3

7) Which of the following compounds is most
acidic?

O

OH

H

2

N

O

OH

O

2

N

O

OH

O

OH

O

OHO

O

2

N

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

H

CH

3

MgBr

H

2

O

O

CH

3

MgBr

H

2

O

O

H

CH

3

MgBr

H

2

O

O

H

OH

OH

OH

O

O

CH

3

MgBr

H

2

O

OH

O

9) Draw N-ethyl-N-propylformamide.

N

O

H

10) What does a positive iodoform test
tell you is present

It indicates the presence of a methyl
ketone.

O

O

Cl

O

O

NaOH

H

2

O

11)

O

O

O

O

O

O

12
)

O

O

O

O

H

3

O

+

H

2

O

HO

O

OH

O

13)

14)

O

OH

O

NaBH

4

, MeOH

H

+

, H

2

O, heat

O

I

2

NaOH

O

O

O

O

+ CHI

3

15
)

16)

O

Br

2

aq. NaOH

O

Br

Br

Br

O

HCl

H

2

O

O

17)

18)

O

O

NaOH

heat

O

O

O

+

O

O

NaOCH

2

CH

3

H

+

, H

2

O, heat

O

19
)

20
)

OCH

3

O

O

NaOCH

2

CH

3

H

3

O

+

O

O

O

+

O

H

3

CO

O

OCH

3

NaOCH

2

CH

3

CH

3

CH

2

OH

O

O

OCH

3

O

H

3

CO

O

O

OCH

2

CH

3

NaOCH

2

CH

3

CHCCH

2

Br

O

O

OCH

2

CH

3

O

H

3

O

+

heat

HOCH

2

CH

2

OH

TsOH

O

O

NaOH

CH

3

I

O

O

O

O

H

2

Pd-C

H

3

O

+

O

O

Br

2

CH

3

COOH

O

Li

2

CO

3

LiBr

DMF

O

Br

NaOCH

2

CH

3

O

O

OCH

2

CH

3

H

2

O

O

O

O

H

3

CH

2

CO

O

O

H

3

O

+

heat

O

O

OCH

2

CH

3

+

O

O

O

1) NaOEt, EtOH

2) H

+

, H

2

O, heat

O

O

OCH

2

CH

3

H

OH

O

O

OCH

2

CH

3

O

O

O

OCH

2

CH

3

O

O

H

H

O

O

OCH

2

CH

3

O

O

O

OCH

2

CH

3

O

H

OH

2

O HO

OCH

2

CH

3

O

H

2

O

O HO

OCH

2

CH

3

O

OH

H

H

2

O

O HO

OCH

2

CH

3

O

OH

O HO

OCH

2

CH

3

O

OH

H

3

O

+

O HO

O

CH

2

CH

3

O

OH

H

O HO

O

OH

H

2

O

O

O

O

OH

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

C

13

H

8

2(13) + 2 – 8 = 20/2
= 10

C

12

H

20

2(12) + 2 – 20 = 6/2
= 3

C

11

H

16

O

2(11) + 2 – 16 = 8/2
= 4

C

10

H

12

O

2

2(10) + 2 – 12 = 10/2
= 5

Looking at this you have sp

2

and sp

3

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