CHCH3

CH3

H3C Cl

heat (∆)

(MCPBA)

ClO

O

OH

OH SOCl2

OH

AlCl3

1) CH3MgBr

2) H3O+

CrO3, H2SO4

H2O, Acetone

1) NaBH4

2) H3O+

excess KMnO4

OH

OH

NBS

peroxides:(PhCO2)2

CH3CH2O-K+

CH3CH2OH

O

Cl

3) PBr3

CH3H3C

NaOH

EtOH

O

OH

O OH

OH

O

H3C

(+ some o- isomer)

CO

OH

C

O

HO

Br

OOH

OCH2CH3

Cl

OH

CH3H3C

O

AlCl3

Br CH3

How many ways can you come up with to make the following molecules? (For practice, think of some other examples on your own and see how many ways you can come upwith to make them.)

OH

OH

BrCH3

OH

O

H

O

OMgBr

NaOHNBS, peroxides

KMnO4LiAlH4

NaBH4

HX

O

H H

Here are a few possibilities. Next work on the precursors.

O

HO

MgBr

O1)

2) H3O+

H3C H

O

POCl3pyridine

1) Hg(OAc)2 H2O NaBH4

H3C Cl

O

AlCl3

Again, these are some of the possibilities. Figuring out schemes like these for different target molecules with help you when you are faced with a more complex molecule.

2) NaBH4

Explain the following reactions with detailed mechanisms.

NH2

HNO3

H2SO4

NH2

NO2

only product, slow reaction, low yield

2

H2SO4

NH2

H+

NH3 Direct attachment of positive charge to ring gives inductive electron-withdrawing effect, and thereby 1) deactivates the ring and 2) gives the meta-substituted product.Draw the resonance structures of the possible intermediates to check this out.When chemists need to nitrate aminobenzenes, they usually use other non-acidic nitrating reagents like CH3CH2ONO2 (but you don’t need to know this) or they first convert the amine to an amide by reaction with an acid chloride or an anhydride (you don’t need to now this yet).

H+

H

HSO4-

H3C

=

Two electrophilic addition steps followed by one electrophilic aromatic substitution step.Redrawing the intermediate helpsyou see where the last substitution occurs.

ClO

NO2

Prepare the following compounds from benzene, inorganic reagents, and organic compounds with 4 or fewercarbons.

OCH3

H O

OCH3

H OHH

OCH3

H BrH

OCH3

MgBr

OCH3

CH3

OCH3

Br

OCH3 OH

OH

CH3

SO3H

SO3H

CH3 CH3

a) PCC; b) PBr3; c) NBS,(PhCO2)2; d) NaH, MeBr; e) NaOH, heat; f) H2SO4/SO3; g) CH3Cl, AlCl3; h) H2C=O; i) Mg, Ether; j) Br2, FeBr3

a)b) c) d) e) f)

g)h)

i) j) d) e) f)

Cl

NO2

Cl

NO2

OH

Cl

NO2

Br

Cl

NO2

O

Cl

NO2

Any way you go about it, it looks like seven steps. If you make the ether at an early stage of the procedure,you avoid problems with using NaH in the presence of the aldehyde or the alkyl bromide.Addition of formaldehyde to a grignard reagent is a nice way to make a primary alcohol with an extra carbon. Remember that the ArSO3H to ArOH conversion can only tolerate alkyl substituents on the aryl ring

g)g)

Cl

Cl

O O

Cl

NO2

MgBr BrCl

NO2

a)b)

c)

d)

e)

f)

g)

h)

a) MCPBA

b) NaOH/EtOH (E2)

c) NBS/peroxides

d) HNO3/H2SO4

e) H2/Pd

f) Cl2, FeCl3g) (CH3)2CHCOCl,

AlCl3h) POCl3, pyridine

i) (CH3)2CHCHO

(then H3O+)

j) Mg, Ether

k) O3(you don’t

have to know this

one)

l) Br2/FeBr3

m) Cl2/FeCl3n) H2/Pd

o) PCC

i)

j) BrCl

NH2

Cl

NH2

NH2NO2

k) l)

m)

n)

d)

Figure out the epoxide group and the side chain first (i.e., as the first precursor to the product). Remember that epoxides open easily under acidic conditions and that most of the electrophilic aromatic subst. steps we know use strong Lewis acids. Note that the branching on the side chain rules out simple Friedel-Crafts alkylation. The top route takes seven steps and looks pretty good. The bottom route is longer, and steps l) and m) will probably give mixtures. Step j) might also give problems with competing formation of the ArMgCl Grignard. Step o) probably takes us to a dead end.

o)