Organic Chemistry II
CHEM 2325
Packet #6
1.
Draw the arenium ion intermediates when either toluene or α,α,α-trifluorotoluene are
reacted with Cl2 and AlCl3 at both the para and meta positions.
CH3
CH3
Cl2
AlCl3
H
H
Cl
H
CH3
Cl
CH3
Cl
Cl2
AlCl3
H
H
CF3
H
Cl
Cl
CH3
H
Cl
CF3
H
Cl
CF3
H
Cl
CF3
H
CH3
H
Cl
Cl
CF3
CH3
Cl
CF3
H
CF3
Cl
Which resonance structure is the most stable?
When the positive charge is placed adjacent to the electron donating methyl group, the most
stable resonance form is obtained. This is why toluene is an ortho/para director.
Which resonance structure is the least stable?
When the positive charge is placed adjacent to the electron withdrawing trifluoromethyl group,
the least stable resonance form is obtained. This is why this compound is a meta director.
Predict the rate of reaction for all four scenarios (toluene at para, toluene at meta,
trifluorotoluene at para and trifluorotoluene at meta) relative to benzene.
Toluene at para is the largest, greater than 1 – toluene at meta is still greater than 1, but less than
the para position – trifluorotoluene at para is less than 1 – trifluorotoluene at meta is also less
than 1, but greater than the para which has the slowest rate.
List in order the relative rate versus benzene for the four scenarios.
Toluene (para) > toluene (meta) > trifluorotoluene (meta) > trifluorotoluene (para)
2.
Consider the following reaction scheme.
1) Mg
O
Br
Br2, FeBr3
O
2) H
NH2NH2
KOH
3) CrO3
Indicate reagents to cause each interconversion.
Draw the product obtained if each of the four structures shown are reacted with HNO3 and
H2SO4.
Br
NO2
HNO3
H2SO4
O
HNO3
O2N
O 2N
HNO3
H2SO4
deactivator
meta
H2SO4
deactivator
o/p
O
Br
HNO3
H2SO4
activator
o/p
O 2N
Which structure would react the fastest with HNO3/H2SO4?
Only the compound with a n-propyl substituent is an activator, thus it reacts the fastest.
Indicate a method to synthesize the compound shown from benzene.
H2N
Consider aspects of the target molecule: 1) both substituents are o/p directors, therefore they
must have been a meta director at some point of synthesis, 2) an n-propyl substituent cannot be
added in a Friedel-Crafts alkylation because rearrangements would occur, therefore probably
attach carbon framework with a Freidel-Crafts acylation, 3) Friedel-Crafts acylations cannot
occur on deactivated rings (outside of halogens), 4) amines typically are added to rings by
reducing nitro groups. Considering these aspects a potential synthesis is as follows.
O
O
O
HNO3
Cl
AlCl3
H2SO4
O 2N
Zn(Hg)
HCl !
H 2N
3.
Consider the following reaction.
H3CO
H3CO
AlCl3
O
Cl
Draw the product obtained when 1 equivalent of AlCl3 is used.
H3CO
H3CO
AlCl3
O
Cl
H3CO
H3CO
H3CO
O
Cl
AlCl3
H3CO
O
With one equivalent, the AlCl3 binds to the carbonyl activating the acid chloride for a FriedelCrafts acylation. Reaction then occurs on the more activated methoxy substituted aromatic ring.
When 3 equivalents of AlCl3 are used, however, a different product is obtained. Draw this
product.
H3CO
H3CO
AlCl3
O
Cl
AlCl3
H3CO
H3CO
AlCl3
H3CO
O
Cl
AlCl3
H3CO
O
With 3 equivalents of AlCl3 the extra Lewis acid will bind to the methoxy oxygen groups. This
binding causes the oxygens to have a partial positive charge, thus deactivating the ring for
aromatic substitution. The Friedel-Crafts reaction will thus react with the benzene ring without
the methoxy substituents as it is now a more activated ring than the other.
Explain why two different products are obtained with different amounts of AlCl3.
See explanations above.