Some Answers to Hour Examination #1, Chemistry 302/302A, 2004

1. In this variation of the Diels-Alder reaction, both the diene and the dienophile are cyclic compounds. The first complication is to decide whether the reaction will pass over an endo or exo transition state. In fact, the cycloadduct has the endo stereochemistry, and this reaction is normal in that respect

+

O

O

H

H

O

O

C11H10O2

note:endo

You may have noticed that the product of this Diels-Alder reaction still has a double bond with two attached carbonyl groups. This carbon-carbon double bond is also a quite reactive dienophile. When this reaction is run in the presence of two equivalents of 1,3-cyclopentadiene, a 2:1 adduct is readily formed:

H

H

O

O

H

H

O

O

H H

C16H16O2

double endo

2. Adamantene is unstable because the π bond is a very poor one.The orbitals do not overlap at all well.

very poor overlap

=

reverse

Diels-Alder

1 2

The reaction succeeds, but the unstable adamantene undergoes a reverse Diels-Alder reaction. Remember: all cyclohexenes are related to a diene and a dienophile through the Diels-Alder reaction.

3. Solvolysis gives pentadienyl cation I that closes to a cyclic ion. Addition of water, followed by deprotonation gives 4 .

OTs

H3C CH 3

HH

H H

H2O

H3C CH3

HH

H HCH 3

H3C

H2O

I

H3O+ H3O+

++

3

4

Intermediate I is a pentadienyl cation, a four-electron system. Analysis of the HOMO shows that closure must take place in conrotatory fashion to give the compound with trans methyl groups. If you can’t figure out the MOs, you do know that all thermal 4n electrocyclic reactions are conrotatory.

HOMO = psi 2, top view + + 0 – –

H3CCH 3

CH 3

H3C

H2O

H3O+

+

4H

H

4. Compound A is formed through a straightforward Friedel-Crafts alkylation reaction.

CH3

CH3

(CH3)2CH Cl AlCl3CH3

CH3

CH(CH3)2

(+)

(+) H

AlCl4

CH3

CH3

CH(CH3)2

+ HClA

+ AlCl3

What can HCl do? The hint tells you to focus on this product. If the product A is protonated, we can generate an intermediate in which an alkyl shift moves the relatively large alkyl groups apart. That’s the key to why this rearrangement happens - the relief of steric strain when the thermodynamically more stable 1,3,5-trisubstituted benzene is formed from a less stable 1,2,4-trialkylbenzene.

(+)

CH3

CH3

CH(CH3)2

HCl

H

CH3

CH(CH3)2

H3C

(+)

H

CH3

CH(CH3)2H3C

H

(+)(+)

CH3

CH(CH3)2H3C

B

5a. Straightforward. Friedel-Crafts acylation, sulfonation, reduction.

CH3CH2CH2

O

Cl

AlCl3O

SO3

H2SO4O

SO2OHZn/HCl

SO2OH

CH3CH2CH2

O

ClCH3CH2CH2CH2OH CH3CH2CH2

O

OHCrO3

H2O

SOCl2

5b. Harder: Indirection must be used because Friedel-Crafts alkylation will not be effective on the strongly deactivated nitrobenzene. Of course, the tert-butyl group is an ortho/para director and will not nitrate in the meta position. So, we use the strongly activating amino group to force nitration into the position adjacent to it, and then remove the amine. In the answer, we acylate the amine to avoid substitution in the meta position directed by the protonated amine, but we will let you get away (partially at least) with an answer without the protection-deprotection sequence.

C(CH 3)3C(CH 3)3

NH2

C(CH 3)3

NHAc

C(CH 3)3

NHAc

NO2

C(CH 3)3

NH2

NO2

C(CH 3)3

NO2

(CH 3)3COH

CH 3COOH

SOCl 2

AlCl3

(CH 3)3CCl

SOCl 2

H2SO4

AcCl

AcCl

HNO3

H2SO4

1. HNO3

2. H2/Pd/C(protection)

KOH/H2O

(deprotection)

1. HONO/HCl

2. H3PO2

Alternatively,NH2 NHAc

NHAc

C(CH 3)3

AcCl

AlCl 3

(CH3)3C-Cl

1. HNO3/H2SO4

2. H2/Pd/C pyridine

as above

6. Compound 3 is aromatic (4n + 2, n = 4), whereas compound 4 is not (4n, n = 4). Accordingly, the 12 outside hydrogens in 3 appear far downfield of the normal alkene position [ Bnet = Bo + Bi], and the six inside hydrogens are shifted far upfield [Bnet =

Bo – Bi]. No such effect is apparent in 4 , of course.

HinHout

B induced

B induced

aromatic ring

Bapplied

7. Here is a set of structures:

O O

HOOC

O

O

Cl Cl

Cl Cl Cl

Cl

A

D

B

Cl

O

C

Friedel-Craf ts withthe acyl chlo ride

Photochlo rinate allbenzyl positions

Double E2

ozono lysis/oxida tionof the double bond

acid ch loride formationand another Friedel-Craf ts

Double Wo lff-Kishnerreduces C=O groups

E

F =