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Solutions for Condensations and Alpha Substitutions of Carbonyl Compounds
71 continued
EtO
O O
EtO
O O
O
O
EtOOC
NaOEt
O
OO
EtO
OO
EtO
O
H3O+
H3O+
O
O
O NaOH O
+ CO2 + EtOH
∆2) Br(CH2)4Br
1) 2 NaOEt(b)
make by aldol
cyclization/dehydration
make by conjugate
addition
The single bond to this substituted acetone can be made by the acetoacetic ester synthesis. How can we 
make the α,β double bond? Aldol condensation!
substituted acetone
∆
+
(c) The acetoacetic ester synthesis makes substituted acetone, so where is the acetone in this product?
Ph C
O
H
CHPh
OH
PhCH2CH2
C Ph
O
H3C
Ph
O
O
OCH3
O
NaOH
CHPhCH C Ph
O
CH C
O
PhPhCH
O
C HPh
OCH3
O
O
Ph
CH3 C
O
Ph
PhCH2CH2 CHPh
OH
forward synthesis
reverse
aldol
+
H2, Pt
(a) Where is the possible α,β-unsaturated carbonyl in this skeleton?
72 These compounds are made by aldol condensations followed by other reactions. The key is to find the 
skeleton made by the aldol.
AHA! The aldol product reveals itself. (See the 
solution to 71 (c).)
+
β-ketoester
(b) The aldol skeleton is not immediately apparent in this formidable product. What can we see from it? 
Most obvious is the β-dicarbonyl (β-ketoester), which we know to be a good nucleophile, capable of 
substitution or Michael addition. In this case, Michael addition is most likely as the site of attack is β to 
another carbonyl.
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+
O
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