Chapter 17 Problem Set Answers April 30, 2013 17.37 How would you prepare the following ketones using an acetoacetic ester synthesis? In each case, think about where the α carbon is. That is the site that was the key enolate in the original reaction. Add an ester back to that position, then disconnect bonds between the α and β carbons to get to the alkyl electrophiles of interest. 17.38 How would you prepare the following compounds using either an acetoacetic ester synthesis or a malonic ester synthesis? Here also, you have to reinsert the other carbonyl group (for compounds b‐d). After that, you can disconnect the bond between the α and β carbons to get to the desired starting materials. 17.39 How might you synthesize each of the following compounds using an aldol reaction? In each case, show the structure of the starting aldehyde(s) or ketone(s) you would use. An α,β‐unsaturated carbonyl can be made from an aldol reaction. The first step is to break that double bond and put a hydroxyl group on the beta position. Then disconnect the bond between the alpha and beta carbon to get an enolate nucleophile – that is the alpha carbon, and a carbonyl electrophile – that is the beta carbon which now has a hydroxyl group on it. Replace the hydroxyl group with a carbonyl group retrosynthetically. 17.41 Base treatment of the following α,β‐unsaturated carbonyl compounds yields an anion by removal of H+ from the gamma carbon. Why are hydrogens on the gamma carbon atom acidic? That anion is in resonance with the α,β‐unsaturated carbonyl system – acts like an extended enolate: 17.43 When optically active (R)‐2‐methylcyclohexanone is treated with either aqueous base or acid, racemization occurs. Explain. Acid generates a planar enol species, and base generates a planar enolate species. Either way, the alpha carbon becomes sp2 hybridized during the course of that reaction, and therefore loses any stereochemical information. 17.45 Intramolecular aldol cyclization of heptane‐2,5‐dione with aqueous NaOH yields a mixture of two enone products in the approximate ratio 9:1. Write their structures and show how each is formed. You have 4 different alpha carbons in this molecule. However, deprotonating on carbons 3 or 4 (between the two ketones) is going to give you cyclized products with highly strained ring systems. The two productive enolates are the ones that form on carbon 1 and on carbon 6. Enolate on carbon 1: O-
O
base
H3C
OH
O-
elimination
H2C
O
O
O
O
O
Enolate on carbon 6: O
O
H H
base
H3C
H3C
O-
O
H3C
H
H3C
O- CH3
H3C
OH CH3
O
O
CH3
O
The major product is the enolate formed from carbon 6, because the double bond is more substituted. 17.46 The major product formed by intramolecular aldol cyclization of heptane‐2,5‐dione has two singlet absorptions in the 1H NMR spectrum at 1.65 and 1.90 ppm and has no absorptions in the range 3 to 10 ppm. What is its structure? The one from the carbon 6 enolate, because it is more substituted. The singlets at 1.65 and 1.90 ppm are from the two methyl groups. 17.49 Butan‐1‐ol is prepared commercially by a route that begins with an aldol reaction. What ssteps are likely to be involved? To get an aldol reaction product, you need to start with an αβ‐unsaturated carbonyl: 17.52 Fill in the reagents a‐c that are missing from the following scheme: O
CO2CH3
O
O
O
CO2CH3
CH3
a
b
CH3
H3C
c
CH3
17.56 Cinnamaldehyde, the aromatic constituent of cinnamon oil, can be synthesized by a mixed aldol condensation between two different carbonyl compounds. Show the starting materials you would use, and write the reaction.