CHEM 345 Problem Set 18 Key

1.) Write the mechanism for the following reactions.

O a.) KCN EtOH O

OH racemic

1.) Write the mechanism for the following reactions. b.) KCN AcOH O NC

OH racemic

O c.) N S R O

NEt3 OH racemic

2.) What is the structure of AcOH?Why does changing the solvent from EtOH to AcOH make such a big difference? O

OH AcOH acetic acid

The pKa of acetic acid is approximately 5. The pKa of ethanol is approximately 15. When you take a proton off of ethanol, you generate ethoxide which is about 1010 times stronger of a base than acetate.

3.) Give two instances when you need to use the thiazolium salt and triethylamine rather than KCN and EtOH.

If the aldehydes contain an enolizable proton then you cannot use KCN/EtOH, instead you must use the thiazolium. Also, if the electrophile is a Michael acceptor to give a 1,4 dicarbonyl, then the thiazolium catalyst should be used.

4.) Break the following compound down as far as you can using Aldol, Michael, and Claisen reactions. Above each retrosynthetic arrow, write the name of the reaction.

O HO O HO Aldol

O

Michael

HO O O HO

O Aldol

O O

Aldol

HO O O HO Michael

O O O Aldol There are other possibilities for order.

O O HO

Aldol

O O

5.) Synthesize the following molecules. All carbons in the molecules must come from benzene or compounds with 5C’s or less. a.) O H2SO4 O HNO3 O2N

AlCl3

O O SOCl2

HO Cl

H2CrO4

1.) BuLi, Et2O + O 2.) H3O HO b.) O O

O

Cl

+ H3O

NaOEt, EtOH

O O

O O

Cl

O

AlCl3 Cl

Cl

AlCl3 Cl2

c.) O

OMe

NaOMe MeOH

O O

1.) POCl3, DMF 2.) H2O

OMe OMe O

AlCl3 MeI

Cl

ONa

HCl NaOH ZnHg

1.) NaOH + mcpba O 2.) H3O

O OH

O O AlCl3

Cl

6.) Write the mechanism for the following reactions. a.) S

N O + O R NEt O 3

O

b.) O O

cat. AcOH O + + H NH NH2

Ph

7.) The following molecule can be broken down two ways. Show both ways. Then, decide which will be the best way synthetically to make the target compound. Route A

O O Claisen O O 2 1 3 O 2 1 O 3 O

Route B

O O Claisen O O 2 3 1 O 2 1 O 3 O

Route A vs. Route B Route A is tempting as only one of the two reactants can be turned into a nucleophile as only one of them (the ketone) has an enolizable proton. Unfortunately, it has two different types of enolizable protons. They have similar acidities and similar sterics, so even LDA cannot differentiate them. As a result, you would get a mixture of nucleophiles.

O

O LDA

O Et2O

The starting materials for route B both have enolizable protons, so they both have the potential to be nucleophiles. Therefore, reversible conditions like sodium methoxide in methanol will not work as a complex mixture of products would be obtained. However, LDA can be used to designate the nucleophile. O O O O O O + LDA H3O O O Et O O 2 Et2O

8.) Breakdown the following molecules. One step only. Write the name of the reaction above the arrow. (Aldol, Michael, Mannich, Claisen, Umpolung)

a.) O O Umpolung 3 4 1 O 2 O b.) O Aldol O

1 3 O 2

c.) O OH Aldol O

3 1 O 2

d.) O O Michael O O

H2N H N 1 3 5 2 2 4

d.) O O Michael O O

H N H2N 2 3 5 1 4 2

e.) Mannich NH 2 3 OH H2N 1 OH 3 O O O

f.) O OH Umpolung

1 2

OH O OH O Chem 345 Reaction Sheets:

Stetter Reaction: S O O + N R NEt 3 O O Mechanism: S O + N R

NEt3 O

Breakdown: O O 3 4 1 2 H O O

Summary (Key words): Product is a 1,4 dicarbonyl. All steps are reversible. You must use a thiazolium catalyst. Cyanide will not work. Michael acceptors are the electrophiles in this umpolung reaction. Breakdown: Break 1,2 bond. Carbon 1 gets a H. A new pi bond is placed between carbons 2 and 3. Be careful that a five bonded carbon is not formed when drawing in the pi bond. Chem 345 Reaction Sheets:

Benzoin Condensation:

Cl Cl O KCN OH Cl EtOH O

Mechanism: Cl O KCN

EtOH

Breakdown: O O OH

1 2

O

Summary (Key words): Product is a 2-hydroxy carbonyl. All steps are reversible. The R groups attached to the 2-hydroxy carbonyl must be the same, otherwise dithiane chemistry must be used. The R groups must not have an enolizable proton. In those cases, the thiazolium must be used instead. Breakdown: Break 1,2 bond. Carbon 1 gets a hydrogen. Carbon 2 becomes a carbonyl. Chem 345 Reaction Sheets: Thiazolium Benzoin Condensation: S O

N O R NEt 3 OH

Mechanism: S

N O R

NEt3

Breakdown: O O OH

1 2

O

Summary (Key words): Product is a 2-hydroxy carbonyl. All steps are reversible. The R groups attached to the 2-hydroxy carbonyl must be the same, otherwise dithiane chemistry must be used. Breakdown: Break 1,2 bond. Carbon 1 gets a hydrogen. Carbon 2 becomes a carbonyl.