Synthesis of Organolithium reagents II

Question 1

Lithiation α to a oxygen heteroatom can also occur at other C(sp²)-H bonds like alkenes
How does this reaction occur

Answer 1

Similarly we have to use a directing group to do so

Question 2

We can replace copper with lithium to make it a softer nucleophile and react it with the following electrophile to form

Answer 2

Question 3

The species formed from the copper nucleophile continues to react with the with the acidic workup
What is formed as the final product?

Answer 3

• MeO is caused to be a leaving group
• Allows for the reformation of the ketone

Question 4

We can also deprotonate an sp³ C-H which is alpha to oxygen if there is a strong electron-withdrawing group on the oxygen
What is the purpose of the highlighted Me groups

Answer 4

The Me groups stop the attack at C=O

Question 5

What is the benefit of the EWG group in this reaction?

Answer 5

• Coordinated to the base bringing it closer to the proton that we want to deprotonate
• Stabilising the negative charge that we form when we deprotonate the proton
• (this allows us to deprotonate an sp³ carbon)

Question 6

What is the mechanism for this deprotonation at a sp³ centre

Answer 6

• Last step involves hydroloysis using a hydroxide base

Question 7

We can also deprotonate sp³ C-H alpha to nitrogen if there is an electron-withdrawing group on the nitrogen
In this case the N-Boc protecting/directing group work well

Answer 7

• The nitrogen coordinates to the base to bring it close to the acidic proton where we deprotonate and stabilises the negative charge that we form upon deprotonation of the C-H centre

Question 8

In order to deprotonate an acidic proton, you need a base that has a …… pKaH than the pKa of the proton you are trying to deprotonate

Answer 8

Higher pKaH

Question 8

Carbonyls are another example of the groups we can use as an electron withdrawing group to stabilise the negative charge that we’re forming
How?

Answer 8

• Two EWG which can stabilise the negative charge

Question 9

Where does the carbonyl deprotonate when reacted with NaH and BuLi

Answer 9

• 1st deprotonation occurs at the most acidic proton
• a strong base (BuLi) is required to deprotonate the less reactive/acidic proton

Question 10

Which centre is more reactive after the two deprotonations?

Answer 10

The centre on the left is more reactive because it is less stable
Hence it would react with any subsequent electrophile to produce a more stable product

Question 11

The more reactive organometallic centre reacts with our electrophile allowing for the installation of our new C-C bond
what is the mechanism?

Answer 11

Question 12

Schlosser’s base (superbase) is formed from the combination of two bases (an alkyllithium base and a group I metal alkoxide)
What does this enable

Answer 12

• deprotonation at very weakly acidic positions
• Schlosser’s base is one of the most powerful bases known, and will abstract protons from allylic or benzylic positions and will even deprotonate benzene

Question 13

How could the following intermediate formed from Schlosser’s base react with carbon dioxide

Answer 13

Question 14

How could the following intermediate formed from Schlosser’s base react with furan

Answer 14