I recently got an email from a reader with an advanced practice problem they were struggling with.
Here’s the problem. I really hope this wasn’t for a take home exam or anything. I want to show an example of the thought process that goes into solving a synthesis problem by applying the “3 questions” method I teach.
The first thing I’d notice is that all the groups on carbons 1 through 4 (that you’ve labelled) are the same. What’s changing is C-5 and C-6.
– new methyl group on C-5
– c-6 attached to an amine (NH2)
– C6 attached to a carboxylic acid.
What reactions do we know that can do each of the following
– form a C-C bond (such as C-CH3)
– form a C-NH2 bond
– form a C- CO2H bond.
Now let’s think:
CO2H can come from hydrolysis of CN
C-NH2 can come from reduction of C=N
[together these 2 steps are called the Strecker synthesis]
C-CH3 can come from alkylation of an enolate
this means C-6 used to be C=O…
because we can easily turn C=O into C=N with NH2 [typo: should have written NH3]
and we can add CN to C=N (this is the Strecker synthesis)
we can also form the enolate of C=O by treating with LDA and then CH3-Br
So we need to turn that C5-C6 double bond so that there’s a C=O at C-6.
This requires anti-markovnikov hydroboration of an alkene (BH3 then H2O2)
Then oxidize that alcohol to an aldehyde (PCC)
Part 3 is putting all those reactions together in the right order.
I’m omitting one final detail: acetone is used here too [any idea why?! ] But this shows the thought process
I hope this helps? James