[deezykem]

I am taking a summer course on organic chemistry right now (bad idea) and it has been 8 years since my last chem experience.  I don't have a solid understanding of reactions yet, but we are getting close to the end.  I have a homework problem that is boggling me where to start (and finish), and would love some guidance on where my disconnects should be and what reagents do them.  Thanks for the help,  I have enjoyed reading other posts on here.

[Dan]

Consider the functional groups in our molecule.

How are these functional groups made?

Consider your starting materials.

What functional groups do you have, and what reactions can they take part in?

Try making some key disconnections first.

You must show you have attempted the question, this is a Forum Rule.

[deezykem]

My bad, I forgot to upload my work so far.  I feel it is incomplete though

[salteen]

This isn't a bad attempt, except for the fact that LiAlH₄ + H₂O = Fire ..

If I were correcting this, I'd probably take a mark off for that.  What other solvent could you use instead?

[magician4]

transformation #1 looks good to me, sofar

nevertheless, as the direct reaction of an amine and an acid (first step of seq. #3) is most difficult, i would propose to move on to the p-nitro benzoic acid chloride here (i.e: further reaction with SOCl₂)

transformation seq. #2 : ok
(pls know that other setups are available for this transformation also, like for example Gabriel-synthesis and thatlike )

seq. #3. (see #1 for starters)
besides that, the final aldol reaction is possible, yes, but enolate addition to the acetaldehyde ( i.e 1 eq. LDA as pre-treatment for the ketone) and a very carefull workup (preventing of aldol condensation) would be my choice instead

regards

Ingo

[orgopete]

Progress! The mixed aldol is the challenge here. How and when to do it.

In route, can you hydrolyze the ketal without hydrolyzing the amide?

[TwistedConf]

nevertheless, as the direct reaction of an amine and an acid (first step of seq. #3) is most difficult,

Besides this fact that the reaction won't work--- your acetal would be gone before you even get the hot plate turned on.

You really need to make that amide under neutral or basic conditions.

[deezykem]

This isn't a bad attempt, except for the fact that LiAlH₄ + H₂O = Fire ..

If I were correcting this, I'd probably take a mark off for that.  What other solvent could you use instead?

Doesn't the presence of ether stop the violent reaction of lithium aluminum hydride and water?

[deezykem]

transformation #1 looks good to me, sofar

nevertheless, as the direct reaction of an amine and an acid (first step of seq. #3) is most difficult, i would propose to move on to the p-nitro benzoic acid chloride here (i.e: further reaction with SOCl₂)

transformation seq. #2 : ok
(pls know that other setups are available for this transformation also, like for example Gabriel-synthesis and thatlike )

seq. #3. (see #1 for starters)
besides that, the final aldol reaction is possible, yes, but enolate addition to the acetaldehyde ( i.e 1 eq. LDA as pre-treatment for the ketone) and a very carefull workup (preventing of aldol condensation) would be my choice instead

regards

Ingo

Thanks!

[Babcock_Hall]

@OP, Can you think of other ways to form the amide bond?

[deezykem]

We are stuck on that final part, I don't know how the LDA and SOCl2/NH3 works

[Babcock_Hall]

I don't see SOCl₂/NH₃ above, but I do see just thionyl chloride.  Let me ask this question in another way.  Do you need a good leaving group, and if so, do you have a good leaving group when you have the free acid?

[magician4]

We are stuck on that final part, I don't know how the LDA and SOCl2/NH3 works

SOCl₂ would be needed to transform an carbon acid to the respective acid chloride:

RCOOH + SOCl₂  RCOCl + HCl + SO₂
the acid chloride can / should be isolated

acid chlorides will react with amines to form amides  (without the problem of salt formation which otherwise would occur, if we'd use the acid instead with the amine, like in your original proposal):

RCOCl + H₂N-R' --(NEt₃₎--> RCONHR' ( + NEt₃ x HCl )

as a side effect, your acetal would remain intact, as you always will remain on the alkaline side due to the triethylamine excess


"LDA" ( LiNEt₂) is what you would gain from reaction of butyllithium and diethylamine:
C₄H₉Li + HNEt₂  C₄H₁₀ + LiNEt₂
normally, it's no nucleophile worthy to mention, but a very powerfull deprotonator nevertheless, and will activate your ketone instantly:
R-CO-CH₂-R' + LDA  [R-CO-CH^--R'  R-C(O^-)=CH-R' ]Li  (+ HNEt₃)
the resulting enolate/HNEt₂ complex is much better than the OH^- you originally proposed as a catalyst for the aldol reaction, esp. if it came to "mixed" aldol reactions ketone/aldehyde, both with respect to steering directions and with respect to yield


regards

Ingo

[Babcock_Hall]

@OP, How does the use of thionyl chloride relate to the issue of leaving groups I mentioned?