This synthesis problem is giving me a lot of trouble. The main problem is that the four steps I have down I think are the only ones I need but there is another blank. Anyone have any thoughts?enter image description here

  • 2
    $\begingroup$ With so many options, I wouldn't be surprised if there was another equally valid path you were expected to find. Nevertheless, if you insist on finding something "wrong" with your perfectly valid path, I would say that you might want to have a H2O step after the LiAlH4 reduction. (Usually when writing chemical schemes, aqueous workup after every step is implied so you are not supposed to have to write H2O after every step.) $\endgroup$
    – orthocresol
    Oct 18, 2016 at 23:25
  • $\begingroup$ That worked, thank you. I'm just so frustrated because how was I supposed to know that? I can't find a strict h2o step anywhere near this section of my notes $\endgroup$
    – 5antoro
    Oct 18, 2016 at 23:27
  • 2
    $\begingroup$ In the lab, there are quite a few reactions where you could probably get away with not performing an aqueous workup. However, with the LiAlH4 step, it is crucial: 1) You need to protonate the alkoxide 2) You need to destroy any remaining LiAlH4 in solution. So, I figured that if you had to add a H2O step, it was probably after the LiAlH4. $\endgroup$
    – orthocresol
    Oct 18, 2016 at 23:29
  • $\begingroup$ And you need to remove all of the lithium salts. $\endgroup$
    – Zhe
    Oct 19, 2016 at 0:39
  • 2
    $\begingroup$ How are you supposed to know that? I understand your frustration. Most reactions have a 'workup' step which for the most part is about neutralizing strong, reactive reagents. In this case, the strong reagent being neutralized is the reasonably strong base alkoxide. This step is normally considered 'trivial' by organic chemists, and is often not dwelled on in lecture discussion. The only comfort I can offer is this will get easier with exposure to more synthetic pathways. $\endgroup$
    – Lighthart
    Oct 19, 2016 at 2:35


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.