1
$\begingroup$

enter image description here

If anyone can give me some guidance here, it'd be much appreciated. I know the NaNH2 would dissociate and the NH2- base would deprotonate the terminal acetylene. I'm assuming the resultant product would act as some sort of nucleophile which would open the ring of the epoxide. The Na+ would form an ionic bond with the negatively polar oxygen after which the acid would protonate said oxygen, but I'm not entirely sure.

$\endgroup$
1
  • 5
    $\begingroup$ The final structure is wrong. Count carbons. Missing a CH2. $\endgroup$
    – user55119
    Sep 20, 2021 at 23:08

1 Answer 1

4
$\begingroup$

Basically, you are on the right track and answered the question. First, NaNH$_2$ is used as a base to deprotonate the terminal alkyne (see this Reagent Friday). Second, the generated nucleophile opens the epoxide in an S$_\mathrm{N}$2 fashion, usually adding to the lesser substituted carbon which in your case yields a quaternary alcohol after acidic workup (see also Opening of Epoxides With Acid).

(Note: Indeed you are missing the CH$_\mathrm{2}$-group where the nucleophile attacks in the final structure)

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.