2
$\begingroup$

enter image description here

I dont understand why they are strong nucleophiles when the negative charge is resonance stabilised. Also it is bulky, is there any chance of elimination?

$\endgroup$
4
  • $\begingroup$ You may find useful these links for text formatting ( not to be applied to titles ): notation , text + formula formatting and upright vs italic $\endgroup$
    – Poutnik
    Dec 8, 2021 at 7:24
  • 1
    $\begingroup$ What exactly is a "strong" nucleophile? IMHO, for the purposes of organic chemistry, it suffices to know that it is a nucleophile. Nucleophilicity scales do exist, but aren't really needed if you just want to draw a curly arrow mechanism. Also, if the charge weren't resonance stabilised, then lots of bad things happen. Alkyllithiums will react in all sorts of conceivable ways (as a base, as a nucleophile, lithium–halogen exchange). Tempering the negative charge helps to keep the reactivity under control. $\endgroup$ Dec 8, 2021 at 10:35
  • $\begingroup$ I'm in high school and I've been taught while comparing nucleophilic strengths(I'm really sorry) , nucleophiles with delocalized charges are poorer nucleophiles because they cannot donate electrons as effectively as the ones with localized charges. I haven't really dealt with Sn1,Sn2 questions involving resonance stabilized nucleophiles before this.(Pardon me for any mistakes, English is not my first language) $\endgroup$
    – jen
    Dec 8, 2021 at 13:20
  • 1
    $\begingroup$ Hi Jen, welcome to ChemSE. Your question is quite insightful for a high school student. Whether or not the nucleophile/base is resonance stabilized is not the main issue. What counts is the relative rate of the irreversible SN2 vs. E2 reactions. I did a quick Chemical Abstracts search of this reaction and found yields of 16%, 21% and 40% for the SN2 product. These yields are modest at best. They may indicate substantial E2 elimination is competing with displacement. I will dismiss poor technique in isolating the product. ;) $\endgroup$
    – user55119
    Dec 9, 2021 at 17:04

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.