# Multiple leaving groups and SN2/E2

In the first stage of the reaction i am confused in which place will the SN2 happen (we have aprotic polar solvent and 4 leaving groups). In my opinion SN2 will happen either to the carbon with F- attached or the carbon with $\ce{Br-}$, $\ce{MeSO3}$ attached as these are the primary carbons. The leaving group strength is $\ce{F- < MeSO3 < Br-}$ so I suppose the Br will be substituted. The thing is that the nucleophile must attack from 180 degrees direction and I think steric hindrance in this carbon is bigger than the one with $\ce{F-}$. Another factor is that the oxygen atom from $\ce{MeSO3}$ can stabilize the intermediate with resonance.

$\ce{NaH (H^-)}$ activates the nucleophile by taking the most acidic proton from $\ce{-SH}$. So I think compound X will be:

Then E2 will occur ($\ce{NEt3}$ strong base) and the leaving group will be $\ce{MeSO3}$ (best LG). forming Z :

Is my solution correct?

• Are you sure that the -OH is the most acidic proton? Perhaps you should look up the pKa of thiophenol – Waylander Jul 19 '18 at 14:58
• The question here is more like which group is the most nucleophilic on the aromatic compound after deprotonation by NaH AND which is the best leaving group for an SN2 (F, OMe, Br, Ms). FYI, the best leaving groups are the ones which can distribute a charge in a wider volume. – SteffX Jul 19 '18 at 16:54
• I was wrong -SH group is more nucleophilic – Aris Jul 19 '18 at 16:58
• Ms and Br can stabilize the negative charge better... also Ms can stabilize with resonance this makes it better leaving group from Br. But comparing their pka HBr has -9 and MsOH -1.9 making conjucate Br- less basic than MsO- and therefore the weaker base is better leaving goup. Which criteria wins? – Aris Jul 19 '18 at 17:38
• Are you sure about the E2 step? Where is the most acidic proton in X? – Waylander Jul 19 '18 at 18:02