Question
I am confused between the questions B. And C. The earlier is SN2 and later is E2. But what makes the difference?
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$\begingroup$ Look at the transition state in SN2. How likely (energy barrier?) is an inversion in C)? $\endgroup$ – imalipusram Apr 22 '20 at 5:45
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$\begingroup$ It is pretty clear that you can't perform any kind of elimination in B due to valency constraints, it has to be SN2. For C,you can look at the chair form and decide $\endgroup$ – Yusuf Hasan Apr 22 '20 at 6:45
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1$\begingroup$ With methoxide and benzyl chloride, the substitution is more likely to be bimolecular. Regular solvolysis of benzyl halides is on the border between the two mechanisms. $\endgroup$ – Zhe Apr 22 '20 at 13:46
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1$\begingroup$ I didn't. Solvolysis would be in methanol, without methoxide. Since methoxide is better at SN2, I think we're breaking the tie. $\endgroup$ – Zhe Apr 22 '20 at 19:33
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1$\begingroup$ For B you get a benzylic cation in a protic solvent. Surely SN1 would be favoured. $\endgroup$ – Josh Mitchell Apr 23 '20 at 14:01
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The most important factor in these reactions are nucleophiles. Here, the use of strong base (Sodium methoxide) in both of the reactions favors the bi-molecular mechanisms ($\ce{S_N2}$ and $\ce{E2}$).
Now, the second factor would be degree of halogen-linked carbon. Lower degree carbons ($\ce{\text{deg.} < 2}$) favor $\ce{S_N2}$, while higher degree favors $\ce{E2, E1, S_N1}$. So, $\ce{S_N2}$ is favored in first and $\ce{E2}$ in second.
Temperature conditions and nature of solvent are least important factors, and should be looked up when above two factors are similar.
answered Apr 23 '20 at 11:00
