One base is bulkier than the other, that's my observation. How does it affect the product? Could someone please give proper reasoning and the product? Thanks a lot.
P.S. I guess both follow E2
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asked Jun 22, 2017 at 11:57
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1$\begingroup$ With methoxide there is a reasonable chance of getting nucleophilic substitution as well as elimination generating a mixture of products. With the much bulkier t-butoxide this is pretty well non-existent so you only get elimination. $\endgroup$– WaylanderCommented Jun 22, 2017 at 12:31
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1$\begingroup$ It's also possible that the stronger base will abstract the most accessible proton, producing a less stable (kinetic) product. The Zaitsev (thermodynamic) product comes from the least accessible proton in your case. $\endgroup$– schneiderfelipeCommented Jun 22, 2017 at 12:40
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1$\begingroup$ You might be under E1 conditions and not E2. $\endgroup$– ZheCommented Jun 22, 2017 at 14:55
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1 Answer
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Under $E_2$ reaction conditions, the MeO$^-$ product will be
while the Me$_3$CO$^-$ product will be
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This is due to the bulkiness of Me$_3$CO$^-$ being unable to reach the hydrogen at the substituted carbon so instead it goes to the primary position instead.
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$\begingroup$ Do you have references that support this black/white distinction? $\endgroup$– JanCommented Jun 23, 2017 at 15:37