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I thought $\ce{KOtBu}$ was used to form the less substituted alkene. Why is that not the case here?

enter image description here

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That is because the benzylic hydrogen is more acidic. When deprotonated, the anion is stabilized through resonance in the aromatic ring.

The base will take with the most acidic proton first.

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  • $\begingroup$ Ahh, makes sense. Is this the same thing for allylic hydrogens, so if t-buOK had a choice between a hydrogen attached to a primary carbon or to a secondary but allylic carbon, it would choose the allylic carbon because it would form a more stable carbocation? $\endgroup$ – Sarah Smith Feb 6 '18 at 7:06
  • $\begingroup$ You seem to suggest that an acid base reaction is preferred over the nucleophilic reaction. Is that a correct interpretation? $\endgroup$ – Gaurang Tandon Feb 6 '18 at 7:57
  • $\begingroup$ @SarahSmith That would be a carbanion $\endgroup$ – Raoul Kessels Feb 6 '18 at 8:01
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    $\begingroup$ @GaurangTandon In this case it is. t-BuO is a very strong base but a poor nucleophile due to its large bulk which impedes its approach to the substrate. $\endgroup$ – Raoul Kessels Feb 7 '18 at 22:01

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