39Yep that's it.
this one seemed me to be simple clemmensen reduction
IS IT ???
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39No...it's a clever question. Actually after the =O group has been destroyed, a negative charge appears in place of it to which HCl provides a proton. Not likely to happen here, there's a leaving group next to it. So it(the negative charge) bangs out the leaving group and forms a double bond.
39Depends on the mols of HCl used. Usually we use that many mols which are sufficient for the clemmenson reduction. Your point is right, if HCl is in excess, we will have Markownikov addition across double bond.
A variation of this question...if it were Wolff-Kishner reduction, what would be the product?