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enter image description here

So basically first step will be replacement of $\ce{-CH2 -}$ by a ($\ce{=O}$ bond) making a aldehyde and ketone side by side. Now what should I do after this? I tried attacking the aldehyde first as it is more electrophilic. I think both carbonyl groups will be attacked because of high concentration of hydroxyl ($\ce{OH-}$) ions. I even tried dehydration of the formed product after attacking with $\ce{OH-}$ ions but I always end up with a ketone and a carboxylic acid by its side after this.

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    $\begingroup$ First is Riley oxidation, then is cannizzaro reaction. $\endgroup$ – Zenix May 11 at 2:52
  • $\begingroup$ oh! damn that was simple thanks T-T $\endgroup$ – shreya May 11 at 3:21
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    $\begingroup$ @Zenix Please avoid answering in comments, it is not aligned with our site policy. This is a good example of a homework question, so I don't see why you shouldn't try answering it. ;) $\endgroup$ – William R. Ebenezer May 11 at 10:23
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    $\begingroup$ @WilliamR.Ebenezer Thanks for the encouragement, I am (and was) quite busy, will try to answer within a day. $\endgroup$ – Zenix May 11 at 21:47
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    $\begingroup$ @WilliamR.Ebenezer Answered within 23.30 hours ;) $\endgroup$ – Zenix May 12 at 22:49
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In the first step, Riley Oxidation (selenium dioxide-mediated oxidation of methylene groups adjacent to carbonyls) is taking place. You will get A as: enter image description here

A on reaction with conc. $\ce{OH-}$ (i.e. in strongly alkaline condition) undergoes redox reaction (famous Cannizzaro reaction). And terminal aldehyde will be oxidized to form the metal salt of acid, ketone will be reduced to form the alcohol, note that intramolecular reaction is taking place, to finally form the product B.

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