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Upon reading the given question, I realized that:

The first reaction is an aldol condensation reaction between the given compound and formaldehyde.

So the product after 1st step is:

enter image description here

Now, the second part has thrown me off. The current ketone has no $\alpha$ hydrogen left. Condensation reactions are out of the question. Cannizzaro reaction can not take place as this is a ketone (and not an aldehyde) also the medium is now acidic. However, I saw one possibility of formation of hemi-acetals. (I did not think about full acetals, as such a structure was not present in the given options to this question)

Which gave me the following product:

enter image description here

However, the answer provided seems to be not this. The given answer is this:

enter image description here

Can someone give a suggestion that where I want wrong and/or how could I reach the answer and also what was wrong in my thinking as well?

This question is taken from the JEE(A)-2016 [Paper - 2]

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  • $\begingroup$ If that ketone part would've been an alcohol, an easy intramolecular reaction would take place - giving me a full acetal as well as the answer...if only. $\endgroup$ – McSuperbX1 Mar 27 '19 at 13:17
  • $\begingroup$ @Waylander, thanks for your input. Amazing. However, when my hemi-acetal is formed, I am already on step (ii). At step (ii), an acidic medium is provided instead of a strongly basic one that is needed for Cannizzaro reaction. $\endgroup$ – McSuperbX1 Mar 27 '19 at 14:00
  • $\begingroup$ You have misidentified the intermediate at the end of step i. The formaldehyde/NaOH conditions will deliver the diol (intermediate 9 in the scheme in the linked answer) through aldol followed by crossed Cannizzaro. The conditions for step ii are for forming the acetal only. $\endgroup$ – Waylander Mar 27 '19 at 14:46
  • $\begingroup$ @Waylander, you're correct. Thanks for the explanation. Everything seems to fall in place now. $\endgroup$ – McSuperbX1 Mar 27 '19 at 16:13
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I think a cross Cannizzaro with the Ketone and Formaldehyde takes place. The ketone give the alcohol and Formaldehyde formiate. In a second step a full acetal will be created with formaldehyd and now the Diole. Sorry I cannot add comments here. So use answer function.

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  • $\begingroup$ Isn't it that Cannizzaro reactions happen only between two aldehydes? Sure that there are no $\alpha$ hydrogen present, but they aren't aldehydes either. $\endgroup$ – McSuperbX1 Mar 27 '19 at 13:48
  • $\begingroup$ It is a "special case" Cannizaro. See the answer linked in the comments for the mechanism. $\endgroup$ – Waylander Mar 27 '19 at 13:52

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