I came across the following question :

enter image description here

My reasoning was as follows :

  1. It can't be option (c) because there is no chance of losing a carbon during Cannizzaro's reaction
  2. It can't be option (b) because basically option (c) gives (b) finally
  3. It can't be option (a) or option (d) because we can't have all carb acid groups or all alcohol groups. we need a combination of both.

So, are the options wrong?

  • $\begingroup$ Your reasoning looks correct $\endgroup$
    – Waylander
    May 8 '21 at 6:14
  • $\begingroup$ I too think there is some mistake in the options, none of them seems correct. What is the given answer? $\endgroup$ May 8 '21 at 8:40
  • $\begingroup$ given answer includes 2 cooh groups and 2 ch2oh groups alternately. but that isnt in the options. the key says (b). $\endgroup$ May 8 '21 at 8:49

Your reasoning is correct and the problem is with the publisher as they made a mistake while compiling these previous year questions.

I found out the original 2003 question paper(pdf). The question you mentioned had the following options:


It is clear form the options that the answer is C(which is the product you expected) which can also be confirmed by the key attached at the end of the pdf.


Yes, the reasons you have given for eliminating options seem correct...I think the answer should be the two Ph groups both with COOH and CH2OH (since it has to have carboxylic acid and alcohol groups in the product) so the structure will be similar to option (C) but with CH2OH in place of the OH groups.

  • $\begingroup$ I have a little confusion as to how (c) and (b) are the same thing though, maybe check your calculations once again, I solved twice but messed up somewhere and got varying answers $\endgroup$ May 8 '21 at 12:30
  • $\begingroup$ option (c) will convert spontaneously to (c). look at the stability of the products. there is lots of conjugation and it is cyclic and completed. moreover, alcohol and carboxylic acids undergo intramolecular condensation to esters very fast $\endgroup$ May 9 '21 at 9:49

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