4
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enter image description here

enter image description here I managed to solve the question till D but I'm stuck at the last reaction. I tried to find its solution and the source suggest it asenter image description here

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Upto (D) you are correct From there it's :enter image description here

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    $\begingroup$ I'm not so sure about that four-membered transition state. $\endgroup$ – orthocresol May 13 '18 at 11:36
  • $\begingroup$ @AmanChande Why base don't abstract H+ from -NH2 as in case of Hoffmann bromamide degradation reaction? $\endgroup$ – Anirudh May 13 '18 at 15:32
  • $\begingroup$ In your question NaOD/D2O is given in excess so we have high concentration of OD- which attacks rather than extracting a proton. If it was Dil. NaOD then it will extract proton. THATS THE DIFFERENCE BETWEEN CANNIZARO AND ALDOL CONDENSATION RXN ( conc. NaOH and Dil. NaOH) $\endgroup$ – Aman Chande May 13 '18 at 16:30
  • $\begingroup$ @Aman Chande: What do you mean by "attacks rather than extracting a proton"? Are you implying that under strong basic conditions H/D exchange doesn't occur but it does in dilute base? Not from my perspective. $\endgroup$ – user55119 Dec 27 '18 at 23:05
  • $\begingroup$ It will be through tuotomerism $\endgroup$ – Anirudh Feb 17 at 15:33
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Upto D) you are absolutely correct. Now observe carefully. In D) you have imine groups as well as amide groups. Under basic conditions imines don't hydrolyse, but amides can undergo hydrolysis. So, the amide parts will hydrolyse and as the reagents are in excess, you will get the end product as
enter image description here

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  • $\begingroup$ your answer also makes sense but the book sugest the above one. $\endgroup$ – Anirudh May 12 '18 at 15:44
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    $\begingroup$ But as I know, hydrolysis of imines are possible in acidic medium mainly. $\endgroup$ – Soumik Das May 12 '18 at 16:17
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This is what I reffered frm book and conclude d.. enter image description here

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  • $\begingroup$ nice quote :)... $\endgroup$ – Nilay Ghosh May 12 at 3:26

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