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The question is to find out the final product $T$ in the following series of reactions.

enter image description here

I am provided the following options to choose from

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With alcoholic $\ce{KOH}$ elimination will occur and with $\ce{KOH}$ syn diol will be formed in place of double bonds.I am unable to see what will $\ce{CaO}$ do to the syn diol.I had an idea that it is used for decarboxylation hence the final product should have 1 carbon less but I am unable to get the right answer.Any ideas?Thanks.

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  • $\begingroup$ Why are you forming a syn diol? That too with alcoholic KOH? In my opinion, Q should be Hexan-1,6-dioic acid. For CaO and Ca(OH)2, I guess they are elaborating the decarboxylation mechanism step by step but that isn't leading me to any of the options. However, by seeing the options, I am getting a hint of an aldol condensation reaction taking place in the last step. $\endgroup$ – Reeshabh Ranjan Apr 17 '17 at 3:41
  • $\begingroup$ @ReeshabhRanjan no not with alcoholic KOH but with KMnO4.My mistake it is hot KMn04 so that it will eliminate a molecule of water from syn diol forming hexan-1-one which on aldol condensation gives D as correct answer.Am I right? $\endgroup$ – Pink Apr 17 '17 at 3:53
  • $\begingroup$ I have realised that D cannot be the right answer in aldol condensation.The nearest option if it were via aldol condensation should be B. $\endgroup$ – Pink Apr 17 '17 at 4:09
  • $\begingroup$ P will be Cyclohexene. KMnO4 is not a dehydrating agent. P2O5 or hot and concentrated sulphuric acid are some examples, or Copper at 573 K temperature. KMnO4 should just break the double bonds in Cyclohexene and convert it into Hexan-1,6-dioic acid. I may be wrong as the exact reagent to so this is hot and alkaline KMnO4. Here the KMnO4 is indeed hot, but no alkaline medium is provided. For CaO being used independently, it in my rough guess may be used as a weak base. I am still unsure of that. $\endgroup$ – Reeshabh Ranjan Apr 17 '17 at 4:12
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    $\begingroup$ @ReeshabhRanjan I was wrong in stating that syn diol will be formed because the appropriate reagent for that is cold dilute KMnO4 but in the question it is hot.Alkaline medium is provided by the previous reaction.So you are correct.After forming hexane-1,6-dioic acid it will decarboxylate forming cyclopentanone which will react via aldol condensation followed by dehydration to give B as final product. $\endgroup$ – Pink Apr 17 '17 at 4:53
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Lets see this step-by-step.

Bromocyclohexane is reacted with alcoholic KOH. This, as you would know, results in the formation of an alkene by elimination reaction. Product P would be Cyclohexene.

Cyclohexene is heated with KMnO4. You might think a cis-1,2-cyclohexanediol would form, and indeed it does. But vigorous oxidation provided by the heat cleaves the C-C bond between the two hydroxy groups and essentially converts into 1,6-hexanedioic acid (product Q, a.k.a adipic acid).

Calcium Oxide, being a pretty strong base, reacts with the dicarboxylic acid and forms a saltcalcium adipate (product R)

Heating a calcium salt of any carboxylic acid results in the elimination of a $\ce{CaCO3}$ molecule, and forms a ketone. If you take calcium adipate, upon heating would give cyclopentanone(product S).

Now, cyclopentanone, just like any other carbonyl compound with α-hydrogens, undergoes an aldol reaction, heating which would result in the formation of the compound given in option (B).

Here is the complete reaction mechanisms: enter image description here

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