4
$\begingroup$

hexane‐2,5‐dione and (NH4)2CO3

Which of the following statement is INCORRECT?

(A) The product X has a five membered ring.
(B) The product X is aromatic.
(C) The degree of unsaturation of the product X is 3.
(D) The product X has a six membered ring.

Ammonium carbonate will probably lose a proton and one of the carbonyls may be protonated. But from here onwards I can't find any leads. Does the lone pair of other oxygen atom plays a role?

Usually the heat provided may lead to a cyclic structure, but I am not sure.

Improve this question
$\endgroup$
2
  • $\begingroup$ Yes it is a cyclic compound. First think, what would happen to ammonium nitrate when it is heated? $\endgroup$
    – Mathew Mahindaratne
    Nov 24, 2019 at 5:57
  • $\begingroup$ Ammonia gas , carbon dioxide and water is released...nitrogen lone pair will attack on carbonyl carbon and the oxygen will attack the other carbonyl carbon...is this assumption correct? But what happens next? $\endgroup$
    – raj pattnaik
    Nov 24, 2019 at 6:32

1 Answer 1

Reset to default
7
$\begingroup$

Heating $\ce{(NH4)2CO3}$ will generate $\ce{NH3 + CO2 + H2O}.$ Out of these, $\ce{NH3}$ will act as a nucleophile as it's the most nucleophilic of the products. Addition of $\ce{NH3}$ to the two ketones, followed by elimination of water molecules (due to the high temperature provided by ∆) will generate the final product as follows:

enter image description here

The product is a five-membered aromatic ring called 2,5-dimethylpyrrole.

Improve this answer
$\endgroup$
2
  • 4
    $\begingroup$ Try to avoid "lassos" in mechanisms. Use arrows. ;) $\endgroup$
    – user55119
    Dec 2, 2019 at 0:33
  • $\begingroup$ @user55119 Well,my handmade mechanisms tend to have a lot of lassos,still will keep that in mind later :) $\endgroup$
    – Yusuf Hasan
    Dec 2, 2019 at 0:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.