A compound having formula $\ce{C9H7O2Cl}$ (O = oxygen) exists predominantly in enol form (A) and also in keto form (B). On oxidation with $\ce{KMnO4}$ it gives m-chlorobenzoic acid as one of the products. Identify (A) and (B).

(subjective question; no options were given)

I tried it and found the double bond equivalent as 6, 4 of which were utilized in benzene ring. Also, the benzylic carbon must have at least one hydrogen. Therefore I determined the following structure as (A):

enter image description here

But my book suggests (A) should be:

enter image description here

Please explain whether both are correct or only one product will exist.

Source: Joint Entrance Exam (JEE) 2003 India

  • $\begingroup$ The answer given by the book is correct because of the formation of 6 membered chelation ring. As well as extended conjugation provides additional stability. As the structure drawn by you has only 5 memebered chelation ring as well as there is no active methylene group(acidic H) so it will not form. $\endgroup$ Commented Sep 15, 2021 at 5:19
  • $\begingroup$ Learn to draw structures such as [A] with proper bond angles. $\endgroup$
    – user55119
    Commented Sep 15, 2021 at 15:22

2 Answers 2


The representation from the book is not the only possibility, but likely more stable than the one you drew and thus the more favored answer.

  • There are many constitutional isomers sharing the sum formula $\ce{C9H7ClO2}$. Assuming a disubsituted benzene, one variation is to put the chloro- and the alkyl group in either ortho, meta, or para relationship to each other. (Side note: Hill formulae and sum formulae about organic compounds typically report the molecule's composition in the sequence about the number of atoms of carbon, then hydrogen, then by alphabetic order of the element symbol.)

  • For the sake of clarity, the structure formula depicted in the exam is redrawn below (left hand side):

    enter image description here

    The formula depicts the same molecule as the first of the three formulae depicting the tautomers which may coexist in the thermodynamic equilibrium with the $\beta$ dicarbonyl compound depicted on the right.

    The use of the dashed line between oxygen of the carbonyl group and hydrogen of the enol highlights an intramolecular hydrogen bond in addition to the typical covalent bond between $\ce{O}$ and $\ce{H}$ of the alcohol. If you compare the two with each other, the covalent bond is stronger than the hydrogen bond. The attraction between the carbonyl $\ce{O}$ and the enol $\ce{H}$ is strong enough that a conformation similar to a six-membered ring is energetic favorable. Here, in an analogy you may think of the hydrogen and oxygen atom attracting each other like opposite magnets:

    enter image description here

    (For comparison, the intramolecular $\ce{O-H}$ bond length in water is about $\pu{0.95 Å}$ (ref).)

  • The molecule you drew derives from a $\alpha$-keto aldehyde, a constitutional isomer of the above:

    enter image description here

    Even in a conformation most favorable for the opposition of hydrogen bond donor and hydrogen bond acceptor, the intramolecular distance is greater than in the earlier example of the $\beta$-keto enol:

    enter image description here

    Is the opposition of $\ce{H}$ of the enol and $\ce{O}$ of the carbonyl favorable (i.e., a lower energy conformation than other ones)? Yes. However given the hydrogen bond donor and hydrogen bond acceptor are farther away, the mutual attraction is less strong (remember Coulomb's law, electrostatic forces scale over distance $r$ by $1/r^2$) and the overall stabilization for the molecule is less important, than in the case of the $\beta$-keto enol.

The distances depicted in the images are obtained by running a quick and cheap MMFF86 force field optimization in Jmol. More important than remembering these distances is to retain the overall trend (the closer the partners, the stronger the potential intramolecular hydrogen bond, the stronger the overall stabilization of the molecule).


Both the answers are correct but the most appropriate will be B as the chelation occurs because of hydrogen bonding which makes it more stable. H-bonding is possible in your structure but the ring is less stable .Less stable chelate ring is formed than the other.

  • $\begingroup$ But there will also be H bonding i product 1 $\endgroup$
    – Anirudh
    Commented May 8, 2018 at 13:20
  • $\begingroup$ In your case the ring will be 5 membered which is less stable than 6 membered ring $\endgroup$ Commented May 8, 2018 at 13:54
  • $\begingroup$ Where is there a 5-membered ring? $\endgroup$ Commented May 8, 2018 at 14:14
  • $\begingroup$ What do you mean by 5 membered ring $\endgroup$
    – Anirudh
    Commented May 8, 2018 at 14:40
  • $\begingroup$ There is no actual ring there $\endgroup$
    – Anirudh
    Commented May 8, 2018 at 14:41

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