The following question was asked in JEE Advanced 2017:


The answers given are option A and B. I was able to get B as the correct answer but A look like the wrong option to me.

My approach for option (A) and checking the reactions for Q:


Since (Q) of option (A) is able to give both haloform and aldol reaction, it should be incorrect. The reason I say that aldol reaction makes the option incorrect is that cannizaro reaction is given by aldehydes or ketones which have non enolisable hydrogen atoms. But aldol reaction requires enolisable hydrogen.

Can anyone confirm what should be the correct answer and why is my approach wrong if the answer given is right.

  • $\begingroup$ Q gives Cannizzaro NOT haloform. Check formulas of Q and S. A and B are correct. $\endgroup$
    – user55119
    May 14, 2019 at 12:28

2 Answers 2


First, Cannizzaro reaction is not given by ketones (your statement says ketones give cannizzaro reaction).

Counter to your analysis, the following reaction will not occur since $\ce{H}$ bonded to methyl group is not sufficiently acidic for hydroxide ion to attack. Refer to $\mathrm{p}K_\mathrm{a}$ values given by Mathew Mahindaratne.

enter image description here

Under the given conditions P on ozonnolysis gives formaldehyde and 4-methylbenzaldehyde. Both are nonenolizable aldehydes capable of undergoing Cannizzaro reaction. Both compounds do not undergo haloform reaction.

Scheme 1

enter image description here

Under the given conditions R gives acetophenone and formaldehyde. Acetophenone gives haloform test and it also undergoes aldol condensation, but not cannizaro reaction.

Scheme 2 enter image description here

Choice B also meets the conditions given. Thus choices A & B are correct.

  • 1
    $\begingroup$ Please note followings: (i) R in Scheme 2 is not denoted; (ii) R does not gives acetone upon ozonolysis (I corrected it to formaldehyde in the text); and (iii) On last diagram, iodoform products are idoform and sodium salt of benzoic acid. I'd like you to correct those errors. $\endgroup$ May 15, 2019 at 16:10
  • 1
    $\begingroup$ @MathewMahindaratne agreed R does not give acetone but formaldehyde..It was a TYPING error..I have corrected haloform product as sodium salt of benzoic acid. $\endgroup$ May 15, 2019 at 17:19
  • $\begingroup$ @MathewMahindaratne i was trying to focus on why 4-methyl benzaldehyde does not undergo haloform reaction which Groverkss wasnt convinced. $\endgroup$ May 15, 2019 at 17:27

The Cannizzaro reaction is a chemical reaction that involves the base-induced disproportionation of two molecules of a non-enolizable aldehyde to give a primary alcohol and a carboxylic acid (Wikipedia).

The haloform reaction requires a methyl ketone as the substrate.

Two compounds (P and R, respectively) in choice [A] give you 4-methylbenzaldehyde, $\ce{C8H8O}$ (Q; a non-enolizable aldehyde) and acetophenone, $\ce{C8H8O}$ (S; a methyl ketone) upon ozonolysis (byproduct from each compound is formaldehyde): 4-Methylbenzaldehyde can undergo Cannizzaro reaction but not haloform reaction (your given mechanism for bromform is unacceptable). Acetophenone, an enolizable ketone, easily undergoes haloform reaction but not Cannizzaro reaction (it is not an aldehyde).

Also, two compounds (P and R, respectively) in choice [B] give you 3-methylbenzaldehyde, $\ce{C8H8O}$ (Q; a non-enolizable aldehyde) and acetophenone, $\ce{C8H8O}$ (S; a methyl ketone) upon ozonolysis as well (byproducts from P and R here are acetaldehyde and acetone, respectively): 3-Methylbenzaldehyde can undergo Cannizzaro reaction (it is also a non-enolizable aldehyde) but not haloform reaction. Acetophenone, as pointed out earlier, undergoes haloform reaction but not Cannizzaro reaction (it is not an aldehyde).

Thus, both choices are correct.

Please also note that the compounds representing P and R in choices [C] and [D] will also give you a non-enolizable aldehyde and a methyl ketone upon ozonolysis. Each of these non-enolizable aldehyde and methyl ketone set shows the ideal reactivity towards Cannizzaro and haloform reactions as described earlier. However, they simply don't have the identical molecular formula as required.

  • $\begingroup$ I'm unable to understand why the hydrogen on the methyl part of 4-methylbenzaldehyde are non enolisable. As I see, they act similar to alpha-H in a ketone due to conjugation with carbonyl group. Am I wrong? $\endgroup$
    – Groverkss
    May 14, 2019 at 20:32
  • 2
    $\begingroup$ Note that $\mathrm{p}K_\mathrm{a}$ of methane and toluene is about 48 and 41 (about 56 and 43 in DMSO). That of diphenylmethane is about 34. What I'm trying to say is 4-aldehyde group should be lowering the $\mathrm{p}K_\mathrm{a}$ of tolyl group, but not that much when compared to that of acetophenone ($\mathrm{p}K_\mathrm{a}$ is about 25 in DMSO). Thus, heloform reaction won't work with 4-methylbenzaldehyde. $\endgroup$ May 14, 2019 at 21:07

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