An organic compound A $(\ce{C10H14O})$ exhibits following characteristics.

A) It reacts with metallic sodium to give a colorless odorless gas.
B) It is oxidized by $\ce{KMnO4}$ to benzoic acid.
C) It can be resolved.
D) It does not give precipitate with iodine in presence of $\ce{NaOH}.$
E) It changes the color of acidified potassium dichromate from orange to bluish green, and gives a chiral compound.

How many of the following compounds satisfy all the five criteria from A to E?

The given answer is the compounds 4, 5, 10 satisfy the criteria but I think the compounds 5, 6, 12 satisfy the criteria.

From statement A, we come to know it's an alcohol (which is common to every compound).

From statement B, we come to know that it has a benzylic hydrogen (which rules out the compound 11).

From statement C, we come to know that it has a chiral carbon. (which rules out options 4, 7, 9).

From statement D, we come to know that $\ce{OH}$ group is not attached to a methyl group (which rules out options 3, 10).

For statement E, I guess potassium dichromate will oxidise 2° alcohol to ketone and 1° alcohol to carboxylic acid and the compound still remains chiral (which rules out options 1, 2, 8).

So where's the mistake?


2 Answers 2


The given answer is wrong in many ways:

  • 4 did not satisfy both criteria C and E, because it is not a chiral compound to begin with.
  • 10 did not satisfy the criteria D, because it does give a precipitate with iodine in presence of $\ce{NaOH}$.

That gives only 5 in given answer to satisfy all criteria. Let's eliminate given compounds in systematic order:

eliminate given compounds according to criteria

That leaves only 5 and 6 to satisfy all criteria.

  • $\begingroup$ How do you know that #3 and #10 precipitate? $\endgroup$ Jul 17, 2020 at 21:09
  • 1
    $\begingroup$ @Stanislav Bashkyrtsev: 3 and 10 are oxidized by $\ce{I2}$ to give methyl ketones first, which give positive idoform test. $\endgroup$ Jul 18, 2020 at 3:19
  • $\begingroup$ @ Mathew Mahindaratne, why don't the carbanion form at the benzylic carbon, and react further giving a negative iodoform test $\endgroup$
    – Govind
    Feb 7, 2021 at 20:44
  • $\begingroup$ @Govind: Can you elaborate your question a bit? I didn't understand your angle. $\endgroup$ Feb 7, 2021 at 21:16
  • $\begingroup$ I think 10th should also be in the answer and should satisfy condition D. As the carbanion intermediate formed while iodoform test would be more stable at benzylic carbon $\endgroup$
    – Govind
    Feb 8, 2021 at 2:56

I think that correct answer should be 5 and 6 only, because:

  • 12 will produce terephthalic acid, instead of benzoic acid as it possess two benzylic-H (also given by 7).

terephthalic acid

  • 4 and 10 doesn't satisfy statement C and D, as you've mentioned.
  • 1
    $\begingroup$ I agree; the answer given is clearly incorrect. 4 and 10 fail the criteria $\endgroup$
    – Waylander
    Jul 8, 2020 at 12:11
  • 1
    $\begingroup$ Thanks for the answer. I didn't notice the extra methyl group in 12 :). $\endgroup$
    – V.G
    Jul 8, 2020 at 12:57

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