# Identification of the organic compound

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?

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).

• 4 and 10 doesn't satisfy statement C and D, as you've mentioned.
• I agree; the answer given is clearly incorrect. 4 and 10 fail the criteria – Waylander Jul 8 '20 at 12:11
• Thanks for the answer. I didn't notice the extra methyl group in 12 :). – Light Yagami Jul 8 '20 at 12:57

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:

That leaves only 5 and 6 to satisfy all criteria.

• How do you know that #3 and #10 precipitate? – Stanislav Bashkyrtsev Jul 17 '20 at 21:09
• @Stanislav Bashkyrtsev: 3 and 10 are oxidized by $\ce{I2}$ to give methyl ketones first, which give positive idoform test. – Mathew Mahindaratne Jul 18 '20 at 3:19
• @ Mathew Mahindaratne, why don't the carbanion form at the benzylic carbon, and react further giving a negative iodoform test – Govind Feb 7 at 20:44
• @Govind: Can you elaborate your question a bit? I didn't understand your angle. – Mathew Mahindaratne Feb 7 at 21:16
• 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 – Govind Feb 8 at 2:56