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A neutral compound (A) with molecular formula $ {{C}_{4}}{{H}_{6}}{{O}_{2}} $ reduced Fehling's solution, liberated hydrogen when treated with sodium metal and gave a positive iodoform test. The structure of (A) is:
(A) $ C{{H}_{3}}CH\left( OH \right)C{{H}_{2}}CHO $
(B) $ C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}CHO $
(C) $ C{{H}_{3}}COC{{H}_{2}}CHO $
(D) $ C{{H}_{3}}COC{{H}_{2}}C{{H}_{2}}OH $

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Last updated date: 14th Jul 2024
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Answer
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Hint: We know that Iodoform tests can be used to determine the presence of carbonyl compounds or alcohols. The reaction of Iodine along with the base with methyl ketones results in the appearance of a very pale yellow precipitate of triiodomethane or iodoform.

Complete answer:
Any compounds containing the group or the group give a positive result with the iodoform test. When iodine and Sodium hydroxide is added to a compound containing one of these groups, a pale yellow precipitate of iodoform or triiodomethane is formed. Ethanol is the only primary alcohol to give the triiodomethane (iodoform) reaction. Other primary alcohols don’t show an iodoform test.
The reaction of iodine and base with methyl ketones is so reliable that the iodoform test is used to probe the presence of a methyl ketone. This is also the case when testing for specific secondary alcohols containing at least one methyl group in alpha-position.
Thus, a neutral compound (A) with molecular formula $ {{C}_{4}}{{H}_{6}}{{O}_{2}} $ reduced Fehling's solution, liberated hydrogen when treated with sodium metal and gave a positive iodoform test. The structure of (A) is $ C{{H}_{3}}CH\left( OH \right)C{{H}_{2}}CHO $
Therefore, the correct answer is option A.

Note:
Remember that you may get confused about compound thinking of it as aldehyde or ketone because the Clemmensen reaction is the reduction reaction for both aldehyde and ketone. Since compound A is not giving any reaction with Tollen’s reagent and Fehling’s solution, this cancels out the chances of compound A being an aldehyde.