Solving Organic Deductive Questions
Compound J, C8H6O4, forms a molecule K, C8H10O3, when reacted with LiAlH4 in dry ether. However, when J was reacted with NaBH4 instead, a product of molecular formula C8H8O4 was formed.
J gives an orange precipitate with 2,4-dinitrophenylhydrazine but does not give any precipitate when warmed with Fehling’s reagent. K reacts with bromine water to form L, C8H7O3Br3.
Suggest the structures of J, K and L.
First information is the molecular formula, it has almost the same C to H ratio and has more than six carbons => likely to have benzene ring C6H6
(Note that there is still two carbon atoms not accounted for)
When it undergoes reduction with LiAlH4, 4 hydrogen atoms were added but 1 oxygen atom was removed. => likely to have one COOH group, accounts for 2 oxygen atoms
But when it undergoes reduction with NaBH4, only 2 hydrogen atoms were added.
=> confirmed one of the reduction by LiAlH4 was on COOH.
J undergoes condensation with 2,4-DNPH but does not undergo oxidation with Fehling’s reagent. => carbonyl but not aliphatic aldehyde. Possible to be aromatic aldehyde or ketone.
K undergoes electrophilic substitution (tri-substitution) with bromine water => phenol present and also 2,4,6 positions does not have any substituent groups. This means that the two remaining carbon atoms must be at 3 and 5 positions. Additionally, this means that ketone is no longer possible as we would need two carbon on the same substituent group.
Summary: From the above deductions, J has phenol, aromatic aldehyde and carboxylic acid on positions 1,3,5.
K has phenol and two primary alcohols.
L has tri-substituted Br on positions 2,4,6, phenol, aromatic aldehyde and carboxylic acid on position 1,3,5.
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