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Question: A and B are two functional isomers of compound \[{C_3}{H_6}O\]. On heating with \(NaOH\)and \({I_2}\...

A and B are two functional isomers of compound C3H6O{C_3}{H_6}O. On heating with NaOHNaOHand I2{I_2} , isomer B forms yellow precipitate whereas isomer A does not form any precipitate. What will be the formula of isomers A and B?

Explanation

Solution

As we know that the given compound can be a ketone or an aldehyde as ketone reacts with a halogen like iodine and a base like sodium hydroxide and where the base helps in removing the hydrogen atoms and then the halogen acts as a nucleophile to get attached on the same carbon.

Complete Step by step answer:
Functional isomers are the one that have the same molecular formula but different functional group. The given compounds have three carbon atoms and six hydrogen groups attached to these carbons along with oxygen, so it can be a ketone or an aldehyde. And we also know that any compound containing CH3C=OC{H_3}C = O group or CH3CH(OH)C{H_3}CH(OH) group can react with a base and a hydrogen to undergo the Haloform reaction of Iodoform test and results into a precipitate formation.
Thus, the given compound can be an Acetone possessing CH3C=OC{H_3}C = O group, which would undergo haloform reaction on heating with sodium hydroxide and iodine results into formation of yellow precipitate of iodoform and a sodium salt of acetic acid along with three molecules of sodium iodide and water.

This reaction can be shown as follows:
CH3COCH3+3I2+4NaOH→I3HCyellow+3NaI+CH3COONa+3H2OC{H_3}COC{H_3} + 3{I_2} + 4NaOH \to \mathop {{I_3}HC}\limits_{yellow} + 3NaI + C{H_3}COONa + 3{H_2}O
Next, the given compound can also be a propionaldehyde or simply propanal having CH3CH2HC=OC{H_3}C{H_2}HC = O formula but it does not form any precipitate as it does not undergo the iodoform test.
CH3CH2CHO+I2+NaOH→No  reactionC{H_3}C{H_2}CHO + {I_2} + NaOH \to No\;reaction

Therefore the correct formula of the isomer A is CH3CH2HC=OC{H_3}C{H_2}HC = O and B is CH3−CO−CH3C{H_3} - CO - C{H_3}.

Note: Remember that three hydrogen atoms are replaced by three iodine from ketones and thus the carbon-carbon double bond in ketone breaks due to the electron withdrawing effect of these three Iodine atoms thereby resulting in the formation of iodoform with a yellowish precipitate in the solution.