
Methyl ketone group is identified by which test?
A) Iodoform Test
B) Fehling Solution
C) Tollen’s Reagent
D) Schiff’s Reagent
Answer
572.1k+ views
Hint: Iodoform Reaction: The iodoform test indicates the presence of an aldehyde or ketone in which one of the groups directly attached to the carbonyl carbon is a methyl group. Such a ketone is called a methyl ketone. In the iodoform test, the unknown is allowed to react with a mixture of excess iodine and excess hydroxide.
Complete answer:
Iodoform test is done to detect the presence of the \[C{H_3}CO\] group in organic compounds.
When methyl ketone is heated with iodine in the presence of sodium hydroxide, yellow precipitate of iodoform is obtained. Hence, methyl ketones are characterized by iodoform test. In this reaction, methyl ketone is converted to the sodium salt of carboxylic acid.
Iodoform test is used to check the presence of carbonyl compounds with the structure $R - CO - C{H_3}$ or alcohols with the structure $R - CH(OH) - C{H_3}$ in a given unknown substance.
The reaction of iodine, a base and a methyl ketone gives a yellow precipitate along with an “antiseptic” smell. It also tests positive for a few specific secondary alcohols that contain at least one methyl group in the alpha position.
Reaction Involved:
$R - CO - C{H_3}$ + $3{I_2}$ + $4NaOH$ $ \to $ $R - CO - ONa$ + $CH{I_3}$ + $3NaI$ + $3{H_2}O$
When Iodine and sodium hydroxide are added to a compound that contains either a methyl ketone or a secondary alcohol with a methyl group in the alpha position, a pale yellow precipitate of iodoform or triiodomethane is formed. It can be used to identify aldehydes or ketones. If an aldehyde gives a positive iodoform test, then it must be acetaldehyde since it is the only aldehyde with a $C{H_3}C = O$ group.
So the correct option is (A).
Note: Compounds that give positive iodoform test:
Acetaldehyde, Methyl Ketones, Ethanol, Secondary Alcohols that contain Methyl Groups in Alpha Position.
Methyl ketone gives yellow precipitates of iodoform on reaction with sodium hydroxide and iodine.
Complete answer:
Iodoform test is done to detect the presence of the \[C{H_3}CO\] group in organic compounds.
When methyl ketone is heated with iodine in the presence of sodium hydroxide, yellow precipitate of iodoform is obtained. Hence, methyl ketones are characterized by iodoform test. In this reaction, methyl ketone is converted to the sodium salt of carboxylic acid.
Iodoform test is used to check the presence of carbonyl compounds with the structure $R - CO - C{H_3}$ or alcohols with the structure $R - CH(OH) - C{H_3}$ in a given unknown substance.
The reaction of iodine, a base and a methyl ketone gives a yellow precipitate along with an “antiseptic” smell. It also tests positive for a few specific secondary alcohols that contain at least one methyl group in the alpha position.
Reaction Involved:
$R - CO - C{H_3}$ + $3{I_2}$ + $4NaOH$ $ \to $ $R - CO - ONa$ + $CH{I_3}$ + $3NaI$ + $3{H_2}O$
When Iodine and sodium hydroxide are added to a compound that contains either a methyl ketone or a secondary alcohol with a methyl group in the alpha position, a pale yellow precipitate of iodoform or triiodomethane is formed. It can be used to identify aldehydes or ketones. If an aldehyde gives a positive iodoform test, then it must be acetaldehyde since it is the only aldehyde with a $C{H_3}C = O$ group.
So the correct option is (A).
Note: Compounds that give positive iodoform test:
Acetaldehyde, Methyl Ketones, Ethanol, Secondary Alcohols that contain Methyl Groups in Alpha Position.
Methyl ketone gives yellow precipitates of iodoform on reaction with sodium hydroxide and iodine.
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