Which of the following does not give iodoform?
(A) Acetic acid
(B) Lactic acid
(C) Acetophenone
(D) Propionic acid
Answer
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Hint: Iodoform test is used to check the presence of carbonyl compounds with a structure of \[R-CO-C{{H}_{3}}\] or alcohols with structure \[R-CH\left( OH \right)-C{{H}_{3}}\] in a given unknown substance.
- A positive result is the appearance of a very pale yellow precipitate of triiodomethane or \[CH{{I}_{3}}\] which is commonly known as iodoform. Apart from its colour, this can be recognised by its fairly “medical” or antiseptic smell.
Complete Solution :
Option A: acetic acid- acetic acid is a carboxylic acid, that is, it has the \[-COOH\] bond. Due to delocalisation of electrons, there is presence of \[o\] adjacent to carbonyl. Therefore, acetic acid does not give iodoform test.
Option B: Lactic acid- lactic acid gives iodoform reaction. The equation is given by:
\[C{{H}_{3}}-\overset{\overset{OH}{\mathop{\text{I}}}\,}{\mathop{C}}\,-\overset{\overset{O}{\mathop{\text{II}}}\,}{\mathop{C}}\,-OH\xrightarrow{{}}\underset{\underset{CH{{I}_{3}}+Na-\overset{\overset{O}{\mathop{\text{II}}}\,}{\mathop{C}}\,-\overset{\overset{O}{\mathop{\text{II}}}\,}{\mathop{C}}\,-OH}{\mathop{\downarrow {{I}_{2}}/NaOh/{{H}_{2}}O}}\,}{\mathop{C{{H}_{3}}}}\,-\overset{\overset{O}{\mathop{\text{II}}}\,}{\mathop{C}}\,-\overset{\overset{O}{\mathop{\text{II}}}\,}{\mathop{C}}\,-OH\]
Option C: Acetophenone- This has the formula - \[{{C}_{6}}{{H}_{5}}COC{{H}_{3}}\]. Since there is a presence of \[C{{H}_{3}}CO\] attached to the carbon, Acetophenone gives the iodoform test.
Option D: Propionic acid- As we can see a \[-COOH\] group in which electrons are delocalised, hence, it also does not give the iodoform test.
So, the correct answer is “Option A and D”.
Note: 1. A pale yellow precipitate of iodoform is formed when iodine and sodium hydroxide are added to a compound that contains either a methyl ketone or secondary alcohol with a methyl group in the alpha position. This test can be used to identify aldehydes or ketones.
2. Ethanol is the only aldehyde to give the iodoform test.
- A positive result is the appearance of a very pale yellow precipitate of triiodomethane or \[CH{{I}_{3}}\] which is commonly known as iodoform. Apart from its colour, this can be recognised by its fairly “medical” or antiseptic smell.
Complete Solution :
Option A: acetic acid- acetic acid is a carboxylic acid, that is, it has the \[-COOH\] bond. Due to delocalisation of electrons, there is presence of \[o\] adjacent to carbonyl. Therefore, acetic acid does not give iodoform test.
Option B: Lactic acid- lactic acid gives iodoform reaction. The equation is given by:
\[C{{H}_{3}}-\overset{\overset{OH}{\mathop{\text{I}}}\,}{\mathop{C}}\,-\overset{\overset{O}{\mathop{\text{II}}}\,}{\mathop{C}}\,-OH\xrightarrow{{}}\underset{\underset{CH{{I}_{3}}+Na-\overset{\overset{O}{\mathop{\text{II}}}\,}{\mathop{C}}\,-\overset{\overset{O}{\mathop{\text{II}}}\,}{\mathop{C}}\,-OH}{\mathop{\downarrow {{I}_{2}}/NaOh/{{H}_{2}}O}}\,}{\mathop{C{{H}_{3}}}}\,-\overset{\overset{O}{\mathop{\text{II}}}\,}{\mathop{C}}\,-\overset{\overset{O}{\mathop{\text{II}}}\,}{\mathop{C}}\,-OH\]
Option C: Acetophenone- This has the formula - \[{{C}_{6}}{{H}_{5}}COC{{H}_{3}}\]. Since there is a presence of \[C{{H}_{3}}CO\] attached to the carbon, Acetophenone gives the iodoform test.
Option D: Propionic acid- As we can see a \[-COOH\] group in which electrons are delocalised, hence, it also does not give the iodoform test.
So, the correct answer is “Option A and D”.
Note: 1. A pale yellow precipitate of iodoform is formed when iodine and sodium hydroxide are added to a compound that contains either a methyl ketone or secondary alcohol with a methyl group in the alpha position. This test can be used to identify aldehydes or ketones.
2. Ethanol is the only aldehyde to give the iodoform test.
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