Iodoform can be prepared from all except
(a) ethyl methyl ketone
(b) isopropyl alcohol
(c) 3-methyl-

Q: Why does ethanol give a positive iodoform test even though it's a primary alcohol?

Ethanol ($CH_3CH_2OH$) is the only primary alcohol that gives a positive iodoform test. This is because it contains the required $CH_3-CH(OH)-$ group (where the other attachment is just a hydrogen). During the reaction, it gets oxidized to acetaldehyde ($CH_3CHO$), which has the necessary methyl carbonyl group.

Q: Do all secondary alcohols give the iodoform test?

No, only those secondary alcohols that have a methyl group attached to the carbon with the -OH group (i.e., 2-alkanols) will give a positive test. For example, 3-pentanol will not give a positive test because it lacks the terminal methyl group next to the hydroxyl-bearing carbon.

Q: What is the yellow precipitate formed in this reaction?

The yellow precipitate is iodoform, which has the chemical formula $CHI_3$. It is a solid at room temperature and has a characteristic, somewhat sweet, antiseptic smell.

Q: Why is NaOH used along with iodine?

Sodium hydroxide ($NaOH$) reacts with iodine ($I_2$) to form sodium hypoiodite ($NaOI$). This species acts as both an oxidizing agent (to convert alcohols to ketones) and a halogenating agent (to replace the alpha-hydrogens of the methyl group with iodine atoms).

Q: Can acetic acid give a positive iodoform test?

No. Even though acetic acid ($CH_3COOH$) has a methyl group attached to a carbonyl carbon, the hydroxyl group of the carboxylic acid interferes. The base will simply deprotonate the acid to form an acetate ion, and it will not undergo the haloform reaction.

Question

Iodoform can be prepared from all except

Options:

(A) ethyl methyl ketone

(B) isopropyl alcohol

(C) 3-methyl-2-butanone

(D) isobutyl alcohol

Sigma Percentile · Accepted Answer

### The Magic of the Iodoform Test

The iodoform test is a classic and visually striking reaction in organic chemistry. When a compound containing specific structural features is treated with iodine ($I_2$) and a strong base like sodium hydroxide ($NaOH$), a bright yellow precipitate of iodoform ($CHI_3$) is formed. This precipitate has a very distinct, antiseptic-like smell, making it incredibly easy to identify in a laboratory setting.

### Structural Requirements

Not just any organic compound can perform this magic trick. The iodoform reaction is highly specific. It requires the presence of one of two functional groups:

1.  **Methyl Ketones:** The compound must have a methyl group ($CH_3$) directly attached to a carbonyl carbon ($C=O$). The general structure is **$CH_3-C(=O)-R$**.
2.  **Specific Alcohols:** The compound must have a methyl group attached to a carbon that also bears a hydroxyl group ($-OH$). The general structure is **$CH_3-CH(OH)-R$**. Under the basic and oxidizing conditions of the test, these alcohols are first oxidized to their corresponding methyl ketones, which then undergo the haloform reaction.

### Analyzing the Options

Let's break down the options provided in the question to see which ones meet these strict criteria:

*   **Ethyl methyl ketone:** Its structure is $CH_3-C(=O)-CH_2-CH_3$. As the name suggests, it has a methyl group directly attached to the carbonyl carbon. It perfectly fits the requirement for a methyl ketone and will give a positive test.
*   **Isopropyl alcohol:** Its structure is $CH_3-CH(OH)-CH_3$. This is a secondary alcohol where the carbon holding the $-OH$ group is attached to two methyl groups. It contains the required $CH_3-CH(OH)-$ unit, so it will be oxidized to acetone and give a positive test.
*   **3-methyl-2-butanone:** Its structure is $CH_3-C(=O)-CH(CH_3)_2$. Despite the bulky isopropyl group on one side, the other side of the carbonyl carbon has a simple methyl group. It is a methyl ketone and will yield iodoform.

### The Exception

Now, let's look at **isobutyl alcohol**. Its structure is $(CH_3)_2CH-CH_2OH$. This is a primary alcohol. The carbon atom bearing the hydroxyl group is attached to a hydrogen atom and an isopropyl group, but **not** a methyl group.

Because it lacks both the $CH_3-C(=O)-$ and the $CH_3-CH(OH)-$ structural motifs, isobutyl alcohol cannot undergo the necessary chemical transformations to produce iodoform. Therefore, it is the only compound among the choices that will fail the iodoform test.

Animated Solution for Chemistry - Organic Chemistry: Iodoform can be prepared from all except

Select Answer:

Visualized Solution

$I o d o f or m T es t R e q u i r e m e n t s$

$O pt i o n A : E t h y l m e t h y l k e t o n e$

$O pt i o n B : I so p r o p y l a l co h o l$

$O pt i o n C : 3 - m e t h y l - 2 - b u t an o n e$

$O pt i o n D : I so b u t y l a l co h o l$

$C o n c l u s i o n$

The Sigma Insight: Carbonyl Compounds

The Magic of the Iodoform Test

Structural Requirements

Analyzing the Options

The Exception

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