Iodoform can be prepared from all except :
The iodoform test is a chemical test used to identify methyl ketones (R-CO-CH3) and alcohols that can be oxidized to methyl ketones or acetaldehyde (specifically, ethanol and secondary alcohols with a methyl group adjacent to the carbon with the -OH group). The reaction produces a yellow precipitate of iodoform (CHI3).
For a compound to give a positive iodoform test, it must have one of the following structural features:
Step 1: Analyze Isobutyl Alcohol
Isobutyl alcohol has the structure: (CH3)2CH-CH2OH. It is a primary alcohol. It cannot be oxidized to a methyl ketone or acetaldehyde. Therefore, it will not give a positive iodoform test.
Step 2: Analyze Ethyl Methyl Ketone
Ethyl Methyl Ketone has the structure: CH3CH2COCH3. It is a classic methyl ketone (R-CO-CH3). Therefore, it will give a positive iodoform test.
Step 3: Analyze Isopropyl Alcohol
Isopropyl alcohol has the structure: (CH3)2CHOH. It is a secondary alcohol with a methyl group adjacent to the carbon bearing the -OH group. It is oxidized by the iodine and base to a methyl ketone (acetone), which then undergoes the reaction. Therefore, it will give a positive iodoform test.
Step 4: Analyze 3-Methyl-2-butanone
3-Methyl-2-butanone has the structure: (CH3)2CHCOCH3. This is also a methyl ketone (it has the -CO-CH3 group). Therefore, it will give a positive iodoform test.
Final Answer: The compound that cannot be used to prepare iodoform is Isobutyl alcohol.
Core Concept: Haloform Reaction
The iodoform test is a specific example of the haloform reaction. The general reaction for a methyl ketone is:
Key Theory: The reaction involves three steps:
Compounds Giving Positive Test:
In the above chemical reaction sequence "A" and "B" respectively are