Find the Major Product: Iodoform Test & Cyanohydrins
One of the classic hallmarks of a high-level JEE or NEET Organic Chemistry question is starting you off with a "Mystery Compound". Instead of giving you the starting structure, examiners give you a molecular formula and a Qualitative Analysis hint (like a flame test or precipitate color).
In this brilliant PYQ, we must first crack the identity of a compound using the Iodoform Test, and then push it through a Nucleophilic Addition reaction. Let's decode it!
Video Tutorial: Decoding the Mystery Sequence
Watch Abhishek Sengar sir from CHEMCA deduce the identity of the $C_4H_8O$ compound and correctly execute the IUPAC naming rules for the final product.
Step-by-Step Problem Breakdown
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Crack the Formula (The Iodoform Clue):
We are given the formula C4H8O. This fits the general formula CnH2nO, which tells us the compound has exactly 1 Double Bond Equivalent (DBE). This means it is likely an Aldehyde or a Ketone.
The Iodoform Rule:Therefore, the mystery compound is Butanone (or 2-Butanone): CH3-C(=O)-CH2-CH3.
The problem states it gives a positive Iodoform Test. This test is strictly positive ONLY for compounds containing a Methyl Ketone group (CH3-C(=O)-), or compounds that can oxidize into one. Among aldehydes, only Acetaldehyde works. Since we have 4 carbons, it must be a ketone! -
Nucleophilic Addition (HCN):
Reacting Butanone with Hydrogen Cyanide (HCN) triggers a nucleophilic addition at the carbonyl carbon. The CN- attacks the carbon, and the oxygen grabs the H+.
This forms a Cyanohydrin: A central carbon bonded to a Methyl group, an Ethyl group, an -OH group, and a -CN group. -
Hydrolysis of the Nitrile:
The problem then subjects the Cyanohydrin to hydrolysis (H3O+ / Δ). As we know, complete hydrolysis of a Nitrile (-C≡N) converts it entirely into a Carboxylic Acid (-COOH). The -OH group remains untouched.
Fig: Decoding the structural identity first, then tracing the chemical transformations.
Mastering the IUPAC Naming
Students often mess up the final IUPAC name. Let's apply the rules to our final structure:
- Principal Functional Group: The Carboxylic Acid (-COOH) takes top priority. Its carbon is firmly assigned as Carbon-1.
- Longest Chain: From C1, we trace the longest continuous chain of carbons. It goes through the central carbon (C2) and down the ethyl group (C3, C4). So the parent chain is Butanoic Acid.
- Substituents: At Carbon-2, we are left with two branches: a Hydroxy group (-OH) and a Methyl group (-CH3).
- Alphabetical Order: "Hydroxy" comes before "Methyl".
Practice Questions for JEE & NEET
Test your knowledge on the strict rules of the Iodoform test and Nitrile reactions!
Question 1: A compound has the molecular formula C3H6O. It gives an orange precipitate with 2,4-DNP, but it fails the Iodoform test. What is the identity of this compound?
Answer: Propanal (Propionaldehyde).
Reasoning:
The positive 2,4-DNP test confirms the compound has a carbonyl group (it is an aldehyde or ketone). With 3 carbons, the only options are Propanal (CH3-CH2-CHO) or Propanone/Acetone (CH3-CO-CH3).
Acetone contains the CH3-C(=O)- grouping, so it WOULD give a positive Iodoform test. Since our compound failed the test, it cannot be a methyl ketone. The only remaining answer is Propanal!
Question 2: In Step 2 of our main reaction, we formed Butanone Cyanohydrin. What would be the major product if we treated this Cyanohydrin with Lithium Aluminum Hydride (LiAlH4) instead of acidic water?
Answer: 1-Amino-2-methylbutan-2-ol (A β-amino alcohol).
Reasoning:
While H3O+ causes hydrolysis of a Nitrile into a Carboxylic Acid, LiAlH4 causes strong reduction.
The strong reducing agent will convert the -C≡N triple bond entirely into a primary amine (-CH2-NH2). The hydroxyl (-OH) group on the adjacent carbon will remain unaffected. This is a classic method for synthesizing amino alcohols!
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