Question

Organic compound (P) $\xrightarrow{(i) \text{ excess of} HI, (ii) \text{ Aq. } NaOH} Q + R$. $Q$ and $R$ both gives Iodoform test, Which among the following is (P) from the given organic compound?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

When an ether is cleaved by excess $\text{HI}$, it yields alkyl iodides, which are then hydrolyzed to corresponding alcohols. For both products to pass the iodoform test, both resultant alcohols must contain the $\text{R}-\text{CH}(\text{OH})\text{CH}_3$ functional group (where R can be H).

Answer 1

The Correct Option is C

The problem involves identifying an organic compound 'P' that, when treated with excess HI followed by aqueous NaOH, gives products 'Q' and 'R' which both give a positive iodoform test. Let's analyze each option step-by-step: 

1. The iodoform test is positive for compounds containing the structure CH₃C=O or CH₃CH(OH), typically methyl ketones. Both Q and R should meet this condition.
2. When ethers react with excess HI, they undergo cleavage to form alcohols, which in turn can form iodoform by reaction with iodine and a base if they have a methyl group next to the carbonyl group after oxidation.
3. For Option 3, the compound is 2-methoxy-2-methylpropane:
   • Treatment with excess HI:
     • This will cleave the ether to form 2-methylpropan-2-ol (tert-butyl alcohol) and iodomethane.
   • Treatment with aqueous NaOH:
     • 2-methylpropan-2-ol gives acetone (propane-2-one) after oxidation, which gives a positive iodoform test, since acetone is a methyl ketone.
     • Iodomethane does not give a direct iodoform test but allows further transformation if necessary. In their scenarios, it's more about the alcohol formed in the first step.

Therefore, compound 3 is (P) since it leads to products that can both give a positive result in the iodoform test. The correct answer is:

Option 3