JEE Main 2023 · 25 January · evening · Numerical answer · +4 / −0
Number of compounds giving (i) red colouration with ceric ammonium nitrate and also (ii) positive iodoform test from the following is ___________
Answer and solution
Answer 3
Solution
Chemistry ma’am · 3 steps
Both tests at once means you need an alcohol whose OH sits on a CH₃CH(OH)⁻ carbon.
- Ceric ammonium nitrate. CAN gives a red colour only with an alcoholic -OH group. Primary, secondary and tertiary alcohols all respond. Ketones, aldehydes and ethers have no OH, so they stay negative.
- Iodoform test. A yellow CHI₃ precipitate needs a CH₃CO⁻ group, or a CH₃CH(OH)⁻ group that the hypoiodite can oxidise to one. A tertiary alcohol has no H on the carbinol carbon, so it cannot become a methyl ketone.
Compound CAN (alcoholic OH) Iodoform Both C₆H₅COCH₃ No Yes, CH₃CO⁻ No CH₃CH(OH)CH₂C₆H₅ Yes Yes, CH₃CH(OH)⁻ Yes CH₃CHO No Yes, CH₃CO⁻ No CH₃CH(OH)CH₂CH₃ Yes Yes, CH₃CH(OH)⁻ Yes CH₃OCH₂CH₃ No, ether No No C₆H₅CH₂CH₂OH Yes No, primary OH without CH₃ next to it No CH₃CH(OH)CH₂C₆H₁₁ Yes Yes, CH₃CH(OH)⁻ Yes C₆H₁₁C(CH₃)₂OH Yes No, tertiary alcohol No - So. Only 1-phenylpropan-2-ol, butan-2-ol and 1-cyclohexylpropan-2-ol pass both tests, so the answer is 3.
The common slip: 4. Counting 2-cyclohexylpropan-2-ol as iodoform positive because of its two methyl groups gives 4. A tertiary carbinol carbon has no H, so it cannot be oxidised to a methyl ketone.
The idea to keep. CAN asks one question: is there an alcoholic OH? Iodoform asks another: is there a CH₃CO⁻ or a CH₃CH(OH)⁻? A tertiary alcohol fails iodoform even when it carries methyl groups.
Answer from the NTA final answer key. Solution by PrepWiser.