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Question #43984
The Question
Consider the following reaction sequence: Identify the major organic products P and Q.
Options
- A.
P = 2-(2,4,6-trimethylphenyl)propan-2-ol; Q = No reaction
- B.
P = 1-(2,4,6-trimethylphenyl)ethan-1-one; Q = Sodium 2,4,6-trimethylbenzoate + \ce{CHI3}
- C.
P = 1-(2,4,6-trimethylphenyl)ethan-1-one; Q = 2,2,2-triiodo-1-(2,4,6-trimethylphenyl)ethan-1-one
- D.
P = 1-(2,4,6-trimethylphenyl)ethan-1-one; Q = 2-iodo-1-(2,4,6-trimethylphenyl)ethan-1-one
Check Your Answer
Correct Answer
Step-by-Step Solution
Explanation
How the Examiner Thinks
Mechanism_Trap
Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.
Concepts Behind This Question
Haloform Reaction Logic|Grignard Reaction Pathways
Why This Question Is Tricky
Common Mistakes
This problem tests the limits of two fundamental reactions under extreme steric hindrance.
Trap 1: Students pattern-match 'Acid Chloride + Excess Grignard -> Tertiary Alcohol'.
However, the ortho-methyl groups block the second nucleophilic addition.
Instead, the Grignard reagent acts as a base, enolizing the intermediate ketone, which upon hydrolysis yields the methyl ketone (acetomesitylene).
Trap 2: Students pattern-match 'Methyl Ketone + I2/NaOH -> Carboxylate + Iodoform'.
While base-catalyzed halogenation occurs to form the triiodo intermediate, the final cleavage step requires nucleophilic attack of OH- at the carbonyl carbon.
The same steric hindrance completely blocks this attack, halting the reaction at the triiodo ketone and failing to produce iodoform.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Aldehydes, Ketones, and Carboxylic Acids - Class 12 |
| Concept | Haloform Reaction Logic |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |