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Acetyl bromide reacts with excess of CH3MgI followed by treatment with a saturated solution of NH4Cl gives
  • a)
    2-methyl-2propanol
  • b)
    acetamide
  • c)
    acetone
  • d)
    acetyl iodide
Correct answer is option 'A'. Can you explain this answer?
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Acetyl bromide reacts with excess of CH3MgI followed by treatment with...
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Acetyl bromide reacts with excess of CH3MgI followed by treatment with...
Reaction Overview:
- Acetyl bromide reacts with excess CH3MgI (Grignard reagent) to form a ketone intermediate.
- The ketone intermediate is then treated with a saturated solution of NH4Cl to give 2-methyl-2-propanol.

Reaction Steps:

Step 1: Formation of Grignard reagent
- CH3MgI is prepared by reacting methyl iodide (CH3I) with magnesium (Mg).
- The reaction occurs as follows:
CH3I + Mg → CH3MgI

Step 2: Reaction of Acetyl bromide with CH3MgI
- Acetyl bromide (CH3COBr) reacts with the Grignard reagent (CH3MgI) to form a ketone intermediate.
- The reaction occurs as follows:
CH3COBr + CH3MgI → CH3COCH3

Step 3: Treatment with NH4Cl
- The ketone intermediate (CH3COCH3) is treated with a saturated solution of NH4Cl to give 2-methyl-2-propanol.
- The reaction occurs as follows:
CH3COCH3 + NH4Cl → (CH3)3COH

Explanation:
- Acetyl bromide (CH3COBr) reacts with the Grignard reagent (CH3MgI) to form a ketone intermediate, acetone (CH3COCH3).
- The reaction between acetyl bromide and the Grignard reagent is a nucleophilic addition reaction. The nucleophilic carbon of the Grignard reagent attacks the electrophilic carbonyl carbon of acetyl bromide. This leads to the formation of a tetrahedral intermediate, which subsequently collapses to form the ketone.
- The ketone intermediate, acetone, is then treated with a saturated solution of NH4Cl. This treatment results in the formation of 2-methyl-2-propanol ((CH3)3COH) through a reduction reaction.
- The saturated solution of NH4Cl serves as a source of H+ ions, which act as a reducing agent in the reaction. The H+ ions react with the carbonyl oxygen of the ketone, causing it to be reduced to a hydroxyl group (-OH). This results in the formation of 2-methyl-2-propanol.
- The final product, 2-methyl-2-propanol, is a tertiary alcohol with a branched structure.

Conclusion:
- The reaction of acetyl bromide with excess CH3MgI followed by treatment with a saturated solution of NH4Cl leads to the formation of 2-methyl-2-propanol.
- This reaction demonstrates the use of Grignard reagents and reduction reactions in organic synthesis.
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Acetyl bromide reacts with excess of CH3MgI followed by treatment with a saturated solution of NH4Cl givesa)2-methyl-2propanolb)acetamidec)acetoned)acetyl iodideCorrect answer is option 'A'. Can you explain this answer?
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Acetyl bromide reacts with excess of CH3MgI followed by treatment with a saturated solution of NH4Cl givesa)2-methyl-2propanolb)acetamidec)acetoned)acetyl iodideCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Acetyl bromide reacts with excess of CH3MgI followed by treatment with a saturated solution of NH4Cl givesa)2-methyl-2propanolb)acetamidec)acetoned)acetyl iodideCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Acetyl bromide reacts with excess of CH3MgI followed by treatment with a saturated solution of NH4Cl givesa)2-methyl-2propanolb)acetamidec)acetoned)acetyl iodideCorrect answer is option 'A'. Can you explain this answer?.
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