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Alcohols Phenols and Ethers question

2004 · Shift 0 · Q18
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Alcohols Phenols and Ethers question

2004 · Shift 0 · Q18

JEE MainChemistryAlcohols Phenols and EthersMCQ+4 / −1
Acetyl bromide reacts with excess of CH3MgICH_3MgICH3​MgI followed by treatment with a saturated solution of NH4ClNH_4ClNH4​Cl given
  1. A
    acetone
  2. B
    acetyl iodide
  3. C
    2- methyl -2- propanol
  4. D
    acetamide
View written solutionFree

Correct answer: C

  1. Identify the reaction type

Acetyl bromide is an acid halide: CH3COBrCH_3COBrCH3​COBr

Grignard reagent is methyl magnesium iodide: CH3MgICH_3MgICH3​MgI

Acid halides react with Grignard reagents twice when the Grignard reagent is in excess.

  1. First addition of CH3MgICH_3MgICH3​MgI

The methyl group from the Grignard attacks the carbonyl carbon of acetyl bromide:

CH3COBr+CH3MgI→CH3COCH3CH_3COBr + CH_3MgI \rightarrow CH_3COCH_3CH3​COBr+CH3​MgI→CH3​COCH3​

So, the first step gives acetone after elimination of Br−Br^-Br−.

  1. Second addition because Grignard reagent is in excess

The acetone formed also reacts with another mole of CH3MgICH_3MgICH3​MgI:

CH3COCH3+CH3MgI→(CH3)3COMgICH_3COCH_3 + CH_3MgI \rightarrow (CH_3)_3COMgICH3​COCH3​+CH3​MgI→(CH3​)3​COMgI

This is a tertiary alkoxide.

  1. Hydrolysis with saturated NH4ClNH_4ClNH4​Cl

On treatment with saturated NH4ClNH_4ClNH4​Cl, the alkoxide is protonated:

(CH3)3COMgI→NH4Cl(CH3)3COH(CH_3)_3COMgI \xrightarrow{NH_4Cl} (CH_3)_3COH(CH3​)3​COMgINH4​Cl​(CH3​)3​COH

Thus, the final product is

(CH3)3COH(CH_3)_3COH(CH3​)3​COH

which is 2-methyl-2-propanol (tert-butanol).

  1. Check options
  • A: acetone — intermediate only, not final product.
  • B: acetyl iodide — not formed.
  • C: 2-methyl-2-propanol — correct.
  • D: acetamide — not formed.
  1. Final answer

The correct option is: C\boxed{\text{C}}C​

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