Acetyl bromide reacts with excess of  CH_3MgI followed by treatment with a saturated solution of  NH_4Cl  gives

  • Option 1)

    acetone

  • Option 2)

    acetamide

  • Option 3)

    2-­methyl­-2-­propanol

  • Option 4)

    acetyl iodide.

 

Answers (1)

As we learnt in

Reaction of Grignard reagent with Alcohol -

Alkane is obtained.    

- wherein

 

 CH_{3}COBr\frac{(i)excess\:CH_{3}MgI}{(ii)staurated\:solution\:of\:NH_{4}CI}\ 2 methyl-2-propanol 

 

 


Option 1)

acetone

This option is incorrect.

Option 2)

acetamide

This option is incorrect.

Option 3)

2-­methyl­-2-­propanol

This option is correct.

Option 4)

acetyl iodide.

This option is incorrect.

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