CHCl 3 + MeCOMe → OH − A → excess KOH B → NaOH − CaO C → KOBr D

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Question: Answered & Verified by Expert

{CHCl}_{3} + MeCOMe \overset{{OH}^{-}}{\to} A \overset{excess KOH}{\to} B \overset{NaOH - CaO}{\to} C \overset{KOBr}{\to} D

ChemistryCarboxylic Acid DerivativesNEET

Options:

A $M e C H_{2} O H$
B $M e C O O H$
C $M e_{2} C H O H$
D $C H_{2} C l_{2}$

Solution:

2049 Upvotes Verified Answer

The correct answer is:

M e C O O H

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