How is ethanoic acid prepared from dry ice?

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Solution

$M e t h y l m a g n e s i u m b r o m i d e$ reacts with solid $c a r b o n d i o x i d e$ ( $d r y i c e$ ) to form a $m a g n e s i u m s a l t o f c a r b o x y l i c a c i d$ . This on acidification gives $e t h a n o i c a c i d$ .
$C H_{3} - M g - B r + O = C = O d r y e t h e r - ---- \to [C H_{3} - C O O M g^{+} B r^{-}] H - O H - --- \to d i l H C l C H_{3} - C O O H + M g (B r) O H$

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