PMA (Poly Methyl Acrylate) polymer is formed by methyl acrylate, which is prepared as follows-

R - C \equiv C H C O + R O H - ----- \to

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H C \equiv C H C O + C H_{3} O H - ------- \to N i (C O)_{4}

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H C \equiv C H N i (C O)_{4} - ----- \to C O + H_{2} O

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None of these

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Solution

The correct option is A $H C \equiv C H C O + C H_{3} O H - ------- \to N i (C O)_{4}$
This is oxidation reaction of alkyne in presence of $C O$ and $C H_{3} O H$ .
Reaction:
$H C \equiv C H C O + C H_{3} O H - ------- \to N i (C O)_{4} C H_{2} = C H C O O C H_{3}$ (methyl acrylate).

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$C H_{3} - C H_{3} | C H - C H_{2} - C H = C H - C H_{3}$

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