CH_2=CH-CO-OCH_3 (methyl acrylate,bp 81^circ | vedclass

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Question

(A) The reaction is an acid-catalyzed transesterification. Methyl acrylate reacts with $n$-butyl alcohol in the presence of $TsOH$ ($p$-toluenesulfoni

Vedclass · Accepted Answer

$CH_2=CH-CO-OCH_2-CH_2-CH_2-CH_3$

Vedclass · Answer

(A) The reaction is an acid-catalyzed transesterification. Methyl acrylate reacts with $n$-butyl alcohol in the presence of $TsOH$ ($p$-toluenesulfonic acid) to form $n$-butyl acrylate and methanol. Methanol,having the lowest boiling point $(65^\circ C)$,is removed by distillation to shift the equilibrium toward the formation of the product $(A)$ ($n$-butyl acrylate). $CH_2=CH-CO-OCH_3 + CH_3-CH_2-CH_2-CH_2-OH \xrightarrow{TsOH, \Delta} CH_2=CH-CO-OCH_2-CH_2-CH_2-CH_3 + CH_3OH$

$CH_2=CH-CO-OCH_3$ (methyl acrylate,bp $81^\circ C$) + $CH_3-CH_2-CH_2-CH_2-OH$ (n-butyl alcohol,bp $117^\circ C$) $\xrightarrow{TsOH, \Delta} (A)$ (bp $145^\circ C$) + $CH_3OH$ (bp $65^\circ C$). Product $(A)$ of the above reaction is:

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