CH_3-C(ONa)=CH_2 xrightarrow{HCequiv CH} xrig | vedclass

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(A) $1$. The reaction starts with the nucleophilic attack of the enolate $CH_3-C(O^-)=CH_2$ on acetylene $(HC\equiv CH)$,followed by protonation to fo

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$H_2C=C(CH_3)-CH=CH_2$

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(A) $1$. The reaction starts with the nucleophilic attack of the enolate $CH_3-C(O^-)=CH_2$ on acetylene $(HC\equiv CH)$,followed by protonation to form $CH_3-C(OH)(CH_3)-C\equiv CH$ (after tautomerization and rearrangement steps as shown in the mechanism).$2$. The partial hydrogenation of the alkyne using Lindlar's catalyst $(H_2/Pd-BaSO_4)$ reduces the triple bond to a double bond,yielding $CH_3-C(OH)(CH_3)-CH=CH_2$.$3$. Finally,dehydration using $Al_2O_3$ at high temperature removes a water molecule to form the conjugated diene,$H_2C=C(CH_3)-CH=CH_2$ (Isoprene).

$CH_3-C(ONa)=CH_2$ $\xrightarrow{HC\equiv CH}$ $\xrightarrow{H^{+}}$ $\xrightarrow[Pd-BaSO_4]{H_2}$ $\xrightarrow[\Delta]{Al_2O_3}$
End product of the reaction is

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$CH_3-C(ONa)=CH_2$ $\xrightarrow{HC\equiv CH}$ $\xrightarrow{H^{+}}$ $\xrightarrow[Pd-BaSO_4]{H_2}$ $\xrightarrow[\Delta]{Al_2O_3}$End product of the reaction is