What is (P) in the reaction given below?(P) x | vedclass

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(C) The final product is $4	ext{-methylpent-3-en-2-one}$,which is an $\alpha,\beta	ext{-unsaturated ketone}$ formed by the aldol condensation of $2$

Vedclass · Accepted Answer

Ethane nitrile

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(C) The final product is $4	ext{-methylpent-3-en-2-one}$,which is an $\alpha,\beta	ext{-unsaturated ketone}$ formed by the aldol condensation of $2$ moles of acetone $(propanone)$.Thus,$(R)$ must be $propanone$ $(CH_3COCH_3)$.The reaction of $(P)$ with $CH_3MgBr$ followed by $H_3O^{+}$ yields $propanone$. This is a characteristic reaction of nitriles.$CH_3CN + CH_3MgBr$ $ightarrow CH_3C(CH_3)=NMgBr$ $\xrightarrow{H_3O^{+}} CH_3COCH_3 + NH_3 + Mg(OH)Br$.Therefore,$(P)$ is $CH_3CN$ $(Ethane\ nitrile)$.

What is $(P)$ in the reaction given below?
$(P)$ $\xrightarrow[\text{(ii) } H_3O^{+}]{\text{(i) } CH_3MgBr} (R)$ $\xrightarrow[\text{(ii) } \Delta]{\text{(i) dil. } NaOH} \text{4-methylpent-3-en-2-one}$

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What is $(P)$ in the reaction given below?$(P)$ $\xrightarrow[\text{(ii) } H_3O^{+}]{\text{(i) } CH_3MgBr} (R)$ $\xrightarrow[\text{(ii) } \Delta]{\text{(i) dil. } NaOH} \text{4-methylpent-3-en-2-one}$