The compound A in the following reaction is:A | vedclass

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(A) The product formed from $D$ by the action of $Ba(OH)_2$ and $\Delta$ is mesityl oxide $(CH_3-C(CH_3)=CH-CO-CH_3)$,which is a self-aldol condensati

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$C_6H_5-CH_2-CO-CH_3$

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(A) The product formed from $D$ by the action of $Ba(OH)_2$ and $\Delta$ is mesityl oxide $(CH_3-C(CH_3)=CH-CO-CH_3)$,which is a self-aldol condensation product of acetone. Therefore,$D$ is acetone $(CH_3-CO-CH_3)$.Since ozonolysis of $B$ gives $C$ and $D$ (acetone),$B$ must contain the $(CH_3)_2C=$ group.When $A$ $(C_6H_5-CH_2-CO-CH_3)$ reacts with $CH_3MgBr$ followed by dehydration with $Conc. H_2SO_4$,it forms $B$ $(C_6H_5-CH=C(CH_3)_2)$.$C_6H_5-CH_2-CO-CH_3$ $\xrightarrow{CH_3MgBr} C_6H_5-CH_2-C(OH)(CH_3)_2$ $\xrightarrow{H^{+}, \Delta} C_6H_5-CH=C(CH_3)_2 (B)$.Ozonolysis of $B$: $C_6H_5-CH=C(CH_3)_2 \xrightarrow{O_3/Zn, H_2O} C_6H_5-CHO (C) + CH_3-CO-CH_3 (D)$.

The compound $A$ in the following reaction is:
$A$ $\xrightarrow[(ii) Conc. H_2SO_4 / \Delta]{(i) CH_3MgBr / H_2O} B$ $\xrightarrow[(ii) Zn / H_2O]{(i) O_3} C + D$
$D \xrightarrow[Ba(OH)_2, \Delta]{} CH_3-C(CH_3)=CH-CO-CH_3$

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The compound $A$ in the following reaction is:$A$ $\xrightarrow[(ii) Conc. H_2SO_4 / \Delta]{(i) CH_3MgBr / H_2O} B$ $\xrightarrow[(ii) Zn / H_2O]{(i) O_3} C + D$$D \xrightarrow[Ba(OH)_2, \Delta]{} CH_3-C(CH_3)=CH-CO-CH_3$