What will be the main product in the followin | vedclass

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(B) The given reaction is a $Cross-Aldol$ condensation reaction between $benzaldehyde$ $(C_6H_5CHO)$ and $acetaldehyde$ $(CH_3CHO)$ in the presence of

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$C_6H_5CH=CHCHO$

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(B) The given reaction is a $Cross-Aldol$ condensation reaction between $benzaldehyde$ $(C_6H_5CHO)$ and $acetaldehyde$ $(CH_3CHO)$ in the presence of a dilute base $(OH^-)$ and heat $(\Delta)$.$1.$ $Acetaldehyde$ contains $\alpha$-hydrogens,so it forms an enolate ion in the presence of a base.$2.$ This enolate ion attacks the carbonyl carbon of $benzaldehyde$ (which has no $\alpha$-hydrogens) to form a $\beta$-hydroxy aldehyde intermediate $(C_6H_5CH(OH)CH_2CHO)$.$3.$ Upon heating $(\Delta)$,this intermediate undergoes dehydration (loss of $H_2O$) to form the stable $\alpha,\beta$-unsaturated aldehyde,which is $cinnamaldehyde$ $(C_6H_5CH=CHCHO)$.Therefore,the main product is $C_6H_5CH=CHCHO$.

List-$I$ (Reactions)	List-$II$ (Reagents)
$(A)$ $CH_3(CH_2)_5COOC_2H_5 \rightarrow CH_3(CH_2)_5CHO$	$(I)$ $CH_3MgBr, H_2O$
$(B)$ $C_6H_5COC_6H_5 \rightarrow C_6H_5CH_2C_6H_5$	$(II)$ $Zn(Hg)$ and conc. $HCl$
$(C)$ $C_6H_5CHO \rightarrow C_6H_5CH(OH)CH_3$	$(III)$ $NaBH_4, H^+$
$(D)$ $CH_3COCH_2COOC_2H_5 \rightarrow CH_3CH(OH)CH_2COOC_2H_5$	$(IV)$ $DIBAL-H, H_2O$

What will be the main product in the following reaction?
$C_6H_5CHO + CH_3CHO \xrightarrow{OH^-, \Delta} ?$

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What will be the main product in the following reaction?$C_6H_5CHO + CH_3CHO \xrightarrow{OH^-, \Delta} ?$