Consider the following reaction:text{Propanal | vedclass

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(C) $1$. The reaction of $	ext{Propanal} (CH_3CH_2CHO)$ and $	ext{Methanal} (HCHO)$ in the presence of $	ext{dil. NaOH}$ followed by heating $(\Del

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racemic mixture and gives a gas with saturated $	ext{NaHCO}_3$ solution

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(C) $1$. The reaction of $	ext{Propanal} (CH_3CH_2CHO)$ and $	ext{Methanal} (HCHO)$ in the presence of $	ext{dil. NaOH}$ followed by heating $(\Delta)$ undergoes a cross-aldol condensation to form $2-	ext{methylacrolein}$ $(CH_2=C(CH_3)CHO)$.$2$. The subsequent reaction with $	ext{NaCN}$ followed by acid hydrolysis $(	ext{H}_3	ext{O}^ )$ is a cyanohydrin formation reaction,which converts the aldehyde group $(-CHO)$ into a carboxylic acid group $(-COOH)$ with an adjacent hydroxyl group $(-OH)$.$3$. The final product $B$ is $2-	ext{hydroxy}-2-	ext{methylbut}-3-	ext{enoic acid}$ (or a similar structure depending on the specific pathway,but the key is the presence of the $-COOH$ group and the chiral center).$4$. The product contains a chiral center (marked with $*$ in the mechanism),and since the reaction conditions do not involve chiral catalysts,it is formed as a racemic mixture.$5$. The presence of the carboxylic acid group $(-COOH)$ ensures that it reacts with $	ext{NaHCO}_3$ to release $	ext{CO}_2$ gas.

List-$I$ (Test)	List-$II$ (Identification)
$A$. Bayer's test	$I$. Phenol
$B$. Ceric ammonium nitrate test	$II$. Aldehyde
$C$. Phthalein dye test	$III$. Alcoholic-$OH$ group
$D$. Schiff's test	$IV$. Unsaturation

Consider the following reaction:
$\text{Propanal} + \text{Methanal}$ $\xrightarrow[\text{(ii)}\Delta, \text{(iii)}\text{NaCN}, \text{(iv)}\text{H}_3\text{O}^+]{\text{(i)}\text{dil. NaOH}} \text{Product } B (C_5H_8O_3)$
The correct statement for product $B$ is:

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Consider the following reaction:$\text{Propanal} + \text{Methanal}$ $\xrightarrow[\text{(ii)}\Delta, \text{(iii)}\text{NaCN}, \text{(iv)}\text{H}_3\text{O}^+]{\text{(i)}\text{dil. NaOH}} \text{Product } B (C_5H_8O_3)$The correct statement for product $B$ is: