$CH_3-CH(OH)-CH_2-CHO$
$3HCHO + A \xrightarrow[40^o C]{Na_2CO_3} (B) (82\%)$
Product $(B)$ of the above reaction is

$A$ compound $A$ with molecular formula $C_{10}H_{13}Cl$ gives a white precipitate on adding silver nitrate solution. $A$ on reacting with alcoholic $KOH$ gives compound $B$ as the main product. $B$ on ozonolysis gives $C$ and $D$. $C$ gives Cannizzaro reaction but not aldol condensation. $D$ gives aldol condensation but not Cannizzaro reaction. $A$ is:

What will be the compound $X$ in the given reaction?
$MeO-C_6H_4-CHO + (X) \xrightarrow{CH_3COONa, H_3O^+} MeO-C_6H_4-CH=CHCOOH$

The product $(Q)$ will be:

$CH_3CH=CHCHO$ is oxidized to $CH_3CH=CHCOOH$ using

Which of the following does not give iodoform reaction?