An alkene $A$ (molecular formula $C_{5}H_{10}$) on ozonolysis gives a mixture of two compounds $B$ and $C$. Compound $B$ gives a positive Fehling's test and also forms iodoform on treatment with $I_{2}$ and $NaOH$. Compound $C$ does not give Fehling's test but forms iodoform. Identify the compounds $A, B$ and $C$. Write the reaction for ozonolysis and the formation of iodoform from $B$ and $C$.

(A) The molecular formula $C_{5}H_{10}$ corresponds to an alkene. Ozonolysis of an alkene $R_{1}R_{2}C=CR_{3}R_{4}$ yields carbonyl compounds.
$B$ gives a positive Fehling's test,so it must be an aldehyde. Since it also gives the iodoform test,it must be acetaldehyde $(CH_{3}CHO)$.
$C$ does not give Fehling's test,so it is a ketone. Since it gives the iodoform test,it must be a methyl ketone $(CH_{3}COR)$. Given the total carbon count is $5$,$C$ is propanone $(CH_{3}COCH_{3})$.
Thus,the alkene $A$ is $2$-methylbut-$2$-ene $(CH_{3}-C(CH_{3})=CH-CH_{3})$.
Ozonolysis reaction:
$CH_{3}-C(CH_{3})=CH-CH_{3} \xrightarrow{O_{3}, Zn/H_{2}O} CH_{3}COCH_{3} (C) + CH_{3}CHO (B)$
Iodoform test:
$CH_{3}CHO + 3I_{2} + 4NaOH \rightarrow CHI_{3} + HCOONa + 3NaI + 3H_{2}O$
$CH_{3}COCH_{3} + 3I_{2} + 4NaOH \rightarrow CHI_{3} + CH_{3}COONa + 3NaI + 3H_{2}O$