Solution:
There are two isomers of $C_3H_6O$. They are:
Propanal and Propanone
To distinguish these isomers, we can take Tollen's Test into account.
Propanal is an aldehyde. Thus, it reduces Tollen’s reagent. But, propanone being a ketone does not reduce Tollen’s reagent.Reaction: $CH_3CH_2CHO$ + 2[Ag(NH3)2]+
+ 3OH-à $CH_3CH_2COO^-$
+ Ag + 4NH3 + 2H2O
Lactic Acid can be prepared from ethanol (aldehyde). When ethanol is treated with HCN it gives Cyanohydrin on further hydrolysis it gives lactic Acid.
Reaction: Conversion of Propanol to Lactic Acid:
Firstly, We convert Propanol to Ethanol then on further addition of hydrogen cyanide to carbonyl group of aldehyde gives cyanohydrin and on hydrolysis of that cyanohydrin we get lactic acid.
$C{H_3} - C{H_2} - \mathop C\limits^{\mathop {||}\limits^O } - H\mathop {\mathop \to \limits_{[O]} }\limits^{KMn{O_4}/{H^ + }} C{H_3} - C{H_2} - COOH$
$C{H_3} - C{H_2} - COOH\mathop \to \limits^{NaOH} C{H_3} - C{H_2} - COONa$
$C{H_3} - C{H_2} - COONa\mathop \to \limits^{CaO + NaOH} C{H_3} - C{H_3} + N{a_2}C{O_3}$
$C{H_3} - C{H_3} + C{l_2}\mathop \to \limits^{Sunlight} C{H_3} - C{H_2} - Cl + HCl$
$C{H_3} - C{H_2} - Cl\mathop \to \limits^{aq.NaOH} C{H_3} - C{H_2} - OH + NaCl$
$C{H_3} - C{H_2} - OH\mathop \to \limits^{KMn{O_4}} C{H_3} - CHO$
$C{H_3} - CHO + HCN \to C{H_3} - \mathop {\mathop {{\rm{ }}C}\limits_{\mathop {{\rm{ }}|}\limits_{CN} } }\limits^{\mathop {{\rm{ }}|}\limits^{OH} } - H$
$C{H_3} - \mathop {\mathop {{\rm{ }}C}\limits_{\mathop {{\rm{ }}|}\limits_{CN} } }\limits^{\mathop {{\rm{ }}|}\limits^{OH} } - H\mathop \to \limits^{{H_2}O/{H^ + }} C{H_3} - \mathop {\mathop C\limits_{\mathop |\limits_H } }\limits^{\mathop |\limits^{OH} } - COOH + NH_4^ + $