Question

Formaldehyde can be distinguished from acetaldehyde by:
A.Schiff’s reagent
B.Tollen’s reagent
C.Fehling’s solution
D.Iodine in presence of base

Answer 1

In this question, to distinguish one from the other, use the reactions to observe some physical change. These physical changes may include change of color of solution, formation of a precipitate, or the generation of some kind of effervescence, etc. Also, check both the compounds as the two compounds may generate a similar kind of result.

Complete answer: On treatment of formaldehyde and acetaldehyde with iodine in the presence of base, acetaldehyde gives off yellow color precipitation while formaldehyde does not react with it. This is known as iodoform reaction and the test is called iodoform test.
Hence, it can be said that acetaldehyde $(C{H_3}CHO)$ gives a positive iodoform test while formaldehyde gives a negative iodoform test.
$C{H_3}CHO + 3{I_2} + 4NaOH \to C{H_3}CO{O^ - }N{a^ + } + 3NaI + 3{H_2}O + CH{I_3}$ (yellow ppt)
$HCHO + 3{I_2} + 4NaOH \to$ No reaction
Both formaldehyde and acetaldehyde give positive Schiff’s test, Tollen’s test and Fehling’s test. Therefore, the correct option is D.
Iodoform reaction is used to prove the presence of $R - CO - C{H_3}$ a group of carbonyls and $R - CH(OH) - C{H_3}$ alcohol. In this reaction, oxidation of carbonyl and alcohol groups takes place. By oxidation, they form a carboxylic acid and a yellow precipitate. This yellow precipitate indicates that a reaction has occurred.

Note:
If in place of iodine, chlorine is used, then it is called a chloroform test. In this reaction, precipitation of white chloroform is observed. If bromine is used instead of iodine or chlorine, then it is called a bromoform test. In this reaction, precipitation of pale yellow bromoform is formed. These types of reactions are collectively called haloform reactions.