Question

(i) Give a chemical test to distinguish between ethanal and propanal.

(ii) Which of the following acids is stronger and why ?

\( \mathrm{CH_2FCOOH} \) and \( \mathrm{CH_2ClCOOH} \)

(iii) Complete the given equation :

\( \mathrm{C_6H_5-C(=O)-O-C(=O)-CH_3} \xrightarrow{\mathrm{H_2O}} \)

Hint:

Only ethanal (a methyl carbonyl compound) gives the iodoform test. Fluorine has the stronger -I effect, so the fluoro acid is stronger, and hydrolysis of the anhydride gives benzoic and ethanoic acids.

Answer 1

Step 1: Part (i), find the structural difference:
Ethanal is \( \mathrm{CH_3CHO} \). Propanal is \( \mathrm{CH_3CH_2CHO} \).
Ethanal has a \( \mathrm{CH_3CO-} \) group, so it is a methyl carbonyl compound. Propanal does not have it.

Step 2: Part (i), the iodoform test:
Warm each compound with iodine and sodium hydroxide solution.
Ethanal gives a yellow precipitate of iodoform, \( \mathrm{CHI_3} \). Propanal gives no yellow precipitate.
\[ \mathrm{CH_3CHO} + 3\,\mathrm{I_2} + 4\,\mathrm{NaOH} \rightarrow \mathrm{CHI_3}\downarrow + \mathrm{HCOONa} + 3\,\mathrm{NaI} + 3\,\mathrm{H_2O} \]

Step 3: Part (ii), compare the two acids:
Fluorine is more electronegative than chlorine. It pulls electrons through the sigma bonds (the -I effect) more strongly.
This spreads the negative charge on the carboxylate ion and makes it more stable.
A more stable anion means a stronger acid.
So \( \mathrm{CH_2FCOOH} \) is the stronger acid.

Step 4: Part (iii), hydrolyse the anhydride:
The given compound is benzoic ethanoic anhydride (a mixed anhydride).
Water breaks the C-O-C link and gives the two parent acids.
\[ \mathrm{C_6H_5CO-O-COCH_3} + \mathrm{H_2O} \rightarrow \mathrm{C_6H_5COOH} + \mathrm{CH_3COOH} \]

Final Answer:
(i) Iodoform test: only ethanal gives yellow \( \mathrm{CHI_3} \).
(ii) \( \mathrm{CH_2FCOOH} \) is stronger because of the stronger -I effect of fluorine.
(iii) Benzoic acid and ethanoic acid are formed.