Question

(i) Give a chemical test to distinguish between Propanal and Propanone.

(ii) Which of the given acids is stronger and why ? \( \mathrm{F_3C{-}C_6H_4{-}COOH} \) or \( \mathrm{H_3C{-}C_6H_4{-}COOH} \)

(iii) Complete the given reaction sequence :

\[ \mathrm{CH_3CH_2CH_2COOC_2H_5} \xrightarrow{\mathrm{NaOH}} \mathrm{A + C_2H_5OH} \]

\[ \mathrm{A} \xrightarrow{\mathrm{H_3O^+}} \mathrm{B} \]

Hint:

Use Tollens test for the aldehyde, the electron withdrawing CF3 group for acid strength, and saponification followed by acidification for the sequence.

Answer 1

Step 1: Understanding the Concept:
The question has three short parts. They test aldehyde versus ketone tests, the effect of substituents on acid strength, and basic hydrolysis of an ester.
In (ii), both acids have the \(\mathrm{-COOH}\) group on a benzene ring, with the other group at the para position.

Step 2: Part (i), Tollens test:
Propanal is an aldehyde. It is oxidised easily by Tollens reagent, an ammoniacal silver nitrate solution.
Warm propanal with it and a bright silver mirror forms on the walls of the tube.
\[ \mathrm{CH_3CH_2CHO + 2[Ag(NH_3)_2]^+ + 3OH^- \rightarrow CH_3CH_2COO^- + 2Ag + 4NH_3 + 2H_2O} \]
Propanone is a ketone. It is not oxidised by this mild reagent, so no silver mirror forms. (Fehling solution test also works: only propanal gives a red-brown precipitate.)

Step 3: Part (ii), compare the two acids:
An acid is stronger when its conjugate base (the carboxylate ion) is more stable.
The \(\mathrm{-CF_3}\) group has a strong electron withdrawing inductive effect (-I). It pulls electron density away from the ring and the carboxylate ion. This spreads out the negative charge and stabilises the ion.
The \(\mathrm{-CH_3}\) group has an electron releasing effect (+I). It pushes charge towards the carboxylate and makes it less stable.
So \(\mathrm{F_3C{-}C_6H_4{-}COOH}\) is the stronger acid.

Step 4: Part (iii), find A:
NaOH hydrolyses the ester (saponification). The acid part becomes its sodium salt and the alcohol part leaves as ethanol.
\[ \mathrm{CH_3CH_2CH_2COOC_2H_5 + NaOH \rightarrow CH_3CH_2CH_2COONa + C_2H_5OH} \]
So A is sodium butanoate, \(\mathrm{CH_3CH_2CH_2COONa}\).

Step 5: Part (iii), find B:
A strong acid (\(\mathrm{H_3O^+}\)) protonates the carboxylate salt and releases the free carboxylic acid.
\[ \mathrm{CH_3CH_2CH_2COONa + H_3O^+ \rightarrow CH_3CH_2CH_2COOH + Na^+ + H_2O} \]
So B is butanoic acid, \(\mathrm{CH_3CH_2CH_2COOH}\).

Final Answer:
(i) Tollens test: propanal gives a silver mirror, propanone does not. (ii) The \(\mathrm{CF_3}\) acid is stronger because of the -I effect. (iii) A is sodium butanoate and B is butanoic acid.
\[ \boxed{\mathrm{A = CH_3CH_2CH_2COONa,\ B = CH_3CH_2CH_2COOH}} \]