Aldehydes, Ketones and Carboxylic Acids

The reaction is a dehydrohalogenation E2-elimination reaction. Elimination takes place in single step and proceed by formation of transition state from anti position.

CH₃COOK is most basic among the given salts.

$C{H}_{3}COOK + H_{2}O \stackrel{ }{\to } \underset{weak Acid}{C{H}_{3}COOH }+ \underset{Strong Base}{KOH}$

Among the given list of acids, the acidic strength increases on-

(i) Presence of electron with drawing group (i.e. - I group)

(ii) Closeness of electron withdrawing group from the -COOH group.

(iii) Number of electron with drawing groups (i.e. -I group)

Hence, the correct order among the given options will be I < III < II < IV

Alkaline KMnO₄ is called Baeyer's reagent. It is used to test unsaturation in an organic compound, when it decolorises KMnO₄.

CH₂ = CH₂ + $\underset{\mathrm{Alk}. {\mathrm{KMnO}}_4}{\underbrace{{\mathrm{H}}_2\mathrm{O} + \mathrm{O}}}$ → $\underset{\mathrm{OH}}{\underset{|}{\mathrm{C}}}{\mathrm{H}}_2-\underset{\mathrm{OH}}{\underset{|}{\mathrm{C}}}{\mathrm{H}}_2$

When two moles of benzaldehyde react with one mole of acetone. It shows aldol condensation as follows-

Therefore, among all the given options, option c is the correct one.