Distinguish between pairs — aldehyde/ketone, 1°/2°/3° alcohol, acid/ester

Question

How do we chemically distinguish between: (a) aldehyde and ketone, (b) primary, secondary, and tertiary alcohols, (c) carboxylic acid and ester? List the test, reagent, and expected observation.

(CBSE 12 + JEE Main + NEET — asked almost every year)

Solution — Step by Step

Test	Reagent	Aldehyde	Ketone
Tollens’ test	$\text{Ag(NH}_3\text{)}_2^+$ (ammoniacal silver nitrate)	Silver mirror forms	No reaction
Fehling’s test	$\text{Cu}^{2+}$ in alkaline tartrate	Red precipitate ( $\text{Cu}_2\text{O}$ )	No reaction
Schiff’s test	Schiff’s reagent (fuchsin + $\text{SO}_2$ )	Pink/magenta colour	No colour

Aldehydes are easier to oxidise — they have an H on the carbonyl carbon that ketones lack.

Alcohol type	Lucas reagent ( $\text{ZnCl}_2$ + conc. HCl)	Time for turbidity
Tertiary (3°)	Immediate turbidity	Within 5 minutes
Secondary (2°)	Turbidity in 5-10 minutes	Moderate wait
Primary (1°)	No turbidity at room temperature	Requires heating

Turbidity appears because the alkyl chloride formed is insoluble. 3° alcohols react fastest because they form the most stable carbocation.

Test	Acid	Ester
$\text{NaHCO}_3$ test	Brisk effervescence ( $\text{CO}_2$ )	No reaction
Litmus test	Turns blue litmus red	No effect
Smell	Sharp, pungent	Fruity, pleasant

The $\text{NaHCO}_3$ test is definitive — only carboxylic acids (and some very strong phenols) release $\text{CO}_2$ .

flowchart TD
    A["Unknown Organic Compound"] --> B{"Contains C=O group?"}
    B -- Yes --> C["Tollens' or Fehling's test"]
    C -- "Silver mirror / red ppt" --> D["ALDEHYDE"]
    C -- "No reaction" --> E["KETONE"]
    B -- No --> F{"Contains -OH group?"}
    F -- Yes --> G["Lucas Test with ZnCl₂/HCl"]
    G -- "Immediate turbidity" --> H["3° ALCOHOL"]
    G -- "5-10 min turbidity" --> I["2° ALCOHOL"]
    G -- "No turbidity" --> J["1° ALCOHOL"]
    F -- No --> K{"Acidic or neutral?"}
    K --> L["NaHCO₃ test"]
    L -- "Effervescence" --> M["CARBOXYLIC ACID"]
    L -- "No reaction" --> N["ESTER"]

Why This Works

These tests exploit fundamental reactivity differences. Aldehydes are mild reducing agents (they get oxidised to acids), so they reduce $\text{Ag}^+$ to Ag (Tollens’) and $\text{Cu}^{2+}$ to $\text{Cu}^+$ (Fehling’s). Ketones cannot do this because oxidising a ketone would require breaking a C-C bond.

In the Lucas test, the $\text{S}_N1$ mechanism is faster for 3° alcohols (stable carbocation intermediate) than for 2° or 1°. The rate of reaction directly corresponds to carbocation stability: 3° > 2° > 1°.

Carboxylic acids donate $\text{H}^+$ to $\text{NaHCO}_3$ , producing $\text{CO}_2$ . Esters do not have an acidic proton, so no gas is released.

Alternative Method

For aldehyde vs ketone, you can also use the iodoform test — but this distinguishes methyl ketones from other ketones, not all aldehydes from ketones. Acetaldehyde ( $\text{CH}_3\text{CHO}$ ) and methyl ketones both give a positive iodoform test (yellow precipitate of $\text{CHI}_3$ ).

For NEET, the Tollens’ and Fehling’s tests are the most frequently tested. Remember: Tollens’ uses silver (silver mirror), Fehling’s uses copper (red precipitate). Both are positive for aldehydes, negative for ketones. Know the reagent compositions cold.

Common Mistake

Students write that Fehling’s test works for all aldehydes. It does NOT work for aromatic aldehydes like benzaldehyde — benzaldehyde gives a negative Fehling’s test (but positive Tollens’). This is because Fehling’s reagent is a weaker oxidising agent. For board exams, always specify “aliphatic aldehydes” when discussing Fehling’s test.