Q35

5 marksLong AnswerSection E

(a) (i) An organic compound (X) has the molecular formula $C_5H_{10}O$ . Draw structures for (X) if it: (3 + 2)

(I) does not give Tollen's test but gives a positive iodoform test.

(II) does not give Tollen's test and iodoform test but undergoes Aldol condensation.

(III) undergoes Cannizzaro's reaction.

(ii) Show how each of the following compounds can be converted to benzoic acid:

(I) Acetophenone (II) Ethyl benzene

OR

(b) Answer the following questions: $(5 \times 1)$

(i) Draw structure of the 2, 4-dinitrophenyl hydrazone derivative of benzaldehyde.

(ii) Arrange the following in increasing order of their reactivity towards HCN: Di-tert. butyl ketone, Acetaldehyde, Acetone

(iii) Give a simple chemical test to distinguish between benzoic acid and ethyl benzoate.

(iv) Write the name of the reagent to convert Ethanenitrile to Ethanal.

(v) Draw the structure of 'X' in the following reaction: Cyclohexanol —( $CrO_3$ )→ 'X'

Q: (a) (i) An organic compound (X) has the molecular formula $C_5H_{10}O$. Draw structures for (X) if it: (3 + 2) (I) does not give Tollen's test but gives a positive iodoform test. (II) does not give Tollen's test and iodoform test but undergoes Aldol condensation. (III) undergoes Cannizzaro's reaction. (ii) Show how each of the following compounds can be converted to benzoic acid: (I) Acetophenone (II) Ethyl benzene OR (b) Answer the following questions: $(5 \times 1)$ (i) Draw structure of the 2, 4-dinitrophenyl hydrazone derivative of benzaldehyde. (ii) Arrange the following in increasing order of their reactivity towards HCN: Di-tert. butyl ketone, Acetaldehyde, Acetone (iii) Give a simple chemical test to distinguish between benzoic acid and ethyl benzoate. (iv) Write the name of the reagent to convert Ethanenitrile to Ethanal. (v) Draw the structure of 'X' in the following reaction: Cyclohexanol —($CrO_3$)→ 'X'

**Primary answer — part (a)** **(a)(i) Structures of $C_5H_{10}O$** (one degree of unsaturation = one $C=O$ group) - **(I) No Tollens', positive iodoform → a methyl ketone:** **Pentan-2-one, CH₃–CO–CH₂–CH₂–CH₃** (also acceptable: 3-methylbutan-2-one, CH₃–CO–CH(CH₃)₂). It has the CH₃CO– unit needed for iodoform but, being a ketone, gives no Tollens' test. - **(II) No Tollens', no iodoform, undergoes Aldol → a non-methyl ketone with α-hydrogen:** **Pentan-3-one, CH₃CH₂–CO–CH₂CH₃**. Not a methyl ketone (no iodoform), a ketone (no Tollens'), but has α-H for aldol. - **(III) Undergoes Cannizzaro → an aldehyde with NO α-hydrogen:** **2,2-Dimethylpropanal (pivaldehyde), (CH₃)₃C–CHO**. **(a)(ii) Conversions to benzoic acid** - **(I) Acetophenone:** $C_6H_5COCH_3$ —(alkaline $KMnO_4$, Δ; then $H_3O^+$)→ **$C_6H_5COOH$**. The oxidant degrades the –COCH₃ side chain to –COOH. - **(II) Ethylbenzene:** $C_6H_5CH_2CH_3$ —(alkaline $KMnO_4$, Δ; then $H_3O^+$)→ **$C_6H_5COOH$**. Any alkyl side chain with a benzylic H is oxidised fully to –COOH. **OR — part (b) (brief)** 1. **(i)** 2,4-DNP of benzaldehyde: **C₆H₅–CH=N–NH–C₆H₃(NO₂)₂** (the N is attached to a 2,4-dinitrophenyl ring). 2. **(ii)** Increasing reactivity towards $HCN$: **Di-tert-butyl ketone < Acetone < Acetaldehyde** (steric + electronic effects; aldehyde most reactive). 3. **(iii)** Add **$NaHCO_3$**: benzoic acid gives **brisk effervescence ($CO_2$)**; ethyl benzoate does not. 4. **(iv)** Reagent: **DIBAL-H** (di-isobutylaluminium hydride), then $H_2O$ — (Stephen reduction, $SnCl_2$/HCl, also acceptable). 5. **(v)** Cyclohexanol —($CrO_3$)→ **Cyclohexanone** (X).

Q: Why is 2,2-dimethylpropanal the answer for the Cannizzaro clue?

Cannizzaro reaction occurs only for aldehydes that lack an α-hydrogen. Among $C_5H_{10}O$ aldehydes, (CH₃)₃C–CHO (2,2-dimethylpropanal) has no α-H, so it disproportionates instead of undergoing aldol.

Q: How do you distinguish benzoic acid from ethyl benzoate?

Add aqueous sodium bicarbonate ($NaHCO_3$): benzoic acid, being a carboxylic acid, gives brisk effervescence of $CO_2$, while the ester ethyl benzoate gives no reaction.

Q: Which reagent converts ethanenitrile to ethanal?

DIBAL-H (di-isobutylaluminium hydride) followed by hydrolysis reduces $CH_3CN$ to $CH_3CHO$; the Stephen reduction ($SnCl_2$/HCl, then $H_3O^+$) is an alternative.

Aldehydes, Ketones and Carboxylic Acids

Structure elucidation of C₅H₁₀O and conversions to benzoic acid

Official Answer

Primary answer — part (a)

(a)(i) Structures of $C_5H_{10}O$ (one degree of unsaturation = one $C=O$ group)

(I) No Tollens', positive iodoform → a methyl ketone:

Pentan-2-one, CH₃–CO–CH₂–CH₂–CH₃ (also acceptable: 3-methylbutan-2-one, CH₃–CO–CH(CH₃)₂). It has the CH₃CO– unit needed for iodoform but, being a ketone, gives no Tollens' test.

(II) No Tollens', no iodoform, undergoes Aldol → a non-methyl ketone with α-hydrogen:

Pentan-3-one, CH₃CH₂–CO–CH₂CH₃. Not a methyl ketone (no iodoform), a ketone (no Tollens'), but has α-H for aldol.

(III) Undergoes Cannizzaro → an aldehyde with NO α-hydrogen:

2,2-Dimethylpropanal (pivaldehyde), (CH₃)₃C–CHO.

(a)(ii) Conversions to benzoic acid

(I) Acetophenone: $C_6H_5COCH_3$ —(alkaline $KMnO_4$ , Δ; then $H_3O^+$ )→ $C_6H_5COOH$ . The oxidant degrades the –COCH₃ side chain to –COOH.
(II) Ethylbenzene: $C_6H_5CH_2CH_3$ —(alkaline $KMnO_4$ , Δ; then $H_3O^+$ )→ $C_6H_5COOH$ . Any alkyl side chain with a benzylic H is oxidised fully to –COOH.

OR — part (b) (brief)

(i) 2,4-DNP of benzaldehyde: C₆H₅–CH=N–NH–C₆H₃(NO₂)₂ (the N is attached to a 2,4-dinitrophenyl ring).
(ii) Increasing reactivity towards $HCN$ : Di-tert-butyl ketone < Acetone < Acetaldehyde (steric + electronic effects; aldehyde most reactive).
(iii) Add $NaHCO_3$ : benzoic acid gives brisk effervescence ( $CO_2$ ); ethyl benzoate does not.
(iv) Reagent: DIBAL-H (di-isobutylaluminium hydride), then $H_2O$ — (Stephen reduction, $SnCl_2$ /HCl, also acceptable).
(v) Cyclohexanol —( $CrO_3$ )→ Cyclohexanone (X).

C₅H₁₀O structureiodoform test methyl ketoneTollens testaldol condensation alpha hydrogenCannizzaro reactionpentan-2-one2,2-dimethylpropanalalkaline KMnO₄ benzoic acid

Marking Scheme

1(a)(i) 3 marks: 1 mark each for correct structures — (I) pentan-2-one (methyl ketone), (II) pentan-3-one (non-methyl ketone with α-H), (III) 2,2-dimethylpropanal (α-H-free aldehyde).
2(a)(ii) 2 marks: 1 mark each — acetophenone → benzoic acid using alkaline $KMnO_4$ /Δ; ethylbenzene → benzoic acid using alkaline $KMnO_4$ /Δ (reagent and product both required).
3OR (b) $5 \times 1$ : (i) correct 2,4-DNP structure; (ii) di-tert-butyl ketone < acetone < acetaldehyde; (iii) $NaHCO_3$ effervescence test; (iv) DIBAL-H (or $SnCl_2$ /HCl, Stephen reduction); (v) cyclohexanone.

Hint

Decode each test: no Tollens' = ketone (or α-H-free aldehyde for Cannizzaro); iodoform = methyl ketone; aldol = needs α-H; Cannizzaro = aldehyde with no α-H. For benzoic acid, oxidise any benzylic side chain with alkaline $KMnO_4$ .

Quick Oral Answer

Since $C_5H_{10}O$ has one C=O, I use the tests as filters: pentan-2-one is the iodoform-positive methyl ketone, pentan-3-one is the non-methyl ketone that still does aldol, and 2,2-dimethylpropanal is the α-hydrogen-free aldehyde that gives Cannizzaro; and alkaline $KMnO_4$ oxidises both acetophenone and ethylbenzene straight to benzoic acid.

Analysis & Explanation

Concept — reading the tests as clues

Each chemical test narrows the structure of $C_5H_{10}O$ (which must contain one $C=O$ since the degree of unsaturation is 1).

Tollens' distinguishes aldehyde (positive) from ketone (negative).
Iodoform is positive only for CH₃CO– (methyl ketone) or CH₃CH(OH)– groups.
Aldol needs at least one α-hydrogen.
Cannizzaro is given only by aldehydes with NO α-hydrogen.

Combining these: a methyl ketone (pentan-2-one) for (I); a symmetrical non-methyl ketone with α-H (pentan-3-one) for (II); an α-H-free aldehyde (2,2-dimethylpropanal) for (III).

Exam trap

For (III) students often draw pentanal, but pentanal has α-hydrogens and would undergo aldol, not Cannizzaro. Only a branched aldehyde like (CH₃)₃C–CHO has no α-H.
In (a)(ii), remember the side chain length does not matter — oxidation always trims a benzylic alkyl group down to a single –COOH.

Real-world

Side-chain oxidation with $KMnO_4$ is the industrial route from toluene/ethylbenzene feedstocks to benzoic acid, a common food preservative (E210) and precursor of phenol and caprolactam.

Common Mistakes

1Drawing pentanal for the Cannizzaro case — pentanal has α-hydrogens, so it undergoes aldol, not Cannizzaro; the correct answer needs an α-H-free aldehyde like (CH₃)₃C–CHO.
2Choosing an aldehyde for the 'no Tollens'' clues — parts (I) and (II) must be ketones.
3Forgetting the oxidising conditions (alkaline/acidic $KMnO_4$ , Δ) when converting acetophenone or ethylbenzene to benzoic acid, or thinking a longer side chain gives a longer acid.

Interesting Facts

The Cannizzaro reaction (S. Cannizzaro, 1853) is a disproportionation: one aldehyde molecule is oxidised to the acid while another is reduced to the alcohol.

The iodoform test gives a pale-yellow precipitate of $CHI_3$ with a characteristic antiseptic smell and was historically used to detect ethanol and acetone in body fluids.

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Frequently Asked Questions

Why is 2,2-dimethylpropanal the answer for the Cannizzaro clue?

Cannizzaro reaction occurs only for aldehydes that lack an α-hydrogen. Among $C_5H_{10}O$ aldehydes, (CH₃)₃C–CHO (2,2-dimethylpropanal) has no α-H, so it disproportionates instead of undergoing aldol.

How do you distinguish benzoic acid from ethyl benzoate?

Add aqueous sodium bicarbonate ( $NaHCO_3$ ): benzoic acid, being a carboxylic acid, gives brisk effervescence of $CO_2$ , while the ester ethyl benzoate gives no reaction.

Which reagent converts ethanenitrile to ethanal?

DIBAL-H (di-isobutylaluminium hydride) followed by hydrolysis reduces $CH_3CN$ to $CH_3CHO$ ; the Stephen reduction ( $SnCl_2$ /HCl, then $H_3O^+$ ) is an alternative.