(i) STRUCTURES OF (X), MOLECULAR FORMULA $C_5H_{10}O$

Preliminary - degree of unsaturation.
For $C_5H_{10}O$: $\text{DoU} = \dfrac{2(5)+2-10}{2} = 1$, which corresponds to ONE $C=O$
group. So (X) is a saturated aldehyde or ketone.

(I) DOES NOT GIVE TOLLEN'S TEST BUT GIVES A POSITIVE IODOFORM TEST

Step 1 - Interpret 'no Tollen's test'.
Tollen's reagent (ammoniacal $AgNO_3$) oxidises ALDEHYDES to acids, giving a silver
mirror. A negative test means (X) is NOT an aldehyde $\Rightarrow$ it must be a KETONE.

Step 2 - Interpret 'positive iodoform test'.
The iodoform test ($I_2$ + NaOH) is positive only for a METHYL KETONE, i.e. a compound
containing the $CH_3CO-$ group (or $CH_3CH(OH)-$).

Step 3 - Combine.
(X) must be a $C_5$ methyl ketone.
$$\textbf{(X)} = CH_3-CO-CH_2-CH_2-CH_3 \qquad \textbf{PENTAN-2-ONE}$$
$$CH_3COCH_2CH_2CH_3 + 3I_2 + 4NaOH \rightarrow CHI_3\downarrow + CH_3CH_2CH_2COONa + 3NaI + 3H_2O$$
(3-methylbutan-2-one, $CH_3COCH(CH_3)_2$, is also acceptable.)

(II) NO TOLLEN'S TEST, NO IODOFORM TEST, BUT UNDERGOES ALDOL CONDENSATION

Step 1 - No Tollen's test $\Rightarrow$ it is a KETONE, not an aldehyde.
Step 2 - No iodoform test $\Rightarrow$ it must NOT contain a $CH_3CO-$ group.
Step 3 - Undergoes Aldol condensation $\Rightarrow$ it MUST have at least one
ALPHA-HYDROGEN.
Step 4 - Combine: a symmetrical $C_5$ ketone with ethyl groups on both sides.
$$\textbf{(X)} = CH_3-CH_2-CO-CH_2-CH_3 \qquad \textbf{PENTAN-3-ONE}$$
It has alpha-hydrogens on both sides, so aldol condensation is possible, but there is no
$CH_3CO-$ group, so the iodoform test is negative.

(III) UNDERGOES CANNIZZARO'S REACTION

Step 1 - Recall the requirement.
Cannizzaro's reaction (disproportionation with concentrated alkali into an alcohol and a
carboxylate salt) is given ONLY by aldehydes that have NO ALPHA-HYDROGEN.

Step 2 - Build a $C_5H_{10}O$ aldehyde with no alpha-hydrogen.
The carbon next to $-CHO$ must carry no hydrogen $\Rightarrow$ it must be a quaternary
carbon.
$$\textbf{(X)} = (CH_3)_3C-CHO \qquad \textbf{2,2-DIMETHYLPROPANAL (trimethylacetaldehyde)}$$

Step 3 - Write its Cannizzaro reaction.
$$2(CH_3)_3C-CHO \xrightarrow{\text{conc. NaOH}} (CH_3)_3C-CH_2OH + (CH_3)_3C-COO^-Na^+$$
(one molecule is reduced to the alcohol, the other oxidised to the acid salt)

(ii) CONVERSIONS TO BENZOIC ACID

(I) ACETOPHENONE $\rightarrow$ BENZOIC ACID

Method 1 - Haloform (iodoform) reaction.
Acetophenone contains a $CH_3CO-$ group attached to the ring, so it responds to the
haloform reaction. The three hydrogens of the methyl group are replaced by iodine and
the $C-C$ bond is then cleaved by hydroxide.
$$C_6H_5COCH_3 + 3I_2 + 4NaOH \longrightarrow C_6H_5COO^-Na^+ + CHI_3\downarrow + 3NaI + 3H_2O$$
$$C_6H_5COO^-Na^+ \xrightarrow{H_3O^+} C_6H_5COOH$$

Method 2 - Oxidation.
$$C_6H_5COCH_3 \xrightarrow[\Delta]{\text{alk. } KMnO_4} C_6H_5COO^-K^+ \xrightarrow{H_3O^+} C_6H_5COOH$$

(II) ETHYLBENZENE $\rightarrow$ BENZOIC ACID

Step 1 - Principle.
Any alkylbenzene that has at least one BENZYLIC HYDROGEN is oxidised by a strong
oxidising agent. No matter how long the side chain is, the WHOLE SIDE CHAIN is oxidised
down to a single $-COOH$ group attached to the ring.

Step 2 - Write the reaction.
$$C_6H_5CH_2CH_3 \xrightarrow[\Delta]{\text{alkaline } KMnO_4} C_6H_5COO^-K^+ \xrightarrow{H_3O^+} C_6H_5COOH$$
(Acidic $KMnO_4$ or chromic acid, $K_2Cr_2O_7/H_2SO_4$, may also be used.)

Step 3 - Note.
The benzene ring itself is not attacked because it is aromatic and very stable; only the
side chain is oxidised.