Question

Complete each synthesis by giving missing starting material, reagent or products:
(I) Ethylbenzene \(\xrightarrow{\text{(i) KMnO}_4,\ \text{KOH};\ \text{(ii) H}_3\text{O}^{+}}\ ?\)
(II) Methylenecyclohexane \(\xrightarrow{\ ?\ }\) Cyclohexanecarbaldehyde
(III) 4-Oxocyclohexane-1-carbaldehyde \(\xrightarrow{[\text{Ag(NH}_3)_2]^{+}}\ ?\)

Hint:

(I) Alkaline KMnO₄ turns the whole alkyl chain into -COOH.
(II) Hydroboration-oxidation, then PCC.
(III) Tollens reagent oxidises only the -CHO group.

Answer 1

Step 1: Part (I), oxidise the side chain of ethylbenzene:
Hot alkaline \(\mathrm{KMnO_4}\) oxidises any alkyl side chain that has at least one benzylic hydrogen. The whole chain is cut down to a -COOH group, whatever its length. So the -\(\mathrm{CH_2CH_3}\) group becomes -COOK, and acid then frees the acid.
\[ \mathrm{C_6H_5CH_2CH_3 \xrightarrow{KMnO_4,\ KOH,\ heat} C_6H_5COOK \xrightarrow{H_3O^+} C_6H_5COOH} \]
Product of (I): benzoic acid.

Step 2: Part (II), see what must change:
Methylenecyclohexane has a ring carbon doubly bonded to \(\mathrm{CH_2}\). The target, cyclohexanecarbaldehyde, has the ring carbon joined to a -CHO group. The carbon count is the same, so we must add H to the ring carbon and turn the \(\mathrm{CH_2}\) end into -CHO.

Step 3: Part (II), choose the reagents:
First do hydroboration-oxidation. Boron adds to the less crowded \(\mathrm{CH_2}\) carbon, so the OH ends up on the terminal carbon (anti-Markovnikov). This gives a primary alcohol, cyclohexylmethanol.
\[ \text{Methylenecyclohexane} \xrightarrow{\text{(i) } B_2H_6;\ \text{(ii) } H_2O_2,\ OH^-} \text{Cyclohexylmethanol} \]
Then oxidise the primary alcohol with PCC in \(\mathrm{CH_2Cl_2}\). PCC is mild, so it stops at the aldehyde and does not go on to the acid.
\[ \text{Cyclohexylmethanol} \xrightarrow{\text{PCC, } CH_2Cl_2} \text{Cyclohexanecarbaldehyde} \]
Missing reagents of (II): (i) \(\mathrm{B_2H_6}\); (ii) \(\mathrm{H_2O_2/OH^-}\); (iii) PCC.

Step 4: Part (III), oxidise the aldehyde group only:
Tollens reagent, \([\mathrm{Ag(NH_3)_2}]^+\), is a mild oxidising agent. It oxidises an aldehyde to a carboxylate and is itself reduced to silver metal (silver mirror). A ketone is not touched by it. So only the -CHO group changes and the ring C=O stays.
\[ \mathrm{OHC{-}C_6H_{9}{=}O \xrightarrow{[Ag(NH_3)_2]^+} {}^{-}OOC{-}C_6H_{9}{=}O + Ag\downarrow} \]
On acidifying we get the free acid. Product of (III): 4-oxocyclohexane-1-carboxylic acid.

Final Answer:
(I) benzoic acid; (II) (i) \(\mathrm{B_2H_6}\), (ii) \(\mathrm{H_2O_2/OH^-}\), (iii) PCC; (III) 4-oxocyclohexane-1-carboxylic acid (with silver mirror).
\[ \boxed{\text{(I) } C_6H_5COOH;\ \text{(II) } B_2H_6,\ H_2O_2/OH^-,\ \text{PCC};\ \text{(III) 4-oxocyclohexane-1-carboxylic acid}} \]