Carboxylic acids preparation

Oxidation-state routes (no new carbon added). A primary alcohol sits three oxidation steps below the acid; an aldehyde sits just one step below it. Strong, often hot, oxidants drive the alcohol all the way to the acid, while the aldehyde — already close — needs only a mild oxidant:

where: $\text{R}$ is any alkyl or aryl-attached carbon chain; the reagent above the alcohol arrow must be strong and usually heated, while the reagent below the aldehyde arrow only needs to be mildly oxidising, since the aldehyde starts one step higher already. For the aromatic case, the same strong oxidant attacks the benzylic side chain of an alkylbenzene — never the ring itself — and degrades it, carbon by carbon, until only the ring-attached carbon survives as $\text{–COOH}$:

Carbon-adding routes (chain grows by one carbon). A nitrile hydrolyses, through an amide intermediate, into the acid — the nitrile carbon itself becomes the new carboxyl carbon:

A Grignard reagent adds across the carbon of dry carbon dioxide, and acid workup liberates the acid; this route also works for aromatic acids from an aryl Grignard reagent:

Every one of the four routes can be traced back to one bookkeeping question — did this step raise the oxidation state of an existing carbon, or did it splice in a brand-new carbon — and getting that distinction right is the difference between correctly predicting the product and miscounting carbons on the acid.

Step 1. Fix the oxidation-state picture. A primary alcohol carbon (bonded to one O and two H, alongside R) is at a lower oxidation level than an aldehyde carbon (one O double bond, one H), which is in turn lower than a carboxylic acid carbon (two bonds to oxygen, no H left on that carbon). Climbing from alcohol to acid therefore crosses two full oxidation steps, which is why it needs a strong, often hot, excess oxidant; climbing from aldehyde to acid crosses only one step, so a mild oxidant such as Tollens' reagent suffices — the same reagent that is too weak to touch the alcohol directly.

Step 2. Fix why the aromatic side chain always collapses to the same product. Hot alkaline permanganate attacks the weakest C–H bonds in the molecule, and the benzylic hydrogens (attached to the carbon nearest the ring) are weakened by conjugation with the aromatic ring, so the oxidant strips the chain back one carbon at a time. It cannot stop partway and it cannot touch the ring itself (the ring is far more oxidation-resistant than an open chain), so the degradation runs all the way down to the single carbon still bonded to the ring, which becomes $\text{–COOH}$. This is why toluene, ethylbenzene, or any longer alkylbenzene all give the same product, benzoic acid.

Step 3. Fix the carbon-adding routes. Both the nitrile route and the Grignard route start from an alkyl or aryl halide and insert one new carbon supplied entirely by the reagent — the cyanide carbon in $\text{KCN}$, or the carbon of $\text{CO}_2$ in the Grignard route — before that new carbon is converted to the carboxyl carbon (by hydrolysis, or by acid workup of the magnesium carboxylate salt). Neither route touches the carbons already present in R; they only append one more. That is the entire reason working backward from a target acid to a starting halide always means subtracting one carbon, never zero and never two.

Problem: Propanoic acid is required from bromoethane, $\text{CH}_3\text{CH}_2\text{Br}$, using two different routes: (a) via the nitrile, (b) via the Grignard reagent. Write both sequences and confirm the final carbon count in each case.

Step 1. Nitrile route, step one — displace bromide with cyanide in alcoholic $\text{KCN}$: $$CH_3CH_2Br + KCN \xrightarrow{\text{ethanol}} CH_3CH_2CN + KBr$$ This adds one carbon (the nitrile carbon) onto the original two-carbon ethyl group, giving a three-carbon nitrile skeleton.

Step 2. Nitrile route, step two — hydrolyse with aqueous acid and heat: $$CH_3CH_2CN + 2H_2O \xrightarrow{H^+,\ \Delta} CH_3CH_2COOH + NH_4^+$$ The nitrile carbon becomes the carboxyl carbon, and the product is the three-carbon acid, propanoic acid.

Step 3. Grignard route, step one — form the Grignard reagent in dry ether, keeping the two-carbon skeleton unchanged: $$CH_3CH_2Br + Mg \xrightarrow{\text{dry ether}} CH_3CH_2MgBr$$

Step 4. Grignard route, step two — carbonate with dry $\text{CO}_2$, then hydrolyse: $$CH_3CH_2MgBr + CO_2 \rightarrow CH_3CH_2COOMgBr \xrightarrow{H_3O^+} CH_3CH_2COOH$$ The $\text{CO}_2$ carbon becomes the new carboxyl carbon, again yielding the three-carbon acid.

Answer: both routes add exactly one carbon to the two-carbon starting halide, and both correctly deliver propanoic acid, $CH_3CH_2COOH$.

• Trying to make a longer-chain aromatic acid by oxidising a longer alkylbenzene — hot $\text{KMnO}_4$ degrades the whole side chain down to $\text{–COOH}$ on the ring regardless of chain length, so this route only ever gives benzoic acid.
• Using a mild oxidant (like PCC) where the target is the acid, not the aldehyde — mild oxidants stop at the aldehyde; reaching the acid from an alcohol needs a strong, usually hot, excess oxidant.
• Letting water or a protic solvent near the Grignard reagent — even a trace destroys it before it can react with $\text{CO}_2$, so the carbonation step must be done with dry $\text{CO}_2$ in dry ether.
• Miscounting carbons when working backward from an acid to a starting halide — both the nitrile and Grignard routes add exactly one carbon, so the correct starting halide always has one carbon fewer than the target acid.

Ask the tutor: try a few retrosynthesis problems — given a target acid, work out the correct starting halide and route, then check your answer with the tutor.