Organic chemistry reaction map — alkane to alcohol to aldehyde to acid chain

Question

Map the interconversion chain from alkane to alcohol to aldehyde to carboxylic acid. What reagents are needed at each step, and how can we reverse these conversions?

(JEE Main, NEET, CBSE 12 — functional group interconversion is one of the highest-weightage organic chemistry topics)

Solution — Step by Step

Halogenation then hydrolysis:

\text{R-H} \xrightarrow{Cl_2/h\nu} \text{R-Cl} \xrightarrow{\text{aq. NaOH}} \text{R-OH}

Alternatively, for alkenes: acid-catalysed hydration or hydroboration-oxidation gives the alcohol directly.

\text{R-CH=CH}_2 \xrightarrow{H_2O/H^+} \text{R-CH(OH)-CH}_3 \quad \text{(Markovnikov)}

Primary alcohol to aldehyde: Use a mild oxidising agent — PCC (pyridinium chlorochromate) or Collins reagent in anhydrous conditions.

\text{R-CH}_2\text{OH} \xrightarrow{PCC} \text{R-CHO}

Aldehyde to carboxylic acid: Further oxidation with KMnO4 or K2Cr2O7 in acidic medium.

\text{R-CHO} \xrightarrow{KMnO_4} \text{R-COOH}

If you use a strong oxidising agent (KMnO4) directly on a primary alcohol, it goes all the way to the carboxylic acid — you cannot stop at the aldehyde.

Carboxylic acid to aldehyde: Reduce with LiAlH(OtBu)3 or use Rosenmund reduction on the acid chloride.

\text{R-COOH} \xrightarrow{SOCl_2} \text{R-COCl} \xrightarrow{H_2/Pd-BaSO_4} \text{R-CHO}

Aldehyde to alcohol: Reduce with NaBH4 or LiAlH4.

\text{R-CHO} \xrightarrow{NaBH_4} \text{R-CH}_2\text{OH}

graph LR
    A["Alkane R-H"] -->|"Cl2/hv then NaOH"| B["Alcohol R-OH"]
    B -->|"PCC"| C["Aldehyde R-CHO"]
    C -->|"KMnO4"| D["Carboxylic Acid R-COOH"]
    D -->|"SOCl2 then H2/Pd-BaSO4"| C
    C -->|"NaBH4"| B
    B -->|"KMnO4 strong"| D
    D -->|"LiAlH4"| B

Why This Works

The reaction chain follows the oxidation state of carbon: alkane (most reduced) to alcohol to aldehyde to carboxylic acid (most oxidised). Each step is an oxidation, and each reverse step is a reduction.

The reagent choice is critical:

Mild oxidant (PCC): Stops at aldehyde — cannot over-oxidise
Strong oxidant (KMnO4, K2Cr2O7): Goes all the way to carboxylic acid
Strong reductant (LiAlH4): Reduces acid all the way to alcohol
Mild reductant (NaBH4): Reduces aldehyde/ketone to alcohol but does not touch carboxylic acids

Alternative Method

For JEE Main, the Rosenmund reduction (acid chloride + H2/Pd-BaSO4 to aldehyde) and Stephen’s reduction (nitrile + SnCl2/HCl to aldehyde) are frequently tested as methods to prepare aldehydes. Remember: BaSO4 is the poison that prevents over-reduction in Rosenmund reduction.

Common Mistake

The most common error: using KMnO4 to prepare an aldehyde from a primary alcohol. KMnO4 is too strong — it oxidises the alcohol past the aldehyde stage directly to the carboxylic acid. To stop at the aldehyde, you must use PCC or Collins reagent (mild, anhydrous oxidants). This reagent selection is tested in JEE every year.

Also, NaBH4 does not reduce carboxylic acids — only LiAlH4 can do that. NaBH4 is selective for aldehydes and ketones. This selectivity difference is a JEE Advanced favourite.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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