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

Question

How do we convert an alkane step-by-step through alcohol, aldehyde, and finally to a carboxylic acid? What reagents are needed at each stage, and what controls selective oxidation?

(JEE Main, NEET, CBSE 12 — functional group interconversion is the backbone of organic chemistry problems)

Solution — Step by Step

Alkanes are unreactive under normal conditions. We need to first functionalize them.

Free radical halogenation:

CH_4 + Cl_2 \xrightarrow{h\nu} CH_3Cl + HCl

UV light or heat initiates homolytic fission of $Cl_2$ , generating $Cl\cdot$ radicals. The selectivity follows: 3-degree H > 2-degree H > 1-degree H (due to stability of the carbon radical formed).

CH_3Cl + NaOH_{(aq)} \xrightarrow{\Delta} CH_3OH + NaCl

This is an $S_N2$ reaction for primary halides — $OH^-$ attacks the carbon bearing the halogen. For tertiary halides, the mechanism shifts to $S_N1$ .

Alternative route: Hydroboration-oxidation of alkenes (anti-Markovnikov addition) or acid-catalysed hydration of alkenes (Markovnikov addition).

This step requires mild, selective oxidation to stop at the aldehyde stage without over-oxidizing to carboxylic acid.

Best reagent: PCC (Pyridinium Chlorochromate, $C_5H_5NH^+ CrO_3Cl^-$ ) in $CH_2Cl_2$

RCH_2OH \xrightarrow{PCC/CH_2Cl_2} RCHO

PCC works in anhydrous conditions — the absence of water prevents further oxidation. Using $KMnO_4$ or $K_2Cr_2O_7$ would push the oxidation all the way to the carboxylic acid.

RCHO \xrightarrow{K_2Cr_2O_7/H^+} RCOOH

Strong oxidising agents like acidified potassium dichromate or $KMnO_4$ complete the oxidation. Even mild oxidants like Tollens’ reagent ( $Ag(NH_3)_2^+$ ) can oxidize aldehydes — this is the basis of the silver mirror test.

RCHO + 2Ag(NH_3)_2^+ + 3OH^- \rightarrow RCOO^- + 2Ag\downarrow + 4NH_3 + 2H_2O

flowchart LR
    A["Alkane<br/>RCH₃"] -->|"Cl₂/hν<br/>(free radical halogenation)"| B["Alkyl halide<br/>RCH₂Cl"]
    B -->|"NaOH(aq)/Δ<br/>(SN2)"| C["Alcohol<br/>RCH₂OH"]
    C -->|"PCC/CH₂Cl₂<br/>(mild oxidation)"| D["Aldehyde<br/>RCHO"]
    D -->|"K₂Cr₂O₇/H⁺<br/>(strong oxidation)"| E["Carboxylic acid<br/>RCOOH"]
    C -->|"K₂Cr₂O₇/H⁺<br/>(direct strong oxidation)"| E

Why This Works

Each step changes the oxidation state of the carbon atom:

Alkane: C is in the lowest oxidation state (surrounded by C-H bonds)
Alcohol: one C-O bond formed (oxidation state increases)
Aldehyde: C=O bond (further increase)
Carboxylic acid: C bonded to two oxygens (highest common oxidation state)

The whole chain is about progressively breaking C-H bonds and forming C-O bonds. PCC is the hero here because it stops at the aldehyde — the anhydrous medium prevents the aldehyde from forming a geminal diol, which is the intermediate that strong aqueous oxidants attack.

Alternative Method

Direct route — alkane to carboxylic acid using strong conditions:

RCH_3 \xrightarrow{KMnO_4/H^+/\Delta} RCOOH

This skips all intermediate steps but gives no control over which product you get. Useful when you only need the final acid.

Reverse direction (acid to aldehyde): Use Rosenmund reduction — $RCOCl + H_2/Pd-BaSO_4 \rightarrow RCHO$ . The poisoned catalyst (Pd on BaSO_4 with quinoline) prevents over-reduction to alcohol.

Common Mistake

The number one error: using $KMnO_4$ or $K_2Cr_2O_7$ when the question asks for aldehyde as the product. These strong oxidants will blow right past the aldehyde to give carboxylic acid. If a question says “convert ethanol to ethanal”, the answer MUST use PCC or Collins’ reagent — not $KMnO_4$ . JEE Main 2023 penalised this specifically.

Memorise this hierarchy: PCC = stop at aldehyde, Jones’ reagent ( $CrO_3/H_2SO_4$ ) = go to acid, $KMnO_4$ = go to acid. For secondary alcohols, all oxidants give the same product (ketone) since there is no further oxidation possible.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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