Preparation methods of alcohols — complete map from different starting materials

medium CBSE JEE-MAIN NEET 3 min read

Question

List all the methods to prepare alcohols, organized by starting material. Which method gives which type of alcohol (primary, secondary, tertiary)?

(CBSE 12 + JEE Main + NEET)


Solution — Step by Step

R-CH=CH2+H2OH+R-CH(OH)-CH3\text{R-CH=CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{R-CH(OH)-CH}_3

Follows Markovnikov’s rule: OH goes to the more substituted carbon. Gives secondary/tertiary alcohols.

For anti-Markovnikov (primary alcohol): use hydroboration-oxidation (BH3/THF\text{BH}_3/\text{THF}, then H2O2/NaOH\text{H}_2\text{O}_2/\text{NaOH}).

Starting materialReagentProduct
Aldehyde (RCHO)NaBH4\text{NaBH}_4 or LiAlH4\text{LiAlH}_4 or H2/Ni\text{H}_2/\text{Ni}Primary alcohol
Ketone (RCOR’)NaBH4\text{NaBH}_4 or LiAlH4\text{LiAlH}_4 or H2/Ni\text{H}_2/\text{Ni}Secondary alcohol
Ester (RCOOR’)LiAlH4\text{LiAlH}_4 (strong)Primary alcohol
Carboxylic acid (RCOOH)LiAlH4\text{LiAlH}_4 (strong)Primary alcohol

NaBH4\text{NaBH}_4 is a mild reducer — works for aldehydes and ketones only. LiAlH4\text{LiAlH}_4 is powerful — reduces everything including esters and acids.

RMgX+R’CHOdry etherR-CH(OH)-R’H3O+secondary alcohol\text{RMgX} + \text{R'CHO} \xrightarrow{\text{dry ether}} \text{R-CH(OH)-R'} \xrightarrow{\text{H}_3\text{O}^+} \text{secondary alcohol}
Carbonyl usedGrignard gives
HCHO (formaldehyde)Primary alcohol
RCHO (aldehyde)Secondary alcohol
R₂CO (ketone)Tertiary alcohol

This is the most versatile method — by choosing the right Grignard + carbonyl pair, we can make any type of alcohol.

R-X+NaOH (aq)R-OH+NaX\text{R-X} + \text{NaOH (aq)} \to \text{R-OH} + \text{NaX}

Works best for primary halides (SN2\text{S}_N2). Tertiary halides prefer elimination — you get alkenes instead.

flowchart TD
    A["Need to prepare an alcohol"] --> B{"What starting material?"}
    B -- Alkene --> C["Hydration: H₂O/H⁺ or BH₃ then H₂O₂"]
    B -- Aldehyde --> D["Reduce with NaBH₄ → 1° alcohol"]
    B -- Ketone --> E["Reduce with NaBH₄ → 2° alcohol"]
    B -- Ester/Acid --> F["Reduce with LiAlH₄ → 1° alcohol"]
    B -- Grignard + HCHO --> G["1° alcohol"]
    B -- Grignard + RCHO --> H["2° alcohol"]
    B -- Grignard + Ketone --> I["3° alcohol"]
    B -- Alkyl halide --> J["NaOH(aq) → alcohol via SN2"]

Why This Works

Each method exploits a different type of reaction: hydration adds water across a double bond, reduction adds hydrogen to a C=O bond, Grignard forms a new C-C bond, and substitution replaces a halide with OH. The product type (1°/2°/3°) depends on the degree of substitution at the carbon bearing the OH group.

The Grignard reaction is the most powerful because it simultaneously forms a new C-C bond and introduces the OH group. This makes it invaluable for synthesizing complex alcohols that cannot be made by simple reduction.


Alternative Method

For NEET, focus on the reduction of carbonyl compounds — this is the most frequently tested preparation method. Remember: LiAlH4\text{LiAlH}_4 reduces everything (even esters and acids to alcohols), while NaBH4\text{NaBH}_4 only reduces aldehydes and ketones. H2/Ni\text{H}_2/\text{Ni} (catalytic hydrogenation) reduces C=O to CH-OH but also reduces C=C bonds — use it only when there are no double bonds you want to preserve.


Common Mistake

Students forget that Grignard reagents react violently with water and acids. The reaction must be done in anhydrous conditions (dry ether). Only AFTER the Grignard has reacted with the carbonyl do we add dilute acid (H3O+\text{H}_3\text{O}^+) to protonate the product. If water is present during the Grignard reaction, it destroys the reagent: RMgX+H2ORH+Mg(OH)X\text{RMgX} + \text{H}_2\text{O} \to \text{RH} + \text{Mg(OH)X}.

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