Preparation of Alcohols

The Five Routes at a Glance

An alcohol is defined by an –OH group bonded to an sp³ carbon. The strategic question in synthesis is always the same: where does that –OH go, and what class of alcohol — primary, secondary or tertiary — results? NCERT groups the laboratory and industrial preparations into three families (from alkenes, from carbonyl compounds, and from Grignard reagents), while NIOS adds the simplest one-step route of all, the hydrolysis of haloalkanes.

Before we open the mechanisms, the table below fixes the destination of each route. Read it as a lookup table: the starting material on the left determines both the regiochemistry and the product class on the right.

From Alkenes: Acid-Catalysed Hydration

Alkenes add water across the double bond in the presence of a mineral acid catalyst. The reaction is the industrial route to ethanol from ethene, and for unsymmetrical alkenes it obeys Markovnikov's rule: the hydrogen of water attaches to the carbon already bearing more hydrogens, and the –OH ends up on the more substituted carbon. Propene therefore gives propan-2-ol, not propan-1-ol.

$$\ce{CH3-CH=CH2 + H2O ->[H^+] CH3-CH(OH)-CH3}$$

The selectivity is a direct consequence of carbocation stability. NCERT lays out the mechanism in three steps.

Because step 1 builds the more stable carbocation (3° > 2° > 1°), the product is the more substituted alcohol. This is exactly why the route is excellent for tertiary and secondary alcohols but a poor choice when a primary alcohol is the target.

From Alkenes: Hydroboration–Oxidation

This is the complementary, anti-Markovnikov route, first reported by H. C. Brown in 1959 (for which he shared the 1979 Nobel Prize in Chemistry). Diborane, $\ce{(BH3)2}$, adds to the alkene to give a trialkylborane; oxidation of that borane with alkaline hydrogen peroxide then liberates the alcohol and boric acid.

$$\ce{3 CH3-CH=CH2 + (BH3)2 -> 2 (CH3CH2CH2)3B}$$

$$\ce{(CH3CH2CH2)3B ->[H2O2,\ OH^-] 3 CH3CH2CH2OH + B(OH)3}$$

The key feature is the orientation of addition. The boron atom — being electron-deficient — attaches to the sp² carbon carrying the greater number of hydrogen atoms (the less substituted, terminal carbon). When boron is later replaced by –OH, the hydroxyl ends up on that terminal carbon. The net effect is addition of water in the opposite sense to Markovnikov's rule, and the addition is syn (H and OH add to the same face). A terminal alkene therefore yields a primary alcohol in excellent yield.

From Carbonyl Compounds: Reduction

Carbonyl compounds — aldehydes, ketones, carboxylic acids and esters — all contain a C=O group that can be reduced to the corresponding alcohol. Reduction simply delivers hydrogen across the carbonyl, and the class of alcohol produced depends entirely on the starting carbonyl.

Catalytic hydrogenation uses a finely divided metal (platinum, palladium or nickel) to add dihydrogen:

$$\ce{CH3CHO ->[H2/\ Ni] CH3CH2OH}\qquad \ce{CH3COCH3 ->[H2/\ Ni] CH3CH(OH)CH3}$$

Hydride reagents are the laboratory workhorses. Lithium aluminium hydride is the strong reagent that reduces every carbonyl class, including the otherwise sluggish acids and esters:

$$\ce{RCOOH ->[(i)\ LiAlH4][(ii)\ H2O] RCH2OH}$$

NCERT notes that LiAlH₄ is expensive and reserved for special chemicals; commercially, acids are first converted to esters and then hydrogenated over a catalyst. NIOS adds the crucial selectivity point: NaBH₄ reduces only aldehydes and ketones and leaves carboxylic acid and ester groups untouched, so it can chemoselectively reduce a carbonyl in a molecule that also carries an acid or ester function.

From Grignard Reagents

The Grignard route is the most versatile carbon–carbon bond-forming method for alcohols, and it is the single most heavily examined sub-topic here in NEET. A Grignard reagent, R–MgX, has a strongly nucleophilic carbon. It adds to the electrophilic carbon of a carbonyl group to form a magnesium alkoxide adduct, which on hydrolysis gives the alcohol.

$$\ce{R-MgX + R'2C=O -> R'2C(OMgX)R ->[H3O^+] R'2C(OH)R}$$

The decisive insight is that the carbonyl carbon becomes the carbinol (C–OH) carbon. The number of carbon groups already on that carbon, plus the new group from the Grignard reagent, fixes whether the product is primary, secondary or tertiary. NCERT states the result plainly: methanal gives a primary alcohol, any other aldehyde gives a secondary alcohol, and a ketone gives a tertiary alcohol.

Mechanism and work-up

The reaction proceeds in two distinct stages. First, the nucleophilic carbon of the Grignard reagent adds across the polar C=O bond, with the carbonyl carbon becoming bonded to the new alkyl group and the oxygen taking the negative charge as a magnesium halide alkoxide. Second, this adduct is hydrolysed — the work-up — to protonate the alkoxide oxygen and release the free alcohol.

$$\ce{C2H5MgBr + (CH3)2C=O -> (CH3)2C(OMgBr)C2H5}$$

$$\ce{(CH3)2C(OMgBr)C2H5 ->[H2O,\ H^+] (CH3)2C(OH)C2H5}$$

The product above is 2-methylbutan-2-ol, a tertiary alcohol, because acetone (a ketone) already supplied two methyl groups on the carbonyl carbon and the Grignard reagent added a third carbon group.

From Haloalkanes

The most direct preparation is the hydrolysis of a haloalkane. Heating an alkyl halide with aqueous sodium or potassium hydroxide lets hydroxide ion act as a nucleophile, displacing the halogen and installing –OH on the same carbon skeleton.

$$\ce{CH3CH2Cl + NaOH(aq) -> CH3CH2OH + NaCl}$$

This route preserves the carbon framework of the halide exactly, so the alcohol class matches the halide class: a primary halide gives a primary alcohol, and so on. Aqueous (not alcoholic) conditions are essential — alcoholic KOH favours elimination to an alkene rather than substitution to an alcohol.

Choosing the Right Method

NEET conversion questions reward a decision rule rather than rote recall. When asked to convert a given starting material into a target alcohol, work backwards from the carbinol carbon and its class.