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Chemistry · Ch 7 — Alcohols, Phenols and Ethers

Preparation of Alcohols

7.4.1

Preparation of Alcohols

Alcohols can be built up from three broad starting points: alkenes, carbonyl compounds, and Grignard reagents. Each route installs the –OH group by a different mechanism, and the position of the –OH in the product (which carbon gets oxidised) depends heavily on which route is chosen.

1. From Alkenes

(i) Acid-Catalysed Hydration

An alkene adds water across its double bond when an acid catalyst is present, giving an alcohol. For an unsymmetrical alkene the addition follows Markovnikov's rule — the –OH ends up on the more substituted carbon (the one that can better stabilise a positive charge), while H adds to the less substituted carbon.

>C=C<  +  H2O    ⇌H+    >C∣H–C∣OH<{>}\text{C=C}{<} \; + \; \text{H}_2\text{O} \;\; \overset{\text{H}^+}{\rightleftharpoons} \;\; {>}\underset{\underset{\displaystyle \text{H}}{|}}{\text{C}}\text{–}\underset{\underset{\displaystyle \text{OH}}{|}}{\text{C}}{<}

CH3–CH=CH2  +  H2O    ⇌H+    CH3–CH∣OH–CH3\text{CH}_3\text{–CH=CH}_2 \; + \; \text{H}_2\text{O} \;\; \overset{\text{H}^+}{\rightleftharpoons} \;\; \text{CH}_3\text{–}\underset{\underset{\displaystyle \text{OH}}{|}}{\text{CH}}\text{–CH}_3

Mechanism — three steps

  1. Protonation of the alkene. The acid first reacts with water to generate the active electrophile, H3O+\text{H}_3\text{O}^+. The alkene's π electrons then attack this proton, forming the more stable carbocation (Markovnikov orientation) and releasing a water molecule. >C=C<  +  H−O+∣H−H  ⇌  −C∣H−C+<  +  H2O¨{>}\text{C=C}{<} \; + \; \text{H}-\overset{\overset{\displaystyle \text{H}}{|}}{\overset{+}{\text{O}}}-\text{H} \;\rightleftharpoons\; -\overset{\overset{\displaystyle \text{H}}{|}}{\text{C}}-\overset{+}{\text{C}}{<} \; + \; \text{H}_2\ddot{\text{O}}
  2. Nucleophilic attack by water. A fresh molecule of water, acting as a nucleophile through its lone pair, attacks the electron-deficient carbocation, giving a protonated alcohol (an oxonium ion). −C∣H−C+<  +  H2O¨  ⇌  −C∣H−C−O+∣H−H-\overset{\overset{\displaystyle \text{H}}{|}}{\text{C}}-\overset{+}{\text{C}}{<} \; + \; \text{H}_2\ddot{\text{O}} \;\rightleftharpoons\; -\overset{\overset{\displaystyle \text{H}}{|}}{\text{C}}-\text{C}-\overset{\overset{\displaystyle \text{H}}{|}}{\overset{+}{\text{O}}}-\text{H}
  3. Deprotonation. Another water molecule removes the extra proton from the oxonium intermediate, regenerating H3O+\text{H}_3\text{O}^+ and releasing the neutral alcohol as the final product. −C∣H−C−O+∣H−H  +  H2O¨  ⟶  −C∣H−C∣:OH−  +  H3O+-\overset{\overset{\displaystyle \text{H}}{|}}{\text{C}}-\text{C}-\overset{\overset{\displaystyle \text{H}}{|}}{\overset{+}{\text{O}}}-\text{H} \; + \; \text{H}_2\ddot{\text{O}} \;\longrightarrow\; -\overset{\overset{\displaystyle \text{H}}{|}}{\text{C}}-\overset{\overset{\displaystyle :\text{OH}}{|}}{\text{C}}- \; + \; \text{H}_3\text{O}^{+}

Because every step is reversible, this hydration is an equilibrium reaction — the same acid-catalysed pathway run in reverse is exactly how alcohols are dehydrated back to alkenes.

(ii) Hydroboration–Oxidation

Diborane, (BH3)2(\text{BH}_3)_2, adds across an alkene's double bond to give a trialkylborane, which is then oxidised to an alcohol using hydrogen peroxide in the presence of aqueous sodium hydroxide.

CH3–CH=CH2  +  (H–BH2)2  ⟶  CH3–CH∣H–CH2∣BH2    ⟶CH3–CH=CH2    (CH3–CH2–CH2)2BH    ⟶CH3–CH=CH2    (CH3–CH2–CH2)3B\text{CH}_3\text{–CH=CH}_2 \; + \; (\text{H–BH}_2)_2 \;\longrightarrow\; \text{CH}_3\text{–}\underset{\underset{\displaystyle \text{H}}{|}}{\text{CH}}\text{–}\underset{\underset{\displaystyle \text{BH}_2}{|}}{\text{CH}_2} \;\; \overset{\text{CH}_3\text{–CH=CH}_2}{\longrightarrow} \;\; (\text{CH}_3\text{–CH}_2\text{–CH}_2)_2\text{BH} \;\; \overset{\text{CH}_3\text{–CH=CH}_2}{\longrightarrow} \;\; (\text{CH}_3\text{–CH}_2\text{–CH}_2)_3\text{B}

(CH3–CH2–CH2)3B    ⟶H2O,  3H2O2,  OˉH    3 CH3–CH2–CH2–OHPropan-1-ol  +  B(OH)3(\text{CH}_3\text{–CH}_2\text{–CH}_2)_3\text{B} \;\; \overset{\text{H}_2\text{O},\; 3\text{H}_2\text{O}_2,\; \bar{\text{O}}\text{H}}{\longrightarrow} \;\; 3\,\underset{\text{Propan-1-ol}}{\text{CH}_3\text{–CH}_2\text{–CH}_2\text{–OH}} \; + \; \text{B(OH)}_3

The boron atom always attaches to the carbon carrying the greater number of hydrogens — the less substituted carbon of the double bond — so the addition is anti-Markovnikov. The two new bonds (C–H and C–B) also form on the same face of the double bond, i.e. the addition is syn. Because the alkyl group migrates from boron to oxygen with retention of configuration during the peroxide oxidation, the net result looks exactly as if water had added across the alkene the "wrong way round" relative to Markovnikov's rule, and the alcohol is obtained in excellent yield.

2. From Carbonyl Compounds

(i) Reduction of Aldehydes and Ketones

Aldehydes and ketones are reduced to the corresponding alcohols by addition of hydrogen across the carbonyl double bond. This can be done either by catalytic hydrogenation, using a finely divided metal such as platinum, palladium or nickel, or by treatment with a hydride reducing agent such as sodium borohydride (NaBH4\text{NaBH}_4) or lithium aluminium hydride (LiAlH4\text{LiAlH}_4).

RCHO  +  H2    ⟶Pd    RCH2OH\text{RCHO} \; + \; \text{H}_2 \;\; \overset{\text{Pd}}{\longrightarrow} \;\; \text{RCH}_2\text{OH}

RCOR′    ⟶NaBH4    R–CH∣OH–R′\text{RCOR}' \;\; \overset{\text{NaBH}_4}{\longrightarrow} \;\; \text{R–}\underset{\underset{\displaystyle \text{OH}}{|}}{\text{CH}}\text{–R}'

An aldehyde, having only one carbon substituent on the carbonyl carbon, is reduced to a primary alcohol; a ketone, with two carbon substituents, is reduced to a secondary alcohol.

(ii) Reduction of Carboxylic Acids and Esters

Carboxylic acids are reduced all the way to primary alcohols in excellent yield by lithium aluminium hydride, a powerful reducing agent.

RCOOH    ⟶(ii) H2O(i) LiAlH4    RCH2OH\text{RCOOH} \;\; \overset{\text{(i) LiAlH}_4}{\underset{\text{(ii) H}_2\text{O}}{\longrightarrow}} \;\; \text{RCH}_2\text{OH}

Because LiAlH4\text{LiAlH}_4 is expensive, it is reserved for small-scale preparation of specialty chemicals. On a commercial scale, acids are instead first converted to their esters (acid-catalysed esterification with an alcohol, discussed later under the reactions of alcohols), and the ester is then reduced by catalytic hydrogenation, which is far cheaper.

RCOOH    ⟶H+R′OH    RCOOR′    ⟶CatalystH2    RCH2OH  +  R′OH\text{RCOOH} \;\; \overset{\text{R}'\text{OH}}{\underset{\text{H}^+}{\longrightarrow}} \;\; \text{RCOOR}' \;\; \overset{\text{H}_2}{\underset{\text{Catalyst}}{\longrightarrow}} \;\; \text{RCH}_2\text{OH} \; + \; \text{R}'\text{OH}

3. From Grignard Reagents

A Grignard reagent, R–MgX\text{R–MgX}, adds to the carbonyl carbon of an aldehyde or ketone. The carbon–magnesium bond is polarised so that the alkyl/aryl carbon carries partial negative character; this carbon behaves as a nucleophile and attacks the electrophilic carbonyl carbon, pushing the C=O π electrons onto oxygen. The resulting magnesium alkoxide adduct is then hydrolysed (with water or dilute acid) to liberate the alcohol.

>C=O  +  Rδ−–Mgδ+–X  ⟶  [ >C∣R–Oˉ  Mg+–X ]Adduct    ⟶H2O    >C∣R–OH  +  Mg(OH)X{>}\text{C=O} \; + \; \overset{\delta-}{\text{R}}\text{–}\overset{\delta+}{\text{Mg}}\text{–X} \;\longrightarrow\; \Big[\, {>}\underset{\underset{\displaystyle \text{R}}{|}}{\text{C}}\text{–}\bar{\text{O}}\;\overset{+}{\text{Mg}}\text{–X} \,\Big]_{\text{Adduct}} \;\; \overset{\text{H}_2\text{O}}{\longrightarrow} \;\; {>}\underset{\underset{\displaystyle \text{R}}{|}}{\text{C}}\text{–OH} \; + \; \text{Mg(OH)X}

The identity of the alcohol formed depends entirely on which carbonyl compound the Grignard reagent attacks:

HCHO  +  RMgX  ⟶  RCH2OMgX    ⟶H2O    RCH2OH  +  Mg(OH)X\text{HCHO} \; + \; \text{RMgX} \;\longrightarrow\; \text{RCH}_2\text{OMgX} \;\; \overset{\text{H}_2\text{O}}{\longrightarrow} \;\; \text{RCH}_2\text{OH} \; + \; \text{Mg(OH)X} …