Preparing alcohols

Which route gives which alcohol?

A chemist who wants an alcohol asks two questions. Which carbon will hold the OH? How many carbons does the product need? The routes below answer both. All three families are in the syllabus.

From alkenes

Acid-catalysed hydration. An alkene reacts with water in the presence of an acid catalyst such as dilute sulphuric acid. The acid adds a proton to the carbon of the double bond that already has more hydrogen atoms. This forms the more stable carbocation. Water attacks the carbocation, and the oxonium ion gives up its proton. Markovnikov's rule describes the result: the OH ends up on the carbon with more alkyl groups.

So propene gives propan-2-ol, a secondary alcohol. Only ethene gives a primary alcohol, ethanol, by this route. Ethanol is made this way at a large scale, with phosphoric acid on a solid support at high temperature and pressure. Because a carbocation is involved, it can shift, and some alkenes give a mixture of alcohols.

Hydroboration then oxidation. Borane, $\text{BH}_{3}$ (used as diborane, $\text{B}_{2}\text{H}_{6}$, in tetrahydrofuran), adds across the double bond. The boron goes to the carbon with fewer alkyl groups, and the hydrogen to the other one. Three alkene molecules fill the three B-H bonds and make a trialkylborane. Then hydrogen peroxide with aqueous sodium hydroxide replaces boron by OH. The result is anti-Markovnikov: the OH ends up on the less substituted carbon.

Propene gives propan-1-ol this way, and no carbocation forms, so there is no shift. The two routes complete each other.

From carbonyl compounds by reduction

An aldehyde or a ketone has a C=O group. Adding hydrogen to C=O gives an alcohol.

• An aldehyde $\text{RCHO}$ gives a primary alcohol $\text{RCH}_{2}\text{OH}$.
• A ketone $\text{R}_{2}\text{CO}$ gives a secondary alcohol $\text{R}_{2}\text{CHOH}$.

Three reducing agents are common. Sodium borohydride, $\text{NaBH}_{4}$, in ethanol or water is mild and reduces aldehydes and ketones. Lithium aluminium hydride, $\text{LiAlH}_{4}$, in dry ether, then water, is stronger. Hydrogen gas with a nickel, palladium or platinum catalyst also works.

A carboxylic acid $\text{RCOOH}$ is harder to reduce. $\text{LiAlH}_{4}$ reduces it to the primary alcohol $\text{RCH}_{2}\text{OH}$, and $\text{NaBH}_{4}$ does not. An ester $\text{RCOOR}'$ gives two alcohols with $\text{LiAlH}_{4}$: $\text{RCH}_{2}\text{OH}$ and $\text{R}'\text{OH}$. $\text{LiAlH}_{4}$ is costly, so industry reduces an acid by converting it to an ester and then adding hydrogen over a catalyst.

From Grignard reagents

A Grignard reagent $\text{RMgBr}$ is made from a haloalkane and magnesium in dry ether. The carbon bonded to magnesium behaves as a carbanion and attacks the carbon of C=O. Water (with a little acid) then turns the alkoxide into the alcohol.

• Methanal $\text{HCHO}$ gives a primary alcohol $\text{RCH}_{2}\text{OH}$. One carbon is added to R.
• Another aldehyde $\text{R}'\text{CHO}$ gives a secondary alcohol $\text{RCH}(\text{OH})\text{R}'$.
• A ketone $\text{R}'_{2}\text{CO}$ gives a tertiary alcohol $\text{RR}'_{2}\text{COH}$.
• Ethylene oxide (an epoxide) gives a primary alcohol $\text{RCH}_{2}\text{CH}_{2}\text{OH}$. Two carbons are added to R.

An ester $\text{R}'\text{COOR}''$ needs two molecules of the Grignard reagent and gives a tertiary alcohol with two identical R groups.

Water must be absent. A Grignard reagent reacts at once with water to give an alkane, $\text{RMgBr} + \text{H}_{2}\text{O} \to \text{RH} + \text{Mg}(\text{OH})\text{Br}$, and the reagent is lost.

Other sources

A haloalkane reacts with aqueous potassium hydroxide to give an alcohol by substitution. Sugar fermentation gives ethanol: the enzyme zymase in yeast splits glucose into ethanol and carbon dioxide. Methanol is made from water gas over a zinc oxide and chromium oxide catalyst at high pressure.

Practise with new numbers

Try it in Python

The program uses the rule for each route to decide which carbon of the double bond takes the OH.

Common mistakes

• Expecting a primary alcohol from acid hydration. Markovnikov's rule puts OH on the more substituted carbon. Only ethene gives a primary alcohol.
• Stopping the hydroboration after the first step. The trialkylborane is not an alcohol. Hydrogen peroxide and sodium hydroxide are needed.
• Using $\text{NaBH}_{4}$ on a carboxylic acid. It is too mild. Use $\text{LiAlH}_{4}$.
• Making a Grignard reagent in water. It is destroyed. Use dry ether.
• Counting carbons wrongly. Methanal adds one carbon to R. Ethylene oxide adds two.

Remember these

• Acid hydration: Markovnikov, carbocation, may shift. Hydroboration then oxidation: anti-Markovnikov, no shift.
• Aldehyde to primary alcohol, ketone to secondary alcohol, acid to primary alcohol ($\text{LiAlH}_{4}$ only).
• Grignard reagent plus methanal gives a primary alcohol, plus another aldehyde a secondary, plus a ketone a tertiary alcohol.
• Dry ether for every Grignard step.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which alkene and which method give 2-methylpropan-1-ol?

2. Give two different Grignard routes to butan-2-ol.

3. Why does a carboxylic acid need $\text{LiAlH}_{4}$ and not $\text{NaBH}_{4}$?

4. What does ethylmagnesium bromide give, after water is added, with (a) methanal and (b) ethylene oxide?

References

• NCERT Chemistry Part II (Class 12), Chapter 7, the section on the preparation of alcohols.
• OpenStax Chemistry 2e, Chapter 20, section 20.2 on alcohols and ethers.
• Morrison and Boyd, Organic Chemistry, the chapter on the preparation of alcohols.