Pentan-1-ol, CH3CH2CH2CH2CH2OH, can be dehydrated to form an alkene. (a) State two different sets of conditions that could each be used to dehydrate pentan-1-ol to an alkene. [2] (b) Name the type of reaction occurring, and write an equation, using structural formulas, for the d…

Answer: ((a)) Passing the alcohol vapour over a heated aluminium oxide, Al2O3, catalyst; OR heating with an excess of concentrated sulfuric acid, H2SO4. ((b)) Elimination; CH3CH2CH2CH2CH2OH → CH3CH2CH2CH=CH2 + H2O. ((c)) Pentan-1-ol has the -OH on the terminal C1, with only one neighbouring carbon (C2) bearing hydrogens available for elimination, so only one alkene (pent-1-ene) can form. Pentan-2-ol has the -OH on C2, with two different neighbouring carbons (C1 and C3) each bearing hydrogens, so…. Part (a): Conditions for dehydration An alcohol can be dehydrated to an alkene in either of two standard ways: - Passing the alcohol vapour over a heated aluminium oxide (Al2O3) catalyst. - Heating the alcohol with an excess of concentrate…

AS Chemistry 9701: Dehydration of Pentan-1-ol to an Alkene

Part (a): Conditions for dehydration

An alcohol can be dehydrated to an alkene in either of two standard ways:

Passing the alcohol vapour over a heated aluminium oxide ( $\text{Al}_2\text{O}_3$ ) catalyst.
Heating the alcohol with an excess of concentrated sulfuric acid ( $\text{H}_2\text{SO}_4$ ).

Part (b): Type of reaction and equation

Dehydration removes an $-\text{OH}$ group from one carbon and a hydrogen atom from an adjacent carbon, forming a carbon–carbon double bond and a molecule of water. This is an elimination reaction.

Since pentan-1-ol has its $-\text{OH}$ group on the terminal carbon (C1), the only adjacent carbon that can lose a hydrogen is C2, so the double bond must form between C1 and C2:

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2 + \text{H}_2\text{O}$

The alkene formed is pent-1-ene.

Part (c): Why the number of products differs

In pentan-1-ol, the $-\text{OH}$ -bearing carbon (C1) has only one neighbouring carbon, C2, which carries hydrogen atoms available for elimination. There is therefore only one possible position for the new double bond, so dehydration gives a single product, pent-1-ene.

In pentan-2-ol, the $-\text{OH}$ -bearing carbon (C2) has two different neighbouring carbons, C1 and C3, each of which carries hydrogen atoms that could be lost during elimination. Removing a hydrogen from C1 gives pent-1-ene, while removing a hydrogen from C3 gives pent-2-ene, so dehydrating pentan-2-ol produces a mixture of these two structurally isomeric alkenes.

Final answers

(a) Heated $\text{Al}_2\text{O}_3$ catalyst (vapour passed over it), OR excess concentrated $\text{H}_2\text{SO}_4$ .
(b) Elimination; $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}=\text{CH}_2 + \text{H}_2\text{O}$ .
(c) Pentan-1-ol has only one carbon (C2) adjacent to the $-\text{OH}$ -bearing carbon that can lose a hydrogen, giving a single product; pentan-2-ol has two such neighbouring carbons (C1 and C3), giving a mixture of pent-1-ene and pent-2-ene.