2-Methylpropene, CH2=C(CH3)2, reacts with steam in the presence of a catalyst of concentrated phosphoric(V) acid, H3PO4, to give a mixture of two isomeric alcohols, with one isomer strongly predominating. (a) State the type of mechanism for this reaction. [1] (b) Describe, in wo…

Answer: ((a)) Electrophilic addition. ((b)) A proton, H^+, donated by the acid catalyst, acts as the electrophile; the alkene's π electrons attack it, forming a new C-H bond and leaving a carbocation on the other carbon of the former double bond. If H^+ bonds to the CH2 carbon, the tertiary carbocation (CH3)3C^+ forms; if… ((c)) (CH3)3C^+ + H2O → (CH3)3C- +OH2, followed by loss of a proton to give (CH3)3C-OH + H^+. Because the H^+ used in the first step of the mechanism is exactly regenerated in this final step, the phosphoric(V) acid is not consumed overall and acts as a true catalyst.. Part (a): Naming the mechanism Addition of steam across the C=C double bond of 2-methylpropene is an example of electrophilic addition. Part (b): The mechanism and Markovnikov's rule The H3PO4 catalyst first donates a proton, H^+, which ac…

AS Chemistry 9701: Electrophilic Addition of Steam to 2-Methylpropene

2-Methylpropene, $\text{CH}_2=\text{C}(\text{CH}_3)_2$ , reacts with steam in the presence of a catalyst of concentrated phosphoric(V) acid, $\text{H}_3\text{PO}_4$ , to give a mixture of two isomeric alcohols, with one isomer strongly predominating.

(a) State the type of mechanism for this reaction. [1]

(b) Describe, in words, how this mechanism proceeds, including the structure of any intermediate(s) formed, and use it to explain which of the two possible alcohol products is obtained as the major product. [4]

(c) Write an equation for the reaction between the major carbocation intermediate and water, give the further step needed to form the neutral major alcohol product, and explain why the $\text{H}_3\text{PO}_4$ catalyst is not used up overall. [3]

Part (a): Naming the mechanism

Addition of steam across the C=C double bond of 2-methylpropene is an example of electrophilic addition.

Part (b): The mechanism and Markovnikov’s rule

The $\text{H}_3\text{PO}_4$ catalyst first donates a proton, $\text{H}^+$ , which acts as the electrophile. The electron-rich $\pi$ bond of 2-methylpropene attacks this electrophilic proton, forming a new C-H bond and leaving a positively charged carbon (a carbocation) at whichever carbon did not bond to the incoming hydrogen atom.

Two outcomes are possible, depending on which carbon of the double bond bonds to the incoming $\text{H}^+$ :

If $\text{H}^+$ bonds to the $\text{CH}_2$ carbon (which already carries two hydrogens), the positive charge is left on the carbon bearing the two methyl groups, giving the tertiary carbocation $(\text{CH}_3)_3\text{C}^+$ .
If $\text{H}^+$ bonds to the carbon bearing the two methyl groups (which carries no hydrogens directly), the positive charge is left on the $\text{CH}_2$ carbon, giving the primary carbocation $^+\text{CH}_2\text{CH}(\text{CH}_3)_2$ .

The tertiary carbocation is far more stable than the primary one, because three alkyl groups attached to the positively charged carbon donate electron density towards it (an inductive, electron-releasing effect), spreading out and reducing the positive charge, compared with only one such alkyl group near the primary carbocation. Since the tertiary carbocation forms far more readily and is much lower in energy, it is produced in much greater amount, and this is the intermediate that goes on to react with water.

Part (c): Completing the mechanism and the role of the catalyst

Water acts as a nucleophile, using a lone pair on its oxygen atom to attack the tertiary carbocation:

$(\text{CH}_3)_3\text{C}^+ + \text{H}_2\text{O} \rightarrow (\text{CH}_3)_3\text{C}-\overset{+}{\text{O}}\text{H}_2$

This protonated alcohol then loses a proton (to a base such as $\text{H}_2\text{PO}_4^-$ present in the mixture) to form the neutral major product, 2-methylpropan-2-ol, and regenerate a hydrogen ion:

$(\text{CH}_3)_3\text{C}-\overset{+}{\text{O}}\text{H}_2 \rightarrow (\text{CH}_3)_3\text{C}-\text{OH} + \text{H}^+$

Because the $\text{H}^+$ used in the first step is exactly regenerated in this final step, the phosphoric(V) acid is not consumed overall, it acts as a true catalyst. (Water attacking the minor primary carbocation, $^+\text{CH}_2\text{CH}(\text{CH}_3)_2$ , would similarly give the minor product, 2-methylpropan-1-ol, but in a much smaller amount.)

Final answers

(a) Electrophilic addition.
(b) Major product is 2-methylpropan-2-ol, via the more stable tertiary carbocation $(\text{CH}_3)_3\text{C}^+$ (Markovnikov’s rule); minor product is 2-methylpropan-1-ol, via the less stable primary carbocation.
(c) $(\text{CH}_3)_3\text{C}^+ + \text{H}_2\text{O} \rightarrow (\text{CH}_3)_3\text{C}-\overset{+}{\text{O}}\text{H}_2$ , then loss of $\text{H}^+$ gives $(\text{CH}_3)_3\text{C}-\text{OH}$ ; $\text{H}^+$ is regenerated, so the catalyst is unchanged overall.