Sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\) adds to the double bond of alkenes as \(\mathrm{H}^{+}\) and \(-\mathrm{OSO}_{3} \mathrm{H}\). Predict the products when sulfuric acid reacts with (a) ethylene and (b) propene.

Short Answer

Expert verified
The product when sulfuric acid reacts with (a) ethylene is ethyl hydrogen sulfate, and (b) with propene is propyl hydrogen sulfate.

Step by step solution

01

Understand sulfuric acid reaction mechanisms

To solve this problem, we should first understand that sulfuric acid can add to the double bond of alkenes. It adds as \(H^+\) and \(-OSO_3H\), which implies that one hydrogen atom and one hydrogen sulfate group will interact with the carbon atoms in the alkene to break the double bond and create new single bonds.
02

Apply mechanism to ethylene

Ethylene has the molecular formula \(C_2H_4\). Reacting with sulfuric acid, we get: \(C_2H_4 + H_2SO_4 \rightarrow C_2H_5HSO_4\). This means that the ethylene molecule now has an extra \(H\) bonded to one carbon and \(-OSO_3H\) bonded to the other carbon. Thus product is ethyl hydrogen sulfate.
03

Apply mechanism to propene

Propene has the molecular formula \(C_3H_6\). Reacting with sulfuric acid, we get: \(C_3H_6 + H_2SO_4 \rightarrow C_3H_7HSO_4\). This means that the propene molecule now has an extra \(H\) bonded to one carbon and \(-OSO_3H\) bonded to the other carbon. Thus product is propyl hydrogen sulfate.

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