Question

Draw the neutral organic product that results from the following reaction. Include all hydrogen atoms. Hint: we are looking for a 3-carbon product.

Answer 1

Concepts and reason

The problem is based on the concept of dehydration reaction. Alcohol can be dehydrated in the presence of concentrated sulfuric acid to form alkene.

Fundamentals

Concentrated Sulfuric acid acts as dehydrating agent. On the action of concentrated sulfuric acid on alcohol it removes the hydroxyl group to form respective alkene. Although the $^ - {\rm{OH}}$ group of the alcohol is a poor leaving group, ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ acts as a dehydrating agent and remove $^ - {\rm{OH}}$ by protonating it as ${\rm{O}}{{\rm{H}}_{\rm{2}}}^ +$ which is a better leaving group.

Sulfuric acid, ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ protonates the $^ - {\rm{OH}}$ group (a poor leaving group) of the propanol to ${\rm{O}}{{\rm{H}}_{\rm{2}}}^ +$ (a better leaving group) group. The ${\rm{O}}{{\rm{H}}_{\rm{2}}}^ +$ group can easily leave the propanol leaving behind the carbocation as an intermediate. The reaction of propanol with ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ is as follows:

The primary carbocation rearranges itself to form the more stable secondary carbocation which can further proceed in the reaction. The rearrangement carried out in the primary carbocation is as follows:

The removal of a proton from the $\beta$ position of the secondary carbocation will lead in the formation of propene and the released proton will satisfy the proton deficiency of the ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ . The reaction of removal of $\beta$ proton is as follows:

Ans:

The neutral organic product results by reaction of concentrated ${{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ on propanol is as follows: