Q16-17P - Organic Chemistry (Mcmurry)

Question

Treatment of p-bromotoluene with NaOH at $\mathbf{300}\mathbf{°}\mathbf{C}$ yields a mixture of two products, but treatment of m-bromotoluene with NaOH yields a mixture of three products. Explain.

Step-by-Step Solution

Answer

The reaction of p-bromotoulene with NaOH at $300°\mathrm{C}$ produces a mixture of two products since the reaction occurs through the formation of a benzyne intermediate.

 The reaction of m-bromotoulene with NaOH at $300°\mathrm{C}$ produces a mixture of three products since the reaction occurs through the formation of two different benzyne intermediates.

Step 1: Nucleophilic substitution on halobenzene

Nucleophlilic substitution on halobenzene happens at higher temperature and pressure through the formation of a benzyne intermediate. The benzyne intermediate leads to two products if the reactant is a substituted halobenzene.

Step 2: Reaction of p-bromotoulene and m-bromotoulene with NaOH at 300 ° C

When p-bromotoulene is treated with NaOH at $300°\mathrm{C}$ , the base picks up a proton from a carbon near to that which comprise a bromine atom. This results in the elimination of  ${\mathrm{Br}}^{-}$forming a benzyne intermediate. The addition of water happens into any of the two carbons in the triple bond resulting in two different phenols.

                              Reaction of p-bromotoulene with NaOH at $300 °\mathrm{C}$

The similar mechanism is seen in m-bromotoulene. But here the adjacent proton is removed from any of the two carbons near to that which comprise the bromine atom. This forms different benzyne intermediates. Each can produce two phenols on adding water. Out of the four products, two are identical. This three phenols are obtained as products.

                                     Reaction of m-bromotoulene with NaOH at  $300°\mathrm{C}$