Reaction of Anisole with Hydrogen Bromide

To determine the products of the reaction of anisole ($\text{C}_6\text{H}_5\text{OCH}_3$) with hydrogen bromide ($\text{HBr}$), we analyze the reaction mechanism step by step:

1. Protonation of the Ether:
   The lone pair of electrons on the oxygen atom of anisole acts as a Lewis base and accepts a proton from $\text{HBr}$ to form an oxonium ion intermediate:
   $\text{C}_6\text{H}_5-\text{O}-\text{CH}_3 + \text{HBr} \rightleftharpoons \text{C}_6\text{H}_5-\overset{+}{\underset{\text{H}}{\text{O}}}-\text{CH}_3 + \text{Br}^-$

2. Nucleophilic Attack by Bromide Ion ($\text{Br}^-$):
   In the protonated ether, there are two potential carbon atoms for nucleophilic attack by the bromide ion:

   • The phenyl carbon ($\text{C}_{\text{aromatic}}$): The bond between the phenyl carbon and the oxygen atom has partial double-bond character due to resonance with the benzene ring. It is stronger and shorter than a typical single bond. Moreover, $\text{S}_\text{N}2$ displacement at an $\text{sp}^2$-hybridized aromatic carbon is energetically unfavorable and hindered by electronic repulsion from the $\pi$-cloud of the ring.
   • The methyl carbon ($\text{C}_{\text{aliphatic}}$): The bond between the methyl carbon and the oxygen atom is a normal single bond at an unhindered, $\text{sp}^3$-hybridized carbon.

   Therefore, the nucleophile $\text{Br}^-$ selectively attacks the methyl carbon via an $\text{S}_\text{N}2$ pathway:
   $\text{Br}^- + \text{CH}_3-\overset{+}{\underset{\text{H}}{\text{O}}}-\text{C}_6\text{H}_5 \longrightarrow \text{CH}_3\text{Br} + \text{C}_6\text{H}_5\text{OH}$

Thus, the reaction produces phenol ($\text{C}_6\text{H}_5\text{OH}$) and bromomethane ($\text{CH}_3\text{Br}$).

Therefore, the correct option is D.