\n $\underbrace{\text{CH}_2 = \text{CH}}_{\substack{ \text{Богатый} \\ \text{Бедный} }} - \text{CH}_2 - \text{CH}_3$\n
+
\n $\text{H-OH}$\n
$\xrightarrow{\text{H}^{+}}$\
\n $\underbrace{\text{CH}_3 - \text{CH(OH)} - \text{CH}_2 - \text{CH}_3}_{\text{Бутанол-2}}$\n