Animated Solution for Chemistry - Organic Chemistry: Following reaction, $(C{H}_3{)}_{3}CBr+H_{2}O⟶(C{H}_3{)}_{3}COH+HBr$ is an example of

Let's embark on a journey to understand the fundamental nature of a classic organic reaction. We are given the following chemical transformation:

At first glance, this might just look like a jumble of letters and numbers, but if we look closely, a beautiful story of molecular exchange is unfolding. Our starting material is tert-butyl bromide, a tertiary alkyl halide. It reacts with water to form tert-butyl alcohol.

To classify any organic reaction, the first step is to play a game of "spot the difference" between the reactants and the products.

In our reactant, the central carbon atom is bonded to three methyl groups and one bromine ($Br$) atom. Now, look at the product. The central carbon is still bonded to the three methyl groups, but the bromine atom has vanished! In its place, a hydroxyl ($-OH$) group has appeared.

When one atom or group of atoms in a molecule is replaced by another atom or group, the reaction is fundamentally classified as a Substitution Reaction.

Now that we know it's a substitution, we need to determine what kind of substitution it is. This depends entirely on the nature of the incoming group.

The new group is $-OH$, which originates from our reagent, water ($H_{2}O$). Let's think about the structure of water. The oxygen atom in water possesses two lone pairs of electrons. This makes water an electron-rich species.

In the language of chemistry, an electron-rich species that seeks out a positive center (a nucleus) is called a Nucleophile (nucleus-loving). Therefore, because a nucleophile is doing the substituting, this reaction is a Nucleophilic Substitution.

While the question only asks for the reaction type, understanding the mechanism gives us a deeper appreciation. Because our substrate is a tertiary alkyl halide, it is highly sterically hindered, making a direct backside attack (the $S_{N}2$ pathway) impossible.

Instead, the reaction proceeds via the $S_{N}1$ mechanism (Substitution Nucleophilic Unimolecular).

1. Step 1 (Slow): The carbon-bromine bond breaks heterolytically. The bromine takes the bonding electrons and leaves as a bromide ion ($B{r}^{-}$), leaving behind a positively charged, highly stable tertiary carbocation. 2. Step 2 (Fast): The nucleophile ($H_{2}O$) rapidly attacks the electron-deficient carbocation. After a quick loss of a proton ($H^{+}$), we get our final product, tert-butyl alcohol.

By carefully observing the exchange of groups and identifying the electron-rich nature of the attacking reagent, we can confidently conclude that this is a Nucleophilic Substitution reaction.