Animated Solution for Chemistry - Organic Chemistry: Which of the following reactions will yield 2, 2-dibromopropane?

Welcome to a fascinating journey through the world of organic synthesis! Today, we are tasked with a specific mission: to synthesize 2,2-dibromopropane. Imagine this molecule in your mind's eye. It is a three-carbon chain, a propane backbone, with two heavy bromine atoms attached directly to the central carbon. This specific arrangement is our target, and we must carefully evaluate our starting materials to find the perfect match.

The options provided to us are a mix of alkenes and alkynes, all reacting with hydrogen bromide ($HBr$). The reaction of unsaturated hydrocarbons with hydrogen halides is a classic example of electrophilic addition. But to get two bromine atoms on the same carbon, we need a starting material that can undergo addition twice. This immediately points us towards alkynes, which have a triple bond and can react with two equivalents of $HBr$.

Markovnikov's Rule in Action

Let us dive deep into option (a), which features propyne ($C{H}_3-C\equiv CH$). Propyne is an asymmetric alkyne. When the first molecule of $HBr$ approaches, it faces a choice. Which carbon should the hydrogen atom attack? This is where the legendary Markovnikov's rule comes into play. The rule states that in the addition of a protic acid to an asymmetric alkene or alkyne, the hydrogen atom will attach to the carbon that already holds a greater number of hydrogen atoms.

In propyne, the terminal carbon has one hydrogen, while the central carbon has none. Therefore, the electrophilic proton ($H^{+}$) will attack the terminal carbon. This attack breaks one of the pi bonds and leaves a positive charge on the central carbon, creating a vinylic carbocation. The bromide ion ($B{r}^{-}$) then swiftly attacks this carbocation, resulting in the formation of 2-bromopropene ($C{H}_3-C(Br)=C{H}_2$).

The Magic of Resonance

Now, the stage is set for the second act. We have 2-bromopropene, and another molecule of $HBr$ is ready to react. Once again, the proton ($H^{+}$) must choose its target. Following Markovnikov's rule, it attacks the terminal $C{H}_2$ group, converting it into a $C{H}_3$ group. This leaves a positive charge on the central carbon, forming a new carbocation: $C{H}_3-C^{+}(Br)-C{H}_3$.

But wait, there is a catch here! You might wonder, doesn't the highly electronegative bromine atom destabilize this positive charge through its electron-withdrawing inductive effect (-I)? This is where the magic of organic chemistry reveals itself. While bromine does pull electrons through the sigma bond, it also possesses lone pairs of electrons in its p-orbitals.

These lone pairs can overlap with the empty p-orbital of the adjacent carbocation. This phenomenon, known as resonance or the +M effect, allows the bromine atom to share its electron density, effectively spreading the positive charge over both the carbon and the bromine atoms. This back-bonding provides extraordinary stability to the intermediate, far outweighing the destabilizing -I effect.

Because this resonance-stabilized carbocation is so incredibly stable, the reaction proceeds rapidly and exclusively through this pathway. The intermediate is now primed for the final strike. The remaining bromide ion ($B{r}^{-}$) in the solution acts as a nucleophile, seeking out the positive charge.

It attacks the central carbon, forming a new carbon-bromine bond. The dust settles, and we are left with our glorious final product: 2,2-dibromopropane ($C{H}_3-C(Br{)}_2-C{H}_3$). We have successfully achieved our target!

For the sake of completeness, let us briefly evaluate why the other options fail our mission. Option (b) features 1-bromopropene. The addition of $HBr$ here would place the hydrogen on the carbon with more hydrogens, yielding 1,1-dibromopropane. Option (c) is ethyne, a two-carbon alkyne. Adding two equivalents of $HBr$ would give 1,1-dibromoethane, which lacks the necessary three-carbon chain. Finally, option (d) is propene, an alkene. It can only react with one equivalent of $HBr$, yielding 2-bromopropane.

Therefore, our journey confirms that option (a) is the one and only correct path to our desired destination. The interplay of Markovnikov's rule and resonance stabilization makes this a beautiful and elegant chemical transformation.