Animated Solution for Chemistry - Hydrocarbons: The reaction of propene with $\text{HOCl}$ (${\text{Cl}}_2+{\text{H}}_2\text{O}$) proceeds through the intermediate

The reaction between propene and hypochlorous acid ($\text{HOCl}$) is a classic example of an electrophilic addition reaction. To predict the correct intermediate, we must carefully analyze the properties of our reagents and the stability of the possible carbocations formed during the reaction.

Understanding Polarity

Before we look at the alkene, we must understand the attacking reagent, $\text{HOCl}$. In the $\text{HO-Cl}$ bond, oxygen is significantly more electronegative than chlorine. This difference in electronegativity means that oxygen pulls the shared electron density towards itself.

As a result, the oxygen atom acquires a partial negative charge (${\delta }^{-}$), leaving the chlorine atom electron-deficient with a partial positive charge (${\delta }^{+}$). This polarization is crucial because it makes the chlorine atom the electrophile in this reaction.

Pi Electrons in Action

Propene (${\text{CH}}_3-\text{CH}={\text{CH}}_2$) contains a carbon-carbon double bond, which consists of a strong $\sigma$ bond and a weaker $\pi$ bond. The electrons in the $\pi$ bond are loosely held and create a region of high electron density, making the alkene a good nucleophile.

When propene and $\text{HOCl}$ interact, the electron-rich $\pi$ bond attacks the electron-deficient chlorine atom (${\text{Cl}}^{\delta +}$). As the new carbon-chlorine bond begins to form, the oxygen-chlorine bond breaks heterolytically, pushing both electrons onto the oxygen atom to form a hydroxide ion (${\text{OH}}^{-}$).

Carbocation Stability

When the $\pi$ bond breaks to bond with the chlorine atom, the chlorine can attach to either of the two carbons that originally shared the double bond. This leads to two possible carbocation intermediates:

1. Path A: If chlorine attaches to the terminal carbon (${\text{CH}}_2$), the middle carbon is left with a positive charge, forming a secondary ($2^{\circ }$) carbocation: ${\text{CH}}_3-{\text{CH}}^{+}-{\text{CH}}_2-\text{Cl}$. 2. Path B: If chlorine attaches to the middle carbon ($\text{CH}$), the terminal carbon gets the positive charge, forming a primary ($1^{\circ }$) carbocation: ${\text{CH}}_3-\text{CHCl}-{\text{CH}}_2^{+}$.

According to Markovnikov's rule, the reaction will proceed via the most stable carbocation intermediate. A secondary carbocation is significantly more stable than a primary carbocation due to two main factors: - Inductive Effect ($+\text{I}$): The adjacent alkyl groups push electron density towards the positively charged carbon, stabilizing it. - Hyperconjugation: The $\sigma$ electrons of adjacent $\text{C-H}$ bonds can partially delocalize into the empty p-orbital of the carbocation.

Because the secondary carbocation (${\text{CH}}_3-{\text{CH}}^{+}-{\text{CH}}_2-\text{Cl}$) is much more stable, it forms at a faster rate and is the major intermediate of the reaction.

Eventually, the nucleophilic hydroxide ion (${\text{OH}}^{-}$) will attack this secondary carbocation to form the final product, propylene chlorohydrin (${\text{CH}}_3-\text{CH}(\text{OH})-{\text{CH}}_2-\text{Cl}$). However, since the question specifically asks for the intermediate, the correct answer is the secondary carbocation.