Q.The intermediate carbocation formed in the reactions of HI, HBr and HCl with propene is the same and the bond energy of HCl, HBr and HI is 430.5 kJ mol^-1, 363.7 kJ mol^-1 and 296.8 kJ mol^-1 respectively. What will be the order of reactivity of these halogen acids?

The reaction of halogen acids (HX) with propene is an electrophilic addition reaction, following Markovnikov's rule, where the hydrogen atom adds to the carbon atom of the double bond that already has more hydrogen atoms. The rate-determining step in this reaction is the initial attack of the electrophile ($H^+$) on the alkene to form a carbocation.

1. The ease of formation of the electrophile ($H^+$) from HX depends on the strength of the H-X bond. A weaker bond requires less energy to break, making $H^+$ more readily available.
2. The given bond energies are: HCl ($430.5 \text{ kJ mol}^{-1}$), HBr ($363.7 \text{ kJ mol}^{-1}$), and HI ($296.8 \text{ kJ mol}^{-1}$).
3. Comparing these values, HI has the weakest H-I bond, followed by HBr, and then HCl with the strongest H-Cl bond. …

The reaction of propene with halogen acids (HCl, HBr, HI) is an example of electrophilic addition. In this type of reaction, the double bond of the alkene acts as a nucleophile, attacking the electrophilic hydrogen of the halogen acid. The rate of this reaction is primarily governed by the stability of the intermediate carbocation formed and the ease with which the H-X bond breaks.

The problem statement simplifies one aspect for us: it explicitly states that the intermediate carbocation formed is the same for all three halogen acids. This means we do not need to consider differences in carbocation stability when comparing reactivity. Instead, we must focus on the other factor influencing the reaction rate: the energy required to break the H-X bond.

Here's a step-by-step breakdown:

1. Mechanism of Electrophilic Addition: The reaction proceeds in two main steps. The first step involves the protonation of the alkene by the halogen acid, forming a carbocation. According to Markovnikov's rule, the hydrogen atom adds to the carbon atom of the double bond that already has more hydrogen atoms, leading to the formation of the more stable carbocation. For propene ($CH_3-CH=CH_2$), this results in the formation of a secondary carbocation (isopropyl carbocation):

The second step is the rapid attack of the halide ion ($X^-$) on the carbocation to form the final product.

2. Identifying the Rate-Determining Step (RDS):

The first step, the formation of the carbocation, is typically the slow and rate-determining step (RDS) of the reaction. This step involves the breaking of the H-X bond and the formation of a new C-H bond, leading to a high-energy carbocation intermediate. The energy required for this step, known as the activation energy, dictates the overall reaction rate.

3. Relating Bond Energy to Reaction Rate: For the H-X bond to break and the carbocation to form, energy must be supplied to overcome the bond energy of H-X.
   • A lower bond energy for H-X means that less energy is required to break that bond.
   • If less energy is required to break the H-X bond, the activation energy for the rate-determining step will be lower. …