Question

The correct decreasing order of reactivity of the halides for \(\mathrm{S}_{\mathrm{N}} 1\) reaction is (A) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}\) (B) \(\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}\) (c) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}\) (d) \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\)

Short answer

Question: Arrange the given halides in the decreasing order of reactivity towards Sₙ1 reaction: 1. PhCH2Br 2. CH2=CHCH2Br 3. CH2=CHBr 4. CH3CH2Br Answer: The correct order of reactivity for Sₙ1 reaction is: PhCH2Br > CH2=CHCH2Br > CH2=CHBr > CH3CH2Br.

Step-by-step solution

Identify the carbocations

The following carbocations are formed from the given halides: 1. \(\mathrm{PhCH}_{2}^+\) 2. \(\mathrm{CH}_{2}=\mathrm{CH}^+\) 3. \(\mathrm{CH}_{3}\mathrm{CH}_{2}^+\) 4. \(\mathrm{CH}_{2}=\mathrm{CH}\mathrm{CH}_{2}^+\)

Determine the stability of carbocations

The stability order for carbocations is: 1. Allylic and benzylic carbocations, 2. 3° (tertiary) carbocations, 3. 2° (secondary) carbocations, 4. 1° (primary) carbocations. In this case, we have: 1. \(\mathrm{PhCH}_{2}^+\) is a benzylic carbocation. 2. \(\mathrm{CH}_{2}=\mathrm{CH}^+\) and \(\mathrm{CH}_{2}=\mathrm{CH}\mathrm{CH}_{2}^+\) are allylic carbocations. 3. \(\mathrm{CH}_{3}\mathrm{CH}_{2}^+\) is a 1° (primary) carbocation.

Arrange the carbocations in decreasing order of stability

Based on the stability order, we can arrange the carbocations as follows: 1. \(\mathrm{PhCH}_{2}^+\) 2. \(\mathrm{CH}_{2}=\mathrm{CH}^+\) and \(\mathrm{CH}_{2}=\mathrm{CH}\mathrm{CH}_{2}^+\) 3. \(\mathrm{CH}_{3}\mathrm{CH}_{2}^+\)

Determine the correct option

Based on the decreasing order of stability of the carbocations, the correct order of reactivity for Sₙ1 reaction is: \(\mathrm{PhCH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CH}-\mathrm{CH}_{2} \mathrm{Br}>\mathrm{CH}_{2}=\mathrm{CHBr}>\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Br}\) So, the correct answer is (d).

Key concepts

Carbocation Stability

Understanding carbocation stability is crucial for predicting chemical reaction outcomes, especially in substitution reactions like the SN1 mechanism. A carbocation is an intermediate with a positive charge on a carbon atom, making it a very reactive entity. The stability of carbocations is hierarchically ordered based on the number of alkyl groups attached to the positively charged carbon. The most stable configuration has three alkyl groups, termed a tertiary carbocation, followed by secondary and primary carbocations, respectively.

But why are tertiary carbocations more stable? It's because the alkyl groups surrounding the positively charged carbon can donate electron density through what's called inductive effect, helping to delocalize the positive charge. This delocalization reduces the energy of the carbocation, thus stabilizing it. Additionally, hyperconjugation, where neighboring C-H bonds can overlap with the empty p-orbital of the carbocation, further stabilizes the positively charged carbon atom.

In the context of the provided exercise, identifying the potential carbocations helps in determining the reactivity order of the halides in the SN1 reaction. The more stable the carbocation can form, the more reactive the halide will be, as it's more likely to reach that intermediate step in the pathway of the reaction.

Benzylic and Allylic Carbocations

Benzylic and allylic carbocations occupy a special place in the reactivity order due to their unique stability, often surpassing even tertiary carbocations. Benzylic carbocations form when the positive charge is on the carbon atom directly attached to an aromatic ring, as seen in the carbocation \(\mathrm{PhCH}_{2}^{+}\). The stability comes from resonance stabilization, where the positive charge can be delocalized over the aromatic system. This delocalization allows the charge to be spread out over a larger volume, significantly reducing the energy of the carbocation.

Similarly, allylic carbocations, such as \(\mathrm{CH}_{2}=\mathrm{CH}^+\) and \(\mathrm{CH}_{2}=\mathrm{CH}\mathrm{CH}_{2}^+\), are stabilized by resonance. The positive charge can resonate along the double bond system of an alkene, which allows the charge to be shared by more than one carbon atom, enhancing stability. In fact, resonance stabilization is so powerful that it places allylic and benzylic carbocations at the top of the carbocation stability chart, explaining their high reactivity in SN1 reactions.

Halides Reactivity

When it comes to SN1 reactions, the reactivity of halides is a direct reflection of the stability of the corresponding carbocations they can form. This is because SN1 reactions proceed through a carbocation intermediate after the halide ion leaves. The ease with which a halide ion can leave is also influenced by its bond strength and the stability of the halide ion itself.

For example, iodide is a better leaving group than bromide, which is in turn better than chloride and fluoride, because the larger the halide ion, the more stable it is when it takes on a negative charge. But in the context of an SN1 reaction, the nature of the carbon atom the halide is attached to is even more important. If the halide is attached to a benzylic or allylic position, or a tertiary carbon, then the SN1 reaction is more favorable, due to the relatively stable carbocations formed upon ionization.

The exercise correctly identifies the reactivity order based on the ability of different halides to form stable carbocations. This illustrates the overarching concept that the easier it is to form a stable carbocation, the higher the reactivity of the halide will be in an SN1 reaction, which corresponds nicely with the provided order of reactivity for halides in the context of SN1.