Radical Stability and Halogenation Selectivity in Organic Chemistry

1. Tertiary ($3^\circ$) Hydrogens: There is one hydrogen on the C3 carbon (the carbon with the methyl branch). Replacing this leads to a $3^\circ$ radical.
2. Secondary ($2^\circ$) Hydrogens: There are two different sets of $CH_2$ groups. One set is at C2 and the other is at C4. These provide four $2^\circ$ hydrogens in total.
3. Primary ($1^\circ$) Hydrogens: There are three methyl groups ($CH_3$). One is the branch at C3, and the other two are at the ends of the pentane chain. This gives us a total of nine $1^\circ$ hydrogens.

When we perform monochlorination or bromination, each of these distinct sites can produce a different constitutional isomer.

The Competition: Chlorination vs. Bromination

The "selectivity" of a halogen refers to its ability to "choose" the most stable radical path over the statistically more likely path.

Monochlorination: The Unselective Path

Chlorine is highly reactive. In the transition state of a chlorination reaction, the bond to the hydrogen is only partially broken. Because the transition state does not look very much like the final radical, the energy difference between forming a $1^\circ$, $2^\circ$, or $3^\circ$ radical is relatively small.

Because chlorine is "impatient," it often reacts with whatever hydrogen it hits first. While it still prefers more stable radicals, the sheer number of $1^\circ$ hydrogens (nine) compared to the single $3^\circ$ hydrogen (one) means that chlorination often produces a complex mixture of many different products.

Bromination: The Selective Path

Bromine is much less reactive and far more "picky." The transition state for bromine abstraction occurs much later in the reaction coordinate. This means the transition state closely resembles the actual radical intermediate.

Because the transition state "feels" the stability of the radical, the energy barrier to form a $3^\circ$ radical is much lower than the barrier for a $1^\circ$ radical. Consequently, bromination is highly regioselective. Even if there are nine $1^\circ$ hydrogens and only one $3^\circ$ hydrogen, bromine will almost exclusively target the $3^\circ$ position.

Exam Tip: If a question asks which reagent is better for synthesizing a specific tertiary halide, the answer is almost always Bromine due to its high selectivity.

Quantitative Selectivity and Relative Reactivity

To move from qualitative predictions ("more stable is better") to quantitative ones ("what percentage of product X will I get?"), we use relative reactivity values. These values represent the rate at which a specific type of hydrogen reacts compared to a standard (usually methyl hydrogens).

To calculate the predicted percentage of a product, use the following formula: $\text{Relative Yield} = (\text{Number of equivalent hydrogens}) \times (\text{Relative reactivity of that type})$

Example Calculation Workflow

If you were provided with a relative reactivity table (for example: $1^\circ = 1$, $2^\circ = 3.8$, $3^\circ = 5$), you would follow these steps for 3-methylpentane:

1. Calculate the weighted value for each type:
   • $1^\circ$ contribution: $9 \text{ hydrogens} \times 1 = 9$
   • $2^\circ$ contribution: $4 \text{ hydrogens} \times 3.8 = 15.2$
   • $3^\circ$ contribution: $1 \text{ hydrogen} \times 5 = 5$
2. Find the total sum: $\text{Total} = 9 + 15.2 + 5 = 29.2$
3. Calculate the percentage for the $3^\circ$ product: $\% 3^\circ = \left( \frac{5}{29.2} \right) \times 100 \approx 17.1\%$

Note that in bromination, the relative reactivity values for $3^\circ$ hydrogens are much higher (often $>1600$), which is why the $3^\circ$ product dominates the yield regardless of the low number of hydrogens available.

Summary Table