Question

$\frac{Br_2}{dark} \longrightarrow P$

Answer 1

Product P is 1,3–dibromo–2–methylpropane.

Answer 2

Solution:

The reaction is an electrophilic “opening” of a cyclopropane. Under Br₂ in the dark the electrophile (Br⁺) attacks the strained cyclopropane (which, though “saturated,” has bent bonds that are unusually reactive). When the electrophile adds (via formation of a cyclic bromonium‐ion intermediate) the ring opens. In our substrate the cyclopropane has a CH₃ substituent at one carbon and a Br at another. The bromonium–ion is formed over the more electron‐rich bond so that when Br⁻ attacks the intermediate (anti–attack common for such additions) the ring opens to give an open–chain dibromide.

Careful analysis of the regioselectivity (the CH₃ group helps stabilize any developing positive character) shows that the resulting product is

$$\textbf{1,3–dibromo–2–methylpropane.}$$

Thus, product \textbf{P} is 1,3–dibromo–2–methylpropane.

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Minimal Explanation (Core Steps):

1. Br₂ adds electrophilically to the cyclopropane (which is reactive due to ring strain) forming a bromonium ion.

2. The ring opens via backside attack by Br⁻.

3. Regioselectivity is governed by stabilization from the CH₃ group → product: 1,3–dibromo–2–methylpropane.