1-methyl ethylene oxide when treated with an excess of HBr produces

1. Step 1: Epoxide Ring Opening under Acidic Conditions ($$\text{HBr}$$, $$\text{SN}_2$$)

   When 1-methyl ethylene oxide is treated with $$\text{HBr}$$, the oxygen atom of the epoxide ring is first protonated by the strong acid to form a highly reactive oxonium ion intermediate. This protonation increases the electrophilicity of both ring carbon atoms.

   Although the reaction follows an $$\text{SN}_2$$-like pathway where the bromide ion ($$\text{Br}^\ominus$$) attacks as a nucleophile, the bond-breaking in a protonated epoxide precedes bond-making. This builds up significant partial positive charge ($$\delta^+$$) on the ring carbons. Because the more substituted secondary ($$2^\circ$$) carbon stabilizes this partial positive charge much better than the primary ($1^\circ$$) carbon, the bromide ion selectively attacks the more substituted secondary carbon:

   $$\text{CH}_3\text{--(CH--O--CH}_2\text{)} \xrightarrow{\text{H}^\oplus} \text{CH}_3\text{--(CH--OH}^\oplus\text{--CH}_2\text{)} \xrightarrow{\text{Br}^\ominus} \text{CH}_3\text{--CH(Br)--CH}_2\text{OH}$$

   This regioselective ring opening yields 1-bromopropan-2-ol as the intermediate.




2. Step 2: Substitution of the Hydroxyl Group with Excess $$\text{HBr}$$

   Because the reaction mixture contains an excess of $$\text{HBr}$$, the remaining secondary alcohol group ($$\text{--OH}$$) in 1-bromopropan-2-ol undergoes a subsequent nucleophilic substitution reaction:

   • The hydroxyl group is protonated by another equivalent of $$\text{HBr}$$ to form an excellent leaving group ($$\text{--OH}_2^\oplus$$).
   • A second bromide ion ($$\text{Br}^\ominus$$) then attacks the carbon, displacing the water molecule.
   $$\text{CH}_3\text{--CH(Br)--CH}_2\text{OH} \xrightarrow{\text{H}^\oplus} \text{CH}_3\text{--CH(Br)--CH}_2\text{OH}_2^\oplus \xrightarrow{\text{Br}^\ominus} \text{CH}_3\text{--CH(Br)--CH}_2\text{Br} + \text{H}_2\text{O}$$

   This converts the secondary alcohol into a vicinal dibromide: 1,2-dibromopropane.

Why Other Options Are Incorrect:

• Halohydrin Intermediates (e.g., 1-bromopropan-2-ol or 2-bromopropan-1-ol):

  Structures that contain a remaining hydroxyl group ($$\text{--OH}$$) alongside a single bromine atom only represent the intermediate species formed after the initial ring opening. They fail to account for the secondary substitution step driven by the specified excess reagent conditions.




• Geminal Dibromides (e.g., 1,1-dibromopropane or 2,2-dibromopropane):

  Geminal dibromides require both halogens to be added to the exact same carbon atom. The epoxide ring opening naturally distributes functional groups across adjacent carbons (carbons 1 and 2), making geminal placement chemically impossible here.

Conclusion:

The combination of regioselective acidic epoxide cleavage followed by acid-catalyzed substitution of the remaining alcohol transforms 1-methyl ethylene oxide completely into 1,2-dibromopropane.

Answer: Option B (1,2-dibromopropane)