Question

$1-\mathrm{methyl}\mathrm{ethylene}\mathrm{oxide}$ when treated with an $\mathrm{excess}\mathrm{of}\mathrm{HBr}$ produces:

• A.
• B.

  

• C.

  

• D.

  

Answer 1

The correct option is C

Explanation for the correct option:

C)

Step 1: Draw the structures of $1-\mathrm{methyl}\mathrm{ethylene}\mathrm{oxide}$ and $\mathrm{HBr}$:

and

Step 2: Releasing of $\mathit{B}{\mathit{r}}^{\mathbf{-}}$ ion:

1. Due to the addition of ${\mathrm{H}}^{+}$ ion, oxygen gets a positive charge on it.
2. 

Step 3: Ring opening due to the attack of $\mathit{B}{\mathit{r}}^{\mathbf{-}}$ ion:

1. The released $B{r}^{-}$ ion gets attacked on the ring carbon-containing $\mathrm{methyl}\left(-{\mathrm{CH}}_3\right)$group.
2. Due to this, the ring-opening takes place.
3. This results in the formation of $2-\mathrm{Bromoprop}-1-\mathrm{ol}$.
4. 

Step 4: Releasing of $\mathit{B}{\mathit{r}}^{\mathbf{-}}$ ion:

1. The lone pair of electrons on oxygen attack on ${\mathrm{H}}^{+}$ ion of $\mathrm{HBr}$ and becomes a positively charged ion.
2. 

Step 5: Attack of $\mathit{B}{\mathit{r}}^{\mathbf{-}}$ ion on $\mathbf{C}\mathbf{1}$ carbon:

1. The released $B{r}^{-}$ ion from $\mathrm{HBr}$ attacks on the $\mathrm{C}1$ carbon.
2. Due to this, the ${\mathrm{H}}_2\mathrm{O}$ molecule gets eliminated.
3. This results in the formation of $\mathrm{trans}-1,2-\mathrm{dibromopropane}$.
4. 

Hence, option C is correct. The product obtained when $1-\mathrm{methyl}\mathrm{ethylene}\mathrm{oxide}$ treated with an $\mathrm{excess}\mathrm{of}\mathrm{HBr}$ is $\mathrm{trans}-1,2-\mathrm{dibromopropane}$.