Propylene oxide is a chiral molecule. Hydrolysis of propylene oxide gives propylene glycol, another chiral molecule.

(a) Draw the enantiomers of propylene oxide.

(b) Propose a mechanism for the acid-catalyzed hydrolysis of pure $\mathbf{\text{(}}\mathit{R}\mathbf{\text{)-propyleneoxide}}$.

(c) Propose a mechanism for the base-catalyzed hydrolysis of pure $\mathbf{\text{(}}\mathit{R}\mathbf{\text{)-propyleneoxide}}$.

(d) Explain why the acid-catalyzed hydrolysis of optically active propylene oxide gives a product with lower enantiomeric excess and a rotation opposite that of the product of the base- catalyzed hydrolysis.

Molecules having exact mirror images of itself and has non- super imposable mirror image is known as a chiral molecule. Enantiomers or optical isomers are two mirror images of a chiral molecule.

Epoxides react under both acidic and basic conditions. The products of the acid-catalyzed ring opening depend primarily on the solvent used.

Epoxides are considered to be more reactive since opening the epoxide ring relieves the strain of the three-membered ring. Strong bases can attack and open epoxides, even though the leaving group is an alkoxide. Amines can also open epoxides.

(a) The enantiomers of propylene oxide are $\text{(}S)\text{-2-methyoxirane}$ and$\text{(}R)\text{-2-methyoxirane}$ .

$\text{enantiomers\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}propylene\hspace{0.17em}\hspace{0.17em}oxide}$

(b)The acid-catalyzed hydrolysis of pure $\text{(}R\text{)-propyleneoxide}$ gives $\left(S\right)\text{-propane-1,2-diol}$.

$\text{acid-catalyzedhydrolysisof(}\mathrm{R}\text{)-propyleneoxide}$

(c) The base-catalyzed hydrolysis of pure$\text{(}R\text{)-propyleneoxide}$ gives $\left(\mathrm{R}\right)\text{-propane-1,2-diol}$.

$\text{base-catalyzedhydrolysisof(}R\text{)-propyleneoxide}$

(d) Acid catalyzed hydrolysis of epoxides gives the diols with anti-stereochemistry. The mechanism for this hydrolysis process involves protonation of oxygen (epoxide) that is followed by${{\text{S}}_{\text{N}}}^{\text{2}}$ attack by a water molecule. Anti-stereochemistry results since there is a back side attack of water on the protonated epoxide. Hence, we get the inversion of configuration. This is the reason for low enantiomeric excess of the product formed by acid-catalyzed hydrolysis.