Question

$\beta$-d-Glucose, a hemiacetal, can be converted to a mixture of acetals on treatment with in the presence of acid. Draw a stepwise mechanism for this reaction. Explain why two acetals are formed from a single starting material.

Short answer

The stepwise mechanism is shown here under

Attack of HCl

Formation of acetals

The two acetals are formed owing to the trigonal planar symmetry of the carbocationic center.

Step-by-step solution

Dehydroxylation

The $\beta$-d-Glucose undergoes dehydroxylation due to the attack of a strong acid HCl. The OH group next to the hetero-atom in the ring is eliminated, leaving a carbocationic center at its position.

Nucleophilic addition

The oxygen in the CH₃OH group donates its lone pair of electrons to the carbocationic center produced upon dehydroxylation in the ring.

Mechanism

The $\beta$-d-Glucose ends up forming two acetals, one with an up oxy methyl group and another one with a down oxy methyl group.

In the first step, HCl attacks the OH group to form a H₂O⁺ group on the ring.

Attack of HCl

The obtained carbocation then reacts with the CH₃OHgroup to form two acetals. The two acetals are formed owing to the trigonal planar symmetry of the carbocationic center.

Formation of acetals