Q4-44E - Organic Chemistry (Mcmurry)

Question

Galactose, a sugar related to glucose, contains a six-membered ring in which all the substituents except the $\mathbf{-}\mathbf{OH}$ group, indicated below in red, are equatorial. Draw galactose in its more stable chair conformation.

Step-by-Step Solution

Answer

$\mathrm{\beta }$ Galactose is more stable than $\mathrm{\alpha }$ galactose.

Step 1: Chair form of cyclohexane

Four conformations are possible for cyclohexane: a chair, twist boat, boat, and half chair. Among all the four conformations chair form of cyclohexane is considered a very stable one. Chair conformation of cyclohexane is as follows;

Chair form of cyclohexane

We can say if the relation is up-up, then cis; if down–down relationship, then trans.

If the up-down or down-up relation between the bond, then trans.

Step 2: Chair confirmation of galactose

The two possible structures of galactose are as follows;

In the case of  $\mathrm{\alpha }$galactose, there are two $-\mathrm{OH}$groups that occupy the axial position, and hence due to axial strain, their stability decreases.

In the $\mathrm{\beta }$ galactose, there is only one $-\mathrm{OH}$group that occupies the axial position; hence it faces less 1, 3 diaxial strain as compared to $\mathrm{\alpha }$ galactose.

Therefore  $\mathrm{\beta }$ galactose is more stable than that of  $\alpha$ galactose.