Formation of $\alpha$ and $\beta$ methyl glucosides by glucose on heating with ${\mathrm{CH}}_3\mathrm{OH}$ in presence of dry $\mathrm{HCl}$ gas indicates the presence of a/ an

The reaction discussed above is-

$${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+{\mathrm{CH}}_3\mathrm{OH}\to \alpha -{\mathrm{C}}_6{\mathrm{H}}_{11}{\mathrm{O}}_6{\mathrm{CH}}_3+\beta -{\mathrm{C}}_6{\mathrm{H}}_{11}{\mathrm{O}}_6{\mathrm{CH}}_3$$

The structure of glucose is cyclic i.e

Here the shown number 1 carbon in the structure of glucose is chiral carbon ( carbon which has all 4 different valencies) and it is chiral due to the presence of an aldehydic group due to which we get a product with two different configurations otherwise the hydroxy group is there at many carbons so aldehyde group is responsible for configurations.

Hence option A is the correct answer.