(a). Name the smallest aldose which forms cyclic hemiacetal and the functional groups are involved in its formation.
(b). What is invert sugar ?
(c ). Calculate the specific rotation of invert sugar. Given, `|alpha|_(D)"of" D-glucose=52.7^(@)`
`|alpha|_(D) "of" D-fructose=-92.4^(@)`
(d). Give the mechanism of mutarotaion of `beta-D-` gluoco-pyranose in (i) `aq. H^(oplus)` and (ii) `overset(Θ)(O)H`.
(e). Why is the mutaraotation faster in the presence of `2-`pyridinol?

Explanation

a. The smallest aldose is a tetraldose which has four `C` atoms and an `O` atom to form a five-membered ring. The `CHO` group and `1^(@) OH` group of tetrose are involved in the formation of ring. Both erythrose and threose form cyclic hemiacetals.
i.
ii.
b. Equimolar mixture of `D-`glucose and `D-`fructose obtained but the hydrolysis of sucrose is called invert sugar. since the specific rotation of sucrose is positive and after hydrolysis it changes to negative value and this process is called inversion of cane sugar.
c. The specific rotation is half the sun of specific rotation of glucose and fructose.
:. Specific rotaiion of invert sugar `=(1)/(2)[+52.7^(@)+(-92.4^(@))]=-19.9^(@)`
d. i. In acidic medium:

ii. In basic medium:

The rate depends on the conversion of cyclic hemiacetal to open-chain aldehyde `(I)`.
e. `2-`Pyridinol is an acid-base catalyst (containing both basic `(N)` and acidic `(OH)` groups] which gives `H^(o+)` ion to hemiacetalic `O` atom and simultaneously removes `H^(o+)` ion from the `OH` group of the hemiacetal by the formation a cyclic intermediate transition state.