The incorrect statement among the following is

To discover which of the four statements is incorrect, let us recall the precise stereochemical meanings of the terms “enantiomers”, “anomers”, and the structural facts about glucose derivatives and cellulose.

For any chiral molecule having several stereogenic centers, an enantiomer is defined as the non-superposable mirror image of the molecule as a whole. That means every chiral centre is inverted: for every $$\mathrm{R}$$ configuration we obtain $$\mathrm{S}$$ and vice-versa. In contrast, an anomer is a special kind of diastereomer found only in cyclic hemiacetals or hemiketals of sugars. Anomers differ in configuration only at the new stereogenic centre created when the carbonyl carbon becomes the anomeric carbon (C-1 in aldoses, C-2 in ketoses). All the other chiral centres remain unchanged.

Consider $$\alpha \text{-D-glucose}$$ and $$\beta \text{-D-glucose}$$.

Write the Fischer projection for open-chain D-glucose:

$$ \begin{array}{c} \mathrm{CHO}\\ |\\ \mathrm{H}\!\!-\!\!\mathrm{C}\!\!\!\!\! \begin{array}{c}OH\end{array}\\ |\\ \mathrm{HO}\!\!-\!\!\mathrm{C}\!\!\!\!\! \begin{array}{c}H\end{array}\\ |\\ \mathrm{H}\!\!-\!\!\mathrm{C}\!\!\!\!\! \begin{array}{c}OH\end{array}\\ |\\ \mathrm{H}\!\!-\!\!\mathrm{C}\!\!\!\!\! \begin{array}{c}OH\end{array}\\ |\\ \mathrm{CH}_2OH \end{array} $$

When this aldehyde cyclises, the carbonyl carbon (C-1) becomes a new stereocentre. If, in the Haworth representation, the -OH on C-1 is drawn down (trans to the CH$$_2$$OH substituent at C-5), the configuration is named $$\alpha$$; if the -OH is drawn up (cis to CH$$_2$$OH), the configuration is named $$\beta$$. All the other chiral centres at C-2, C-3, C-4, and C-5 remain identical in both forms. Therefore:

$$\alpha \text{-D-glucose} \quad \text{and} \quad \beta \text{-D-glucose}$$

differ only at C-1, so they are anomers. They are not related as mirror images, because only one of the five stereocentres has opposite configuration while the rest are the same. Thus they cannot be enantiomers.

Consequently, Option A, which states $$\alpha$$ $$-$$ D $$-$$ glucose and $$\beta$$ $$-$$ D $$-$$ glucose are enantiomers, is false.

Now consider Option B. When glucose is converted into its penta-acetate, every hydroxyl group, including the anomeric -OH at C-1, becomes acetylated ($$-OCOCH_3$$). Hydroxylamine reacts with free carbonyl groups or free anomeric hydroxyls to form oximes; but in the penta-acetate, there is no free -OH or free carbonyl. Therefore the penta-acetate does indeed fail to react with hydroxylamine, so Option B is true.

Option C explicitly states that $$\alpha$$ $$-$$ D $$-$$ glucose and $$\beta$$ $$-$$ D $$-$$ glucose are anomers. As shown above, this is correct, so Option C is true.

Option D claims that cellulose is a straight-chain polysaccharide made exclusively of $$\beta$$ $$-$$ D $$-$$ glucose units. In cellulose, each glucose residue is linked by $$\beta(1 \rightarrow 4)$$ glycosidic bonds, producing an unbranched, ribbon-like chain. Hence the statement is accurate, so Option D is true.

Only Option A does not align with chemical facts.

Hence, the correct answer is Option A.