Match items of Row I with those of Row II.
Row I :
Row II :
(i) α-D-(-)-Fructofuranose
(ii) β-D-(-)-Fructofuranose
(iii) α-D-(-) Glucopyranose
(iv) β-D-(-)-Glucopyranose
Correct match is

Solution

Chemistry ma’am · 3 steps

Read ring size first, then compare the anomeric OH with the CH₂OH that sits on top.

1. Ring size. A and B have six-membered rings (five C and one O), so they are pyranoses, the glucose forms. C and D have five-membered rings (four C and one O), so they are furanoses, the fructose forms.
2. Alpha or beta. In a D-sugar Haworth structure, the CH₂OH on the last ring carbon points up. At the anomeric carbon (C1 of glucose, C2 of fructose), an OH pointing down gives α. An OH pointing up, on the same side as that CH₂OH, gives β.

   Structure	Ring	Anomeric OH	Name
   A	pyranose	C1, down	α-D-glucopyranose (iii)
   B	pyranose	C1, up	β-D-glucopyranose (iv)
   C	furanose	C2, down (CH₂OH up)	α-D-fructofuranose (i)
   D	furanose	C2, up (CH₂OH down)	β-D-fructofuranose (ii)

3. So. The match is A → iii, B → iv, C → i, D → ii, which is option (a).

The idea to keep. In a D-sugar Haworth ring, the terminal CH₂OH points up. Anomeric OH on the opposite side (down) is α, on the same side (up) is β. Six-membered ring is pyranose, five-membered is furanose.

Answer from the NTA final answer key. Solution by PrepWiser.