The number of chiral carbons present in sucrose is..........

We first recall that sucrose is a disaccharide obtained by joining an $$\alpha$$-D-glucopyranose unit with a $$\beta$$-D-fructofuranose unit through an $$\alpha(1\rightarrow2)\beta$$ glycosidic linkage. In other words, the anomeric carbon $$C_1$$ of glucose is linked to the anomeric carbon $$C_2$$ of fructose. Our task is to count how many carbon atoms in the whole sucrose molecule are chiral, i.e. attached to four different substituents.

Let us examine the glucose part first. In the six-membered ring (pyranose form) of D-glucose we number the ring atoms clockwise starting from the anomeric centre. The atoms and their substituents are:

$$ \begin{aligned} C_1 &:~\text{attached to } H,\;O\!-\!\text{fructose},\;O(\text{ring}),\;C_2 \\ C_2 &:~\text{attached to } OH,\;H,\;C_1,\;C_3 \\ C_3 &:~\text{attached to } OH,\;H,\;C_2,\;C_4 \\ C_4 &:~\text{attached to } OH,\;H,\;C_3,\;C_5 \\ C_5 &:~\text{attached to } OH,\;CH_2OH,\;C_4,\;O(\text{ring}) \end{aligned} $$

At every one of $$C_1,\,C_2,\,C_3,\,C_4,$$ and $$C_5$$ the four groups listed are all different, so each of these five carbons is chiral. Thus the glucose moiety contributes

$$ 5 \text{ chiral carbons.} $$

Now we consider the fructose part. In the five-membered furanose ring of D-fructose we number the ring atoms starting at the anomeric carbon $$C_2$$ (because $$C_1$$ is the exocyclic $$CH_2OH$$ group). The relevant atoms are:

$$ \begin{aligned} C_2 &:~\text{attached to } H,\;O\!-\!\text{glucose},\;O(\text{ring}),\;C_3 \\ C_3 &:~\text{attached to } OH,\;H,\;C_2,\;C_4 \\ C_4 &:~\text{attached to } OH,\;H,\;C_3,\;C_5 \\ C_5 &:~\text{attached to } CH_2OH,\;OH,\;C_4,\;O(\text{ring}) \end{aligned} $$

Again, each of $$C_2,\,C_3,\,C_4,$$ and $$C_5$$ is bonded to four distinct groups, and each is therefore chiral. Hence the fructose moiety contributes

$$ 4 \text{ chiral carbons.} $$

Adding the contributions from both halves of the molecule, we obtain

$$ \text{Total chiral carbons} = 5 \;(\text{glucose}) + 4 \;(\text{fructose}) = 9. $$

So, the answer is $$9$$.