Question

The number of chiral centers in open-chain structures of glucose is
A.3
B.4
C.5
D.6

Answer 1

Hint: We know that the chiral center is a point at which all the four groups surrounding the point are of different atoms. An example of it is $C(F)(Cl)(Br)H$. The molecular formula of Glucose is
${{{C}}_{{6}}}{{{H}}_{{{12}}}}{{{O}}_{{6}}}$. From the open chain structure of glucose, the number of chiral centers present in the glucose molecule can be found out.

Complete step by step answer:
As we know that here, the chiral center is a carbon that is attached to four different groups of atoms.
The open-chain structure of glucose is drawn as -

           .

Here we can see that there is no internal plane of symmetry.
In ${C^1}$, there is a double bond to Oxygen, so we can consider it a 2 oxygen atom. So it is not a chiral carbon.
On checking the ${C^2},{C^3},{C^4},{C^5}$ , the valency of carbon is satisfied by 4 different groups of atoms. Hence they are chiral.
Therefore the total number of chiral centers present in the open-chain structure of glucose is four.

Therefore, the atomic masses of the element X and Y are 25.6 and 42.6

So, the correct answer is Option B.

Note: The open-chain structure of glucose has 4 chiral centers. The cyclic structure of glucose resembles pyran and due to the formation of cyclic hemiacetal, there are five chiral centers present in the cyclic form of glucose.