Question

The number of stereoisomers for n-carbon aldoses and ketoses respectively are:

• A. $2^{(n-2)}$ and $2^{(n-3)}$
• B. 0
• C. $2^{(n-1)}$ and $2^{(n-2)}$
• D. $2^{(n-3)}$ and $2^{(n-2)}$

Hint:

For a six-carbon aldohexose (like glucose, $n=6$), the number of stereoisomers is $2^{6-2} = 2^4 = 16$. For a six-carbon ketohexose (like fructose, $n=6$), the number of stereoisomers is $2^{6-3} = 2^3 = 8$.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Stereoisomers are molecules that share the same chemical formula and connectivity but differ in the spatial arrangement of their constituent atoms.
The number of stereoisomers for an organic molecule with asymmetric (chiral) carbon centers is determined by the number of those chiral centers.
Key Formula or Approach:
According to van't Hoff's rule, the maximum number of stereoisomers for a molecule with $k$ different chiral centers is given by:
\[ \text{Number of stereoisomers} = 2^k \]

Step 2: Detailed Explanation:
Let us determine the number of chiral centers ($k$) for an $n$-carbon aldose and an $n$-carbon ketose:
1. Aldoses: An aldose has an aldehyde group ($-\text{CHO}$) at carbon 1 and a primary alcohol group ($-\text{CH}_2\text{OH}$) at the terminal carbon ($n$).
Neither of these two terminal carbons is chiral because carbon 1 is double-bonded to oxygen, and the terminal carbon has two identical hydrogen atoms.
All the intermediate carbons ($n - 2$ carbons) are asymmetric, serving as chiral centers.
Therefore, for an $n$-carbon aldose, $k = n - 2$.
The number of stereoisomers is:
\[ \text{Stereoisomers} = 2^{(n-2)} \]
2. Ketoses: A ketose contains a ketone carbonyl group ($-\text{C}=\text{O}$) typically at carbon 2, and primary alcohol groups ($-\text{CH}_2\text{OH}$) at both carbon 1 and the terminal carbon ($n$).
These three carbons are non-chiral.
The remaining intermediate carbons are asymmetric, serving as chiral centers.
The number of chiral centers is $n - 3$.
Therefore, for an $n$-carbon ketose, $k = n - 3$.
The number of stereoisomers is:
\[ \text{Stereoisomers} = 2^{(n-3)} \]

Step 3: Final Answer:
The number of stereoisomers for $n$-carbon aldoses and ketoses are $2^{(n-2)}$ and $2^{(n-3)}$ respectively, corresponding to option (A).