L-Glucose is the enantiomer of D-Glucose, which means it has the same structure as D-Glucose but with opposite configurations at each chiral center.
In L-Glucose, the -OH groups are positioned in such a way that it is a mirror image of D-Glucose.
For L-Glucose, the configuration of the hydroxyl (-OH) groups should be as follows from top to bottom:
Based on the given configurations, Option (1) matches the structure of L-Glucose.
The correct structure of L-Glucose is represented by Option (1).
Match the following List-I with List-II and choose the correct option: List-I (Compounds) | List-II (Shape and Hybridisation) (A) PF\(_{3}\) (I) Tetrahedral and sp\(^3\) (B) SF\(_{6}\) (III) Octahedral and sp\(^3\)d\(^2\) (C) Ni(CO)\(_{4}\) (I) Tetrahedral and sp\(^3\) (D) [PtCl\(_{4}\)]\(^{2-}\) (II) Square planar and dsp\(^2\)
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to: