Write the product obtained when D-glucose reacts with \( H_2N - OH \).
When D-glucose reacts with hydroxylamine (\( H_2N - OH \)), the product obtained is D-glucose oxime. The reaction involves the formation of an oxime by the condensation of the aldehyde group (at C-1) of D-glucose with hydroxylamine, eliminating a molecule of water (\( H_2O \)).
Reaction:
\[
\text{D-Glucose} + H_2N-OH \rightarrow \text{D-Glucose Oxime} + H_2O
\]
Structure of D-Glucose Oxime:
The oxime forms at the C-1 position of D-glucose, replacing the aldehyde group (\( -CHO \)) with the oxime group (\( C=N-OH \)). The rest of the glucose structure remains unchanged.
Key Points:
Final Answer:
The product is D-glucose oxime (\( C_6H_{12}O_5N-OH \)), formed at the C-1 position.
Identify A in the following reaction. 
For the reaction, \(N_{2}O_{4} \rightleftharpoons 2NO_{2}\) graph is plotted as shown below. Identify correct statements.
A. Standard free energy change for the reaction is 5.40 kJ \(mol^{-1}\).
B. As \(\Delta G\) in graph is positive, \(N_{2}O_{4}\) will not dissociate into \(NO_{2}\) at all.
C. Reverse reaction will go to completion.
D. When 1 mole of \(N_{2}O_{4}\) changes into equilibrium mixture, value of \(\Delta G = -0.84 \text{ kJ mol}^{-1}\).
E. When 2 mole of \(NO_{2}\) changes into equilibrium mixture, \(\Delta G\) for equilibrium mixture is \(-6.24 \text{ kJ mol}^{-1}\).
Choose the correct answer from the following.
