Advertisements
Advertisements
प्रश्न
Account for the following:
There are 5 OH groups in glucose
Advertisements
उत्तर
Acetylation of glucose with acetic anhydride gives glucose pentaacetate which confirms the presence of five –OH groups. Since it exists as a stable compound, five –OH groups should be attached to different carbon atoms.
\[\begin{array}{cc}
\phantom{...}\ce{CHO}\phantom{...................}\ce{CHO}\phantom{....}\phantom{}\ce{O}\phantom{.........}\\
\phantom{..}|\phantom{.......................}|\phantom{}\phantom{........}||\phantom{........}\\
\phantom{}\ce{(CHOH)4 ->[Acetic anhydride] (CH-O-C-CH3)4}\\
\phantom{..........................}\ce{O}\\
\phantom{..}|\phantom{.......................}|\phantom{.}\phantom{........}||\phantom{........}\\
\phantom{}\ce{CH2OH}\phantom{................}\ce{CH2-O-C-CH3}\phantom{}
\end{array}\]
APPEARS IN
संबंधित प्रश्न
Differentiable between the following:
Amylose and Amylopectin
What do you observe when glucose is treated with bromine water?
The symbols D and L represents ____________.
Which is the least stable form of glucose?
Which one is correct?
In the following reaction, identify A and B:
\[\begin{array}{cc}
\ce{C6H12O6 ->[Acetic anhydride] A}\\
\downarrow \text{Conc. nitric acid}\phantom{...}\\
\ce{B}\phantom{.................}\end{array}\]
Why does compound (A) given below not form an oxime?
(A)
How will you distinguish 1° and 2° hydroxyl groups present in glucose? Explain with reactions.
On the basis of which evidences D-glucose was assigned the following structure?
\[\begin{array}{cc}
\ce{CHO}\\
|\phantom{....}\\
\phantom{..}\ce{(CHOH)4}\\
|\phantom{....}\\
\phantom{..}\ce{CH2OH}
\end{array}\]
Consider the following reactions:
(i) \[\ce{Glucose + R-OH ->[Conc. HNO3] [A] ->[X eq of][(CH3CO)2O] Acetyl derivative}\]
(ii) \[\ce{Glucose ->[Ni/H2] [A] ->[Y eq of][(CH3CO)2O] Acetyl derivative}\]
(iii) \[\ce{Glucose ->[Z eq of][(CH3CO)2O] Acetyl derivative}\]
'X, 'Y' and 'Z' in these reactions are respectively:
