Sr.No |
Aldehyde(R1) |
M.F. and M.W. |
Ketone(R2) |
M.F. And M.W> |
Chalcone M.F. and M.W. |
||
1 |
-CH3 |
C8H8O(120) |
-CH3 |
C9H10O(134) |
C17H16O(236) |
||
2 |
-CH3 |
C8H8O(120) |
-OCH3 |
C9H10O2(150) |
C17H16O2(252) |
||
3 |
-CH3 |
C8H8O(120) |
-F |
C8H7OF(138) |
C16H13OF(240) |
||
4 |
-Cl |
C7H5OCl(140.5) |
-CH3 |
C9H10O(134) |
C16H13OCl(256.5) |
||
5 |
-Cl |
C7H5OCl(140.5) |
-OCH3 |
C9H10O2(150) |
C16H13O2Cl(172.5) |
||
6 |
-Cl |
C7H5OCl(140.5) |
F |
C8H7OF(138) |
C15H10OClF(260.5) |
||
7 |
-OCH3 |
C8H8O2(136) |
-OCH3 |
C9H10O2(150) |
C17H16O3(268) |
||
8 |
-O-CH2-C6H5 |
C14H12O2(212) |
-CH3 |
C9H10O(134) |
C23H20O2(328) |
||
9 |
-O-CH2-C6H5 |
C14H12O2(212) |
-OCH3 |
C9H10O2(150) |
C23H20O3(354) |
2)Conversion of mmol (milimole) into gram :
We know that the molecular weight of given compound expressed in gram corresponds to 1 mole of substance.
And 1 mmol = 1 x 10-3 mole ----(similar to gram to milligram)
We have 1 mole of compound = molar mass of that compound in gram.
1 mmol of compound = molar mass of that compound in gram x 10-3.
Hence,
3.5 mmol of compound = 3.5 x molar mass of that compound in gram x 10-3.
e.g.
For 4-Methylbenzaldehyde molar mass = 120 g
It means 1 mole 4-Methylbenzaldehyde = 120 g 4-Methylbenzaldehyde.
Hence, using above eq.(1)
3.5 mmol 4-Methylbenzaldehyde = 3.5 x molar mass of 4-Methylbenzaldehyde in gram x 10-3
= 3.5 x 120 x 10-3.
= 0.420 g
3.5 mmol 4-Methylbenzaldehyde = 0.420 g or
3.5 mmol 4-Methylbenzaldehyde = 420 mg.
Hence the conversion of mmol to gram.
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