Cistus L is an aromatic, erect branched shrub and is a significant element of Greek flora. It can be — Organic Chemistry Chemistry Question
The Chemistry and Identification of Flavonoids
Cistus L is an aromatic, erect branched shrub and is a significant element of Greek flora. It can be found in stony slopes and hills and it can also be found in pinewoods. In folk medicine the flower branches of cistus monospeliensis have been used for asthma, while the leaves may replace tea. Flavonoids are widely distributed in plants as glycosides or as free aglycons. They are known to exhibit a broad spectrum of pharmacological properties including antimicrobial, antitumor, antiviral, as well as enzyme inhibition and central vascular system activity.
Apigenin is a very widely distributed flavonoid. Its structure is shown below:
[VISUAL]
In the following reactions draw the structures of products B and C.
Apigenin
1. NaH/DMF
2. excess MeI
⟶ B
Apigenin
acetic anhydride(excess)
pyridine
⟶ C
Model Answer
Product B is the trimethyl ether derivative of apigenin (5,7,4'-trimethoxyflavone), where all three phenolic hydroxyl groups (at positions 5, 7, and 4') are converted to methyl ethers (-OCH3). Product C is the triacetate derivative of apigenin (5,7,4'-triacetoxyflavone), where all three phenolic hydroxyl groups are converted to acetyl esters (-OCOCH3).
Apigenin can form a hydrogen bond between the phenolic hydroxyl group attached to C–5 and the carbonyl group at C–4. The 1 H–NMR resonance of the phenolic proton at C–5 will be shifted relative to the phenolic protons at C–7 and C–4':
a) down field, b) up field, c) not shifted.
Model Answer
a) down field. The 1H-NMR resonance of phenolic proton involvement in hydrogen bonding will be observed at very low magnetic field (~ 12ppm).
When treated with aqueous solution of NaOH (c = 2 mol dm–3), apigenin gives (among other products) D and E
NaOH, c = 2 mol dm-3 Apigenin ⟶ D + E
Compound D (C6H6O3) gives a positive test with FeCl3 and its 1 H–NMR spectrum consists of only one aromatic singlet peak (spectrum I). Compound E (C9H12O2) also gives a positive test with FeCl3. In the 1 H-NMR spectrum the aliphatic region shows one multiplet and two triplet peaks, while the aromatic region consists of two doublets (spectrum II). Draw the structure of compounds D and E.
[VISUAL]
Model Answer
Compound D is phloroglucinol (benzene-1,3,5-triol), which gives a single aromatic singlet peak at 6.15 ppm in its 1H-NMR spectrum because of its high symmetry. Compound E is 3-(4-hydroxyphenyl)propan-1-ol, which shows a para-substituted pattern (two doublets) in the aromatic region of its 1H-NMR spectrum and a propyl-1-ol side chain (one multiplet and two triplets) in the aliphatic region.
Indicate with arrows the three carbon atoms in structure C that will give rise to characteristic peaks in 13 C-NMR which distinguish structure C from B.
Model Answer
The three carbon atoms in structure C are the carbonyl carbons of the three acetyl groups (-OCOCH3). These carbonyl carbons are part of ester groups and show characteristic signals in the 13C-NMR spectrum at high chemical shifts (~165-175 ppm), distinguishing them from the methoxy-substituted B. Note: In the solution key, the explanation of the three characteristic carbonyl peaks is incorrectly printed under 27.3, but it corresponds to 27.4.