Citric acid (2-hydroxy-1,2,3-propanetricarboxylic acid) is the primary acid of citrus fruits, which — Analytical Chemistry Chemistry Question
Structure elucidation with stereochemistry
Citric acid (2-hydroxy-1,2,3-propanetricarboxylic acid) is the primary acid of citrus fruits, which contributes to their sour taste. Commercial manufacturing of citric acid involves fermentation of molasses or starch using the fungus Aspergillus niger at pH 3.5. It is widely used in food, soft drinks and as a mordant in dyeing. It is also an important biochemical intermediate.
What transformation will citric acid undergo when warmed with concentrated sulfuric acid at 45 – 50 °C? Give the structure and IUPAC name of the product obtained. Which type of organic acids would undergo a similar reaction?
Model Answer
Citric acid undergoes decarbonylation and dehydration when warmed with concentrated sulfuric acid at 45 – 50 °C to yield 3-oxo-1,3-pentanedioic acid (acetonedicarboxylic acid):
[VISUAL]
3-oxo-1,3-pentanedioic acid
-Hydroxy carboxylic acids undergo similar reaction.
After warming citric acid with sulfuric acid, anisole (methoxybenzene) is added to the reaction mixture and product A (C12H12O5) is obtained. On heating with acetic anhydride, A forms an anhydride. 118 mg of A requires 20 cm3 of KOH solution (c = 0.05 mol dm-3) for neutralisation.
Reaction with bromine indicates that the same amount of compound A requires 80 mg of bromine to give an addition product.
Deduce the structure of A.
Model Answer
Molar mass of A = 236 g mol -1.
20 cm3 of KOH solution (c = 0.05 mol dm-3) <=> 118 mg A
1000 cm3 of KOH solution (c = 1 mol dm-3) <=> 118 g A
Thus, the acid is dibasic.
Molar mass of A = 236 g mol -1
80 mg Br2 <=> 118 mg A
160 mg Br2 <=> 236 g A
A contains one double bond C=C.
It has anisole ring in the molecule (OCH3) and is formed from HOOCCH2COCH2COOH. It has molecular formula C12H12O5.
Due to steric hindrance the attachment of the aliphatic portion on the anisole ring will be para with respect to -OCH3. Hence the structure will be:
[VISUAL]
As A forms anhydride the two COOH groups should be on the same side of the double bond.
Identify the possible isomers of A in this reaction and give their structures, absolute configurations and the IUPAC names.
Model Answer
The possible isomers of A are:
[VISUAL]
(E) 3-( 2-methoxyphenyl )-2-pentenedioic acid
[VISUAL]
(Z) 3-( 2-methoxyphenyl )-2-pentenedioic acid
[VISUAL]
(Z) 3-( 4-methoxyphenyl )-2-pentenedioic acid
How many stereoisomers of A will be obtained in the bromination reaction? Draw their Fischer projections.
Model Answer
Two products are possible when compound A reacts with bromine:
[VISUAL]
Structures 1 and 2 are enantiomers.
Assign absolute configurations to the stereocentres in all the stereoisomers formed in 16.4.
Model Answer
The absolute configurations of the stereocentres are assigned as:
[VISUAL]
Structure [1]: R, S
Structure [2]: S, R
If phenol and resorcinol are separately added to the reaction mixture instead of anisole, compounds B and C are obtained, respectively. B does not give any coloration with neutral FeCl3 solution, but C does. Under identical reaction conditions, the yield of compound C is much higher than that of B.
Give appropriate structures for B and C.
Model Answer
The structures for B and C are:
[VISUAL]
Structure B (Product obtained by reaction with phenol)
[VISUAL]
Structure C (Product obtained by reaction with resorcinol)
What is the difference between the reactions leading to the formations of A and B, respectively?
Model Answer
In the formation of compound A from anisole, the attack takes place at the p-position of the OCH3 group. However, when compound B is formed from phenol, the attack takes place at the o-position of the OH group. Steric hindrance of OCH3 group favours the attack at the para position. Steric hindrance of the OH group is comparatively less. Thus, the attack is possible at the ortho or para positions. However, addition at ortho position is favoured as it leads to cyclization of the intermediate acid to stable B.
Why is the yield of C higher than that of B?
Model Answer
Phenol has only one OH group on the phenyl ring whereas resorcinol has two OH groups on the phenyl ring at the m-positions. Hence, position 4 is considerably more activated (i.e, electron rich) in the case of resorcinol.
[VISUAL]
Therefore, under identical reaction conditions, the yield of compound C is much higher than that of B.