Since ancient times, the bark of willow trees has been used for pain relief - VCE - SSCE Chemistry - Question 8 - 2010 - Paper 1
Question 8
Since ancient times, the bark of willow trees has been used for pain relief. In the 19th century, chemists isolated the active compound, salicin, from the bark. This... show full transcript
Worked Solution & Example Answer:Since ancient times, the bark of willow trees has been used for pain relief - VCE - SSCE Chemistry - Question 8 - 2010 - Paper 1
Step 1
What type of linkage is circled in the structure of salicin?
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Answer
The linkage circled in the structure of salicin is an ester linkage. This type of linkage is characterized by the formation of an ester bond between a hydroxyl group and a carboxyl group.
Step 2
In step 1, salicyl alcohol and another compound is produced.
- i. What group of biomolecules does this other compound belong to?
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The other compound produced in step 1 belongs to the group of carbohydrates. Specifically, it can be categorized as a sugar or monosaccharide.
Step 3
- ii. The structure of this other compound is not complete. Write the formula of the atom or group of atoms represented by A in the reaction scheme above.
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The atom or group of atoms represented by A in the reaction scheme above is the hydroxyl group (-OH), which is characteristic of alcohols.
Step 4
Step 2 involves the conversion of salicyl alcohol into salicylic acid.
- i. What type of reaction is step 2?
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Step 2 is classified as an oxidation reaction. In this reaction, the salicyl alcohol undergoes oxidation to form salicylic acid.
Step 5
- ii. Suggest a suitable reagent to carry out the reaction.
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A suitable reagent to facilitate the oxidation of salicyl alcohol to salicylic acid is chromic acid (H2CrO4).
Step 6
Step 3 requires sulfuric acid catalyst and another reagent. Name this reagent.
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The other reagent required in step 3 is acetic anhydride, which is often used in the acetylation process.
Step 7
Draw the structure of product B.
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Product B, as a result of the reaction with a strong base, would be the deprotonated form of aspirins, often referred to as the aspirin anion, represented as:
O
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C6H4 – O–
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CH3
This indicates the loss of the hydrogen from the hydroxyl group in the presence of a strong base.
