This question is from Organic chemistry section of JEE Advance 2019. First time when you see this question, it looks a bit difficult. Don’t lose your confidence. I’ll tell you how easy it is. We have to apply the basic organic reactions and named reactions, that’s it!!
In this question, set of reactions are given and we are asked to find out the products.
Explanation
This complete reaction includes:
- In first step, reaction of alkyne with HgSO4 and dilute H2SO4, the product formed will be a ketone at triple bond. In second step, the aldehyde group present at the end of the chain show tollen’s reagent test. In third step, Clemmenson reduction occurs, which finally form the product Q.
- Now in product -COOH group is present which reacts with SOCl2 in pyridine to form acid chloride. Then, Friedel Craft acylation reaction occurs to form the product R.
- In last step, again Clemmenson reduction occurs to form the final product S.
The step-wise reaction is given below:
For complete explanation, go through this video:
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