Pent-1-yne, pent-2-yne and 3-methylbut-1-yne are each treated with water, mercury(II) sulphate and…
Contributed by StemPy LabTry it on paper first. A hint, the key idea and the worked solution are here when you need them.
Pent-1-yne, pent-2-yne and 3-methylbut-1-yne are each treated with water, mercury(II) sulphate and dilute sulphuric acid. Name the carbonyl product or products of each. How many of the three alkynes give a single product?
Need a reminder? The key ideaShow
A primary alcohol gives an aldehyde with PCC, or with copper at 573 K. A secondary alcohol gives a ketone. Terminal alkynes give methyl ketones with Hg and dilute acid. Rosenmund: acyl chloride, H, Pd-BaSO. Stephen: nitrile, SnCl and HCl, then water. The percentage yield is .
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Pent-1-yne. The terminal carbon holds the hydrogen. Water adds to the inner carbon, so the product is pentan-2-one only.
Step 2. Pent-2-yne. The two carbons of the triple bond are about equally substituted: one has a methyl group and the other an ethyl group. Water can add to either one. The product is a mixture of pentan-2-one and pentan-3-one.
Step 3. 3-Methylbut-1-yne. Water adds to the inner carbon of the terminal alkyne. The product is 3-methylbutan-2-one only.
Answer: two of the three give a single product.
Check. Every terminal alkyne except ethyne gives a methyl ketone, and a symmetrical alkyne gives one ketone.
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