
The compound with formula ${C_6}{H_10}O$:
A.
B.
C.
D. None of these




Answer
516k+ views
Hint: We know that the carbonyl compounds on reaction with iodine in the presence of an alkali results in the formation of iodoform which is a yellow color solid and has chemical formula $\mathrm{CHI}_{3}$ and sodium salt of carboxylic acid.
Complete step by step answer:
When we react 4-methoxypent-2-yne with HI then it results in the formation of pent-3-yne-2-ol. Now, when we react pent-3-yne-2-ol with sodium hydroxide and iodine, the formation of iodoform takes place and this solid precipitate of yellow color is formed.
The whole chemical equation which is explained above is shown as follows.
Thus, we can say that the unknown compound with the chemical formula $\mathrm{C}_{6}\mathrm{H}_{10}\mathrm{O}$ is 4-methoxypent-2-yne.
Hence, we can conclude that the correct option is C.
Note:
We can also say that the reaction which is carried out in the question is also given by other halogens and commonly referred to as haloform reaction. Therefore, it is also true that the reaction with iodine is used in the quantitative analysis of organic compounds.
Complete step by step answer:
When we react 4-methoxypent-2-yne with HI then it results in the formation of pent-3-yne-2-ol. Now, when we react pent-3-yne-2-ol with sodium hydroxide and iodine, the formation of iodoform takes place and this solid precipitate of yellow color is formed.
The whole chemical equation which is explained above is shown as follows.

Thus, we can say that the unknown compound with the chemical formula $\mathrm{C}_{6}\mathrm{H}_{10}\mathrm{O}$ is 4-methoxypent-2-yne.
Hence, we can conclude that the correct option is C.
Note:
We can also say that the reaction which is carried out in the question is also given by other halogens and commonly referred to as haloform reaction. Therefore, it is also true that the reaction with iodine is used in the quantitative analysis of organic compounds.
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