Draw the major product for the reaction of 1-butyne with water in the presence of catalytic TfOH (i.e., CF3SO3H). Then answer th
1 answer:
Answer:
2-Butanone
Explanation:
From the given information:
The presence of mercury as an acid catalyst brings about the addition of water to the triple bond which yields enol. Then, according to Markownikov's rule and after tautomerism has occurred, we have a methyl ketone ( 2- Butanone) as the product.
The answer regarding the transformation is addition and hydration.
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At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.
The melting point is the temperature at which a substance passes from solid to liquid. Below the melting point, a substance is in the solid state. Above the melting point, a substance is in the liquid or gas state.
The boiling point is the temperature at which a substance passes from liquid to gas. Below the boiling point, a substance is solid or liquid. Above the boiling point, a substance is in the gas state.
At -25 °C, methanol is above the melting point (-97.6 °C) and below the boiling point (64.7 °C). Thus, it is in the liquid state.
At -25 °C, methanol, whose boiling point is 64.7 °C and its melting point is -97.6 °C, is in the liquid state.
You can learn more about the melting and boiling points here: brainly.com/question/5753603?referrer=searchResults
