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.
You might be interested in
Answer:
The vapor pressure of the solution is 23.636 torr
Explanation:
Where;
is the vapor pressure of the solution
is the mole fraction of the solvent
is the vapor pressure of the pure solvent
Thus,
15.27 g of NaCl = [(15.27)/(58.5)]moles = 0.261 moles of NaCl
0.67 kg of water = [(0.67*1000)/(18)]moles = 37.222 moles of H₂O
Mole fraction of solvent (water) = (number of moles of water)/(total number of moles present in solution)
Mole fraction of solvent (water) = (37.222)/(37.222+0.261)
Mole fraction of solvent (water) = 0.993
<u>Note:</u> the vapor pressure of water at 25°C is 0.0313 atm
Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm
the vapor pressure of the solution = 0.0311 atm = 23.636 torr