What alcohol is formed when the alkene is treated with h2o in the presence of h2so4? qs14?
1 answer:
Here we have to get the alcohols which are formed in the reaction of an alkene with H₂SO₄ and H₂O.
The alcohols which are formed in the reaction between alkene and H₂SO₄ and H₂O are majorly secondary and tertiary alcohol. As primary alcohol only ethanol can be produced in this reaction.
The addition of H₂SO₄/H₂O generates alcohol from alkene. The general reaction is shown in the figure. In this reaction methanol (CH₃OH) cannot be prepared and in primary alcohol only ethanol (C₂H₅OH) can be prepared from ethene.
In case of propene and other higher alkene always secondary or tertiary alcohols are produced as the addition of sulphuric acid and water depends on the Markonikov's rule of addition to the less hindered position.
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Answer:
Explanation:
From similar question, equation (b) is
Molecular Equation:
- Cu(s) + 2AgNO₃(aq) → Cu(NO₃)₂(aq) + 2Ag(s)
Net Ionic Equation:
- Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s)
<u>1. Mole ratio:</u>
<u>2. Convert 1.147 g of Cu(s) to moles:</u>
- Atomic mass of Cu: 63.546g/mol
- Number of moles = mass in grams / atomic mass
- Number of moles = 1.147 g / 63.546 g/mol = 0.01805 mol Cu(s)
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<u>3. Calculate the moles of Ag(s):</u>
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<u>4. Convert 0.03610 mol Ag(s) to grams:</u>
- Atomic mass of Ag(s) = 107.868g/mol
- Mass = 0.03610mol × 107.868g/mol = 3.894g