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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<u>Answer:</u> The isomers are shown in the image below.
<u>Explanation:</u>
Isomers are defined as the chemical compounds having the same number and kinds of atoms but arrangement are different.
For the alkane having four carbon atoms and 1 bromine atom, the IUPAC name of the haloalkane is bromobutane
There are 4 possible isomers for the given haloalkane compound:
- 1-bromobutane
- 2-bromobutane
- 1-bromo-2-methylpropane
- 2-bromo-2-methylpropane
The isomers of the given organic compound is shown in the image below.
Answer:
35.7%
Explanation:
The percent yield is calculated by the formula:
<em>[(actual yield) / (theoretical yield)] * 100</em>
In this case, the actual yield is 0.15 grams, and the theoretical yield is 0.42 grams. So, putting these values into the equation, we have:
Thus, the percent yield of is 35.7%.
Hope this helps!