Answer:
Structures are given below.
Explanation:
- Treatment of 2-bromo-2-methylbutane with KOH in ethanol will give elimination of HBr through E2 mechanism.
- H atoms adjacent to Br will be eliminated.
- 2-bromo-2-methylbutane has two possible adjacent H atoms that can be eliminated giving mixture of products.
- Product of this elimination reaction is alkene. Here saytzeff fule is followed during elimination. So most substituted alkene will be major product.
- Structure of alkenes are given below.
The reaction has had a heat that is enthalpy of -22 kJ/mol. The exothermic process has been signaled by the negative sign.
The amount of energy that the system absorbs or releases to create the products is described as the heat of reaction.
The source of the reaction's heat is
H is equal to 3(413 Kj/mol) + 358 Kj/mol + 467 Kj/mol + 1070 Kj/mol = 3134 Kj/mol.
H prod equals 3(413 kj/mol) plus 347 kj/mol plus 358 kj/mol plus 467 kj/mol plus 745 kj/mol, or 3156 kj/mol.
H=3134 kj/mol - 3156 kj/mol = -22 Kj/mol
Negative findings point to an exothermic response.
A chemical process known as an exothermic reaction releases energy in the form of heat or light.
Learn more about exothermic reaction here-
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