Cation and an anion the differences in charge make them attracted to each other
Answer:

Explanation:
Hess's Law of Constant Heat Summation states that if a chemical equation can be written as the sum of several other chemical equations, the enthalpy change of the first chemical equation is equal to the sum of the enthalpy changes of the other chemical equations. Thus, the reaction that involves the conversion of reactant A to B, for example, has the same enthalpy change even if you convert A to C, before converting it to B. Regardless of how many steps it takes for the reactant to be converted to the product, the enthalpy change of the overall reaction is constant.
With Hess's Law in mind, let's see how A can be converted to 2C +E.
(Δ
) -----(1)
Since we have 2B, multiply the whole of II. by 2:
(2Δ
) -----(2)
This step converts all the B intermediates to 2C +2D. This means that the overall reaction at this stage is
.
Reversing III. gives us a negative enthalpy change as such:
(-Δ
) -----(3)
This step converts all the D intermediates formed from step (2) to E. This results in the overall equation of
, which is also the equation of interest.
Adding all three together:
(
)
Thus, the first option is the correct answer.
Supplementary:
To learn more about Hess's Law, do check out: brainly.com/question/26491956
To name this Alkyne, simply count from the direction that will give the lowest starting number to appear at the beginning of the carbon triple bond.
If you were to count from the top of the chain, the position of the carbon next to the triple bond would be 4. Yet if you count from the bottom chain going left to right and above the chain, the position of the carbon next to the triple bond would be 3.
Then identify the groups that are connected off the parent chain, here we have a methyl group on carbon 2.
Thus the name would be 2 - methyl - 3 - heptyne. I believe.
Those reactions in which Alkyl Halide reacts with the solvent without the involvement of any acid or base is called as
Solvolysis. In given problem <em>tert</em>-Butyl Bromide is a tertiary Alkyl Halide and we know well that tertiary alkyl halides undergo
SN¹ and
E¹ elimination reaction due to the formation of
stable tertiary carbocation. In given example after the formation of carbocation when Isopropyl act as
nucleophile it will produce
ether and when it acts as a
base it will produce
unsaturated compound. The reaction along with both products is shown below,