Answer: (Structure attached).
Explanation:
This type of reaction is an aromatic electrophilic substitution. The overall reaction is the replacement of a proton (H +) with an electrophile (E +) in the aromatic ring.
The aromatic ring in p-fluoroanisole has two sustituents, an <u>halogen</u> and a <u>methoxy group</u>, which are <em>ortho-para</em> directing substituents.
Aryl sulfonic acids are easily synthesized by an electrophilic substitution reaction aromatic using <u>sulfur trioxide as an electrophile</u> (very reactive).
The reaction occurs in three steps:
- The attack on the electrophile forms the sigma complex.
- The loss of a proton regenerates an aromatic ring.
- The sulfonate group can be protonated in the presence of a strong acid (H₂SO₄).
Normally, a mixture of <em>ortho-para</em> substituted products would be obtained. However, since both <em>para</em> positions are occupied, only the <em>ortho </em>substituted product is obtained here.
Based on the data given in this question, the statement that shows a correct interpretation of the chemical reactions is as follows: reaction A was exothermic and reaction B was endothermic.
<h3>What are endothermic and exothermic reactions?</h3>
Endothermic reaction is a chemical reaction that absorbs heat energy from its surroundings while exothermic reaction is a reaction that releases energy in the form of heat.
Endothermic reactions leave their surroundings cooler while exothermic reactions leave their surroundings hotter.
According to this question, the initial and final temperatures of two reactions are given as follows:
- Reaction A: 25.1°C and 30.2°C
- Reaction B: 25.1°C and 20.0°C
From the above data, reaction A was exothermic because it increased the surrounding temperature and reaction B was endothermic because it reduced the surrounding's temperature.
Learn more about endothermic and exothermic at: brainly.com/question/23184814
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Answer:
Explanation:
In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).
A and C reacts with two differents reagents and conditions, however both of them gives the same product.
Let's analyze each reaction.
First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.
Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.
Explanation:
Potential energy is the the relationship between work done height mass and acceleration due to gravity, because of this some objects also experience kinetic energy due to the factors mentioned above
Answer:
17202.6 years
Explanation:
Activity of the living sample (Ao) = 160 counts per minute
Activity of the wood sample (A) = 20 counts per minute
Half life of carbon-14 = 5730 years
t= age of the artifact
From;
0.693/t1/2= 2.303/t log Ao/A
Then;
0.693/ 5730= 2.303/t log Ao/A
Substituting values;
0.693/5730= 2.303/t log (160/20)
Then we obtain;
1.209×10^-4 = 2.0798/t
t= 2.0798/1.209×10^-4
Thus;
t= 17202.6 years
Therefore the artifact is 17202.6 years old.