Substituents on an aromatic ring can have several effects on electrophilic aromatic substitution reactions. Substituents can act
ivate or deactivate the ring to substitution, donate or withdraw electrons inductively, donate or withdraw electrons through resonance, and direct substitution either to the ortho/para or to the meta positions.
1. From the following list, select the substituents that have the indicated property. The substituents are written as -XY, where X is the atom directly bound to the aromatic ring.
a. Activation of the ring towards substitution
i. -CF3
ii. -NO2
iii. -OCOCH3
iv. -NH2
1. From the following list, select the substituents that have the indicated property. The substituents are written as -XY, where X is the atom directly bound to the aromatic ring.
a. Activation of the ring towards substitution
i. -CF3
ii. -NO2
iii. -OCOCH3
iv. -NH2
1 answer:
Answer:
iv. -NH₂
Explanation:
The amine group is an activating substituent and provides an <em>ortho-para </em>orientation.
The free electrons in the nitrogen atom provide <u>stabilization by resonance </u>in the sigma complex if the attack by the electrophilic reactant occurs in the positions <em>ortho</em> or <em>para </em>with regards to the nitrogen atom.
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Answer:
See explanation below
Explanation:
In this case, let's see both molecules per separate:
In the case of SeO₂ the central atom would be the Se. The Se has oxidation states of 2+, and 4+. In this molecule it's working with the 4+, while oxygen is working with the 2- state. Now, how do we know that Se is working with that state?, simply, let's do an equation for it. We know that this molecule has a formal charge of 0, so:
Se = x
O = -2
x + (-2)*2 = 0
x - 4 = 0
x = +4.
Therefore, Selenium is working with +4 state, the only way to bond this molecule is with a covalent bond, and in the case of the oxygen will be with double bond. See picture below.
In the case of CO₂ happens something similar. Carbon is working with +4 state, so in order to stabilize the charges, it has to be bonded with double bonds with both oxygens. The picture below shows.
