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weeeeeb [17]
3 years ago
13

Any groups present on a benzene ring can impact the success and regioselectivity of an electrophilic aromatic substitution. Dete

rmine which group from the list best fits each activation and directing description.
Strongly deactivating meta- director __________
Moderately deactivating meta- director ________
Strongly activating ortho-/para- director _________
Weakly deactivating ortho-/para- director _________
Weakly activating ortho-/para- director ________
Chemistry
1 answer:
r-ruslan [8.4K]3 years ago
4 0

Answer:

Explanation:

In electrophilic aromatic substitution, the substituent(s) that favors electrophilic attack is said to be ortho or para to the substituent. Meta directors are deactivators. Altogether, these directors have an integral role to play in the success and regioselectivity of electrophilic aromatic substitution reactions.

From the information given: The;

Strongly deactivating meta- director  ⇒ NO₂ (nitro group)

Moderately deactivating meta- director  ⇒ CN (cyano group)

Strongly activating ortho-/para- director ⇒ OH (hydroxy; group)

Weakly deactivating ortho-/para- director ⇒ -X (halide group)

Weakly activating ortho-/para- director ⇒ - Ph  (phenyl group)

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The equation used to calculate enthalpy change is of a reaction is:

\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]

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The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(SO_3(g))})]-[(2\times \Delta H_f_{(SO_2(g))})+(1\times \Delta H_f_{(O_2(g))})]

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Putting values in above equation, we get:

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To calculate the number of moles, we use ideal gas equation, which is:

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P = pressure of the gas = 1.00 bar

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T = temperature of the mixture = 25^oC=[25+273]K=298K

Putting values in above equation, we get:

1.00bar\times 2.67L=n\times 0.0831\text{ L. bar }mol^{-1}K^{-1}\times 298K\\\\n=\frac{1\times 2.67}{0.0831\times 298}=0.108mol

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Putting values in above equation, we get:

-197.8kJ/mol=\frac{q}{0.108mol}\\\\q=(-197.8kJ/mol\times 0.108mol)=-21.36kJ

Hence, the amount of heat produced by the reaction is -21.36 kJ

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