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3 years ago

Classify each substituent as ortho, para activators; ortho, para deactivators; or meta deactivators.

1 answer:

8 0

The activating groups are; -NH2, -NR2 > -OH, -OR> -NHCOR> -CH3. while the deactivating groups are; -NO2, -CF3> -COR, -CO2R, -SO3H > Halogens.

<h3>Electrophillic substitution</h3>

Owing to the unusual stability of the benzene ring, it does not undergo addition like other unsaturated species, rather benzene does undergo electrophillic substitution.

The substituent groups determine whether the molecule will be activated or deactivated towards electrophilic substitution, the following groups decativate benzene towards electrophilic substitution; -NO2, -CF3> -COR, -CO2R, -SO3H > Halogens. The following groups activate benzene ring towrads electrophilic substitution; -NH2, -NR2 > -OH, -OR> -NHCOR> -CH3.

Learn more about activators and decativators: brainly.com/question/9904853

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Diethyl ether is produced from ethanol according to the following equation:

juin [17]

The balanced chemical equation is given as:

2CH3CH2OH(l) → CH3CH2OCH2CH3(l) + H2O(l)

We are given the yield of CH3CH2OCH2CH3 and the amount of ethanol to be used for the reaction. These values will be the starting point for the calculations.

Theoretical amount of product produced:
329 g CH3CH2OH ( 1 mol / 46.07 g ) ( 1 mol CH3CH2OCH2CH3 / 2 mol CH3CH2OH ) (74.12 g / mol ) = 264.66 g CH3CH2OCH2CH3

% yield = .775 = actual yield / 264.66

actual yield = 205.11 g CH3CH2OCH2CH3

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3 years ago

Is the reaction mg²⁺(aq) + h₂o(l) → mgo(s) + 2h⁺(aq) endothermic or exothermic?

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<span>If energy is released, the reaction is exothermic. If energy is absorbed, the reaction is endothermic. Since heat is being absorbed in this reaction (to break down H2O into H2 and O), the reaction is endothermic.</span>

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3 years ago

Which of the following describes a synthesis reaction?

Grace [21]

Answer:

D. Two reactants combine to form one product

Explanation:

The definition of synthesis in chemistry is "the production of chemical compounds by reaction from simpler materials."

A is decomposition, B is DOUBLE replacement, and C is SINGLE replacement.

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4 years ago

Balance the following equations:

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<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2>_____________________________________</h2>

(I) $N_{2} + 3H_{2} --> 2NH_{3}$

(II) $P_{4} + 5O_{2} --> 2P_{2} O_{5}$

(III) $2NaF + Br_{2} -->2NaBr + F_{2}$

(IV) $2ZnS + 3O_{2} --> 2ZnO + 2SO_2$

(V) $Pb(NO_3)_2 + 2NaCl --> 2NaNO_3 + PbCl_2$

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7 0

3 years ago

A 3.4 g sample of an unknown monoprotic organic acid composed of C,H, and O is burned in air to produce 8.58 grams of carbon dio

Pavlova-9 [17]

Answer:

$C_7H_6O_2$

Explanation:

Hello there!

In this case, we can divide the problem in three stages: (1) determine the empirical formula with the combustion analysis, (2) compute the molar mass of acid via the moles of the acid in the neutralization and (3) determine the molecular formula.

(1) In this case, since 8.58 g of carbon dioxide are released, we can first compute the moles of carbon in the compound:

$n_C=8.58gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.195molC$

And the moles of hydrogen due to the produced 1.50 grams of water:

$n_H=1.50gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O} =0.166molH$

Next, to compute the mass and moles of oxygen, we need to use the initial 3.4 g of the acid:

$m_O=3.4g-0.195molC*\frac{12.01gC}{1molC}-0.166molH*\frac{1.01gH}{1molH} =0.89gO\\\\n_O=0.89gO*\frac{1molO}{16.0gO}=0.0556molO$

Thus, the subscripts in the empirical formula are:

$C=\frac{0.195}{0.0556}=3.5 \\\\H=\frac{0.166}{0.0556}=3\\\\O=\frac{0.0556}{0.0556}=1\\\\C_7H_6O_2$

As they cannot be fractions.

(2) In this case, since the acid is monoprotic, we can compute the moles by multiplying the concentration and volume of KOH:

$n_{KOH}=0.279L*0.1mol/L\\\\n_{KOH}=0.0279mol$

Which are equal to the moles of the acid:

$n_{acid}=0.0279mol$

And the molar mass:

$MM_{acid}=\frac{3.4g}{0.0279mol} =121.86g/mol$

(3) Finally, since the molar mass of the empirical formula is:

7*12.01 + 6*1.01 + 2*16.00 = 122.13 g/mol

Thus, since the ratio of molar masses is 122.86/122.13 = 1, we infer that the empirical formula equals the molecular one:

$C_7H_6O_2$

Best regards!

8 0

3 years ago