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3 years ago

Write the mechanism of the reaction of methyl benzoate to form methyl 2-nitrobenzoate.

Chemistry

2 answers:

Galina-37 [17]3 years ago

8 0

The question is partially incorrect, because nitration of <span> methyl benzoate results in generation of methyl 3-nitrobenzoate, and not methyl 2-benzoate.

This a because of the present of ester group, which deactivated benzene ring at ortho and para position. Due to this, the electrophile (NO2+) attackes on meta position.

The detailed mechanism is attached below.</span>

mylen [45]3 years ago

3 0

Answer:

See the attached picture.

Explanation:

Hello,

Nitration of alkyl benzenes is carried out adding both nitric and sulfuric acid to the organic reactant in order to promote the inclusion of the nitro group into the ring. Nonetheless, due to the steric hindrance such inclusion is not at the second carbon but at the third one. The addition of nitric acid promotes the attaching of the nitro group and the sulfuric acid shifts the equilibrium towards the nitration is it ionizes the nitric acid to nitronium ion NO₂⁺. The mechanism is shown on the attached picture.

Best regards.

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Answers and Explanation:

a)- The chemical equation for the corresponden equilibrium of Ka1 is:

2. HNO2(aq)⇌H+(aq)+NO−2

Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.

b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:

ΔG= ΔGº + RT ln Q

Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)

At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:

⇒ 0 = ΔGº + RT ln Ka

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ΔGº= 19092.8 J/mol

c)- According to the previous demonstation, at equilibrium ΔG= 0.

d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:

ΔG= ΔGº + RT ln Q

Q= ((H⁺) (NO₂⁻))/(HNO₂)

Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)

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We know that ΔGº= 19092.8 J/mol, so:

ΔG= ΔGº + RT ln Q

ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)

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Notice that ΔG<0, so the process is spontaneous in that direction.

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