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iragen [17]
2 years ago
9

When N,N-Dimethylaniline is treated with bromine, ortho and para products are observed. However, when N,N-Dimethylaniline is tre

ated with a mixture of nitric and sulfuric acid, only the meta product is observed. Explain these results. (Hint: what can happen to the nitrogen atom in the presence of the strong acids
Chemistry
1 answer:
lilavasa [31]2 years ago
6 0

Answer:

See explanation below

Explanation:

To get a better understanding watch the picture attached.

In the case of the reaction with Bromine, the -N(CH₃)₂ is a strong ring activator, therefore, it promotes a electrophilic aromatic sustitution, so, in the mechanism of reaction, the lone pair of the Nitrogen, will move to the ring by resonance and activate the ortho and para positions. That's why the bromine wil go to the ortho and para positions, mostly the para position, because the -N(CH₃)₂ cause a steric hindrance in the ortho position.

In the case of the reaction with HNO₃/H₂SO₄, the acid transform the -N(CH₃)₂ in a protonated form, the anilinium ion, which is a deactivating of the ring, and also a strong electron withdrawing, so, the electrophile will go to the meta position instead.

Hope this helps.

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Answer:

\%m/m=14\%

Explanation:

Hello!

In this case, since the molarity of magnesium chloride (molar mass = 95.211 g/mol) is 1.672 mol/L and we know the density of the solution, we can first compute the concentration in g/L as shown below:

[MgCl_2]=1.672\frac{molMgCl_2}{L}*\frac{95.211gMgCl_2}{1molMgCl_2}=159.2\frac{gMgCl_2}{L}

Next, since the density of the solution is 1.137 g/mL, we can compute the concentration in g/g as shown below:

[MgCl_2]=159.2\frac{gMgCl_2}{L}*\frac{1L}{1000mL}*\frac{1mL}{1.137g}=0.14

Which is also the by-mass fraction and in percent it turns out:

\%m/m=0.14*100\%\\\\\%m/m=14\%

Best regards!

6 0
3 years ago
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The reaction will move to the left.

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H⁺ will combine with OH⁻ to form water.

<em>So, the concentration of OH⁻ will decrease and the equilibrium is disturbed.</em>

<em />

<em>According to Le Châtelier's principle: </em>when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

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  • The base will dissociate to form more OH⁻ and thus, the quantity of Ba(OH)₂ will decrease.

<em>So, the right choice is: the reaction will move to the left, is the choice that will not happen to the equilibrium.</em>

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