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iVinArrow [24]
3 years ago
13

Draw the structure of the product of each step in the following three-step synthesis. Show the formal charges, if applicable. As

a start, the benzene ring is drawn for you in each product. Although the first step produces a mixture of isomers, the para isomer is isolated as the sole product of interest and used in the second step. Give only the major product for the second and third steps.

Chemistry
1 answer:
Dvinal [7]3 years ago
5 0

Answer:

Second step: 4-bromo-1-methyl-2-nitrobenzene.

Third step: 1.5-dibromo-2-methyl-3-nitrobenzene.

Explanation:

To solve this exercise I will use the concepts of electrophilic substitution. In these reactions, a functional group is displaced by an electrophile. In the attached image are the two main products.

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Read 2 more answers
Calculate the mass of MgCO3 (84.31 g/mol) precipitated by mixing 10.0 mL of a 0.300 M Na2CO3 solution with 6.00 mL of 0.0400 M M
I am Lyosha [343]

Answer:

m_{MgCO_3}=0.0202molMgCO_3

Explanation:

Hello,

In this case, for this purpose we first have to write the undergoing chemical reaction:

Na_2CO_3+Mg(NO_3)_2\rightarrow MgCO_3+2NaNO_3

Thus, since the mole ratio between the reactants is 1:1, we next identify the limiting reactant by computing the available moles of sodium carbonate and those moles of the same reactant consumed by the magnesium nitrate considering the given solutions:

n_{Na_2CO_3}=0.010L*0.300\frac{molNa_2CO_3}{1L}=0.003molNa_2CO_3 \\\\n_{Na_2CO_3}^{consumed}=0.006L*0.0400\frac{molMg(NO_3)_2}{1L}*\frac{1molNa_2CO_3}{1molMg(NO_3)_2} =0.00024molNa_2CO_3

In such a way, since less moles are consumed, we can say that the sodium carbonate is excess whereas the magnesium nitrate is the limiting one, therefore, the yielded mass of magnesium carbonate turns out:

m_{MgCO_3}=0.00024molMg(NO_3)_2*\frac{1molMgCO_3}{1molMg(NO_3)_2}*\frac{84.31gMgCO_3}{1molMgCO_3}  \\\\m_{MgCO_3}=0.0202molMgCO_3

Regards.

7 0
3 years ago
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