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Luba_88 [7]
4 years ago
14

Does pyridine undergo nucleophilic aromatic substitution reactions? If no, why not. If yes, on what position does the substituti

on occur? Use resonance contributors to justify your answer.(15 points) b) Draw the tautomers of 2-hydroxypyridine and 4-hydroxypyridine (5 points)

Chemistry
1 answer:
sergij07 [2.7K]4 years ago
5 0

Explanation:

Part a

Pyridine is an aromatic heterocyclic compound and undergoes nucleophilic aromatic substitution reaction at 2 and 4 positions.

This is because, when a  nucleophile attack pyridine at 2 and 4 positions, the anoin formed is stabilized. In the anion formed is tabilized by resonance and also negative charge is present at electronegative atom, nitrogen.

If the nucleophile attack at 3 position, the anion formed is not as stable as anions formed when nucleophile attach at 2 and 4 positions. This is because negative charge is not present at the nitrogen atom.

Part b:

Tautomers are structural isomers of each others formed by migration atom within the molecule. Tautomers exist in equilibrium with each other.

Tautomers of 2-Hydroxypyridine and 4-Hydroxypyrine are given in the attachment.

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Which equation represents the reaction of a weak acid with water?
shusha [124]
Acid is anything which will give H+ to other species.

Have a look at the equations given.

In A) HCl changes to Cl- by giving its H+ to H2O. So HCl is an acid.

In B) HCO3- changes to H2CO3^2- by accepting H+. It did Not give its H+ rather it takes from other species. So it is not an acid at all.

In C) H2O is just breaking to H+ and OH-. It is not giving H+ to other species. So it is also not an acid in this reaction.

In D) HCOOH is giving its H+ to H2O. So it is also an acid.

So out of all reactions with water. The only two species are acting as acid with water namely HCl and HCOOH.

Out of these two HCl is very strong acid but HCOOH is a weak acid.

So the answer is D
6 0
3 years ago
Read 2 more answers
A chemist adds 26.5g of ammonium chloride to 10g of sodium hydroxide, which follows the reaction below. Assuming the reaction go
4vir4ik [10]

Answer :  The amount of reactant left in excess is 13.1075 grams.

Explanation : Given,

Mass of NH_4Cl = 26.5 g

Mass of NaOH = 10 g

Molar mass of NH_4Cl = 53.5 g/mole

Molar mass of NaOH = 40 g/mole

First we have to calculate the moles of NH_4Cl and NaOH.

\text{Moles of }NH_4Cl=\frac{\text{Mass of }NH_4Cl}{\text{Molar mass of }NH_4Cl}=\frac{26.5g}{53.5g/mole}=0.495moles

\text{Moles of }NaOH=\frac{\text{Mass of }NaOH}{\text{Molar mass of }NaOH}=\frac{10g}{40g/mole}=0.25moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

NH_4Cl+NaOH\rightarrow NH_4OH+NaCl

From the balanced reaction we conclude that

As, 1 mole of NaOH react with 1 mole of NH_4Cl

So, 0.25 moles of NaOH react with 0.25 moles of NH_4Cl

From this we conclude that, NH_4Cl is an excess reagent because the given moles are greater than the required moles and NaOH is a limiting reagent and it limits the formation of product.

Moles of remaining excess reactant = 0.495 - 0.25 = 0.245 moles

Now we have to calculate the mass of NH_4Cl.

\text{Mass of }NH_4Cl=\text{Moles of }NH_4Cl\times \text{Molar mass of }NH_4Cl

\text{Mass of }NH_4Cl=(0.245mole)\times (53.5g/mole)=13.1075g

Therefore, the amount of reactant left in excess is 13.1075 grams.

5 0
3 years ago
A chemist makes 940. mL of sodium carbonate (Na_2CO_3) working solution by adding distilled water to 200. mL of a 0.171 m stock
Alexxandr [17]

Answer:

The answer is 0.0698 M

Explanation:

The concentration was prepared by a serial dilution method.

The formula for the preparation I M1V1 = M2V2

M1= the concentration of the stock solution = 0.171 M

V1= volume of the stock solution taken = 200 mL

M2 = the concentration produced

V2 = the volume of the solution produced = 940 mL

Substitute these values in the formula

0.171 × 200 = 490 × M2

34.2 = 490 × M2

Make M2 the subject of the formula

M2 = 34.2/490

M2 = 0.069795

M2 = 0.0698 M ( 3 s.f)

The concentration of the Chemist's working solution to 3 significant figures is 0.0698M

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3 years ago
Calculate the mass percent 250g of salt solution that has 3.50g salt residue
Anna35 [415]

Answer:

1.4% salt residue

Explanation:

Mass percent can be found using the formula:

mass of solute (grams)
------------------------------------  x 100%
mass of solution (grams)

Therefore, since you have been given the weights of both the solute and solution, you can use the formula to solve for the mass percent.

3.50 g salt residue
-----------------------------  =  0.014 x 100%  = 1.4% salt residue
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