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3 years ago

Which is a stronger nucleophile in an Sn2 raction, pyridine or pyrole?

Chemistry

1 answer:

atroni [7]3 years ago

3 0

Answer:

Pyridine

Explanation:

Pyridine is a stronger nucleophile than pyrole ,

Since , both have a lone pair of electron on its nitrogen atom , and both the structure are aromatic in nature .

But , in case of pyrole , the lone pair on nitrogen participates in the delocalization and makes the ring aromatic in nature . hence the lone pair losses its nuleophilic character .

But , in case of pyridine , the lone are free to act as a nucleophile , since they are not involved in the delocalization of the ring .

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3 years ago

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Roman55 [17]

Answer:

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Explanation:

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In this case, according to the reaction:

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3 years ago

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4 years ago

Read 2 more answers

Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant o

sattari [20]

Answer:

4.26 %

Explanation:

There is some info missing. I think this is the original question.

<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10 ⁻⁴.</em>

<em />

Step 1: Given data

Initial concentration of the acid (Ca): 0.249 M

Acid dissociation constant (Ka): 4.50 × 10 ⁻⁴

Step 2: Write the ionization reaction for nitrous acid

HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)

Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])

We will use the following expression.

$[A^{-} ] = \sqrt{Ca \times Ka } = \sqrt{0.249 \times 4.50 \times 10^{-4} } = 0.0106 M$

Step 4: Calculate the percent ionization of nitrous acid

We will use the following expression.

$\alpha = \frac{[A^{-} ]}{[HA]} \times 100\% = \frac{0.0106M}{0.249} \times 100\% = 4.26\%$

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4 years ago