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Nimfa-mama [501]
3 years ago
12

What is the major driving force for losing a proton as the last step in an electrophilic aromatic substitution reaction?

Chemistry
1 answer:
Fofino [41]3 years ago
4 0

Answer:

Explanation:

In the following reaction we have shown an example of aromatic substitution reaction .

C₆H₆ + RCl = C₆H₅R  + HCl

This  reaction takes place in the presence of catalyst like AlCl₃ which is a lewis acid .

First of all formation of carbocation is made as follows .

RCl + AlCl₃ = R⁺ + AlCl₄⁻

This R⁺ is carbocation which is also called electrophile . It attacks the ring to get attached with it .

C₆H₆ + R⁺  =   C₆H₅R⁺H.

The complex formed is unstable , though it is stabilized by resonance effect . In the last step H⁺ is kicked out of the ring . The driving force that does it is the steric hindrance due to presence of two adjacent group of H and R⁺ at the same place . Second driving force is attack by the base  AlCl₄⁻ that had been formed earlier . It acts as base and it extracts proton ( H⁺ ) from the ring .

C₆H₅R⁺H + AlCl₄⁻  = C₆H₆ + AlCl₃ + HCl .

The formation of a stable product C₆H₆ also drives the reaction to form this product .

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Which separation method would be most successful in separating the components of a homogeneous mixture?
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Match each of the following descriptions to thecorrect element listed
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Answer: Please see explanation for answer

Explanation:Matching the  descriptions to the correct element listed gives

Element Group Name Group No Metal/ Nonmetal/ Metalloid Description

Neon    Noble gases       18           Non metal Colorless, tasteless, and        odorless gas, unreactive, nonconductive.

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The breaking buffer that we use this week contains 10mM Tris, pH 8.0, 150mM NaCl. The elution buffer is breaking buffer that als
olasank [31]

Answer:

A breakdown of the breaking buffer was first listed with its respective component and their corresponding value; then a table was made for the stock concentrations in which the volume that is being added was determined by using the formula M_1*V_1 = M_2*V_2. It was the addition of these volumes altogether that make up the 0.25 L (i.e 250 mL)  with water

Explanation:

Given data includes:

Tris= 10mM

pH = 8.0

NaCl = 150 mM

Imidazole = 300 mM

In order to make 0.25 L solution buffer ; i.e (250 mL); we have the following component.

Stock Concentration             Volume to be             Final Concentration

                                               added            

1 M Tris                                     2.5 mL                         10 mM

5 M NaCl                                  7.5 mL                        150 mM

1 M Imidazole                           75 mL                         300 mM    

M_1*V_1 = M_2*V_2. is the formula that is used to determine the corresponding volume that is added for each stock concentration

The stock concentration of Tris ( 1 M ) is as follows:

M_1*V_1 = M_2*V_2.

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The stock concentration of NaCl (5 M ) is as follows:

M_1*V_1 = M_2*V_2.

1*V_1= 0.15 M *250mL\\V_1 = 7.5mL

The stock concentration of Imidazole (1 M ) is as follows:

M_1*V_1 = M_2*V_2.

1*V_1= 0.03 M *250mL\\V_1 = 75mL

Hence, it is the addition of all the volumes altogether that make up 0.25L (i.e 250 mL) with water.

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