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erastovalidia [21]
2 years ago
15

Why is an amide less reactive to nucleophilic acyl substitution than an acid chloride?

Chemistry
1 answer:
Olin [163]2 years ago
7 0

An amide is less reactive to nucleophilic acyl substitution than an acid chloride because more electron density is donated to the carbonyl by nitrogen.

<h3>What is electron density?</h3>

In quantum chemistry, electron density, also known as electronic density, is a metric for the likelihood that an electron will be found at a microscopic portion of space surrounding a specific point. The likelihood of finding an electron at a particular position near an atom or molecule is represented by electron density. In general, areas with a high electron density are where the electron is most likely to be located. The attractive forces between the electrons and the nuclei in a molecule are what hold the nuclei together in molecule, hence the electron density is essential to the bonding and geometry of a molecule. The repulsions between the nuclei and the electrons function as an antagonist to these attractive forces.

To learn more about electron density, visit:

brainly.com/question/13152866

#SPJ4

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3 years ago
At a high temperature, equal concentrations of 0.160 mol/L of H2(g) and I2(g) are initially present in a flask. The H2 and I2 re
Nikitich [7]

Explanation:

The given reaction is as follows.

                                H_{2} + I_{2} \rightarrow 2HI

Initial :                  0.160    0.160          0  

Change :                  -x           -x              2x

Equilibrium:        0.160 - x    0.160 - x       x

It is given that [H_{2}] = [0.160 - x] = 0.036 M

and,                [I_{2}] = [0.160 - x] = 0.036 M      

so,                             x = (0.160 - 0.036) M

                                    = 0.124 M

As, [HI] = 2x.

So,           [HI] = 2 \times 0.124

                       = 0.248 M

As it is known that expression for equilibrium constant is as follows.

               K_{eq} = \frac{[HI]^{2}}{[H_{2}][I_{2}]}

                                  = \frac{(0.248)^{2}}{(0.036)(0.036)}

                                  = 47.46

Thus, we can conclude that the equilibrium constant, Kc, for the given reaction is 47.46.

                               

7 0
3 years ago
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