Answer:
The order of reactivity towards electrophilic susbtitution is shown below:
a. anisole > ethylbenzene>benzene>chlorobenzene>nitrobenzene
b. p-cresol>p-xylene>toluene>benzene
c.Phenol>propylbenzene>benzene>benzoic acid
d.p-chloromethylbenzene>p-methylnitrobenzene> 2-chloro-1-methyl-4-nitrobenzene> 1-methyl-2,4-dinitrobenzene
Explanation:
Electron donating groups favor the electrophilic substitution reactions at ortho and para positions of the benzene ring.
For example: -OH, -OCH3, -NH2, Alkyl groups favor electrophilic aromatic substitution in benzene.
The -I (negative inductive effect) groups, electron-withdrawing groups deactivate the benzene ring towards electrophilic aromatic substitution.
Examples: -NO2, -SO3H, halide groups, Carboxylic acid groups, carbonyl gropus.
Given:
M = 0.0150 mol/L HF solution
T = 26°C = 299.15 K
π = 0.449 atm
Required:
percent ionization
Solution:
First, we get the van't Hoff factor using this equation:
π = i MRT
0.449 atm = i (0.0150 mol/L) (0.08206 L atm / mol K) (299.15 K)
i = 1.219367
Next, calculate the concentration of the ions and the acid.
We let x = [H+] = [F-]
[HF] = 0.0150 - x
Adding all the concentration and equating to iM
x +x + 0.0150 - x = <span>1.219367 (0.0150)
x = 3.2905 x 10^-3
percent dissociation = (x/M) (100) = (3.2905 x 10-3/0.0150) (100) = 21.94%
Also,
percent dissociation = (i -1) (100) = (</span><span>1.219367 * 1) (100) = 21.94%</span>
The answer to the question is b
Answer:
Its B
Explanation: I did the test passed btw
