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blondinia [14]
2 years ago
11

In electrophilic aromatic substitution reactions the hydroxyl group is an o p-director because?

Chemistry
1 answer:
Natalka [10]2 years ago
5 0

In electrophilic aromatic substitution reactions the hydroxyl group is an o,p-director because: hydroxyl group donates the electron density to the ring by induction and destabilizes the meta sigma complex and  by resonance and it  stabilizes the ortho and para sigma complexes of aromatic ring .

Most ring activators have atoms with unshared electron pairs directly attached to a carbon atom of the benzene ring . For example, the — OH group has two pairs of unshared electrons on the oxygen atom , which will form a bond to a carbon atom of the benzene ring . Thus , the — OH group will be an activating group in electrophilic aromatic substitution reactions .

to learn more about electrophilic aromatic substitution reactions please click here ,

brainly.com/question/28286554

#SPJ4

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The standard reduction potentials for the Ag+|Ag(s) and Zn2+| Zn(s) half-cell reactions are +0.799 V and -0.762 V, respectively.
mihalych1998 [28]

<u>Answer:</u> The potential of the given cell is 1.551 V

<u>Explanation:</u>

The given chemical cell follows:

Zn(s)|Zn^{2+}(0.125M)||Ag^{+}(0.240M)|Ag(s)

<u>Oxidation half reaction:</u> Zn(s)\rightarrow Zn^{2+}(0.125M)+2e^-;E^o_{Zn^{2+}/Zn}=-0.762V

<u>Reduction half reaction:</u> Ag^{+}(0.240M)+e^-\rightarrow Ag(s);E^o_{Ag^{+}/Ag}=0.799V       ( × 2)

<u>Net cell reaction:</u> Zn(s)+2Ag^{+}(0.240M)\rightarrow Zn^{2+}(0.125M)+2Ag(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

Putting values in above equation, we get:

E^o_{cell}=0.799-(-0.762)=1.561V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Zn^{2+}]}{[Ag^{+}]^2}

where,

E_{cell} = electrode potential of the cell = ? V

E^o_{cell} = standard electrode potential of the cell = +1.561 V

n = number of electrons exchanged = 2

[Zn^{2+}]=0.125M

[Ag^{+}]=0.240M

Putting values in above equation, we get:

E_{cell}=1.561-\frac{0.059}{2}\times \log(\frac{(0.125)}{(0.240)^2})

E_{cell}=1.551V

Hence, the potential of the given cell is 1.551 V

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