When optically active (S)-2-methylcyclopentanone is treated with aqueous base, the compound loses its optical activity. Explain
1 answer:
Answer:
See explanation and image attached
Explanation:
The first step in the reaction is the attack of the hydroxide ion from the base abstracting a proton from (S)-2-methylcyclopentanone.
The abstraction of a proton from water by the substrate yields the enol form in equilibrium with the keto form.
The product formed is racemic hence the optical activity of (S)-2-methylcyclopentanone is lost.
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Answer :
(a) The concentration of is, 0.0337 M
(b) The concentration of is,
Solution :
<u>(a) As per question, lead is oxidized and copper is reduced.</u>
The oxidation-reduction half cell reaction will be,
Oxidation half reaction:
Reduction half reaction:
The balanced cell reaction will be,
Here lead (Pb) undergoes oxidation by loss of electrons, thus act as anode. Copper (Cu) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
Now we have to calculate the concentration of .
Using Nernest equation :
where,
n = number of electrons in oxidation-reduction reaction = 2
= 0.507 V
Now put all the given values in the above equation, we get:
Therefore, the concentration of is, 0.0337 M
<u>(b) As per question, lead is reduced and copper is oxidized.</u>
The oxidation-reduction half cell reaction will be,
Oxidation half reaction:
Reduction half reaction:
The balanced cell reaction will be,
Here Copper (Cu) undergoes oxidation by loss of electrons, thus act as anode. Lead (Pb) undergoes reduction by gain of electrons and thus act as cathode.
First we have to calculate the standard electrode potential of the cell.
Now we have to calculate the concentration of .
Using Nernest equation :
where,
n = number of electrons in oxidation-reduction reaction = 2
= 0.507 V
Now put all the given values in the above equation, we get:
Therefore, the concentration of is,
The fraction of acetic acid that is dissociated is 0.18
Why?
The chemical equation for the dissociation of acetic acid (HAc) is the following:
HAc(aq) + H₂O(l) ⇄ H₃O⁺(aq) + Ac⁻(aq)
To find the fraction of acetic acid that is in the dissociated form (f), we apply the following equation (Ka for acetic acid is 1.76*10⁻⁵). This equation comes from solving the equation of the equilibrium constant for the dissociated fraction of HAc:
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