Q.47P Substituent effects: resonance a... [FREE SOLUTION]

Chapter 20: Q.47P (page 1078)

Substituent effects: resonance and induction.
(a)When the methoxy group is in the meta position inB, is the acid stronger or weaker than inA? Is the methoxy group in the meta position electron-donating or withdrawing?
(b)When the methoxy group is in the para position inC, is the acid stronger or weaker than inA? Is the methoxy group in the para position electron-donating or withdrawing?
(c)How can this apparent contradiction be explained? Which effect is stronger for methoxy?
(d)When the fluoro group is in the meta position inE, is the acid stronger or weaker than inD? Is the fluoro group in the meta position electron-donating or withdrawing?
(e)When the fluoro group is in the para position inF, is the acid stronger or weaker than inD? Is the fluoro group in the para position electron-donating or withdrawing?
(f)Compare the results of the fluoro group with those of the above methoxy group. What must be different about the relative strength of the resonance and inductive effects for fluoro compared with methoxy?

Short Answer

Expert verified

(a) The methoxy group present in the meta position act as an electron-withdrawing group. So B is a stronger acid than A.

(b) The methoxy group present in the para position is electron-donating so C is a weaker acid than A.

(c) The meta substituent exerts inductive effect so it is electron-withdrawing and para position exerts resonance effect, so it is electron-donating. Thus, the resonance is stronger than induction.

(d) The fluoro group in meta position is electron-withdrawing so e is a stronger acid than D.

(e) The fluoro group is present in a para position slightly withdrawing so F is a stronger acid than D.

(f) The meta position exerts an inductive effect while the para position has a resonance effect. Thus, the resonance effect results in electron-donating but for the fluoro group, the inductive effect becomes stronger than the resonance effect.

Step by step solution

About the resonance

The effect that explains the polarity present in a molecule due to the interaction taking place between pi bonds and electron pairs or between the pi bonds of adjacent atoms is called resonance.

There will be a change in the properties of chemical compounds when the resonance occurs.

(a) About the stronger acid

There is a methoxy group present at the meta position in structure B. So, this acts as an electron-withdrawing group. Hence, compound B is a stronger acid than compound A.

(b) About the weaker acid

There is a methoxy group at the para position in structure C. This acts as an electron-donating group. Hence, compound C is a weaker acid than compound B.

(c) Explanation

The meta position supports only the inductive effect but not the resonance effect. This indicates that electron-withdrawing takes place by induction because of the electronegative oxygen.

The para position supports only the resonance effect. So, there must be electron-donating. The resonance effect will always be stronger than the induction but in resonance, the electron donation is stronger so it overpowers the inductive effect.

(d) About the stronger acid

There is a fluoro (F) group present at meta position in structure E. So, this acts as an electron-withdrawing group. Hence, compound E is a stronger acid than compound D.

(e) About the stronger acid

There is a fluoro group at the para position in structure F. This acts as an electron-withdrawing. Hence, compound F is a stronger acid than compound E.

(f) Explanation

The meta position supports only the inductive effect but not the resonance effect. By the induction concept, the fluoro group should be electron-withdrawing because of its higher electronegativity.

The para position supports only the resonance effect. So, there must be electron-donating. But due to the higher electronegativity of the fluoro group, the inductive effect becomes stronger than the resonance effect.