Question

2. Not only does an aromatic ring itself affect the reactivity of an acid derivative (an ester, on last year's quiz), substituents on the aromatic ring can further modify the reactivity. First, we will consider a phenyl amide with a para-acetyl substituent: CH, (a) What is the n-effect (resonance) of the para-acetyl substituent? How will that affect the reactivity of the other (amide) carbonyl in the molecule below (increase or decrease, compared to the same amide but with no p- acetyl)? Draw resonance structures to support your answer. (4 points) (b) Would a meta-acetyl substituent (not shown) have the same effect on the reactivity of the amide carbonyl? Why or why not? Draw resonance structures to support your answer. (3 points) (c) What is the o-effect (inductive) of the acetyl substituent? How does it affect the reactivity of the amide carbonyl? How does its effect vary with location (para vs. meta)? (3 points) (d) The two isomers below can isomerize by way of an acid-catalyzed mechanism in which the acyl group is transferred directly from one -N to the other it is an intramolecular mechanism, not intermolecular). Draw this mechanism, using a general H-A/A catalyst. (5 points) CH (e) Based on your analysis above, do you expect this equilibrium to favor the reactant or product or is AG-0? Why? (3 points)

Answer 1

2.a. Acetyl group is Electron-withdrawing group.

• The Electron withdrawing effect decreases the reactivity of the Amide carbonyl than an amide without para Acetyl group.

cl 3 一。

2.b. The meta acetyl group would not be such suitably substituted(the acetyl cannot take part n the conjugation) as it was in the case of para.So,para substitution will be less deactivating.

3 Cut cit 3

c. Sigma effect of the substituent is the Inductive effect due to its local polarity .The sigma effect of Acetyl group is shown below-

So inductively also,the Acetyl group withdraw electrons in the bond due to the bond polarity.

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• This Inductive effect( -I) decreases the reactivity of amide carbonyl more when it is near to the amide carbonyl. So the Acetyl in Meta- position will be more deactivating than the acetyl in the Para- position.

d. The isomerization reaction is shown below-

e. In the product ,the Amine group i.e. -NH2 is in the para position(it is meta in the reactant). As -NH2 is more effective electron pair donor here,the product will be more deactivated than the reactant (as the Lone pair of -NH2 is involved in resonance in the product due to the para- substitution of NH2-,  but not in the reactant,as here it is meta- substituted). As the resonance effect is more powerful than the sigma(inductive effect),so the REACTANT will be more favorable here or $\Delta G=+ve$ for the reaction.