Two reactions which are examples of nucleophilic attack are given as below.
Reaction I: $ {{R}_{1}}-C\left( {{R}_{2}} \right)=O+HCN\xrightarrow{pH=x} $
Reaction II: $ {{R}_{1}}-C\left( {{R}_{2}} \right)=O+N{{H}_{2}}-C\left( =O \right)-NH-N{{H}_{2}}\xrightarrow{pH=y} $
1. Value of x is:
(A) $ x\le 4.5 $
(B) $ x=6 $
(C) $ x>7 $
(D) Can’t decide.
2. Value of y is:
(A) $ x=4.5 $
(B) $ x=1.5 $
(C) $ x=7 $
(D) $ x=9 $
Answer
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Hint: We know that the substitution of the existing nucleophile by the incoming and more reactive nucleophile is known as the nucleophilic substitution reaction. The nucleophiles are the electron-rich species, and have an affinity towards the positive charge.
Complete answer:
The nucleophilic substitution may proceed through the bimolecular or unimolecular pathways. The substitution of one nucleophile by the more reactive nucleophile is the nucleophilic substitution reaction. A nucleophilic aromatic substitution reaction, the nucleophile replaces the leaving group in the reagent structure. The leaving capacity of a leaving group or element depends on the stability of that group or element after leaving the reagent structure. One can think about the options and which can form a stable anion after it leaves.
$ {{R}_{1}}-C\left( {{R}_{2}} \right)=O+HCN\xrightarrow{pH=4.5}{{R}_{1}}{{R}_{2}}-C\left( OH \right)-CN $
The leaving atom or group ability is dependent on the stability of the anion formed by the leaving group. The stability of an anion depends on the electronegativity of that element. The more the electronegativity of an element is, the more will be its capacity to hold the anionic negative charge on itself. The more the capacity of an element to hold the charge on the element, the more will be its stability and it will act as a better leaving group.
$ {{R}_{1}}-C\left( {{R}_{2}} \right)=O+N{{H}_{2}}-C\left( =O \right)-NH-N{{H}_{2}}\xrightarrow{pH=9}{{R}_{1}}{{R}_{2}}-C=N-C\left( =O \right)-NH-N{{H}_{2}} $
Therefore, correct answer is option A and D i.e. $ \le 4.5 $ and $ 9. $
Note:
Remember that the rate of the reaction depends only on the concentration of reactant (substrate) thus it is first-order kinetics. The extent of the reaction of the replacement of the one nucleophile by the other depends on the leaving group's ability to knock itself out and the reactivity of the incoming nucleophile.
Complete answer:
The nucleophilic substitution may proceed through the bimolecular or unimolecular pathways. The substitution of one nucleophile by the more reactive nucleophile is the nucleophilic substitution reaction. A nucleophilic aromatic substitution reaction, the nucleophile replaces the leaving group in the reagent structure. The leaving capacity of a leaving group or element depends on the stability of that group or element after leaving the reagent structure. One can think about the options and which can form a stable anion after it leaves.
$ {{R}_{1}}-C\left( {{R}_{2}} \right)=O+HCN\xrightarrow{pH=4.5}{{R}_{1}}{{R}_{2}}-C\left( OH \right)-CN $
The leaving atom or group ability is dependent on the stability of the anion formed by the leaving group. The stability of an anion depends on the electronegativity of that element. The more the electronegativity of an element is, the more will be its capacity to hold the anionic negative charge on itself. The more the capacity of an element to hold the charge on the element, the more will be its stability and it will act as a better leaving group.
$ {{R}_{1}}-C\left( {{R}_{2}} \right)=O+N{{H}_{2}}-C\left( =O \right)-NH-N{{H}_{2}}\xrightarrow{pH=9}{{R}_{1}}{{R}_{2}}-C=N-C\left( =O \right)-NH-N{{H}_{2}} $
Therefore, correct answer is option A and D i.e. $ \le 4.5 $ and $ 9. $
Note:
Remember that the rate of the reaction depends only on the concentration of reactant (substrate) thus it is first-order kinetics. The extent of the reaction of the replacement of the one nucleophile by the other depends on the leaving group's ability to knock itself out and the reactivity of the incoming nucleophile.
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