
The reaction is fastest when X is: \[{{H}_{3}}C-\overset{-}{\mathop{S}}\,\overset{\oplus }{\mathop{Na}}\,+C{{H}_{3}}C{{H}_{2}}-X\to \]
(A) -OH
(B) -F
(C) $ -O-\overset{\overset{O}{\mathop{\parallel }}\,}{\mathop{C}}\,-C{{F}_{3}} $
(D) $ -O-\overset{\overset{O}{\mathop{\parallel }}\,}{\mathop{C}}\,-C{{H}_{3}} $
Answer
462.9k+ views
Hint To solve this question, first we have to understand the reaction conditions as in the questions related to organic chemistry it is very important to take care of the reaction condition otherwise it will affect the formation of the product.
Complete step by step solution:
A leaving group is an anion or a neutral molecule fragment which departs with the pair of electrons in the heterolytic bond cleavage. The leaving group can be positive or negative or neutral charge. In a nucleophilic substitution reaction, the leaving group that gets displaced according to the predefined stoichiometry.
In the given reaction, the better leaving group fastest will be the reaction, the given reaction is $ S{{N}_{2}} $ . the rate of $ S{{N}_{2}} $ reaction depends on the substrate and leaving group. More stable the leaving group, the more the speed of the reaction. Hence, the most stable leaving group among the following is $ -O-\overset{\overset{O}{\mathop{\parallel }}\,}{\mathop{C}}\,-C{{F}_{3}} $ , due to -I effect because of the presence of electron withdrawing group.
Hence, the correct answer is option (C) i.e. the reaction will be fastest when X is $ -O-\overset{\overset{O}{\mathop{\parallel }}\,}{\mathop{C}}\,-C{{F}_{3}} $ .
Note: Don’t get confused between the basicity and nucleophilicity. Remember that basicity is the subset of nucleophilicity. All the nucleophiles are Lewis bases because they donate a pair of electrons. A base is just the name which we give to a nucleophile while it forms a bond with the proton. It is not important that strong bases are good nucleophiles. There are four factors that affect nucleophilicity, there are charges, electronegativity, solvent and the steric factor.
The conjugate bases are always a better nucleophile. Hydroxide ions are always a better nucleophile than water.
Complete step by step solution:
A leaving group is an anion or a neutral molecule fragment which departs with the pair of electrons in the heterolytic bond cleavage. The leaving group can be positive or negative or neutral charge. In a nucleophilic substitution reaction, the leaving group that gets displaced according to the predefined stoichiometry.
In the given reaction, the better leaving group fastest will be the reaction, the given reaction is $ S{{N}_{2}} $ . the rate of $ S{{N}_{2}} $ reaction depends on the substrate and leaving group. More stable the leaving group, the more the speed of the reaction. Hence, the most stable leaving group among the following is $ -O-\overset{\overset{O}{\mathop{\parallel }}\,}{\mathop{C}}\,-C{{F}_{3}} $ , due to -I effect because of the presence of electron withdrawing group.
Hence, the correct answer is option (C) i.e. the reaction will be fastest when X is $ -O-\overset{\overset{O}{\mathop{\parallel }}\,}{\mathop{C}}\,-C{{F}_{3}} $ .
Note: Don’t get confused between the basicity and nucleophilicity. Remember that basicity is the subset of nucleophilicity. All the nucleophiles are Lewis bases because they donate a pair of electrons. A base is just the name which we give to a nucleophile while it forms a bond with the proton. It is not important that strong bases are good nucleophiles. There are four factors that affect nucleophilicity, there are charges, electronegativity, solvent and the steric factor.
The conjugate bases are always a better nucleophile. Hydroxide ions are always a better nucleophile than water.
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