The order of reactivity of various alkyl halides towards the nucleophilic substitution follows the order:
A. \[RI>RBr>RCl>RF\]
B. \[RF>RCl>RBr>RI\]
C. \[RCl>RBr>RI>RF\]
D. \[RBr>RI>RCl>RF\]
Answer
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Hint: We know that alkyl halides are the chain of hydrocarbons in which hydrogen are replaced by halogens. The reactivity of the alkyl halide is decided by the ease with which the halide leaves the substrate.
Complete step-by-step answer:
Nucleophilic substitution is a fundamental class of reactions in which a leaving group (nucleophile) is replaced by an electron rich compound (nucleophile).
The electron pair from the nucleophile (nuc) attacks the substrate(R-LG) forming a new bond, while the leaving group (LG) departs with an electron pair. The principal product in this case is R-nuc.
And the nucleophile may be electrically neutral or negatively charged, whereas the substrate is typically neutral and positively charged. Anions of halogens are good nucleophiles. In polar, protic solvents, \[{{F}^{-}}\]is the weakest nucleophile, and \[{{I}^{-}}\] the strongest.
The alkyl halides are very reactive due to highly polarized \[CX\] bonds with a large difference in electronegativities of carbon and halogen atoms.
As per the leaving ability, the order is \[I>Br>Cl>F\]. So, the order of \[S{{N}^{2}}\] reaction is \[I>Br>Cl>F\]. Because as the atoms get bigger from \[F\] to \[I\], bonds get longer, so weaker and react faster. Length and strength dominate, not polarity of the carbon-halogen bond, which would give the opposite order.
From the above points we can conclude that the order of reactivity in polar protic solvent is \[RI>RBr>RCl>RF\].
So, the correct option is “A”.
Note: The reactivity order also depends on solvent. The above mentioned order is in polar protic solvent and if the solvent is polar aprotic then the order gets reversed.
Complete step-by-step answer:
Nucleophilic substitution is a fundamental class of reactions in which a leaving group (nucleophile) is replaced by an electron rich compound (nucleophile).
The electron pair from the nucleophile (nuc) attacks the substrate(R-LG) forming a new bond, while the leaving group (LG) departs with an electron pair. The principal product in this case is R-nuc.
And the nucleophile may be electrically neutral or negatively charged, whereas the substrate is typically neutral and positively charged. Anions of halogens are good nucleophiles. In polar, protic solvents, \[{{F}^{-}}\]is the weakest nucleophile, and \[{{I}^{-}}\] the strongest.
The alkyl halides are very reactive due to highly polarized \[CX\] bonds with a large difference in electronegativities of carbon and halogen atoms.
As per the leaving ability, the order is \[I>Br>Cl>F\]. So, the order of \[S{{N}^{2}}\] reaction is \[I>Br>Cl>F\]. Because as the atoms get bigger from \[F\] to \[I\], bonds get longer, so weaker and react faster. Length and strength dominate, not polarity of the carbon-halogen bond, which would give the opposite order.
From the above points we can conclude that the order of reactivity in polar protic solvent is \[RI>RBr>RCl>RF\].
So, the correct option is “A”.
Note: The reactivity order also depends on solvent. The above mentioned order is in polar protic solvent and if the solvent is polar aprotic then the order gets reversed.
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