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Which reaction in each of the following pairs will take place more rapidly?
A. \[C{H_3}Br + H{O^ - } \to C{H_3}OH + Br\], \[C{H_3}Br + {H_2}O \to C{H_3}OH + HBr\]
B. \[C{H_3}Br + H{O^ - } \to C{H_3}OH + {I^ - }\], \[C{H_3}Br + H{O^ - } \to C{H_3}OH + C{l^ - }\]
C. \[C{H_3}Br + N{H_3} \to C{H_3}{N^ + }H + B{r^ - }\], \[C{H_3}Br + {H_2}O \to C{H_3}OH + HBr\]
D. \[C{H_3}Br + O{H^ - }\xrightarrow{{DMSO}}C{H_3}OH + B{r^ - }\], \[C{H_3}Br + {H_2}O\xrightarrow{{EtOH}}C{H_3}OH + B{r^ - }\]
E. \[C{H_3}Br + N{H_3}\xrightarrow{{E{t_2}O}}C{H_3}{N^ + }H + B{r^ - }\], \[C{H_3}Br + N{H_3}\xrightarrow{{EtOH}}C{H_3}{N^ + }H + B{r^ - }\]

Last updated date: 20th Jun 2024
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Hint: Nucleophile is a chemical species that donates its electron pair to form a bond with other atoms and thus, the nucleophile is also called an electron-rich species. When nucleophile undergoes substitution, based on the nature of reactant, it comes under two categories such as Unimolecular Nucleophilic substitution (\[{S_N}^1\]) and Bimolecular Nucleophilic substitution (\[{S_N}^2\]) reaction.

Complete step by step answer:
\[{S_N}^2\] reaction is nothing but bimolecular nucleophilic substitution reaction. It is a single-step process and thus, no carbocation is formed as an intermediate in the reaction. Since \[{S_N}^2\] is a single-step process, the rate of the reactions depends on both reactant and also nucleophile. The reaction will be influenced by the steric effect and thus, the primary alkyl group is more reactive towards \[{S_N}^2\] reaction than secondary and tertiary alkyl groups. The order of reactivity of alkyl order for \[{S_N}^2\] reaction can be written as,
Primary alkyl \[ > \] Secondary alkyl \[ > \] Tertiary alkyl
The leaving group in \[{S_N}^2\] reaction will influence the rate of the reaction. Among alkyl halides, the order of reactivity can be written as,
\[R - I > R - Br > R - Cl > R - F\]
Thus, iodine ion will act as a better leaving group among all halogens.
\[C{H_3}Br + H{O^ - } \to C{H_3}OH + Br\] Will take place faster than \[C{H_3}Br + {H_2}O \to C{H_3}OH + HBr\]. This is because of the nature of nucleophiles. \[{S_N}^2\] reaction favours strong nucleophile and alkoxide ion (\[H{O^ - }\]) is stronger nucleophile than water.
\[C{H_3}Br + H{O^ - } \to C{H_3}OH + {I^ - }\] will take place faster than \[C{H_3}Br + H{O^ - } \to C{H_3}OH + C{l^ - }\] since iodine ion is better leaving group than chlorine ion and it makes the reaction to occur fast.
\[C{H_3}Br + N{H_3} \to C{H_3}{N^ + }H + B{r^ - }\] will take place faster than \[C{H_3}Br + {H_2}O \to C{H_3}OH + HBr\] since ammonia is strong nucleophile than water. Nitrogen is less electronegative than oxygen and thus, it tends to donate its lone pair of electrons.
\[C{H_3}Br + O{H^ - }\xrightarrow{{DMSO}}C{H_3}OH + B{r^ - }\] will take place faster than \[C{H_3}Br + {H_2}O\xrightarrow{{EtOH}}C{H_3}OH + B{r^ - }\] since alkoxide ion is stronger nucleophile than water. And also DMSO is less polar solvent than ethanol. \[{S_N}^2\] reactions will be favoured by solvents with low polarity.
\[C{H_3}Br + N{H_3}\xrightarrow{{E{t_2}O}}C{H_3}{N^ + }H + B{r^ - }\] will take place faster than \[C{H_3}Br + N{H_3}\xrightarrow{{EtOH}}C{H_3}{N^ + }H + B{r^ - }\] since ether is nonpolar and generally \[{S_N}^2\] reactions will be favoured by solvents with low polarity.
In the following pairs, the reaction which will take place rapidly
\[C{H_3}Br + H{O^ - } \to C{H_3}OH + Br\]
\[C{H_3}Br + H{O^ - } \to C{H_3}OH + {I^ - }\]
\[C{H_3}Br + N{H_3} \to C{H_3}{N^ + }H + B{r^ - }\]
\[C{H_3}Br + O{H^ - }\xrightarrow{{DMSO}}C{H_3}OH + B{r^ - }\]
\[C{H_3}Br + N{H_3}\xrightarrow{{E{t_2}O}}C{H_3}{N^ + }H + B{r^ - }\]

Note: Backside attack is possible in \[{S_N}^2\] reaction and stereochemically, inversion product is obtained from \[{S_N}^2\] reaction. No rearrangement is possible in \[{S_N}^2\] reaction since no carbocation is formed as an intermediate.