
What is chlorination reaction?
Answer
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Hint:In a chlorination reaction, the chlorine atom gets attached to one position, in alkyl halides. This reaction can proceed in two methods, depending on the condition given.
Complete step-by-step answer:In order to answer the question, let us learn about substitution reactions in organic chemistry. Nucleophilic substitution bimolecular or ${{S}_{N}}2$ is a single step bimolecular reaction in which the incoming nucleophile attacks the C-atom of substrate in a direction opposite to the outgoing nucleophile.. In the transition state, the C-atom is $s{{p}^{2}}$ hybridised with a p-orbital whose one lobe overlaps with an orbital of incoming nucleophile and the other lobe overlaps with an orbital of outgoing nucleophile. The reaction is completed when the outgoing nucleophile leaves with the bond pair of electrons and simultaneously the incoming nucleophile binds to the C-atom. As the reaction progresses, the configuration of the C-atom under attack is inverted.
Now, let us come to ${{S}_{N}}1$ reaction. Nucleophilic substitution unimolecular or ${{S}_{N}}1$ is a two step unimolecular reaction. Slow ionisation of substrate and is the rate-determining step. The second step consists of a rapid reaction between the carbocation (formed in the first step) and the nucleophile. ${{S}_{N}}1$ reactions generally proceed in polar protic solvents. Since the first step is the rate-determining step, the rate of reaction depends upon the concentration of alkyl halide only and is independent of the concentration of nucleophile.
Both the ${{S}_{N}}1,{{S}_{N}}2$ reactions are examples of chlorination reactions. In a chlorination reaction, a chlorine atom gets attached to the alkane to form alkyl chloride. For example: \[{{C}_{6}}{{H}_{6}}C{{H}_{3}}+C{{l}_{2}}\to {{C}_{6}}{{H}_{6}}C{{H}_{2}}\mathbf{Cl}\]
Note:It is to be noted that in ${{S}_{N}}1$ reaction, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction.
Complete step-by-step answer:In order to answer the question, let us learn about substitution reactions in organic chemistry. Nucleophilic substitution bimolecular or ${{S}_{N}}2$ is a single step bimolecular reaction in which the incoming nucleophile attacks the C-atom of substrate in a direction opposite to the outgoing nucleophile.. In the transition state, the C-atom is $s{{p}^{2}}$ hybridised with a p-orbital whose one lobe overlaps with an orbital of incoming nucleophile and the other lobe overlaps with an orbital of outgoing nucleophile. The reaction is completed when the outgoing nucleophile leaves with the bond pair of electrons and simultaneously the incoming nucleophile binds to the C-atom. As the reaction progresses, the configuration of the C-atom under attack is inverted.
Now, let us come to ${{S}_{N}}1$ reaction. Nucleophilic substitution unimolecular or ${{S}_{N}}1$ is a two step unimolecular reaction. Slow ionisation of substrate and is the rate-determining step. The second step consists of a rapid reaction between the carbocation (formed in the first step) and the nucleophile. ${{S}_{N}}1$ reactions generally proceed in polar protic solvents. Since the first step is the rate-determining step, the rate of reaction depends upon the concentration of alkyl halide only and is independent of the concentration of nucleophile.
Both the ${{S}_{N}}1,{{S}_{N}}2$ reactions are examples of chlorination reactions. In a chlorination reaction, a chlorine atom gets attached to the alkane to form alkyl chloride. For example: \[{{C}_{6}}{{H}_{6}}C{{H}_{3}}+C{{l}_{2}}\to {{C}_{6}}{{H}_{6}}C{{H}_{2}}\mathbf{Cl}\]
Note:It is to be noted that in ${{S}_{N}}1$ reaction, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction.
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