
Alkyl halides can be obtained by all methods except:
(A) $C{{H}_{3}}C{{H}_{2}}OH+HCl\xrightarrow{ZnC{{l}_{2}}}$
(B) $C{{H}_{3}}CH=C{{H}_{2}}\xrightarrow{S{{O}_{2}}C{{l}_{2}}/475K}$
(C) ${{C}_{2}}{{H}_{5}}OH+NaCl\to $
(D) $C{{H}_{3}}COOAg+B{{r}_{2}}\xrightarrow[\text{Reflux}]{CC{{l}_{4}}}$
Answer
583.2k+ views
Hint: In presence of lewis acid, and chloride ions, alcohol can give halides. Thionyl chloride can add a halogen group to the $\alpha $ position of alkenes. Borodine-Hundsdicker reaction gives alkyl halides as a product.
Complete step by step solution:
Let’s see all the given reactions in the options and find which one is incorrect.
i) $C{{H}_{3}}C{{H}_{2}}OH+HCl\xrightarrow{ZnC{{l}_{2}}}C{{H}_{3}}C{{H}_{2}}Cl+{{H}_{2}}O$
In this reaction, we can see that alkyl halide is obtained as a product. Here, in presence of lewis acid zinc chloride, hydrochloric acid can react with alcohol to give alkyl halides. The name of the reaction is Groove’s reaction.
ii) $C{{H}_{3}}CH=C{{H}_{2}}\xrightarrow{S{{O}_{2}}C{{l}_{2}}/475K}Cl-C{{H}_{2}}CH=C{{H}_{2}}$
In this reaction, we will also get an alkyl halide as a product. Thionyl chloride can add halogen to the $\alpha $ carbon atom if any hydrogen atom is present there. So, we will get alkyl halide as a product here as well.
iii) ${{C}_{2}}{{H}_{5}}OH+NaCl\to \text{no reaction}$
Here, we can see that reaction is between alcohol and sodium chloride. But chloride ion here cannot displace hydroxyl groups because the C-O bond is stronger enough that chloride ions from its salt cannot displace it. Thus, this reaction will not proceed further.
iv) $C{{H}_{3}}COOAg+B{{r}_{2}}\xrightarrow[\text{Reflux}]{CC{{l}_{4}}}C{{H}_{2}}Br+C{{O}_{2}}$
This is an example of the Borodine-Hunsdiecker reaction. Here, bromine reacts with a silver salt of an ester in carbon tetrachloride to give carbon dioxide and an alkyl bromide. So, an alkyl halide is also formed here.
Thus, we can conclude that option (C) is the correct answer for this question.
Note: Keep in mind that chloride ions from any salt cannot displace hydroxyl groups of alcohol groups. Remember that with thionyl chloride, it is necessary that the compound has C-C double bond in order to undergo reaction to give alkyl halides.
Complete step by step solution:
Let’s see all the given reactions in the options and find which one is incorrect.
i) $C{{H}_{3}}C{{H}_{2}}OH+HCl\xrightarrow{ZnC{{l}_{2}}}C{{H}_{3}}C{{H}_{2}}Cl+{{H}_{2}}O$
In this reaction, we can see that alkyl halide is obtained as a product. Here, in presence of lewis acid zinc chloride, hydrochloric acid can react with alcohol to give alkyl halides. The name of the reaction is Groove’s reaction.
ii) $C{{H}_{3}}CH=C{{H}_{2}}\xrightarrow{S{{O}_{2}}C{{l}_{2}}/475K}Cl-C{{H}_{2}}CH=C{{H}_{2}}$
In this reaction, we will also get an alkyl halide as a product. Thionyl chloride can add halogen to the $\alpha $ carbon atom if any hydrogen atom is present there. So, we will get alkyl halide as a product here as well.
iii) ${{C}_{2}}{{H}_{5}}OH+NaCl\to \text{no reaction}$
Here, we can see that reaction is between alcohol and sodium chloride. But chloride ion here cannot displace hydroxyl groups because the C-O bond is stronger enough that chloride ions from its salt cannot displace it. Thus, this reaction will not proceed further.
iv) $C{{H}_{3}}COOAg+B{{r}_{2}}\xrightarrow[\text{Reflux}]{CC{{l}_{4}}}C{{H}_{2}}Br+C{{O}_{2}}$
This is an example of the Borodine-Hunsdiecker reaction. Here, bromine reacts with a silver salt of an ester in carbon tetrachloride to give carbon dioxide and an alkyl bromide. So, an alkyl halide is also formed here.
Thus, we can conclude that option (C) is the correct answer for this question.
Note: Keep in mind that chloride ions from any salt cannot displace hydroxyl groups of alcohol groups. Remember that with thionyl chloride, it is necessary that the compound has C-C double bond in order to undergo reaction to give alkyl halides.
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