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Question: \( C{{H}_{3}}-C{{H}_{2}}-Br\xrightarrow[dry\text{ }ether]{Mg}A\xrightarrow{{{H}_{2}}O}B \) Identi...

CH3−CH2−Br→dry etherMgA→H2OBC{{H}_{3}}-C{{H}_{2}}-Br\xrightarrow[dry\text{ }ether]{Mg}A\xrightarrow{{{H}_{2}}O}B
Identify A & B compounds.

Explanation

Solution

Hint : We know that in the given question firstly we have to define what exactly is the Sodium thiosulfate. Then we have to give a proper overview of the processes and then the given reaction that will take place that is Na2S2O3+4Cl2+5H2O→2NaHSO4+8HClNa{{}_{2}}{{S}_{2}}{{O}_{3}}+4C{{l}_{2}}+5{{H}_{2}}O\to 2NaHS{{O}_{4}}+8HCl . This will give us a solution to the problem.

Complete Step By Step Answer:
The given question statement asks about the kind of arrow through which we would be able to get and explain the proper and correct explanation of all the steps involved in the reaction. So we have to explain what exactly are the reversible reactions and the correct answer of the problem.
In the given problem the compound A formed is of ethyl magnesium bromide which is a grignard reagent whereas the compound B is ethane.
The chemical reaction for the given process would be as follows :
CH3−CH2−Br→dry etherMgAC{{H}_{3}}-C{{H}_{2}}-Br\xrightarrow[dry\text{ }ether]{Mg}A
CH3−CH2−Br→dry etherMgCH3−CH2−MgBrC{{H}_{3}}-C{{H}_{2}}-Br\xrightarrow[dry\text{ }ether]{Mg}C{{H}_{3}}-C{{H}_{2}}-MgBr
Therefore that makes A=CH3−CH2−MgBrA=C{{H}_{3}}-C{{H}_{2}}-MgBr
This is a Grignard Reagent. Grignard reagents or the Grignard compounds are very common and popular reagents in organic synthesis for creating new carbon-carbon bonds.
The chemical reaction for the given process would be as follows :
CH3−CH2−MgBr→H2OBC{{H}_{3}}-C{{H}_{2}}-MgBr\xrightarrow{{{H}_{2}}O}B
CH3−CH2−MgBr→H2OCH3−CH2−Mg(OH)BrC{{H}_{3}}-C{{H}_{2}}-MgBr\xrightarrow{{{H}_{2}}O}C{{H}_{3}}-C{{H}_{2}}-Mg(OH)Br
Therefore that makes as B=CH3−CH2−Mg(OH)BrB=C{{H}_{3}}-C{{H}_{2}}-Mg(OH)Br
Now by this reaction we can clearly observe that the product which is yielded at the products end of the reaction are of A=CH3−CH2−MgBrA=C{{H}_{3}}-C{{H}_{2}}-MgBr and B=CH3−CH2−Mg(OH)BrB=C{{H}_{3}}-C{{H}_{2}}-Mg(OH)Br .

Note :
Remember that for the example for Grignard Reagents, when reacted with another halogenated compound R′−X′R'-X' in the presence of a suitable catalyst, they typically yield R−R′R-R' and the magnesium halide MgXX′MgXX' as a byproduct; and the latter is insoluble in the solvents normally used. In this aspect, they are similar to organolithium reagents.