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Question: The correct set of reagents for the following conversion is: ![](https://www.vedantu.com/question-...

The correct set of reagents for the following conversion is:

A. P2O5/Δ,P4/I2,Na{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}/\Delta ,{P_{\text{4}}}{\text{/}}\,{{\text{I}}_{\text{2}}}{\text{,Na}}
B. P4/I2,Na, P2O5/Δ{P_{\text{4}}}{\text{/}}\,{{\text{I}}_{\text{2}}}{\text{,Na, }}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}/\Delta
C. P2O5,NaBH4{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}{\text{,}}\,{\text{NaB}}{{\text{H}}_{\text{4}}}
D. P4/I2,Na,conc.H2SO4{P_{\text{4}}}{\text{/}}\,{{\text{I}}_{\text{2}}}{\text{,Na,}}\,{\text{conc}}{\text{.}}\,{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}

Explanation

Solution

We can synthesize the product by ionizing the reactant in the first step then we will condense the iodized reactant and the condensed product will be treated with the reagent which can cause anhydride formation.

Complete step by step answer:
The reaction of (CH3)2CHCOOH{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHCOOH}} with P2O5/Δ{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}/\Delta lead to the formation of anhydride but not cyclic anhydride then treating with P4/I2,Na{P_{\text{4}}}{\text{/}}\,{{\text{I}}_{\text{2}}}{\text{,Na}} leads to a different product so, option (A) is incorrect.
P4/I2{P_{\text{4}}}{\text{/}}\,{{\text{I}}_{\text{2}}} is an oxidizing reagent. This reagent gives a nucleophilic substitution reaction. The nucleophile iodide substitutes the hydrogen.
The reaction is shown as follows:


The two iodized reactants can be condensed by using sodium in presence of dry ether.
The reaction is as follows:


The condensed product is tetramethyl succinic acid.
When two carboxylic groups are condensed they form anhydride. The P2O5/Δ{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}/\Delta reagent causes the anhydride formation.
The reaction is as follows:


So, the reactant can be converted into the product by using the P4/I2,Na, P2O5/Δ{P_{\text{4}}}{\text{/}}\,{{\text{I}}_{\text{2}}}{\text{,Na, }}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}/\Delta .
The reaction of (CH3)2CHCOOH{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHCOOH}} with P2O5/Δ{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}/\Delta lead to the formation of anhydride but not cyclic anhydride then treating with NaBH4{\text{NaB}}{{\text{H}}_{\text{4}}} will give alcohol as a product so, option (C) is incorrect.
The reaction of (CH3)2CHCOOH{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHCOOH}} with P4/I2,Na{P_{\text{4}}}{\text{/}}\,{{\text{I}}_{\text{2}}}{\text{,Na}} lead to the formation of tetramethyl succinic acid then treating with conc.H2SO4{\text{conc}}{\text{.}}\,{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} leads to a different product so, option (D) is incorrect.

Therefore, option (B) P4/I2,Na, P2O5/Δ{P_{\text{4}}}{\text{/}}\,{{\text{I}}_{\text{2}}}{\text{,Na, }}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}/\Delta is correct.

Note: The reagent P4/I2{P_{\text{4}}}{\text{/}}\,{{\text{I}}_{\text{2}}} introduced iodine into the reactant. conc.H2SO4{\text{conc}}{\text{.}}\,{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} is an oxidizing and dehydrating reagent. NaBH4{\text{NaB}}{{\text{H}}_{\text{4}}} is a reducing agent. P2O5/Δ{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{5}}}/\Delta cause anhydride formation. Sodium in dry ether case the joining of aryl halide and alkyl halide or two aryl or two alkyl halides. The reaction is known as Wurtz fittig reaction.