Question

The most common method for the synthesis of unsymmetrical ethers is the Williamson synthesis, a reaction (SN2) of an alkoxide ion with an alkyl halide. Two pathways are possible, but often one is preferred. Construct the preferred pathway for the synthesis of 2-propoxypropane from propene, with propene-derived alkyl halide and alkoxide intermediates, by dragging the appropriate intermediates and reagents into their bins. Not every given reagent or intermediate will be used.

The most common method for the synthesis of unsymmetrical ethers is the Williamson synthesis, a reaction (SN2) of an alkoxide ion with an alkyl halide. Two pathways are possible, but often one is preferred. Construct the preferred pathway for the synthesis of 2-propoxypropane from propene, with propene-derived alkyl halide and alkoxide intermediates, by dragging the appropriate intermediates and reagents into their bins. Not every given reagent or intermediate will be used.

Answer 1

Concepts and reason

The concept used in the problem is of Williamson’s Synthesis. Williamson’s Synthesis or Williamson’s ether synthesis is very important and convenient method for the synthesis of simple and mixed ethers. As the unsymmetrical ether synthesis requires selectivity of the reactant to give the desired product. Several pathways are possible for the synthesis.

Fundamentals

The Williamson’s ether synthesis involves the reaction of alkoxides and phenoxides with alkyl halides, which undergo bimolecular substitution reaction (S_N2 reaction). In the asymmetric synthesis, there are two possibilities for the choice of reactants, which is more available and reactive to form desired product.

Since the preliminary requirement for the synthesis is the selection of correct reactant. In which alkoxide (deprotonated alcohol) or the phenoxide may be primary, secondary or tertiary depending upon the alcohol taken. Whereas the alkyl halide, are taken most preferably primary, because the reaction undergoes S_N2 reaction in which the primary halides are more reactive.

As the alkoxide are very reactive in nature, so they are just prepared from the alcohol immediately before performing Williamson’s synthesis, by the use of strong base such as metal hydride, or carbonate base.

Alkyl halide can also be prepared from the hydro halogenation of alkene.

To obtain primary halide and the secondary alkoxide, propene must undergo different types of electrophilic addition reactions.

Firstly, the Markovnikov’s addition of ${{\rm{H}}_{\rm{2}}}{\rm{O/}}{{\rm{H}}_{\rm{2}}}{\rm{S}}{{\rm{O}}_{\rm{4}}}$ (Reagent 1) to propene to get 2-propanol is done (Secondary alkoxide as Step 1 product).

To obtain 2-propoxypropane as the final desired product there must be a primary halide and secondary alkoxide as the reactant.

The primary halide can be obtained by Anti-Markovnikov addition of ${\rm{HBr}}/ROOR(peroxide)$ (Reagent 2) to propene to get 1-bromopropane (primary halide as Step 2 product).

This secondary alcohol is further treated with metal base ${\rm{NaH}}$ (Reagent 3) to give the Sodium isopropoxide (secondary alkoxide as Step 3 product).

The secondary alkoxide ion attacks in the ${{\rm{S}}_{\rm{N}}}{\rm{2}}$ fashion on the ${\sigma ^ * }$ orbital of the C-Br bond resulting in the breakage of the sigma bond between C-Br and simultaneously formation of the sigma bond between carbon and the upcoming nucleophile, in a concerted manner.

Resulting the formation of 2-propxypropane as the Williamson’s unsymmetrical ether.

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