Starting with acetylene and bromoethane, show how you would use reagents from the table to synthesize 4-methyl-3-hexanone. (enter your choices as a string of letters without punctuation, i. e. ace.)

I have provided the full reaction scheme for the synthesis of 4-methyl-3-hexanone from the reaction of acetylene and bromoethane. Acetylene is initially reacted with NaNH₂ which is a strong base that deprotonates the C-H of the acetylene which creates a carbon nucleophile which will then attack the electrophilic carbon containing the bromo in bromoethane. This is a simple sn2 substitution. Essentially an ethyl group is added to each side of the triple bond in acetylene.

With the 3-hexyne in hand, the triple bond is reduced using Lindlar's catalyst which will hydrogenate only to the alkene and stop. The 3-hexene is then reacted with a peroxycarboxylic acid which is used to epoxidize the alkene, to give the epoxide.

The epoxide is reacted with the grignard reagent which treats the methyl as a strong nucleophile. The methyl adds to one carbon of the epoxide and opens the ring. The acid is added at the end to protonate the alcohol.

Finally, the alcohol is oxidized with chromic acid which will oxidize a secondary alcohol to the ketone. The final product is 4-methyl-3-hexanone.

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

Answer(false);