Starting with acetylene and bromoethane, show how you would use reagents from the table to synthesize 4-methyl-3-hexanone. (ente
r your choices as a string of letters without punctuation, i.e. ace.)
1 answer:
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.
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.
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Answer: polar molecule.
Explanation:
The boiling point is the temperature at which the vapor pressure of a liquid equals the external pressure surrounding the liquid. The boiling point is dependent on the type of forces present.
Iodine monochloride (ICl) is a polar molecule due to the difference in electronegativities of iodine and chlorine. Thus the molecules are bonded by strong dipole dipole forces. Thus a higher temperature is needed to generate enough vapor pressure.
Bromine is a non polar molecule as there is no electronegativity difference between two bromine atoms. The molecules are bonded by weak vanderwaal forces and thus has low boiling point.