SYNTHESIS OF   ALCOHOLS

(Synthesis Tree Method)

Propose a synthesis plan for the indicated target molecules from the indicated single molecules (MOb 30 -41). To do this, use the reagents and reaction conditions that you think are necessary:

MOb 30 solution.

Strategy: It is observed that the starting molecule has been dehydrated and in the initially unsubstituted allylic position, a hydrogen has been displaced by the cyano or nitrile group. This last reaction can occur only if the precursor molecule is an allylic halide, which is why it is proposed as a precursor of the mob 30.

The Br is introduced in the desired position with the NBS and the alkene is the product of the dehydration of the starting molecule.

Solution MOb 31.

Strategy : It is a uncle ether, the necessary precursor molecule will be a 1,3-cyclopentadiene halide.

This halide is prepared by the action of NBS on the dienic cycloalkene, which in turn is prepared by the dehydrobromination of the precursor molecule, which is reached by the action of NBS on the cycloalkene formed.   previously by dehydrohalogenation of the starting molecule brominated by radicals

Mob 32 solution.

Strategy : It is similar to the one used in obtaining the mob 30

Solution MOb 33.

Strategy : the MOb 33 has an increase of two alkyl groups (methyl and ethyl) in relation to the starting molecule, which are introduced in different stages, for which a ketone is proposed as a precursor, carrying the ethyl group, which is introduced into the aldehyde. obtained from the starting alcohol.

Solution MOb 34.

Strategy : the MOb 34, shows that an initial methyl group was transformed into an aldehyde, which will obviously be formed from an alcohol, and the latter by hydration of an alkene, which is reached by dehydrobromination of a brominated molecule by radicals, of the starting material.

MOb 35 solution.

Strategy : An alkene is proposed as a precursor molecule, which is brominated under antimarkovnikov conditions. The alkene was prepared by dehydration of the alcohol formed, by the action of a Grignard on a ketone, which is prepared by oxidation of cyclohexanol.

Solution MOb 36.

Strategy: the MOb ,   it has twice as many C atoms as the starting one. Then an alcohol is proposed as a precursor, the same one that will be prepared between a Grignard and an aldehyde formed from the same starting molecule.

Solution MOb 37.

Strategy: The methyl ketone of the mob 37, can be prepared by oxidation of an acetylide group with Hg(II) salts in an acidic medium, the alcohol is not affected by this reagent.

The alcohol was formed by the action of sodium acetylide on a ketone derived from the starting alcohol by oxidation.

Mob solution 38.

Strategy: The starting material is oxidized to aldehyde with disiamilborane, which attacked by a suitable Grignard to form the mob 38

Mob solution 39.

Strategy :   As the mob 39 is an aldehyde, an alcohol is proposed as the precursor molecule, which, in turn, is prepared by opening an epoxide with a Grignard formed from the bromide of the starting molecule. This is brominated by radicals, to have the necessary compound.

MOb 40 solution.

Strategy : The halides are obtained from alkenes or alcohols, the latter is the best option, because the action of a Grignard group that contains the phenyl group allows the alcohol required to be formed as a precursor molecule.

The following two syntheses require a little more effort from the chemist, because the defined starting materials require the use of some reactions that are not the most common, despite the fact that the structures of the target molecules are relatively simple.

So how is the formation of molecules 41 and 42 to from the materials listed?

the MOb 41, can be obtained by reduction of the carbonyl group of the alpha, beta-unsaturated ketone substrate. Which, as we well know, is formed by the dehydration of an aldol, the result of the intramolecular aldol condensation of a 1,5-dicarbonyl precursor. Here we could use the Michael reaction to form it, but we are going to resort to a variant that uses 2,4-dichloro-2-butene, on an amide carbanion, which is only formed with very strong bases such as Et ₂ N ^- Li ⁺ in highly polar media such as hexamethylphosphorotriamide (HMPT).

On the other hand, methyl lithium in stoichiometric amounts leads to methyl ketone only when the substrate is an amide or a free acid (J. Organometal Chem., 9, 165 (1967). Alternatively, the use of methyl magnesium iodide is also suggested in place of methyllithium.

Mob 42 solution.

The opening of the cyclopropane molecule, a precursor to the mob 42, is carried out with a catalytic hydrogenation and it can be formed through a variant of the Simmons-Smith reaction, called the Furukawa reaction. The unsaturation necessary for cyclopropanation is formed from the elimination, with heat, of a xanthate formed   Starting with a secondary alcohol, the rest of the reactions are sufficiently known for further explanation.