Propose an efficient synthesis for the following transformation.

Question 1

Propose an efficient synthesis for the following transformation shown on the attached image.

The transformation shown on the attached image can be performed with some reagent or combination of the reagents listed below. Give the necessary reagent(s) in the correct order, as a string of letters (without spaces or punctuation, such as “EBF”). If there is more than one correct solution, provide just one answer.



2) H3O+H3O+










2) H2O2,H2OH2O2,H2O









Question 2

Propose an efficient synthesis for the following reaction transformation.

Question 3

Answer to question 1

The most correct answer is FG.

The alkane to be transformed has the standard IUPAC name 2-methylbutane, meaning a four-carbon chain with a one-carbon side branch attached one carbon from the end. Carbon has four electrons in its outer valence shell, meaning a neutral carbon atom is only stable when participating in four covalent bonds. In molecular shorthand, only the carbon-carbon bonds are shown for this molecule, but wherever a carbon (represented by a kink, intersection, or line ending) is drawn with less than four bonds, the remaining bonds are implied to be with hydrogen, which has a single electron in its neutral state and participates in one bond at a time.

Opposite the methyl group on carbon two is a single hydrogen surrounded by three carbons. Considering that hydrogen has an electronegativity value of 2.1 and carbon one of 2.5, the carbons have slightly more pull on electrons than hydrogen, making this C-H bond the weakest in the molecule and the easiest for a reagent to attack. Next door, carbon three of the butane chain has two bonds to carbon atoms and two bonds to hydrogen atoms. These two C-H bonds are the next weakest in the molecule.

Reagent F, tert-butoxypotassiumide, is an organic ionic compound and weak base. The anion is a central carbon atom with covalent bonds to three more and and oxygen atom, the last carrying the negative charge. The cation is Potassium, with an electronegativity value of 0.8. Although oxygen has a stronger electronegativity value (3.5) than carbon, it has two electrons in the outer valence shell, giving it a preference for covalent bonds. The hydrogen opposite the methyl group in 2-methyl-butane is easily shed, make the compound a weak acid. The potassium ion will take that hydrogen atom’s place, freeing it to form an alcohol group (-OH) with the oxygen in tert-butoxypotassiumide. The process is then repeated at carbon three in the butane chain.

Reagent G is molecular bromine with hv (light) as a catalyst. Bromine has an electronegativity value of 2.8 and seven electrons in the outer valence shell, one less than a complete set of eight. Potassium-bromine compounds are much more stable than potassium-carbon bonds. Light energy is absorbed by the bromine molecule, dividing it into two bromine free radicals. The carbon with the methyl group donates one electron and its negative charge to one of these free radicals, turining it into a Br- anion that forms an ionic compound with K+. This leaves carbon two of the chain as a free radical, pulling even more on carbon three and further weakening its C-H bonds. One hydrogen is sacrificed to form a double C-C bond. The result is the alkene 2-methylbut-2,3-ene The remaining bromine free radical can then attack the outcast hydrogen atom.

Answer to question 2

The given transformation can be achieved efficiently in the following steps as described and shown below:

Step 1: Chlorination of the given substrate.

Step 2: Formation of a less substituted alkene.

Step 3: Hydroboration-oxidation of the double bond.

Step 4: Oxidation of secondary alcohol into a ketone.

Step 5: Incorporation of one methyl group into a ketone.

Step 6: Nucleophilic addition to the keto group.

Answer to question 3

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