1.

H2SO4 1) BH3, THF - 2) OH , H2O2, H2O OH OH

PBr3 Pyridine

O Mg HO 1) Ether 2) HCl Br MgBr

1. In the first step, acidic elimination of an alcohol always results in the most-substituted alkene. 2. Be prepared for them to throw in a reaction from Ochem 1 in one of the synthesis steps such as the second step of this synthesis. Since the alcohol is on the least substituted carbon, you know you have to use hydroboration in order to get the desired product. 3. Also, be aware of stereochemistry. Here, the question decides the stereochemistry for you for the alcohol in the third step. When PBr3 reacts with a secondary alcohol, you must remember that this is a Sn2 reaction and therefore inverts the stereochemistry. 4. It then reacts to become a Grignard reagent and the stereochemistry remains the same when preparing a Grignard reagent. 5. The next part is tricky and requires you to carefully count the carbons. Remember that Grignard reagents attack the less substituted carbon of an epoxide (the right side of the epoxide in this picture). When the Grignard reagent attacks the epoxide it loses its stereochemistry. 6. In the final product, the alcohol group is a chiral center, but it will be racemic (equal amounts of R and S), so you can either write it without stereochemistry (as shown) or draw both the R and S structures of it. Both ways are correct.

2. O-

Cl2 hv

Cl

H2SO4,

H2O

1.Ts-Cl, pyridine

N 2. NaCN OH

1. In the first step, although this would not be the major product in alkane halogenation, it is still a possible product, and frankly, the only way to obtain the given compound. 2. Then, remember the tert-butoxy ion always indicates an elimination reaction. It eliminates a hydrogen forming an alkene and the chlorine leaves. 3. Then, this synthesis calls on a ch 6 reaction of turning an alkene into an alcohol (a common step in these synthesis problems and should be kept in mind). 4. Finally, the compound is first turned into a tosylate, meaning that it will bond with the oxygen from the alcohol group. The -OTs group on the cyclohexane is a good leaving group. Since CN is a strong nucleophile, an SN2 reaction will occur, eliminating the -OTs group and replacing it with the CN group.

3. OH 1) MCPBA H2SO4 O 2) RCOOH

1) (CH2=CHCH3)2CuLi

2) HCl

OH

Seeing the stereochemistry may throw you off at first. I recommend finding the correct steps that get you the product first, disregarding stereochemistry, and then seeing how those can be modified to bring about the correct stereochemistry. So, when looking at your starting material, I would see that the only group we know how to manipulate is the alcohol group. There is very often a synthesis which involves a Gilman or Grignard reagent attacking an epoxide, so you should always be on the lookout for this in a synthesis. In the first step, the alcohol can be turned into an alkene, which is needed to create an epoxide. Then once it is transformed into an epoxide, you can see that you need a reagent that will attack the least substituted side with three carbons which includes a C-C double bond. We know that organocuprates attack the carbon of an epoxide that is the least substituted. This will give you the final product. The stereochemistry in the beginning is diminished when an alkene is formed, so the final product is just an example of one of the stereoisomers that this synthesis will create. Technically, both R and S are formed.

Extra tips: • Always compare the starting reagent and final product and write out what has changed/ been added • Think of all the relevant reactions that could help you get to the final product and try to find the correct order to put them in • Trial and error is a common route to go for synthesis problems; you are going to have to play around to get the right answer, but your mistakes will eventually lead you to the correct answer! • Always remember the Ochem 1 reactions because they may be very useful in a synthesis problem

4. O

OH

O

MCPBA

RCOOH H2CrO4 OR Na2Cr2O7

O O

CH3OH OH

H2SO4

Once again, when looking at the starting material and the ending material, try to decipher what all has changed.

1. In this synthesis it is easy to see that a methoxy group was added, there is no longer a double bond, and a carboxylic acid group was added. 2. You should remember that the only way you have learned to add a methoxy group was either acidic or basic addition to an epoxide. So, now you know you need to first transform the double bond into an epoxide, which we know is using MCPBA. 3. In order to add a methoxy group we need to decide which side of the epoxide we want it to attack so we know if we should use acidic or basic conditions. When looking at the final product, we can see the methoxy group was added as a result of attacking the more substituted carbon. 4. Therefore, we need to use acidic conditions. This will give us one methoxy group and one alcohol group. Now, when looking at the final product we can see the only thing left to do is turn the alcohol group into a carboxylic acid group. You should recognize that this is an oxidation of a primary alcohol. We only know that ����or ����� are capable of doing this. So, either of those reagents will give you the correct final product.

5. SH 2 6 SH 4 5 3 1

(R)-hexane-1,4-dithiol

The systematic naming rules apply the same to sulfides. However, keep in mind they are a priority group. Since the sulfide is the only priority group present, you will use the suffix -thiol. In compliance with the systematic naming rules, you always want to give priority groups the lowest number possible. So, you would number the compound as shown above. Remember because there are two sulfides that the suffix is -dithiol.

6 SH 5 1 2 4 SH 3

1-methyl-1,2-cyclohexanedithiol

This is similar to the preceding example. When deciding which sulfide group to number first since both could possibly be labeled as number one, remember the rule that you want all the substituents to have the lowest number possible. So, you would number the compound as shown above allowing the methyl substituent to be numbered as 1 also.

6 1 O O 2 3 5 3 2 1 4

2,3-epoxy-5-methylhexane 2-isobutyl-3-methyloxirane

There are two different ways you can name epoxides which requires different numbering systems. For the epoxide on the left, you number the compound based on the longest continuous carbon chain and then treat the oxygen of the epoxide as a substituent using the prefix -epoxy. So, in this case you number in order to allow the epoxide group to have the lowest number possible and then add the prefix -2,3-epoxy and name the rest using the normal systematic naming rules.

Using the naming method on the right, you always number the oxygen on the ring as 1 and then number only the two carbons attached, while naming the rest like substituents. So, in this example the carbon on the right side of the epoxide ring gets the lower number because its substituent is first in the alphabet. Lastly, add the suffix -oxirane. 3 4 2 O 1 5 6

1,2-epoxy-1-methylcyclohexane

In this example, only this method of naming is possible. So, you number the ring as the parent chain and then number the carbons of the epoxide as 1 and 2. Make sure the carbon with the lower number has the substituent so that the substituent also has the lower number.

6.

7.

8.

9.

2 1 3 4

The rate of dehydration for alcohols goes primary < secondary < tertiary. However, the molecule given the fastest ranking is a secondary alcohol. This is because the carbocation formed would be an allylic carbocation. Because it would be a secondary allylic carbocation, it is more stable than even a tertiary carbocation and would form the most quickly. The other 3 molecules follow the simple rule stated in the first sentence.

10.

** by using ���/���/���� with pyridine allows for reaction without carbocation rearrangement ** the first reaction occurs through a secondary alcohol (SN1 mechanism) and thus give a racemic mixture of R/S products, second reaction is always SN2 and thus gives an inversion of stereochemistry

11.

** step 1 activates the alcohol turning it into a good leaving group (retention of stereochemistry), negatively charged oxygen in step 2 acts as nucleophile displacing the OTsCl with a back side attach leading to an inversion of stereochemistry

12.

** when using acid catalyzed dehydration, the major product will be the most stable alkene (most stable alkene = most substituted). This can occur through a carbocation rearrangement so be careful ** use ���� and pyridine at 0°C to avoid carbocation rearrangement (this will lead to the most stable/most substituted alkene possible without rearrangement)

13.

**weak oxidant (PCC, ���� or NaOCl, ������) will oxidize a primary alcohol to an aldehyde **strong oxidant ( ���� or ����) will oxidize a primary alcohol to a carboxylic acid ** secondary alcohols will ALWAYS be oxidized to a ketone ** tertiary alcohols CANNOT be oxidized using these reagents

14.

**Reaction occurs through SN1 when the leaving group makes a stable carbocation. Most substituted side of the of the ether becomes the alkyl halide, and least substituted side becomes the alcohol **Reaction occurs through an SN2 when the ether doesn’t form a stable carbocation. Pay careful attention because the nucleophile attacks the least sterically hindered of the 2 alkyl groups (forming least substituted alkyl halide) and the alcohol forms on the most sterically hindered side of the alkyl group

15.

** Williamson ether synthesis (halide is good leaving group which is displaced by the nucleophile)

16.

** In acidic conditions the nucleophile attacks the MOST SUBSTITUTED side of the epoxide ** In basic conditions the nucleophile attacks the LEAST SUBSTITUTED side of the epoxide

17.

** trans diols formed when cyclohexene oxide reacts with a hydroxide ion and then hydrochloric acid

18.

** Hoffman Elimination where a quaternary ammonium ion can undergo E2 elimination with a strong base, producing a tertiary amine and the Anti-Zaitsev double bond (least substituted) this occurs as a hydrogen is plucked from the beta carbon with the most hydrogens

19.

** First must add 2 equivalents of lithium to an alkyl halide in a non-polar solvent (hexane) ** The organolithium compound then acts as a nucleophile that react in the same manner as a SN2 reaction would

20.

**palladium in a basic solution replaces the halogen (iodine/bromine preferred) with a carbon containing group of an organoboron in this case ** if reacting with vinyl halide conserve the stereochemistry of the double bond

21.

**For the heck reaction a vinylic or aryl halide must be used in the presence of a base (triethylamine) and the palladium catalyst (in this reaction the R group of the halide replaces the least sterically hindered vinyl hydrogen) **stereochemistry of the vinyl halide is conserved, if there is a substituent bonded to the alkene (Z), the new group will be trans to Z

22.

**For monochlorination resulting from radical chlorination the reaction will for an alkyl halide for every unique carbon in the compound. This compound has 7 unique carbons that lack symmetry therefore there will be 7 monochlorination products 23.

**For determining the major monochlorination products one must calculate the probable product percentages. This is done by determining the amount of substitutable hydrogens at each carbon (look out for planes of symmetry) and multiply by the radical of formation number given in the Relative Rates of Formation table (see chapter 12 packet). Then divide each product by the sum of all the possible hydrogens to get the relative percentage of halogenating product. Example for answer shown above: 3.8 (value from table) x 4 (4 hydrogens at those 2 secondary positions) = 15.2 / 18 = 0.84 3.8 x 4 =15.2 / 18 = 0.84 = 84% This was the highest yield out of all possible monochlorination products

24.

** alkenes are halogenated at the anti-Markovnikov position with HBr via radical reaction (initiated by peroxide along with heat or light). If peroxide is not used the Markovnikov product will form