1.

2.

H2SO4 H2SO4 H2SO41) BH3, THF 1) BH3, THF 1) BH3, THF - - - 2) OH , H2O2, H2O2) OH , H2O2, H2O 2) OH , H2O2, H2O OH OH OH

3. OH OH OH

H2SO4 1) BH3, THF - 2) OH , H2O2, H2O OH PBr3 PBr3 PBr3 Pyridine Pyridine Pyridine OH

PBr3 Pyridine O O O Mg Mg Mg HO HO 1)HO 1) 1) Ether Ether Ether 2) HCl 2) HCl 2) HCl Br Br Br MgBr MgBr MgBr O Mg HO 1) Ether 2) HCl Br 4. MgBr

**This reaction reduces an aldehyde to a primary alcohol and a ketone to a secondary alcohol. Since only one EQ of reagent is given the most reactive functional group will be reduced (in this case the aldehyde)

**ketones are less reactive than aldehydes and the more steric hindrance surrounding the ketone the less reactive it is.

Note: NaBH4 reduces ketones but is too weak to reduce an ester/double/triple bonds

5.

** With these reagents a tertiary amide will be reduced to a tertiary amine, a secondary amide will be reduced to a secondary amine, and a primary amide will be reduced to a primary amine 6.

** Recall from chapter 7, OChem 1 H2 and Lindlars catalyst always reduces an alkyne to a CIS ALKENE

** Li/Na and liquid NH3 always reduces an alkyne to a TRANS ALKENE

7.

**A ketone/aldehyde reacts with primary amines in trace acid to produce an imine, and reacts with a secondary amine in trace acid to produce an enamine (double bond forms between the alpha and beta carbons)

**Sodium cyanoborohydride (NaBH3CN) is used to reduce imines/enamines back to their respective secondary/tertiary amines

8.

** H2, Raney Nickle reduces nitriles to primary amides

** H2, Raney Nickle also reduces aldehydes to PRIMARRY ALCOHOLS, and reduces ketones to SECONDARRY ALCOHOLS 9.

** with the Wittig reaction the alkyl group of the ylide replaces the carbonyl oxygen forming an alkene

** to prepare triphenylphosphate (the wittig reactant an aldehyde/ketone reacts with phosponium ylide to produce an alkene (to prepare the ylide, triphenylphosphate reacts with an alkyl bromide via SN2 reaction

10.

** When an aldehyde reacts with a Grignard a secondary alcohol forms, with the Grignard acting as a nucleophile attacking the carbonyl carbon.

11.

** Then Cyanide ion acts as a nucleophile to form a cyanohydrin

Note: must use EXCESS CYANIDE ION and HCl

12.

** Aldehydes (and ketones) react with water to form a gem-diol (2 alcohols bonded to the same carbon).

Note: to hydrolyze back to the original aldehyde/ketone react acid (HCl) in water. 13.

** This an alpha/beta unsaturated ketone (double bond between the alpha and beta carbons) and can form 2 products the direct addition product (nucleophile attacks carbonyl carbon) and the conjugate addition product (nucleophile attacks the beta carbon.

** the direct addition product is formed from strong nucleophiles such as Grignard and NaBH4 (hydride reduction) shown above.

** the conjugate addition product is formed from weak nucleophiles such as cyanide ion, HBr, thiols, amines, and the gilman reagent.

14.

1. DIBAL-H, -78℃ 2. H2O

DIBAL-H is a mild reducing agent that stops the reduction at the aldehyde rather than converting it all the way to a primary alcohol

15.

**Grignard reagents are synthesized from adding magnesium to an alkyl halide in diethyl ether or tetrahydrofuran (THF). One the Grignard reagent is synthesized a carbon chain can be added.

** magnesium is inserted between the carbon and the halogen

** Grignard reagents react as nucleophiles and follow SN2 rules.

16.

** LDA is a strong base (poor bulky nucleophile) it leads to enolization via removal of an alpha hydrogen to form a double bond, the reaction can then go through a SN2 substitution with a primary/methyl alkyl halide

** Kinetic product: formed at low temperatures (-78°C) and extracts the least hindered alpha hydrogen resulting in the least substituted double bond

** Thermodynamic product: forms at high temperatures, and prefers forming the most substituted double bond (most stable)

17.

**This is a base catalyzed alpha carbon halogenation. There is only one alpha carbon in this compound. Also take note that in base catalyzed alpha carbon halogenation every available alpha hydrogen is replaced with the given halogen.

18.

**The top (kinetic product) which is formed in low temperatures with LDA and prefers the least sterically hindered enolate.

**The bottom (thermodynamic product) is formed with high temperatures and heat and prefers the most substituted/sterically hindered double bond. 19.

** Robinson Ring Annulation: this reaction includes a ring formation via a micheal reaction (steps 1-3). Then in goes through an aldol formation (intermolecularly in this case (step 4)) and finally a dehydration (step 5) ** Remember to always number your compound in order to more accurately predict the structure in ring formation reactions as I have shown above.

20.

**Dieckmann cyclization mechanism: this is an intramolecular Claisen condensation which occurs between 2 esters or a di ester to form a beta-keto ester compound in this case. Only 5 or 6 member rings are formed due to stability. Remember it is key to remove a second alpha hydrogen after product formation to drive the reaction to completion.

** Remember to number your compound before beginning to avoid confusion. 21.

** This is the mechanism for a Michael reaction (steps 1-3) where a carbon nucleophile is formed first to attack the alpha/beta-unsaturated ketone and then again to form the ring structure. Finally step 4 involves the formation of an aldol after the ring is formed using hydrochloric acid as a proton donor.

** Remember to number your compound before beginning to avoid confusion.

22.

A strong base such as NaOH will remove every alpha hydrogen and replace it with a Bromine. This contrasts the base LDA which only removes one alpha hydrogen.

23.

24.

I2, H2O2, H2SO4

Br2, FeBr2

HNO , H SO 3 2 4 This is a matter of memorization of reagents for electrophilic aromatic substitution. Make flashcards! 25.

Remember that carbocation rearrangement is possible in Friedel Crafts Alkylation!! Always look for this first when doing a Friedel Crafts Alkylation problem. This rearrangement allowed for there to be a secondary carbocation instead of a primary carbocation. They WILL ask you a question that involves a carbocation rearrangement.

26.

Br NaNO2, HCl 0o C

The first set of reagents create the arenediazonium salt which then allow us to do a Sandmeyer reaction. Once you have your diazonium salt, when you add CuBr, it adds bromine to the benzene ring.

NO2 NO2

Br

This is a matter of ortho/meta/para and their directing effects. First, the ring is alkylated and the methyl group is added to the ring. Then, we know a methyl group is a weak activator, so it either attaches at the ortho or para position. However, if possible, the substituent would rather go to the ortho position, so NO2 attaches to the ortho position respective to the methyl group. Now, NO2 is strongly deactivating, making it a meta director, and as stated before, the methyl group is an ortho/ para director, so we try to see if there is a position in which they both direct to, because this would be where bromine adds to. The position bromine is shown in is para to the methyl group AND meta to the NO2 group. 27.

OH NO2 Br

28. Br 3 1 2

The more strongly activating a substituent is, the more reactive towards EAS it will be. For example, Bromine is weakly deactivating, NO2 is strongly deactivating, and OH is strongly activating, which leads to the rankings above. Remember that more activating means the substituent will donate a lone pair in to the pi system, making it more negative, and therefore a better nucleophile to electrophilic addition.

29. O

Cl Br AlCl3 2 hv H2O

O

H2NNH2 -OH,

Remember that you cannot add the alkyl group onto the benzene ring using Friedel- Crafts Alkylation because it would rearrange so a better way is to use Friedel- Crafts Acylation, and then reduce it using a Wolff-Kishner to get the alkyl group. Then you can use a reaction for ch 12, halogenation of an alkane by using Br2 and hv. 30.

** parent chain has to contain the ketone (because it has the highest priority), giving it the lowest locant (#2), next number down the chain giving the LONGEST POSSIBLE PARENT NAME.

** complex substituent on carbon #3 main chain of this substituent is cyclohexyl methyl with a hydroxy.

31.

** put ester chain name out in front of name (butyl and isopropyl in that order based on alphabetization). Malonate is the common name for this diester

32.

** give the aldehyde lowest locant because it’s the highest priory next number down the ring to give the bromine the lowest locant possible (#3).

33.

** Alcohol has highest priority thus is numbered first on the parent chain, then the alkyl chain because it gives the next lowest locant on the ring vs going around the ring to chloro first. Name the alkyl substituent as a complex substituent because there is no common name for it.

34.

** Styrene is a common name for this compound. You need to memorize the common names in the chapter 18 packet, they will most likely show up on the test.

35.

** benzoic acid is another common name to know, next number to give the bromine the lowest locant (#3) then number down the chain.

**The methyl is wedges to trick you this is not a chiral center because it only has 3 groups bonded to it. 36.

** The first step converts the alcohol to a bromide and then creates the Grignard reagent.

** Attacking carbon dioxide with Grignard and using hydronium as proton donor results in the formation of a carboxylic acid

** Then convert to a ester using ethanol (any primary alcohol would work as well) and HCl (proton donor)

** Finally use two equivalents of Grignard to with an acid work up to create the desired tertiary alcohol.

37.

This is a Wolff-Kishner Reduction. It reduces all ketone carbonyl groups, not just those adjacent to benzene rings. The carbonyl group is just reduced to an alkane. Make sure the number of carbons remains constant. Essentially, you just “remove” the carbonyl group.

H2, Raney Ni

Raney Nickel must be used in order to reduce a Carbon-Nitrogen triple bond, but remember the ONLY thing that can reduce the double bonds of a benzene ring is H2, Ni, 25 atm, and 250 degrees Celsius (extreme conditions)

no reaction

Normally, this would turn the attached alkane groups into carboxylic acids, BUT since there are no benzylic hydrogens available to oxidize it, there is no reaction.

OCH3

NO 2

Since the incoming group is a stronger base than Cl, then nucleophilic aromatic substitution occurs and the methoxy group replaces Chlorine. Also because Cl is a really good leaving group. NO2 is a strong base itself and the chloride group is a way better leaving group (weaker base) so NO2 remains in place.

38. N NH H2, Pd/C 2 N + NaNO2 H3O HCl, 0o

Hydrogen and Pd/C can reduce NO2 into NH2, which we then can use to make a diazonium ion. Once we have our diazonium ion, it is very reactive and there are many reactions we can choose from. Since we are adding an alcohol group, we are going to use H30+ and heat.

39.

NH2 Br Br NH2

N OCH3 N

This is just a nucleophilic substitution reaction so the NH2 is a strong base and Br is the best leaving group so NH2 attacks and then the electrons are put back on nitrogen. Then nitrogen’s lone pair will go back in the ring and eventually kick Br off and leave NH2.

40.

CO, HCl, AlCl3

These are the reagents used to formylate a benzene ring (add an aldehyde to a benzene ring). The electrons in the benzene ring will attack the Carbon in the Carbon-Oxygen triple bond of CO. Then, a pair of electrons from the triple bond will go back on oxygen forming the carbonyl.

41.

Solution: This reaction acts the same as Wolff Kishner and reduces a carbonyl group into carbon-carbon single bonds as shown below (essentially just take the oxygen and double bond away).