organic reaction guide beauchamp 1 chem 316 ...psbeauchamp/pdf/org_rxns_study_list.pdforganic...

Organic Reaction Guide Beauchamp 1 Chem 316 / Beauchamp Reactions Review Sheet Name

SN2 Reactionsspecial features: biomolecular kinetics Rate = kSN2[RX][Nu-], single step concerted reaction, E2 is a competing reaction relative order of reactivity: CH3X > 1oRX > 2o RX >> 3oRX (based on steric hinderance, no SN2 at 3o RX) allylic & benzylic RX are very reactive, adjacent pi bonds help stabilize transition state and lower TS energy (Ea) complete substitution at Cα (3o RX) or Cβ (neopentyl pattern) almost completely inhibits SN2 reactions vinyl & phenyl are very unreactive, bonds are stronger and poor backside approach leaving group ability: OTs = I > Br > Cl in neutral or basic conditions (just like E2, SN1 adn E1), and neutral molecule leaving groups are good from protonated, cationic intermediates in acid conditions, -OH2

+, -ORH+, -OR2+, -NR3

+, etc. we will consider all anions, ammonia, amines, thiols and sulfides to be strong nucleophiles (favors SN2 and E2 reactions) in our course some electron pair donors are mainly nucleophiles (sulfur, azide, cyanide, carboxylates) and some are mainly bases (t-BuO - K+, Na+ H2N -, Na+ H -) polar, aprotic solvents work best for SN2 reactions because nucleophiles are relatively unencombered for electron doantion (dimethyl sulofoxide = DMSO, dimethylformamide = DMF, acetonitrile = AN, acetone, etc.) in our course some electron pair donors are mainly nucleophiles (sulfur, azide, cyanide, carboxylates) and we will consider neutral solvent molecules such as water, alcohols and acids to be weak nucleophiles (favors SN1 and E1) stereoselectivity: 100% inversion of configuration from backside atack

regioselectivity: reacts at carbon with leaving group, completely unambiguous

chemoselectivity: N/A

The following list is designed to emphasize SN2 reactions. Other possibilities (E2) are not listed. a. primary RX (X = Cl, Br, I, OTs)

X CN

nitrile

CN

Possible additional steps1. make amide (HCl/H2O)2. make acid (H2SO4/∆)3. make aldehyde (DIBALH)4. make ketone (RMgBr)5. make 1o amine (LiAlH4)LimitationsSN2 at Me, 1o and 2o RX

X

Possible additional steps1. make cis alkene (Pd/H2/quinoline)2. make trans alkene (Na/NH3)3. make alkane (Pd/H2)4. make ketone (H2SO4/Η2Ο)5. make aldehyde (a.R2BH, b.H2O2)6. zipper reaction (NaNR2)LimitationsSN2 at Me and 1o RX

CCRR

alkyne (terminal or internal)

(from alkyne + NaNH2)

X

Possible additional steps1. make RX (SOCl2,PBr3,HI)2. make tosylate (TsCl/py)3. make aldehyde (PCC/no H2O)4. make acid (Jones/Η2Ο)5. make alkoxide (NaH)LimitationsSN2 at Me and 1o RX

OHOH

alcohol

X

Possible additional steps1. protonate in acid2. stable in base

LimitationsSN2 at Me and 1o RX

OR OR

ether

(from alcohol + NaH)

C:\Documents and Settings\butterfly\My Documents\classes\316\special handouts\org_rxns_study_list_for_web.doc

Organic Reaction Guide Beauchamp 2

X

Possible additional steps1. hydrolyze in acid or base2. reduce with LiAlH43. react twice with organometallics

(from acid + NaOH)

O

O

RR

O

O

ester LimitationsSN2 at Me, 1o and 2o RX

X

Possible additional steps1. make thiolate with NaOH (good nucleophile)

(from NaSH)

SH

thiol LimitationsSN2 at Me, 1o and 2o RX

SH

X

Possible additional steps1. can oxidize ro sulfoxide or sulfone

(from thiol + NaOH)

SR

LimitationsSN2 at Me, 1o and 2o RX

SR

sulfide

X

Possible additional steps1. hydrolyze to 1o amine with NaOH/H2O (or hydrazine, H2NNH2)


N

O

O

N

O

O

imide

(from phthalimide + NaOH)

NaOH/H2O

NH2

1o amine

X

Possible additional steps1. azides can by hydrogenated (reduced) to a 1o amines with Pd/H2


N3

azide NN N

NN N2

(from NaN3)sodium azide

Pd/H2

NH2

1o amine

X

Possible additional steps1. make carbanion nucleophile with n-BuLi and react with aldehydes and ketones to make very specific alkenesLimitationsSN2 at Me, 1o and 2o RX

P

= Ph3P

triphenylphosphine

P

PhPhPh X

alkyltriphenylphosphonium halide, this salt is used in Wittig reactions



Possible additional steps1. makes RX center into alkane functionality


XLiHAl

H

H

H

lithium aluminium hydride (LAH)

Possible additional steps1. makes RX center into alkane functionality


NaHB

H

H

HX

X

Possible additional steps1. Couples two "R" parts from two different RX starting structures, one is made into an alkyl lithium, then a cuprate and coupled to another RX compound, cuprates are needed because this reaction does not work with Mg or Li reagents.LimitationsSN2 at Me, 1o and 2o RX

Li

Cu

organocuprate (from organolithium from RX compound)

We will view cuprate + RX as an SN2 reaction, eventhoughfree radicals may be involved

X

Possible additional steps1. LDA is made from diisopropyl amine and n-BuLi, usually in THF at room temperature and the alkylation reaction run at -78oCLimitationsSN2 at Me, 1o and 2o RX

Li

enolates from carbonyl compounds + lithium diisopropyl amide (LDA), at very low temperatures,many variations possible

R

O

CH2

O

R

special RX (allyl, benzyl, vinyl, phenyl, neopentyl)

XX

XX X

exceptionally good electrophiles in SN2 reactions very poor electrophiles in SN2 reactions

allyl RX benzyl RX vinyl RX phenyl RX neopentyl RX

A good exercise would be to write out each reaction above with allyl and benzyl RX compounds.


Organic Reaction Guide Beauchamp 4 b. secondary RX (X = Cl, Br, I, OTs)

Possible additional steps1. see 1o RX, cyanide is not too basic for mainly SN2 at 2o RX, acid's pKa = 9


nitrile

CNXC

N

Possible additional steps1. not useful at 2o RX, see 1o RX reactions, acdtylides are too basic for mainly SN2 at 2o RX, acid's pKa = 25

X CCR

mainly E2 reaction

2-E and 2-Z and 1- alkenes LimitationsSN2 at Me and 1o RX

Possible additional steps1. not useful at 2o RX, see 1o RX reactions, messy product mixture (SN2 and E2), acid's pKa = 16

X


OH alcohol

OH

2-E and 2-Z and 1- alkenes

X


alcohol

OR

2-E and 2-Z and 1- alkenes

OR

Possible additional steps1. not useful at 2o RX, see 1o RX reactions, messy product mixture (SN2 and E2), acid's pKa = 16-18

X O

O

RR

O

O

ester LimitationsSN2 at Me, 1o and 2o RX

Possible additional steps1. see 1o RX, less basic carboxylates are better behaved nucleophiles and give good yields for SN2 at 2o RX centers, conjugate acid's pKa = 9

Possible additional steps1. see 1o RX reactions

X


SH

thiol

SH




X


SR SR

sulfide


X


N

O

O

N

O

ONaOH/H2O

NH2

1o amine


X


N3

azide NN N

NN N2

(from NaN3)sodium azide

Pd/H2

1o amine

NH2


X


triphenylphosphine alkyltriphenylphosphonium halide, used in the Wittig reaction,

P

PhPhPh XPhPPh

Ph

Possible additional steps1. see 1o RX reactions, We use LAH as nucleophilic hydride in this book. If we need basic hydride, we'll use sodium hydride, NaH.


X LiHAl

H

H

H

lithium aluminium hydride (LAH)


NaHB

H

H

HX

Possible additional steps1. see 1o RX reactions, We use NaBH4 as nucleophilic hydride in this book. If we need basic hydride, we'll use sodium hydride, NaH.

sodium borohydride





X Li

Cu

organocuprate



XLi

enolate chemistry

R

O

CH2

O

R

Intramolecular SN2 reaction.Br

O

O

O

O

special RX (allyl, benzyl, vinyl, phenyl, neopentyl)

XX

X

exceptionally good electrophiles in SN2 reactions very poor electrophiles in SN2 reactions

allyl RX benzyl RX vinyl RX phenyl RX neopentyl RX

XX

SN reactions of epoxide electrophiles are shown in a later table.


Organic Reaction Guide Beauchamp 7 E2 Reactions are emphasized in this sectionspecial features: biomolecular kinetics (Rate = kE2[RX][B-], single step concerted reaction, competing reaction is SN2 favored reactivity: 3oRX > 2o RX > 1oRX (none at CH3X, need Cβ-H), 1oRX will produce mainly SN2 product excet for mostly E2 with the sterically hindered and highly basic potassium t-butoxide and generally more E2 occurs from the less hindered side of the RX allylic & benzylic RX are very reactive if a conjugated pi bond can form complete substitution at Cα (3o RX) shuts down SN2 and makes E2 the only choice, but there maay be many possible E2 products a completely substituted Cβ makes E2 impossible from that position, but if other Cβ's are present with a hydrogen present, then E2 can occur from those atoms vinyl & phenyl are fairly unreactive, but with really stong bases (R2N ) E2 can form an alkyne, or if the alkyne pi bond becomes conjugated, the reaction can occur more easily with less basic RO- leaving group ability: OTs = I > Br > Cl in neutral or basic conditions (just like SN2/E2 reactions) anions whose conjugate acids have high pKa's (weaker acids have stronger bases) generally produce more E2 relative to SN2, the two examples we will emphasize at 2o RX centers are carboxylates (SN2 > E2) vs hydroxide and alkoxides (E2 > SN2) and cyanide (SN2 > E2) vs terminal acetylides (E2 > SN2) we will consider neutral solvent molecules such as water, alcohols and acids to be weak nucleophiles (favors SN1 and E1) stereoselectivity: mainly anti Cβ-H and Cα-X elimination since parallel orbital overlap of the favored staggered conformation allows formation of pi bonds with lower Ea, syn elimination can occur in rigid systems that lock in the required ecliplsed conformation, there can be a lot of possibilities to consider with up to three beta atoms with hydrogen atoms, also each hydrogen of a C-beta CH2 will often be different, producing E or Z stereoisomer alkenes depending on the anti conformations present, also a chiral 3o RX Cβ-H may have R (E or Z) and S (Z or E).

regioselectivity: anti C-beta atoms (or syn in rigid systems) having a hydrogen are required relative to the C-alpha with the leaving group

chemoselectivity: N/A

Cα

Cβ

HH X

H

Nu

NuCα

Cβ

Cβ

H X

H

HB

BCα

CβHH

X

H

Nu

B

Cα

CβCβH

X

HHNu

B

Two different perspectives to show either SN2 or E2 reactions. Additional features need to be drawn in. The three templates canwork for 1o, 2o and 3o RX compounds. A cyclohexane template is also provided. The anti requirement for E2 reactions requires that X be in an axial position in cyclohexanes, which also works better for SN2 reactions.

1o RX template

perspective 1 perspective 2

2o RX template


NuCα

Cβ

Cβ

CβX

H

HB

Cα

CβCβCβ

X

HHNu

B

3o RX template


H

H

E2 target

SN2 target

At 1o RX SN2 is usually favored over E2, except if the sterically large and very basic potassium t-butoxide is used.

At 2o RX SN2 and E2, are in competition, less basic electron pair donors tend to favor SN2 and more basic electron pair donors tend to favor E2 reactions, any feature that adds sterically large groups pushes the reaction towards E2.

Only E2 reactions are expected at 3o RX when reactied with strong electron pair donors.

Nu B=

Nu B=

Nu B=



interconvert infast equilibrium

C

C

12

3

45

6

possiblereactions possible

reactions an axial "X" is necessary for a succesful E2 reaction and also works better for SN2

Templates for both cyclohexane chair possibilities with a carbon substituent present. The leaving group, X, can be added at anyblank bond position and needs to be in an axial position to be anti in the ring (not true for carbon branch).

? ?

An example of some possible choices for SN2 versus E2 in reactions at a secondary RX center having a chiral Cα, chiral Cβ and a beta CH2.

Nu

B

Cα

Cβ

Cβ

H I

HOCH3

Ha

CH3

Hb

CH2CH3

Cα

Cβ

Cβ

H I

HOCH3

Ha

CH3

Hb

CH2CH3

Cα

Cβ

Cβ

H I

HOCH3

Hb

Ha

CH3

CH2CH3

Cα

Cβ

Cβ

H I

HOCH3

Ha

CH3

Hb

CH2CH3

B

C

C

CH3CH2

H CH2CH3

C

C

H

CH3

C

H

H CH2CH3

OCH3

OCH3

B C

C

H

H

C

CH3

H CH2CH3

OCH3

(2Z,4S)-3-methoxy-2-hexene

(2E,4S)-3-methoxy-2-hexene

CNu

Cβ

HOCH3

CH2CH3

HCβHa

CH3

Hb

(3S,4S) (3R,4S)

(3Z)-3-methoxy-3-hexene

SN2 E2

E2E2

Contrast with the reaction below.

OH

OR

X

a. primary RX (X = Cl, Br, I, OTs), typically see mostly SN2 and do not consider the E2 product, unless the base is potassium t-butoxide, then E2 is the major product

or

OH

mainly SN2

ORor

Contrast with the reaction above.

X

BUT

mainly E2

KO

a big, bulky, very strong base



CCRCN

more basic, mainly E2 reaction

b. secondary RX (X = Cl, Br, I, OTs), SN2 and E2 products are both competitive at 2oRX, less basic anions are often good nucleophiles and produce more SN2 product, while more basic anions are better at plucking off a Cβ−Η producing more E2 product, any feature that introduces steric hindrance will favor E2 product (hindrance at Cα, Cβ or in the electron pair donor = base/nucleophile)

similar looking base/nucleophiles (used in this book) that react differently with 2oRX structures

less basic, mainly SN2 reaction

pKa of conjugate acid = 9 pKa of conjugate acid = 16-18

OH ORRO

OpKa of conjugate acid = 5pKa of conjugate acid = 25

less basic, mainly SN2 reaction more basic, mainly E2 reaction

E2 reactions of 2o RX (and a few 3o RX) compounds

E2 requires Cβ-H and Cα-X bonds to be in anti conformation

Possible additional steps1. many additional alkene reactions, are possible although in this reaction these would not be productive because too many different products are obtainedLimitationsSN2 and E2 products obtained

OHSN2 product

X OHE2 products

or OR

(or OR)

only one hydrogen can be antior three Cβ-H's

Possible additional steps1. alkene reactions

Limitationsmainly E2 product

OH or OR

X

C

CH3

H3C

CH3

Oonly E2 reaction, t-butoxide is too big and bulky for SN2 reactions, an anti Cβ-H is possible on either side when the Br is axial

Br

K

enantiomers

High pKa , sterically bulky base, should be only E2, but Br cannot significantly rotate to an axial position since the very large t-butyl group locks the ring into confromation having t-but yl equatorial.

C

CH3

H3C

CH3

O

No reaction

K

Br

Only E2 at a 3o RX with a strong base/nucleophile. There are three Cβ's but only C6 and the methyl carbons allow the necessary anti conformation for E2 reactions. C2 cannot rotate its hydrogen anti.

R3NBr1

2

6

(DBU or DBN)



Only E2 at a 3o RX with a strong base/nucleophile. All three Cβ's can rotate a hydrogen anti and form E2 product.Br1

3

6

F

C N A Cβ carbon is fully substituted so SN2 reaction is greatly inhibited. There is an anti Cβ-H possibility on the other side so E2 can occur there.

Br

SN2 and E2 reactions possible, this is a good example for showingwhy you need o be able to draw and understand 3D drawings, this example is comprehensively viewed below.

CH3O

BrCH3

H

H

H

H

See choices below.(3S,4S)

Br

CH3

H

H2

3 14

56

CH3OOCH3

H

H

R

S S

S BrBr

H

H

H

H

SN2

Br

HH

H

CH3

H

C-C rotation

C-C rotationH

Br

HH

CH3

H

CH3

Br

HH

HH

E2a E2b E2c

C-C rotation

H

CH3

H

H

H

H

CH3

H

CH3

CH3

H

H

H"Z" configuration "E" configuration

"E" configuration

CH3OCH3O CH3O

SN2 reaction

E2 reactions

S

S Br

H

H

H

H


Organic Reaction Guide Beauchamp 11 SN1 and E1 Reactions (they share a common carbocation intermediate)special features: unimolecular kinetics ( Rate = kSN1[RX] or Rate = kE1[RX], the relative rates depend on the k's ), multi step reaction, SN1 and E1 are competing reactions (usually SN1 dominates) favored reactivity: 3oRX > 2o RX and we say none at a 1o RX or CH3X because those carbocations are too unstable to form allylic & benzylic RX are reactive because of resonance stabilization of carbocation (even if 1o RX) the number of R gourps at Cβ is not especially important in our course, because the key step in either SN1 or E1 is the first step involving ionization of the Cα-X bond, but highly substituted Cβ positions mean there will probably be rearrangements there is no anti Cβ-H requirement for E1 because the carobcation is so reactive and any Cβ-H can be rotated parallel to the empty p orb ital, generally the more substituted, more stable alkene forms to the greatest extent vinyl & phenyl are completely unreactive because the sp2 bonds are generally too strong to break and the sp carbocation is too unstable in the case of vinyl and the empty sp2 orbital is too unstable in the case of phenyl leaving group ability: OTs = I > Br > Cl in neutral or basic conditions (same for all of the reactions), and neutral molecule leaving groups are good from protonated, cationic intermediates in acid conditions, -OH2

+, -ORH+, -OR2+, -NR3

+, etc. only weak base/nucleophiles (usually the same molecule: H-B: = H-Nu:) will be used in these reactions, usually it is the solvent molecule (H2O, ROH or RCO2H in this book) the solvent is usually a polar, protic solvent that is capable of stabilizing charged intermediates the only synthetically useful E1 reaction in this book will be dehydration of alcohols, ROH, in concentrated H2SO4 with heating (∆) which distills out the alkene and shifts the equilibrium towards E1 whenever carbocations are formed, rearrangements must be considered and are likely if similar or more stable carboncations can form (we will usually only emphasize rearrangements to more stable carbocations), but the ultimate two leading to stable products reactions are add a nucleophile or lose a beta hydrogen stereoselectivity: any Cβ-H can be lost in an E1 reaction because any Cβ-H can be rotated parallel to the empty p orbital allowing formation of pi bonds, generally the most stable, more substituted (or E over Z) alkene forms to the greatest extent, SN1 reactions lead to racemization of chiral RX centers

regioselectivity: any Cβ hydrogen atom can be lost in E1 reactions, in SN1 reactions the nucleophile will add to either face of the Cα carbon unless there rearrangement occurschemoselectivity: N/A

In this book SN1 reactions attack will occur from either side of the flat Cα carbon. This will result in racimization of configuration at chiral centers or cis/trans products in rings.

CNu

2

34C X

2

34

NuH

X

"S" configuration "R" configuration

R

X

HNu H

R Nu

trans ringcis ring

X

i. (R racemization) or (S racemization)

C

2

34

Nu H

attack can occur from eitherface of the flat carbocation

C Nu

2

34

"S" configurationleads to racemization (50/50 mixture for our course)

R

Nu

R

H

Nu H

protontransfer

trans ringattack can occur from top or bottom of carbocation

ii. (cis ring cis/trans ring) or (trans ring cis/trans ring)

Br

H

CH3

H

H

HHCH3

H

HHBr

=

only one anti Cβ-H "IF" E2 reaction.

first step for SN1/E1 reactions

CH3

H

H

H

H

1. rearrange 2. add Nu 3. lose beta-H

conformation changes of two chairs allows any beta hydrogen to be lost from the carbocation

(likely in this reaction because a 3o R+ can form)

ROH

"IF" RO

SN2/E2

protontransfer

X

NuH

NuH



Br H2O

via

OH

SN1 (major)

E1 (minor)

Our general rules are SN1 > E1, except for high temperature, acid dehydration of alcohols. Rearrnagement here is not a problem because nothing changes. No stereochemistry or regiochemistry to consider.

Br ROH

via

OR

SN1 (major)

E1 (minor)


Br

O

OH

via

OO

SN1 (major)

E1 (minor)


Rearrangement of 2o carbocation is possible, but not shown in this problem.

H2O

via

OH SN1 (major)diastereomers

E1 (minor)

Brenantiomers

OH

ROHBr

Rearrangement of 2o carbocation is possible, but not shown in this problem.

via

OR SN1 (major)diastereomers

E1 (minor)

enantiomers

OR

O

OHBrRearrangement of 2o carbocation is possible, but not shown in this problem.

via

OO

SN1 (major)diastereomers

E1 (minor)

enantiomers

ester on top and bottom



Rearrangement is likely because a 3o

carbocation can form from the originally formed 2o carbocation.

H2O

via

OHSN1 (major)

E1 (minor)Br

Rearrangement is likely because a 3o carbocation can form from the originally formed 2o carbocation.

ROH

via

ORSN1 (major)

E1 (minor)Br

Rearrangement is likely because a 3o carbocation can form from the originally formed 2o carbocation.

O

OH

via

O

O

SN1 (major)

E1 (minor)Br

Br H2O

via

OH

SN1 E1

Rearangement is likely since a 3o carbocation can form, normally SN1 dominates over E1, with racimization of a chiral center (stereogenic centers are drawn with a wiggly lines), however significant E1 is expected because the alkene is tetrasubstituted.

Br ROH

via

OR

SN1 E1

Rearangement is likely since a 3o carbocation can form, SN1 will dominate over E1, racimization of chiral center is expected (stereogenic centers are drawn with a wiggly line, only the most stable E1 alkene is shown.

Br

O

OH

via

O

OSN1

E1

Rearangement is likely since a 3o carbocation can form, normally SN1 dominates over E1, with racimization of a chiral center (stereogenic centers are drawn with a wiggly lines), however significant E1 is expected because the alkene is tetrasubstituted.



H2O

SN1, very good at benzylic even though primary can still occur because of resonance stabilization of carbocation, E1 is not possible here because there is no Cβ-H to lose a hydrogen.

Br OH

O

OH Phenyl carbocation is VERY difficult to form in sp2 orbital.Br No reaction

ROH

SN1, very good at allylic R+ common intermediate via resonance, two different products formed from a commonintermediate having partial positive charge at two sites as shown by the resonance structures

Br

Bror

OROR

thermocynamic product kinetic

product

Vinyl carbocation is VERY difficult to form in p orbital of sp hybridized carbon. Carbocations that we typically see are empty p orbitals of a sp2 hybridized carbon.

No reactionBr

H2O

HI

SN1, acid protonates OH and makes -OH2

+. Water is a good leaving group and when at 2o or 3o forms a carbocation. Mainly SN1 with HX acids. Nucleophile adds from top and bottom.

OH I

cis or trans cis and trans

SO2 leaving group, HCl also formed. View reaction as SN1 at 2o and 3o ROH.

OH SOCl2 Cl


PBr3OH

Oxygen first does SN2 at phosphorous and then HOPBr2 is the leaving group in second step (repeated two more times). View reaction as SN1 at 2o and 3o ROH and SN2 at methyl or 1o ROH.

Br


Substitution reacton is at the sulfur and represents another way to make the OH into a good leaving group, as a tosylate (inorganic sulfur ester). Pyridine added to sponge up (neutralize) the HCl generated in the reaction.

= TsCl

OH

S

O

O

Cl OTs

= pyridineN cis goes to cis and trans goes to trans



Dehydration of alcohols in strong acid, with heat, represents our main E1 conditions. Rearrangement is possible, but not shown.

OH H2SO4 / ∆

enantiomerscis or trans

HBr OH would protonate, but -OH2

+ cannot leave from a phenyl carbon. the empty sp2 orbital (NOT a p orbital!) is too unstable.

OH No Reaction

HBr

Just a reminder at 1o ROH, OH would protonate, but -OH2

+ cannot leave from a 1o carbon. It needs to be pushed off via SN2 by the bromide.

OH Br

E1, initial carbocation rearranges and then eliminates at high temperature. The alkene(s) distill out. Tetrasubstitution is the most stable alkene shown.

OH H2SO4 / ∆ major E1

E1 product. The most stable alkene is shown due to more substituted, trans and conjugated.

H2SO4 / ∆

HO

PBr3

Oxygen first does SN2 at phosphorous and then HOPBr2 is the leaving group in second step (repeated two more times). View reaction as SN1 at 2o and3o ROH and SN2 at methyl or 1o ROH.

OH Br

1. makes tosylate (good leaving group) reaction occurs at sulfur, not carbon so no change in chiral center2. SN2 by bromide at carbon, chiral center inverts

1. TsCl, pyridine2. NaBr

OH Br

SO2 leaving group, HCl also formed. View reaction as SN1 at 2o and 3o ROH and SN2 at methyl and 1o ROH.

SOCl2

OH Cl


Organic Reaction Guide Beauchamp 16 Reactions of Alkenes (and Alkynes) -

Stereoselectivity (cis vs trans, E vs Z, R vs S) and Regioselectivity (reacts at C1 vs C2), it is sometimes possible to see these features in the product structures which can be achiral, enantiomers, diasteromers and/or meso. Reagents generally can attack from either face. Sometimes the approaches are equivalent and sometimes one face is preferred over the other.

In the reactions below, if "Stereo = Y" is written, then the reaction can be stereoselective even if not all products will show it. If "Regio = Y" is written, the reaction can be regioselective even if not all products will show it.

Markovnikov addition,intermediate carbocation (can rearrange, add a nucleophile or lose a beta hydrogen.

HBr

Br

racemic

1 Stereo = NRegio = Y

a. Hg(OAc)2 H2O (or ROH)b. NaBH4

Markovnikov addition,intermediate carbocation does not rearrange because of mercury bridge, NaBH4 reduces off mercury to hydrogen

OH

racemic


OH

racemic

3

H2SO4 / H2O

Markovnikov addition, intermediate carbocation can rearrange if more stable carbocation can form, if heated the E1 alkene is the product

Stereo = NRegio = Y

a. BH3b. H2O2 / HO

4 anit-Markovnikov addition, borane forms trialkylborane,second step oxidizes boron position to an OH

OHStereo = Y (syn)Regio = Y

a. BH3b. Br2 / CH3O

5Same as above, but instead of boron becoming an OH, it becomes a Br

BrStereo = Y (syn)Regio = Y

Br2

6 Bromonium bridge prevents rearrangement, bromide adds anti to first bromine at more partial postive carbon.

Br

Br

racemic

Stereo = Y (anti)Regio = N

Br2 / H2O

7OH

Brracemic

Bromonium bridge preventsrearrangement, hydroxide adds anti to first bromine at more partial positive carbon. Similar toabove reaction.

Stereo = Y (anti)Regio = Y

a. Br2 / H2Ob. NaOH

8 Product from #7 forms an epoxide. Oxgen attacks from backside in anti conformation. Can also be made directly from alkene with mCPBA.

O

racemic

Stereo = Y(anti, SN2)

Regio = N (overall)


Organic Reaction Guide Beauchamp 17 9

mCPBAReaction occurs in a single step via concerted mechanism.

O

racemic

Stereo = Y (syn)Regio = N

10 Net result of epoxidation followed

by opening the epoxide in acid or base is anti addition of two vicinal alcohol groups. This is opposite to the next reaction (syn addition).

OH

OH

racemic

a. mCPBAb. H3O / H2O orb. HO / H2O

Stereo = Y (anti)Regio = N

OsO4 orKMnO4

11OH

OH

racemic

Stereo = Y (syn)Regio = N

Dioxygenation occurs in a singleconcerted step via syn addition, followed by aqueous hydrolysis to form a vicinal diol. This is oppositeto the above reaction (anti addition).

H2 / Pd

The metal activates the hydrogen and two hydrogen atoms add to the pi bond in a cis or syn manner.achiral

12 Stereo = Y (syn)Regio = N

Ozone cuts pi bond in two and workup leaves a carbonyl bond at each carbon (aldehydes and/or ketones are the products).

131. O3, -78oC2. CH3SCH3 or Zn

Stereo = NRegio = NH

OO H

H

HOCH3

14

1. O3, -78oC2. NaBH4

Ozone cuts pi bond in two and workup reduces carbonyl bonds at each carbon to alcohol groups.

Stereo = NRegio = N

OH

OH

O

O OH

OH

151. O3, -78oC2. H2O2

Ozone cuts pi bond in two and workup oxidizes carbonyl bondsat each carbon to carboxylic acids or ketones if the alkene carbon is geminally substituted.

Stereo = NRegio = N

= CO2

1. Br22. NaNH23. WK

16 Bromine adds as in reaction 6 above.Sodium amide performs two E2 reactions and then deprotonates the alkyne. Workup with acid protonates the sp carbanion to form an alkyne.

Stereo = NRegio = N

H2 / Pd17 The metal activates the hydrogen

and two hydrogen atoms add to the first pi bond in a cis or syn manner to form an alkene, which then reduces to the alkane.

Stereo = YRegio = N

H2 / Pd / quinoline

18 The metal activates the hydrogenand two hydrogen atoms add to the first orbital in a cis or syn manner. The quinoline poisons the catalyst so that alkenes do not react further.

Stereo = YRegio = N



Na / NH3

An electron from sodium adds tothe LUMO orbital. the resulting anion protonates. Another electron adds and the resulting anion protonates to form the more stable E alkene.

19 Stereo = YRegio = N

Markovnikov addition, intermediatecarbocation (can rearrange, add a nucleophile or lose a beta hydrogen.achiral, but cis/trans diastereomers

HBr

20

Br

Stereo = NRegio = Y

a. Hg(OAc)2 H2O (or ROH)b. NaBH4

Markovnikov addition,intermediate carbocation does not rearrange because of mercury bridge, NaBH4 reduces off mercury to hydrogen


OH

22

H2SO4 / H2O

Markovnikov addition, intermediate carbocation can rearrange if more stable carbocation can form, if heated the E1 alkene is the product

Stereo = NRegio = Y

OH

a. BH3b. H2O2 / HO

23 anit-Markovnikov addition, boron and hydrogen add syn,borane forms trialkylborane,second step oxidizes boron position to an OH

OH

OH

Stereo = YRegio = Y

a. BH3b. Br2 / CH3O

24Same as above, but instead of boron becoming an OH, it becomes a Br

Br

Br

Stereo = YRegio = Y

Br2

25

Bromonium bridge prevents rearrangement, bromide adds anti to first bromine.

Br

Br

Br

Br

Stereo = YRegio = N

Br2 / H2O

26 Bromonium bridge preventsrearrangement, hydroxide adds anti to first bromine at more partial positive carbon. Similar to above reaction.

Br

OH

Br

OHStereo = Y, Regio = Y

a. Br2 / H2Ob. NaOH

27 Product from #7 forms anepoxide. Oxgen attacks from backside in anti conformation. Can also be made directly from alkene with mCPBA.

O O

Stereo = Y, Regio = N 28

mCPBA

Reaction occurs in a single step via concerted mechanism.Epoxide oxygen adds syn.

O O

Stereo = Y, Regio = N




29 Net result of epoxidation followed by opening the epoxide in acid or base is anti addition of two vicinal alcohol groups. This is opposite tothe next reaction (syn addition).

a. mCPBAb. H3O / H2O orb. HO / H2O

OH

OH

OH

OH

OsO4 orKMnO4

30 OH

OH

OH

OH

Dioxygenation occurs in a single concerted step via syn addition, followed by aqueous hydrolysis to form a vicinal diol. This is opposite to the above reaction (anti addition).Stereo = Y, Regio = N

H2 / Pd

The metal activates the hydrogen and two hydrogen atoms add to the pi bond in a cis or syn manner.

31


Ozone cuts pi bond in two and workup leaves a carbonyl bond at each carbon (aldehydes and/or ketones are the products).

321. O3, -78oC2. CH3SCH3 or Zn

O

HO

33

1. O3, -78oC2. NaBH4

Ozone cuts pi bond in two and workup reduces carbonyl bonds at each carbon to alcohol groups.

OH

OH

341. O3, -78oC2. H2O2

Ozone cuts pi bond in two andworkup oxidizes carbonyl bondsat each carbon to carboxylic acids or ketone if the alkene carbon is geminally substituted.

O

OHO

1. Br22. NaNH23. WK

35 Bromine adds as in reaction 6 above.Sodium amide performs two E2 reactions and then deprotonates the alkyne. Workup with acid protonatesthe sp carbanion to form an alkyne.

H2 / Pd36 The metal activates the hydrogen

and two hydrogen atoms add to the first pi bond in a cis or syn manner to form an alkene, which then reduces to the alkane.

H2 / Pd / quinoline

37 The metal activates the hydrogenand two hydrogen atoms add to the first pi bond in a cis or syn manner.The quinoline poisons the catalyst so that alkenes do not react further.



Na / NH3

An electron from sodium adds tothe LUMO pi bond. the resulting anionprotonates. Another electron adds and the resulting anion protonatesto form the more stable E alkene.

38

Zipper reaction, tautomer-likeproton exchanges that move pi bonds to the terminal position where anion is most stable (in sp orbital)as:

39

R

Na NR21. 2. workup

Zipper reaction, opens epoxide, workup protonates the alkoxide.

OH

40 RNa NR21. 2.

3. workup

O

Zipper reaction, does SN2 on R-Br.

41R

Na NR21. 2.

3. workupBr

Step one forms terminal sp carbanion nucleophile. Step 2 adds methanal (formaldehyde) as a one carbon electrophile. Workup protonates the alkoxide anion.

42 Na NR21. 2. H2C=O3. workupH

OH

Similar to above except a generic aldehyde is used as the carbon electrophile. Workup protonates the alkoxide anion.

43 Na NR21.

2.

3. workup

HOHO

H

44

H2SO4 / H2O HgSO4

HO Markovnicov addition of water (H

and OH) via most stable carbocation. Forms enol which tautomerizes to methyl ketone when a terminal alkynereacts.

If a nonterminal alkyne is used, reaction could occur from either side, possibly producing different ketones probably in similar amounts.

45H2SO4 / H2O HgSO4

O

O2-hexanone

3-hexanone

O46

H2SO4 / H2O HgSO4

If some special feature makes one side preferred (e.g. resonance) then a single ketone might be preferred.

Anti-Markovnikov addition of dialkylborane toa terminal alkyne. Two large R groups are used (9-BBN is common) so the addition only occurs once to the skinny alkyne. Workup with H2O2 oxidizes carbon with boron to an enol-like structure which when released protonates to form an aldehyde

47a. R2BHb. H2O2 / HO

H

H

O



a. R2BHb. H2O2 / HO

If a nonterminal alkyne is used, reaction could occur from either side, possibly producing different ketones probably in similar amounts.

O

O2-hexanone

3-hexanone

Miscellaneous alkenes and alkynes to consider.


Organic Reaction Guide Beauchamp 22 Reactions of Alcohols (SN, E, oxidation, esterification, acetal/ketal formation, acid/base...)

There is methanol and there are primary, secondary, tertiary, allylic, benzylic alcohols. Phenols (aromatic alcohols) are considered separately.

OHH3C CR

H

H

OH CR

R

H

OH CR

R

R

OH OH OHOH

methanol 1o ROH 2o ROH 3o ROH allylic alcohols benzylic alcohols phenols

HI1

OHNo carbocations at primary carbon. Mechanism is SN2.I

2

OHSOCl2

Carbocations form at secondary carbon. Mechanism is SN1.Cl

3

OH PBr3Carbocations form at secondary carbon. Mechanism is SN1.

Br

4 E1 conditions, alkene would

distill out and shift equilibrium towards products. Rearrangements are expected.

H2SO4 / ∆OH major

5

H2SO4 / ∆OHE1 conditions, alkene would distill out and shift equilibrium towards products. Rearrangements are expected.

6

H2SO4 / ∆OH

E1 conditions, alkene would distill out and shift equilibrium towards products. Rearrangements are expected.major minor

7 Sodium hydride is only basic in

our course. It is very useful for pulling off very weakly acidic protons. LAH and NaBH4 can be nucleophilic in our course.

Na HOH O

a strong base/nucleophile

Na

Na(s) metal can also be used 8

Na HOH

Sodium hydride is only basic inour course. It is very useful for pulling off very weakly acidic protons. LAH and NaBH4 can be nucleophilic in our course.

O


Na

Na(s) metal can also be used



Na H OSodium hydride is only basic inour course. It is very useful for pulling off very weakly acidic protons. LAH and NaBH4 can be nucleophilic in our course.a strong base/nucleophile

OH Na

Na(s) metal can also be used

The first reaction is acid/baseand the second reaction is SN2.This reaction would work in either direction of alcohol and RX compound.

10OH

1. NaH

2. Br O

11

OH1. NaH

2. BrO

The first reaction is acid/base and the second reaction is SN2.If the reaction were tried the other way around there would be consideralbe E2 product

12

OH 1. NaH

2. Br

OThe first reaction is acid/baseand the second reaction is SN2.If the reaction were tried the other way around there would only be E2 product

The OH of an alcohol can be made into a toslyate (an inorganic ester). Tosyl group is a very good leaving group in SN and E chemistry (similar to iodides).

13S

O

O

Cl tosyl chloride

pyridine = proton sponge

OH OTs

1o tosylate, good leaving group 14

OHS

O

O

Cl tosyl chloride


OTs

2o tosylate, good leaving group

The OH of an alcohol can be made into a toslyate (an inorganic ester). Tosyl group is a very good leaving group in SN and E chemistry (similar to iodides).

15

NOH

Ts-Cl = tosyl chloride


OTs

3o tosylate, good leaving group

The OH of an alcohol can be made into a toslyate (an inorganic ester). Tosyl group is a very good leaving group in SN and E chemistry (similar to iodides). Susceptible to E1.

16 Removing water shifts equilibrium

to the right and adding water shifts equilibrium to the left. Toluene sulfonic acid is a common catalyst.

OH

Fischer esterification

O

OHTsOH (cat.)

(remove H2O) O

O

17OH


O

OHTsOH (cat.)

(remove H2O)

Removing water shifts equilibrium to the right and adding water shifts equilibrium to the left. Toluene sulfonic acid is a common catalyst.O

O



18OH


O

OHTsOH (cat.)

(remove H2O)

Removing water shifts equilibrium to the right and adding water shifts equilibrium to the left. Toluene sulfonic acid is a common catalyst.E1 possible at 3oRX, reducing yields.

O

O

Need OH and H on the same carbonatom. Highly oxidized chromium strips electrons from oxygen and base removes proton in E2 reaction to form pi bond between the carbon and the oxygen.

19OH

CrO3 / pyridine / no H2O PCC

O

Haldehyde at 1o ROH

20OH

CrO3 / pyridine / no H2O PCC


O

ketone at 2o ROH 21

OH CrO3 / pyridine / no H2O PCC


No reaction at 3o ROH

CrO3 / H2O / acid Jones reagent

22OH

In acid the aldehyde hydrates and OHand H are on the same carbon atom again. The second oxidation forms a carboxylic acid.

O

OHacid at 1o ROH


23OH O

ketone at 2o ROH

No further reaction is possible on a ketone because there is no hydrogen atom to allow the E2 reaction to occur.


24OH No reaction at 3o ROH

Tertiary alchols do not react because there is not hydrogen atom to allow the E2 reaction to occur.

25

toluene sulfonic acid = TsOH

OHHO

(remove H2O)

O

H H

OO

aldehyde

diol = ethylene glycol

acetal

Acetals are used to protect aldehydes. They are stable in neutral and basic solution, but reactive in acid solution. Removing water shifts equilibrium to the right, adding it shifts to the left.

Ketals are used to protect ketones.They are stable in neutral and basic solution, but reactive in acid solution. Removing water shifts equilibrium to the right, adding it shifts to the left.

26

toluene sulfonic acid = TsOH

OHHO

(remove H2O)

diol = ethylene glycolO

ketone

OO

ketal

27Alcohols can be protected with DHPforming a THP acetal. There is a disquised carbonyl and second OH hidden in the DHP and THP groups. THP acetals are stable in neutral and basic solution, but reactive in acid solution. Removing water shifts equilibrium to the right, adding it shifts to the left.

OH O

TsOH

(remove H2O)

O O

DHP = dihydropyran

OTHP

THP = tetrahydropyran

enol ether

acetal


Organic Reaction Guide Beauchamp 25 Reactions of Epoxides (mainly SN2-like reactions, E2-like reactions are possible but not emphasized in our course)

Epoxides are unusual ethers. Because of their large ring strain (26 kcal/mole) they can open in acid or base conditions. In base the attack of the strong nucleophile is at the less hindered position as one would expect in an SN2-like reaction. But in acid the attack of the weak nucleophile is at the more hindered position because it carries more of the partial positive charge which more strongly attracks the weak nucleophile. In both reaction attack is forced to occur from the opposite side of the epoxide bridge. In the second compound below, no inversion is observed at the more hindered position in base and inversionis observed there in acid.

O OH

2RO O

ethylene oxideethenoxirane

(2R)-propylene oxide(2R)-propenoxirane

cyclohexene oxidecyclohexenoxirane

(1R,2S,4S)-methylcyclohexene oxide(1R,2S,4S)-methylcyclohexenoxirane

1R

2S4S O

(1R,2S,4S)-2,4-dimethyl- cyclohexenoxirane

1R

2S4S

1

H2O / H3O+Opens trans or anti from backside attack. Cannot tell in this simple epoxide.

OHO

OH

2

H2O / H3O+

Opens trans or anti frombackside attack at the more substituted, more partial positive carbon. Chiral center does invert, becomes S.

OH

2R 2SHOOH

H

3

O H2O / H3O+OH

OH50/50 mixture of enantiomers

Similar to #1. Opens trans or anti from backside attack. Enantiomers are formed in equal amounts (a racemicmixture).

OH

OH

4

O H2O / H3O+OH

OHunequal mixture of diasteromers

Similar to #1. Opens trans or anti from backside attack. Diasteromers are formed in unequal amounts.

OH

OH

5


O HO / H2O

In strong base/nucleophile conditionsattack is at the less hindered positon as expected in SN2-type reactions.A chiral center at the more substituted position is not inverted.

OHHO

6

HO / H2OOH

2R

In strong base/nucleophile conditionsattack is at the less hindered positon as expected in SN2-type reactions.A chiral center at the more substituted position is not inverted.

2R

H OH

HO 7

HO / H2OO

OH

OHracemic mixture of enantiomers

Similar to #5. Opens trans or anti from backside attack. Enantiomers are formed in a50/50 racemic mixture.



HO / H2OO

Similar to #5. Opens trans oranti from backside attack. Diasteromers are formed in unequal amounts.

OH

OHunequal mixture of diasteromers

OH

OH

9 Similar to #5. Mg and Li reagents are

formed from the metals and an RX compound. Cuprates are made from lithium reagents and cuprous bromide (CuBr). Workup is necessary

O2. workup

(MgBr)CH2

Grignard, lithium or cuprate organometallic

HO

10

OH

2R

Similar to #6.2. workup

(MgBr)CH2


2R

H OH

11

OSimilar to #7.

2. workup

(MgBr)CH2


OH

racemic mixture of enantiomers 12

OSimilar to #8.

2. workup

(MgBr)CH2


OH

unequal mixture of diasteromersOH

13

O2. workup

terminal acetylides

H NaSN2 reaction at less hindered center. The acetylide is made from a terminal alkyne + NaNH2.HO

14

OH

2R2. workup

terminal acetylides

H Na Similar to #6.HHO

2R

15O

2. workupterminal acetylides

H Na Similar to #7.OH

racemic mixture of enantiomers 16

O2. workup

terminal acetylides

H Na Similar to #8.OH

unequal mixture of diasteromersOH


Organic Reaction Guide Beauchamp 27 Reactions of Aldehydes and Ketones (mainly carbonyl addition reactions in strongly acidic or strongly basic conditions)

1. Carbonyl groups in strong acid (weak base/nucleophile conditions) a. Always begin by protonating the carbonyl oxygen lone pair. b. Protonation on oxygen generates a resonance stabilized carbocation with reactions of carbocations 1. add nucleophile = first step of carbonyl addition and substitution reactions 2. lose beta hydrogen = tautomer reactions 3. no rearrangements because of resonance.

C OH

CαH

C OH

CαH

X C

X

OH

CαH

The addition is totally regioselective. Use a lone pair for electron donation. In our course, begin every carbonyl reaction in acid this way.

top/bottom or syn/anti addition is not relevant in our course, but could lead to R/S stereochemistry (loss of "β H" is also possible)resonance shows that the positive charge is spread over the carbon and oxygen atoms, the first resonace structure is better with full octets and an extra bond, but the second resonance structure is very informative about the ultimate fate of the intermediate

addition could produce enantiomers, if the carbon with the OH group is the only chiral center...or could lead to diasteromers, if there is one or more other chiral centers

additionproduct

H X

C OH

CαH

competing pathway, redrawn from above (keto/enol tautomerization)

X

C OH

Cαenol structure

C O

CαH

2. Carbonyl groups in strong nucleophile/base conditions (weak acid or nonacidic conditions). Two sites of attack are possible by strong electron pair donation (recall SN2 at carbon and E2 at hydrogen).

a. Nucleophilic attack at carbon (C=O) or b. Basic attack at an adjacent hydrogen (Cα-H)

Remember a similar competition about where to donate the electrons in SN (carbon) versus E (hydrogen) reactions.

a. Nucleophilic attack is possible at electrophilic carbon using strong electron pair donation. Often all acidic protons are excluded to avoid quenching the strong electron pair donor.

CCα

O

H

Nu CNu

Cα

O

H

H X

neutralize, often as a second workup step

CNu

Cα

O H

Hadditionproduct

often all acidic protons are excluded to avoid protonating the nucleophile


Organic Reaction Guide Beauchamp 28 b. Basic attack is possible at the “relatively acidic” Cα-H. Often all acidic protons are avoided to avoid quenching the strong electron pair donor.

CCα

O

H

"keto"tautomer

further reactions at C or O are possible

reaction with an electrophile,usually at carbon

reaction with a proton, can occur at carbon or oxygen (tautomers)C

Cα

O

Cα-H's to carbonyl groups are moreacidic than typical C-H bonds due toresonance stabilization by oxygen inthe enolate conjugate base.

enolateintermediate

XH

CCα

O H

CCα

O

EEδ

carbonyl with electrophile

"enol"tautomer

pKa (CH in alkane) = 50pKa (CH α to C=O) = 20

B

CCα

O

tautomerization

carbon electrophilesoften add at Cα

hydrogen usually equilibrates amongall basic positions,preferred locationdepends on thermo-dynamics thoughless stable contributorsmay be more reactive

Reactions of Aldehydes and Ketones (mainly carbonyl addition reactions)

O

H

OO

propene 2-butanone 4-methylcyclohexan-1-one

O

H

2S-methylbutanal

O

2R-methylpentan-2-one

O

3R-methylcyclohexan-1-one 1

H2O / H3O+

Hydration of a carbonyl.Equilibrium favors the keto form. Very fast in acid or base and very slow in neutral water. Keto/enol tautomeriztion is also possible.

O

H H

OHHO

2

H2O / H3O+O HO OH

Same as #1.

3

H2O / H3O+OOHOH

Same as #1.

4HO / H2O

O

H H

OHHO

Hydration of a carbonyl.Equilibrium favors the keto form. Very fast in acid or base and very slow in neutral water. Keto/enol tautomeriztionis also possible.

5

HO / H2OO HO OH Same as #2.



O HO / H2OOHOH

Same as #2.

7

O

H

Li1.

2. workup

OH Organometallic (from RX compound) adds to carbonyl electrophile. Racemic (R/S) 2o benzylic ROH forms in this reaction.

8 O (MgBr)CH3CH21.

2. workup

OH Organometallic (from RX compound) adds to carbonyl electrophile. Achiral 3o ROH forms in this reaction.

9O

Na1.

2. workupOH

"OH" on top and bottom

Terminal acetylide adds to top and bottom of carbonyl face.Cis/trans diastereomers form.

10

O

H

CNNa1.

2. workup

OH

CN

Cyanide nucleophile formscyanohydrin with carbonyls and slow addition of acid. Aldehydes and less substituted ketones work best. R/S is possible here.

11O H2N

TsOHpH = 5 (-H2O)

1o RNH2 + aldehyde or ketone forms imines. Removing water shifts equilibrium to right and adding water shifts it to left. Many 1o RNH2 derivatives react similarly. E/Z stereochemistry is possible but not shown.

N

12

O

TsOHpH = 5 (-H2O)

NHN

Pyrolidine (2o amine) forms enamines with carbonyl compounds. Removing water shifts equilibrium to right and adding water shifts it to left. Makes Cαinto a neutral nucleophilic carbon.

13 O

H

CrO3, H2O, H3O+

(Jones)O

OH

Oxidizes via carbonyl hydrate.OH

OH

H

14 O Zn, HCl

Clemmenson ReductionReduces C=O to CH2 in acid.Zn supplies the electrons and HCl supplies the protons



O

Reduces C=O to CH2 in base.A hydrazone forms, there are a number of proton transfers, some resonance structures and loss of nitrogen.

H2NNH2, KOH, ROH(Wolff-Kishner Reduction)

16 O

H

HOOH

TsOH (-H2O) H

OO

Acetals form (via hemiacetals). Removingwater shifts equilibrium to right and adding water shifts it to left. Makes carbonyls unreactive under neutral and basic conditions (protects them), but are reactive in acid to break down back to the carbonyl.

17 O

HOOH

TsOH (-H2O)

OO

Similar reaction with ketones.Ketals form (via hemiketals) and are used to protect ketones. Removing water shifts equilibrium to right and adding water shifts it to left.

18 O

H

1. NaBH42. workup

Sodium borohydride only reduces aldehydes and ketones and opens epoxides. Workup protonates intermediate anion. An aldehyde forms a 1o ROH.

OH

19O 1. LiAlH4

2. workup

Lithium aluminium hydride reduces all carbonyl compounds, nitriles and opens epoxides. Workup protonates intermediate anion. A ketone forms a 2o ROH. Cis/trans diastereomers form in this example.

H

OH

20O

HP

PhPh

Ph

1.

2. workup

Wittig reactions form really specific alkenes. The Wittig salt is formed via RX + Ph3P. Nucleophilic carban is generated with nBuLi.

21O P

PhPh

Ph

1.

2. workup

Wittig reaction. Wittig salt formed via RX + Ph3P. Carbanion generated with nBuLi.

22 O HO

HOTsOH, (-H2O)

(diol)OO

Ketal formation (remove water).Hydrolyze ketal to ketone under same conditions, but add water. Protecting group for carbonyls.

23

TsOHpH = 5 (-H2O)

NHPyrolidine (2o amine) forms enamines with carbonyl compounds. Removing water shifts equilibrium to right and adding water shifts it to left. Makes Cαinto a neutral nucleophilic carbon.

O

H N



H2NpH = 5 (-H2O)

1o amines form imines with carbonyl compounds, which can be reduced to 2o amines by NaBH3CN. 2o amines react in an analogous way to form 3o amines (see next example).

ON

H

H

1.

2. Na H3BCN 25 pH = 5 (-H2O) 2o amines form iminium ions with

carbonyl compounds, which can be reduced to 3o amines by NaBH3CN. (See examples above.)

O N

1.

2. Na H3BCN

NH

26 O

Hα,β-unsaturated carbonyl

Li and Mg organometallics prefer1,2 attack of α,β-unsatruated carbonyls. Cuprates prefer attack at Cβ, called conjugate addition or 1,4-addition. Allylic/benzylic OH probably unstablein this example.

OHLi

2. workup

1.

αβ

27 O

Li2. workup

1.Cu

O Li and Mg organometallics prefer 1,2 attack of α,β-unsatruated carbonyls. Cuprates prefer attack at Cβ, called conjugate addition or 1,4-addition. Carbonyl group is retained in this example.α,β-unsaturated carbonyl

βα

28O

α,β-unsaturated carbonyl

Na CN

H2O/ROH

O

CN

More stable nucleophiles, likecyanide, prefer attack at Cβ, forming the more stable, thermodynamic product, also called conjugate addition or 1,4-addition. Stabilized enolates, discussed later, react in a similar manner in the Michael reaction.

αβ


Organic Reaction Guide Beauchamp 32 Reactions of Carboxylic Acids and their derivatives (typically acyl substitution reactions, because there is aleaving group at the acyl carbon, as opposed to typically addition reactions at aldehydes and ketones)

O

OH

O

Clpropanoic acid propanoyl chloride

O

O

O

propanoic anhydride

O

Oethyl propanoate

O

NH2propanamide

CN

propanenitrile

Which C=O groups are most reactive? Why? (The answers are found in steric, inductive and resonance effects.) Thiol esters (structure C, below) are not usually emphasized in organic chemistry, but are very important for living organisms (biochemistry). Acetyl CoA is a well known example. One might say that thioesters are nature's acid chlorides. In carboxylic acids and their derivatives, the third resonance structure is a strong contributor when the contributing lone pair comes from a 2p orbital (oxygen and nitrogen) that overlaps well with the 2p orbitals of the C=O pi bond (the third resonance structure is more important than the second resonance structure). However, resonance donation from chlorine and sulfur is not as good becasue the 3p orbital is larger and electron delocalization is not as efficient. Because chlorine and sulfur are somewhat electronegative there is increased partial positive character on the carbonyl carbons from an inductive effect, with little return of electron density via resonance. Additionally, chloride and sulfides are stable anions and good leaving groups which leads to high reactivity in acyl substitution reactions. The middle oxygen of anhydrides has good overlap with the carbonyl carbons, but is split between two carbonyls, which reduces its resonance effect, while the electron withdrawing inductive effect is even larger than that found in an ester. The carboxylate group of an anhydride is also a good leaving group. Aldehydes and ketones do not have a stabilizing third resonace contributor which makes them more reactive than those functional groups that do, where resonance donation is important (esters and amides). Aldehydes are more reactive than ketones because they have a sterically small hydrogen substituent and aldehydes do not have the extra "R" group of a ketone which is inductively donating and reduces the parital positive of the carbonyl carbon and thereby reducing their reactivity with nucleophiles. Esters (we will consider carboxylic acids and esters equivalently) and amides are less reactive than any of the previously discussed groups, because the third resonance structure below significantly reduces the partial positive at the carbonyl carbon. Amides are less reactive than esters because nitrogen is many orders of magnitude better at donating electrons than oxygen on the basis of electronegativity. Normally, we would never even think of a negatively charged carboxylate as being an electrophile, however, even the carboxylate will react that way with an excess of organolithium compounds, the most pushy electron pair donors that we enounter. Thus, the typical order of reactivity is that shown below (A>B>C>D>E>E>F>G>H)

CR

O

Cl

A B C D E F G H

CR

O

OC

O

RC

R

O

SRC

R

O

HC

R

O

RC

R

O

ORC

R

O

NR2C

R

O

O

The first number represents the pKa of LG part of acyl group when protonated (the corresponding ∆G in parentheses in kcal/mole). This provides a measure of how stable the leaving group is on its own.

-7 (-10) +5 (+7) +7 (+10) +40 (+56) +50 (+70) +18 (+25) +37 (+52) +25 (+35)

1

2

3

pKa (∆G) of HCl

CR Cl

CR

O

OC

RC

R SRC

R HC

R RC

R ORC

R NR2C

R O

O O O O O O O O

CR Cl

CR

O

OC

RC

R SRC

R ORC

R NR2C

R O

O O O O O O

no additional resonance



O

Cl

O

OH

There are a variety of ways to transform a carboxylic acid into an acid chloride. Two of these are shown and use thionyl chloride or oxalyl chloride.

S

O

Cl Cl

thionyl chloride

2O

Cl

O

Cl oxalyl chloride

O

OH

O

Cl

See #1. Acid chlorides are at the top of the reactivity hill and all of the other acid derivatives can be made from them by using the appropriate nucleophile.

3

O

Cl

O

O

O

anhydrideHO

OCarboxylates are even better nucelophiles and are easily formed by neutralizing the acid.

4 O

ClO

OH

O

O

O

anhydride

5

O

Cl

This would probably be a pretty stinky reaction.

SH

ethanethiol

O

S

6O

Cl

Ester synthesis. Adding a 3o amine will neutrialize the HCl that also forms and protect an organic molecule that has other sensitive functionality.

HOO

O

pyridine (proton sponge) 7

O

Cl

O

OHO

pyridine (proton sponge)

Ester synthesis.

8 O

Cl H2O

O

OH

Generally, we are doing everything in our power to avoid this reaction. It undoes the first two reactions above that made the acid chlorides.

9 O

Cl NH3

O

NH2

Reaction with ammonia will form 1o amides. An extra equivalent is needed to neutralize the HCl formed.



O

Cl

O

N

H

H2NReaction with 1o amines will form 2o amides. An extra equivalent is needed to neutralize the HCl formed.

11 O

Cl

O

N

Reaction with 2o amines will form 3o amides. An extra equivalent is needed to neutralizethe HCl formed.

HN

13 O

Cl

O

HDIBALH

-78oCAlH

2. workup

1.Reaction with diisobutyl aluminiumhydride (DIBALH) at very low temperature will form aldehydes, after acidic workup. Nitriles and esters also form aldehydes with DIBALH.

14

O

Cl

Cu

Licuprates

OReaction with cuprates at very lowtemperature will form ketones.

-78oC 15

HO

O

O

OO A more complex anhydride is preparedfrom a simpler anhydride. Ethanoic anhydride (acetic anhydride) is readily available and commonly used in this manner.

O

O

O

16 Ester synthesis. A carboxylic acid

also forms. If the molecule has other sensitive functionality then an amine base may be needed to neutrialize the acid.

O

O

O HO O

O

17

Ester synthesis. DMAP is acommon catalyst in these reactions.O

O

O HO O

O N(CH3)2N

N,N-dimethylaminopyridine (DMAP) 18

H2O

O

OHGenerally, we are doing everything in our power to avoid this reaction.

O

O

O

2

O

N

H

H2N

pyridine (proton sponge)

1o, 2o or 3o amide synthesis is possible in a manner similar to the acid chlorides above from ammonia, 1o amines or 2o amines. Excess amine is needed to neutralize the carboxylic acid.

19 O

O

O


Organic Reaction Guide Beauchamp 35 20 OH

Lithium aluminium hyhdride, (LAH)reduces all carbonyl functional groups, as well as nitriles. LAH supplies nucleophilic hydride and workup supplies acidic (electrophilic) hydrogen.

1. LiAlH42. workup

O

O

O

OH

isolated

discarded in workup

Fischer ester synthesis. Uses a catalyticamount of acid and water is removed, which shifts the equilibrium to the ester side. Adding water under the same conditions would hydrolyze the ester back to an alcohol and a carboxylic acid.

O

OH

HO

TsOH (cat.), (-H2O)

O

O

21

22

OH Lithium aluminium hyhdride reduces all carbonyl functional groups, as well as nitriles.

1. LiAlH42. workup

O

OH

Hydroxide neutralizes the carboxylicacid. The carboxylate acts as a well behaved nucleophile to do SN2 reactions at methyl, 1o and 2o RX compounds.

O

OH

Br O

O

232.1. NaOH

25

24 O

N

H

Amides are pretty hardy and requirepretty harsh acid or base conditions to hydrolyze them to carboxylic acids.In base the formation of a carboxylate drives the reaction and in acid complete protonation of the amine drives the reaction (no longer nucleophilic). The target molecule could be either part (the acid or the amine) or both of them. To extract the neutral acid into an ether layer a low pH extraction would be necessary, while to get the amine into an ether layer a high pH extraction would be needed.

O

N

H2SO4 / H2O / ∆

O

OH

H3N

O

OHN

H

HNaOH / H2O / ∆1.

2. neutralize with acid

after acidic workup

26 O

N1. LiAlH42. workup N

A 3o amide generally forms a 3o amine when reduced with LAH.

26 O

NH

H

1. LiAlH42. workup N

H

H

A 1o, 2o or 3o amide generally forms a 1o, 2o or 3o amine when reduced with LAH, followed by acidic workup.

27 O

N

H

1. LiAlH42. workup N

H



28O

N1. LiAlH42. workup

N

2. workup

29 O3o amindes + Grignard reagents lead to ketones after hydrolytic workup.

O

N (MgBr)1.

30

CN 1o amindes can be dehydrated

with thionyl chloride to form nitriles.

O

NH

H

S

O

Cl Cl

thionyl chloride

33

32

31 As with amides, nitriles require harsh acid or base conditions to hydrolyze them to first amides, then carboxylic acids. In base the formation of a carboxylate drives the reaction and in acid complete protonation of the ammonia drives the reaction (no longer nucleophilic).

H2SO4 / H2O / ∆

1. NaOH / H2O2. workup

CN

CN

CN

HCl / H2O / ∆

C

O

OH

C

O

NH2

C

O

OH

34

O

HC

N

Reaction with diisobutyl aluminiumhydride (DIBALH) at very low temperature will form aldehydes, after acidic workup. Acid chlorides and esters also form aldehydes with DIBALH.

DIBALH

-78oCAlH

Esters should have been placed up above carboxylic acid reactions.

35Esters are hydrolyzed back to acarboxylic acid and an alcohol in aqueous acid. Either compound could be the desired target. Overall this is the reverse of Fischer ester synthesis.

H2SO4 / H2O / ∆O

O

OH

O

HO

36Esters can also be hydrolyzed backto a carboxylic acid and an alcohol in aqueous base. Either compound could be the desired target. This is sometimes called saponification (soap making).

O

O

OH

O

HO

NaOH / H2O / ∆



371. LiAlH42. workup

LAH reduces esters to 1o alcohols after acidic workup. Either product alcohol could be the desired target. While NaBH4 will reduce aldehydes and ketones, it will not reduce esters (no reaction).

O

O OH

HO

from reduced carbonyl,desired target

discardedin workup

38

O

O

from reduced carbonyl,discarded in workup

OH1. LiAlH42. workup

HO

desired target

LAH reduces esters to 1o alcohols after acidic workup. Either product alcohol could be the desired target.

39Reaction with diisobutyl aluminiumhydride (DIBALH) at very low temperature will form aldehydes, after acidic workup. Acid chlorides and nitriles also form aldehydes with DIBALH.

DIBALH

-78oCAlH

2. workup

1.

O

O O

H

HO

discardedin workup

desired target

40

O

O

2. workupCH3CH2 Li1.

OH

Esters react twice with Mg and Li organometallicsforming 3o ROH after acidic workup. Two of the R groups of the 3o ROH are the same, being added from the organometallic. Benzylic, 3o ROH would be sensitive to substitution (SN1) or elimination (E1) in the workup in this example.


Organic Reaction Guide Beauchamp 38 Reactions of Aromatic Compounds 1. Electrophilic aromatic substitution reactions

a. (activating substituents)

Common ortho / para directing substituents: any alkyl substituent, any group with a lone pair next to the aromatic ring that can beused in resonance with the intermediate carbocation. This could include the following commonly encountered groups in our course.

R OH OR OO

R

NR2 NR

OR

X

R = alkyl phenols ethers esters amines (except in strong acid where they are protonated)

amides halogen comounds (X = F, Cl, Br, I), while ortho/para directing, these substituents are deactivating

Nitration conditions (HNO3/H2SO4) make nitroaromatic compounds. Here with an ortho/para activating substituent. Ortho product expected, but not shown. A small amount of meta product is also likely.

Sulfonation conditions (H2SO4/SO3, oleum), makes aromatic sulfonic acids. Mainly ortho/para product with methyl substituent.

Halogenation (FeX3/X2), (Cl2 and Br2), halogenates (chlorine or bromine) aromatic compounds. Mainly ortho/para product with methyl substituent.

Friedel Crafts alkylation (RX/AlX3), forms carbocations and rearrangements are likely, adds alkyl groups to aromatic ring, which makes the aromatic ring more activated and likely to react again.

Friedel Crafts acylation (RCOCl/AlCl3), forms a resonance stabilize carbocation (an acylium ion) so no rearrangement is expected, makes aromatic ketones, which deactivate the ring and make it less likely that another reaction will occur

HNO3 / H2SO4 NO2faster than benzene

SO3 / H2SO4SO3Hfaster than

benzene

Clfaster than benzene

AlCl3 / faster than benzene

FeCl3 / Cl2

Cl

AlCl3 / faster than benzene

Cl

O

O

1

2

3

4

5


Organic Reaction Guide Beauchamp 39 b. Electrophilic aromatic substitution reactions (deactivating substituents) Common meta directing substituents: any strongly electron withdrawing substituent by resonance or inductive effect. They often have anatom with a double bond to oxygen next to the aromatic ring (can be carbon, nitrogen, sulfur and others). The attached group is destabilizing to the positively charged aromatic substitution intermediate when they are directly facing one another which occurs with ortho/para attack, so electrophiles prefer the meta position for attack in substitution reactions and the reaction is always slower than benzene. Some commonly encountered groups in our course are shown below.

CO

H

NR3 C

X

X

X

aldehydes ketones acids esters amides

nitro groupssulfonic acids, amides, esters, X = OH, NR2, OR

sp3 carbon with strongly withdrawing groups attached

positively charged groups, like ammonium ions

CO

RC

O

OHC

O

ORC

O

NR2

S

O

O

X NO

O

Nitration conditions (HNO3/H2SO4) make nitroaromatic compounds. Here with an meta deactivating substituent. Meta is the main product expected and a slow reaction is likely.

Sulfonation conditions (H2SO4/SO3, oleum), makes aromatic sulfonic acids. Mainly meta product with a nitro substituent.

Halogenation (FeX3/X2), (Cl2 and Br2), halogenates (chlorine or bromine) aromatic compounds. Mainly metal product with a nitro substituent.

Friedel Crafts alkylation (RX/AlX3), forms carbocations and rearrangements are likely, but the reaction is rarely successful with only deactivating substituents present.

Friedel Crafts acylation (RCOCl/AlCl3), forms a resonance stabilize carbocation so no rearrangement is expected, but the reaction is rarely successful with only deactivating substituents present.

O2NHNO3 / H2SO4

O2N

NO2

slower than benzene

SO3 / H2SO4

O2N

SO3H

O2N

Br

AlCl3 / does not typically work with deactivated aromatic rings

FeCl3 / Cl2

Cl No reaction

No reactionAlCl3 / Cl

O

6

7

8

9

10

O2N

O2N

O2N

O2N

slower than benzene

slower than benzene

does not typically work with deactivated aromatic rings


Organic Reaction Guide Beauchamp 40 2. Nucleophilic reactions (addition/elimination),

diazonium chemistry (Ar-N2+) from following sequence: nitro amino diazonium salts

O2N

HNO3 H2SO4

reduce nitro group

Fe / HClH2N

possible reactions at meta position(s)

reactions are more controlable with less activating amide than amine, possible reactions at ortho/para position(s), can hydrolyze amide back to the amineto continue the reaction sequence towards the diazonium chemistry

O

Cl

N

O

H

NaNO2 / HCl(makes HONO)

NN

diazonium salt, stable below 5oC, decomposes to carbocation (or free radical ) above 5oC,see reactions below

T > 5oC

aromatic ring is the electrophile !

First step to diazonium salt is nitration of an aromatic ring.

Meta groups might be added at this stage before the nitro group is reduced to an amine. A metal in a reduce state donates electrons and an acid donates the protons.

If additional substitution at this stage is desired, the amine group is usually protected as an amide to reduce its basicity and activating power. All available positions (o + p) might be substituted or the amine might be protonated and turned into a meta director. If the amine is protected as an amide it must be hydrolyzed in acid or base to get back the amine functionality

HNO3 / H2SO4

O2N

1

2

3

4

5

O2N

CuCl

reduce nitro group

Fe / HCl orSnCl2/HCl H2N

O

Cl N

O

H

NaNO2 / HCl(makes HO-N=O) NN

diazonium salt is stable below 5oC, it decomposes to an unusual carbocation in that the empty orbital is sp2, (some reactions may proceed by a free radical intermediate), a variety of nucleophiles can be added at this point, see reactions below

Cl

Ipso substitution (same position), Copper reactions are called Sandmeyer reactions, the chlorine put on in this reaction is backwards to the way it was put on above, nucleophilic chloride adds here.

T > 5oCaromatic ring is the electrophile !

H2N

N

O

H

H2NH3O+ / H2O orNaOH / H2O

H2N

6

NN

Nitrous acid is the electrophile and the aromatic amine group is the nucleophile, which joins two nitrogen atoms together,followed by some proton transfers, loss of water and resonance.

7

NNT > 5oC



CuBr

CuCN

H2O

H3PO2

KI

Br

NC

HO

H

I

FBF4

Ipso substitution (same position), Copper reactions are called Sandmeyer reactions, the bromine put on in this reaction is backwards to the way it was put on above, nucleophilic bromide adds here.

8

NNT > 5oC

Ipso substitution (same position), Copper reactions are called Sandmeyer reactions, cyanide is nucleophilic in this reaction, the aromatic ring is the electrophile.

9

NNT > 5oC

Ipso substitution (same position), iodide is the nucleophile.

10

NNT > 5oC

Ipso substitution (same position), water is the nucleophile.

11

NNT > 5oC

Ipso substitution (same position), a fluoride from tetrafluoroborate is the nucleophile.

12

NNT > 5oC

Ipso substitution (same position), hypohphosphorous acid is a very unusual ACID hydride donor. A possible reaction sequence is shown below.

13

NNT > 5oC

ElectrophileHOH2

PO

OH

HPO

OH

HH

Electrophile PO

OH

OH H

lose proton

Hypohphosphorous acid is a very unusual ACID hydride donor. The hydride reduces something and the phosphorous gets oxidized.


Organic Reaction Guide Beauchamp 42 Nucleophilic reactions (addition / elimination), similar to conjugate substitution at an α,β-unsaturated carbonyl having a leaving group, except followed by elimination to reform the aromatic compound

1

2

O2N Cl

Electron poor aromatic rings with a good leaving group attached allow nucleophiles to add at the carbon with the leaving group,followed by rearomatization when the good leaving group leaves. The leaving group has to be ortho or para to the electron withdrawing group so that the substituent can stabilize the negative charge in the intermediate. It is similar to the diazonium intermediate in that the aromatic ring is the electrophile.

H2OOH

OCH3

CH3OH

O2N OH

O2N

Br

O2N

CH3O

The nitro substituent is electron withdrawing and can stabilize negative charge by resonance and chloride is a good leaving group, attack by a nucleophile can displace the chloride, the reaction is somewhat like a substitution reaction on a vinylogous acid halide.

O

Cl

N

Br

O

O

Nu

Nu

O

ClNu

NO

O

BrNu

N

Nu

O

O

O

Nu

analogous nucleophilic substitution reaction on a vinylogous acid chloride

3. Benzyne (elimination / addition), uses a very strong base, sodium amide (NaNH2)to force a 1,2

elimination beta to good leaving group forming highly reactive benzyne. A nucleophile can add to the “pseudo” triple bond on either side, after which the aromatic ring protonates at the other position.

1

2

Benzyne forms in an E2-like reaction requirering a very strong base (often some sort of sodium amide, NaNR2), except because ofthe rigid, flat nature of the aromatic ring, no real pi bond can form. The parallel sp2 orbitals are angled away from one another making for a very unstable and highly reactive arrangement of orbitals. An electron pair donor in the vacinity of either sp2 ortibal will add its electrons to form a sigma bond and isolate the pseudo pi electrons in a highly basic sp2 orbital which quickly protonates in an acid/base reaction to regenerate a neutral aromatic ring. If an unchanged substituent is present on the ring, it is easily seen that either carbon of thepseudo pi bond can be attacked by a nucleophile, because isomeric products are obtained.

Br

Cl

N(CH3)2

HN(CH3)2benzyne intermediate

N N

N(CH3)2

HN(CH3)2

NN

Nu: adds at either position

Nu: adds at either position

meta and para products obtained

ortho and metaproducts obtained


Organic Reaction Guide Beauchamp 43 4. Miscellaneous side chain reactions

1

2

Clemmenson reduction (HCl/Zn), reduces aromatic ketone to methylene carbon (CH2) under acid conditions. The Zn supplies the electrons and the HCl supplies the protons.

Pd/H2 reduction, reduces aromatic ketone to methylene carbon (CH2) because it's benzylic, reduces, reaction occurs under neutral conditions unlike Clemmenson's (acidic) or Wolff-Kishner (basic).

CrO3/∆ or KMnO4/∆ oxidations, very harsh conditions, no sensitive groups tolerated, oxidizes any carbon side chain with a benzylic hydrogen to a carboxylic acid, a quaternary benzylic carbon will either not react or if really pushed the aromatic ring is oxidized away, leaving a carboxylic acid in place of the ring.

Free radical substitution (Cl2 or Br2 and light, hν), chain reaction mechanism prefers benzylic position because of weaker C-H bond

O

R HCl / ZnCH2 R

O

RCH2 R

O

R Pd / H2CH2 R

H2NNH2/RO /∆

KMnO4/∆

CrO3/∆ HO2C CO2H

HO2C

BrBr2 / hν + HBr

3

4

6

Wolff Kishner reduction (H2NNH2/RO /∆), reduces aromatic ketone to methylene carbon (CH2), under base conditions. Imine-like structure forms, acid/base proton transfers, tautomer-like changes and ultimately loss of nitrogen gas.

KMnO4/∆

7

forcing conditions

HO2C

8


Organic Reaction Guide Beauchamp 44 Synthesis of Functional Groups – most of the reactions listed below have been listed above, but are listed here by the common theme of functional group preparation so you can consider a variety of possibilities when considering the synthesis of a particular functional group.

1. Synthesis of RX compounds RX compounds from free radicals substitution of sp3 C-H bonds - the weakest C-H bond is attacked fastest

H3C CH3CH4

Typical range of sp3 C-H bonds

Typical free radical substitution mechanism

Br Br

H Br

1. initiation

2 propagationa. abstraction of H - bromine atom abstracts hydrogen atom from weakest C-H bond fastest (3o > 2o > 1o > methyl).

b. abstraction of Br - carbon free radical abstracts Br from Br2 molecule (very weak bond).

3 termination - two free radicals diffuse near one another and quench each other in bond formation.

hν and/or ∆BrBr

Br

H

H

2o > 1o

H ∆Hrxn depends on difference in C-H and H-Br bond energies.

H

Br BrH

BrBr

∆Hrxn is always favorable because Br-Br bond is so weak.

H H



1

2

3

SOCl2

PBr3

OH Cl

OH Br

Use of HX acids (HCl, HBr, HI).

Use of thionyl chloride.

OHHBr

Br

Use of phosphorous trichloride(PCl3, PBr3 or P/I2).

a. Synthesis of RX compounds from alcohols

4 OHIOTs

N

S

O

O

ClNa I

Make tosylate followed by an SN2 reaction (Me, 1o, 2o RX) with a sodium halide salt

a. b. a. b.

HBr Br

b. Synthesis of RX compounds from alkenes

HX acids (Markovnikov addition)

1. BH3 2. Br2 / CH3OCH3

H

Br

5

6a. b.

a.b.CH3

H

B

R

Ri. BH3 ii. H2O2/HO (anti-Markovnikov & "syn" addition)

2. Synthesis of alcohols

NaOH

a. Synthesis of alcohol compounds from RX compounds

SN2 at Me, 1o with NaOH

SN1 at 2o, 3o RX with H2O, possible rearrangements, reasonable if rearrangement is not a problem

Br OH

OHBrH2O

SN2 at Me, 1o , 2o with CH3CO2Na, followed by base hydrolysis to form the alcohol and acetate which is discarded.

Br1. CH3CO2Na2. NaOH OH

(via ester)

1

2

3


Organic Reaction Guide Beauchamp 46 b. Synthesis of alcohol compounds from alkenes

1. HgX2 / H2O2. NaBH4

1

2

3

OH Mercury bridge of cation intermediate minimizes rearrangement. Borohydride reduces mercury off and substitutes hydrogen on.

H3O+ / H2O

OH Aqueous acid forms carbocation intermediate which can rearrange

OH

1. O3 / -78oC2. NaBH4

CH3OHlost in aqueousworkup

Ozonolysis cuts double bond in two forming carbonyls and sodium borohydride workup reduces carbonyls to alcohols.

4OH

Anti-Markovnikov and "syn" addition of borane, BH3, followed by oxidation with perioxide to form an alcohol,

1. BH3 2. H2O2 / HO

1

2

3

c. Synthesis of alcohol compounds from RMgX & RLi organometallics reacted with aldehydes, ketones, esters (twice) and epoxides (all followed by acid workup)

1. Mg2.

3. WK

O

H

O

OCH3

1. Mg (2 eqs.)2.

3. WK

O1. Li2.

3. WK

Br

Br

OH

Br OH

Organomethallics + aldehydes and ketones makes 1o (from CH2=O), 2o (from RCH=O) or 3o (from R2C=O) alcohols.

Organomethallics + esters (react twice) makes 3o alcohols.

Organomethallics + epoxides, SN2-like reaction at less hindered side of the epoxide, workup protonates the alkoxide.

OH



1

2

3

4

d. Synthesis of alcohol compounds from metal hyrides, LiAlH4 or NaBH4 reacted with aldehydes, ketones, esters and acids (twice & only with LAH) and epoxides (all followed by acid workup)

Sodium borohydride (or LAH) + aldehydes and ketones makes 1o or 2o alcohols after workup.

Only lithium aluminium hydride works on esters, reacts twice to make 1o alcohols after workup.

NaBH4 or LiAlH4 + epoxides, SN2-like reaction at less hindered side of the epoxide, workup protonates the alkoxide.

O

H

1. NaBH42. workup

1. LiAlH42. workup

O

OCH3

OH

OH

(+ CH3OH, discarded)

O 1. NaBH42. workup

OH

(R) (R)

O

OH1. LiAlH42. workup OH

Only lithium aluminium hydride works on esters, reacts twice to make 1o alcohols after workup.

1

2

Base hydrolysis of esters is often called saponification (soap making)

e. Acid or base hydrolysis of esters forms a carboxylic acid and an alcohol (either or both could be the desired result).

O

OCH31. H2O / HO2. WK

O

OHCH3OH

H3O+ / H2O O

O

O

OHHO

(discard ?)

(discard ?)

Acid hydrolysis of esters is the reverse of Fischer ester synthesis.Water is added instead of removed.

3. Synthesis of ethers from alcohols and alkenes

1

2

3

An SN1 reaction at 2o adn 3o ROH (could have E1 complications), and an SN2 reaction at 1o ROH.

Markovnikov addition of an alcohol at an alkene. Rearrangement is minimized by bridging mercury atom, which gets reduce off with hydride (free radical intermediate). Note that the alcohol used in this reaction could have been made by a similar procedure between and alkene and water.

conc. H2SO4 cold

OH 1. NaH2.

O

1. HgX2

2. NaBH4

O

Br

OH

OH O

Make an alcohol into a stronger nucleophile by removing proton with hydride (strong base). SN2 works OK at methyl and primary RX centers.


Organic Reaction Guide Beauchamp 48 4. Synthesis of epoxides from alkenes

1

2

3

In a second reaction, the alkoxide is formed and does an SN2 on the vicinal bromide to form the epoxide. Products in this example are enantiomers.

meta-chloroperoxybenzoic acid (mCPBA) accomplished the same transformation in a single step. The products in this example is meso (achiral with chiral centers).

First, a halohydrin is made from an alkene. Products in this example are enantiomers.

1. Br2 H2O

BrH

OH

O

H

CH3

O

H

H

mCPBA

(d / l)

(d / l)

NaOHBrH

OH

5. Synthesis of alkenes

1

2 This represents our only productive conditions for E2 at a primary center and is due to the sterically bulky and very basic potassium t-butoxide.

These are E1 conditions. Rearrangements are possible.

OHH2SO4 / ∆

(major)(dehydration)

BrO K

3

4Wittig reaction, generally the best bet to get the exact alkene you are looking for.

Strong base/nucleophile and a 3o RX mean an E2 reaction. Only the more stable alkene is shown.

5

6

Poisoned Pd catalyst stops at the cis alkene.

Birch reduction of alkyne mostly forms E (trans) alkene.

Br NaOH

O 1. Ph3P=CH2 (from CH3X)2. WK

Pd / H2quinoline

(Lindlar's cat.)

Na / NH3(l)

(Birch)



7Birch reduction of aromatic ring putstwo pi bonds opposite one another. If an electron donating substituent is present, it will be at one of the sp2 positions.

8Complete reduction of an alkyne to an alkane for comparison.

Na / NH3(l) ROH

(Birch)

Pd / H2

6. Synthesis of alkynes

1

2

3

Double elimination product (plus loss of terminal sp C-H under the reaction conditions. Workup protonates or an electrophile could be added. The starting dibromide can be made from an alkene + Br2.

The Zipper reaction moves a triple bond through any number of CH2's until the termial position is found, where the sp C-H is lost, forming the most stable anion in the pot.

Terminal acetylide carbanion works well at methyl and primary RX.

NaNH21.

2.Br

Br

Br

excessNaNH2

1.

2. WK

excessNaNH2

1.

2. WK

SN2

7. Synthesis of amines

1

2

3

4

1.

2.N

O

O

H

Gabriel amine synthesis, starts with phthalimide, removes proton on nitrogen, does an SN2 reaction on an RX and hydrolyzes off the two carbonyl portions to obtain a 1o RNH2.

NaOH

SN2

Br N

O

O

NaOH NH21o aminesN

O

O

H

ONH2

1.

2. NaBH3CN3. WK H

NH Reductive alkylation of imine with

sodium cyanoborohydride to make 2o or 3o amines.

H

O NH

1.

2. NaBH3CN3. WK H

N

1o amines

3o amines

3o amines

Reductive alkylation of imine with sodium cyanoborohydride to make 2o or 3o amines.



6

NH2

O1. LiAlH42. WK NH2

LAH reduction of 1o, 2o and 3o amidesmakes 1o, 2o and 3o amines.

CN 1. LiAlH4

2. WK NH2

LAH reduction of nitriles makes 1o amines after workup.

8. Synthesis of ketones

1

2

3

OH CrO3 (no H2O or H2O)

OPCC or Jones at 2o ROH.

Nitriles + RLi compounds, followed by hydrolysis form ketones.

CN Li

2. WK

O

H3O+ /Hg+2

H2O

O Hydration of an alkyne. Markovnikovadditon, forms enol, which tautomerizes to ketone.

4

5

6

7

8

O1. O3, -78o

2. CH3SCH3 2

Ozonolysis of alkene, DMS workup. Many other workup conditions are possible. For best results here, it would be nice to have a symmentrical alkene.

Cl

O (CH3)2Cu

Li

O Cuprates + acid chlorides form ketones. Cuprates come from organolithium compounds which come from RX compounds.

S

S 1. nBuLi2. CH3Br3. WK

S

S1. nBuLi2. CH3CH2Br3. WK

S

S Dithiane alkylation (twice), then hydrolysis to the carbonyl compound. If hydrolyzed after one alkylation, then an aldehyde is obtained.

S

S

Hg+2 / H2OO

S

S



10

OH

OCH3 Li

2 eqs.

2. WK

O 2 eqs.RLi + carboxylic acid, forms a double alkoxide because of the power of the organolithium nucleophile, which hydrolyzes to a ketone in the workup.

Cl

O Friedel Crafts Acylation, makes aromatic ketones, deactivating substituents inhibit the reaction. Only reacts one time because keto group is a deactivating, meta director.

AlCl3

O

9. Synthesis of aldehydes

1

2

3

Nitriles + DIBALH, followed by hydrolysis form aldehydes.

CN

CrO3 (no H2O)

PCC OHH

O

H

O1. DIBAH -78oC2. WK

H

Ohydroboration of an alkyne, then H2O2/HO

1. BH32. H2O2./ HO

4

5

6

7

8

Ozonolysis of alkene, DMS workup. Many other workup conditions are possible. For best results here, it would be nice to have a symmentrical alkene.

Cuprates + acid chlorides form ketones. Cuprates come from organolithium compounds which come from RX compounds.

Dithiane alkylation (once) , then hydrolysis to the carbonyl compound. An aldehyde is obtained.

1. O3, -78o

2. CH3SCH3 2 H

O

Cl

O

H

O1. DIBAH -78oC2. WK

S

S 1. nBuLi2. CH3CH2Br3. WK

S

S

Hg+2 / H2OH

O

S

S

O

O

H

O DIBAH + ester at low temperature makes aldehydes after hydrolysis.

1. DIBAH -78oC2. WK


Organic Reaction Guide Beauchamp 52 9 Vilsmeier reaction on activated aromatics

(many variations) to make aromatic aldehydes.AlCl3

O

HHClOC

10. Protection of aldehydes and ketones

1

2

3

C

ON HO

OHTsOH (-H2O) C

OON

1. CH3Li2. mild WK OOO

Protection of aldehyde or ketonewith ethylene glycol. Acid catalysis and removal or water forms acetal or ketal. Carbonyl becomes inert to many strong nucleophiles that would otherwise react with it.

COON

H3O+ / H2O OOO O O

Run the desired reaction, an organo-metallic reaction here. It is possible to do a mild workup and not hydrolyze the ketal, or a more vigorous workup, as in the next frame could deptrotect the ketal at the same time as the other reaction is worked up.

11. Synthesis of acids

1

2

3

4

5

CrO3 (H2O)

Jones conditions oxidize 1o ROH to carboxylic acids and 2o ROH to ketones.

OH OH

O

i. O3 / -78oC ii. H2O2 /HO

OH

O2Ozonolsis with oxidative workup, using H2O2, forms acid if alkene carbon has a hydrogen and ketones if alkene carbon is geminally disubstituted.

CN

OH

O H2SO4H2O / ∆

Full acid hydrolysis of a nitrile.

CN

OH

O1. H2O/HO ∆2. WK

Full base hydrolysis of a nitrile.

OR

O aqueousacid or base

OH

OAqueous hydrolyisis of acid derivatives.



7

8

9

Cl

O

O

O O

aqueousacid or base

aqueousacid or base

OH

O

OH

O

Aqueous hydrolyisis of acid derivatives.

NH2

O H2SO4H2O / ∆ OH

O Full acid hydrolysis of a amide.

Br1. Mg2. CO23. WK OH

OGrignard reagent reacted with carbon dioxide, then workup.

Aqueous hydrolyisis of acid derivatives.

12. Synthesis of acid chlorides

1

2

3

A few ways to make an acid chloride from a carboxylic acid: thionl chloride,oxalyl chloride and phosphorous trichloride.

SOCl2

OH

O

Cl

O

OH

O

PCl3

thionyl chloride

oxalyl chloride

ClCl

O

O

OH

O

Cl

O

Cl

O

phosphorous trichloride

13. Synthesis of anhydrides

1

2

A couple ways to make an unsymmetrical anhydride from a carboxylic acid and either another anhydride or an acid chloride.

Cl

OOH

O

O

O O

OH

O

O

O O

O

O O


Organic Reaction Guide Beauchamp 54 14. Synthesis of esters

1

2

3

Make a good carboxylate nucleophile then ract with a methyl, 1o or 2o RXcompound.

O

OH

1. NaOH2. Br

O

O

O

OH

OH TsOH(-H2O)

O

O

Fischer ester synthesis, acid catalysis and remove water to from ester. Use opposite conditions to hydrolyze ester(acid catalysis and lots of water).

O

Cl

OH O

O

A tertiary amine is sometimes used to make the reaction work faster and better.

4

5

O

O O OH O

O

A tertiary amine is sometimes used to make the reaction work faster and better.

OO

O

Cl OH

mCPBA

O

O

Another use of mCPBA is to oxidize ketones to esters. Called the Baeyer Villegar Rxn.

15. Synthesis of amides

1

2

3

4

Milder conditions hydrolyze nitriles to amides, harsher conditions hydrolyze nitriles to carboxylic acids.

Acid chlorides + ammonia, 1o or 2o amines makes 1o, 2o or 3o amides.

O

O O

CN HCl

H2O

O

NH2

O

ClNH3

O

NH2

O

Cl

O

N

NH

Anhydrides + ammonia, 1o or 2o amines makes 1o, 2o or 3o amides.

Anhydrides + ammonia, 1o or 2o amines makes 1o, 2o or 3o amides.

NH

O

N


Organic Reaction Guide Beauchamp 55 16. Synthesis of nitriles

1

2 O

NH2

Br

CNNa C

N

CNSOCl2

(-H2O)

SN2 reactions of cyanide withmethyl, 1o or 2o RX compounds

Dehydration of primary amide with thionyl chloride.

17. Enamine Chemistry

1

2

ON H

pH = 5(-H2O)

N

Br

Hydrolyze alkylation product via this intermediate

WK = aq. hydrolysis H2O

ON

1.

2. N

Start with carbonyl, make enamine,alkylate enamine, hydrolyze back to carbonyl compound with extra R group added (methyl, allyl or benzyl in our course).

18. Synthesis of β-hydroxycarbonyl and α,β-unsaturated carbonyl

1

2

O

H

Na ROO

H

Aldol reaction of two carbonyls, usually the same one. Base is used to make an enolate which then attacks another carbonyl to form a β-hydroxy carbonyl structure that can be isolated or dehydrated to an α,β-unsatruated carbonyl shown in the next frame. this often done with acid, but in this book we will indicate the next step with the symbol for heat, ∆.

acid. hydrolysisH3O+ / H2O / ∆

O

H

α,β-unsatruated carbonyl

O

H

OH

α

β

O

H

OH

α

β

β

α

β-hydroxy carbonyl


Organic Reaction Guide Beauchamp 56 19. Malonic ester synthesis (produces mono and disubstituted acetic acids)

1

2

3

4

O

O O

O

Br

1. CH3CH2O2.

3. WKO

O O

O

1. CH3CH2O2.

3. WK

BrO

O O

O

H3O+ / H2O (-CO2)

H3O+ / H2O (-CO2)

O

OH

O

OH

O

O O

O

product from 1

product from 1

O

O O

O

O

O O

O

product from 3

Malonic ester synthesis. Remove acidic proton with alkoxide base then alkylate with electrophile (RX, epoxide or another carbonyl). Can hydrolyze ester at this point, decarboxylate (-CO2) to obtain a monoalkylated ethanoic acid (acetic acid), or repeat the reaction a second time and then decarboxylate to obtain a dialkylated ethanoic acid (acetic acid). It's also possible to do similar chemistry by using the dianion of ethanoic acid or using a simple ethanoate ester and LDA to generate the ester enolate and perform an alkylation at low temperature (-78oC)

monoalkylated acetic acid

dialkylated acetic acid

20. Ethyl acetoacetate synthesis (produces mono and disubstituted acetones)

1

2

3

4

Br

1. CH3CH2O2.

3. WK

1. CH3CH2O2.

3. WK

BrO

O O

H3O+ / H2O (-CO2)

H3O+ / H2O (-CO2)

O

O

product from 1

product from 1

product from 3

Acetoacetic ester synthesis. Remove acidic proton with alkoxide base then alkylate with electrophile (RX, epoxide or another carbonyl). Can hydrolyze ester at this point, decarboxylate (-CO2) to obtain a monoalkylated 2-propanone (acetone), or repeat the reaction a second time and then decarboxylate to obtain a dialkylated 2-propanone (acetone). It's also possible to do similar chemistry by using a simple ketone and LDA to generate the ketone enolate and perform an alkylation at low temperature (-78oC)

monoalkylated acetone

dialkylated acetone

O

O O

O

O O

O

O O

O

O O

O

O O


Organic Reaction Guide Beauchamp 57 21. Cuprate chemistry

1

2

3

4

Li

0.5 eq CuBr

Li

Cu

Li

O

2. Cuprate coupling reaction with an RX compound. Remember the cuprate comes from an organolithium, which comes from another RX compound. It hard to tell which bond was formed and in what direction the atoms were used because there are a lot of possibilities

Br

Cuprates are prepared from organo-lithium reagents + CuBr (cuprous salt) in a 2 to 1 ratio.

There are three choices for a cuprate in our course.

RX compound

Li

5

organolithium reagent

organocuprate

Cu

Liorganocuprate

Cu

Liorganocuprate

Cu

Liorganocuprate

O

α,β-unsaturate carbonyl

Br

RX compound

O

ClO

acid chloride

α

β

1. conjugate addition to an α,β-unsaturate carbonyl

3. Cuprate substitution of chlorine in an acid chloride to make a ketone. There are two possible ways you could consider joining the acid chloride and cuprate.

Organolithium reagents come from RX compounds + Li metal.

22. Conjugate addition – stable anions often add at the C-β carbon

O

CNNa

O

CN

O

O OO

H3O+ / H2O (-CO2)

O

O

23. Dithiane Chemistry – see aldehydes and ketones above 1

2

Dithiane alkylation (once) , then hydrolysis to the carbonyl compound. An aldehyde is obtained.

S

S 1. nBuLi2. CH3CH2Br3. WK

S

S

Hg+2 / H2OH

O

S

S

3

S

S 1. nBuLi2. CH3Br3. WK

S

S Dithiane alkylation (twice), then hydrolysis to the carbonyl compound. If hydrolyzed after one alkylation, then an aldehyde is obtained.



5

S

S1. nBuLi2. CH3CH2Br3. WK

S

S

Hg+2 / H2OO

S

S

24. Dianion Chemistry

1

2The reaction can be run once, worked up and decarboxylated (shown in the next frame). In this case the same product could have been obtained using the normal ethyl acetoacetate synthesis.

To make dianion requires very strongbases. To simplify our reaction we will write 2 eqs of LDA to show formation of dianion. The second acidic site is the more reactive site in the alkylation.

O

O O

O

O O 1. NaH or LDA2. nBuLi

Br

O

O O

O

O O

O

O O workup

3

4 A second alkylation can be performed using the normal ethyl acetoacetate synthesis, alcylating the position in between the two carbonyl groups. After ester hydrolysis and decarboxylation a disubstituted acetone is obtained with a alkylation on both sides of the carbonyl.

5

6

7

H3O+ / H2O (-CO2)

O

O O O

O

O O O

CH3Br

1.

2. O

O O

H3O+ / H2O (-CO2)O

O O O

The same product as using the normal ethyl acetoacetate synthesis.

1. NaOH 2. LDA

O

OH

O

O

Dianions from carboxylic acid can be formed using a strong, non-nucleophilic base for the Cα-H position. The carbanionic site is themore reactive site and alkylation occurs there. The product will be an alkylated carboxylic acid, which can be esterified by making the carboxylate and doing an SN2 on an RX compound, as discussed earlier.

O

OBr1.

2. workup

O

OH



8 O

OH

O

OHBr

1. NaOH

2.

25. Robinson Annelation

1

2

3

i. Base makes enolateO O

O HO

O O O

O

ii. conjugate addition to α,ß-unsaturated carbonyl O O

O 2. workup

O O

O

O

O

iii. aldol condensation (-H2O), 1,6 atoms join together

HO

O

O

OH

O

Oβ-hydroxycarbonyl (aldol product)

O

O

O

O

12

34

56

1

2

3

4

56

4

5

6

-H2O

OO

O α,β-unsaturatedcarbonyl

β-hydroxycarbonylO

OH

O

O

If ester group is hydrolyzed the acid will decarboxylate (-CO2) forming a cyclic ketone. (annelation = ring forming)

-H2OOO

O

∆

H3O

OO

HO

- CO2 followed by tautomerization of enol

OOO

HO

α,β-unsaturatedcarbonyl
