Nucleophilic Acyl Substitution Reactions of Carboxylic Derivatives

Question 1

What are the major steps of all nucleophilic acyl substitution reactions?

Answer 1

1st major step = Nucleophilic attack of the carbonyl grp carbon to produce the alkoxide intermediate

2nd step = Elimintation of the leaving grp; producing new carbonyl compound

NOTE: There may be a step before the nucleophilic attack to alter the carboxy. acid derivative to be more electrophilic (reactive)

Question 2

How does NU acyl substitution rxn differ from NU add. rxn?

Answer 2

1st major step = SAME (nucleophilic attack to produce alkoxide)

Following steps = DIFFERENT
–> NU add. rxn = Alkoxide ion protonates to yield alcohol

–> NU acyl sub. rxn. = Alkoxide ion undergoes elimination of the LG to yield new carbonyl compound

Question 3

What determines reactivity of a compound in NU acyl sub. rxn?

Answer 3

The reactivity of the carbonyl carbon! Influenced by:

1) Steric hindrance around the carbonyl carbon (branching on the alpha carbons)

2) Electronics; Electrophilic nature of the carbonyl carbon

Question 4

What is the rate limiting step of the NU acyl sub. rxn?

Answer 4

The nucleophilic attack of the carbonyl carbon

Question 5

Greater branching on the alpha carbons of carboxylic acid derivatives =…

Answer 5

LESS REACTIVE in Nu acyl sub rxns.

Question 6

Relationship between alpha-carbon branching and reactivity in Nu. acyl sub rxns?

Answer 6

Quarternary alpha-carbon (LEAST reactive)
<
Tertiary alpha-carbon
<
Secondary alpha-carbon
<
Primary alpha-carbon (MOST reactive)

Question 7

What affects electrophilic character of the carbonyl carbon?

What is the relationship to reactivity in Nu. acyl sub rxns?

Answer 7

The groups bonded to the carbonyl carbon

–> EDGs will decrease E+ character = Decreases reactivity
–> EWGs will increase E+ character = Increases reactivity

Question 8

Relative ranking of reactivity of carboxy. acid derivatives in Nu. acyl sub rxns

Answer 8

MOST Reactive to LEAST reactive:

1) Acyl Halides (most)
2) Acid Anhydrides
3) Thioesters
4) Esters
5) Amides (least)

Question 9

ALL carboxy. acid derivatives can. be produced from carboxylic acids via…

BUT what must occur first?

Answer 9

Nucleophilic Acyl Substitution Reactions

BUT the OH grp is a poor LG so alterations must be made to make the carboxy. acid MORE reactive to nucleophile

Question 10

What are the two main types of alterations to carboxy. acids to increase reactivity to nucleophile?

Answer 10

1) PROTONATION of the carbonyl oxygen to give a (+) charge to the carbonyl grp = increases E+ character!

–> Protonation typically mediated by a strong acid catalyst

2) Conversion of OH to a BETTER LG

Question 11

Carboxylic Acid to Acyl Halide: Reactants

Answer 11

For Acyl Chloride:
SOCl₂
———->
CHCl₃

For Acyl Bromide:
PBr₃
———->
ether

Question 12

Carboxylic Acid to Acyl Chloride: General mechanism steps:

Answer 12

1) SOCl₂ binds to the oxygen on the OH grp of the carboxylic acid, releasing Cl^-

== SO₂Cl bound to the carbonyl grp which is a better LG than the original OH

2) Free Cl^- does nucleophilic attack on the carbonyl carbon

3) Oxygen electrons push down to eliminate SO₂ from the carbonyl grp

== Acyl chloride!

Question 13

Carboxylic Acid to Ester: Reactants

Answer 13

FISCHER ESTERIFICATION:

H₂SO₄ or HCl
————————->
ROH

Question 14

Fischer Esterification (carboxylic acid to ester) PROCESS

Answer 14

PATED!

1) Protonation of the carbonyl oxygen by the acid

2) Attack of the nucleophile (ALCOHOL) to the carbonyl carbon

3) Transfer of a proton (H+) from (+) ROH to the OH grp to form OH₂+

4) Elimination of OH₂ (alleviating the + charge on the OH₂)
–> Does so bY pushing electrons from the previously carbonyl oxygen (now an OH), reforming the carbonyl double bond

5) Deprotonation of the carbonyl oxygen by the acid catalyst conjugate base to alleviate the (+) charge

== Yields the ester! (and reforms the acid catalyst)

Question 15

What are the two potential methods for:

Carboxylic Acid —> AMIDE

Answer 15

1) Form ACYL HALIDE and then react with an AMINE

2) React carboxylic acid with DCC and then with an AMINE

–> BOTH methods involve converting the OH grp to a better LG before reacting with the amine!

Question 16

Carboxylic Acid to AMIDE via acyl halide: Reactants

Answer 16

1. SOCl₂, CHCl₃
——————————->
2. AMINE (< tertiary)

Question 17

Carboxylic Acid to AMIDE via acyl halide: Process

Answer 17

1) Carboxylic acid reacts with SOCl₂ to yield an acyl chloride

2) The amine (that is less than tertiary) attacks the carbonyl carbon of the acyl chloride

3) Free amine in solution then deprotonates the added amine to get rid of the (+) charge

4) Electrons from the alkoxide (-) push down to reform the carbonyl double bond AND eliminate (push off) the chlorine group

== AMIDE formed!

Question 18

Carboxylic Acid to AMIDE with DCC: Reactants

Answer 18

1. DCC
—————————–>
2. AMINE (< tertiary)

Question 19

Carboxylic Acid to AMIDE with DCC: Process

Answer 19

1) Carboxylic acid reacts with DCC; bonds to the oxygen of the OH grp to form a NEW and better LG!

2) Amine attacks the carbonyl carbon on the DCC sub. compound

3) Alkoxide electrons push down to reform carbonyl double bond and push off the DCC LG as DCU!

4) The pushed off DCC derivative deprotonates the added amine to get rid of the (+) charge on it

== AMIDE formed!

Question 20

What are the reactions that acyl halides undergo (not including making them)?

Answer 20

SIX main reactions:

1) Hydrolysis: Acyl halide —-> Carboxylic Acid

2) Alcoholysis: Acyl halide —-> ESTER

3) Aminolysis: Acyl halide —-> AMIDE

4) Grignard: Acyl halide —> 3° alcohol!

5) FULL Reduction: Acyl halide —> 1° alcohol

6) PARTIAL Reduction: Acyl halide —> ALDEHYDE

Question 21

Acyl Halide Hydrolysis: Reactants

Answer 21

Acyl Halide (RCOX) —-> Carboxylic Acid (RCOOH)

H₂O
———————–>
Base? (to prevent excessive halide acid formation)

Question 22

Acyl Halide Hydrolysis: Process

Answer 22

1) Acyl halide carbonyl carbon is attacked by WATER = sub. alkoxide

2) Alkoxide electrons push down to reform carbonyl double bond AND push off the halide LG (X)!

3) Base or X^- deprotonates the H₂O⁺ on the carbonyl

== carboxylic acid!

Question 23

Acyl Halide Alcoholysis: Reactants

Answer 23

Acyl halide —-> ESTER

ROH (alcohol)
———————————>
Et₃N (tertiary amine base!)

Question 24

Acyl Halide Alcoholysis: Process

Answer 24

1) Alcohol (ROH) acts as nucleophile and attacks the carbonyl carbon of the acyl halide = alkoxide forms

2) The alkoxide pushes electrons down to reform carbonyl double bond AND push off the halide LG

3) Et₃N deprotonates the added ROH⁺ grp on the carbonyl

== ESTER forms!

Question 25

How does substitution of the alcohol affect acyl halide alcoholysis?

Answer 25

The LESS SUBSTITUTED the alcohol, the more reactive it is! Ex: If a molecule has a 2° and a 1° alcohol, the 1° OH will undergo alcoholysis to form an ester because it is more reactive! Most reactive = 1° ROH -----> 3°ROH (LEAST reactive!)

Question 26

Acyl halide Aminolysis: Reactants

Answer 26

Two potential methods: **2 eq. (AMINE, <3°)** --------------------------------> or **AMINE (< 3°)** ----------------------------> **Et₃N**

Question 27

Acyl halide Aminolysis: Process (for both methods)

Answer 27

1) Amine acts as nucleophile and attacks the carbonyl carbon 2) Alkoxide electrons push down to reform carbonyl DB and to push off (eliminate) halide LG 3) Deprotonation: mechanism depends on base present! --> If 2 eq. amine were used: second equivalent of amine will come and deprotonate the first added amine to resolve the (+) charge and produce the amide --> If 1 eq. of amine + Et₃N were used: Et₃N will deprotonate the added amine to resolve the (+) charge and produce the amide

Question 28

Acyl halide + Grignard: Reactants

Answer 28

Acyl halide ----> 3° Alcohol! **1. 2 eq. GRIGNARD (RMgX)** ----------------------------------------> **2. H₃O⁺**

Question 29

Acyl halide + Grignard: Process

Answer 29

Reaction happens TWICE! **Rxn 1:** 1) ONE grignard reagent does Nu attack to the carbonyl carbon of the acyl halide = Alkoxide forms with R grp from grignard 2) Alkoxide electrons push down to reform carbonyl DB and kick off the halide LG == KETONE is formed! (Intermediate) **Rxn 2:** 3) Formed ketone is immediately attacked by the SECOND grignard reagent at the carbonyl carbon = Alkoxide forms with a second R grp from grignard 4) Alkoxide is protonated by the H₃O⁺ (CANT push down to form DB as the carbon is now 4°!) == 3° alcohol (with two identical R groups from the grignards!)

Question 30

Acyl halide FULL reduction: Reactants

Answer 30

Acyl halide ----> 1° alcohol! **1. LiAlH₄** --------------------------> **2.H₃O⁺**

Question 31

Acyl halide FULL reduction: Process

Answer 31

Rxn. occurs TWICE: **Rxn 1:** 1) H^- of Li⁺AlH₄^- attacks the carbonyl carbon = alkoxide forms 2) Alkoxide electrons push down to reform carbonyl DB and push off the halide LG == ALDEHYDE forms! (Intermediate) **Rxn 2:** 3) Aldehyde immediately gets attacked at carbonyl carbon by H^- of more Li⁺AlH₄^- = forms alkoxide 4) Alkoxide is PROTONATED by H₃O⁺ (Alkoxide electrons CANT push down to reform carbonyl DB because the carbon is now 4°!) == 1° alcohol forms! (with two Hs on previously carbonyl carbon from LiAlH₄)

Question 32

Acyl Halide PARTIAL Reduction: Reactants

Answer 32

Acyl halide ----> KETONE **R₂CuLi** -----------------> Dialkylcopper lithium reagent = GILMAN reagent! May also see as: **(CH₃CH₂)₂CuLi**

Question 33

Acyl Halide PARTIAL Reduction: Process

Answer 33

1) The copper of the dialkyl copper (R₂Cu) portion of the gilman reagent replaces halide on the acyl halide 2) RCu leaves and one of the R groups stays behind, bound to the carbonyl carbon == KETONE is formed!

Question 34

Reactions of acid anhydrides:

Answer 34

Undergo almost ALL the same reactions as acyl halides BUT just slower (because they are less reactive): 1) Hydrolysis (Anhydride --> 2x carboxylic acid) 2) Aminolysis (Anhydride --> Amide + carboxylate ion) 3) Alcoholysis (Anhydride ---> Ester + carboxylate ion) 4) Grignard (Anhydride ---> 2x 3° alcohol!) 5) FULL reduction (Anhydride ---> 2x 1° alcohol!) (DOES NOT do partial reduction! Gilman reagents only work with acyl halides!)

Question 35

Anhydride Hydrolysis: Reactants + Process

Answer 35

Anhydride ---> 2x Carboxylic Acids **H₂O** --------------> 1) H₂O acts as nucleophile and attacks ONE of the carbonyl carbons = Alkoxide forms 2) Electrons of alkoxide push down to reform carbonyl grp and kick off a carboxylate ion 3) Carboxylate ion deprotonates H₂O⁺ on the other carbonyl carbon == Forms TWO carboxylic acids (will be identical if the anhydride was symmetrical)

Question 36

Anhydride FULL reduction: Reactants + Process

Answer 36

Anhydride ---> 2x 1° alcohols **1. LiAlH₄** --------------------------> **2.H₃O⁺** 1) First reduction occurs to yield an aldehyde and a carboxylate ion (which can protonate to carboxylic acid) 2) BOTH the carboxylic acid and the aldehyde are immediately reduced and then the alkoxide is protonated on both to yield TWO 1° alcohols!

Question 37

Anhydride + Grignard: Reactants + Process

Answer 37

Anhydride ---> 2x 3° alcohols **1. 4x GRIGNARD (RMgX)** ----------------------------------------> **2. H₃O⁺** 1) First grignard reaction produces ONE ketone and one carboxylate ion 2) Second round of grignard reactions produces ONE 3° alcohol and ONE ketone 3) Last grignard reaction turns the remaining ketone into a 3° alcohol == TWO 3° alcohols!

Question 38

Anhydride Alcoholysis: Reactants + Process

Answer 38

Anhydride ---> ONE ester + ONE carboxylate ion **ROH (alcohol)** ---------------------------------> **Et₃N** (tertiary amine base!) 1) Alcohol does Nu attack at ONE of the carbonyl carbons = alkoxide forms 2) Formed alkoxide pushed electrons down to reform carbonyl DB and kick off a carboxylate ion 3) Et₃N deprotonates the ROH⁺ to give the ESTER == ESTER + carboxylate ion

Question 39

Anhydride Aminolysis: Reactants + Process

Answer 39

Anhydride ---> ONE amide + ONE carboxylate ion **Amine, <3°** ------------------------------> **Et₃N** (tertiary amine base!)** 1) Amine does Nu attack at ONE of the carbonyl carbons = alkoxide forms 2) Alkoxide electrons push down to reform carbonyl DB and kick off a carboxylate ion 3) Et₃N deprotonates the added Amine⁺ to give the AMIDE == AMIDE + carboxylate ion

Question 40

What are the methods in which esters can be MADE?

Answer 40

1) FISCHER ESTERIFICATION (carboxylic acid + ROH + ACID CAT) 2) Acyl halide alcoholysis (acyl halide + ROH + Et₃N) --------------------------------------------- NOT usually used: 3) Base hydrolysis followed by alkyl halide attack (Carboxylic acid + NaOH + 1°RX) --> Carboxylic acid is converted to carboxylate ion via hydroxide (base) --> RCOO- then attacks a PRIMARY alkyl halide to form ester (SN2 like mechanism)

Question 41

Why is base hydrolysis + alkyl halide attack not preferred for ester preparation?

Answer 41

Because it is limited in the esters it can make! The R grp that gets attached to the bridge carbon can only be PRIMARY as the alkyl halide used to make it must be primary!

Question 42

Ester Reactions:

Answer 42

Can undergo 6 reactions: 1A) Acid-catalyzed hydrolysis (RCOOR ---> Carboxylic acid + ROH) 1B) Base-induced hydrolysis (RCOOR ---> Carboxylic acid + ROH) 2) Aminolysis (RCOOR ---> Amide + ROH) 3) FULL Reduction (RCOOR ----> 1° alcohol + ROH) 4) PARTIAL Reduction (RCOOR ---> Aldehyde + ROH) 5) Grignard rxn. (RCOOR ---> 3° alcohol + ROH

Question 43

Ester Acid-Catalyzed Hydrolysis: Reactants

Answer 43

RCOOR ---> Carboxylic acid + byproduct ROH **H₂SO₄** ---------------------> **H₂O

Question 44

Ester Acid-Catalyzed Hydrolysis: Process

Answer 44

1) H₂SO₄ protonates the carbonyl oxygen = gives (+) charge to the ester 2) H₂O acts as nucleophile and attacks carbonyl carbon (DB electrons push up to form stable OH grp) 3) Proton transfer: H⁺ transfers from the added H₂O⁺ to the bridge oxygen (-OR grp) to create ROH⁺ LG! 4) Electrons from oxygen in one of the two OH grps push down to reform carbonyl DB and pushes off ROH! 5) HSO₄^- deprotonates the protonated carbonyl oxygen == Carboxylic Acid + ROH byproduct!

Question 45

Ester Base-Induced Hydrolysis: Reactants

Answer 45

RCOOR ---> Carboxylic acid + byproduct ROH **NaOH** -----------------> **H₂O** (just there to solubilize NaOH)

Question 46

Ester Base-Induced Hydrolysis: Process

Answer 46

1) OH- of NaOH attacks carbonyl carbon as a nucleophile = alkoxide forms 2) Alkoxide electrons push down to reform carbonyl DB and push off OR^- = Carboxylic acid is formed! 3) Free OR^- is very basic and deprotonates the acidic OH on the formed carboxylic acid (acid-base rxn) == Carboxylate ion + ROH byproduct NOTE: Must hit with acid to reprotonate the carboxylate ion to get the carboxylic acid

Question 47

Ester Aminolysis: Reactants

Answer 47

Ester ---> AMIDE + ROH **Amine, <3°** ----------------->

Question 48

Ester Aminolysis: Process

Answer 48

1) Amine acts as nucleophile and attacks carbonyl carbon = alkoxide forms 3) Alkoxide electrons push down to reform carbonyl DB and pushes off OR^- 3) Free OR^- deprotonates the added amine⁺ to remove the (+) charge ==AMIDE formed! + ROH byproduct

Question 49

Ester FULL Reduction: Reactants

Answer 49

Ester ---> 1° alcohol + ROH **1. LiAlH₄** --------------------------> **2.H₃O⁺**

Question 50

Ester FULL reduction: Process

Answer 50

TWO reductions occur: **Rxn 1:** 1) AlH₄^- attacks the carbonyl carbon of the ester = alkoxide forms 2) Alkoxide pushes down electrons to reform carbonyl DB and pushes off OR^- == ALDEHYDE forms + OR^- byproduct (gets protonated later to produce ROH **Rxn. 2:** 3) Aldehyde gets immediately reduced by LiAlH₄ again to yield an alkoxide 4) Alkoxide is protonated by H₃O⁺ == 1° alcohol + ROH byproduct

Question 51

Ester PARTIAL Reduction: Reactants

Answer 51

Ester ----> Aldehyde + ROH **1. DIBAH** or **i-Bu₂Al-H**, **-78C in hexane** ----------------------------------------------------------> **2.H₃O⁺** DIBAH = Diisobutylaluminum hydride

Question 52

Ester PARTIAL Reduction: Process

Answer 52

1) DIBAH acts as H- source to form the aldehyde 2) Acid is used to protonate the OR- to give the byproduct ROH == ALDEHYDE + byproduct ROH

Question 53

Ester + Grignard Reagent: Reactants

Answer 53

Ester ---> 3° alcohol + ROH **1. 2x GRIGNARD (RMgX)** ----------------------------------------> **2. H₃O⁺**

Question 54

What conditions are needed for partial reduction using DIBAH in esters?

Answer 54

1) Organic solvent (Like toluene or hexane) 2) -78°C

Question 55

Ester + Grignard Reagent: Process

Answer 55

Rxn happens TWICE: 1) One grignard reacts with the ester to produce a KETONE + OR- 2) Ketone immediately reacts with second grignard to produce an alkoxide compound 3) Alkoxide is protonated by acid to yield the 3° alcohol (and OR- from the first grignard reaction is protonated to yield ROH byproduct) == 3° alcohol + ROH byproduct

Question 56

What are the methods of MAKING amides?

Answer 56

TWO main methods: 1) **Acyl halide aminolysis** (acyl halide + Amine (<3°) + Et₃N) 2) **Carboxylic acid and DCC!** (carboxy acid + DCC + Amine (<3°)) (No extra base needed)

Question 57

Amide Reactions:

Answer 57

Undergo THREE main reactions: 1A) Acid-Catalyzed Hydrolysis (Preferred) (amide --> RCOOH + amine) 1B) Base-Induced Hydrolysis (amide ---> RCOO- + amine) 2) FULL Reduction (Amide --> AMINE) (NO mechanism for partial reduction) (NO reaction with grignard!)

Question 58

Amide Acid-Catalyzed Hydrolysis: Reactants

Answer 58

Amide ---> Carboxylic acid + Amine **H₃O⁺** ----------------> **H₂O**

Question 59

Amide Acid-Catalyzed Hydrolysis: Process

Answer 59

1) Carbonly oxygen of the amide protonates from the acid = gives amide a (+) charge (attracts Nu) 2) H₂O acts as nucleophile and attacks the carbonyl carbon and pushes the DB electrons up to form a stable OH grp 3) PROTON TRANSFER: Nitrogen of the nitro grp. gets protonated by taking H+ from the added H₂O⁺ 4) Electrons from OH grp push down to reform carbonyl DB and push off the protonated nitro grp = releases an AMINE 5) H₂O deprotonates the carbonyl oxygen to get rid of the (+) charge and produce the carboxylic acid! == Carboxylic acid + AMINE

Question 60

Amide Basic Hydrolysis: Reactants

Answer 60

Amide ---> Carboxylate ion + Amine **NaOH** ----------------> **H₂O** (just to solubilize NaOH)

Question 61

Amide Basic Hydrolysis: Process

Answer 61

1) Hydroxide acts as nucleophile and attacks the carbonyl carbon = forms alkoxide 2) Alkoxide pushes electrons down to reform carbonyl DB and pushes off a (-) amine grp == Forms carboxylic acid! 3) The (-) amine grp comes back and deprotonates the OH grp of the carboxylic acid (acid-base rxn) == Carboxylate ion + AMINE

Question 62

Amide FULL Reduction: Reactants

Answer 62

Amide ----> Amine **1. LiAlH₄** --------------------------> **2.H₃O⁺**

Question 63

Amide FULL reduction: Process

Answer 63

Involves two reduction steps! **Rxn 1:** 1) H- attacks the carbonyl carbon to form alkoxide --> which immediately reacts with AlH₃ to form an OAlH₃ leaving group! 2) Electrons from nitrogen of the nitro grp push down to form DB with carbonyl carbon and in this process pushes off the OAlH₃ leaving group ==IMMINIUM ION forms (nitro group has a + charge) **Rxn 2:** 3) H- attacks imminium ion at the nitro CARBON, pushing the ntiro DB electrons UP to the nitro group = gets rid of the (+) charge and the amine is formed! == AMINE formed!