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OChem2 > Exam 3 > Flashcards

Exam 3 Flashcards

1
Q

When does nucleophilic addition occur on a carbonyl C?

A

When there is no electronegative atom Z on the carbonyl group

  • Ex: Aldehydes have an H group
  • Ex: Ketones have a CH group of some sort
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2
Q

When does nucleophilic acyl substitution occur on a carbonyl C?

A

When there is an electronegative atom Z on the carbonyl group

-Ex: Carboxylic acids, acid chlorides, etc

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3
Q

What is substituted on the alpha C (R group) of a carbonyl compound?

A

One of the hydrogens on the alpha C is replaced by the electrophilic group attacking the compound

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4
Q

Tautomerization (definition)

A

The process of converting one tautomer to another (Ex: keto to enol)

  • Catalyzed by both an acid and a base
  • Requires two steps (protonation and deprotonation)
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5
Q

Describe tautomerization in acid

A

-Protonation PRECEDES deprotonation

  • STEP 1: Carbonyl O deprotonates water (and is protonated itself)
  • STEP 2: -OH Deprotonates the alpha C-H, forming a double bond between the alpha and carbonyl C
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6
Q

Describe tautomerization in base

A

-Deprotonation PRECEEDS protonation

  • STEP 1: The -OH deprotonates the alpha C, forming a double bond between the carbonyl C and the alpha C, which pushes the pi bond in the carbonyl group up onto the O
  • STEP 2: Carbonyl O deprotonates H2O (therefore becoming protonated itself)
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7
Q

Does equillibirum favor the side with a weaker acidity or stronger?

A

Favors the side with a weaker acidity (higher pKa)

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8
Q

What does LDA stand for? What is it used for?

A

It is a very strong base (weak nucleophile due to steric hindrance)

It stands for Li(+)N(-)[CH(CH3)2]2

It is used to form enolate in 100% yield

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9
Q

What is an enolate?

A

A carbonyl with a negatively charged alpha C (in the R group)

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10
Q

How would you substitute an electrophile onto a carbonyl compound?

A

STEP 1: A strong base deprotonates the alpha carbon (in the R group)– this makes the enolate (the H is eliminated)

STEP 2: The negatively charged alpha carbon (in the enolate R group) attacks the electrophile, adding it to the alpha C

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11
Q

What happens when an unsymmetrical carbonyl compound is treated with a base?

A

Two enolates are possible

One is formed by removal of a 2’ carbon (first R group), the other formed by the removal of a 3’ carbon (other R group)

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12
Q

Does the removal of a 2’ hydrogen make a kinetic or thermodynamic product?

A
  • Removal of a 2’ hydrogen makes a kinetic product

- Faster, but less stable / higher in energy

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13
Q

Does the removal of a 3’ hydrogen make a kinetic or thermodynamic product?

A
  • Removal of a 3’ hydrogen makes a thermodynamic product

- Slower, but more stable / lower in energy

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14
Q

How do you favor the kinetic enolate?

A
  • Use a strong nucleophilic base
  • Polar aprotic solvent
  • Low temperature

Ex: Use LDA in THF at -78C

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15
Q

How do you favor the thermodynamic enolate?

A
  • Use a strong base
  • Protic solvent
  • Room temperature

Ex: Use -OH in ROH at room temp (25 C)

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16
Q

What solvent would you use for an acid halogenation?

A

X2
———–>
CH3COOH

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17
Q

Describe acid-catalyzed halogenation at the alpha carbon

A

i) Carbonyl O deprotonates the acid catalyst– The acid catalyst then deprotonates the R group on the carbonyl group (making a double bond b/w carbonyl C and alpha C)
ii) One of the lone pairs on the carbonyl O comes down, forcing the double bond b/w carbonyl C and alpha C attacks the X2
- –The negatively charged X(-) by itself now deprotonates the carbonyl O, making the final product

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18
Q

Describe base-catalyzed halogenation at the alpha carbon

A

i) The base deprotonates the alpha C, making it negatively charged
ii) The negatively charged alpha C attacks the X2 , finishing the reaction

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19
Q

Describe the formation of a haloform

A

**Only occurs with methyl ketones (ketones with a methyl R group)

i) A base (-OH) deprotonates the alpha C, then the negatively charged alpha C attacks the X2
- –This happens one more time
ii) OH adds to the carbonyl C, pushing the pi bond up to the O
iii) The pi bond comes back down and pushes off the R group (the CX situation)
iv) The CX is negatively charged, and it deprotonates the OH group, finishing the reaction

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20
Q

How would you add a pi bond into a carbonyl hexanone?

A

i) Substitute Br to the alpha C (Br replaces an H)
- –Br2
- ———->
- –CH3COOH

ii) Eliminate the Br to make a pi bond
- –Li2CO3
- ——->
- –LiBr, DMF

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21
Q

How would you add two aldehydes together to form an aldol product? How would you make this aldol product a condensation product?

A

i) Take two identical aldehydes, add -OH
ii) This results in ONE of the two carbonyl compounds becoming protonated, and the OTHER aldehyde loses an H in order to attach to the 1st aldehyde

Ex: Aldehyde + Aldehyde (Me-C(H)=O) –> Me-C(OH)(H)-CH2-C(H)=O

THEN, Add more OH and heat to get rid of the previously protonated OH, leaving an H in its place, and making a double bond b/w the first aldehyde and the C group bonding the two aldehyde groups

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22
Q

Mechanism of aldol synthesis

A

i) Base (OH) deprotonates methyl group on Aldehyde– this creates a double bond b/w the carbonyl C and the methyl C & pushes the C=O bond up to the O
ii) Pi bond of C=O comes back down, forces double bond b/w the carbonyl C and methyl C to attack the carbonyl C of the second aldehyde (which pushes the 2nd aldehyde’s pi bond up to its O)
iii) The negatively charged O on the second aldehyde deprotonates water

That’s it

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23
Q

Mechanism of aldol condensation

A

i) OH deprotonates the CH2 holding the two aldeyde groups together, makes the C there negatively charged
ii) That negative charge moves over to become a pi bond b/w the carbonyl C of the first aldehyde and the CH – This entirely pushes off the OH group on the first aldehyde

That’s it

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24
Q

What must you have in order to create an aldol from an aldehyde?

A

There needs to be an alpha Hydrogen on the R group of the aldehyde

—AKA there needs to be a hydrogen directly attached to the alpha carbon

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25
Q

Acetoacetic acid ester synthesis definition

A

A stepwise method for converting ethyl acetoacetate into a ketone having one or two alkyl groups on the alpha carbon

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26
Q

What does ethyl acetoacetate look like?

A

A carbonyl with the following R groups:

  • Methyl
  • C(H2)-COOEt
27
Q

What reacts in a Claisen reaction?

A

Two molecules of an ester react with each other in the presence of an alkoxide base (such as 1) NaOHEt 2) H3O+)

  • Claisen reactions are SUBSTITUTIONS
  • Only esters with 2 or 3 alpha hydrogens on the alpha carbon undergo this reaction
28
Q

Claisen Reaction Mechanism

A

i) Base deprotonates the CH3 (R-group) to make negatively charged CH2

ii) The negatively charged CH2 attacks the carbonyl C of the second ester, pushing the pi bond of the second ester up to its O
iii) The negative charge comes back down from O and pushes off the OR group of the second ester

iv) The base deprotonates the CH2 group holding together the two esters, making a negatively charged CH
v) Addition of a strong acid re-protonates the CH2

29
Q

What is formed in a Dieckmann reaction?

A
  • A five or six membered ring
  • It is an intramolecular Claisen Rxn
  • 1, 6 Diesters form 5-membered rings
  • 1,7 Diesters form 6-membered rings
30
Q

What is the very basic mechanism of a Dieckman reaction?

A

Ex: Using a 1,7 Diester

-The 1 and 6 carbons fuse, the non-CH2 R group is booted off of the 1 carbon, the R group remains on the 6 carbon

31
Q

Dieckman rxn using a 1,7 Diester Mechanism

A

i) The -OEt base deprotonates the 6-Carbon, making it negatively charged
ii) This negatively charged C attacks the carbonyl 1-Carbon, pushing the carbonyl C=O pi bond up to the O
iii) The pi bond comes back down and pushes off the OR group on the 1-Carbon
iv) To complete the reaction, the remaining H on the 6-C is removed by base, and then re-protonated by H3O+

32
Q

Michael Reaction Mechanism

A

i) The -OEt base deprotonates the alpha carbon of the carbon (r group) which contains multiple O groups
ii) This now negatively-charged alpha C attacks the Beta carbon of the second a-B-unsaturated carbon compound (ex: a benzene ring with a pi bond attached to the Beta carbon); This pushes the pi bond up to the alpha carbon
iii) The now-negatively charged alpha carbon of the unsaturated carbon compound will deprotonate the base which had deprotonated in the first step, this will complete the reaction

33
Q

Robinson Annulation Mechanism - Part 1

A

i) Base removes the most acidic hydrogen (the one between the two carbonyl groups) making that carbon negatively charged
ii) This negatively charged carbon attacks the unsaturated (pi bond) Beta carbon of a carbonyl compound, making the alpha carbon of the unsaturated compound negatively charged (the pi bond next to it bumps onto it)
iii) That negatively charged carbon deprotonates a water to become neutral

34
Q

Robinson Annulation Mechanism - Part 2 (Intramolecular)

A

i) -OH deprotonates the H on the only carbon remaining unattached at the end of the molecule, making that C negatively charged
ii) This negatively charged C attacks the carbonyl C of the other half of the compound, pushing the pi bond of that carbonyl up to its O
iii) The now negatively charged O deprotonates a water to become OH

iv) The -OH deprotonates the alpha C adjacent to the C which contains the OH group (the C that was deprotonated at the very beginning of the reaction), making it negatively charged
v) The negative charge becomes a pi bond between this carbon and the carbon containing the OH group, which pushes the OH group off, ending the reaction

35
Q

Are primary and secondary amines H-bond donors or acceptors?

A

1’ and 2’ amines are both donors and acceptors of H-bonds

36
Q

Are 3’ amines H-bond donors or acceptors?

A

3’ amines are only H-bond acceptors

37
Q

What four elements can undergo H-bonding?

A

F, Cl, O, and N

No other elements are e-negative enough

38
Q

How can an amine act as a base?

A

The lone pair on N will deprotonate an acid

39
Q

What three acids can protonate an amine?

A
  • HCl
  • H2SO4
  • Carboxylic acids

—Must be strong acids

40
Q

Is a more stable compound more acidic or more alkaline?

A

A more stable compound is more alkaline (less acidic)

41
Q

What two compounds are good Sn2 electrophiles?

A
  • 1’ Halide

- CH3I

42
Q

Can you mix a 1:1 ratio of amine and Sn2 electrophile?

A

NO!!

-You must have one or the other in excess in order to drive the reaction

43
Q

What does the Gabriel synthesis make?

A

Synthesizes 1’ amines via nucleophilic substitution

44
Q

What three reagents would you use to do a Hoffman Elimination starting with an amine?

A

1) CH3I
- ———->
2) Ag2O
3) Heat

45
Q

What two sets of reagents can you use to reduce a nitro group (NO2)?

A

H2, Pd/C
————->

OR

Fe, HCl
———–>

Makes 1’ amine

46
Q

What set of reagents can you use to reduce a nitrile?

A

1) LiAlH4
- ————->
2) H2O

Makes 1’ amine

47
Q

What set of reagents can you use to reduce an amine?

A

1) LiAlH4
- ————>
2) H2O

**Can be used for 1’, 2’, and 3’ amines

48
Q

What reagent do you use in gabriel synthesis?

A

Pthalamide

It’s a 5-membered ring with a nitrogen at the top (as part of the ring), which has a =O group on each C to either side of the N, then that ring is attached to a benzene ring

49
Q

Is the N-H bond of an imide acidic or basic?

A

Super acidic

50
Q

Are amines lewis acids or lewis bases?

A

Amines are lewis bases

–They cannot undergo Friedel-Crafts reactions

–Must add protecting group to let it undergo F.C.

51
Q

What reagents would you use to substitute an NH2 group into an OH group on a benzene ring?

A

H2O, Heat

—————>

52
Q

What reagents would you use to substitute out an NH2 group for a halide on a benzene ring?

A

CuCl

or

CuBr

or

KI

53
Q

What reagent would you use to substitute out an NH2 for an H on a benzene ring?

A

H3PO2

54
Q

What reagents do you use to create a diazonium salt from an NH2 group?

A

NaNO2
————>
HCl

55
Q

What does a diazonium salt look like?

A

N(triplebond)N(+) Cl(-)

56
Q

Basic mechanism of a Michael Reaction

A

Adds an H to the alpha carbon of a carbonyl

Eliminates an H from the Beta carbon of a carbonyl, and adds the deprotonated carbonyl from step 1 in its place

57
Q

Hydrolysis of a Nitrile in Base - Mechanism

A

i) Base (-OH) attacks C in R-C(triplebond)N – This pushes one of the pi bonds between C and N onto the N
ii) The now-negatively charged N deprotonates water to become neutrally charged (forms an imidic acid)

iii) Base deprotonates the OH on the C, which results in a negative charge on that O
iv) The negative charge on O comes down to form a pi bond b/w C and O, pushing the remaining pi bond b/w C and N onto the N and making it negatively charged
v) The negatively charged N deprotonates water to become neutrally charged

vi) H2O and -OH are added and this changes out the NH2 group for a negatively charged O group (so you end up with a carbonyl with one of its R groups as a negatively charged O)

58
Q

Reduction of a Nitrile with LiAlH4 and H2O - Mechanism

A

i) H3LiAl-H protonates the C in the nitrile group (R-C(triplebond)N), shoving one of the pi bonds b/w C and N onto the N and negatively charging it
ii) Negatively charged N bonds to AlH3
- —–Molecule: RHC=N-Al(-)H3

iii) H3LiAl-H protonates the C again, causing the remaining pi bond b/w C and C to attack another equivalent of LiAlH3
v) The addition of water will leave the molecule as follows:

RH2C-NH2

OVERALL: Both the C and the N have two hydrogens added to them and only a sigma bond remains between them

59
Q

When LDA is added to a methyl ketone and then a second compound is added to the mixture, what does it do?

A

i) LDA deprotonates the methyl R group on the ketone, leaving a -CH2
ii) The new group attaches at that point

60
Q

What reagents do you need for a Micheal Synthesis?

A

1) OR’ or OH
- —————–>
2) H2O

61
Q

What reagents do you need for a Claisen Synthesis?

A

1) NaOR’
- ————>
2) H3O+

62
Q

What reagents do you need for an Aldol Synthesis?

A

1) OH, H2O
- —————->
2) H3O+

63
Q

What reagents do you need for a Robinson Synthesis?

A

OH
——->
H2O

OChem2 (3 decks)

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