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Organic Chemistry > Chemistry Chapter 11: Organic Reaction Mechanisms > Flashcards

Chemistry Chapter 11: Organic Reaction Mechanisms Flashcards

(64 cards)

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1
Q

SN1 is what rate order

A

First rate order, depends on substrate concentration only

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2
Q

What are the steps of the SN1 mechanism?

A

Step 1: formation of a carbocation (slow)
Step 2: nucleophilic attack (fast)

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3
Q

Anything that increases the stability of the _________ in an SN1 reaction will increase the rate of reaction

A

Carbocation

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4
Q

In SN1, what is the substitutent on carbon?

A

Tertiary > secondary (primary won’t work)

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5
Q

How many steps in SN1?

A

2

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6
Q

Key mechanism in SN1

A

Carbocation stability

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7
Q

Strength on nucleophile for SN1

A

Doesn’t matter- can be weak

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8
Q

Solvent for SN1

A

Polar protic

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9
Q

Stereochemistry for SN1

A

Not preserved (racemic)

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10
Q

Key mechanism in SN2

A

Steric hindrance

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11
Q

Rate law dependence for SN2

A

Bimolecular (substrate and nucleophile concentrations)

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12
Q

Number of steps for SN2

A

1

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13
Q

Substituents on carbon (SN2)

A

Methyl (best) > primary > secondary

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14
Q

Strength of nucleophile for SN2

A

Strong (and non-bulky)

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15
Q

Solvent for SN2

A

Polar aprotic

Polar aprotic solvents are polar solvents that lack acidic protons and cannot form hydrogen bonds

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16
Q

Stereochemistry for SN2

A

Inverted (backside attack)

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17
Q

Solvents that can donate a proton

A

Polar protic (ie OH or NH) water, ethanol etc. polar protic engage in hydrogen bonding to the leaving group which stabilizes the carbocation

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18
Q

Solvent that contain a permanent dipole due to a polar bond (typically C=O) but can’t act as hydrogen bond donors

A

Polar aprotic solvents. Used in SN2 reactions, such as acetone and DMSO

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19
Q

In carboxylic acids, what is typically the electrophile and why?

A

C=O carbonyl carbons (because the oxygen has a partial negative charge and the C has a partial positive charge). The OH group makes the carbonyl carbons even more subject to nucleophile attack

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20
Q

Fischer Esterification

A

An example of nucleophile substitution at a carboxylic acid group.

An acid-catalyzed technique for turning a carboxylic acid into an ester.

You replace the OH group of the COOH carboxylic acid with the OR functional group characteristic of an ester

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21
Q

How to perform an Fischer Esterification

A

Carboxylic acid is mixed with an alcohol under acidic conditions

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22
Q

Steps of an Fischer Esterification

A

1) pronation of carbonyl O
2) nucleophilic addition by alcohol
3) Proton transfer x 2
4) OH2 group leaves
6) deprotonation by base
7) ester formed

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23
Q

Fischer Esterification can be reversed by

A

Hydrolysis

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24
Q

Imine formation is what type of reaction

A

Nucleophile substitution

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25
Describe Imine formation
Occurs at carbonyl carbon (NOT a carboxylic acid group) 1) protonation of carbonyl oxygen 2) nucleophilic attack of the nitrogen atom of the amine and joins carbonyl carbon and amine to form tetrahedral intermediate 3) Proton transfer from amine to OH group 4) N lone pair attacks to form double bond and H2O leaves 5) N is deprotonated to remove positive charge
26
Schiff’s base
An Imine with an additional carbon substituent (C=N-R)
27
Hemiacetals formed from
Nucleophillic addition reactions at carbonyl carbon (aldehydes)
28
Hemiketals are formed how?
Nucleophillic addition at carbonyl carbon of ketones
29
How to form acetals and ketals?
Repeat reaction with excess alcohol
30
Nucleophillic addition reactions
Nucleophilic attacks carbonyl carbon without a leaving group
31
Keto-enol tautomerism can be catalyzed by either
Acid or base, alpha hydrogen is critical in both
32
When is the alpha hydrogen removed in a base catalyzed mechanism vs acid catalyzed mechanism for keto-enol tautomerism?
First step for base, second step for acid
33
What is enolate chemistry?
Resonance stabilized negative charge on alpha carbon allows carbon to be a nucleophile
34
What are 4 reactions that involve enolates
Aldol condensation Retro-aldol Michael addition Robinson Annulation
35
Aldol Condensation
Nucleophilic alpha carbon attacks electrophilic carbonyl C to form a new C-C bond, forming aldol which then continues to react to form a alpha-beta unsaturated ketone
36
Retro-Aldol
Reverse of the Aldol condensation process
37
Michael Addition
An enolate attacks the beta carbon of a alpha, beta unsaturated aldehyde/ketone resulting a compound containing a1,5 di carbonyl
38
Robinson Annulation
Michael addition followed by an Aldol condensation
39
Steps of a base-catalyzed aldol condensation
1) Enolate attack 2) deprotonation 3) Reduction 4) Reprotonation
40
The substrate for SN2 reactions occur BEST if it’s
Methyl, primary or secondary
41
Secondary substrates would undergo
SN1 OR SN2
42
A double bond to an oxygen becomes a double bond to a nitrogen following a nucleophilic attack on a carbonyl carbon creates what
Imine
43
What is DMSO and what reactions is it used for?
DMSO is a polar aprotic solvent used in SN2 reactions
44
What is LDA and what is it used for?
LDA (lithium diisopropylamide) is a strong, bulky base. It is used because its steric hindrance minimizes unwanted side reactions, making it ideal for reactions requiring precise control, such as Kinetic enolate formation (favoring the less-substituted alpha-carbon)
45
How do you create a kinetic enolate?
-formed most quickly, less stable long term -rapid, irreversible -low temp ** use a strong bulky base like LDA -double bond formed on the less substituted
46
What are the characteristics of a thermodynamic enolate?
-slower to form but more stable long term -need higher temp -use a weak, less bulky base -reversible -double bond forms on the more substituted carbon
47
Methoxide (NaOCH3) is a strong or weak base?
A strong base
48
Base-catalyzed Aldol condensation involves an enol or enolate?
Enolate
49
Acid catalyzed adol condensation involves an enol or enolate?
Enol
50
If using a ketone for adol condensation, what do you have to look out for?
Multiple products form because the alpha hydrogens are not identical
51
What does a Michael addition result in?
Enolate attacks a alpha beta unsaturated ketone to form a compound with a 1,5- dicarbonyl group
52
What does the Robinson Annulation result in?
Michael addition followed by an Aldol condensation results in a alpha beta unsaturated ketone which is cyclic
53
Robinson Annulation is important in the formation of what
Steroid hormones because this reaction forms a ring structure
54
How do you perform a cross-Aldol reaction?
1) start with 1 aldehyde or ketone to create an enolate 2) add a strong hindered base such as LDA 3) add the second aldehyde or ketone
55
This is a mixed aldol condensation reaction The catalytic base will take THE MOST ACIDIC hydrogen (found on the alpha carbon between the 2 ketones- most acidic because the conjugate base is the most stable) The carbonyl carbon of the cyclopentanone will act as an electrophile, and undergo nucleophilic attack from the alpha carbanion forming a double bond after removal of the water from the intermediate. This gives us a cyclopentane double bonded to the carbon between the two carbonyls groups.
56
57
In aldol condensations, a bond is formed between which two carbons?
the alpha carbon of one molecule (bearing an enolate ion) reacts with the carbonyl carbon of another molecule to form a new carbon-carbon bond.
58
59
Enols vs enolates
Enolates have a charge
60
SN1 uses what kind of solvent and why?
Polar protic solvents because they stabilize the carbocation intermediate and solvate the leaving group, facilitating its departure.
61
SN2 uses what kind of solvent and why?
polar aprotic solvents because they do not solvate the nucleophile, keeping it strong and reactive for the backside attack.
62
When can an OH- group attack a double bond to O?
How OH⁻ Affects Different Carbonyls: Lactones, Esters, and Amides (Have Leaving Groups) → Break Up OH⁻ attacks the carbonyl → forms a tetrahedral intermediate. The carbonyl reforms, kicking out a leaving group (RO⁻ or NH₂⁻). Final Products: Carboxylate (-COO⁻) + Alcohol (for esters/lactones) or Amine (for amides). Ketones and Aldehydes (No Leaving Groups) → Stay Intact OH⁻ can attack, but no leaving group = carbonyl stays intact. Can form alcohols (if reduced) or enolates (if deprotonated). Key Rule: If a leaving group exists (esters, lactones, amides), OH⁻ breaks up the carbonyl. If not (ketones, aldehydes), the carbonyl remains.
63
is a bi-substrate enzyme reaction where the first substrate modifies the enzyme and leaves before the second substrate binds, preventing a ternary complex; examples include aspartate aminotransferase and serine proteases.
Ping pong mechanism
64
In this question Asp14 acts as a nucleophile. Why is the answer A?
Asp14 = acts as a nucleophile (therefore wants to attack an electrophile) phosphorous has a partial positive charge, therefore it is the electrophile Therefore, Asp attacked phosphorous

Organic Chemistry (4 decks)

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