Ochem 2 Flashcards

Question 1

Q

Aldehydes & Ketones
- Formation of Acetals/Hemiacetals and Ketals/Hemiketals
  - Describe the 4 steps of formation

Answer

A

An alcohol acts as the nucleophile
- Attacking the electrophilic carbonyl carbon and
- pushing the pi electrons from the C=O bond up onto the oxygen
The negatively charged oxygen is protonated to form an alcohol and

the original alcohol is deprotonated to form an ether
- This yields:
  - a hemiacetal if it was originally an aldehyde, or
  - a hemiketal if it was a ketone

The alcohol is protonated again to form the good leaving group water
* a second equivalent of alcohol attacks the central carbon
Deprotonation of the second alcohol results in another ether,

yielding:
- an acetal if it was originally an aldehyde
- or a ketal if it was a ketone

Question 2

Q

Acid (Acyl) Chloride
- Acid Chloride Formation
  - Formula=?
  - What 3 reagents readily produce acid chlorides when added to carboxylic acids?

Answer

A

RCOOH + PCl₃⇒ RCOCl + H₂O

Three reagents readily produce acid chlorides when added to carboxylic acids:
- PCl₃
- PCl₅
- SOCl₂

Question 3

Q

Acid Chloride
- Acid Chloride Formation
  - Addition of chloride ion (Cl-) to a carboxylic acid does NOT produce an acid chloride
  - Provide a possible explanation for WHY

Answer

A

A chloride ion IS capable of attacking the carbonyl carbon of a carboxylic acid
- However, when the electrons in the carbonyl bond are kicked up onto the oxygen, and then collapse back down, the substituent that is the best leaving group will leave
  - …regardless of the original structure of the molecule
Chloride ion is more stable (i.e., it is a weaker base) than hydroxide ion
- so the chlorine will be kicked off to reform the acid

Question 4

Q

Acid Chlorides (a.k.a.___)
- Definition
- Nomenclature
  - Common Names (3)

Answer

A

Definition:

An acid chloride (a.k.a. “acyl chloride”) is any compound containing:
- a carbonyl
  - with a chlorine substituent on the carbonyl carbon

Nomenclature:

Acid chlorides are named with the “–oyl chloride” suffix
- Ex: propanoyl chloride

Common Names:

Know the same three non-IUPAC names we’ve been highlighting for each functional group
1. Formyl chloride
2. Acetyl chloride
3. BenzOYL chloride

Take note that benZYL chloride is NOT an acid chloride

It is a chlorine attached to a benzyl group

Question 5

Q

Acid Chloride
- What property of Acid Chlorides make them IMPORTANT?
- What 2 things are the cause of this property?

Answer

A

Acid Chlorides are important because:
- they are the most reactive of the carboxylic acid derivatives
Their reactivity is due to:
1. The withdrawing power of the chlorine
  - which makes the partial positive charge on the carbonyl larger than normal
2. The fact that chloride ion is a superb LG

Question 6

Q

Acid Chlorides
- Provide a reactant that will form each of the following when reacted with an acid chloride
  1. An ester
  2. An amide
  3. An anhydride
  4. A carboxylic acid

Answer

A

ROH
RNH₂
RCOOH
H₂O

Question 7

Q

Aldehydes & Ketones
- General Characteristics
  - Why don’t aldehydes and ketones undergo substitution reactions?
    - ex: SN1, SN2

Answer

A

In order for a substitution to occur, there must be a LEAVING GROUP

Acid derivatives all have leaving groups:
- Cl^-in the case of acid chlorides
- -OH in the case of carboxylic acids
- a Carboxylate ion in the case of anhydrides, etc.
The stability of these groups after they leave varies widely, but in the case of an aldehyde or ketone:
- there are NO groups that would be reasonably stable,
- and therefore NO candidates to act as leaving groups
The aldehyde hydrogen will not leave as H:^- , nor will an R group leave from a ketone as a carbanion R:^-
In both cases, the leaving group would become a strong base
- Recall that strong bases** **NEVER** **make good leaving groups
  - Good leaving groups must be WEAK bases that are stable AFTER they leave
- For this reason, aldehydes and ketones only undergo addition reactions—lacking a suitable leaving group to undergo substitution

Question 8

Q

Aldehydes & Ketones
- General Characteristics
  - Aldehydes and ketones can also function as Lewis ___s, accepting electrons when WHAT happens?

Answer

A

Aldehydes and ketones can also function as Lewis ACIDS, accepting electrons when
- a base abstracts an alpha hydrogen

Question 9

Q

Aldehydes & Ketones
- Aldol Condensation
  - Describe
  - Explain the 3 steps

Answer

A

Aldol Condensation

The condensation of one aldehyde or ketone with another aldehyde or ketone

STEPS:

A base abstracts an alpha hydrogen, creating a carbanion
The carbanion will attack any carbonyl carbon in the solution
The oxygen is protonated to form an alcohol

Question 10

Q

Aldehydes & Ketones
- Aldol Condensation
  - What’s a common mistake students make with aldol condensations?
  - How can you avoid this mistake?

THE MCAT LOVES ALDOL CONDENSATIONS!

Answer

A

The common mistake seems to be for students to lose track of the carbanion carbon—often drawing a product with the two original carbonyl carbons adjacent to one another

To avoid this:

Always draw out the product
- Do NOT try to predict it in your head
Draw the carbanion nearby the carbonyl and then immediately draw a bond between them

In the product there should always be one carbon in between:

the carbonyl carbon
the carbon bearing the hydroxyl group

This could also be avoided by counting the total carbons before and after the reaction

Question 11

Q

Aldehydes & Ketones
- α-β Unsaturated Carbonyls
  - Describe
  - Is an α-β unsaturated carbonyl a base, a nucleophile, or an electrophile?

Answer

A

An aldehyde or ketone with a double bond between the alpha and beta carbons

In terms of the MCAT, you should think of an α-β-unsaturated carbonyl as an ELECTROPHILE
It might be tempting to think of the double bond between the alpha and beta carbons as a nucleophile that will undergo electrophilic addition
- However, the withdrawing effect of the carbonyl decreases the electron density of the double bond deactivating it toward electrophilic addition

Question 12

Q

Anhydrides
- Definition
- Nomenclature
- Common Names (3)

Answer

A

Definition:

An anhydride is a compound with two acyl groups connected to one another by a single oxygen
Viewed another way, an anhydride is an ester where the –R group is a carbonyl

Nomenclature:

Named by replacing the “-oic” ending of the corresponding carboxylic acid with “-oic anhydride”
- i.e., benzoic acid⇒benzoic anhydride
Mixed acid anhydrides are named alphabetically
- i.e., ethanoic methanoic anhydride

Common Names:

The MCAT will expect you to recognize:
1. formic anhydride
2. acetic anhydride
3. acetic formic anhydride
If it is a mixed anhydride made from:
- ethanoic acid and methanoic acid

Question 13

Q

Aldehydes & Ketones
- General Characteristics
  - Which is more acidic, the alpha hydrogen of a ketone, or the alpha hydrogen of an aldehyde?
  - Provide a possible explanation

Answer

A

The alpha hydrogen of an aldehyde is more acidic than a comparable alpha hydrogen on a ketone

because the conjugate base in the case of the aldehyde is more stable

In an aldehyde, a hydrogen is attached to the carbonyl carbon

Hydrogen is defined as neither a withdrawing group nor a donating group

However, in the case of a ketone, an –R group is attached to that same carbonyl carbon

and –R groups are weakly electron donating
- This will decrease the magnitude of the partial positive charge on the carbonyl carbon in the ketone
  - making it less able to stabilize the negative charge of the carbanion in the conjugate base

Question 14

Q

Aldehydes & Ketones
- General Characteristics
  - Substitution vs. Addition
    - Aldehydes & Ketones undergo…?
    - What 4 FG’s undergo nucleophilic SUBSTITUTION?

Answer

A

Aldehydes and Ketones undergo:

nucleophilic ADDITION

Carboxylic Acids
Amides
Esters
Anhydrides

undergo. ..
* nucleophilic SUBSTITUTION

Question 15

Q

Aldehydes & Ketones
- General Characteristics
  - MAJOR FUNCTION=?

Answer

A

ELECTROPHILES!

with their carbonyl carbon being attacked by Nu:’s

Question 16

Q

Aldehydes & Ketones

Halogenation of an Aldehyde or Ketone
- Describe
- List the 2 steps

Answer

A

Substitution of a Br, Cl or I for one of the alpha hydrogens on an aldehyde or ketone
Multiple halogenations often occur

STEPS:

A base abstracts an alpha hydrogen
- leaving a carbanion
The carbanion attacks a diatomic halogen (Br₂)

Question 17

Q

Aldehydes & Ketones
- Keto-Enol Tautomerization
  - Aldehydes and ketones cannot act as H-bond donors
    - An exception to this rule is 1,3-dicarbonyl compounds
      - They can act as hydrogen bond donors
  - Draw out a 1,3 dicarbonyl compound and propose an explanation

Answer

A

A 1,3-dicarbonyl can undergo an intramolecular hydrogen bond when:

one of the carbonyls is in the keto form and
the other is in the enol form
- This significantly stabilizes the enol compared to a stand-alone enol
In this condition:
- the enol is acting as the hydrogen-bond donor
- the carbonyl as the hydrogen bond acceptor

The MCAT loves alpha hydrogens so much, it wouldn’t be right to mention 1,3-dicarbonyls without also pointing out that they have ULTRA ACIDIC alpha protons on the carbon between the two carbonyl carbons

b/c there’s DOUBLE resonance stabilization!

Question 18

Q

Aldehydes & Ketones
- Keto-Enol Tautomerization
  - Draw a step-wise mechanism for the tautomerization

Question 19

Q

Aldehydes & Ketones
- Keto-Enol Tautomerization
  - Which is more stable, the keto or the enol tautomer?
  - Why?

Answer

A

The keto and enol forms are in an equilibrium with one another that strongly favors the keto form at room temperature

The keto form is more stable
- because the sum of its bond energies is greater than the sum of the bond energies in the enol form
The keto form has a C=O bond, a C-C bond, and a C-H bond
- that are replaced by a C-O bond, a C=C bond, and an O-H bond in the enol form
C-H and O-H bonds are quite close in bond energy
C=C has about 250 kJ/mol more bond energy than a C-C bond (almost double)
The real difference comes in the difference between a C-O bond and a C=O bond
- A C=O bond has about 450 kJ/mole more bond energy!
What you should know is that carbonyl bonds are much shorter and stronger than alkene bonds
- That is the most significant difference between the two forms and is the reason the keto form is favored.

Question 20

Q

Aldehydes & Ketones
- Nomenclature
  - There are a few common aldehydes and ketones for which the MCAT will use non-IUPAC names
    - Name the 4 we need to know
    - Hint: FABA

Answer

A

There are a few common aldehydes and ketones for which the MCAT will use non-IUPAC names
- These include:
  1. formaldehyde
    - HCOH
  2. acetaldehyde
    - CH₃COH
  3. benzaldehyde
    - C₆H₅COH
  4. acetone
    - CH₃COCH₃

Question 21

Q

Aldehydes & Ketones
- Nomenclature
  - What suffixes are they associated with each?
  - If a ketone must be named as a FG, what suffix does it use?
  - What must the parent chain contain?

Answer

A

Aldehydes

are named with the “–al” ending
Aldehyde carbons are always considered carbon #1 for numbering purposes

Ketones

are named with the “–one” ending
If a ketone MUST be named as a substituent, it is called an “-oxo” group
- as in 4-oxopentanal

In either case, the parent chain must be the longest chain that includes the carbonyl

Question 22

Q

Aldehydes & Ketones

Ketones are given the name “-oxo” as substituents
- What is an aldehyde named if it must be labeled as a substituent?
Aldehydes and ketones can ONLY be substituents when WHAT is present?
If the aldehyde or ketone is the TOP priority functional group:
- Which Carbon is labeled as C-1?

Answer

A

Surprisingly, substituent aldehydes are given the SAME “-oxo” name as ketone substituents!

There really should not be any confusion, however, because:

if the identified carbon is TERMINAL:
- it MUST be an aldehyde
  - and cannot be a ketone
and if it is SECONDARY:
- it MUST be a ketone
  - and cannot be an aldehyde

Remember that aldehydes and ketones can ONLY be substituents when:

there is a HIGHER priority functional group present*
such as a carboxylic acid

If the aldehyde or ketone is the *TOP* priority functional group:

then the carbonyl carbon is always labeled as C-1

Question 23

Q

Aldehydes & Ketones
- Physical Properties
  - Solubility & BP trends
    - Aldeyhdes and ketones can act as _____recipients, but NOT as____donors

Answer

A

Aldeyhdes and ketones can act as H-bond recipients, but NOT as H-bond donors

Question 24

Q

Aldehydes & Ketones
- Physical Properties
  - Solubility & BP trends
    - Use your knowledge of structure and function to predict the relative water solubility of aldehydes and ketones compared to comparable alkanes or alcohols
    - How will boiling point differ between these same species?

Answer

A

SOLUBILITY:

Alkanes

Alkanes are non-polar and therefore insoluble in water
Aldehydes and ketones can act as hydrogen-bond acceptors when dissolved in water, with water acting as the hydrogen bond donor
- Therefore, aldehydes and ketones will be far more soluble than alkanes
Finally, alcohols can hydrogen bond as both a donor and acceptor with water, so they will be the most soluble
- These trends assume comparable molecular weight, chain length, etc
  - This is important because a small alcohol such as methanol is water soluble, but dodecanol (big) is considered insoluble

BOILING POINTS

Alkanes

The boiling point of alkanes will be the lowest
- because their only intermolecular attraction would be van der Waals forces
Aldehydes and ketones do NOT hydrogen bond with one another
- but they are both polar and will therefore have much higher boiling points than alkanes
Finally, alcohols will have the highest boiling points
- due to intramolecular hydrogen bonding (Strong IMFs)

Question 25

Q

Aldehydes & Ketones
- Keto-Enol Tautomerization

Answer

A

Keto-Enol Tautomerization:

This is the process by which an alpha hydrogen adjacent to an aldehyde or ketone becomes bonded to the carbonyl oxygen,
……while the double bond is switched from the carbonoxygen bond to the bond between the carbonyl carbon and the alpha carbon

Question 26

Q

Aldehydes & Ketones
- α-β Unsaturated Carbonyls
  - What are the 2 possible ways to visualize this mechanism?

Answer

A

STEPS:

There are two possible ways to visualize this mechanism, based on which resonance form you start with:

With the double bond between the alpha and beta carbons, the nucleophile attacks the beta carbon
- pushing the double bond over one carbon and forcing the C=O electrons up onto the oxygen
With a carbocation on the beta carbon, the nucleophile simply attacks the beta carbon directly

Starting with either resonance form, the oxygen will get protonated to form an alcohol
Note that the protonated oxygen is really just the enol form of a keto-enol tautomer

Question 27

Q

Aldehydes & Ketones
- α-β Unsaturated Carbonyls
  - Draw two possible resonance structures for an α-β unsaturated carbonyl
  - Which one will be the major contributor to the resonance hybrid?

Answer

A

The one on the left is clearly the more significant contributor to the actual structure
- because it has no formal charges, compared to a charge separation in the structure on the right

Question 28

Q

Aldehydres & Ketones
- Define

Answer

A

Aldehyde

is any compound containing a carbonyl…
- with one or more hydrogen substituents on the carbonyl carbon

Ketone

is any compound containing a carbonyl…
- with two carbon substituents on the carbonyl carbon

Question 29

Q

Amide
- Properties
  - Among acid derivatives, amides are…?
  - Describe the reactivity of an amide’s carbonyl carbon

Answer

A

Amides are theMOST STABLE of all acid derivatives
- Their carbonyl carbons are UNreactive
  - This is because –NH₂ is NOT a good leaving group

Question 30

Q

Amide
- Properties
  - 1° and 2º amides can do what?
  - What about 3º amides?
  - What is the BIOCHEMISTRY connection here?

Answer

A

Primary and secondary amides can HYDROGEN BOND
- ∴ amides are water soluble as long as:
  - they lack long alkyl chains
Tertiary amides cannot H-bond

Biochemistry Connection:

Amide hydrogen bonding is perfectly illustrated in the secondary structure of proteins
- In an alpha helix every amine hydrogen forms an H-bond with the carbonyl four residues previous to it in the chain

Question 31

Q

Amides

Hoffman Degradation
- What happens to the Carbon Chain in this reaction?
- What does the mechanism include?
- What do you reach with what, to produce what?

Why is this reaction important?

Answer

A

Primary amides (amides with only hydrogens on the nitrogen) react in strong, basic solutions of Cl₂ or Br₂ to form primary amines
- The mechanism includes decarboxylation,
  - and thus SHORTENS (!!) the length of the carbon chain

This reaction is important because it allows you to ADD AN AMIDE TO A TERTIARY CARBON

Question 32

Q

Amides
- Physical Properties
  - The nitrogen of an amINE is normally sp³ hybridized
  - What is the expected hybridization of the nitrogen in an amIDE?

Answer

A

Because the nitrogen in an amide donates its lone pair via resonance, both the C-O and the C-N bond have double-bond character
- Therefore the hydbridization of the nitrogen will be closer to sp² than to sp³

Question 33

Q

Amides
- Physical Properties
  - What does Resonance (aka “Double Bond character”) do to amides?

Answer

A

Resonance (Double Bond Character) LIMITS ROTATION:

The lone pair on the amide nitrogen resonates with the carbonyl double bond
- …giving both the C-O and the C-N bonds double bond character
  - This prevents rotation

Question 34

Q

Amides
- Properties
  - Would you expect amIDES to be more or less basic than comparable amINES?
  - Why?

Answer

A

Nitrogen has less electron density in an amide than it would in a normal amine
- b/c of donation of the lone pair on the nitrogen into the conjugated system
  - Therefore, it will be LESS BASIC than comparable amines

One could also consider the effect of induction

The carbonyl carbon has a strong partial positive charge
- ….which will withdraw electron density from the amine through the sigma bond

Question 35

Q

Amides
- Definition
- Nomenclature

Answer

A

Definition

An amide is any compound containing a carbonyl with an amine substituent on the carbonyl carbon

Nomenclature

Named by replacing the “-oic” ending of the corresponding carboxylic acid with “amide”
- i.e., benzoic acid⇒benzamide

Question 36

Q

Amines
- Addition of Amines to Carbonyls
  - aka, the Formation of ___s and ___s
    - Amines add to ___ and ___ to form these ^^^
- Describe the 3 steps

Answer

A

Addition of Amines to Carbonyls….

aka Formation of ENAMINES and IMINES

Amines add to aldehydes and ketones to form imines and enamines

STEPS:

The amine acts as a nucleophile
- …attacking the electrophilic carbonyl carbon
The oxygen is protonated twice
- creating the good LG water
A base abstracts a hydrogen from the nitrogen and kicks off water in an E2 mechanism
- This forms either an imine or an enamine
  - (depending on the substitution pattern of the nitrogen)

Question 37

Q

Amines
- Addition of Amines to Carbonyls (Formation of Enamines and Imines)
  - What do 1°, 2°, and 3° amines yield, respectively?

Answer

A

Primary (1°) amines:
- yield IMINES
Secondary (2°) amines:
- yield ENAMINES
Tertiary (3°) amines
- DO NOT REACT

Question 38

Q

Amines
- Definition
- Properties
  - Amines can act as either…?
    - 1° and 2° amines usually act as?
    - 3° amines always act as?
      - Why?
  - Amines with 4 R groups act as…?
    - as long as…?
  - Amines are capable of…?

Answer

A

Definition:

An amine is any organic compound that contains a basic nitrogen atom

Properties

Amines can act as either bases or nucleophiles
- Primary or secondary amines usually act as nucleophiles
- Tertiary amines ALWAYS act as bases
  - (because they are too sterically hindered to act as nucleophiles)
Amine Basicity:
- Basicity decreases from tertiary to secondary to primary to ammonia due to the electron donating effects of the R- groups
- Amines attached to aromatic rings are significantly less basic than standard amines
Amines with four substituents act as *ELECTROPHILES*
- as long as they have at least one hydrogen
- ex: Ammonium, NH₄⁺
Amines are capable of *hydrogen bonding*

Question 39

Q

Amines
- Gabriel Synthesis
  - Describe the steps
  - Draw the e^- pushing mechanism

Answer

A

STEPS:

The phthalimide ion (a reactive species with a full negative charge on the nitrogen) acts as a nucleophile
- …attacking the alkyl halide
  - via SN₂

Question 40

Q

Amines
- Gabriel Synthesis
  - = the formation of what from what?
  - What does this rxn AVOID?

Answer

A

Formation of a 1° amine from a 1° alkyl halide

Avoids the side products of alkyl amine synthesis

Question 41

Q

Amines
- Properties
  - Which functional group forms stronger hydrogen bonds, alcohols or amines?

Answer

A

ALCOHOLS will form stronger hydrogen bonds
- because there is a greater difference in electronegativity between oxygen and hydrogen than there is between nitrogen and hydrogen
This greater dipole will create a stronger electrostatic attraction
- …and therefore a stronger hydrogen bond

Question 42

Q

Amines
- Nomenclature
  - It’s UNIQUE for amines
  - Describe the 4 steps for naming amines
    - Hint: It does matter whether the amine is primary, secondary, etc.

Answer

A

Naming amines is a bit unique, so be sure you understand the system:

Name the alkane to which the N is attached e.g., propane)
Add “amine” in place of the “e” on the end of “ane” (e.g., propanamine)
- It is also acceptable to separate the substituent name (e.g., propyl amine)
If the amine is secondary, the longest chain is included in the name as indicated above
- The other chain is added at the beginning, proceeded by the letter “N-“
  - e.g., N-ethylpropanamine
If the amine is tertiary or quaternary, add additional substituents to the front of the name in alphabetical order, all with the prefix N- included

e.g., N,N-diethylpropanamine, or N-ethyl-Nmethylpropanamine, or N,N-dimethyl-N-ethylpropanamine (ATTACHED)

Question 43

Q

Amines
- Reduction Synthesis of Amines
  - Describe the reduction of Nitro and Nitrile groups
  - What do they both get REDUCED TO?
  - What do most O-Chem books focus on these groups being reduced by?

Answer

A

Nitro groups

can be reduced to the associated primary amine
- via all of the listed reducing agents
  - e.g., LiAlH₄, NaBH₄, and H₂/catalyst with pressure
Most O-Chem books focus on nitro groups being reduced by:
- metals in HCl M•HCl
  - M=Fe, Zn, Sn

Nitrile groups

can be reduced to the associated primary amine
- via all of the above listed reducing agents
Most O-Chem books focus on nitriles being reduced by:
- LiAlH₄

Question 44

Q

Amines
- Reduction Synthesis of Amines
  - Describe the reduction of Imine and AmIDE groups
    - What are both reduced TO?
    - What do most O-chem books focus on these groups being reduced BY?

Hint: Amides can only be reduced with ONE reducing agent….

Answer

A

Imines

can be reduced to the associated 1° amine
- via all of the above listed reducing agents
- O-Chem books focus on Imines being reduced by:
  - BH₃
  - ^-:CN or
  - H₂ / catalyst

Amides

reduce to the associated 1° amine
- via LiAlH₄ ONLY

Question 45

Q

Amines
- Tautomerization
  - Define
    - what is it analogous to?
  - Draw

Answer

A

Tautomerization

An enamine and imine interchange
- via a proton shift

Analogous to the keto-enol tautomerization

Question 46

Q

Amines
- Amine Basicity
  - Offer a plausible explanation for the decreased basicity of aromatic amines

Answer

A

Aromatic amines are less basic because they donate their electron pair into the ring
- forming a conjugated system with the ring

Question 47

Q

Amines
- Addition of Amines to Carbonyls (Formation of Enamines and Imines)
  - Provide a possible explanation for why tertiary amines DO NOT REACT with carbonyls
    - Hint: Attempt to draw out a mechanism

Answer

A

Tertiary amines are not good nucleophiles
- because they are TOO STERICALLY HINDERED
They are more likely to act as a base
Even if they were to attack the carbonyl carbon, this would form:
- an unstable quaternary amine
  - with a full positive (+) formal charge
- This would be a better LG than the water formed by protonation of the carbonyl
  - …and would therefore be kicked back off anyway

Question 48

Q

Amines
- “Reduction Synthesis of Amines”
  - …Is the reduction of what 4 groups?
    - using common reducing agents, such as? (3)

Answer

A

Reduction of:
1. AmIDES
2. Imines
3. Nitriles
4. Nitro groups
using common reducing agents such as:
- LiAlH₄
- NaBH₄
- H₂/catalyst with pressure.

Question 49

Q

Amines
- Synthesis of Alkyl Amines
  - Describe
  - Formula=?
  - Describe the 2 Steps
  - Remember that this rxn results in many side products
    - Why?

Answer

A

Formation of an alkylamine from an amine and an alkyl halide

NH₃ + CH₃Br ⇒ NH₂CH₃ + HBr

STEPS:

Ammonia acts as a nucleophile, attacking the alkyl halide via SN₂
- …and kicking off the halide ion
The halide ion acts as a base, abstracting a hydrogen
- This quenches the charge on the nitrogen

NOTE:

This reaction results in many side products

because the resultant amine is still a good nucleophile and can react again

Question 50

Q

Amines
- _____synthesis is the most common human-body example of an amine acting as a nucleophile

Answer

A

Protein Synthesis

the amine acts as a nucleophile
- attacking the carbonyl carbon of another amino acid to form a peptide bond

Question 51

Q

Anhydrides
- Properties
  - Anhydrides are excellent _____

Answer

A

Anhydrides are excellent ELECTROPHILES!

The two carbonyl carbons are highly reactive to nucleophiles
- because the leaving group is a resonance-stabilized carboxylate ion

Question 52

Q

Carboxylic Acids
- Decarboxylation
  - Draw a mechanism for the base-catalyzed decarboxylation of a β-keto acid

Answer

A

Without a base catalyst, this mechanism can also be visualized as a 6- member, concerted, “ring-like” intramolecular reaction
- ….that does not require protonation of the enolate ion

Question 53

Q

Carboxylic Acids
- Describe “Decarboxylation”
  - What is lost, what is left behind?
  - What does the process usually require?

Answer

A

Decarboxylation

The loss of a CO₂ molecule from a beta-keto carboxylic acid
- leaving behind a resonance-stabilized carbanion
The process usually requires catalysis by a base
The carboxylate ion usually retakes the hydrogen from the base, forming a keto-enol tautomer

Question 54

Q

Carboxylic Acids
- Esterification
  - Define
  - Formula=?
  - What is required 1^stin order for this rxn to proceed?
  - How can you get higher yields?
  - Esterification is how ____s are formed

Answer

A

Reaction of an alcohol with a carboxylic acid to form an ester (ROR)

RCOOH + ROH ⇒ RCOOR + H₂O

The hydroxyl group will never leave without being protonated first
- ….to form the “good leaving group water”
  - thus this reaction requires an acid catalyst
Higher yields can be obtained by reacting an anhydride with an alcohol

This is how triacylglycerols are formed:

A glycerol undergoes esterification with three fatty acids

Question 55

Q

Carboxylic Acids
- Nucleophilic Attack of Carbonyls
  - Formula=?

(ex: __+__⇒__)

Answer

A

RCOOH + H₂O ⇒ RCOOH₂⁺ + Nu:^-⇒ RCONu + H₂O

Question 56

Q

Carboxylic Acids
- Physical Properties
  - Surprisingly, short-chain carboxylic acids are also soluble in many relatively non-polar solvents
    - Such as chloroform (even though they are clearly polar)
  - Provide a possible explanation for this observation

Answer

A

Theoretically, the carboxylic acid dimer (pictured below) would have no net dipole moment

This explains its solubility in non-polar solvents

Question 57

Q

Carboxylic Acids
- Describe the “3 Key Features of Carboxylic Acids”

Answer

A

Resonance Stabilization
- The carboxylate ion is uniquely stable due to resonance stabilization
Induction
- Alpha substituents can either donate or withdraw from the carboxylate ion
  - increasing or decreasing acidity
    - REMEMBER: To predict acidity examine the stability of the conjugate base!
      - (A principle applicable to any acid)
Hydrogen Bonding
- It is worth mentioning twice; don’t forget that carboxylic acids can not only hydrogen bond, but do it twice—to form dimers

Question 58

Q

Carboxylic Acids
- Definition
- Nomenclature
- What are some common acids whose non-IUPAC names you should know? (3)

Answer

A

A carboxylic acid is any compound containing a carbonyl with a hydroxyl substituent on the carbonyl carbon

NOMENCLATURE

Carboxylic acids are named with the “-oic acid” ending
A carboxylate ion is the result of abstraction of a proton
- leaving a negative charge on the oxygen
Carboxylate ions are named with an “-ate” ending
- e.g., formic acid ⇒formate
If it is a salt formed between the carboxylate ion and a metal:
- name the metal first
- then the ion
- e.g., benzoic acid⇒benzoate⇒sodium benzoate

Common Names:

There are few common acids for which the MCAT will use non-IUPAC names
These include:
1. Formic acid (HCOOH)
2. Acetic acid (CH₃COOH)
3. Benzoic acid (C₆H₅COOH)

Question 59

Q

Carboxylic Acids
- Physical Properties
  - Describe BP & Solubility properties

Answer

A

Carboxylic acids have very high boiling points
- due to their ability to form strong dimers involving two hydrogen bonds
Without long alkyl chains, they are soluble in water
Surprisingly, short-chain carboxylic acids are also soluble in many relatively non-polar solvents
- Such as chloroform
  - (even though they are clearly polar)

Question 60

Q

Define “Electrophiles”

Answer

A

“Electron lovers”

is attracted to e^-s / e^- rich centers
Usually POSITIVELY CHARGED

Question 61

Q

Describe Nucleophilic Addition
- What reacts with what?
- What happens as a result? (2)

Answer

A

Is an addition reaction
Here, a compound with a double or triple bond (aka one that has 1 or 2 π bonds)
- …reacts with electron-rich reactant (“nucleophile”)

As a result:

Disappearance of the double bond
Creation of two new single (“σ”) bonds

Question 62

Q

Describe Nucleophilic Substitution (both kinds)

Answer

A

Here, an electron nucleophile selectively bonds with (or attacks) the positive or partially positive charge of an electrophile or a group of atoms

…to REPLACE** a so-called **“leaving group”

Question 63

Q

Esters
- Describe “Saponification”
  - How does it utilize an ester?
  - What 2 things does it yield?
  - Describe the 3 steps

Answer

A

Saponification *(HYDROLYSIS OF AN ESTER)*

Hydrolysis of an ester to yield:
1. an alcohol
2. the salt of a carboxylic acid

STEPS:

The hydroxide ion (NaOH or KOH) attacks the carbonyl carbon and pushes the C=O electrons up onto the oxygen
The electrons collapse back down and kick off the –OR group
Either the –OR group, or hydroxide ion, abstracts the carboxylic acid hydrogen, yielding a carboxylate ion
- This associates with the Na+ or K+ in the solution to form “soap”

Question 64

Q

Esters
- INorganic Esters
  - Familiar Examples
    - ATP, GTP and UTP are examples of inorganic ______ esters
    - FADH₂ and NADH are examples of ______ esters
    - FMN, DNA and RNA are examples of ______ esters

Answer

A

Familiar Examples:

ATP, GTP, UTP, etc. are examples of inorganic triphosphate esters
FADH₂ and NADH are examples of inorganic diphosphate esters
FMN, DNA and RNA are examples of inorganicmonophosphate esters

Answer 64

A

Esters act as H-bond recipients, but NOT donors

Without long alkyl chains:
- they are SLIGHTLY soluble in water
- but LESS soluble than acids or alcohols

Answer 65

A

Reaction of an existing ester with an alcohol, creating a different ester
Also requires acid catalysis

RCOOR_a + R_bOH⇒RCOOR_b + R_aOH

Answer 66

A

Formation of a ketone from a β-keto ester

STEPS:

A base abstracts the acidic alpha hydrogen, leaving a carbanion
The carbanion attacks an alkyl halide (R-X), resulting in addition of the –R group to the alpha carbon
Hot acid during workup causes loss of the entire –COOR group

Answer 67

A

Acetals/ketals

have two –OR substituents and

Hemiacetals/hemiketals

have one –OR substituent plus one alcohol substituent (-OH group)

The MCAT will sometimes refer to both hemiacetals and acetals as simply “acetals”

On other occasions they have differentiated the two
Be aware of both conventions

Answer 68

A

Definition:

An ester is any compound containing a carbonyl with an –OR group substituted on the carbonyl carbon

Nomenclature:

Esters are named with the “–oate” ending, with the R portion of the –OR group named and placed in front of the name
- ex: “methyl” pentanoate

Common Names:

The common name applies to the second half of the name
The first portion of the name will be different for different R groups
- We will use methyl for all three examples:
  1. methyl formate
  2. methyl acetate
  3. methyl benzoate

Answer 69

A

Ketones & Enols

Answer 70

A

the LEAST POLAR

Answer 71

A

Used to isolate a specific molecule or product based on:

a VERY SPECIFIC affinity or binding interaction

To elute, the bound target molecules must disrupt the binding interaction

This could be accomplished via:

a SALT SOLUTION, but often requires
a chemical REVERSAL OF THE RXN that BOUND the target TO the column

Answer 72

A

The mixture to be separated is passed through a column packed with charged glass beads
- or some other polar matrix
The solution is collected in fractions at the bottom of the column
- i.e., the collecting tubes are changed at regular intervals

In column chromatography, assuming a polar matrix, the MOST NON-POLAR substances will elute first

followed by increasingly more polar substances

Answer 73

A

Distillation:

To separate two substances by simple distillation they must have boiling points that are at least 25˚C apart

Simple Distillation
Fractional Distillation
Vacuum Distillation

Answer 74

A

How it Works:

The desired product (which still contains impurities) is dissolved in the minimum amount of hot solvent necessary to create a saturated solution
The solution is then cooled as slowly as possible
Because pure substances usually crystallize at a higher temperature than impure substances,
- if the temperature is dropped just below the melting point of the product:
  - the crystals that form will be product
  - the impurities will remain in solution

This is rarely a perfect process, but repeated cycles can produce an increasingly pure product

Answer 75

A

The column or stationary phase is coated with cations or anions
The mixture is passed through and oppositely-charged ions adhere to the column
The target molecules can then be eluted
- by washing with a salt solution

Answer 76

A

Heat the mixture in a flask
- the liquid with the lower boiling point evaporates first,
  - enters a collecting arm,
  - cools,
  - and drops into a collecting flask

To improve separation process:

Cool the condensing arm or
Place the collecting flask on ice

Answer 77

A

A fractionating column is placed between the heating flask and the condensing arm
The mixture is heated to slightly above the boiling point of the more volatile liquid
The gas rises through a column of glass beads or metal shards
- This causes any impurities in the vapor (i.e., molecules from the liquid with the higher boiling point) to condense and fall back into the flask
  - resulting in a better separation
The more volatile component will be above its boiling point
- and therefore will NOT condense
This approach allows separation of compounds with boiling points less than 25˚C apart

Answer 78

A

How it Works:

The two solvents used will ideally have widely different polarities and densities, and will therefore create a distinct line of separation
- If the target product is already dissolved in a solvent that fits these requirements, it can serve as one of the two immiscible liquids
- If not, the mixture containing the target will be added to the separatory funnel along with two other immiscible solvents

The contents of the funnel are gently mixed or swirled
The mixture is then allowed to sit until the two solvents are fully separated
- If the target is a polar compound that was synthesized in a non-polar solvent then the target should easily move into the polar layer during mixing (like dissolves like)
- If the target is a non-polar product it should readily move into the non-polar solvent
After the two layers are fully separated, the bottom layer (usually aqueous) is drained out through the stopcock
The solvent layer containing the product is then evaporated to obtain the target

Answer 79

A

The separation can be aided by one or more of the following:

Repetition:
- Rarely is a full separation obtained after only one cycle
Fractional Extraction:
- Extracting 5mL ten separate times will produce a much better separation than extracting 50mL all at once
Addition of an acid
- to protonate the product
Addition of a base
- to deprotonate the product

Answer 80

A

An acid would protonate bases in the mixture,
- potentially adding a formal charge and dramatically increasing solubility in the aqueous layer
  - i.e., protonated amines
Bases could similarly abstract protons to create a negatively charged species—with the identical impact upon solubility
- i.e. deprotonated carboxylic acids

Answer 81

A

Mixing TOO vigorously
- This can result in the formation of emulsions that are difficult to separate
Reactive solvents
- The chosen solvents should not be reactive with:
  1. Each other
  2. The targeted product
High boiling point
- If a solvent with a high boiling point is used, it will be difficult to evaporate off the solvent to obtain the product

Answer 82

A

Purpose:

Used to:
1. Separate two compounds
2. To remove a desired product from a reaction mixture

Two immiscible (“un-mixable”) liquids, a polar (aqueous) and a non-polar (organic) are used

The separation depends on the target molecule having differential solubility in the two solvents

Answer 83

A

Gas Chromatography:

A liquid is used as the stationary phase
The mixture is dissolved into a heated gas and then passed through the liquid
Various components reach the exit port at different rates based on
1. Boiling point, and
2. Polarity
- Only consider polarity if the two substances have almost identical boiling points

On a gas chromatograph, there will be one peak for each unique compound in the mixture

The height of the peak in a gas chromatograph is relative to the ABUNDANCE of that component

Answer 84

A

Physical separation of a solid (either crystallized product or solid impurities) from a liquid by passing it through filtration paper
- Fluted filter paper is recommended

Fluted filter paper increases the surface area for filtration

Answer 85

A

Whether you would want to filter hot or cold would be determined by the conditions

If you are filtering out a solid impurity, filtering hot could prevent your product from crystallizing on the filter paper
However, if you are close to the freezing point of the solid then filtering hot could bring impurities back into solution
In other cases you may be trying to capture product that crystallized upon cooling of your reaction mixture
In that case you would certainly want to fitler cold or else you would redissolve your product

Answer 86

A

Notice how the O-H and C=O absorbances OVERLAP

Answer 87

A

The bond roughly follows Hooke’s Law
- and either atom can be compared to the mass
A STRONGER bond
- is analogous to a spring with a larger spring constant (k)
- (∴ a weaker bond would be analogous to a spring with a smaller spring constant)
The MW of the atom
- is analogous to the size of the mass on the spring

PER HOOKE’S LAW:

Stronger spring (larger k) increases oscillation frequency according to:

f = ½π √k/m*
A weaker bond would therefore vibrate at a lower frequency

A smaller atom *(lower MW)* has ahigher** vibrational frequency

∴ a larger atom has a lower vibrational frequency

Answer 88

A

The Vibrational Frequency is determined by:

1) Strength of the bond
2) MW of the bonded atoms

IR absorbances do NOT represent the entire molecule

They only represent a single polar bond in the molecule

Answer 89

A

Notice how shallow the amine absorption is

Answer 90

A

Notice how sharp and deep the absorbance for the C-N triple bond (nitrile) is

Answer 91

A

Carbonyl, C=O
- 1700 cm^-1
- sharp, deep
Alcohol, OH
- 3300 cm^-1
- broad, separate from CH
Saturated Alkane, CH
- 2800 cm^-1
- sharp, deep
Carboxylic Acid, OH
- 3000 cm^-1
- broad, overlaps CH
Amine, NH
- 3300 cm^-1
- broad, shallow
Amide, NH
- 3300 cm-1
- broad, deep
Nitriles, C≡N
- 2250 cm^-1
- sharp, deep

Answer 92

A

notice the OH stretch

Answer 93

A

IR (Infrared) Spectroscopy

How it Works:

If a bond has a DIPOLE, exposing that bond to an external electric field will cause the atoms to move within that field
- much like a charged particle attached to a spring
Infrared radiation is used to create an oscillating electric field
- which in turn causes dipolar bonds to oscillate at a specific “vibrational frequency”
To produce an IR spectrum, we slowly vary the frequency of the IR radiation.
When the IR radiation exactly matches the frequency of vibration of a particular bond, that bond is said to be in “resonance”
- and it will absorb some of the IR energy
  - This absorbance is what is picked up by the detector
A bond with no dipole will NOT be detected by IR

Answer 94

A

The sample is usually a mixture of components which is spotted onto the paper or TLC plate near its bottom edge
The top of the solvent should always be below this line where the sample is spotted
- so that the mixture does not simply dissolve into the solvent

Answer 95

A

R_f is a ratio of the distance traveled by the component over the distance traveled by the solvent

So a number close to one means that the polarity of the component is similar to that of the solvent
- Because non-polar solvents are normally used, the component is therefore relatively non-polar

Answer 96

A

Smears are usually caused when the solvent is unable to dissolve and carry the components as it moves up the paper via capillary action
- This can be because too much of the sample was spotted onto the paper and the solvent could not dissolve ALL of the sample
It can also occur if there is too big of a difference in polarity between the sample and the solvent
- If the polarities are too distant the solvent cannot dissolve the sample well enough to carry it
- By a similar token, if they are too similar, the components may never come out of solution to form a spot and be carried all the way to the top of the paper

Answer 97

A

Higher and sharper melting points indicate better purity
Impurities lower melting point
- They also broaden the range across which the crystals melt

Answer 98

A

How it Works:

The energy difference (in Joules) between two adjacent molecular orbitals, on average, just happens to be about the same amount of energy created by electromagnetic radiation in the UV spectrum
- Thus, when UV radiation is shone on a molecule, electrons within that molecule will often absorb that energy and “excite” to the next highest energy level
- This absorbance is recorded on a UV spectrum

Answer 99

A

Molecules containing only single bonds show low or no UV absorbance
Double and triple bonds absorb UV strongly
- double
Conjugated systems absorb UV even more strongly than do isolated double or triple bonds.
- The greater the degree of conjugation, the farther to the right the species will absorb
  - i.e., at a higher wavelength
A UV spectrum is a graph of absorbance vs. wavelength (nm)

Answer 100

A

Vacuum filtration is performed with a Hirsch or Buchner funnel
A vacuum is created inside of the flask which creates suction to pull the filtrate through the filter paper
- The filter usually has holes in it which are covered by the filter paper

The primary advantage is that it is FASTER than gravity filtration

Answer 101

A

Paper or Thin-Layer Chromatography (TLC):

Paper chromatography uses paper as the stationary phase.
Thin-Layer Chromatography (TLC) is nearly identical
- …but uses manufactured glass or plastic sheets coated with silica, alumina, etc.

R_f =

distance traveled by component, OR
distance traveled by solvent

Answer 102

A

Vacuum Distillation

The air inside the apparatus is evacuated to create a vacuum

Vacuum filtration is used because it dramatically lowers the boiling point,
- allowing you to work at more manageable temperatures with substances that have much higher boiling points at atmospheric pressure
This observation is supported by the fact that liquids can be said to boil at the point where vapor pressure equals atmospheric pressure
- By creating the vacuum, we essentially drive down the atmospheric pressure to meet the vapor pressure

Answer 103

A

Separation of one or more compounds by dissolving them in a “mobile phase” and then passing that phase through or across a “stationary phase”

The substances in the mobile phase interact to varying degrees with the stationary phase based on their polarity
The more interaction that occurs between the two phases
- the SLOWER the substance will move
Substances that do not interact with the stationary phase (or do so to the least degree), will move the FASTEST

Answer 104

A

A linear increase in the magnetic field will create a linear increase in the force exerted on the particles according to:

F = qvBsinθ

For a curved flight tube with a detector at the far end, in order for any particle to strike the detector:
- It must experience exactly the right centripetal acceleration to trace the precise curvature that results in striking the detector
  - As the centripetal force increases, this will occur for increasingly massive particles

The graph below shows the linear relationship between magnitude of the B field and mass of particles at the detector

Answer 105

A

The Height of each peak
- gives the relative abundance of that fragment
The Parent peak
- represents the original molecule minus one electron
  - a.k.a. “Molecular Ion Peak”
The Base peak
- is the most common fragment
  - is usually the most stable fragment generated
  - It is defined as existing at 100% relative abundance
    - In other words, it will be the highest peak on the spectra and the height of all other peaks will be a function of how abundant THAT fragment is compared to the base peak

Answer 106

A

C¹³-NMR Absorbances you should know:

C – C
- 0-50ppm
C – O
- 50-100ppm
C = C
- 100-150ppm
C = O
- 150-200ppm

Somewhat Analogous to IR:

Whereas H-NMR peaks represent unique hydrogen environments, C¹³-NMR peaks represent carbon functional groups
- This is more similar to IR, where the number associated with the absorbance helps you predict the functional group
In H-NMR we do not memorize numbers because although they DO tell us the relative degree of shielding, they DON’T directly correlate with specific functional groups

Answer 107

A

MAGNETIC FIELD, B

Answer 108

A

All nuclei with an odd atomic or mass number have what is called “nuclear spin”
- —a concept analogous to electron spin
When an external magnetic field is applied to a group of nuclei they will align their own magnetic fields with the direction of the external field
Then, if photons are shone onto the nuclei, some of the nuclei will be able to absorb this photon energy and flip their orientation
- so that they are lined up AGAINST the external field
  - It is THIS absorbance of energy that is picked up by NMR
- Different nuclei will require a different frequency photon (i.e., different amount of energy) to cause this flip in orientation
  - These differences are caused by the degree to which each nucleus is shielded by neighboring hydrogens

Answer 109

A

Absorbance Range: 0 – 12 ppm
- 12 ppm is Downfield = Deshielded
- 0 ppm is upfield = shielded
The reference compound tetramethyl silane
- Si(CH₃)⁴, or “TMS”
- is defined as 0 ppm

Answer 110

A

Important Differences:

NO spin-spin splitting
- i.e., all peaks are singlets
NO integration
- i.e., area under the curve does NOT indicate the relative number of carbons

Absorbance Range: 0 – 220 ppm
- 220 ppm is Downfield = Deshielded
- 0 ppm is upfield = shielded

Reference compound= Tetramethyl silane

“TMS,” or Si(CH₃)⁴
is defined as 0 ppm

Answer 111

A

FALSE.

C¹³-NMR on a large steroid would NOT detect every single carbon
- because it only detects the carbon-13 isotope
  - which has a relative abundance of only about 1%

Answer 112

A

Peaks:

Each peak represents all of the hydrogens in a molecule that share an indistinguishable chemical environment
- These are called “equivalent hydrogens”

Spin-spin splitting:

The presence of “neighbors” (non-equivalent hydrogens attached to a neighboring carbon) causes splitting of the peak
The peak for a set of equivalent hydrogens will be split into exactly n + 1 sub-peaks
- where n is the number of non-equivalent hydrogen neighbors
The number of these sub-peaks tells you how many hydrogens are represented by that peak
- (# of sub-peaks – 1)
- This will assist you in deducing which parts of a molecule are represented by each peak

Answer 113

A

This technique is used to differentiate molecules based on the differing chemical environments of:
- their hydrogen nuclei (H-NMR), or
- their carbon nuclei (C 13 -NMR)
On the MCAT, however, H-NMR will be tested about ninety percent of the time
An atom must have either:
- an odd atomic number or
- an odd mass number

… to register on an NMR

For the MCAT, think of MRIs as an NMR of the human body

Answer 114

A

Ketones or aldehydes can be prevented from reaction with a nucleophile or base by conversion to an acetal or ketal (which are UNreactive in all but acidic conditions)
- Any terminal diol with at least two carbons will work

STEPS:

One end of the diol acts as the nucleophile
1. An alcohol acts as the nucleophile, attacking the electrophilic carbonyl carbon and pushing the pi electrons from the C=O bond up onto the oxygen
2. The negatively charged oxygen is protonated to form an alcohol and the original alcohol is deprotonated to form an ether. This yields a hemiacetal if it was originally an aldehyde, or a hemiketal if it was a ketone.
The other end of the diol acts as the “second equivalent of alcohol”
1. The alcohol is protonated again to form the good leaving group water, and a second equivalent of alcohol attacks the central carbon as shown below
2. Deprotonation of the second alcohol results in another ether, yielding an acetal if it was originally an aldehyde or a ketal if it was a ketone.
Acidic conditions will return the acetal/ketal to the original aldehyde/ketone

Answer 115

A

Recall that the intermediate is:
- an oxygen anion that collapses down to re-form the carbonyl
Which substituent will leave as a result of this collapse depends solely on:
- its quality as a leaving group
Often, the attacking nucleophile is the better LG, so the original acid derivative is simply reformed

Answer 116

A

A nucleophile can be added to any carboxylic acid or acid derivative and it will attack the carbonyl carbon
Only sometimes, however, will the result be substitution of that nucleophile for the existing substituent

Answer 117

A

(Best LG) Cl^- > ^-OCOR > ^-OH > ^-OR > ^-NH₂ (Worst LG)

Chlorine is the best leaving group because it is a *weak base*

…that is remarkably stable while holding a formal negative charge
This ability to hold a charge is due to the size of chlorine’s electron butt
The negative charge is spread out across a much larger distance than it would be for a smaller atom
For this reason, bromine would be an even better leaving group, and iodine better yet
The –OCOR group is the leaving group portion of an anhydride—the portion kicked off when a nucleophile attacks one of the carbonyl carbons
- It is a carboxylate ion stabilized by resonance
  - That stabilization makes it much more stable than other leaving groups that place a negative charge on an oxygen

Hydroxide and alkoxide are best compared together

It should be noted that under acidic conditions, hydroxide gets the nod as the better leaving group
- because –R groups are weak donating groups
Hydrogen, by comparison is considered neither donating, nor withdrawing
Therefore, the –R group on an alkoxide donates more electron density
- …to an oxygen that already bears a full negative charge
Under basic conditions the acidic hydroxide gets deprotonated and does not act as a leaving group, therefore only the alkoxide RO can act as a leaving group

Finally, an amine makes a relatively awful leaving group because when it leaves it forms the strong base NH2 -

This is a stronger base than hydroxide or alkoxide and therefore is the most unstable bearing a full negative charge
This is the reason that amides are the LEAST REACTIVE of the acid derivatives

Answer 118

A

ALKENE

An alkene involves a sigma and a pi bond between two carbons

CARBONYL

A carbonyl involves a sigma and a pi bond between a carbon and an oxygen

Whenever you think about pi bonds, one of the first things you should consider is the amount of overlap of the p orbitals

Compared to two carbons, a carbon and an oxygen can get closer to one another
- because oxygen has a smaller atomic radius
- This allows for more pi overlap and therefore a stronger bond

Answer 119

A

Area Under the Peak:

This is a relative representation of the number of hydrogens accounted for by that peak
An “Integral trace” (a step-wise line superimposed across the spectrum) makes this easier to determine
- The relative area under each curve is given by the height of each step.
  - By “relative” we mean that if one peak is 2 units high and the other is 6 units high, this tells you only that the latter is accounted for by 3 times the number of hydrogens—NOT that the first peak accounts for 2 hydrogens and the second for 6 hydrogens

Answer 120

A

There are two resonance structures for the conjugate base formed by removal of an alpha hydrogen
Remember that neither form actually exists
- —the actual structure is a permanent hybrid of the two
In this case, the hybrid will look slightly more like the second structure
- because the negative charge is on the more electronegative oxygen

Answer 121

A

3

These structures differ in the substituents attached to the carbonyl carbon
- An electron donating group on the carbonyl carbon will decrease its partial positive charge and thereby make it less able to stabilize the conjugate base
- An electron withdrawing group will make the carbonyl better at stabilizing the conjugate base and therefore the strongest electron withdrawing group would indicate the strongest acid
  - because it produces the most stable conjugate base
#1
- is the only electron withdrawing group, so it will have the most acidic alpha hydrogens
- Bromines are BIG!
  - “Suck away” electrons

The weakest donating group of the other three is the methyl group on #4

Amines (#3) and hydroxyl groups (#2) are both electron donating
- but amines are better donating groups
  - because nitrogen is less electronegative than oxygen

Answer 122

A

This scenario would occur if the carbonyl was of the form R₁COR₂
…and the nucleophile was something such as a Grignard Reagent (R-MgBr)
- …that added either R₁ OR R₂ to the carbonyl carbon
  - Because two of the substituents would be identical, it could not be chiral
This could also happen if water or hydroxide were added to the bond, creating two –OH substituents

Answer 123

A

Partial positive charge on the carbonyl carbon
- This makes the carbon a good electrophile
α hydrogens
- Hydrogens on the α carbon are surprisingly acidic
  - –especially given that alkane hydrogens normally CANNOT be removed
Electron donating/withdrawing groups
- The reactivity of a carbonyl with a nucleophile is dramatically affected by the presence of electron donating or electron withdrawing groups on the carbonyl carbon
  - Donating groups:
    - decrease the reactivity of the carbonyl carbon
  - Withdrawing groups:
    - increase its reactivity
Steric hindrance
- Bulky substituents attached to the carbonyl carbon decrease its reactivity
Planar stereochemistry
- The sp² hybridized carbonyl carbon is planar
  - and can therefore be attacked from either side
- When the two substituents on the carbonyl carbon are NOT identical, an addition reaction could therefore create both R and S enantiomers in a racemic mixture

Answer 124

A

Hydrogens on a carbon adjacent to a carbonyl carbon are ACIDIC
- due to resonance stabilization of the conjugate base
When there are two carbonyls separated by a single carbon (i.e., 1,3-dicarbonyls), hydrogens on the middle carbon are even more acidic
The greater the partial positive charge on the carbonyl carbon…
- the more acidic its alpha hydrogens will be
- For this reason, be on the lookout for electron donating and withdrawing groups on the carbonyl carbon

Answer 125

A

A carbonyl is a *carbon double bonded to an oxygen*

For the MCAT, you can treat most carbonyl analogues, such as S=O or N=O, as you would carbonyls

However, it is quite possible that the MCAT would require you to use your basic knowledge of electronegativity, bond polarity, atomic radius, etc., to predict difference in reactivity between an analogue and a carbonyl

Answer 126

A

How it works:

The molecules of the sample are bombarded with electrons
- causing them to both break apart into smaller pieces, and ionize
This will happen in a random way
- producing fragments with different masses and charges
These fragments are accelerated through a narrow curved magnet called a “flight tube”.
Only particles with a certain mass-to-charge ratio (m/z) will follow the exact curved path necessary to NOT hit the walls and exit onto the detector at the end of the flight tube.
The strength of the magnetic field is varied from low to high
- …changing the curvature of each fragment until all of the fragments have struck the detector

Answer 127

A

It will turn it into an ALCOHOL

Answer 128

A

Most SUBSTITUTED

(Think of alkenes)

Answer 129

A

The formation of an additional C-C bond

Grignard Reagent:

RMgX

Answer 130

A

Ketones

is NOT oxidized further

Answer 131

A

First gets oxidized to ALDEHYDES

With 2+ equiv of oxidizing agent, gets oxidized again into CARBOXYLIC ACIDS

Answer 132

A

Strong reducing agents such as Lithium Aluminum Hydride essentially produce basic hydride ions that can attack electrophiles
The carboxylic acid will be attacked by the hydride–
- pushing the double bond up onto the oxygen as a lone pair
In any protic solvent this will be protonated to form a hydroxyl group

A gem diol is an exact description of the intermediate, two hydroxyl groups attached to the same carbon

Recall that two hydroxyl groups on neighboring carbons is called a vic diol

Answer 133

A

Pinacol rearrangement

Requires a VIC-diol
- OH’s on neighboring C’s

Involves:

the spontaneous disassociation of one of the protonated alcohols
followed by a methyl shift
Following the methyl shift, the other alcohol’s hydrogen is abstracted
and the electrons condense to form a carbonyl
- …and quench the carbocation

Answer 134

A

Aldehydes are LESS acidic than either water OR alcohols!

Water and alcohols are similar in acidity, but you can compare them by looking at the conjugate bases
The alcohol’s conjugate base has an electron donating group (“R”) destabilizing the negatively charged oxygen
- making it less stable
  - therefore less acidic
Water doesn’t have an e’ donating group, therefore the negatively charged Oxygen is more stable (remember: Oxygen LOVES having a (-) charge!)
- thus more acidic

Answer 135

A

Tosyl chloride is traditionally used to protect alcohols or other nucleophiles

The alcohol attacks the tosyl chloride and becomes “tosylated”

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