4. Carbonyl Chemistry Flashcards

1
Q

Why are halogens typically good leaving groups?

A

They are weak bases that can stabilize a negative charge (EN) well. Additionally there larger size (except F) helps stabilize this charge.

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2
Q

What is an SN2 reaction?

A

SN2 reactions are nucleophilic substitution reactions that occur in one step. The nucleophile (lone pair or negative atom) attacks the electrophile at the SAME time as the leaving group departs.

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3
Q

Why do SN2 reactions require back-side attack? What is the consequence of this?

A

back-side attack occurs because the nucleophile must attack from the opposite side of the leaving group (because in SN2 reactions, this occurs at the same time). The consequence of this is inversion of stereochemistry.

R chiral centers become S chiral centers and vice versa.

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4
Q

What is an SN2 reaction called an SN2 reaction

A

Substitution reaction
Nucleophilic
2 indicates it is bimolecular. The reaction rate is determined by the [N] and [E]

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5
Q

In terms of the (a) substrate and (b) solvent environment, how do you favor an SN2 reaction?

A

The SN2 mechanism goes through a pentavalent TS. Thus, electrophiles with less bulky substituent groups (less steric hindrance) react faster than bulky electrophiles.

Polar protic (H-bonding) solvents will solvate the nucleophile and make it less reactive. Thus, polar, aprotic solvents are best. These will dissolve the E and N without affecting reactivity.

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6
Q

DMF and DMSO are examples of what?

A

polar aprotic solvents (best for SN2 reactions)

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7
Q

Explain the relative stability of carbocations?

A

Carbocations are most stable when they are surrounded by electron-donating groups (alkyl groups).

Therefore,
Greatest stability on the left
tertiary > secondary > primary > methyl

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8
Q

Explain the process of an SN1 reaction.

A

SN1 reactions occur in two steps.

  1. The LG leaves creating a trigonal planar carbocation
  2. The nucleophile attacks the carbocation and forms a bond with it.
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9
Q

Why are SN1 reactions uni-molecular? What is the rate limiting step?

A

SN1 reactions occur in two steps. SN1 reaction rate is dependent on the rate-limiting step (uni-molecular), which is the LG leaving.

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10
Q

Why does SN1 reactions cause racemization?

A

Because once the LG leaves, the nucleophile can attack both sides of the carbocation with equal affinity (since it is a planar, symmetrical molecule). This creates a racemic mixture of R and S.

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11
Q

t or f, reaction rate of
SN2 = k[E][N]
SN1 = k[E]

A

True,
SN2 = bimolecular
SN1 = uni-molecular

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12
Q

How do we speed up an SN1 reaction in terms of the (a) substrate and (b) solvent environment?

A

Substituted carbocations are more favorable. Therefore electrophiles with more substituents (tertiary) are better
tertiary > secondary > primary >

polar protic solvents (H-bonding) are better for SN1 reactions because they stabilize the carbocation and the leaving group. i.e. they facilitate the rate-limiting step of the reaction.

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13
Q

What is solvolysis?

A

Solvolysis is a form of an SN1 reaction in which the solvent is also the nucleophile. An example is water which can attack the electrophile with its free electrons on O2.

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14
Q

What is a carbocation rearrangement? Is this possible with SN1 and SN2 reactions?

A

If a carbocation is formed near a carbon that is more substituted, the (+) charge may shift to the more substituted (i.e. more stable) carbon atom. In the following step, the nucleophile is now attacking a new carbon atom that differs from the LG’s carbon atom. This is a rearrangement.

SN1 reactions can do this. SN2 reactions don’t have carbocations (and only occur in one step) and thus cannot do this.

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15
Q

What kind of nucleophile favors (a) SN1 rxn and (b) SN2 rxn?

A

SN1 –> weaker, non-basic nucleophile. Weaker nucleophiles used because we have a stronger electrophile (full + charge)

SN2 –> strong, non-bulky (small) nucleophile. Needs to be strong so it can attack a delta positive center.

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16
Q

What defines a good leaving group? Is OH a good LG?

A

Anything that can hold a negative charge well. The best LG’s are conjugate bases of strong acids (i.e. weak bases). OH is a strong base and bad LG.

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17
Q

What do you create when you oxidize a primary, secondary, and tertiary alcohol?

A

In an oxidation reaction, a hydrogen is lost at the alpha carbon.

primary OH –> this creates an aldehyde.
secondary OH –> this creates a ketone
tertiary OH –> No reaction because there is no free H to be removed.

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18
Q

PCC, chromic acid (H2CrO4), permanganate (MnO4-) are all examples of what?

A

Oxidizing agents (IN GENERAL, these will oxidize primary alcohols to aldehydes and secondary alcohols to ketones)

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19
Q

What is the only oxidizing agent that will NOT over-oxidize primary alcohols to carboxylic acids?

A

The only oxidizing agent that will stop at the aldehyde stage is PCC.

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20
Q

What is an acidic hydrogen?

A

Hydrogen’s bonded to carbons near the carbonyl group are prone to leaving because the electrons they leave behind can stabilize the delta positive charge on the alpha carbon. They can also become de-localized in the pi system (resonance).

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21
Q

ketones and aldehydes: What is an enolate ion? How is it formed, what are its properties?

A

Once a ketone or aldehyde is made by oxidizing an alcohol, they can be reactive due to their acidic hydrogen’s.

A strong base (OH or OR-) may deprotonate an alpha hydrogen. The resulting carbanion may delocalize with the pi system (C=O). This resonance hybrid is called the enolate ion which is nucleophillic on the carbanion formed.

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22
Q

What is keto-enol tautomerism? How does it work? What are tautomers?

A

keto-enol tautomerism occurs when we create an enolate ion (deprotonation of alpha hydrogen of a ketone). Instead of the carbanion acting as a nucleophile, the oxygen can become protonated creating an alcohol group and an alkene group.

tautomers are constitutional isomers that can readily convert between each other.

Here, the ketone and the enol readily inter-convert.

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23
Q

true or false, keto-enol tautomerism causes racemization.

A

true, since the enol formation is sp2 hybridized and therefore planar. Protonation of the alpha carbon can come from any direction causes a racemic mix.

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24
Q

Sodium borohydride (NaBH4) and lithium aluminum hydride (LiAlH4) are examples of what?

A

NaBH4 and LiAlH4 are examples of reducing agents. These can be used to reduce aldehydes and ketones back into alcohols.

Note: reducing agents (want to be oxidizes, want to gain O, lose H) often have many H atoms bonded to a non EN atom.

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25
Q

What are organometallic reagents? What are Grignard reagents?

A

organometallic reagents come in the form R-M+. They create anionic carbons they can act as a strong base or nucleophile. They are often used in nucleophillic addition reactions with carbonyl centers.

Grignard reagents are typically an alkyl halide reacting with magnesium to form (e.g.) ethyl-MgBr. These reactions take place in aprotic solvents so that the very basic reagent remains active.

26
Q

What occurs when a Grignard reagent reacts with a aldehyde or ketone?

A

Grignard reagents are strong nucleophiles (and bases). It will attack the delta positive carbon center and kick off the C=O bond. In acid workup, the O- is protonated.

This reaction allows you to create an alcohol with a specified R group (whatever R-group is attached to the Grignard reagent during nucleophillic attack).

27
Q

Grignard reagents are strong bases: what is necessary when using these to add R groups?

A

First we need to use an aprotic solvent. Secondly, no carboxylic acid groups (or any acid groups) can be present, otherwise the reagent will react with them instead.

28
Q

How do you protect alcohol groups from reacting?

A

mesylate or tosylate them. These groups can also protect amino groups (NH2)

These are also very good LGs. We can make an OH a better leaving group by first mesylating it.

29
Q

What are hemiacetals and acetals? What molecule exists as these?

A

Both are created by the nucleophillic addition of an alcohol on ketones or aldehydes in the presence of acid.

acetal - two ether groups bonded to the same carbon
hemiacetal - one ether, one alcohol bonded to the same carbon

sugars (glucose) exist as these.

30
Q

describe, in general, the mechanism of acetal formation.

A

It occurs in the presence of acid. First the acid protonates the O group of an aldehyde making the carbonyl more electrophilic. Now the alcohol nucleophile attacks the carbonyl center. Loss of an H gives you a hemiacetal. The alcohol group is now protonated creating water. Water leaves and a c=o bond is formed. Alcohol now attacks again giving an acetal.

31
Q

t or f, acetals, hemiacetals, and k/a are all in equilibrium with each other.

A

true

32
Q

What is cyanide? What is a cyanohydrin?

A

cyanide –> C tripe-bond to N. C is negatively charged and thus, very nucleophilic.

cyanohydride occurs when cyanide attack a ketone or aldehyde.

33
Q

What is the difference between alkyl and aryl amines

A

alkyl amine = NH2 bonded to an sp3 hybridized carbon

aryl amine = NH2 bonded to an aromatic carbon (sp2 hybridized)

34
Q

What occurs when an aldehyde or ketone reacts with a primary amine? What is formed? Explain the reaction mechanism.

A

This forms an imine.

In weakly acidic conditions, a primary amine attacks the carbonyl center of an aldehyde. Through protonation, water leaves as the nitrogen group forms a double bond. A base or water deprotonates the N group and forms an imine.

R2C = NR

35
Q

What occurs when an aldehyde or ketone reacts with a secondary amine? What is formed?

A

Primary amine = imine

secondary amine = enamine.
R2C – NR2

The mechanism is the same as an imine formation. However, no iminum ion is formed (NHR2+). Thus, an alpha acidic hydrogen is deprotonated forming a C=C bond instead.

36
Q

Why are enamines nucleophilic?

A

enamines exist in resonance with imimuim ions (NHR2+). When the nitrogens free electrons form a double bond, the C=C double bone of the enamine jumps onto an alpha carbon creating an carbanion. This is a nucleophilic carbanion.

37
Q

true or false, both imines and enamines are formed in mildly acidic conditions (pH = 4-6).

A

True.

38
Q

What is an aldol condensation? What is the product?

A
  1. A carbonyl group (a/k) forms an enolate ion through deprotonation of an alpha hydrogen.
  2. The enolate ion attacks the carbonyl center of another carbonyl species (a/k).

The product is a beta-hydroxyl carbonyl compound

39
Q

What is necessary for the aldol condensation to occur?

A

You need a strong base (OH, SH, RO) to be able to deprotonate an alpha hydrogen. Weak bases are not sufficient.

40
Q

What is a crossed aldol condensation?

A

when two different carbonyl compounds are used in the reaction. To avoid a mixture of products, one of the carbonyl compounds will typically not have any alpha hydrogen’s and therefore, cannot be the enolate ion (nucleophile).

41
Q

Aldol condensation: When ketones have more than one alpha proton to deprotonate, two different enolate ions can form (and thus, two different products). How do you favor the kinetic product? How do you favor the equilibrium product?

A

A more substituted alkene is more stable. Therefore, if you proceed under regular temperature, the more sterically hindered hydrogen will be removed. It also helps to use a small unrestricted base (NaOH).

The kinetic product is favoured at very low temperatures (-78C) because there is not enough energy to overcome the steric hindrance of the substituted carbon. Also, using a large base such as LDA will allow the kinetic product to form.

42
Q

After an aldol condensation forms a beta-hydroxyl carbonyl compound, what occurs if you heat up the product?

A

The addition of heat causes a dehydration (elimination) reaction.

the hydroxyl group leaves as water and the alpha-beta carbons form an alkene. This product is called an alpha-beta-unsaturated carbonyl compound.

43
Q

What happens if you treat a beta-hydroxyl carbonyl compound with a strong base?

A

The strong base will split the beta-hydroxyl carbonyl compound back into ketones / aldehydes. This is called retro-aldol reaction

44
Q

Why do carboxylic acids have high melting points and boiling points?

A

They can inter-molecular hydrogen bond harboring both H-bond acceptors and donors.

45
Q

Carboxylic acids may be reduced to form ketones and aldehydes. We can also reduce carboxylic acids all the way down to primary alcohols, however, only one reducing agent is capable of this. What is it?

A

sodium borohydride (NaBH4) and lithium aluminum hydride (LiAlH4) are both strong reducing agents. However, ONLY LiAlH4 is capable of reducing carboxlyic acids all the way down to alcohols. NaBH4 will stop at the aldehyde stage.

This is opposite to PCC which is the only oxidizing agent that will stop at the aldehyde stage and not oxidize alcohols all the way to carboxylic acids.

46
Q

What is decarboxylation? What molecules does it typically occur on?

A

Carboxylic acids that have a beta carbonyl group are subject to decarboxylation which is the loss of CO2.

47
Q

t or f, unlike aldehydes and ketones, carboxylic acid derivatives undergo nucleophilic attack, followed by an elimination reaction.

A

True, since the derivative (e.g. acid halide, amide, etc) is a good leaving group.

These are called nucleophilic addition-elimination reactions

48
Q

What is an esterification reaction? What is ester hydrolysis?

A

Esterification = When a carboxylic acid reacts with an alcohol, typically in the presence of an acid. This forms an ester.

Ester hydrolysis = splitting an ester back into a carboxylic acid and an alcohol. The mechanism varies depending on whether you are using a base or acid.

49
Q

What is trans-esterification?

A

In trans-esterification, an alcohol is used as the nucleophile instead of water (ester hydrolysis), in acidic conditions.

The nucleophilic attack of an alcohol results in the replacement of the current OR group on the ester.

Here, the products are the new ester, and an R-OH group that was originally the esters OR group.

50
Q

Explain breifly the difference between acid-catalyzed and base-catalyzed ester hydrolysis.

A

Acid –> The acid protonates the carbonyl oxygen. Water attacks as the nucleophile. The ether group reacts with acid to make it a good leaving group (ROH > RO). The group leaves as an alcohol leaving behind the original carboxylic acid.

Base –> The strong base (OH) attacks the carbonyl center. When the carbonyl C=O reforms, the RO group is kicked off which immediately reacts with the carboxylic acid to form ROH and COO-. Acid work up is needed to recover the carboxylic acid completely.

51
Q

What is saponifcation?

A

The base-catalyzed hydrolysis of triglyceride to form glycerol and three fatty acids. Done to make soap.

52
Q

What is an amphipathic molecule?

A

amphipathic = hydrophobic and hydrophillic

amphoteric = acts as an acid and a base

53
Q

How do you form an acid halide from a carboxylic acid?

A

ROOH reacts with PX3 or SOCl2

X = Cl or Br

54
Q

How do you form an acid anhydride? (n=2)

A

React two carboxylic acids together in heat.

We can also react a carboxylic acid with an acid halide.

55
Q

How do you form an ester? (n=3)

A

esterification by reacting alcohol and ROOH. They may also be formed from an acid halide or acid anhydride and the desired alcohol.

56
Q

How do you form an amide? (n=3)

A

They can be made using the corresponding acid halide, anhydride, or ester and the desired amine.

57
Q

t or f, amides can be synthesized from carboxylic acids.

A

False

they are the only acid derivative that cannot be directly made from ROOH. ROOH must first react to form an acid halide, anhydride, or ester which then can react with an amine to form an amide.

58
Q

t or f, carboxylic acids can be formed from any one of its derivatives simply by heating it in acidic conditions.

A

TRUE

59
Q

explain the reactivity of ROOH derivatives.

A

COOH > AH > AA > ES > AM

Anthing to the left can make anything below it, except for COOH which cannot make amides.

60
Q

t or f, acid halides and anhydrides are readily hydrozyled in water, esters and amides are much more stable.

A

True. this is because of the LG stability

X- and COO- are relatively weak bases and thus, good leaving groups

RO- and NR2 are strong bases and bad LG’s