Aldehydes and Ketones

Question 1

Define:

aldehyde

Answer 1

It is an organic carbonyl compound. It consists of an alkane chain with a carbon-oxygen double bond on its terminal carbon.

Aldehydes are named by removing the “-e” from the end of the name of the parent alkane and replacing it with “-al.” The three-carbon chain above is called propanal.

Question 2

What is the IUPAC name of the molecule below?

Answer 2

It is known as methanal, commonly referred to as formaldehyde.

Aldehydes are named by removing the “-e” from the end of the name of the parent alkane and replacing it with “-al.” For the MCAT, know both the IUPAC and common name for this compound.

Question 3

What is the IUPAC name of the molecule below?

Answer 3

pentanal

To name an aldehyde, remove the “-e” from the end of the name of the parent alkane and replace it with “-al.” Since an aldehyde is a terminal group by definition, it is not necessary to write “1-pentanal.”

Question 4

Define:

ketone

Answer 4

It is an organic carbonyl compound. It consists of an alkane chain with a carbon-oxygen double bond on one of the chain’s internal carbons.

Ketones are named by removing the “-e” from the end of the name of the parent alkane and replacing it with “-one.” The three-carbon chain above is called propanone, commonly known as acetone. For the MCAT, both names are valuable to know for this molecule.

Question 5

What is the IUPAC name of the molecule below?

Answer 5

butanone

Ketones are named by removing the “-e” from the end of the name of the parent alkane and replacing it with “-one.”

Question 6

What is the IUPAC name of the molecule below?

Answer 6

2-pentanone

To name an ketone, remove the “-e” from the end of the name of the parent alkane and replace it with “-one.” Since numbering begins from the end closest to the carbonyl, this structure is called 2-pentanone, not 4-pentanone.

Question 7

How do the boiling points of ketones and aldehydes compare to those of simple hydrocarbons?

Assume molecular masses of all compounds are similar.

Answer 7

Ketones and aldehydes have higher boiling points due to their larger intermolecular forces.

These molecules exert dipole-dipole forces due to their carbonyl groups, while hydrocarbons have only dispersion forces.

For example, propane has a boiling point of -42 °C, compared to propanal at 50 °C and propanone at 56 °C.

Question 8

How do the boiling points of ketones and aldehydes compare to those of alcohols?

Assume all compounds have similar molecular masses, chain length, and branching.

Answer 8

Ketones and aldehydes have lower boiling points due to their smaller intermolecular forces.

Alcohols can hydrogen bond, while ketones and aldehydes exhibit only the comparatively weaker dipole-dipole interactions.

For example, propanol has a boiling point of 97 °C, compared to propanal at 50 °C and propanone at 56 °C.

Question 9

OH^- tends to donate electrons and is attracted to electropositive atoms. What name is given to molecules that share these traits?

Answer 9

nucleophiles

Good nucleophiles have lone pairs or negative charges and are generally basic.

Most reactions in MCAT organic chemistry involve a nucleophile attacking a carbonyl carbon (which possesses a partial positive charge).

Question 10

Which atom in a ketone or aldehyde is most susceptible to nucleophilic attack?

Answer 10

The C=O (carbonyl) carbon is most easily attacked.

Oxygen is very electronegative and pulls electron density away from the carbon. This leaves the carbon with a partial positive charge, which is exactly what attracts nucleophiles.

Question 11

What product is formed when an aldehyde and a single alcohol are reacted in the presence of an acid catalyst?

Answer 11

• Hemiacetal is formed.
• Water is formed as a side product.

A hemiacetal, or “half acetal,” is a functional group that contains one -OR and one -OH group bound to the same carbon. Since they are derived from aldehydes, that carbon is also bound to a hydrogen atom.

Question 12

What product is formed when a ketone and a single alcohol are reacted in the presence of an acid catalyst?

Answer 12

• Hemiketal is formed.
• Water is formed as a side product.

A hemiketal, or “half ketal,” is a functional group that contains one -OR and one -OH group bound to the same carbon. Since they are derived from ketones, that carbon is also bound to two carbon chains.

Question 13

Define:

acetal

Answer 13

It is an aldehyde derivative in which the carbonyl has been replaced by two -OR groups. The “R” can be identical or different.

An acetal is formed via the reaction of an aldehyde molecule with two equivalents of alcohol.

Question 14

What product is formed when an aldehyde and excess alcohol are reacted in the presence of an acid catalyst?

Answer 14

• Acetal is formed.
• Water is formed as the side product.

Note that an acetal is also formed when a hemiacetal is reacted with an additional equivalent of alcohol.

Question 15

Define:

ketal

Answer 15

It is a ketone derivative in which the carbonyl has been replaced by two -OR groups. The “R” can be identical or different.

A ketal is formed via the reaction of a ketone molecule with two equivalents of alcohol.

Question 16

What product forms when a ketone and excess alcohol are reacted in the presence of an acid catalyst?

Answer 16

• Ketal is formed.
• Water is formed as the side product.

Note that a ketal is also formed when a hemiketal is reacted with an additional equivalent of alcohol.

Question 17

What product will this reaction form?

Answer 17

hemiketal

In this reaction, the alcohol’s -OR group is the nucleophile, while the carbonyl carbon on the ketone is the electrophile. After the attack, the oxygen accepts the H instead of reforming the carbonyl. A basic workup keeps the -OH from further protonating and creating a leaving group.

Question 18

What product will this reaction form?

Answer 18

hemiacetal

In this reaction, the alcohol’s -OR group is the nucleophile, while the carbonyl carbon on the aldehyde is the electrophile. After the attack, the oxygen accepts the H instead of reforming the carbonyl. A basic workup keeps the -OH from further protonating and creating a leaving group.

Question 19

What product will this reaction form?

Answer 19

ketal

In this reaction, the alcohol’s -OR group is the nucleophile, while the carbonyl carbon on the ketone is the electrophile. After the attack, the oxygen accepts the H instead of reforming the carbonyl. An acidic workup protonates the -OH into H₂O, creating a leaving group and allowing a second -OR substitution to occur.

Question 20

What product will this reaction form?

Answer 20

acetal

In this reaction, the alcohol’s -OR group is the nucleophile, while the carbonyl carbon on the aldehyde is the electrophile. After the attack, the oxygen accepts the H instead of reforming the carbonyl. An acidic workup protonates the -OH into H₂O, creating a leaving group and allowing a second -OR substitution to occur.

Question 21

What product forms when a ketone or aldehyde and a secondary amine undergo a condensation reaction?

Answer 21

• Enamine is formed.
• Water is formed as a side product.

An enamine consists of a nitrogen group (amine) bound to a carbon that is participating in a C=C double bond.

Question 22

What is the name of the functional group that consists of a carbon double-bonded to a nitrogen atom?

Answer 22

imine

Imine nomenclature is rarely tested on the MCAT, so it is sufficient to simply be able to identify this functional group.

Question 23

What product would form if the oxygen in the C=O bond of a carbonyl was replaced with an N-H group?

Answer 23

An imine would form. Imines are functional groups that include a carbon-nitrogen double bond.

The details of imine formation are not likely to be tested on the MCAT, but it is important to be able to recognize the group.

Question 24

What types of reactants can take part in an aldol condensation, and what final product is formed?

Answer 24

• two aldehydes
• two ketones
• aldehyde and a ketone

The final product is an enone, or a molecule with an alkene adjacent to (and conjugated with) a ketone.

Question 25

What **term** describes an aldol condensation in which the reactants are two equivalents of the same molecule?

Answer 25

self-condensation ## Footnote The aldol condensation is not commonly seen on the MCAT, but when it does appear, it generally references self-condensations.

Question 26

Briefly list the **steps** involved in an **aldol condensation reaction**.

Answer 26

1. A base attracts an alpha hydrogen from one of the reactants, allowing an enolate to form. 2. The enolate attacks the carbonyl carbon of the other reactant. This forms an aldol (a β-hydroxyaldehyde or β-hydroxyketone). 3. Base is used to promote dehydration, or loss of water, which forms a conjugated enone.

Question 27

What **type** of product will be formed from this reaction in the presence of heat?

Answer 27

An **enone** will form in this aldol condensation. ## Footnote First, propanone is deprotonated, forming an enolate. This acts as a nucleophile to attack the intact propanone's carbonyl carbon, then protonates the original carbonyl oxygen. When a beta proton is shifted to protonate that atom again, it becomes H₂O, a good leaving group. Condensation then occurs. The exact product will be 4-methyl-3-penten-2-one, but the MCAT would not ask for such detail. Simply know that this is an aldol condensation.

Question 28

What **name** could be used to describe a reaction in which two carbonyl compounds are formed from a β-hydroxyaldehyde?

Answer 28

retro-aldol reaction ## Footnote As its name implies, a retro-aldol reaction is simply the reverse of the forward reaction, aldol formation. The products will be two smaller carbonyl compounds (such as two aldehydes).

Question 29

In organic chemistry, how can an **oxidation** reaction be easily identified?

Answer 29

It involves the gain of bonds to oxygen or the loss of bonds to hydrogen. ## Footnote This definition differs from the general chemistry idea of oxidation number. While both are valid ways to conceptualize redox reactions, thinking of the gain and loss of bonds can be much easier when dealing with organic compounds.

Question 30

In organic chemistry, how can a **reduction** reaction be easily identified?

Answer 30

It involves the gain of bonds to hydrogen or the loss of bonds to oxygen. ## Footnote This definition differs from the general chemistry idea of oxidation number. While both are valid ways to conceptualize redox reactions, thinking of the gain and loss of bonds can be much easier when dealing with organic compounds.

Question 31

Name at least **three compounds** that are used as oxidizing agents in organic reactions.

Answer 31

* the chromium reagents (CrO₃, CrO₄^2-, Cr₂O₇^2-) * permanganate (MnO^4-) and manganate (MnO₄^2-) * nitric acid (HNO₃) * pyridinium chlorochromate (PCC) ## Footnote For the MCAT, remember that oxidizing agents generally contain many oxygen atoms. Reducing agents usually contain many hydrogen atoms.

Question 32

What **product**, if any, will be formed from the oxidation of an aldehyde group?

Answer 32

carboxylic acid ## Footnote The carbon on an aldehyde already contains two bonds to oxygen. To further oxidize, it would need to gain another one and become a carboxylic acid.

Question 33

What **product**, if any, will be formed from the oxidation of a ketone group?

Answer 33

No product will form. ## Footnote The carbon on a ketone already contains two bonds to oxygen and two bonds to carbon. In other words, it is maximally oxidized; it cannot be reacted further.

Question 34

One ketone is reacted with KMnO₄ (a very strong oxidizing agent), while another identical ketone is reacted with PCC (a considerably weaker oxidant). What **difference** will be seen between the two products?

Answer 34

No difference will exist. ## Footnote Since ketones already contain two bonds to oxygen and two to carbon, they are already fully oxidized. The strength of the oxidizing agent does not matter; both reactants will be unchanged.

Question 35

What is the **main function** of the hydride reagents **NaBH₄** and **LiAlH₄**?

Answer 35

These reagents both act as **reducing agents**. ## Footnote LiAlH₄ is a strong reducing agent, while NaBH₄ is noticeably weaker.

Question 36

What **products** will form from the reduction of a ketone and an aldehyde, respectively? ## Footnote Assume the reducing agent used was either NaBH₄ or LiAlH₄.

Answer 36

* Reduction of a ketone will form a **secondary alcohol**. * Reduction of an aldehyde will form a **primary alcohol**. ## Footnote Reduction is the loss of bonds to oxygen and the gain of bonds to hydrogen. Aldehydes and ketones already contain two carbon-oxygen bonds, and must lose one during reduction.

Question 37

**Name** the two isomeric forms of a carbonyl compound that can interconvert and exist in equilibrium.

Answer 37

* **keto** form * **enol** form ## Footnote The keto form contains a carbonyl group, while the enol form contains an alcohol adjacent to an alkene.

Question 38

What **name** is given to the conversion between a compound's keto and enol forms?

Answer 38

tautomerization ## Footnote Keto-enol tautomerization includes the movement of an alpha hydrogen and the bonding electrons of the pi bond. Because atoms change their positions, it is not a type of resonance. Remember this for the MCAT!

Question 39

Which **form** tends to be more stable in aldehydes and ketones: keto or enol?

Answer 39

The **keto** form is generally more thermodynamically stable than the enol form. For this reason, it tends to predominate. ## Footnote Exceptions include cases in which the enol form is conjugated. One example is phenol, which gains extra stability from conjugation that it would lose as a keto tautomer.

Question 40

What **stable resonance structure** can be formed by a 1,3 dicarbonyl structure?

Answer 40

This molecule can form an **enol** that is stabilized by hydrogen bonding.

Question 41

In addition to the keto and enol forms, what other **functional groups** are involved in tautomerization?

Answer 41

* Imines * Enamines ## Footnote Like the keto form, the imine form tends to predominate due to stability. However, the enamine form can be more prevalent if it allows the compound to be conjugated.

Question 42

What is the **difference** between the kinetic and thermodynamic enolates?

Answer 42

* the **kinetic enolate** is easier to form, with a lower activation energy. It is favored at low temperatures. * the **thermodynamic enolate** is more energetically difficult to form, with a higher activation energy, but is more stable (more substituted). It is favored at high temperatures.

Question 43

Is the circled structure on the right the **kinetic** or **thermodynamic** enolate?

Answer 43

It is the **thermodynamic** enolate. Since it is more substituted, it is more stable; however, its steric hindrance makes it difficult to form at low temperatures. ## Footnote The structure on the left is the kinetic enolate. In contrast to above, it is less substituted, less stable, and easier to form.

Question 44

What is an **organometallic reagent**?

Answer 44

It is a compound that **contains an R group bound to a metal atom**. Examples include n-BuLi and the Grignard reagents. ## Footnote Organometallic reagents allow the R group to act as a nucleophile. They attack the carbonyl carbon, adding the R group to that position and protonating the carbonyl oxygen to an alcohol.

Question 45

What **product** will this reaction form?

Answer 45

2-methyl-butanol ## Footnote A ketone reacting with an organometallic reagent will produce a tertiary alcohol. This occurs because the metal allows the carbon chain to act as a nucleophile, adding to the carbonyl carbon and protonating the carbonyl oxygen.

Question 46

What **atoms** are included in a **Grignard reagent**?

Answer 46

It is a type of **organometallic reagent with the formula R-MgX**. Grignard reagents contain a magnesium halide bound to an R group, or carbon chain. ## Footnote The Grignard reaction is a good way to add an alkyl chain to a carbonyl compound.

Question 47

What **product** will this reaction form?

Answer 47

A Grignard, will form a **secondary alcohol**. ## Footnote The R group acts like a nucleophile and attacks the carbonyl carbon, leading to the protonation of the carbonyl oxygen. The MgBr detatches and moves back into solution.

Question 48

What **product** will this reaction form after an acid work-up?

Answer 48

tertiary alcohol ## Footnote The -OR group on the reactant is a good leaving group, so the overall molecule will react with two equivalents of Grignard reagent: first as a substitution of R' for -OR, and again as an addition of a second R' group.

Question 49

How does the presence of large substituents near the carbonyl carbon **affect** the speed of a nucleophilic attack?

Answer 49

The speed of attack, and reactivity of the compound in general, is **decreased**. ## Footnote Large substituents near the carbonyl carbon cause steric hindrance and block the nucleophile's access to the electrophilic carbon.

Question 50

Scientists are trying to perform a nucleophilic addition on the molecule below. How do the tert-butyl groups likely **affect** the reactivity?

Answer 50

The large, bulky groups **decrease** the reactivity. ## Footnote In other words, the steric hindrance of the tert-butyl groups limits the nucleophile's access to the carbonyl carbon. This slows the reaction.

Question 51

What **chemical property** is displayed by the alpha proton of a carbonyl compound?

Answer 51

Alpha protons are **more acidic** than the other protons in a carbon chain. ## Footnote When the alpha position is deprotonated, the resulting molecule can be stabilized by resonance. In other words, the conjugate base is especially stable, a characteristic that increases acidity.

Question 52

What **product(s)** exist immediately after the deprotonation of a carbonyl compound at the alpha position?

Answer 52

It is a **resonance-stabilized carbanion**, in both the keto and enol forms. ## Footnote While carbanions are extremely reactive, the resonance here allows one to be stable enough to form. This explains the increased acidity of alpha protons.

Question 53

How many **resonance structure(s)** can form from this alpha, beta unsaturated carbonyl?

Answer 53

**Two** resonance structures can form in addition to the original. ## Footnote The structure on the left is an enolate, while that on the right has its double bond shifted to the beta-carbon.

Question 54

What is the **product** of a strong nucleophilic attack on an alpha, beta unsaturated carbonyl (as shown below)?

Answer 54

It will be a **carbonyl** with the nucleophile added to the beta position. ## Footnote The nucleophile attacks at the beta position, causing the alpha, beta-unsaturated carbonyl to convert to the enol form. The enol then tautomerizes back to a carbonyl.

Question 55

What **name** is given to molecules like CH₂-P-Ph₃, and what reaction are they involved in?

Answer 55

CH₂-P-Ph₃ is a ylide, specifically a triphenyl phosphonium ylide. Ylides are used in **Wittig reactions**. ## Footnote The Wittig reaction allows a carbonyl to be converted into an alkene, a common process used in alkene synthesis. For the MCAT, however, you need no outside knowledge about the reaction; just be able to recognize it if you see it in a passage.

Question 56

What is the **name** of the molecule below?

Answer 56

cyanohydrin ## Footnote It can be synthesized by reacting a carbonyl compound with a nitrile, such as CN^-. Cyanohydrins are involved in the Strecker synthesis, a procedure that can synthesize amino acids.