Jones M., Fleming S.A. Organic Chemistry

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15.11 Additional Problems 759

09876543210

(ppm)

Chemical shift (δ)

09 8 7 6 5 4 3 2 1 0

(ppm)Chemical shift (δ)

09 8 7 6 5 4 3 2 1 0

(ppm)

Chemical shift (δ)

PROBLEM 15.82 Compounds A and B, containing only C,

H, and O, gave 63.15% C and 5.30% H upon elemental

analysis. Their mass spectra do not show a molecular ion, but

freezing point depression data indicate a molecular weight of

150 g/mol. Catalytic hydrogenation of both A and B leads to

C. Spectral data for compounds A, B, and C are summarized

below. Assign structures for compounds A, B, and C and

explain your reasoning.

Compound A

IR (Nujol®): 1845 (m), 1770 (s) cm

1

H NMR (CDCl

): δ 2.03–2.90 (m, 2H)

3.15–3.70 (m, 1H)

5.65–6.25 (m, 1H)

Compound B

IR (Nujol®): 1825 (m), 1755 (s) cm

1

H NMR (CDCl

/DMSO-d

): δ 1.50–2.10 (m, 1H)

2.15–2.70 (m, 1H)

Compound C

IR (neat): 1852 (m), 1786 (s) cm

1

H NMR (CDCl

/DMSO-d

): δ 1.10–2.30 (m, 4H)

3.05–3.55 (m, 1H)

PROBLEM 15.83 Spectral data for isomeric compounds A and

B are summarized below. Assign structures for compounds A

and B, and explain your reasoning.

Compound A

Mass spectrum: m/z  148 (M, 7%), 106 (8%), 105 (100%),

77 (29%), 51 (8%)

IR (neat): 1675 (s), 1220 (s), 980 (s), and 702 (s) cm

1

H NMR (CDCl

): δ 1.20 (d, J  7 Hz, 6H)

3.53 (septet, J  7 Hz, 1H)

7.20–7.60 (m, 3H)

7.80–8.08 (m, 2H)

Compound B

IR (neat): 1705 (s), 738 (m), and 698 (s) cm

1

H NMR (CDCl

): δ 0.95 (t, J  7 Hz, 3H)

2.35 (q, J  7 Hz, 2H)

3.60 (s, 2H)

7.20 (s, 5H)

PROBLEM 15.84 Upon direct photolysis or heating at 220 °C,

the dimer of 2,5-dimethyl-3,4-diphenyl-2,4-cyclopentadien-

1-one (A) yields compound B, mp 166–168 °C. Spectral data for

compound B are summarized on the next page. Deduce the struc-

ture of compound B and explain your reasoning. (You may have

to make an educated guess about the actual stereochemistry of B.)

PROBLEM 15.81 Isomeric compounds A and B have the com-

position C

. Spectral data are summarized below.

Deduce structures for A and B and explain your reasoning.

Compound A

IR (KBr): 1720 (s), 1240 (s) cm

1

H NMR (CDCl

): δ 2.46 (s, 3H)

3.94 (s, 6H)

8.05 (d, J  2 Hz, 2H)

8.49 (t, J  2 Hz, 1H)

Compound B

IR (Nujol®): 1720 (s), 1245 (s) cm

1

H NMR (CDCl

): δ 2.63 (s, 3H)

3.91 (s, 6H)

7.28 (d, J  8 Hz, 1H)

8.00 (dd, J  8 Hz, 2 Hz; 1H)

dd  doublet of doublets

8.52 (d, J  2 Hz, 1H)

Compound B

Mass spectrum: m/z  492 (M)

IR (CCl

): 1704 cm

1

H NMR (CDCl

): δ 0.73 (s, 3H)

0.92 (s, 3H)

1.51 (s, 3H)

1.88 (s, 3H)

6.6–7.5 (m, 20H)

760 CHAPTER 15 Analytical Chemistry: Spectroscopy

CD E

116.1

128.8

119.0

126.5

129.5

128.7

134.9

147.7

PROBLEM 15.85 An appreciation of the concepts of elec-

tronegativity (inductive effects) and of electron delocalization

(resonance effects), combined with an understanding of ring

current effects, permits both rationalization and prediction of

many chemical shifts.

(a) The carbons of ethylene have a

C NMR chemical shift of

δ 123.3 ppm. Rationalize the

C NMR chemical shifts of

the β carbons of 2-cyclopenten-1-one (A) and methyl vinyl

ether (B). (Hint: Think resonance!)

B C

MgCl

autoclave

230 ⬚C, 2 h

(b) The

H NMR chemical shifts of hydrogens ortho, meta, and

para to a substituent on a benzene ring can also be correlated

in this way. For reference sake, note that the hydrogens in

benzene appear at about δ 7.3 ppm. Aniline (C) exhibits

three shielded aromatic hydrogens (i.e., δ  7.0), whereas the

other two aromatic hydrogens appear in the normal range.

On the other hand, two of the aromatic hydrogens in ace-

tophenone (D) are strongly deshielded (δ 7.9 ppm), one is

somewhat deshielded (δ 7.5 ppm), and the other two aro-

matic hydrogens appear at δ 7.4. Finally, all five aromatic

hydrogens in chlorobenzene (E) appear at δ 7.3 ppm. Assign

the

H NMR chemical shifts of the aromatic hydrogens for

C, D, and E by considering inductive, resonance, and ring

current effects. Additionally, the

C NMR chemical shifts

for the aromatic carbons of C and E are shown in color in

the next column. (The

C NMR chemical shift for benzene

is δ 128.5 ppm.) Are these

C NMR chemical shifts consis-

tent with the

H NMR chemical shifts for C and E?

PROBLEM 15.86 Compound A was isolated from the bark of

the sweet birch (Betula lenta). Compound A is soluble in 5%

aqueous NaOH solution but not in 5% aqueous NaHCO

solu-

tion. The spectral data for compound A are summarized below.

Deduce the structure of compound A. (Be sure to assign the

aromatic resonances in the

H NMR spectrum. Note that J

meta

and J

para

were not observed.) Hint: The pK

of phenol is about

10, and that of benzoic acid is about 4.2.

Compound A

Mass spectrum: m/z  152 (M, 49%), 121 (29%), 120 (100%),

92 (54%)

IR (neat): 3205 (br), 1675 (s), 1307 (s), 1253 (s), 1220 (s), and

757 (s) cm

1

H NMR (CDCl

, 300 MHz): δ 3.92 (s, 3H)

6.85 (t, J  8 Hz, 1H)

7.00 (d, J  8 Hz, 1H)

7.44 (t, J  8 Hz, 1H)

7.83 (d, J  8 Hz, 1H)

10.8 (s, 1H)

C NMR (CDCl

): δ 52.1 (q), 112.7 (s), 117.7 (d), 119.2 (d),

(hydrogen coupled)

130.1 (d), 135.7 (d), 162.0 (s), 170.7 (s)

PROBLEM 15.87 Dehydration of α-terpineol (A) affords a

mixture of at least two isomers, one of which is B.

(dimer)

hν

or Δ

165.1 84.2

OCH

Isomer B, which exhibits the pleasant odor of lemons and can

be isolated from the essential oils of cardamom, marjoram, and

coriander, reacts with maleic anhydride to yield compound C,

mp 64–65 °C. Spectral data for compound C are summarized

on the next page. Deduce the structures of compounds B and

C. (You may have to make an educated guess about the stereo-

chemistry of compound C.)

Compound C

Mass spectrum: m/z  234 (M)

IR (Nujol®): 1870 (shoulder), 1840 (m), and 1780 (s) cm

1

H NMR (CDCl

): δ 0.98 (d, J  6.8 Hz, 3H)

1.08 (d, J  6.8 Hz, 3H)

1.36 (m, 4H)

1.46 (s, 3H)

2.60 septet ( J  6.8 Hz, 1H)

2.80 (d, J  9 Hz, 1H)

3.20 (d, J  9 Hz, 1H)

6.01 (d, J  8.5 Hz, 1H)

6.10 (d, J  8.5 Hz, 1H)

PROBLEM 15.88 Isoquinoline (1) undergoes the Chichibabin

amination reaction when treated with potassium amide in liquid

ammonia to yield 1-aminoisoquinoline (2) (p. 676).

15.11 Additional Problems 761

2. H

1. KNH

/NH

PROBLEM 15.89 When reactions are run in THF, it is often

possible to isolate traces of an impurity, compound 1.

Compound 1 gives 81.76% C and 10.98% H upon elemental

analysis. Spectral data for 1 are summarized below. Deduce the

structure of compound 1 and explain your reasoning.

Compound 1

Mass spectrum: m/z  220 (M, 23%), 205 (100%), 57 (27%)

IR (Nujol®): 3660 (m, sharp) cm

1

H NMR (CDCl

): δ 1.43 (s, 18H), 2.27 (s, 3H), 5.00 (s, 1H,

shifts with D

O), 6.98 (s, 2H)

C NMR (CDCl

): δ 21.2 (q), 30.4 (q), 34.2 (s), 125.5 (d),

128.2 (s), 135.8 (s), 151.5 (s)

Use Organic Reaction Animations (ORA) to answer the

following questions:

PROBLEM 15.90 Select the reaction “Nucleophilic aromatic

substitution.” Predict the

H NMR spectra of the starting

material (1-chloro-2,4-dinitrobenzene) and the product

(N-methyl-2,4-dinitroaniline). How would the IR data

differ for the starting material and the product?

PROBLEM 15.91 View the “Alkene epoxidation” reaction.

Predict the

H NMR spectra for the starting material

and the product. Compare the

H NMR spectra of the starting

material and the product for the same reaction using

(Z)-2-butene.

PROBLEM 15.92 The animated “Acidic epoxide ring opening”

and “Basic epoxide ring opening” show isomeric products.

What analytical tool (

H NMR,

C NMR, IR, MS) could

you use to distinguish between the 2-methoxy-1-propanol

and the 1-methoxy-2-propanol products? What specific

differences would you look for and what methods would

you use?

When the reaction mixture containing 1, potassium amide, and

liquid ammonia was examined at 10 °C by

H NMR spec-

troscopy, the following signals were observed (solvent hydrogens

and NH

hydrogens not reported):

δ 4.87 (d, J  5.5 Hz, 1H), 5.34 (t, J  7.0 Hz, 1H; signal

collapses to a singlet when amide ion is present in excess),

6.35–7.3 (br m, 5H). Rationalize this low-temperature

H NMR spectrum. Hint: For reference sake, the

H NMR

spectrum of isoquinoline itself (1) is summarized below:

δ 7.3–8.2 (m, H

), 8.61 (d, J  6 Hz, H

9.33 (s, H

Carbonyl Chemistry 1:

Addition Reactions

762

16.1 Preview

16.2 Structure of the

Carbon–Oxygen Double Bond

16.3 Nomenclature of Carbonyl

Compounds

16.4 Physical Properties of

Carbonyl Compounds

16.5 Spectroscopy of Carbonyl

Compounds

16.6 Reactions of Carbonyl

Compounds: Simple Reversible

Additions

16.7 Equilibrium in Addition

Reactions

16.8 Other Addition Reactions:

Additions of Cyanide and

Bisulfite

16.9 Addition Reactions Followed

by Water Loss: Acetal

Formation

16.10 Protecting Groups in Synthesis

16.11 Addition Reactions of Nitrogen

Bases: Imine and Enamine

Formation

16.12 Organometallic Reagents

16.13 Irreversible Addition

Reactions: A General Synthesis

of Alcohols

16.14 Oxidation of Alcohols to

Carbonyl Compounds

16.15 Retrosynthetic Alcohol

Synthesis

16.16 Oxidation of Thiols and Other

Sulfur Compounds

16.17 The Wittig Reaction

16.18 Special Topic: Biological

Oxidation

16.19 Summary

16.20 Additional Problems

RETINAL CRYSTAL This is a polarized light photo of retinal (vitamin A), which is

a nutritionally important aldehyde found in green vegetables and fruit (see p. 533).

16.1 Preview 763

“You have two choices,” Jamf cried, his last lecture of the year: outside were

the flowery strokings of wind, girls in pale colored dresses, oceans of beer,

male choruses intensely, movingly lifted as they sang

Semper sit in flores!

Semper sit in flo-ho-res ...“stay behind with carbon and hydrogen, take

your lunch-bucket in to the works every morning with the faceless droves

who can’t wait to get in out of the sunlight—or move

beyond

. Silicon, boron,

phosphorus—these can replace carbon, and can bond to nitrogen instead of

hydrogen.—” a few snickers here, not unanticipated by the playful old

pedagogue, be he always in flower: his involvement in getting Weimar to

subsidize IG’s Stickstoff Syndikat was well known—“move beyond life,

toward the inorganic. Here is no frailty, no mortality—here is Strength, and

the Timeless.” Then his well-known finale, as he wiped away the scrawled

C—H on his chalkboard and wrote, in enormous letters, Si—N.

—THOMAS PYNCHON,

GRAVITY’S RAINBOW

16.1 Preview

In Chapters 3, 9, and 10 we examined the structure of alkenes and the addition reac-

tions of their carbon–carbon double bonds. Indeed, we have been exploring various

aspects of carbon–carbon double-bond chemistry since Chapter 9. Now it is time to

extend what we know to the chemistry of carbon–oxygen double bonds,the carbonyl

group. Although it bears many similarities to carbon–carbon double bonds, the car-

bon–oxygen double bond is not “just another double bond,” because there are many

important differences.The chemistry of the carbonyl group is the cornerstone of syn-

thetic chemistry. Much of our ability to make molecules depends on manipulation

of carbonyl groups because they allow us to make carbon–carbon bonds. There are

several centers of reactivity in a molecule containing a carbon–oxygen double bond,

and this versatility contributes to the widespread use of carbonyl compounds in syn-

thesis. There is the π bond itself (don’t forget the antibonding π* orbital), and the

nonbonding electrons on oxygen.The centerpiece reaction of this chapter is the acid-

or base-induced addition to the carbon–oxygen double bond. In addition, there is spe-

cial reactivity associated with the hydrogens on the carbon in the position adjacent

(α) to the carbon–oxygen double bond (Fig. 16.1). These hydrogens are called α

hydrogens. A ketone can have α hydrogens on either side of its carbonyl group. The

chemistry of the α position is very important, but will wait until Chapter 19.

ESSENTIAL SKILLS AND DETAILS

1. Almost this entire chapter involves additions of nucleophiles to carbon–oxygen double

bonds. Many of these reactions are reversible, others are not. It is the reversible

reactions that trouble students most, because there seems to be a bewildering array of

additions, proton transfers, and eliminations. So there is, but keep your wits about you,

and you’ll learn to push those protons and waters around properly.

2. In this chapter, you will learn a lot about synthesis, especially alcohol synthesis. It is not hard

to do these syntheses, but you have to learn the basic reaction—the irreversible addition of

organometallic reagents and metal hydrides to carbonyl compounds—very well.

Our first Pynchon quote was in Chapter 9. Same man, same book.

An α hydrogen

π Bond

Nonbonding

electrons

FIGURE 16.1 A generic carbonyl

group has three loci of reactivity: the

nonbonding electrons on the oxygen,

the π bond, and its α hydrogens.

764 CHAPTER 16 Carbonyl Chemistry 1: Addition Reactions

3. A compound does not have to be favored at equilibrium to be the reactive ingredient

in a chemical reaction. Even if it is disfavored, it can still have low barriers to further

reaction. In such a case, it is used up, then regenerated at equilibrium. In such a way, all

of the product can come from the minor partner in an equilibrium.

4. It is very useful to become comfortable with the redox interconversion of carbonyl

compounds (aldehydes and ketones) and the related alcohols.There is a lot of detail

here, and a none-too-simple mechanism as well.

5. Protecting groups are introduced in this chapter. A reactive functional group can be

converted into a nonreactive group and stored in this fashion. Later, when the reactivity

of the functional group is required, it can be regenerated as needed.There are many

protecting groups, and it is very helpful to be able to use them in synthesis.

16.2 Structure of the Carbon–Oxygen Double Bond

By analogy to the carbon–carbon double bond, we might expect sp

hybridization

for the carbon atom in the carbonyl group. The 2p orbitals on carbon and oxygen

form the π bond,and two lone pairs of electrons remain on oxygen.Figure 16.2 gives

a schematic orbital picture of both the σ and π systems for the simplest carbonyl

compound, formaldehyde.

C9 H Bond made

from 1s /sp

overlap

π Bond made from two

overlapping 2p orbitals

σ Bond

Lone-pair

electrons

FIGURE 16.2 One orbital picture of

the simplest carbonyl compound,

formaldehyde.

The structures of some known carbonyl compounds show that this picture is a

reasonable one. Bond angles are close to the 120° required by pure sp

hybridiza-

tion,and of course we would not expect a molecule of less than perfect trigonal sym-

metry to have exactly sp

hybridization. Bond lengths are similar to those in simple

alkenes, although the carbon–oxygen double bond is a bit shorter and stronger than

its carbon–carbon counterpart (Fig. 16.3).

PROBLEM 16.1 Make an orbital interaction diagram for each different kind of

bond in the molecule formaldehyde, shown in Figure 16.2.

Methanal

(formaldehyde)

Ethanal

(acetaldehyde)

Ethene

(ethylene)

118⬚

121⬚

1.06

⬚

1.23 A

⬚

118⬚

121⬚

1.09

⬚

1.22 A

⬚

116.6⬚

121.7⬚

1.08

⬚

1.33 A

⬚

WEB 3D WEB 3D

FIGURE 16.3 A comparison of the

structures of formaldehyde,

acetaldehyde, and ethylene.

16.2 Structure of the Carbon–Oxygen Double Bond 765

Formaldehyde is the smallest and

simplest aldehyde. It can be found

in substantial concentrations in new

buildings as commonly used wood

products age.

The carbonyl π bond is constructed in the same way as the related carbon–

carbon π bond.The 2p orbitals on the adjacent carbon and oxygen atoms are allowed

to interact in a constructive way to form a bonding π molecular orbital, and in a

destructive, antibonding way to form π*. But there is a difference. In an alkene, the

constituent 2p orbitals are of equal energy, and therefore contribute equally to the

formation of π and π*. In a carbonyl group, the two 2p orbitals are not identical.

An electron in the 2p orbital on the more electronegative oxygen atom is lower in

energy than one in the 2p orbital on the carbon atom, so the oxygen atom con-

tributes more to π than the more energetically remote carbon 2p orbital (Fig. 16.4).

The bonding π molecular orbital will be constructed from more than 50% of the

oxygen 2p atomic orbital and less than 50% of the carbon 2p atomic orbital. In a

similar way, we can see that π* will be made up of more than 50% of the energeti-

cally close carbon 2p orbital and less than 50% of the energetically more remote

2p orbital on oxygen.

Energy

C 2p

O 2p

π*

FIGURE 16.4 A graphical

construction of π and π* for the

carbonyl group.The π orbital is

constructed from more than 50% of

the oxygen 2p orbital, and the π*

orbital from more than 50% of the

carbon 2p orbital.

So the carbonyl π and π* orbitals bear a general resemblance to the π and π*

orbitals of the alkenes, but are less symmetrical. The two electrons naturally occu-

py the lower energy bonding molecular orbital. In the π molecular orbital, there is

766 CHAPTER 16 Carbonyl Chemistry 1: Addition Reactions

FIGURE 16.5 Carbonyl compounds

are polar molecules with substantial

dipole moments [measured in debye

(D) units]. Notice the small bond

dipole arrow that points from the

positive end toward the negative end

of the dipole.

a greater probability of finding an electron in the neighborhood of the relatively elec-

tronegative oxygen atom than near the more electropositive carbon, and this unequal

sharing of electrons contributes to the dipole moment in molecules containing

carbonyl groups. As can be seen from Figure 16.5, carbonyl compounds do have

substantial dipole moments and are quite polar molecules.

PROBLEM 16.2 It is not just the π bonds that are unsymmetrical in a carbonyl

group. Use the hybrid orbitals of oxygen and carbon to construct the C–O σ and

σ* orbitals.

PROBLEM 16.3 One way to construct a carbon–oxygen double bond using the

hybridization formalism is to use sp-hybridized oxygen. Make a drawing of this

model in the style of Figure 16.2.

In resonance terms, we can describe the carbonyl group as a combination of

the two major contributing forms shown in Figure 16.6.This picture emphasizes the

Lewis acid (electrophilic) character of the carbon of the carbon–oxygen double bond.

C CO

–

A GENERAL CARBONYL GROUP

FIGURE 16.6 A resonance

formulation of a carbonyl group.

–

WORKED PROBLEM 16.5 Why is the dipole moment in acetone larger than that in

acetaldehyde (Fig. 16.5)?

ANSWER Because the negative end of the dipole is the same for each species,

there can be no answer there. However, the two methyl groups in acetone stabi-

lize the positive charge much better than the methyl and hydrogen of acetalde-

hyde do. The polar form is more stable in acetone than it is in acetaldehyde, and

as a result, acetone has a larger dipole moment.

2.33 D

2.88 D

2.69 D

–

Methanal

(formaldehyde)

Ethanal

(acetaldehyde)

Propanone

(acetone)

WEB 3D WEB 3D

WORKED PROBLEM 16.4 There is another possible polar resonance form in addi-

tion to the one shown in Figure 16.6. What is it, and why is it not considered a

major contributor to the carbonyl group?

ANSWER The other polar resonance form has the positive charge on oxygen and

the negative charge on carbon. Oxygen is more electronegative than carbon and

the negative charge will be more stable on oxygen than on carbon. Similarly, the

relatively electropositive carbon should have the positive charge, not the negative.

For these reasons, this form is not a major contributor to the structure.

16.3 Nomenclature of Carbonyl Compounds 767

16.3 Nomenclature of Carbonyl Compounds

In a generic carbonyl compound R

C O, there is a vast array of possible substi-

tution patterns,and there is a separate name for each one of them. Simple compounds

in which both R groups are hydrocarbons such as alkyl, alkenyl, or aryl are called

ketones. If one R group is hydrogen, the compound is an aldehyde, and the mol-

ecule in which both R groups are hydrogen is called formaldehyde.

Common names are often used for the aldehydes containing four carbons or fewer,

but thereafter the IUPAC system takes over, and the naming process becomes ration-

al. The “e” of the parent alkane’s name is dropped and the suffix “al” is added. In the

IUPAC numbering system the aldehyde group is always number 1. When aldehydes

are attached to cycloalkanes, the term carboxaldehyde is used,as in “cyclopropanecar-

boxaldehyde.” Some aldehydes and their IUPAC names are given in Figure 16.7.

Cyclohexanecarboxaldehyde

3-Chloro-2-hydroxypropanal

Cl OH

2-Methylpentanal

Pentanal

Butanal

(butyraldehyde)

2-Methylpropanal

(isobutyraldehyde)

Propanal

(propionaldehyde)

Ethanal

(acetaldehyde)

FIGURE 16.7 A few aldehydes and their IUPAC names are shown. Common names are given in parentheses.

Compounds containing two aldehydes are dials. In this case, the final “e” of the

parent name is not dropped before adding “dial” (Fig. 16.8).

Propanedial

Butanedial

WEB 3D

FIGURE 16.8 Dialdehydes are named

as dials.

Ketones, too, have common names, but most have been discarded. An exception

is the name for the simplest member of the class, dimethyl ketone,which is still called

acetone. Ketones are named in several ways. For example, acetone is also known as

768 CHAPTER 16 Carbonyl Chemistry 1: Addition Reactions

Propanone

(acetone)

(dimethyl ketone)

Butanone

(ethyl methyl ketone)

3-Methyl-2-pentanone

(sec-butyl methyl ketone)

WEB 3D

FIGURE 16.9 Simple ketones can be

named in many ways. The IUPAC

names are listed first.

PROBLEM 16.6 Why is acetone properly named propanone and not 2-propanone?

The rules are fairly simple: in the presence of a higher priority group the IUPAC

system treats the carbonyl oxygen atom as a substituent that is described by the

prefix “oxo-” and given a number like any other attached group. A great many

common names are retained for aromatic aldehydes and ketones. For example, the

simplest aromatic aldehyde is not generally called “benzenecarboxaldehyde” but

benzaldehyde. That one’s quite easy to figure out, but most are not. A few com-

monly used names are given in Figure 16.11.

Compounds containing two ketones are diones. As with dials, the final “e” of

the parent alkane is retained,and the name begins with the position numbers of the

two carbonyl groups (Fig. 16.10).Ketoaldehydes are named as aldehydes, not ketones,

because the aldehyde group has a higher priority than a keto group (Fig. 16.10).

2-oxopropane, propanone, or dimethyl ketone (Fig. 16.9). In using the IUPAC sys-

tem, the final “e” of the parent alkane is dropped, and the suffix “one” is appended.

The number of the carbon atom bearing the carbon–oxygen double bond is placed

before the parent name.

2,4-Pentanedione

1,4-Cyclohexanedione

3-Oxohexanal

2-Oxopropanal

(pyruvic aldehyde)

FIGURE 16.10 Diketones are named as diones. The final “e” of the parent hydrocarbon is retained.

Ketoaldehydes are named as aldehydes, not ketones.

OCH

Benzaldehyde

4-Methoxybenzaldehyde

( p-anisaldehyde)

4-Methylbenzaldehyde

(p-tolualdehyde)

(E)-3-Phenyl-2-propenal

(trans-cinnamaldehyde)

FIGURE 16.11 Some common names

for aromatic aldehydes are shown.

The IUPAC accepted names

are listed first.