It has been found extremely useful to classify each caibon atom of an alkane with respect to the number of other carbon atoms to which it is attached.
SEC. 3.12 PHYSICAL PROPERTIES 85
A primary (7) carbon atom is attached to only one other carbon atom; a secondary
(2) is attached to two others; and a tertiary (3) to three others. For example:
Each hydrogen atom is similarly classified, being given the same designation of primary, secondary, or tertiary as the carbon atom to which it is attached. We shall make constant use of these designations in our consideration of the
relative reactivities of various parts of an alkane molecule.
Physical properties
The physical properties of the alkanes follow the pattern laid down bymethane, and are consistent with the alkane structure. An alkane molecule is held together entirely by covalent bonds. These bonds either join two atoms of the same kind and hence are non-polar, or join two atoms that differ very little in electronegativity and hence are only slightly polar. Furthermore, these bonds are directed in a very symmetrical way, so that the slight bond polarities tend to cancel out. As a result an alkane molecule is either non-polar or very weakly polar As we have seen ^Sec. 1.19), the forces holding non-polar molecules together (van der Waals forces) arc weak and of very short range; they act only between the portions of different molecules that are in close contact, that is, between the surfaces of molecules. Within a family, therefore, we would expect that the larger the molecule and hence the larger its surface area the stronger the intermolecular forces.
Table 3.3 lists certain physical constants for a number of the w-alkanes. As
we can see, the boiling points and melting points rise as the number of carbons increases. The processes of boiling and melting require overcoming the intermolecular forces of a liquid and a solid; the boiling points and melting points rise because these intermolecular forces increase as the molecules get larger. Except for the very small alkanes, the boiling point rises 20 to 30 degrees for each carbon that is added to the chain ; we shall find that this increment of 20-30 per carbon hofds not only for the alkanes but also for each of the homologous series that we shall study. The increase in melting point is not quite so regular, since the intermolecular forces in a crystal depend not only upon the size of the molecules but also upon how well they fit into a crystal lattice. The first four i-alkanes are gases, but, as a result of the rise in boiling point and melting point with increasing chain length, the next 13 (C$Cn) are liquids, and those- containing 18 carbons or more are solids.
There are somewhat smaller differences among the boiling points of alkanes that have the same carbon number but different structures. On pages 77 and 80 the boiling points of the isomeric butanes, pentanes, and hexanes are given. We see that in every case a branched-chain isomer has a lower boiling point than a straight-chain isomer, and further, that the more numerous the branches, the lower the boiling point. Thus w-butane has a boiling point of and isobutane 12.m w-Pentane has a boiling point of 36, isopentane with a single branch 28, and neopentane with two branches 9.5. This effect of branching on boiling point is observed within all families of organic compounds. That branching should lower the boiling point is reasonable: with branching the shape of the molecule tends to approach that of a sphere; and as this happens the surface area decreases, with the result that the intermolecular forces become weaker and are overcome at a lower temperature. In agreement with the rule of thumb, "like dissolves like," the alkanes are
soluble in non-polar solvents such as benzene, ether, and chloroform, and are insoluble in water and other highly polar solvents. Considered themselves as solvents, the liquid alkanes dissolve compounds of low polarity and do not dissolve compounds of high polarity.
The density increases with size of the alkanes, but tends to level off at about 0.8; thus all alkanes are less dense than water. It is not surprising that nearly all organic compounds are less dense than water since, like the alkanes, they consist chiefly of carbon and hydrogen. In general, to be denser than water a compound must contain a heavy atom like bromine or iodine, or several atoms like chlorine.
Industrial source
The principal source of alkanes is petroleum, together with the accompanying natural gas. Decay and millions of years of geologicaTstresses have transformed the complicated organic compounds that once made up living plants or animals into a mixture of alkanes ranging in size from one carbon to 30 or 40 carbons. Formed along with the alkanes, and particularly abundant in California petroleum, are cycloalkanes (Chap. 9), known to the petroleum industry as naphthenes. The other fossil fuel, coal, is a potential second source of alkanes: processes are being developed to convert coal, through hydrogenation, into gasoline and fuel oil, and into synthetic gas to offset anticipated shortages of natural gas. Natural gas contains, of course, only the more volatile alkanes, that is, those of low molecular weight; it consists chiefly of methane and progressively smaller amounts of ethane, propane, and higher alkanes. For example, a sample taken from a pipeline supplied by a large number of Pennsylvania wells contained methane, ethane, and propane in the ratio of 12:2: 1, with higher alkanes making up only 3% of the total. The propane-butane fraction is separated from the more volatile components by liquefaction, compressed into cylinders, and sold as bottled gas in areas not served by a gas utility. Petroleum is separated by distillation into the various fractions listed in Table 3.4; because of the relationship between boiling point and molecular weight, this amounts to a rough separation according to carbon number. Each fraction is still a very complicated mixture, however, since it contains alkanes of a range of carbon numbers, and since each carbon number is represented by numerous isomers. The use that each fraction is put to depends chiefly upon its volatility or viscosity, and it matters very little whether it is a complicated mixture or a single pure compound. (In gasoline, as we shall see in Sec. 3.30, the structures of the components are of key importance.) The chief use of all but the non-volatile fractions is as fuel. The gas fraction, like natural gas, is used chiefly for heating. Gasoline is used in those internal combustion engines that require a fairly volatile fuel, kerosene is used in tractor and jet engines, and gas oil is used in Diesel engines. Kerosene and gas oil are also used for heating purposes, the latter being the familiar "furnace oil."
The lubricating oil fraction, especially that from Pennsylvania crude oil (paraffin-base petroleum), often contains large amounts of long-chain alkanes (C20-C34) that have fairly high melting points. If these remained in the oil, they might crystallize to waxy solids in an oil line in cold weather. To prevent this, the oil is chilled and the wax is removed by filtration. After purification this is sold as solid paraffin wax (m.p. 50-55) or used in petrolatum jelly (Vaseline). Asphalt is used in roofing and road building. The coke that is obtained from paraflin-base crude oil consists of complex hydrocarbons having a high carbon-to-hydrogen ratio; it is used as a fuel or in the manufacture of carbon electrodes for the electrochemical industries. Petroleum ether and ligroin are useful solvents for many organic materials of low polarity. In addition to being used directly as just described, certain petroleum fractions are converted into other kinds of chemical compounds. Catalytic isomerization
changes straight-chain alkanes into branched-chain ones. The cracking process
(Sec. 3.31) converts higher alkanes into smaller alkanes and alkenes, and thus
increases the gasoline yield; it can even be used for the production of "natural"
gas. In addition, the alkenes thus formed are perhaps the most important raw
materials for the large-scale synthesis of aliphatic compounds. The process of
catalytic reforming (Sec. 12.4)converts alkanes and cycloalkanesinto aromatic hydrocarbons
and thus helps provide the raw material for the large-scale synthesis of
another broad class of compounds.
Industrial source vs. laboratory preparation
We shall generally divide the methods of obtaining a particular kind of
organic compound into two categories: industrial source and laboratory preparation.
We may contrast the two in the following way, although it must be realized that
there are many exceptions to these generalizations.
An industrial source must provide large amounts of the desired material at
the lowest possible cost. A laboratory preparation may be required to produce
only a few hundred grams or even a few grams; cost is usually of less importance
than the time of the investigator.
For many industrial purposes a mixture may be just as suitable as a pure
compound; even when a single compound is required, it may be economically
feasible to separate it from a mixture, particularly when the other components
may also be marketed. In the laboratory a chemist nearly always wants a single
pure compound. Separation of a single compound from a mixture of related
substances is very time-consuming and frequently does not yield material of the
required purity. Furthermore, the raw material for a particular preparation may
well be the hard-won product of a previous preparation or even series of preparations, and hence he wishes to convert it as completely as possible into his desired
compound. On an industrial scale, if a compound cannot be isolated from naturally
occurring material, it may be synthesized along with a number of related
compounds by some inexpensive reaction. In the laboratory, whenever possible,
a reaction is selected that forms a single compound in high yield.
In industry it is frequently worth while to work out a procedure and design
apparatus that may be used in the synthesis of only one member of a chemical
family. In the laboratory a chemist is seldom interested in preparing the same
compound over and over again, and hence he makes use of methods that are
applicable to many or all members of a particular family.
In our study of organic chemistry, we shall concentrate our attention on
versatile laboratory preparations rather than on limited industrial methods. In
learning these we may, for the sake of simplicity, use as examples the preparation
of compounds that may actually never be made by the method shown. We may
discuss the synthesis of ethane by the hydrogenation of ethylene, even though we
can buy all the ethane we need from the petroleum industry. However, if we know
how to convert ethylene into ethane, then, when the need arises, we also know how
to convert 2-methyl-l-hexene into 2-methylhexane, or cholesterol into cholestanol,
or, for that matter, cottonseed oil into oleomargarine.
The Grignard reagent: an organometallic compound
When a solution of an alkyl halide in dry ethyl ether, (C2H5)2O, is allowed to
stand over turnings of metallic magnesium, a vigorous reaction takes place: the
solution turns cloudy, begins to boil, and the magnesium metal gradually disappears.
The resulting solution is known as a Grignard reagent, after Victor Grignard
(of the University of Lyons) who received the Nobel prize in 1912 for its
discovery. It is one of the most useful and versatile reagents known to the organic
chemist.
The Grignard reagent has the general formula RMgX, and the general name
alkylmagnesium halide. The carbon-magnesium bond is covalent but highly polar,
with carbon pulling electrons from electropositive magnesium; the magnesiumhalogen
bond is essentially ionic.
R:Mg+:X:-
Since magnesium becomes bonded to the same carbon that previously held
halogen, the alkyl group remains intact during the preparation of the reagent.
Thus 7i-propyl chloride yields /i-propylmagnesium chloride, and isopropyl chloride
yields isopropylmagnesium chloride.
The Grignard reagent is the best-known member of a broad class of substances,
called organometallic compounds, in which carbon is bonded to a metal:
lithium, potassium, sodium, zinc, mercury, lead, thallium almost any metal
known. Each kind of organometallic compound has, of course, its own set of
properties, and 'its particular uses depend on these. But, whatever the metal, it is
less electronegative than carbon, and the carbon-metal bond like the one in the
Grignard reagent is highly polar. Although the organic group is not a fullfledged
carbanion an anion in which carbon carries negative charge it nevertheless
has considerable carbanion- character. As we shall see, organometallic
compounds owe their enormous usefulness chiefly to one common quality: they
can serve as a source from which carbon is readily transferred with its electrons.
The Grignard reagent is highly reactive. It reacts with numerous inorganic
compounds including water, carbon dioxide, and oxygen, and with most kinds of
organic compounds ; in many of these cases the reaction provides the best way to
make a particular class of organic compound.
The reaction with water to form an alkane is typical of the behavior of the
Grignard reagent and many of the more reactive organometallic compounds
toward acids. In view of the marked carbanion character of the alkyl group, we
may consider the Grignard reagent to be the magnesium salt, RMgX, of the
extremely weak acid, R H. The reaction
is simply the displacement of the weaker acid, R H, from its salt by the stronger
acid, HOH.
An alkane is such a weak acid that it is displaced from the Grignard reagent
by compounds that we might ordinarily consider to be very weak acids themselves,
or possibly not acids at all. Any compound containing hydrogen attached to
oxygen or nitrogen is tremendously more acidic than an alkane, and therefore can
decompose the Grignard reagent: for example, ammonia or methyl alcohol.
For the preparation of an alkane, one acid is as good as another, so we naturally
choose water as the most available and convenient.
Problem 3.9 (a) Which alkane would you expect to get by the action of water
on w-propylmagnesium chloride? (b) On isopropylmagnesium chloride? (c) Answer
(a) and (b) for the action of deuterium oxide ("heavy water," D2O).
Problem 3.10 On conversion into the Grignard reagent followed by treatment
with water, how many alkyl bromides would yield: (a) //-pentane; (b) 2-methylbutane;
(c) 2,3-dimethylbutane; (d) neopentane? Draw the structures in each case.
Non-rearrangement of free radicals. Isotopic tracers
Our interpretation of orientation (Sec. 3.21) was based on an assumption that
we have not yet justified: that the relative amounts of isomeric halides we find in
the product reflect the relative rates at which various free radicals were formed
from the alkane. From isobutane, for example, we obtain twice as much isobutyl
chloride as te/7-butyl chloride, and we assume from this that, by abstraction of
hydrogen, isobutyl radicals are formed twice as fast as /erf-butyl radicals.
Yet how do we know, in this case, that every isobutyl radical that is formed
ultimately yields a molecule of isobutyl chloride? Suppose some isobutyl radicals
were to change- by rearrangement of atoms into /react with chlorine to yield ter/-butyl chloride. This supposition is not so far-
fetched as we, in our present innocence, might think; the doubt Jt raises is a very real one. We shall shortly see that another kind of reactive intermediate particle, the carbonium ion, is \ery prone to rearrange, with less stable ions readil> changing into more stable ones (See. 5.22). H. C. Brown (of Purdue University) and Glen Russell (no\\ of loua State University) decided to test the possibility that free radicals, like carbonium ions, might rearrange, and chose the chlorination of isobutane as a good test case,
because of the large difference in stability between w/-butyl and isobutyl radicals.
If rearrangement of alkyl radicals can indeed lake place, it should certainly happen
here.
What the problem comes down to is this: does every abstraction of primary
hydrogen lead to isobuiyl chloride, and every abstraction of tertiary hydrogen lead
to tert-butyl chloride? This, we might say, we could never know, because all
hydrogen atoms are exactly alike. But are they? Actually, three isotopes of hydrogen
exist: 'H, profhtm, ordinary hydrogen: 2H or D, deuterium* heavy hydrogen;
and 3 H or T, tritium. Protium and deuterium are distributed in nature in
the ratio of 5000: 1. '(Tritium, the unstable, radioactive isotope, is present in
traces, but can be made by neutron bombardment of 6 Li.) Modern methods of
separation of isotopes have made very pure deuterium available, at moderate
prices, in the form of deuterium oxide. D2O, heavy water.
Brown and Russell prepared the deuterium-labeled isobutane I,
photochemically chlorinated it, and analyzed the products. The DClrHCl ratio
(determined by the mass spectrometer) was found to be equal (within experimental
error) to the tert-buiyl chloride: isobutyl chloride ratio. Clearly, every abstraction
of a tertiary hydrogen (deuterium) gave a molecule of tert~buty\ chloride, and every
abstraction of a primary hydrogen (protiwn) gave a molecule of isobutyl chloride.
Rearrangement of the intermediate free radicals did not occur.
All the existing evidence indicates quite strongly that, although rearrangement
of free radicals occasionally happens, it is not very common and does not involve
simple alkyl radicals.
Problem 3.18 (a) What results would have been obtained if some isobutyl
radicals had rearranged to tert-buiyl radicals? (b) Suppose that, instead of rearranging,
isobutyl radicals were, in effect, converted into terr-butyl radicals by the reaction
What results would Brown and Russell have obtained?
Problem 3.19 Keeping in mind the availability of D2O, suggest a way to make I
from ter/~butyl chloride. (Hint: See Sec, 3.16.)
The work of Brown and Russell is just one example of the way in which we can
gain insight into a chemical reaction by using isotopically labeled compounds. We
shall encounter many other examples in which isotopes, used either as tracers, as
in this case, or for the detection of isotope effects (Sec. 11.15), give us information
about reaction mechanisms that we could not get in any other way.
Besides deuterium and tritium, isotopes commonly used in organic chemistry
include: 14C, available a
8O; 15N, as 15NH3 ,
15N(V, and 15NO2
~
;
36
C1, as chlorine or chloride; 131
I, as iodide.
Problem 3.20 Bromination of methane is slowed down by the addition of
HBr (Problem 14, p. 71); this is attributed to the reaction
CH3
- + HBr > CH4 + Brwhich,
as the reverse of one of the chain-carrying steps, slows down bromination.
How might you test whether or not this reaction actually occurs in the bromination
mixture?
Analysis of alkanes
An unknown compound is characterized as an alkane on the basis of negative
evidence.
Upon qualitative elemental analysis, an alkane gives negative tests for all
elements except carbon and hydrogen. A quantitative combustion, if one is carried
out, shows the absence of oxygen; taken with a molecular weight determination,
the combustion gives the molecular formula, -CnH2n + 2 which is that of an alkane.
An alkane is insoluble not only in water but also in dilute acid and base and in
concentrated sulfuric acid. (As we shall see, most kinds of organic compounds
dissolve in one or more of these solvents.)
An alkane is unreactive toward most chemical reagents. Its infrared spectrum
lacks the absorption bands characteristic of groups of atoms present in other families
of organic compounds (like OH, C~O, ,C C, etc.).
Once the unknown has been characterized as an alkane, there remains the
second half of the problem: finding out which alkane.
On the basis of its physical properties boiling point, melting point, density,
refractive index, and, most reliable of all, its infrared and mass spectra it may be
identified as a previously studied alkane of known structure.
If it turns out to be a new alkane, the proof of structure can be a difficult
job. Combustion and molecular weight determination give its molecular formula.
Clues about the arrangement of atoms are given by its infrared and nmr spectra.
(For compounds like alkanes, it may be necessary to lean heavily on x-ray diffraction
and mass spectrometry.)
Final proof lies in synthesis of the unknown by a method that can lead only
to the particular structure assigned.
(The spectroscopic analysis of alkanes will be discussed in Sees. 13.15-13.16.)
Alkena
The functional group
The characteristic feature of the alkene structure is the carbon-carbon double
bond. The characteristic reactions of an alkene are those that take place at the
double bond. The atom or group of atoms that defines the structure of a particular
family of organic compounds and, at the same time, determines their properties is
called the functional group.
In alkyl halides the functional group is the halogen atom, and in alcohols the
OH group; in alkcnes it is the carbon- carbon double bond. We must not forget
that an alkyl halide, alcohol, or alkene has alkyl groups attached to these functional
groups; under the proper conditions, the alkyl portions of these molecules undergo
the reactions typical of alkanes. However, the reactions that are characteristic of
each of these compounds are those that occur at the halogen atom or the hydroxyl
group or the carbon-carbon double bond.
A large part of organic chemistry is therefore the chemistry of the various
functional groups. We shall learn to associate a particular set of properties with a
particular group wherever we may find it. When we encounter a complicated
molecule, which contains a number of different functional groups, we may expect
the properties of this molecule to be roughly a composite of the properties of the
various functional groups. The properties of a particular group may be modified,
of course, by the presence of another group and it is important for us to understand
these modifications, but our point of departure is the chemistry of individual
functional groups.
Analysis of alkencs
The functional group of an alkene is the carbon-carbon double bond. To
characterize an unknown compound as an alkene, therefore, we must show that
it undergoes the reactions typical of the carbon-carbon double bond. Since there
are so many of these reactions, we might at first assume that this is an easy job.
But let us look at the problem more closely.
First of all, which of the many reactions of alkenes do we select? Addition
of hydrogen bromide, for example 9 Hydrogenation ? Let us imagine ourselves
in the laboratory, working with gases and liquids and solids, with flasks and test
tubes and bottles.
We could pass dry hydrogen bromide from a tank through a test tube of an
unknown liquid. But what would we see? How could we tell whether or not a
reaction takes place? A colorless gas bubbles through a colorless liquid; a difierent
colorless liquid may or may not be formed.
We could attempt to hydrogenate the unknown compound. Here, we might
say, we could certainly tell whether or not reaction takes place: a drop in the
hydrogen pressure would show us that addition had occurred. This is true, and
hydrogenalion can be a useful analytical tool. But a catalyst must be prepared,
and a fairly elaborate piece of apparatus must be used ; the whole operation might
take hours.
Whenever possible, we select for a characteritation test a reaction that is
rapidly and conveniently carried out, and that gives rise to an easily observed change.
We select a test that requires a few minutes and a few test tubes, a test in which
a color appears or disappears, or bubbles of gas are evolved, or a precipitate forms
or dissolves.
Experience has shown that an alkene is best characterized, then, by its property
of decolorizing both a solution of bromine in carbon tetrachloride (Sec. 6.5) and
a cold, dilute, neutral permanganate solution (the Baeyer test. Sec. 6.20). Both
tests are easily carried out; in one, a red color disappears, and in the other, a purple
color disappears and is replaced by brown manganese dioxide.
Granting that we have selected the best tests for the characterization of alkenes,
let us go on to another question. We add bromine in carbon tetrachloride to an
unknown organic compound, let us say, and the red color disappears. What
does this tell us? Only that our unknown is a compound that reacts with bromine.
It may be an alkene. But it is not enough merely to know that a particular kind
ofcompound reacts with a given reagent; we must also know what other kinds of
compounds also react with the reagent. In this case, the unknown may equally
well be an alkyne. (It may also be any of a number of compounds that undergo
rapid substitution by bromine; in that case, however, hydrogen bromide would be
evolved and could be detected by the cloud it forms when we blow our breath over
the test tube.)
In the same way, decolorization of permanganate does not prove that a compound
is an alkene, but only that it contains some functional group that can be
oxidized by permanganate. The Compound may be an atttene; but it may instead
be a& atkyne, an aldehyde, or any of a number of easily oxidized compounds.
It may even be a compound that is contaminated with an Impurity that fa oxidised;
alcohols, for example, are not oxidized under tlrtse conditions, but often contain
impurities that ore. We can usually rule out this by making sure that more than a
drop or two of the reagent is decolorized*
By itself, a single characterization test seldom proves that an unknown is
one particular kind of compound. It may limit tte number of possibilities, so
that a final decision can then be made on tbe ba*m of ad4itloaal tests. Of, conversely,
if certain possibilities have already been eHmto*trf, a tingle tort may ponnk
a final choice to be made. Thus, the bromine or pwmaganate test would to
sufficient to differentiate an alkene from an alkane, ot an aJfcene from an alky!
halide, or an alkene from an alcohol.
The tests most used in characterizing alkenes, then, are the following: (a) rapid
decolorization of bromine in carbon tetrachloride without evolution of HBr,
a test also given by alkynes; (b) decolorization of cold, dilute, neutral, aqueous
permanganate solution (the Baeyer test), a test also given by alkynes and aldehydes.
Also helpful is the solubility of alkenes in cold concentrated sulfuric acid, a test
also given by a great many other compounds, including all those containing oxygen
(they form soluble oxonium salts) and compounds that are readily sulfonated
(Sees. 12.11 and 17.8). Alkanes or alkyl halides are not soluble in cold concentrated
sulfuric acid.
Of the compounds we have dealt with so far, alcohols also dissolve in sulfuric
acid. Alcohols can be distinguished from alkenes, however, by the fact that
alcohols give a negative test with bromine in carbon tetrachloride and a negative
Baeyer test so long as we are not misled by impurities. Primary and secondary
alcohols ore oxidized by chromic anhydride, CrO3 , in aqueous sulfuric acid:
within two seconds, the clear orange solution turns blue-green and becomes
opaque.
Tertiary alcohols do not give this test; nor do alkenes.
Problem 6.19 Describe simple chemical tests (if any) that would distinguish j
between: (a) an alkene and an alkane; (b) an alkene and an alkyl halide; (c) an alkenc
"
and a secondary alcohol; (d) an alkene, an alkane, an alkyl halide, and a secondary
alcohol. Tell exactly what you would do and see.
Problem 6.20 Assuming the choice to be limited to alkane, alkene, alkyl halide,
secondary alcohol, and tertiary alcohol, characterize compounds A, B, C, D, and E
on the basis of the following information:
Once characterized as an alkene, an unknown may then be identified as a
previously reported alkene on the basts of its physical properties, including its
infrared spectrum and molecular weight. Proof of structure of a new compound
is best accomplished by degradation: cleavage by ozone or permanganate, followed
by identification of the fragments formed (Sec. 6.29).
(Spectroscopic analysis of alkenes will be discussed in Sees. 13.15-13.16.)
Structure and nomenclature of ethers
Ethers are compounds of the general formula R -O R, Ar O R, or
Ar-O Ar.
To name ethers we usually name the two groups that are attached to oxygen,
and follow these names by the word ether:
Physical properties of ethers
Since the C O C bond angle is not 180, the dipole moments of the two
C O bonds do not cancel each other; consequently, ethers possess a small net
dipole moment (e.g., 1.18 D for ethyl ether).
This weak polarity does not appreciably affect the boiling points of ethers,
which are about the same as those of alkanes having comparable molecular weights,
and much lower than those of isomeric alcohols. Compare, for example, the
boiling points of /i-heptane (98), methyl /i-pentyl ether (100), and w-hexyl alcohol
(157). The hydrogen bonding that holds alcohol molecules strongly together is
not possible for ethers, since they contain hydrogen bonded only to carbon (Sec.
15.4):
On the other hand, ethers show a solubility in water comparable to that of
the alcohols, both ethyl ether and w-butyl alcohol, for example, being soluble to the
extent of about 8 g per 100 g of water. We attributed the water solubility of the
lower alcohols to hydrogen bonding between water molecules and alcohol molecules;
presumably the water solubility of ether arises in the same way.
Cyclic ethers
In their preparation and properties, most cyclic ethers are just like the ethers
we have already studied: the chemistry of the ether linkage is essentially the same
whether it forms part of an open chain or part of an aliphatic ring.
Problem 17.14 1,4-Dioxane is prepared industrially (for use as a water-soluble
solvent) by dehydration of an alcohol. What alcohol is used?
Problem 17.15 The unsaturated cyclic ether fnran can readily be made from sub-
Stances isolated from oat hulls and corncobs; one of its important uses involves its
conversion into (a) tetrahydrofnnin, and (b) 1,4-dichlorobutane. Using your knowledge
of alkene chemistry and ether chemistry, show hens these conversions can be carried
out.
Analysis of ethers
Because of the low reactivity of the functional group, the chemical behavior
of ethers both aliphatic and aromatic resembles that of the hydrocarbons to
which they are related. They are distinguished from hydrocarbons, however,
by their solubility in cold concentrated sulfuric acid through formation of oxonium
salts.
Problem 17.23 Because of their highly reactive benzene rings, aryf ethers may
decolorize bromine in carbon tetrachloride. How coutd this behavior be distinguished
from the usual unsaturation test? (Hint: Sec Sec. 6.30.)
Problem 17.24 Expand the table you made in Problem 16.10, p. 536, to include
ethers.
Problem 17.25 Describe simple chemical tests (if any) that would distinguish
between an aliphatic ether and (a) an aikane; (b) an alkene; (c) an alkyne; (d) an
alkyl halide; (e) a primary or secondary alcohol; (f) a tertiary alcohol; (g) an alkyl
aryl ether
Identification as a previously reported ether is accomplished through the usual
comparison of physical properties. This can be confirmed by cleavage with hot
concentrated hydriodic acid (Sec, 17.7) and identification of one or both products.
Aromatic ethers can be converted into solid bromination or nitration products
whose melting points can then be compared with those of previously reported
derivatives.
Proof of structure of a new ether would involve cleavage by hydriodic acid
and identification of the products formed. Cleavage is used quantitatively in the
Zeisel method to show the number of alkoxyl groups in an alkyl aryl ether.
Problem 17.26 How many methoxyl groups per molecule of papaverine would
be indicated by the following results of a Zeisel analysis ?
Treatment of papaverine (C2oH2|O4N, one of the opium alkaloids) with hot
concentrated hydriodic acid yields CHJ, indicating the presence of the methoxyl
group OCHj. When 4.24 mg of papaverine is treated with hydriodic acid and the
CH3 I thus formed is passed into alcoholic silver nitrate, 11.62 mg of silver iodide is
obtained.
17.17 Spectroscopic analysis of ethers
Infrared. The infrared spectrum of an ether does not, of course, show the
O H band characteristic of alcohols; but the strong band due to C O stretching
PROBLEMS 571
is still present, in the 1060-1300 cm" 1
range, and is the striking feature of the
spectrum. (See Fig. 17.1).
CO stretching, strong, broad
Alkyl ethers 1060-1 150 cm" 1
Aryl and vinyl ethers 1200-1275 cm" 1
(and, weaker, at 1020-1075 cm' 1
)
Carboxylic acids and esters show C O stretching, but show carbonyl absorption
as well. (For a comparison of certain oxygen compounds, see Table 20.3, p. 689.)
Alkohol
Structure
Alcohols are compounds of the general formula ROH, where R is any alkyl
or substituted alkyl group. The group may be primary, secondary, or tertiary;
it may be open-chain or cyclic; it may contain a double bond, a halogen atom,
or an aromatic ring. For example:
All alcohols- contain the hydroxyl (- OH) group, which, as the functional
group, determines the properties characteristic of this family. Variations in
structure of the R group may affect the rate at which the alcohol undergoes certain
reactions, and even, in a few cases, may affect the kind of reaction.
Compounds in which the hydroxyl group is attached directly to an aromatic
ring are not alcohols; they are phenols, and differ so markedly from the alcohols
that we shall consider them in a separate chapter.
Classification
We classify a carbon atom as primary, secondary, or tertiary according to the
number of other carbon atoms attached to it (Sec. 3.1 1). An alcohol is classified
according to the kind of carbon that bears the OH group:
One reaction, oxidation, which directly involves the hydrogen atoms attached
to the carbon bearing the OH group, takes an entirely different course for each
class of alcohol. Usually, however, alcohols of different classes differ only in
rate or mechanism of reaction, and in a way consistent with their structures. Certain
substituents may affect reactivity in such a way as to make an alcohol of one class
resemble the members of a different class; benzyl alcohol, for example, though
formally a primary alcohol, often acts like a tertiary alcohol. We shall find that
these variations, too, are consistent with the structures involved.
Physical properties
The compounds we have studied so far, the various hydrocarbons, are nonpolar or nearly so, and have the physical properties that we might expect of suchcompounds: the relatively low melting points and boiling points that are characteristic
of molecules with weak intermolecular forces; solubility in
non-polar
solvents and insolubility in polar solvents like water.
Alcohols, in contrast, contain the very polar OH group. In particular, this
group contains hydrogen attached to the very electronegative element, oxygen,
and therefore permits hydrogen bonding (Sec. 1.19). The physical properties (Table
15.1) show the effects of this hydrogen bonding.
Let us look first at boiling points. Among hydrocarbons the factors that
determine boiling point seem to be chiefly molecular weight and shape; this is
to be expected of molecules that are held together chiefly by van der Waals forces.
Alcohols, too, show increase in boiling point with increasing carbon number, and
decrease in boiling point with branching. But the unusual thing about alcohols is
that they boil so high: as Table 15.2 shows, much higher than hydrocarbons of
the same molecular weight, and higher, even, than many other compounds of
considerable polarity. How are we to account for this?
The answer is, of course, that alcohols, like water, are associated liquids:
their abnormally high boiling points are due to the greater energy needed to break
the hydrogen bonds that hold the molecules together. Although ethers and aldehydes
contain oxygen, they contain hydrogen that is bonded only to carbon; these
hydrogens are not positive enough to bond appreciably with oxygen.
Infrared spectroscopy (Sec. 13.4) has played a key role in the study of hydrogen
bonding. In dilute solution in a non-polar solvent like carbon tetrachloriue
(or in the gas phase), where association between molecules is minimal, elhanol,
for example, shows an O H stretching band at 3640 cm" 1
. As the concentration
of ethanol is increased, this band is gradually replaced by a broader band at
3350 cm" 1
. The bonding of hydrogen to the second oxygen weakens the O H
bond, and lowers the energy and hence the frequency of vibration.
The solubility behavior of alcohols also reflects their ability to form hydrogen
bonds. In sharp contrast to hydrocarbons, the lower alcohols are miscible with
water. Since alcohol molecules are held together by the same sort of intermodular
forces as water molecules, there can be mixing of the two kinds of molecules: the
energy required to break a hydrogen bond between two water molecules or two
alcohol molecules is provided by formation of a hydrogen bond between a water
molecule and an alcohol molecule.
This is true, however, only for the lower alcohols, where the OH group
constitutes a large portion of the molecule. A long aliphatic chain with a small
OH group at one end is mostly alkane, and its physical properties show this.
The change in solubility with carbon number is a gradual one: the first three
primary alcohols are miscible with water; -butyl alcohol is soluble to the extent
of 8 g per 100 g water; w-pentyl, 2 g; w-hexyl, 1 g; and the higher alcohols still less.
For practical purposes we consider that the borderline between solubility and
insolubility in water occurs at about four to five carbon atoms for normal primary
alcohols.
Polyhydroxy alcohols provide more than one site per molecule for hydrogen
bonding, and their properties reflect this. The simplest glycol, ethylene glycol,
boils at 197. The lower glycols are miscible with water, and those containing as
many as seven carbon atoms show appreciable solubility in water. (Ethylene glycol
owes its use as an antifreeze e.g. Prestone to its high boiling point, low freezing
point, and high solubility in water.)
Chemistry of the OH group
The chemical properties of an alcohol, ROH, are determined by its functional
group, OH, the hydroxyl group. When we have learned the chemistry of the
alcohols, we shall have learned much of the chemistry of the hydroxyl group in
whatever compound it may occur; we shall know, in part at least, what to expect
of hydroxyhalides, hydroxyacids, hydroxyaldehydes, etc.
Reactions of an alcohol can involve the breaking of either of two bonds:
the C OH bond, with removal of the OH group; or the O H bond, with
removal of H. Either kind of reaction can involve substitution, in which a
group replaces the OH or -H, or elimination, in which a double bond is
formed.
Differences in the structure of R cause differences in reactivity, and in a few
cases even profoundly alter the course of the reaction. We shall see what some of
these effects of structure on reactivity are, and how they can be accounted for.
REACTIONS OF ALCOHOLS C-OH BOND CLEAVAGE
Reactivity of ROH: 3 > 2 > 1
Oxidation.
The compound that is formed by oxidation of an alcohol depends upon the
number of hydrogens attached to the carbon bearing the OH group, that is,
upon whether the alcohol is primary, secondary, or tertiary. We have already
encountered these products aldehydes, ketones, and carboxylic acids and should
recognize them from their structures, even though we have not yet discussed much
of their chemistry. They are important compounds, and their preparation by the
oxidation of alcohols is of great value in organic synthesis (Sees. 16.9 and 16.10).
The number of oxidizing agents available to the organic chemist is growing
at a tremendous rate. As with all synthetic methods, emphasis is on the development
of highly selective reagents, which will operate on only one functional group
in a complex molecule, and leave the other functional groups untouched. Of the
many reagents that can be used to oxidize alcohols, we can consider only the most
common ones, those containing Mn(VII) and Cr(VI).
Primary alcohols can be oxidized to carboxylic acids, RCOOH, usually by
heating with aqueous KMnO4 . When reaction is complete, the aqueous solution
of the soluble potassium salt of the carboxylic acid is filtered from MnO2 , and the
acid is liberated by the addition of a stronger mineral acid.
Primary alcohols can be oxidized to aldehydes, RCHO, by the use of K2Cr2O7 .
Since, as we shall see (Sec. 19.9), aldehydes are themselves readily oxidized to
acids, the aldehyde must be removed from the reaction mixture by special techniques
before it is oxidized further.
Secondary alcohols are oxidized to ketones, R2CO, by chromic acid in a form
selected for the job at hand: aqueous K2Cr2O7 , CrO3 in glacial acetic acid, CrO3
in pyridine, etc. Hot permanganate also oxidizes secondary alcohols; it is seldom
used for the synthesis of ketones, however, since oxidation tends to go past the
ketone stage, with breaking of carbon-carbon bonds.
With no hydrogen attached to the carbinol carbon, tertiary alcohols are not
oxidized at all under alkaline conditions. If acid is present, they are rapidly dehydrated
to alkenes, which are then oxidized.
Let us look briefly at the mechanism of just one oxidation reaction, to see
the kind of thing that is involved here. Oxidation of secondary alcohols by Cr(VI) is
believed to involve (1) formation of a chromate ester, which (2) loses a proton and an
HCrO3
~
i n to form the ketone. It is possible that the proton is lost to an oxygen
of the ester group in a cyclic mechanism (2a). Additional alcohol is then oxidized,
evidently by reactions (3)-(5), with chromium finally reaching the Cr(III) state.
The difficult step in ail this is breaking the carbon-hydrogen bond; this is
made possible by the synchronous departure of HCrO3 ~, in what is really an E2
elimination but here with the formation of a carbon-oxygen double bond.
In connection with analysis, we shall encounter two reagents used to oxidize
alcohols of special kinds: (a) hypohalite (Sec. 16.11), and (b) periodic acid (Sec.
Analysis of alcohols. Characterization. lodoform test
Alcohols dissolve in cold concentrated sulfuric acid. This property they share
with alkenes, amines, practically all compounds containing oxygen, and easily
sulfonated compounds. (Alcohols, like other oxygen-containing compounds,
form oxonium salts, which dissolve in the highly polar sulfuric acid.)
Alcohols are not oxidized by cold, dilute, neutral permanganate (although
primary and secondary alcohols are, of course; oxidized by permanganate under
more vigorous conditions). However, as we have seen (Sec. 6.30), alcohols often
contain impurities that are oxidized under these conditions, and so the permanganate
test must be interpreted with caution.
Alcohols do not decolorize bromine in carbon tetrachloride. This property
serves to distinguish them from alkenes and alkynes.
Alcohols are further distinguished from alkenes and alkynes and, indeed,
from nearly every other kind of compound by their oxidation by chromic anhydride,
CrO3 , in aqueous sulfuric acid: within two seconds, the clear orange
solution turns blue-green and becomes opaque.
Tertiary alcohols do not give this test. Aldehydes do, but are easily differentiated
in other ways (Sec. 19.17).
Reaction of alcohols with sodium metal, with the evolution of hydrogen
gas, is of some use in characterization; a wet compound of any kind, of course,
will do the same thing, until the water is used up.
The presence of the OH group in a molecule is often indicated by the formation
of an ester upon treatment with an acid chloride or anhydride (Sec. 18.16).
Some esters are sweet-smelling; others are solids with sharp melting points, and
can be derivatives in identifications. (If the molecular formulas of starting material
and product are determined, it is possible to calculate how many OH groups are
present.)
Make a table to show the response of each kind of compound
we have studied so far toward the following reagents: (a) cold concentrated H2SO4 ;
(b) cold, dilute, neutral KMnO4 ; (c) Br, in CC14 ; (d) CrO3 in H2SO4 ; (e)cold fuming
sulfuric acid; (f) CHC13 and A1C13 ; (g) sodium metal.
Whether an alcohol is primary, secondary, or tertiary is shown by the Lucas
test, which is based upon the difference in reactivity of the three classes toward
hydrogen halides (Sec. 16.4). Alcohols (of not more than six carbons) are soluble
in the Lucas reagent, a mixture of concentrated hydrochloric acid and zinc chloride.
(Why are they more soluble in this than in water?) The corresponding alkyl
chlorides are insoluble. Formation of a chloride from an alcohol is indicated by
the cloudiness that appears when the chloride separates from the solution ; hence, the
time required for cloudiness to appear is a measure of the reactivity of the alcohol.
A tertiary alcohol reacts immediately with the Lucas reagent, and a secondary
alcohol reacts within five minutes; a primary alcohol does not react appreciably at room temperature. As we have seen, benzyl alcohol and allyl alcohol react as
rapidly as tertiary alcohols with the Lucas reagent; allyl chloride, however, is
soluble in the reagent. (Why?)
Whether or not an alcohol contains one particular structural unit is shown
by the iodoform test. The alcohol is treated with iodine and sodium hydroxide
(sodium hypoiodite, NaOI); an alcohol of the structure
yields a yellow precipitate of iodoform (CHI3 , m.p. 119). For example:
The reaction involves oxidation, halogenation, and cleavage.
As would be expected from the equations, a compound of structure
also gives a positive test (Sec. 19.17).
In certain special cases this reaction is used riot as a test, but to synthesize
the carboxylic acid, RCOOH. Here, hypobromite or the cheaper hypochlorite
would probably be used.
Spectroscopic analysis of alcohols
Infrared. In the infrared spectrum of a hydrogen-bonded alcohol and this
is the kind that we commonly see the most conspicuous feature is a strong,
broad band in the 3200-3600 cm- 1
region due to O H stretching (see Fig. 16.1).
(A monomeric alcohol, as discussed in Sec. 15.4, gives a sharp, variable band at
3610-3640 cm-i.)
Another strong, broad band, due to C O stretching, appears in the 1000-
1200 cm" 1 region, the exact frequency depending on the nature of the alcohol:
(Compare the locations of this band in the spectra of Fig. 16.1.)
Phenols (ArOH) also show both these bands, but the C -O stretching appears at somewhat higher frequencies. Ethers show C O stretching, but the O H band is absent. Carboxylic acids and esters show C O stretching, but give absorption characteristic of the carbonyl group, C O, as well. (For a comparison of certain oxygen compounds, see Table 20.3, p. 689.) Nmr. Nmr absorption by a hydroxylic proton (O H) is shifted downtield by hydrogen bonding. The chemical shift that is observed depends, therefore, on the degree of hydrogen bonding, which in turn depends on temperature, concentration, and the nature of the solvent (Sec. 1 5.4). As a result, the signal can
appear anywhere in the range 8 1-5. It may be hidden among the peaks due to
alkyl protons, although its presence there is often revealed through proton counting.
A hydroxyl proton ordinarily gives rise to a singlet in the nmr spectrum:
its signal is not split by nearby protons, nor does it split their signals. Proton
exchange between two (identical) molecules of alcohol
R* -O-H* + R-O-H ^= R*- O-H + R-O-H*
is so fast that the proton now in one molecule and in the next instant in another
cannot see nearby protons in their various combinations of spin alignments, but
in a single average alignment.
Presumably through its inductive effect, the oxygen of an alcohol causes a
downfield shift for nearby protons: a shift of about the same size as other electronegative
atoms (Table 13.4, p. 421).
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