Structure and Bonding

Structure and Bonding 1.1 Introduction Hearing the words “organic chemistry” may strike fear in some students, but this should not be the case at all...

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CHAPTER 1

Structure and Bonding

1.1 Introduction Hearing the words “organic chemistry” may strike fear in some students, but this should not be the case at all. You may have come across complex-looking structures, alien-looking mechanisms and reactions. Exam questions about such may confuse you and heighten anxiety, but the answers to these questions need not be highly sophisticated, as these usually stem from the basic principles of organic chemistry that are covered in this book. Organic chemistry is mainly the study of carbon-containing compounds, excluding those classified as inorganic compounds, such as carbonates and oxides. It is commendable that the carbon element is able to form chains upon chains, resulting in millions of uniquely different compounds. This remarkable feat cannot be reproduced even by its closest group member, silicon, in the periodic table. Although there are numerous organic compounds in existence, one can easily identify various specific combinations and arrangements of atoms that are responsible for the chemical behavior of these compounds. These atoms or group of atoms are known as functional groups, and we will set out to discuss their characteristic properties in this book. In subsequent chapters, we will look into the properties of simple organic compounds grouped into homologous series according to the functional group present. Some of these are presented in Table 1.1. 1

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Common Functional Groups and Homologous Series.

Homologous Series

Alkane

Functional Group Structure



Example

Ethane

Alkene

Ethene

Alkyne

Ethyne

Arene

Benzene

Halogenoalkane

Alcohol

Ether

Aldehyde

Ketone

Carboxylic acid

Bromomethane

Methanol

Dimethyl ether

Ethanal

Propanone

Ethanoic acid

(Continued )

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Structure and Bonding Table 1.1 Homologous Functional Group Series Structure

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(Continued ) Example

Acyl chloride or acid chloride

Ethanoyl chloride

Ester

Methyl propanoate

Acid anhydride

Propanoic anhydride

Amine (Primary amine)

Methylamine

Amide (Primary amide)

Ethanamide

Nitrile

Ethanenitrile

Before we get into the organic chemistry proper, we need to be able to interpret what structures such as those in Table 1.1 represent. The naming of simple organic compounds will be covered under each chapter on the homologous series.

1.2 Structural Formulae of Organic Compounds The molecular formula of a compound informs us of the number and type of atoms present in a molecule. But it may not give us information on how the various atoms are actually connected to one another. For instance, the molecular formula C4 H8 O2 can represent

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more than one compound, and these include the following carboxylic acid and ester:

Butanoic acid

Methyl propanoate

We need to rely on some visuals to depict the arrangement of atoms within a molecule or, in other words, the structural formula of an organic compound. In a displayed formula, otherwise known as the full structural formula, all the atoms and the bonds between them are shown. One of the displayed formulae for a carboxylic acid with the molecular formula C3 H7 COOH looks like this:

A skeletal formula is a simplified drawing in the sense that all C−H bonds are not shown and each C−C single bond is depicted as a line “−” with C atoms present, but not shown, at both ends of the line. If heteroatoms (such as O, N, Cl, etc.) are present, their elemental symbol is included to depict the bonding to the carbon skeletal structure. The skeletal formula for the same carboxylic acid, C3 H7 COOH, that was discussed above is shown as:

There are no hard and fast rules in drawing a structural formula. Basically, a diagram drawn must not be ambiguous. For instance, the following condensed structural formulae are valid in depicting the same carboxylic acid molecule:

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All these structures show the same sequential arrangement of atoms in the molecule when one reads from left to right. However, ambiguity arises when the formula for the acid is written as C3 H7 COOH because it can represent more than one unique compound. The ambiguity resides in the −C3 H7 segment as it can be either one of the following two structures:

With all that said, the various types of structural formulae mentioned so far only show the connectivity of one atom to another. They do not indicate the actual geometry or bond angles in the molecule. To depict the three-dimensional (3-D) geometry about a particular atom, we make use of the 3-D structural formula which utilizes the following conventions:

Q: How can we tell the number of C−H bonds present in a particular carbon atom by the drawing of a skeletal formula? A: Every carbon atom in a stable molecule fulfills the octet rule and can have a maximum of four sigma bonds or combinations of sigma(σ) and pi(π) bonds (for double or triple bonds). Based on this, we can account for the number of C−H bonds

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that are present for each of the carbon atoms in the following example:

1.3 Bonding and Shapes of Organic Molecules Looking at the various molecules in Table 1.1, you will find that a carbon atom has one of the common bonding patterns as shown below:

Each of these bonding patterns corresponds to a different geometry around the carbon atom:

But when we think about the electronic configuration of a carbon atom, 1s2 2s2 2p2 , there is a lack of agreement between the observed experimental data and the idea of atomic orbital overlap during the formation of covalent bond. What do we mean by this? Consider the methane molecule, CH4 .

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The carbon atom is covalently bonded to four hydrogen atoms. But there are only two 2p orbitals, each containing an electron that can be used to form two covalent bonds with other atoms. May be this discrepancy can be resolved by simply unpairing the pair of electrons in the 2s orbital and promoting one of them to the 2p subshell. This process does not really require too much energy as the energy gap between the 2s and 2p subshells is relatively small, thus it can be easily achieved. We now have four unpaired electrons sitting in four orbitals (the 2s orbital and the three 2p orbitals):

These four electrons reside in various types of orbitals, and each of these orbitals comes with a particular shape and orientation in space. If these four orbitals overlap with the 1s orbital of four hydrogen atoms, we are going to get four C−H bonds. However, since the three 2p orbitals are perpendicular to each other, we will get three C−H bonds at 90◦ angles to each other. Furthermore, the four C−H bonds will have different bond lengths since different types of orbitals are used by the carbon atom in bond formation. But experimentally, it has been found that all the bond angles in a CH4 molecule are the same at 109.5◦ , and all the C−H bonds are of the same strength and length! We therefore need a different model to explain what we have observed experimentally. The hybridization model was formulated mathematically for this very purpose. Hybridization is essentially a mathematical linear combination of atomic orbitals to create new orbitals (known as hybrid orbitals) that are then used for bonding. Combinations of a specific number of atomic orbitals will give rise to the same number of hybrid orbitals.

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We will now look at the various types of hybridization and understand their purpose and usage, especially for organic molecules. Q: What is a “linear combination?” A: Linear combination refers to a resultant mathematical function that is obtained by simply adding up other mathematical functions together. But before adding these mathematical functions together, each of these mathematical functions is multiplied by a coefficient, which indicates the loading factor or the “amount of contribution” this particular mathematical function has to the resultant function. An example would be, w = a1 x + a2 y, where both x and y are mathematical functions and a1 and a2 are coefficients. 1.3.1 sp3 Hybridization It can be used to account for the tetrahedral geometry about a carbon atom covalently bonded to four other atoms. Consider the valence orbitals of the C atom:

Four atomic orbitals (one 2s and three 2p orbitals) give rise to four equivalent sp3 hybrid orbitals which are orientated at 109.5◦ apart at the corners of a tetrahedron. In forming methane, each of the four sp3 hybrid orbitals of the C atom overlaps head-on with the 1s orbital of the H atom to form four sigma bonds. This accounts for the tetrahedral geometry of the molecule.

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In the case of ethane, C2 H6 , three of the four sp3 hybrid orbitals of a C atom overlap head-on with the 1s orbital of three H atoms to form three sigma bonds. The same goes for the other C atom. The remaining sp3 hybrid orbitals of both C atoms overlap head-on with each other to form the C−C sigma bond. The geometry about each C atom is tetrahedral, which is in accordance with the experimental data.

1.3.2 sp2 Hybridization To account for the trigonal planar geometry about a carbon atom covalently bonded to three other atoms, such as those in the ethene molecule (C2 H4 ), we have to consider another type of hybridization — the sp2 hybridization. Consider the valence orbitals of the C atom:

This time round, we have three atomic orbitals (one 2s and two 2p orbitals) giving rise to three equivalent sp2 hybrid orbitals which are orientated at 120◦ apart in a triangular plane. The remaining unhybridized p orbital lies perpendicular to this plane. Q: Why is the energy of the sp2 hybrid orbitals lower than that of the p orbitals? A: Since the s orbital has a lower energy than the p orbital, when both s and p orbitals undergo hybridization, the resultant hybrid orbitals would have higher energy than the s orbital but lower than the p orbital.

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In forming ethene, two of the three sp2 hybrid orbitals of each C atom overlap head-on with the 1s orbital of the H atoms forming a total of four C−H sigma bonds. The remaining sp2 hybrid orbitals of both C atoms overlap with each other head-on to form the C−C sigma bond. When this happens, the unhybridized p orbitals are brought close to each other, and they are able to overlap side-on to form the pi bond. As such, the bonding pattern for a sp2 hybridized carbon atom is two single bonds to two atoms and a double bond to a third atom — giving a total of eight valence electrons around the bonded carbon atom. The geometry about each type of hybridized carbon atom is specific — the arrangement of substituents for a sp3 hybridized carbon atom is tetrahedral and that for a sp2 hybridized carbon atom is trigonal planar. Shown here is an example of a molecule that contains both types of hybridized carbon atoms. Can you figure out the geometry about each carbon atom?

Q: Can we talk about the hybridization of the Cl atom in the above molecule? A: What is the purpose of hybridization? It is used to explain shape. It is pointless to describe a hybridization state for the Cl atom in the above molecule as it is only bonded to another atom and the shape is obviously linear. Similarly for the O atom in the C=O functional group. But it is alright to describe the hybridization state of the O atom in the –OH group as it would tell us the shape about this O atom. It is in fact sp3 hybridized with two other atoms (a C and a H atom) bonded to it. In addition, there

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are two lone pairs of electrons and, as a result of these two lone pairs of electrons, the molecular geometry is a bent shape. You may find that it is easier to first account for all the C−H bonds in the molecule before deciding on the hybridization and hence geometry about each carbon atom.

1.3.3 sp Hybridization The sp hybridization is used to account for the linear geometry about a carbon atom covalently bonded to two other atoms, such as those in the ethyne molecule, C2 H2 . Consider the valence orbitals of the C atom:

Two atomic orbitals (one 2s and one 2p orbitals) give rise to two equivalent sp hybrid orbitals which are orientated at 180◦ apart. Two unhybridized p orbitals lie perpendicular to the plane containing the hybrid orbitals. In forming ethyne, one of the two sp hybrid orbitals of each C atom overlaps head-on with the 1s orbital of the H atom forming a total of two C−H sigma bonds altogether. The remaining sp hybrid orbitals of the C atoms overlap with each other head-on to form the C−C sigma bond. Consequently, the two unhybridized p orbitals on each C atom overlap side-on to form two pi bonds.

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Exercise: State the type of hybridization of the carbon atoms, labeled 1 to 5, in the following molecule and hence the geometry about each of them.

Solution: Carbon 1: Carbon 2: Carbon 3: Carbon 4: Carbon 5:

sp2 hybridized; trigonal planar geometry sp3 hybridized; tetrahedral geometry sp2 hybridized; trigonal planar geometry sp2 hybridized; trigonal planar geometry sp hybridized; linear geometry

Q: What is the hybridization state of the carbon atoms in the following molecule?

A: For both C1 and C3, each of them forms two single bonds and one double bond. Hence, they are sp2 hybridized carbon atoms. As for C2, it forms two double bonds which consists a total of two sigma bonds and two pi bonds. In order for C2 to be able to form two pi bonds, it must have two unused p orbitals not involved in hybridization. Therefore, C2 must be sp hybridized. In short, to determine the hybridization state of an atom, we can make use of the following guidelines: • Four single bonds OR four sigma bonds ⇒ atom is sp3 hybridized with tetrahedral geometry; • Two single bonds and one double bond OR three sigma bonds and one pi bond ⇒ atom is sp2 hybridized with trigonal planar geometry;

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• One single bond and one triple bond OR two double bonds OR two sigma bonds and two pi bonds ⇒ atom is sp hybridized with linear geometry. With these simple guidelines, we can determine the hybridization states of the nitrogen atom in the following molecules:

In the molecule of NH3 , there are three sigma bonds and a lone pair of electrons around the nitrogen atom. The lone pair can be considered equivalent to a sigma bond. Hence, the nitrogen atom is sp3 hybridized. In the molecule of N2 H2 , there are two sigma bonds, one pi bond and a lone pair of electrons around the nitrogen atom. Here again, the lone pair can be considered equivalent to a sigma bond. Hence, the nitrogen atom is sp2 hybridized. Q: What happens when the lone pair of electrons on the nitrogen resides in a p orbital instead of in a sp2 hybridized orbital for the NH3 ? A: If the lone pair of electrons resides in a p orbital, this would mean that we cannot use the sp3 hybridization model. Instead, may be we should be using the sp2 hybridization model. But if this is so, then you would find that both sp2 and sp hybridization models cannot be used to account for the experimental geometry around the nitrogen atom of NH3 . In addition, a lone pair of electrons residing in a p orbital that is perpendicular to the nitrogen atom would induce too much inter-electronic repulsion compared to if it were to reside in a hybridized orbital, which is pointing away from the atom. 1.3.4 Delocalized Bonding/Resonance What we have covered thus far are examples that use the hybridization model to account for the shapes and bonding of molecules. But there are certain molecules and ions whose bonding pattern cannot be adequately explained based on this model.

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Take, for instance, the benzene molecule, C6 H6 . It contains six sp2 hybridized carbon atoms with trigonal planar geometry about each of them — as accounted for by the hybridization model. If we are to draw either one of the localized structures using Lewis structures (shown in Fig. 1.1), we would expect to have two different types of bond length — that of the C−C single bond and the C=C double bond. However, experimental data shows that all the carbon–carbon bond lengths in benzene are identical and the bond length is intermediate between the C−C single and C=C double bonds. The only way to be in sync with the observed experimental evidence is to imagine that each p orbital overlaps simultaneously with its two neighboring p orbitals, creating two rings of delocalized electron cloud above and below the planar six-member carbon ring structure (see Fig. 1.2). To describe this phenomenon of electron delocalization, the term resonance is used. As a result, a resonance hybrid, which is described by an “average” of the two equivalent resonance forms (or

Fig. 1.1.

Canonical forms of benzene: Localized structures of benzene.

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Structure and Bonding

Fig. 1.2.

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Resonance hybrid: Delocalized structure of benzene.

canonical structures) is created. Take note that the term resonance DOES NOT mean that the molecule is constantly flipping from one resonance structure to the other. There is only one particular structure for benzene at any one time, and for this, it is represented by the resonance hybrid structure.

1.4 Bonding and Reactivity of Organic Compounds Chemical bonds are electrostatic forces of attraction (positive charge attracting negative charge) that bind particles together to form matter. When different types of particles interact electrostatically, different types of chemical bonds are formed. Since the carbon atoms in an organic molecule are bonded by covalent bonds to either other carbon atoms or some other heteroatoms such as oxygen, nitrogen, halogens, etc., the reactivity of the organic compounds is mostly determined by the relative ease of breaking these covalent bonds. 1.4.1 Nature and Strength of Covalent Bonds Understanding the nature of covalent bonds provides us with deeper insight as to how reactions occur and also the factors that affect the rate of reaction. When we talk about two different species reacting with each other, there must be a reason for them to want to approach each other. For instance, an atom that is electron-rich can “attract” an attack by a complementary species that is electron-deficient and vice versa — a scenario that stems from the all-too-familiar notion of “opposites attract.” As transformation occurs in producing new compounds, a preliminary step consists of the cleaving of bonds within the reactant molecules. Different types of chemical bonds have different strengths.

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Consequently, energy that is needed for bond cleavage varies, and thus the rates of reactions are affected differently. More details on the different types of organic reactions are given in Chapter 3. An electron-rich or electron-deficient atom in a molecule results from the unequal sharing of electrons between the two bonding atoms due to a difference in their electronegativity. Electronegativity denotes the ability of an atom in a molecule to attract electrons to itself. When the two bonding atoms have the same electronegativity, the bonding electrons are shared equally between the two nuclei, forming a pure covalent bond. This normally happens when the bonding atoms consist of identical atoms. A difference in electronegativity creates a permanent separation of charges (dipole) in the bond. The more electronegative atom has a stronger hold onto the bonding electrons. It is thus “electron rich” and gains a partial negative charge (δ−). The other bonding atom partially loses its hold on the bonding electrons and acquires a partial positive charge (δ+), an indication of it being “electron-deficient.” Such covalent bonds with unequal sharing of electrons are termed “polar bonds.”

Bond energy (BE) is a good indicator of the strength of covalent bonds. It is the average amount of energy required to break one mole of a particular bond in any compound in the gas phase. A greater value of bond energy implies a stronger covalent bond. A covalent bond is basically the electrostatic attraction between the localized shared electrons and the two positively charged nuclei. A greater electron density in the inter-nuclei region contributes to a stronger bond. Since an electron is perceived as residing in a region of space known as the orbital, a covalent bond is formed when two partially filled valence orbitals overlap with each other. Hence, the size of the orbital used in bond formation affects bond strength. Example 1.1: Account for the trend of the C−X bond energies whereby X = Cl, Br, and I.

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BE(C−Cl): 340 kJ mol−1 ; BE(C−Br): 280 kJ mol−1 ; BE(C−I): 240 kJ mol−1 . Solution: The orbital used for covalent bonding increases in size from Cl to Br to I. The larger orbitals are more diffuse, and when these orbitals overlap with each other, the accumulation of electron density within the inter-nuclei region is lower. Thus, the orbital overlap is less effective, and this accounts for the weaker bond strength as reflected in the bond energies. Another factor affecting bond strength is the number of bonds between the bonding atoms, as shown in Example 1.2. Example 1.2: Account for the trend of bond energies for the different carbon−carbon bonds: BE(C−C): 350 kJ mol−1 ; BE(C=C): 610 kJ mol−1 ; BE(C≡C): 840 kJ mol−1 . Solution: As the number of shared electrons increases within the inter-nuclei region for multiple bonds, the attractive force for this greater number of electrons increases. Thus, the carbon–carbon covalent bond strength increases from single to double to triple bonds. Q: Why aren’t the bond energies as follows: BE(C−C) = 1/2 × BE(C=C) = 1/3× BE(C≡C)? A: The C=C double bond consists of a sigma bond and a pi bond whereas the C≡C triple bond is made up of one sigma and two pi bonds. The bond energies of the C=C and C≡C bonds are not exact multiples of the C−C single bond as the pi bond is weaker than a sigma bond. Q: Does it mean that all the sigma bonds in each of the carbon−carbon bonds, C−C, C=C and C≡C, are not of the same strength? A: Yes, they are indeed not of the same strength. The presence of a pi bond in the C=C bond would weaken the sigma bond within the double bond itself. Further weakening effect of the sigma bond would be observed in the C≡C bond due to the presence of two pi bonds. Similarly, do not expect the strength of the pi bond in the C=C bond to be of the same strength as each of the two pi bonds in the C≡C bond. In fact, the pi bond in the C=C

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bond should be stronger than each of the two pi bonds in the C≡C bond. Q: So, are the two pi bonds in the C≡C bond of the same strength? A: Yes, they are of the same strength. Since the two pi bonds cannot be really differentiated from one another, and each of the pi bonds would affect the other, on the average, they would have the same strength. The electron density accumulated in the inter-nuclei region of a sigma bond is greater than that in the pi bond. A sigma bond is formed when two orbitals overlap head-on, whereas a pi bond usually occurs when two p orbitals overlap side-on. Therefore, there is more effective overlap of orbitals in forming the sigma bond as compared to the pi bond. The type and strength of covalent bonds influence the chemical properties of organic compounds. These bonds are strong, and some are even comparable in strength to the ionic bonds present in ionic compounds such as sodium chloride. But how does this similarity in bond strength between covalent and ionic compounds account for the difference in the physical properties of these compounds? Why do covalent compounds, organic compounds included, have lower melting and boiling points than ionic compounds? Melting and boiling are phase changes for a substance. When a simple molecular organic compound undergoes a phase change, it still retains its molecular entity. Ethanol, be it in the solid, liquid or vapor phase, still consists of CH3 CH2 OH molecules. No intramolecular bonds are broken in transiting from one physical phase to another. It is the extent of interactions an ethanol molecule has with other ethanol molecules and the degree of freedom of motion of these molecules that causes the lower melting and boiling points for these compounds. So what kinds of interactions are we talking about that exist between ethanol molecules? 1.4.2 Types of Intermolecular Attractive Forces Depending on the type of molecules, one of the following intermolecular forces of attractions, or combinations of these, could exist between molecules:

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• van der Waals forces of attraction ◦ Instantaneous dipole–induced dipole interactions (id–id) ◦ Permanent dipole–permanent dipole interactions (pd–pd) • Hydrogen bonding 1.4.2.1 Instantaneous dipole–induced dipole (id–id) interactions Id–id interactions exist for all types of molecules and it may be used to account for the observed physical properties of matter. As electrons are always in random motion, there is an uneven distribution of electrons in the molecule at any point in time. This separation of charges creates an instantaneous dipole in the molecule. The instantaneous dipole in one molecule can induce the formation of dipoles in nearby unpolarized molecules. As a result, a weak electrostatic attraction forms between these dipoles.

Due to the very fact that electrons are constantly moving, these dipoles are short-lived. However, new dipoles are soon formed again, and the process repeats itself so that on average, there are “permanent” forces of attraction between these molecules that give rise to their observed physical properties such as melting point, boiling point, non-ideal gas behavior and so forth. The strength of id–id interactions depends on: • The number of electrons in the molecule: The greater the number of electrons (i.e. the bigger the electron cloud), the more polarizable would be the electron cloud, and the stronger the id–id interactions. • The surface area for contact of the molecule: The greater the surface area of contact that is possible between the molecules, the greater would be the extent of id–id interactions.

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Q: Why would a greater number of electrons lead to a more polarizable electron cloud? A: A greater number of electrons would imply that more electrons would be further away from the nucleus, hence subjected to weaker attractive force by the nucleus. Thus, the electron cloud would be more polarizable. Q: But a greater number of electrons would also imply a greater number of protons in the nucleus. So, shouldn’t the greater number of protons lead to stronger attractive force by the nucleus? A: Yes, you are right that the greater number of protons would lead to a greater attractive force by the nucleus. Electrostatic force can be perceived as an inverse square law, (F ∝ r12 ), depending on the separation between the electron and nucleus. The effect of an increase in distance leads to a more drastic decrease in the electrostatic force of attraction than an increase in the quantity of charge. In addition, with an increase in the number of electrons, the inter-electronic repulsion would also increase significantly. These two factors coupled together would result in a weaker net attractive force for the electrons in the outermost principal quantum shell. Hence, ionization energy decreases and polarizability increases. 1.4.2.2 Permanent dipole–permanent dipole (pd–pd) interactions Pd–pd interactions exist for polar molecules only. As there is an uneven distribution of electrons in polar bonds, permanent separation of charges (dipole) is found within polar molecules. The permanent dipoles in neighboring polar molecules attract each other. As a result, a weak electrostatic attraction forms between these dipoles, known as pd–pd interactions.

The strength of pd–pd interactions depends on the magnitude of the molecule’s net dipole moment. The greater the magnitude of the dipole moment of a polar bond, the more polar is the bond. The magnitude of the dipole, in turn, depends on the magnitude of the

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electronegativity difference between the bonding atoms. The greater the difference in electronegativity, the greater the dipole moment and the more polar the covalent bond. Dipole moment is a vector quantity that has both magnitude and direction. It is represented , which depicts the δ+ end pointing towards the by the symbol δ− end of the dipole. But one needs to take note that the net polarity of a molecule is also dependent on the geometrical shape of the molecule itself. A molecule may contain polar covalent bonds, but if the dipole moments of these polar bonds cancel each other out due to its geometrical shape, overall this molecule is still a non-polar molecule. Hence, there would not be any pd–pd interaction due to the absence of a net dipole moment. 1.4.2.3 Hydrogen bonding Hydrogen bonding is present between molecules that have one highly electronegative atom (X) – F, O, or N– covalently bonded to an H atom. X, being more electronegative than H, attracts the bonding electrons in the H−X bond closer towards itself. As a result, it is more “electron rich” and gains a partial negative charge (δ−). The H atom, on the other hand, acquires a partial positive charge (δ+). Being “electron deficient,” this H atom is strongly attracted to the lone pair of electrons on another highly electronegative atom (electron-rich region) in other molecules. This electrostatic attraction between the H atom and the lone pair of electrons on the highly electronegative atom, F, O or N, is known as hydrogen bonding.

Hydrogen bonding can be regarded as a more specific type of pd–pd interaction that is applicable to certain types of polar molecules only. So, take note that the terms “hydrogen bonding” and “pd–pd interactions” are non-interchangeable. Examples of compounds that exhibit hydrogen bonding include H2 O, HF, NH3

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and organic compounds such as alcohols, carboxylic acids, amines and amides. The strength of a hydrogen bond depends on: • Dipole moment of the H−X bond, where X is O, F, or N: F−H- - -F−H > O−H- - -O−H > N−H- - -N−H • Ease of donation of a lone pair on Y, where Y is O, F, N: N−H- - -N−H > O−H- - -O−H > F−H- - -F−H Overall, hydrogen bond strength is in the following order: F−H- - -F−H > O−H- - -O−H > N−H- - -N−H This indicates that the dipole moment factor predominates over the ease of donation factor. 1.4.3 Bonding and Physical Properties of Organic Compounds 1.4.3.1 Melting and boiling points

For melting to occur, the crystal lattice has to be broken down so that the discrete molecules are free to move about. Therefore, energy (in the form of heat) is required to overcome the intermolecular forces of attraction among the molecules. Likewise, for boiling to occur, energy is needed to overcome the intermolecular attractions between the mobile molecules so that they can break away from one another and have greater freedom of movement. In trying to relate the type of structure, bonding or intermolecular forces of attraction to the physical properties of substances, it is useful (but NOT necessary always true) to start with the general trend: • The order of decreasing strength of intermolecular forces of attraction: Hydrogen bonding >> pd–pd interactions > id–id interactions (strongest) (weakest)

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Example 1.3: Account for the lower boiling point of ethanal as compared to ethanol.

Solution: Both ethanol and ethanal are polar molecules with simple molecular structures. The lower boiling point of ethanal as compared to ethanol indicates that the pd–pd interactions are weaker than hydrogen bonding. Thus, less energy is required to overcome the attractive forces between ethanal molecules as compared to the hydrogen bonding between ethanol molecules.

1.4.3.2 Miscibility of organic compounds Organic compounds dissolve more readily in non-polar solvents (e.g. benzene and tetrachloromethane) than in polar solvents such as water. Q: Why does an organic compound dissolve in one solvent but not in another? A: For an organic solute to dissolve, energy that is released from the solvation process must be sufficient to offset the energy that is needed to overcome the attractive forces holding the solute molecules together and hence separate them. Q: The solvation energy is the energy released when the solvent and solute molecules interact. We could call this “future” energy, as the energy can only be released after the solute molecules have separated from each other and at the same time, “spaces” are created between the solvent molecules to accommodate the

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incoming solute molecules. The separation of the molecules needs energy. So how can “future” energy be used to compensate for the energy that is currently required for the breaking of bonds? A: One needs to understand how substances actually mix or dissolve. When two substances mix or dissolve, it does not mean that all bonds between the solvent molecules or solute molecules have to be broken first at one go before both the solvent and solute molecules interact. We can imagine the process of an ionic solid dissolving to be like peeling an onion. The positive end of the dipole of a water molecule is attracted to the anions on the surface of the ionic solid, and the negative end of the dipole is attracted towards the cations. The formation of the various ion–dipole interactions releases energy. The energy released is then transferred to the cations and anions, which increases the vibrational energy of these ions. As more ion–dipole interactions occur, releasing more energy, the greater amount of vibrational energy would enable the ions to be freed from the lattice. For the mixing of two liquids, when the boundaries of these two liquids come in contact with each other, the molecules at the two boundaries would interact with each other. If this interaction is favorable, then a sufficient amount of energy would be released, and this energy would be transferred to their neighboring molecules as vibrational energy. With an increase in vibrational energy, the molecules at the boundaries would be able to “break” away from their bulk. This process is similar to the dissolution of ionic compound in water. To ensure that substantial energy is released from solvation, there must be favorable interactions between the solute and solvent molecules that are stronger than or, at the very least, of similar magnitude to, those between the solute molecules themselves. This is the driving force for the solute molecules to be separated. Likewise, to have solvent molecules attracted to the solute molecules, the interactions between them must be stronger than or similar to those between the solvent molecules. Such explanation is the reasoning behind the saying “like dissolves like.” The same reason is used to account for the miscibility, or lack thereof, of two liquids.

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The ability to form hydrogen bonds accounts for the high solubility of substances such as ammonia and short-chain alcohols (e.g. methanol, ethanol) in water. Although the id–id interactions among alkane molecules are relatively weak and easily overcome, you do not find alkanes to be miscible with water. This is because it is not “worthwhile” for both the alkane and water molecules to “give up” the “strong” interactions among their own kind in return for less favorable interactions between the two groups of molecules. We thus expect the solubility of alcohol in water to decrease with increasing carbon chain length due to the increasingly hydrophobic nature of the carbon chain. We will touch upon the physical properties of each homologous series in the corresponding chapters.

My Tutorial 1. (a) (i)

The enthalpies of formation of ethane, ethene, ethyne and benzene are −84.7, +52.3, +227, and +82.9 kJ mol−1 , respectively. With the given bond energies of H−H and C−H as +436 and +412 kJ mol−1 , respectively, calculate values for the carbon−carbon bond energies in the four hydrocarbons. (ii) Account for the differences in C−H bond length in the compounds ethane, ethene, ethyne and benzene. (iii) Draw the dot-and-cross diagrams of ethane, ethene and ethyne. (b) Explain why CH2 =C=CH2 is not a flat molecule and the carbon skeleton of the following molecule is not planar.

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(c) Explain in molecular terms the differences between the boiling points of: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix) (x) (xi) (xii)

methane and n-butane n-butane and methylpropane n-hexane and cyclohexane propene and ethane cis-but-2-ene and trans-but-2-ene n-butane and propanone propanone and propanal propanone and propan-1-ol ethylamine and dimethylamine ethylamine and ethanol ethanol and ethanoic acid ethanoic acid and amino ethanoic acid

(d) Predict whether each of the compounds below will be miscible or immiscible with water. Give reasons for the predictions. (i) (ii) (iii) (iv) (v)

n-butane chloroethane propanone propan-1-ol propylamine

(e) 2,2’-Biquinolyl is an important complexation agent used in the extraction of copper.

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(i)

Identify the type of hybridization of each of the two N atoms. (ii) Circle the atoms that make the compound able to act as a complexation agent. (iii) The N atom of 2,2’-biquinolyl is more basic than the N atom of phenylamine but less basic than the N atom of an aliphatic amine. Explain. (iv) Explain why the carbon skeleton of 2,2’-biquinolyl is planar. (f) Use the following bond energies to account for why the following germinal diol undergoes spontaneous intramolecular dehydration to form the aldehyde functional group.

[Bond energies (in kJ mol−1 ): C−O (358); C−H (413); C=O (736); H−O (464)] (g) The enthalpy changes of hydrogenation of benzene and cyclohexatriene, both to cyclohexane, are −208 and −360 kJ mol−1 , respectively. Discuss the difference between these two values. 2. Morphine is a chemical able to act on the central nervous system to relieve pain, whereas heroin, a synthetic derivative of morphine, is an illegal drug. Both the structures of the compounds are shown below:

(a) Write the condensed structures (i.e. Cw Hx Ny Oz ) for both morphine and heroin. Thus, find the relative molecular mass of these two compounds.

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(b) Identify each type of hybridization used by the carbon atoms in each of the molecule and give the number of carbon atoms exhibiting each type of hybridization. (c) Describe the bonding represented by the circle in the benzene ring of each of the molecule. (d) Identify a pair of atoms involved in a pi bond in each of the molecule. Sketch the pi bond and describe briefly how it is formed. (e) Excluding the benzene ring from consideration, determine the number of pi bonds in each of the molecule. (f) In each part of the molecule linking the benzene ring to the rest of the molecule, (i) name the groups, C−C(O)−O and C−O−C (ii) predict the value of the C−C−O bond angle and explain your reasoning. (g) Identify two different elements in each of the molecules that have lone pairs of electrons and determine the number of such lone pairs in the molecule. (h) Star the chiral carbons in each molecule and hence predict the number of stereoisomers. (i) Predict which of morphine or heroin would be more soluble in water. Explain your prediction.