Organic Chemistry Chapter 2

Watch out: nitro group.

there is one situation where it is not possible to combine charges to form a double bond—this occurs with the nitro group.

2. allylic carbocation.

we are focusing on allylic positions, but this time, we are looking for a positive charge located in an allylic position: we are focusing on allylic positions, but this time, we are looking for a positive charge located in an allylic position: When there is an allylic carbocation, only one curved arrow will be required; this arrow goes from the π bond to form a new π bond: The positive charge is moved to the other end of the system. Never place the tail of a curved arrow on a positive charge (that is a common mistake). If there are multiple double bonds (conjugated), then multiple contributors are possible. They are drawn by moving one electron pair at a time

Localized Lone Pairs general rules

- Don't assume a lone pair is delocalized just because it is next door to a pi bond. - As a general rule, you can assume that when an atom possesses a pi bond and a lone pair, they both will not participate in resonance. - And, as always, if there is no p-orbital next door to the lone pair, then it will be localized

MO theory and Resonnance

- in MO theory, the entire molecule is treated as one entity, and all of the electrons in the entire molecule occupy regions of space called molecular orbitals. - Two electrons are placed in each orbital, starting with the lowest energy orbital, until all electrons occupy orbitals. - According to MO theory, the three p orbitals shown in Figure 2.3 no longer exist. Instead, they have been replaced by three MOs, illustrated in Figure 2.4 in order of increasing energy. Notice that the lowest energy MO, called the bonding molecular orbital, has no vertical nodes. The next higher energy MO, called the nonbonding molecular orbital, has one vertical node. The highest energy MO, called the antibonding molecular orbital, has two vertical nodes. - The π electrons of the allyl system will fill these MOs, starting with the lowest energy MO. How many π electrons will occupy these MOs? The allyl carbocation has only two π electrons, rather than three, because one of the carbon atoms bears a positive formal charge indicating that one electron is missing. The two π electrons of the allyl system will occupy the lowest energy MO (the bonding MO). If the missing electron were to return, it would occupy the next higher energy MO, which is the nonbonding MO. Focus your attention on the nonbonding MO (shown again in Figure 2.5). There should be an electron occupying this nonbonding MO, but the electron is missing. Therefore, the colored lobes are empty and represent regions of space that are electron deficient. In conclusion, MO theory suggests that the positive charge of the allyl carbocation is associated with the two ends of the system, rather than just one end. In a situation like this, any individual bond-line structure that we draw will be inadequate. How can we draw a positive charge that is spread out over two locations, and how can we draw two π electrons that are associated with three carbon atoms?

Resonance hybrid and flipping

- is not flipping back and forth between the different resonance structures - nectarines plum and peach no single drawing adequately describes the nature of the electron density spread out over the molecule The term "resonance" does not describe any real process that is actually happening. Rather, chemists draw multiple resonance structures as a bookkeeping method to overcome the inadequacy of bond-line drawings.

5 general bonding patterns

1. Allylic lone pair 2. Allylic carbocation 3. Lone pair of electrons adjacent to a carbocation 4. A pi bond between two atoms with different electronegativities 5. Conjugated pi bonds in a ring

Drawing Resonance Structures rules

1.Avoid breaking a single bond. By definition, resonance structures must have all the same atoms connected in the same order. Breaking a single bond would change this—hence the first rule: 2 . Never exceed an octet for second-row elements.

RANKING THE SIGNIFICANCE OF RESONANCE STRUCTURES

1.The most significant resonance forms have the greatest number of filled octets. 2. The structure with fewer formal charges is more significant. 3. Other things being equal 4. Resonance forms that have equally good Lewis structures are described as equivalent and contribute equally to the resonance hybrid.

conjugated

A compound in which two π bonds are separated from each other by exactly one σ bond.

Localized Lone Pairs

A lone pair does not participate in resonance if it cannot overlap with an adjacent p-orbital... it is a localized lone pair

How does a resonance stabilize the molecule?

Delocalized electrons are more spread out, thus lower energy, thus more stable. So resonance stabilizes a molecule - Electrons exist in orbitals that span a greater distance giving the electrons more freedom minimizing repulsions - Electrons spend time close to multiple nuclei all at once maximizing attractions

5. Conjugated π bonds enclosed in a ring.

Each atom in the ring has an unhybridized p orbital that can overlap with its neighbors The pi bonds can be pushed over by one position (clockwise or counterclockwise, doesn't matter, the result is the same).

MO and Resonance

From a molecular orbital point of view, the THREE unhybridized p-orbitals overlap to form THREE new MOs The two pi-electrons occupy the lowest energy MO, which is the bonding MO The allyl carbocation has a charge of +1. If it gained an electron it would go to the non-bonding MO The symmetry of the non bonding MO suggests the cationic charge is spread out to both ends of the 3 carbon chain (not just one carbon atom drawn with the positive charge)

Introduction to Resonance

If all of the carbons have unhybridized p orbitals, then all 3 of them overlap side-on-side, and All three overlapping p orbitals allow the electrons to move throughout the overlapping area, and so we say the molecule has resonance (meaning it has delocalized electrons).

neutral allylic atom

If the allylic atom is neutral, then it will become positive, and the atom receiving the lone pair will become negative.

1.The most significant resonance forms have the greatest number of filled octets.

In the following example, the first resonance form exhibits a carbon atom that lacks an octet (C+), while all of the atoms in the second resonance form have filled octets. Therefore, the second resonance form is the major contributor:

Localized electrons vs delocalized electrons

Localized electrons are NOT in resonance Delocalized electrons ARE in resonance (and are more stable)

Resonance MO theory

Once again, this description is consistent with molecular orbital theory for the allylic cation. Positive charges are areas of electron deficiency, and, in the case of an allylic carbocation, these areas are represented by the nonbonding molecular orbital. This molecular orbital indicates that the electron deficiency is delocalized exclusively over the two carbon atoms at the periphery, not the central carbon atom:

3. A lone pair adjacent to a carbocation

Only one arrow is needed The tail of the curved arrow is placed on a lone pair, and the head of the arrow is placed to form a π bond between the lone pair and the positive charge: Charges: Have negative and positive which cancel out or positive charge that stays but mobes

Bond-Line Structures

STEP 1Delete hydrogen atoms, except for those connected to heteroatoms. STEP 2Draw in zigzag format, keeping triple bonds linear. STEP 3Delete carbon atoms.

Ways to represent molecules

The advantage of Lewis structures is that all atoms and bonds are explicitly drawn. However, Lewis structures are only practical for very small molecules. . In partially condensed structures, the bonds are not all drawn explicitly. In condensed structures, single bonds are not drawn. Instead, groups of atoms are clustered together, when possible. The molecular formula of a compound simply shows the number of each type of atom in the compound.

Formal charge in the resonance hybrid

The central carbon atom has no formal charge in the resonance structures, so it must also have no formal charge in the resonance hybrid. The carbon atom on the left and right has no formal charge in one resonance structure and a +1 charge in the other resonance structure. In the resonance hybrid, each of these carbon atoms is assigned a partial positive charge (δ+) to indicate that the actual structure is a combination, or average, of the individual resonance structures. Overall, the resonance hybrid illustrates that the positive charge is delocalized, or spread out, over two of the three carbon atoms in the allylic cation. each of the partial charges is halfway between 0 and +1.

Delocalization of charge

The charge is spread out over more than one atom. The resulting partial charges are more stable than a full +1 charge.

2. The structure with fewer formal charges is more significant.

The first two structures both have full octets, but the first one has fewest formal charges, so it is the most significant resonance contributor. Any resonance form that contains an atom bearing a +2 or −2 charge is highly unlikely

Lone pairs and resonance example

The nitrogen atom is actually sp2 hybridized and trigonal planar in both resonance structures. How? The nitrogen atom has a delocalized lone pair, and it therefore occupies a p orbital (rather than a hybridized orbital), so that it can overlap with the p orbitals of the π bond

4. A π bond between two atoms of differing electronegativity.

The pi electrons will be more attracted to the more electronegative atom, causing the pi bond to be unequally shared These two structure represent the "extreme" descriptions of the pi-bonded electrons (equally shared versus not shared). The actual compound is a hybrid of the two (unequally shared).

Resonance hybrid

The resonance hybrid represents the pi bond of an allylic carbocation as being delocalized over all three carbon atoms This is consistent with MO theory, in that the pi electrons occupy the bonding MO, which has no nodes and is spread out across all three atoms.

delocalization

The spreading of a charge or lone pair as described by resonance theory. is a stabilizing factor. That is, molecules and ions are stabilized by the delocalization of electrons. This stabilization is often referred to as resonance stabilization, and the allyl cation is said to be resonance stabilize

Resonance stabilization

The stabilization associated with the delocalization of electrons via resonance.

Delocalized

To be delocalized, a lone pair of electrons must be adjacent to an atom with an unhybridized p orbital

An allylic lone pair.

We look for allylic lone pairs because they will be resonance delocalized Two curved arrows must be used to show the delocalization of an allylic lone pair.

Vinyl and allyl

When a compound contains a carbon-carbon double bond, the two carbon atoms bearing the double bond are called vinylic positions, while the atoms connected directly to the vinylic positions are called allylic positions:

atom with the allylic lone pair has a negative charge

When the atom with the allylic lone pair has a negative charge, the charge is delocalized with the lone pair, shown here. the atom with the lone pair has a negative charge, then it transfers its negative charge to the atom that ultimately receives a lone pair:

Lone pairs and resonance

Whenever a lone pair participates in resonance, it will occupy a p orbital rather than a hybridized orbital, and this must be taken into account when predicting geometry.

Bottom line... if an atom has a π bond and a lone pair,

Whenever an atom possesses both a π bond and a lone pair, they will not both participate in resonance. In general, only the π bond will participate in resonance, and the lone pair will not.

What is needed for a lone pair to participate in resonance?

a lone pair must occupy a p orbital that can overlap with the neighboring p orbitals, forming a "conduit." Example: the nitrogen atom is already using a p orbital for the π bond. The nitrogen atom can only use one p orbital to join in the conduit shown in and that p orbital is already being utilized by the π bond. As a result, the lone pair cannot join in the conduit, and therefore it cannot participate in resonance. In this case, the lone pair occupies an sp2-hybridized orbital, which is in the plane of the ring.

3. Other things being equal

a structure with a negative charge on the more electronegative element will be more significant, and a positive charge will be more stable on the less electronegative element.

Fisher projections

only used for acyclic compounds

Haworth projection

only used for cyclic compounds

insignificant resonance forms of oxygen

oxygen is much more electronegative than carbon, so you should never draw a resonance form in which an oxygen atom lacks an octet

Resonance

show that the positive charge is spread over two locations (and the π electrons are spread over all three carbon atoms).