Lecture 5 Flashcards

1
Q

What are the 3 types of pericyclic reactions?

A
  1. electrocyclic reactions
  2. cycloaddition reactions
  3. sigmatropic rearrangements (not covered)
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2
Q

What are electrocyclic reactions?

A

An electrocyclic reaction is an intramolecular reaction in which a new
sigma bond is formed between the ends of a conjugated p (pi) system (shown in digital notes). This reaction is easy to recognize—the product is a cyclic compound that has one more ring and one less pi bond than the reactant. These reactions are reversible

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3
Q

What are cycloaddition reactions?

A

In a cycloaddition reaction, two different pi-bond-containing molecules react to form a cyclic compound. Each of the reactants loses a pi bond, and the resulting cyclic product has two new sigma bonds. The Diels–Alder reaction is the best-known
example of a cycloaddition reaction.

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4
Q

What are the common features of all pericyclic reactions?

A

■ They are all concerted reactions—that is, all the electron reorganization takes place in a single step. Therefore, there is one cyclic transition state and no intermediate.

■ They are highly stereoselective.

■ They are generally not affected by catalysts or a change in solvent.

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5
Q

What does the configuration of the product formed in a pericyclic reaction depend on?

A

Before we answer, these have yet to be discussed. Why the product is dependent on them will be discussed in the upcoming flashcards:

■ the configuration of the reactant;

■ the number of conjugated double bonds or pairs of electrons in the reacting system;

■ whether the reaction is a thermal reaction or a photochemical reaction.

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6
Q

What is a thermal and photochemical reactions?

A
  • A thermal reaction is a reaction that takes place without the absorption of light.
  • A photochemical reaction is a reaction that takes place when a reactant absorbs light.

Despite its name, a thermal reaction does not necessarily require more heat than what is available at room temperature. Some thermal reactions do require additional heat in order to take place at a reasonable rate, but others readily occur at, or even below, room temperature

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7
Q

What was so puzzling about pericyclic reactions?

A

The question should be what wasnt so puzzling about this reaction. They broke all the rules that organic chemists set. How come some pericyclic reactions happened under photothermal conditions and others under thermal conditions? And what do you mean that a reaction that can happen under both conditions yields different configurations????

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8
Q

What was the answer that solved the puzzle of pericyclic reactions?

A

Simply it was the introduction of the “conservation of orbital symmetry theory” which states that a pericyclic reaction requires the overlap of in-phase orbitals. This means the symmetry of a molecular orbital controls both the conditions under which a pericyclic reaction takes place and the configuration of the product that is formed.

Because the behaviour of pericyclic reactions is so precise, it is not surprising that everything about their behaviour can be explained by one simple theory

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9
Q

What are the important points to remember about the MO theory?

A

■ When two in-phase p atomic orbitals interact, a covalent bond is formed. The interaction of two out-of-phase p atomic orbitals subtracts from bonding because a node is created between the two nuclei.

■ Electrons fill molecular orbitals according to the same rules that govern how they fill atomic orbitals (Section 1.2)—that is, an electron goes into the available MO with the lowest energy (the aufbau principle); only two electrons can occupy a particular MO and they must be of opposite spin (the Pauli exclusion principle); and an electron will occupy an empty degenerate orbital before it will pair up (Hund’s rule).

■ Each carbon that forms a p bond has a p atomic orbital, and the p atomic orbitals of the carbons combine to produce a set of MOs. Thus, a MO can be described by the linear combination of atomic orbitals (LCAO). In a MO, each electron that previously occupied a p atomic orbital belonging to an individual carbon now occupies the entire part of the molecule that is encompassed by the interacting p orbitals.

■ Orbitals are conserved: two atomic
orbitals combine to produce two
MOs, four atomic orbitals combine
to produce four MOs, six atomic
orbitals combine to produce six
MOs, and so on.)

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10
Q

What is a node again?

A

a node is a place in which there is zero probability of finding an electron

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11
Q

What is the general trend found when the energy of a MO increases?

A

as the energy of the MO increases, the number of bonding interactions decreases and the number of nodes between the nuclei increases

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12
Q

When is a MO bonding and when is it anti-bonding?

A

a MO is bonding if the number of bonding interactions is greater than the number of nodes between the nuclei, and a MO is antibonding if the number of bonding interactions is fewer than the number of nodes between the nuclei.

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13
Q

What happens when light is shown on a molecule, and what does that tell us about how it reacts?

A

If a molecule absorbs light of an appropriate wavelength, the light will promote an electron from its ground-state HOMO to its LUMO. The molecule is
then in an excited state where the HOMO is now the previous LUMO and the LUMO is now the MO above the previous LUMO.

This now tells us that a molecule reacts from its ground state in a thermal reaction
and from its excited state in a photochemical reaction.

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14
Q

What are symmetric and anti-symmetric MO?

A

if the lobes at the ends of the MO are in-phase (both have blue lobes on the top and green lobes on the bottom), then the MO is symmetric; if the two end lobes are out-of-phase, then the MO is antisymmetric

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15
Q

What is the trend observed in symmetry when the MOs increase in energy and what does this tell us about the ground state and excited state HOMO?

A

as the MOs increase in energy, they alternate from being symmetric to being antisymmetric. Therefore, the ground-state HOMO and the excited-state HOMO always have opposite symmetries: if one is symmetric, the other is antisymmetric. Now we can use our brains and see why some pericyclic reactions only happen in photothermal conditions bas the book likes to be nonchalant and wants to explain it slower so gl ;).

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16
Q

What are the frontier orbitals?

A

They are the HOMO and the LUMO and even though a molecule is defined by all of its MOs, examining these two tells us a lot about what will happen in pericyclic reactions

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17
Q

Where does the position of equilibria lie in electrocyclic reactions?

A

It lies towards the product if a six-membered ring is formed, if not it will lie towards the open-chain compound. This is because of the angle strain associated with non-6-membered rings.

a question should have been asked about what happens if a 7-member ring is formed

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18
Q

How does a new sigma bond in electrocyclic reactions (Specifically what happens to the p orbitals at the end of the conjugated system)?

A

For the new sigma bond to form, the p orbitals at the ends of the conjugated system must rotate in order to (in phase) overlap head-to-head as they rehybridize from sp2 to sp3. Rotation can occur in two ways: conrotatory and disrotatory

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19
Q

What is conrotatory and disrotatory ring closure?

A

Conrotatory - If both orbitals rotate in
the same direction (both clockwise or both counterclockwise) RING CLOSURE is conrotatory. (lowkey away from each other as shown in notes)

Disconrotatory - if the orbitals rotate in opposite directions (one clockwise, the other counterclockwise), ring closure
is disrotatory. (lowkey towards each other as shown in notes)

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20
Q

Whether conrotatory or disrotatory ring closure occurs is determined by?

A

Whether conrotatory or disrotatory ring closure occurs depends only on the symmetry of the HOMO of the compound undergoing ring closure. The symmetry of the HOMO determines the course of the reaction because this is where the highest energy electrons are. These are the most loosely held electrons and, therefore, the ones most easily moved during a reaction.

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21
Q

What is a symmetry-allowed pathway and a symmetry-forbidden pathway?

A

A symmetry-allowed pathway is one in which in-phase orbitals overlap where the reaction can take place under relatively mild conditions.

A symmetry-forbidden pathway is one in which out-of-phase orbitals overlap, where a reaction cannot take place by a concerted mechanism (recall that all pericyclic reactions follow a concerted mechanism.

22
Q

When the HOMO is symmetric which ring closure is symmetry-allowed and which is symmetry-forbidden?

A

If the HOMO is symmetric (the end orbitals are identical), rotation will have to be disrotatory to achieve in-phase overlap. In other words, disrotatory ring closure is symmetry-allowed, whereas conrotatory ring closure is symmetry-forbidden. This is shown in digital notes.

23
Q

When the HOMO is asymmetric which ring closure is symmetry-allowed and which is symmetry-forbidden?

A

if the HOMO is antisymmetric, rotation has to be conrotatory in order to achieve in-phase overlap. In other words, conrotatory ring closure is symmetry-allowed, whereas disrotatory ring closure is symmetry-forbidden. This is shown in digital notes.

24
Q

Why is the configuration of the product formed under photochemical conditions the opposite of the configuration of the product formed under thermal conditions?

A

This is because the ground-state and excited-state HOMOs have opposite symmetries—that is, if one is symmetric, the other is antisymmetric. And the stereochemical outcome of an electrocyclic reaction depends only on the symmetry of the HOMO undergoing ring closure

25
Q

What is the general rule that determines whether a ground state HOMO is symmetric or asymmetric?

A

The ground-state HOMO of a compound with an even number of conjugated double bonds is antisymmetric, whereas the ground-state HOMO of a compound with an odd number of conjugated double bonds is symmetric.

Therefore, just knowing the number of conjugated double bonds in a compound, you can tell whether ring closure will be conrotatory (an even number of conjugated double bonds) or disrotatory (an odd number of conjugated double bonds) under thermal conditions. However, if the reaction takes place under photochemical conditions, everything is reversed because the ground-state and excited-state HOMOs have opposite symmetries.

26
Q

What are the Woodward–Hoffmann rules?

A

The rules show that the mode of ring closure depends on the number of conjugated double bonds in the reactant and on whether the reaction is carried out under thermal or photochemical conditions. And once you know the mode of ring closure, you can determine the products of the pericyclic reaction.

Both Electrocyclic Reactions and Cycloaddition reaction have their own Woodward-Hoffmann rules with only the electrocyclic shown in the digital notes since the one for cycloaddition is relevant (we only focus on the Diels–Alder reaction). It can be rather burdensome to memorize these rules (and worrisome if they are forgotten during an exam), but they all can be summarized by the mnemonic TE-AC shown in the next flashcard.

27
Q

What is TE-AC?

A

the selection rules for all pericyclic reactions can be summarized by TE-AC. How to use TE-AC is described below. note that antarafacial/Suprafacial are not important but are still discussed in the future flashcard.

■ If TE (Thermal/Even) describes the reaction, the outcome is given by AC (Antarafacial or Conrotatory).

■ If both of the letters of TE are different (Photochemical/Odd), the outcome is still given by AC (Antarafacial or Conrotatory).

■ If one of the letters of TE is different (the reaction is not Thermal/Even but is Thermal/Odd or Photochemical/Even), the outcome is not given by AC (that is, the outcome is Suprafacial or Disrotatory).

For the Diers-Alders reaction, it is always added that the sum of the number of
bonds in the reacting systems of both reagents is 3. And since the Diers-Alders is thermal that means the addition is Suprafacial. More imp info about configuration in flashcard 49.

28
Q

What are the summarized rules about the configuration of the Product of an Electrocyclic Reaction?

A

This is shown in digital notes

29
Q

What is important to note about electrocyclic reaction when it comes to its configuration?

A
  1. The orbital symmetry rules used for a ring-closure reaction also apply to the reverse ring-opening reaction.
  2. The number of conjugated double bonds we use to determine the mode (dis ot con) of ring opening and ring closure is the number in the compound that is undergoing ring closure.
  3. Generally we can image any compound in an electrocyclic reaction to be in the middle of two equilibria with one under thermal conditions and the other under photothermal conditions and both will have different modes of ring closer
  4. We can play around with the products of the reaction using light and heat as shown in digital notes.
30
Q

How are cycloaddition reactions classified?

A

Cycloaddition reactions are classified according to the number of p electrons that interact to produce the product. The Diels–Alder reaction is a [4 + 2] cycloaddition reaction because one reactant has four interacting p electrons and the other reactant has two. Only the p electrons that participate in the electron rearrangement are counted.

31
Q

What are the different cycloaddition reactions found?

A

Shown in digital notes. Focus only on [4+2] cyclo-addition reaction. Imp note that the [4+2] is controlled by heat and the [2+2] is controlled by light. So the Dies-alder is only available in heat ;). This is because both your reactants must be present in the ground state.

32
Q

What orbitals do we consider in cycloaddition reactions?

A

In a cycloaddition reaction, the new s bonds in the product are formed by donation of electron density from one reactant to the other reactant. Because only an empty orbital can accept electrons, we must consider the HOMO of one of the reactants and the LUMO of the other. It does not matter which reactant’s HOMO is used as long as electron donation occurs between the HOMO of one and the LUMO of the other.

33
Q

What are the two modes of orbital overlap in cycloaddition reaction (not mentioned in the lecture)?

A

There are two modes of orbital overlap for the simultaneous formation of two s bonds, suprafacial and antarafacial. In suprafacial bond formation, both s bonds form on the same side of the p system; in antarafacial bond formation, the two s bonds form on opposite sides of the p system. Suprafacial bond formation is similar to syn addition, whereas antarafacial bond formation resembles anti-addition. This is shown in the notes

Note that A cycloaddition reaction that forms a four-, five-, or six-membered ring must occur by suprafacial bond formation. The geometric constraints of these small rings make the antarafacial approach highly unlikely even if it is symmetry-allowed. (Remember that symmetry-allowed means the loverlapping orbitals are in-phase.) Antarafacial bond formation is more likely in cycloaddition reactions that form larger rings.

34
Q

What is the orbital overlap for [4+2] cycloaddition reactions? (This is imp)

A

This is shown in digital notes. Either the HOMO of the diene (a system with two conjugated double bonds) and the LUMO of the dienophile (a system with one double bond) or the HOMO of the dienophile and the LUMO of the diene can be used to explain the reaction.

No matter which pair of HOMO and LUMO we choose, the overlapping orbitals that form the two new sigma bonds have the same colour. This is due to the symmetries of a conjugated diene and of a single c=c bond (Look at the picture see the colour of the orbital, and identify which homo is symmetric and which isn’t, gain intuition). Thus, a Diels–Alder reaction
occurs with relative ease

35
Q

Reflection point

A

okay, so I am the only person who can make a very straightforward lecture complicated.

Just know that pericyclic reactions are defined by MO (and the rules of MOs mentioned). Know the rules for Electrocyclic reaction. And for cycloaddition, we really just look at Diers shit so the imp configuration shit will come when we start covering the Diers reaction which is now ;).

36
Q

Why is the Diels-Alder reaction important?

A

Because it creates two new carbon-carbon bonds and in the process forms a cyclic compound.

37
Q

What are the reactants in the Diels-Alder reaction?

A

a conjugated diene reacts with a compound containing a carbon–carbon double bond. That compound is called a dienophile because it “loves a diene.” heat is used in this system

38
Q

What is the mechanism for the Diels-alder reaction?

A

Look at the wall.

39
Q

What increases the reactivity of a dienophile?

A

The reactivity of the dienophile is increased when an electron-withdrawing group is attached to one of its sp2 carbons. An electron-withdrawing group, such as a carbonyl group (C=O) or a cyan group (CN), withdraws electrons from the dienophile’s double bond. This puts a partial positive charge on the sp2 carbon that the pi electrons of the conjugated diene add to. Thus, the electron-withdrawing group makes the dienophile a better electrophile. This is shown in the digital notes. Note though:

It is important to say that pericyclic reactions arent polar and they are completely explained by MO. However, the concept that opposite attracts is always present, so the fact that the partial positive charge is increased on the electrophile only attracts the nucleophile of the diene and nothing more.

40
Q

What is an interesting note about Diels-Alder reactions?

A

Compounds containing carbon-carbon triple bonds can also be used as dienophiles in Diels–Alder reactions to prepare compounds with two isolated double bonds. This is shown in the digital notes.

41
Q

What happens if BOTH the diene and the dienophile are unsymmetrically substituted?

A

If both the diene and the dienophile are unsymmetrically substituted two products are possible. The products are constitutional isomers. This is because the reactants can align in two different ways as shown in digital notes.

Note that I said BOTH because if one only is unsymmetrically substituted but the other is symmetrical we still get one product.

42
Q

How can you determine which product is the main one?

A

The product formed in greater yield depends on the charge distribution in each of the reactants. To determine the charge distribution, we need to draw the resonance contributors of the reactants.

In our example, the methoxy group of the diene donates electrons by resonance. As a result, its terminal carbon has a partial negative charge. The carbonyl group of the dienophile, on the other hand, withdraws electrons by resonance, so its terminal carbon has a partial positive charge (this is shown in digital notes). The partially positively charged carbon of the dienophile binds preferentially to the partially
negatively charged carbon of the diene. Therefore, a compound aligns in a way that makes that possible. This is also shown in notes.

43
Q

What are the two conformers that a conjugated double-bond system can have?

A

They can either be a s-cis conformer, where the double bonds are cis about the single bond (s = single), or an s-trans conformer where the double bonds are trans about the single bond (this is shown in digital notes).

44
Q

What is more stable the s-cis conformer or the s-trans conformer?

A

An s-trans conformer is a little
more stable (by 2.3 kcal/mol) because the close proximity of the hydrogens in the s-cis conformer causes some steric strain. The rotational barrier between the s-cis and s-trans conformers is low enough to allow the conformers to interconvert rapidly at room temperature (This is shown in digital notes under flashcard 43)

45
Q

Which conformation does the diene need to be in to react in a Diels-Alder reaction?

A

In order to participate in a Diels–Alder reaction, the conjugated diene must be in an s-cis conformation because when it is in an s-trans conformation, C-1 and C-4 are too far apart to react with the dienophile in a concerted reaction. Therefore, a conjugated diene that is locked in an s-trans conformation cannot undergo a Diels–Alder reaction because it cannot achieve the required s-cis conformation. A conjugated diene that is locked in an s-cis conformation, such as 1,3-cyclopentadiene, is highly reactive in a Diels–Alder reaction. Its structure is shown in the digital notes.

46
Q

What is a bridged bicyclic compound?

A

a compound that contains two rings that
share two nonadjacent carbons. It is a product of the Diels-Alder reaction when the diene is a cyclic compound as shown in digital notes. Comparison of fused and bridged is also shown in digital notes.

47
Q

What are the possible confirmations for bridged bicyclic compounds?

A

There are two possible configurations for substituted bridged bicyclic compounds, because the substituent (R) can either point away from the double bond (the exo configuration) or not point away from the double bond (the endo configuration). This is shown in the digital notes.

48
Q

Is the endo or exo conformation favoured?

A

The endo product is formed faster when the dienophile has a substituent with p electrons. Recent studies suggest that the increased rate of endo product formation is due to interaction between the p electrons of the substituent and the p electrons of the ring, which stabilizes the transition state. These interactions are called secondary orbital interactions and they are so strong that steric hindrance can be ignored. A substituent in the exo position cannot engage in such stabilizing
interactions. This is shown in digital notes

49
Q

What is all that is needed to know for the configuration of the Diers-Alder reaction?

A
  1. The Diels–Alder reaction is a synaddition reaction. One face of the diene adds to one face of the dienophile. Therefore, if the substituents in the dienophile are cis, then they will be cis in the product; if the substituents in the dienophile are trans, then they will be trans in the product. Because each of the following synaddition reactions forms a product with two new asymmetric centres, each reaction forms a pair of enantiomers. 3yani Cis and trans in the diene are conserved in the product. Also, y3ani everything on the right will end up on the right. This is shown in digital notes
  2. The terminal group of the diene will always end up on the same side as the electron-withdrawing group of the dienophile. This is shown in the written notes with the tiles Flashcard 49.
50
Q

Retrosynthesis analysis of Diers-Alder reaction

A

To determine the reactants needed to synthesize a Diels-Alder product:

  1. locate the double bond in the product. The diene that was used to form the cyclic product had double bonds on either side of this double bond, so draw in those double bonds and remove the original double bond.
  2. the new s bonds are now on the other side of these double bonds. Deleting these s bonds and putting a p bond between the two carbons whose s bonds were deleted gives the needed reactants—that is, the diene and the dienophile.

VERY IMP EXAMPLES SHOWN IN NOTES

51
Q

Imp note from the slides

A

Electron donating groups in Diene and
electron-withdrawing groups on the dienophile enhance reaction rate as energy levels come closer together.

52
Q

VERY IMP NOTE ABOUT ENDO AND EXO!!!!!!!!!!!!!!

A

In pericyclic reactions, the endo product forms faster because it is kinetically favored due to secondary orbital interactions, but it is less stable than the exo product. Upon heating, the reaction gains enough energy to overcome the activation barrier, allowing the endo product to convert into the more stable exo product. This happens because the exo product has less steric hindrance and is thermodynamically favored, meaning heating shifts the reaction toward this more stable configuration.