Question about conjugated systems

[Mantis Toboggin]

A conjugated system (always / sometimes / never) involves exactly 4 parallel p orbitals.
A conjugated system (always / sometimes / never) contains at least one alkene.
A conjugated system (always / sometimes / never) involves an sp3 hybridized atom

I honestly need help with this question.

Sometimes
Always
Sometimes

is what I think-- can someone give examples/suggestions for this?

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[gwjib04]

My understanding:
Conjugated systems are an alkene exclusive form of resonance.

A conjugated system involves at least 4 parallel p orbitals
A conjugated system always contains more than one alkene.
A conjugated system sometimes involves an sp3 hybridized atom

Not clear to me if LP-pi bond interactions are called conjugated systems but it seems they are not.

 

[AureatEagle]

A conjugated system (always / sometimes / never) involves exactly 4 parallel p orbitals.
A conjugated system (always / sometimes / never) contains at least one alkene.
A conjugated system (always / sometimes / never) involves an sp3 hybridized atom

I honestly need help with this question.

Sometimes
Always
Sometimes

is what I think-- can someone give examples/suggestions for this?

A conjugated system is defined as a series 3 or more of parallel, consecutive p orbitals. There are many ways to accomplish this.

For the first claim, a conjugated system by definition would sometimes consist of four parallel p orbitals. It could also be fewer than four, or more than four. The p orbitals also have to be consecutive, so four nonconsecutive p orbitals wouldn't be considered conjugated.

For the second claim, this would also be "sometimes." A C=C bond, which contains two consecutive p orbitals, would sometimes be included as part of a conjugated system-- but not always. For example, a triple C-C bond or a C=O bond may comprise the conjugated system instead.

For the last, this claim is never true. An sp3 hybridized atom has no p orbitals, and therefore cannot be part of a conjugated system.

 

[Mantis Toboggin]

A conjugated system is defined as a series 3 or more of parallel, consecutive p orbitals. There are many ways to accomplish this.

For the first claim, a conjugated system by definition would sometimes consist of four parallel p orbitals. It could also be fewer than four, or more than four. The p orbitals also have to be consecutive, so four nonconsecutive p orbitals wouldn't be considered conjugated.

For the second claim, this would also be "sometimes." A C=C bond, which contains two consecutive p orbitals, would sometimes be included as part of a conjugated system-- but not always. For example, a triple C-C bond or a C=O bond may comprise the conjugated system instead.

For the last, this claim is never true. An sp3 hybridized atom has no p orbitals, and therefore cannot be part of a conjugated system.

Thanks a lot; just for further clarification:

In case (a) with the "parallel p orbitals," could you further explain what/how to count as the consecutive p orbitals? For example how many consecutive p orbitals are in a C-C triple bond (I'm guessing a total of 4, 2 per carbon)? Further, these are UNHYBRIDIZED bonding orbitals, right?

In case (c) the particular exception I had in mind was when you have an anion or lone pair on one of the carbons. Wouldn't that constitute a conjugated sp3 bond?

 

[AureatEagle]

Thanks a lot; just for further clarification:

In case (a) with the "parallel p orbitals," could you further explain what/how to count as the consecutive p orbitals? For example how many consecutive p orbitals are in a C-C triple bond (I'm guessing a total of 4, 2 per carbon)? Further, these are UNHYBRIDIZED bonding orbitals, right?

In case (c) the particular exception I had in mind was when you have an anion or lone pair on one of the carbons. Wouldn't that constitute a conjugated sp3 bond?

In a C-C triple bond there would be two pairs of two consecutive p orbitals. The thing to remember here is that the second set is NOT parallel to the first set (they're perpendicular), so only two can be part of a single conjugated system at a time. And yes, these must be unhybridized p orbitals.

Your second question tugs at one of the neat little exceptions to the rule about assigning hybridization to atoms. Consider the example of allyl anion (google it). In this case, the carbon with the lone pair would normally have sp3 character since there are four things attached to it. However, assuming the carbon was sp3 hybridized, the lone pair would not be able to delocalize as part of a conjugated system with the p orbitals from the adjacent double bond (since only electrons in consecutive, adjacent p orbitals can delocalize, and our lone pair is in an sp3 orbital). What we're looking at here is a situation where delocalization wants to happen, but isn't possible due to the carbon's current hybridization.

What ends up happening to fix this is the carbon will spontaneously rehybridize to an sp2 configuration in order to place its lone pair into a p orbital, thus creating a 3 p orbital conjugated system which allows delocalization to take place. So no, an sp3 atom can never participate in a conjugated system. If possible, however, the atom will REHYBRIDIZE itself to sp2 in order to participate.

This is an advanced point. Just let me know if you have further questions. 🙂

 

[Mantis Toboggin]

In a C-C triple bond there would be two pairs of two consecutive p orbitals. The thing to remember here is that the second set is NOT parallel to the first set (they're perpendicular), so only two can be part of a single conjugated system at a time. And yes, these must be unhybridized p orbitals.

Your second question tugs at one of the neat little exceptions to the rule about assigning hybridization to atoms. Consider the example of allyl anion (google it). In this case, the carbon with the lone pair would normally have sp3 character since there are four things attached to it. However, assuming the carbon was sp3 hybridized, the lone pair would not be able to delocalize as part of a conjugated system with the p orbitals from the adjacent double bond (since only electrons in consecutive, adjacent p orbitals can delocalize, and our lone pair is in an sp3 orbital). What we're looking at here is a situation where delocalization wants to happen, but isn't possible due to the carbon's current hybridization.

What ends up happening to fix this is the carbon will spontaneously rehybridize to an sp2 configuration in order to place its lone pair into a p orbital, thus creating a 3 p orbital conjugated system which allows delocalization to take place. So no, an sp3 atom can never participate in a conjugated system. If possible, however, the atom will REHYBRIDIZE itself to sp2 in order to participate.

This is an advanced point. Just let me know if you have further questions. 🙂

Ahhh excellent thanks a lot, makes sense. Basically in order to achieve the stability of a conjugated system it allows for that hybridization. Cool to know, since none of the review books really mention this >_>.

While we're on triple bonds I had some quick questions that were bothering me:

1.) For example, order the following based on bond energy from highest BE to lowest BE

Acetylene, Ethylene, Ethane.

2.) Related to this, what could we say about their reactivity and their stability? My assumption is that a higher BE means it is less reactive, but I'm not sure.

3.) Regarding bond length, a longer bond length signifies what exactly? Less stability (and therefore less BE)?

4.) Related to (3), I ran into a question where the presence of Deuterium on cyclohexane and its shorter bond length made hydrogen the more bulky groups to consider. Does this mean that bond length is directly related to steric hindrance (and if so, when should I consider it a serious factor)?

5.) Also related to (3), why is it that a substance such as benzene is less stable than cyclohexane, even though it has a shorter bond length and resonance stabilization? Is it because of the various reactions with benzene?

 

[AureatEagle]

Ahhh excellent thanks a lot, makes sense. Basically in order to achieve the stability of a conjugated system it allows for that hybridization. Cool to know, since none of the review books really mention this >_>.

While we're on triple bonds I had some quick questions that were bothering me:

1.) For example, order the following based on bond energy from highest BE to lowest BE

Acetylene, Ethylene, Ethane.

2.) Related to this, what could we say about their reactivity and their stability? My assumption is that a higher BE means it is less reactive, but I'm not sure.

3.) Regarding bond length, a longer bond length signifies what exactly? Less stability (and therefore less BE)?

4.) Related to (3), I ran into a question where the presence of Deuterium on cyclohexane and its shorter bond length made hydrogen the more bulky groups to consider. Does this mean that bond length is directly related to steric hindrance (and if so, when should I consider it a serious factor)?

5.) Also related to (3), why is it that a substance such as benzene is less stable than cyclohexane, even though it has a shorter bond length and resonance stabilization? Is it because of the various reactions with benzene?

The books don't mention it because it's an advanced topic, because it confuses many students, and because it's a topic that's on the unlikely side to come up on the MCAT as a result (but not impossible).

Bond dissociation energy refers to the amount of energy it'd take to homolytically cleave the bond (into two equal halves). It stands to reason that the triple bond is harder to break than the double bond, which is harder to break than the single bond. Thus, acetylene has the highest BDE.

A longer bond length implies a weaker bond (i.e. one that is easier to break). Consider the difference between HF and HI. The short bond distance between H and F makes the bond substantially stronger than the bond between H and I, which makes HI the strong acid (completely dissociates) and HF the weak acid.

Stability and reactivity, however, are not necessarily related. Stability is always a "relative" phenomenon-- substances are always more or less stable in relation to something else. "More stable" thermodynamically just means that the molecule overall has stronger bonds than whatever you're comparing it to. Reactivity, on the other hand, refers to how easily a molecule will react. This can change under different circumstances. Stability and reactivity are often correlated (i.e. benzene is very thermodynamically stable relative to a lot of compounds, and very unreactive), but not always. For example, 1,3-butadiene is more stable relative to butane due to conjugation, BUT it's also more reactive than butane since it more easily stabilizes unstable reaction intermediates (thus lowering transition state energy and making it easier for a reaction to occur).

Not 100% sure about the deuterium question, but it seems logical that if the bond length is shorter, the atom is held tighter, and is thus in less of a position to physically interact with other groups connected to neighboring atoms on the ring.

What makes you say that benzene is less stable than cyclohexane? What criteria are you using?