Stability of cycloalkanes | Organic chemistry | Khan Academy

Summary

The video script explores the concept of cycloalkanes, initially believed to be planar structures with varying degrees of angle strain. It delves into the heat of combustion as a measure of stability, revealing that cyclohexane is the most stable, contrary to earlier theories. The script also explains the different types of strain in cycloalkanes, such as angle and torsional strain, and how these affect their reactivity and structure. Finally, it uses the concept to compare the stability and heat of combustion of ethylcyclopropane and methylcyclobutane, concluding that the latter is more stable with a lower heat of combustion.

Takeaways

• 🔍 Initially, cycloalkanes were thought to be planar, with cyclopropane, cyclobutane, cyclopentane, and cyclohexane having bond angles deviating from the ideal tetrahedral angle of 109.5 degrees.
• 📏 Angle strain is the increase in energy due to bond angles deviating from the ideal, causing significant strain in small-ring cycloalkanes like cyclopropane and cyclobutane.
• 🔄 The stability of cycloalkanes was initially thought to be based on bond angles, with cyclopentane being considered the most stable due to its bond angle being closest to 109.5 degrees.
• 🔥 Heat of combustion was used to analyze the stability of cycloalkanes, revealing that cyclopropane has the highest heat of combustion per CH2 group, indicating it is the most unstable.
• ⚖️ To accurately compare cycloalkanes' stability, the heat of combustion should be divided by the number of CH2 groups present in the molecule.
• 📉 Cyclohexane was found to be strain-free and as stable as an open chain alkane, contrary to the earlier theory that suggested it was a flat hexagon.
• 🧬 The structure of cycloalkanes can be stabilized by adopting non-planar conformations, such as the puckered conformation for cyclobutane and the envelope conformation for cyclopentane.
• 🔑 The three-membered ring in cyclopropane is highly reactive due to significant angle and torsional strain, making it susceptible to ring-opening reactions.
• 🔄 The stability of cycloalkanes is inversely related to their heat of combustion; less stable compounds have higher heats of combustion.
• 🔑 Ethylcyclopropane, with a three-membered ring, is less stable than methylcyclobutane, which has a four-membered ring and less strain.
• 🔥 Ethylcyclopropane, being less stable, has a higher heat of combustion compared to the more stable methylcyclobutane.

Q & A

What was the initial belief about the shape of cycloalkanes?

Initially, it was thought that all cycloalkanes were planar, with cyclopropane being a flat triangle, cyclobutane a flat square, cyclopentane a flat pentagon, and cyclohexane a flat hexagon.

What is angle strain and how is it related to the ideal bond angle?

Angle strain is the increase in energy associated with a bond angle that deviates from the ideal bond angle of 109.5 degrees, which is characteristic of a carbon atom in tetrahedral geometry.

Why does cyclopropane have significant angle strain?

Cyclopropane has a bond angle of 60 degrees, which is significantly different from the ideal bond angle of 109.5 degrees, leading to a large amount of angle strain.

Why was cyclopentane initially considered the most stable cycloalkane?

Cyclopentane was considered the most stable because its bond angle of 108 degrees is closest to the ideal bond angle of 109.5 degrees among the cycloalkanes.

How does the heat of combustion relate to the stability of cycloalkanes?

The heat of combustion, when divided by the number of CH2 groups, gives a better indication of the stability of cycloalkanes. A lower value indicates more stability, as it means less energy is required to combust the compound.

Why can't the heats of combustion be used directly to compare cycloalkane stability?

Direct comparison is not possible because the number of carbons increases from cyclopropane to cyclohexane, which inherently increases the heat of combustion regardless of the ring strain.

Which cycloalkane is considered the most stable based on the heat of combustion per CH2 group?

Cyclohexane is considered the most stable cycloalkane based on the heat of combustion per CH2 group, as it is approximately the same as that of a straight chain alkane.

Why does cyclopropane have the highest heat of combustion per CH2 group?

Cyclopropane has the highest heat of combustion per CH2 group due to its significant angle strain, which increases the energy of the molecule, making it more reactive and susceptible to ring-opening reactions.

What is the non-planar confirmation of cyclobutane that helps relieve torsional strain?

Cyclobutane can adopt a puckered confirmation to relieve torsional strain, where the hydrogens in the front are no longer eclipsing the hydrogens in the back.

What is the envelope confirmation of cyclopentane and how does it help relieve torsional strain?

The envelope confirmation of cyclopentane is a non-planar structure where four carbons are in the same plane and the fifth carbon is out of the plane, relieving some of the torsional strain by reducing the eclipsing of hydrogens.

Which isomer, ethylcyclopropane or methylcyclobutane, is more stable and why?

Methylcyclobutane is more stable because it contains a four-membered ring which has less strain compared to the three-membered ring in ethylcyclopropane.

Which isomer has a higher heat of combustion, and what does this indicate about its stability?

Ethylcyclopropane has a higher heat of combustion, indicating that it is less stable compared to methylcyclobutane, which has a lower heat of combustion.

Outlines

00:00

🔍 Cycloalkane Structure and Stability Analysis

This paragraph delves into the historical misconception that cycloalkanes were planar structures, with cyclopropane, cyclobutane, cyclopentane, and cyclohexane being thought of as flat geometric shapes corresponding to their number of carbon atoms. The concept of angle strain is introduced, which is the additional energy associated with bond angles deviating from the ideal tetrahedral angle of 109.5 degrees. The paragraph explains how cyclopropane has the most significant angle strain due to its 60-degree bond angle, while cyclopentane, with a bond angle of 108 degrees, was initially considered the most stable. However, the heat of combustion data shows that cyclohexane is actually the most stable cycloalkane, with cyclopropane being the least stable. The paragraph also discusses the method of comparing cycloalkanes by dividing their heat of combustion by the number of CH2 groups, which provides a more accurate measure of their stability.

05:01

🌀 Strain and Reactivity in Cycloalkanes

This paragraph examines the different types of strain in cycloalkanes, including angle and torsional strain, and how these strains affect their reactivity and stability. Cyclopropane is highlighted as having significant angle strain due to its planar structure and is described as being highly reactive and prone to ring-opening reactions. The paragraph then discusses cyclobutane's ability to relieve torsional strain by adopting a puckered confirmation, cyclopentane's lesser angle strain and its envelope confirmation to alleviate torsional strain, and cyclohexane's strain-free status, making it as stable as an open chain alkane. The paragraph concludes with a problem comparing the stability and heat of combustion between ethylcyclopropane and methylcyclobutane, concluding that the latter is more stable and has a lower heat of combustion due to less strain.

Keywords

💡Cycloalkanes

Cycloalkanes are a class of organic compounds that contain a ring of carbon atoms bonded to each other. In the context of the video, cycloalkanes are discussed in terms of their planarity and stability, with examples of cyclopropane, cyclobutane, cyclopentane, and cyclohexane. The script explores the historical misconceptions about their structures and the actual molecular geometries that contribute to their stability.

💡Planar

Planar refers to a two-dimensional shape where all points lie in the same plane. The script initially describes the old belief that cycloalkanes were planar, with each type having a shape corresponding to a flat polygon (e.g., cyclopropane as a flat triangle). However, it is later clarified that this is not the case, especially for larger rings, which adopt non-planar conformations to reduce strain.

💡Angle Strain

Angle strain is the increase in energy associated with bond angles that deviate from the ideal bond angle for carbon in tetrahedral geometry, which is approximately 109.5 degrees. The video explains how cycloalkanes with bond angles significantly different from this ideal experience angle strain, which affects their stability, as seen in cyclopropane with bond angles of 60 degrees and cyclobutane with 90 degrees.

💡Heat of Combustion

Heat of combustion is the amount of heat released when a substance is burned in oxygen. The video script uses the heat of combustion as a measure of the stability of cycloalkanes, with the understanding that more stable compounds have lower heats of combustion. It is calculated by dividing the total heat released by the number of CH2 groups in the molecule.

💡Torsional Strain

Torsional strain is a type of molecular strain that arises from the eclipsing or overlapping of atomic or group orbitals in a molecule, which can increase the molecule's energy and decrease its stability. The script mentions torsional strain in the context of cyclobutane and cyclopentane, where the strain can be relieved by adopting non-planar conformations.

💡Puckered Conformation

The puckered conformation is a non-planar shape adopted by some cycloalkanes, like cyclobutane, to relieve torsional strain. The video describes how in this conformation, the hydrogens in front are no longer eclipsing the hydrogens in the back, reducing the molecule's overall energy.

💡Envelope Conformation

The envelope conformation is another non-planar shape that cyclopentane can adopt to minimize torsional strain. In this conformation, four carbons lie in the same plane while the fifth carbon is out of the plane, resembling the shape of an envelope, hence the name.

💡Eclipsing

Eclipsing in the context of the video refers to the overlap of hydrogen atoms or groups in a molecule, which contributes to torsional strain. The script explains how eclipsing in cyclopropane and cyclobutane leads to increased strain, and how non-planar conformations can help to alleviate this.

💡Ring-Opening Reactions

Ring-opening reactions are chemical reactions that involve breaking a ring structure in a molecule, often to reduce strain and increase stability. The video script illustrates this concept by showing how breaking the ring in cyclopropane can decrease strain and make the molecule more stable.

💡Isomers

Isomers are compounds with the same molecular formula but different structural arrangements of atoms. The script presents ethylcyclopropane and methylcyclobutane as isomers (C5H10) and discusses their relative stabilities and heats of combustion based on the size of the ring and the associated strain.

💡Stability

Stability in the context of the video refers to the relative energy state of a molecule, with more stable molecules having lower energy and being less reactive. The script uses the concept of stability to compare different cycloalkanes, concluding that cyclohexane is the most stable due to its lack of significant strain.

Highlights

Cycloalkanes were initially thought to be planar structures with specific shapes based on the number of carbon atoms.

Angle strain is the energy increase due to bond angles deviating from the ideal 109.5 degrees in tetrahedral geometry.

Cyclopropane has significant angle strain due to its 60-degree bond angle, deviating greatly from the ideal.

Cyclobutane has a bond angle of 90 degrees, indicating a large amount of angle strain.

Cyclopentane's bond angle is close to ideal at 108 degrees, suggesting it was thought to be the most stable cycloalkane.

Cyclohexane, with a bond angle of 120 degrees, was theorized to be the least stable due to its deviation from the ideal angle.

Heat of combustion analysis shows that cyclopropane releases the most energy per mole when combusted.

The stability of cycloalkanes cannot be determined solely by heats of combustion; it requires normalization by the number of CH2 groups.

Cyclopropane has the highest heat of combustion per CH2 group, indicating it is the most unstable.

Cyclohexane is found to be strain-free and as stable as an open chain alkane, contrary to earlier flat structure theories.

Cyclopropane's high reactivity is attributed to its significant angle and torsional strain.

Cyclobutane adopts a puckered confirmation to relieve torsional strain, indicating a move away from planarity.

Cyclopentane's planar confirmation is destabilized by torsional strain, leading to non-planar conformations like the envelope confirmation.

Ethylcyclopropane and methylcyclobutane, being isomers, differ in stability due to the number of carbon atoms in their rings.

Ethylcyclopropane, with a three-membered ring, is less stable than methylcyclobutane with a four-membered ring.

The more stable compound, methylcyclobutane, has a lower heat of combustion compared to ethylcyclopropane.

Understanding the stability and heat of combustion of cycloalkanes is crucial for predicting their reactivity and behavior in chemical reactions.

Transcripts

- [Voiceover] At one time it was thought that

the cycloalkanes were all planar,

so cyclopropaner was thought to be a flat triangle,

cyclobutane was thought to be a flat square,

cyclopentane was thought to be a flat pentagon

and cyclohexan was thought to be a flat hexagon.

And in terms of analyzing them, the idea of angle strain

was introduced, and angle strain is the increase in energy

that's associated with a bond angle that deviates from

the ideal bond angle of 109.5 degrees,

and this number should sound familiar to you, this was

the bond angle for a carbon of tetrahedral geometry.

So if you go through and you analyze these, the bond angle

in here for this triangle must be 60 degrees,

and 60 degrees is a long ways off from 109.5 degrees

meaning cyclopropane has significant angle strain.

For cyclobutane, this angle would be 90 degrees,

and 90 degrees is still a ways off from 109.5 degrees

so cyclobutane also has a large amount of angle strain,

although not as much as cyclopropane.

For cyclopentane, this bond angle is 108 degrees,

and 108 degrees is pretty close to 109.5 degrees,

closer than it would be for cyclohexane;

This bond angle is 120 degrees.

And so the theory was, cyclopentane is the most stable

out of the cycloalkanes, because this bond angle

is closest to 109.5 degrees.

However that conclusion doesn't hold up

if you look at the heat of combustion of the cycloalkanes.

And first we'll start with cyclopropane.

So if you define the heat of combustion as

the negative change in the enthalpy,

cyclopropane gives off 2,091 kilojoules for every one mole

of cyclopropane that is combusted,

and if you count the number of CH2 groups on cyclopropane,

let's go back here and let's count them up.

So here's one, two, and three on the drawings;

That's why there's a three here.

Now you can't analyze the cycloalkanes

in terms of just the heats of combustion.

So if we look at those we can see that they increase.

2,091 to 2,721,

to 3,291 and then 3,920.

But that's what we expect to happen because as we go from

cyclopropane to cyclobutane, -pentane, and -hexane,

we're increasing in number of carbons and we already know

from the earlier video on heats of combustion,

if you increase the amount of carbons that you have,

you'd expect an increase in the heats of combustion.

So you can't really compare the cycloalkanes directly

in terms of just the heats of combustion, you have to

compare them in terms of their heats of combustion

divided by the number of CH2 groups, and that gives you

a better idea of the stability.

So if you take 2,091, which is the heat of combustion of

cyclopropane, and divide that by the number of CH2 groups,

which is three, you get approximately 697.

So again, this is the heat of combustion divided by

the number of CH2 groups in kilojoules per mole.

And this is a much better way

to compare the stability of the cycloalkanes.

Notice cyclopropane has the highest value here, 697.

Cyclobutane goes down to 680, cyclopentane is 658

and cyclohexane is approximately 653.

If you remember back to the video on heats of combustion,

this number here, 653 kilojoules per mole, is approximately

the same value we got for a straight chain alkane

when you add it on a CH2 group, so each additional CH2 group

increased the heat of combustion by approximately 653,

or 654 kilojoules per mole, and that tells us that

cyclohexane is pretty much strain-free,

cyclohexane is about as stable as an open chain alkane,

and so we know that this idea of cyclohexane being flat

must not be true, so cyclohexane isn't flat

as we'll see in later videos.

So cyclohexane is the most stable out of these cycloalkanes.

Cyclopentane is a little bit higher in energy

and therefore a little bit more unstable,

and cyclobutane even higher than that,

and finally cyclopropane at 697 for a heat of combustion

per CH2 group, this is the most unstable,

this is the highest heat of combustion,

this is the highest in energy, and so let's analyze

why cyclopropane has such a relatively high

heat of combustion per CH2 group.

Here we have a model of the cyclopropane molecule.

If I turn it to the side you can see that

all three carbon atoms are in the same plane.

So cyclopropane is planar.

You can also see that these bonds are bent.

The significant angle strains means

the orbitals don't overlap very well,

which leads to these bent bonds.

You can see the plastic is even bending in the model set.

There's another source of strain associated with

cyclopropane and we can see it if we look down

one of the carbon-carbon bonds, so the front hydrogens

are eclipsing the hydrogens in the back,

and cyclopropane is locked into this eclipse confirmation.

All this increased strain means that a three-membered ring

is very reactive, and highly susceptible

to ring-opening reactions.

So I'm gonna take the ring here, I'm gonna break it and

open up the ring so we can see that decreased the strain,

those bonds even look straight now.

Here we have the cyclobutane molecule,

and you can see there is some angle strain here

although not as much as in cyclopropane.

If we turn it to the side you also see

some torsional strain, the hydrogens in the front

are eclipsing the hydrogens in the back.

To relieve this torsional strain,

cyclobutane can adopt a non-planar confirmation,

this is called the puckered confirmation.

And if you turn it to the side here

so you're staring down one of the carbon-carbon bonds,

you can see how that's relieved some of the torsional strain

so the hydrogens in front are no longer eclipsing

the hydrogens in the back.

Finally we have the cyclopentane molecule, which has

much less angle strain than cyclopropane or cyclobutane

but if you turn it to the side you can see that

the planar confirmation is destabilized by torsional strain,

so we have some eclipsed hydrogens there.

Some of that torsional strain can be relieved

in a non-planar confirmation, so one of the

non-planar confirmations would be to rotate the carbon

up like that, and that's called the envelope confirmation.

And you can see that four carbons are in the same plane.

So this carbon right here, this one, the one in the back

and this one in the back are the same plane

is this fifth one here is up out of the plane.

So this looks a little bit like an envelope,

and so that's why it's called the envelope confirmation.

Now that we understand the ability of cycloalkanes,

let's do a quick problem.

On the left we have ethylcyclopropane,

on the right we have methylcyclobutane.

They're isomers of each other, they both have the

molecular formula C5H10.

The first question is which isomer is more stable.

Well, we're comparing a three-membered ring

to a four-membered ring, and we know that cyclopropane

is higher energy, there's more strain associated with it.

So ethylcyclopropane must be the less-stable isomer.

So this one is less stable, which makes methylcyclobutane

the more stable isomer, so there's not as much strain

in methylcyclobutane.

So I've answered our first question.

Our second question is which one has

the higher heat of combustion.

We know that the more stable compound

has a lower heat of combustion.

We know that from the heat of combustion video.

That must mean methylcyclobutane has the lower

heat of combustion because this one is more stable.

Which means that ethylcyclopropane

must have the higher heat of combustion.

So the higher heat of combustion.

The higher heat of combustion is the one that's less stable.