Molecular Geometry, Hybridization, and Polarity of CH3OCH3 Lewis Structure

The chemical formula for the molecule Dimethyl Ether is CH3OCH3.

Ethers, as we all know, are a class of chemical molecules with the formula R-O-R’, where R and R’ stand for alkyl radicals.

The colourless gas dimethyl ether, commonly known as methoxymethane, has a slight odour. As a gas at 0 C, it has a molar mass of 46.07 g/mol and a density of 2.1146 kg/m3. It is toxic and volatile in liquid form.

CH3OCH3 is used in aerosol goods such as hair sprays as a propellant. It’s a low-temperature solvent that also works as an adhesive and a binding agent. Apart from that, it has uses in oxygen-blown gasifiers, the welding and soldering industries, diesel engines, printing, and polymerization.

The most common way to make dimethyl ether is to dehydrate methanol, which is done as follows:

2CH3OH      ——->      CH3OCH3 + H2O

Because (CH3)2O is highly flammable in nature, it can be dangerous. It can also cause inflammation, blurred vision, and numbness in the eyes.

Let us now move to a detailed study of the nature of chemical bonding within a methoxymethane molecule.

Lewis Structure of CH3OCH3

The Lewis Structure is the first step in determining the chemical bonding in a molecule.

It is concerned with the valence electrons, or those in the outermost shell, which form pairs and establish covalent bonds between atomic elements.

During the sketch of the Lewis Structure, we employ electron dot notations to represent the valence electrons.

Let’s see how to figure out how many valence electrons are in a (CH3)2O molecule:

Carbon has a valency of 4, Oxygen has a valency of 6, while Hydrogen has only one valence electron, according to the periodic table.

In CH3OCH3, the total number of valence electrons is

= 4 + 31 + 6 + 4 + 31 + 4 + 3*1

equals 20.

The valence electrons in a molecule of dimethyl ether are 20.

The core atom of the given molecule must be identified.

One of the most important things to remember while sketching the electron dot structure in organic chemistry is that hydrogen atoms will usually assume the terminal positions regardless of the electronegativity values.

When drawing a CH3OCH3 molecule, keep in mind that the ether (R-O-R’, here R=R’) group is present.

As a result, the skeleton formula for dimethyl ether is as follows:

The centre element in this molecule is oxygen, which is flanked on both sides by two methyl groups.

This is an exception to the typical rule of sketching the Lewis Structure with the most electropositive member in the centre.

The highest electronegativity value of oxygen has taken front stage here.

The electron dot notations will now be placed around the atoms in the diagram:

We started by putting electron pairs between two atoms, such as C and H or C and O, so that bonds might form.

Six C-H and two C-O bonds have been formed using 16 of our valence electrons.

So, where are the other four going to go?

The octet rule is as follows:

Noble gas elements such as Ne, Ar, and others have a tendency to acquire octet valence shell configuration when they are found in groups 1 to 17 of the periodic table.

Hydrogen, on the other hand, tends to reach two electrons in its outermost shell in order to achieve the He configuration.

The octet fulfilment rule is what it’s called.

When we look at the constituents in the dimethyl ether molecule, we can see that all of the hydrogen and carbon atoms have taken on their appropriate positions. Oxygen, on the other hand, has only four valence electrons surrounding it.

We’ve used all 20 valence electrons and followed the octet rule.

We must now examine the formal charge values of these atoms within the molecule.

The following formula is used to determine the formal charge:

In the case of methoxymethane,

Each C atom has a formal charge of 4 – 0.5*8 – 0 = 0.

Each H atom has a formal charge of 1 – 0.5*2 – 0 = 0.

O atom formal charge = 6 – 0.5*4 – 4 = 0.

The molecule’s nine atoms are all at their lowest attainable formal charge levels. As a result, we now have our Lewis Structure diagram:

Molecular Geometry of CH3OCH3

Let’s talk about how we can go even further and have a greater understanding of the inside of the molecule.

We now have a visual representation of a 2-dimensional methoxymethane molecule. We understand valence electrons and the types of bonds that can be formed (six C-H bonds and 2 C-O bonds).

Now we need to figure out how this molecule appears in three dimensions.

To do so, we must first understand the VSEPR model.

The Valence Shell Electron Pair Repulsion theory is abbreviated as VSEPR. Although it has its own limits, this theory is required to forecast the molecular form of numerous substances.

According to VSEPR theory, electrons with similar charges repel each other in a negatively charged environment. Because this force must be minimised in order for the molecule to be stable, atoms tend to stay apart from one another.

These electrons form groups (lone pairs and bound pairs), and in VSEPR theory, we solely consider the centre atom for calculation purposes.

We deal with AXnEx notations in VSEPR theory, where:

The letter ‘A’ represents for the centre atom, which in this case is oxygen.

Because ‘X’ represents the surrounding atoms, the value of ‘n’ is 2. ( if we consider the C atoms or the CH3 groups)

The lone pairs on the centre atom are denoted by the letter ‘E,’ and the number ‘x’ equals two.

So, for the methoxymethane molecule, we have the AX2E2 notation.

As can be seen, we acquire a bent molecule geometry similar to that of water.

Because methyl groups undergo steric repulsion, the bond angle of C-O-C is roughly 110 degrees.

The electrons have a tetrahedral shape. The 3D representation of the form of the CH3OCH3 molecule is shown below.

Hybridization of CH3OCH3

The notion of orbital hybridization is the next one we’ll discuss in this post.

Covalent bonds develop between carbon and hydrogen, as well as carbon and oxygen, in organic molecules like CH3OCH3. We’ll need the help of a model to understand this bond creation phenomena. The hybridization model is used in this case.

Orbital hybridization is the process of combining and mixing atomic orbitals in order to create hybrid orbitals. The atomic orbitals must all belong to the same atom and have the same energy.

Let’s take a peek at the dimethyl ether molecule’s core atom. Oxygen is what it is. With two Carbon atoms on either side, oxygen creates two single bonds.

Always remember that a single bond refers to sigma pairing, or orbital head-on-head overlap. The s and three p orbitals of Oxygen must form sp3 hybridised orbitals for this to happen.

Also, due to the two bonded and two lone or unshared pairs of valence electrons in this molecule, the coordination number of oxygen in this molecule is 4. We can conclude that the oxygen atom possesses sp3 hybridization in either case.

Polarity of CH3OCH3

What exactly do we mean when we say “polarity”? Polarity is a notion or topic in chemistry that refers to the charge distribution within a compound’s molecule.

We know that the dipole moment is utilised to quantify electronegativity, which is defined as the ability of any atomic element to achieve electrons.

When the electronegativity values of two elements forming a bond differ by a specific amount (typically greater than 0.4), we receive charges at the two poles, making the bond polar in nature.

Take a look at the electronegativity chart created by Pauling:

According to the chart, C has an electronegativity of 2.55, H has an electronegativity of 2.20, and O has an electronegativity of 3.44.

Although the difference in C and H values is not considerable, we will have a partial + on C and a – on the O atom in the two C-O bonds.

The difference is about 0.89, resulting in polar bond formation.

The dipole moments of these bonds all point in the same direction, but the dipole of the molecule as a whole is not cancelled. Methoxymethane has a dipole value of 1.3 D.

As a result, we can classify CH3OCH3 as mildly polar.

Conclusion

We addressed Lewis Structure, VSEPR theory for 3D molecular geometry, hybridization, and the concept of polarity in this essay on Methoxymethane or Dimethyl Ether.

Good luck with your studies!

Read more: Hybridization, Molecular Geometry, and Polarity in the IF3 Lewis Structure

Misha Khatri
Misha Khatri is an emeritus professor in the University of Notre Dame's Department of Chemistry and Biochemistry. He graduated from Northern Illinois University with a BSc in Chemistry and Mathematics and a PhD in Physical Analytical Chemistry from the University of Utah.

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