Organic Chemistry – Some Basic Principles and Techniques · Lesson 1 of 15

Carbon Bonding and Molecular Shape

“Connect carbon’s bonding patterns to the shapes, strengths, and reactive regions of organic molecules.”

Learning Objectives

• Explain how catenation and tetravalence generate diverse organic compounds. • Relate carbon hybridisation to shape, bond properties, and electronegativity. • Count sigma and pi bonds in molecules containing single and multiple bonds. • Explain restricted rotation and the accessibility of pi electrons.

Why carbon supports so many structures

A carbon atom can join other carbon atoms while still bonding to hydrogen, oxygen, nitrogen, sulphur, phosphorus, or halogens. Repeating this possibility produces long chains, branches, rings, and combinations of these arrangements. Organic chemistry studies these carbon compounds and the relationships between their structures, properties, and reactions.

Definition

Catenation

The ability of an element to form covalent bonds with other atoms of the same element, creating chains or rings.

Carbon forms four covalent bonds in its familiar neutral compounds: this is tetravalence. A double bond counts as two shared electron pairs and a triple bond as three. Four bonds do not always mean four neighbouring atoms: methane has four neighbours around carbon, whereas carbon in a carbon dioxide molecule has only two neighbours.

Organic compounds include DNA and proteins, as well as fuels, medicines, dyes, clothing fibres, and polymers. Their origin does not determine whether they can be made in a laboratory. The old vital-force idea proposed that living organisms supplied a special force needed to form organic substances. In 1828, Wöhler prepared urea from ammonium cyanate. Later syntheses of acetic acid by Kolbe and methane by Berthelot further established that organic compounds could be prepared from inorganic starting materials.

A historical synthesisLaTeX

N H_{4} CNO heat N H_{2} CON H_{2}

The starting salt is ammonium cyanate; the product is urea. The atoms are rearranged into a different compound.

Hybridisation gives a useful picture of shape

A covalent bond forms when orbitals overlap and share electrons. Hybridisation is a model that combines atomic orbitals into directional orbitals suitable for bonding. For carbon, the three common arrangements are tetrahedral, trigonal planar, and linear. Identify the bonding environment of each carbon separately; a molecule need not contain just one kind of carbon.

Carbon hybridisation	Electron-domain arrangement	Typical example	Ideal angle	s character
$s p^{3}$	Tetrahedral	$C H_{4}$	$109. 5^{\circ}$	25%
$s p^{2}$	Trigonal planar	$C H_{2} = C H_{2}$	$12 0^{\circ}$	About 33%
$s p$	Linear	$HC \equiv CH$	$18 0^{\circ}$	50%

In methane, four $s p^{3}$ orbitals point towards four corners of a tetrahedron. In ethene, each carbon uses three $s p^{2}$ orbitals for sigma bonds in a plane, leaving one unhybridised p orbital. In ethyne, each carbon uses two $s p$ orbitals along a line, leaving two mutually perpendicular p orbitals. A multiple bond counts as one direction when identifying these local arrangements.

Three common carbon arrangements— Follow the bond directions around a single carbon; hybridisation describes the local environment.

The s part of a hybrid orbital holds electron density closer to the nucleus than the p part. Increasing s character therefore tends to produce shorter, stronger bonds in comparable situations. It also increases the electronegativity of the carbon atom: $s p > s p^{2} > s p^{3}$ . These comparisons must be made for comparable bonds, rather than treating every bond in every molecule as identical.

The following comparison emphasises local geometry. In the tetrahedral sketch, the wedge and dashed bond represent different depths; the apparent page angles are not the actual bond angles. A three-dimensional model is useful for seeing the full tetrahedral arrangement.

Sigma and pi bonds explain multiple bonding

A sigma bond is formed by head-on overlap along the line joining two nuclei. A pi bond results from sideways overlap of parallel p orbitals. Every ordinary single bond is a sigma bond. A double bond contains one sigma and one pi bond; a triple bond contains one sigma and two pi bonds. All carbon–hydrogen bonds in the examples here are sigma bonds.

In ethene the two p orbitals must stay parallel for strong sideways overlap. Twisting one end of the double bond disrupts that overlap, so rotation is restricted. The pi electron cloud lies above and below the plane of the atoms. It is relatively exposed to approaching reagents, helping explain why multiple bonds often act as reactive centres.

Counting bonds in an unsaturated chain

Problem
Count sigma and pi bonds in $HC \equiv CCH = CHC H_{3}$ .

1.There are five carbons in one continuous chain, so there are four carbon–carbon connections. Each connection contributes one sigma bond.
2.Count hydrogens carbon by carbon: 1, 0, 1, 1, and 3. Their six carbon–hydrogen bonds are sigma bonds.
3.The triple bond contributes two pi bonds and the double bond contributes one. The total is $10$ sigma bonds and $3$ pi bonds.

Reading different carbon environments

Problem
Identify carbon hybridisation in $C H_{3} CN$ and $C H_{3} CH = CHCN$ .

1.In $C H_{3} CN$ , the methyl carbon makes four sigma bonds and is $s p^{3}$ . The nitrile carbon has a triple bond to nitrogen and a single bond to carbon, so it is $s p$ .
2.In $C H_{3} CH = CHCN$ , follow the chain from the methyl end. The environments are $s p^{3}, s p^{2}, s p^{2}, s p$ . Each carbon must be assessed locally.

Predicting the shape around carbon

Problem
Compare $H_{2} C = O$ , $C H_{3} F$ , and $HC \equiv N$ .

1.The carbonyl carbon has three bonding directions: two C–H bonds and one C=O direction. It is $s p^{2}$ , with a trigonal planar arrangement.
2.The carbon in fluoromethane has four single bonds, so it is $s p^{3}$ and approximately tetrahedral.
3.The carbon in hydrogen cyanide has two bonding directions, so it is $s p$ and the molecule is linear.

Common mistake

A double bond is not two sigma bonds, and a triple bond is not three sigma bonds. Count one sigma bond for each connected pair of atoms, then add the pi bonds. Hybridisation describes each atom’s local environment, not the molecule as a single label.

Parallel p orbitals in ethene— Both carbon atoms and the attached hydrogens lie in a plane. The two p orbitals overlap sideways above and below that plane.

Quick check

Which property directly allows carbon atoms to form long chains?

Quick check

Which carbon hybrid orbital has the greatest s character?

Quick check

What does a carbon–carbon triple bond contain?

Quick check

Why is rotation about an ordinary carbon–carbon double bond restricted?

Quick check

What is the carbon hybridisation in $H_{2} C = O$ ?

Practice Problems

Count all sigma and pi bonds in $C H_{2} = C = CHC H_{3}$ and show how you obtained each count.
Assign the hybridisation of every carbon in $C H_{3} Cl$ , $(C H_{3})_{2} CO$ , and $HCON H_{2}$ .
Compare the shapes around the carbon atoms in $C H_{2} = C = O$ and explain why they differ.
Explain why increasing s character can affect bond length, strength, and electronegativity.
Explain how Wöhler’s synthesis challenged the vital-force idea.
Sketch the p orbitals in ethene and explain what happens if one end is twisted.

Key Takeaways

• Catenation and tetravalence make carbon frameworks diverse. • Local hybridisation connects bonding directions to molecular shape. • A double bond has one pi bond; a triple bond has two. • Pi overlap explains restricted rotation and exposes electron density to reagents.

Start of chapter

Next · Lesson 2

Drawing Organic Molecules

Why carbon supports so many structures

Hybridisation gives a useful picture of shape

Sigma and pi bonds explain multiple bonding

Quiz

Practice Problems

Key Takeaways