Structure and Properties of Organic Compounds

1. Introduction

Organic chemistry mainly deals with carbon compounds. Carbon forms a very large number of compounds because of its ability to form strong covalent bonds with itself and with many other elements.

To understand the properties and reactions of organic compounds, it is not enough to memorize their formulas. We also need to understand how atoms are bonded, how electrons are distributed and how electron density changes inside a molecule.

Important factors controlling the properties of organic compounds include:

Hybridization
Bond type
Bond polarity
Electronegativity
Resonance
Inductive effect
Mesomeric effect
Hyperconjugation
Electron donation and withdrawal
Steric effects

2. Bonding in Organic Compounds

Most organic compounds contain covalent bonds. A covalent bond is formed by sharing electrons between atoms.

Carbon has four valence electrons and usually forms four covalent bonds to achieve a stable electronic arrangement.

Covalent bond

A covalent bond is formed when two atoms share one or more pairs of electrons.

For example, methane is:

CH₄

Carbon forms four C-H sigma bonds.

Single bond

A single bond contains one shared electron pair.

C — C

Double bond

A double bond contains one sigma bond and one pi bond.

C = C

Triple bond

A triple bond contains one sigma bond and two pi bonds.

C ≡ C

3. Hybridization

Hybridization is the mixing of atomic orbitals of similar energy to form new hybrid orbitals having equivalent energy and definite orientation.

sp³ hybridization

One s orbital and three p orbitals combine:

1s + 3p → 4sp³

The four sp³ orbitals are directed approximately toward the corners of a tetrahedron.

Approximate bond angle:

109.5°

Example: methane, CH₄.

sp² hybridization

One s orbital combines with two p orbitals:

1s + 2p → 3sp²

One p orbital remains unhybridized.

The three sp² orbitals lie in one plane.

Bond angle ≈ 120°

Example: ethene, CH₂=CH₂.

sp hybridization

One s orbital combines with one p orbital:

1s + 1p → 2sp

Two p orbitals remain unhybridized.

The geometry is linear.

Bond angle = 180°

Example: ethyne, HC≡CH.

Hybridization	Geometry	Approx. angle	Example
sp	Linear	180°	Ethyne
sp²	Trigonal planar	120°	Ethene
sp³	Tetrahedral	109.5°	Methane

4. Sigma and Pi Bonds

Sigma bond

A sigma bond is formed by head-on overlap of orbitals along the internuclear axis.

Sigma bonds are generally stronger than pi bonds.

Rotation around a simple sigma bond is generally possible unless restricted by other structural factors.

Pi bond

A pi bond is formed by sideways overlap of parallel unhybridized p orbitals.

Pi bonds are generally weaker than sigma bonds.

A double bond contains:

1 σ + 1 π

A triple bond contains:

1 σ + 2 π

Feature	Sigma bond	Pi bond
Overlap	Head-on	Sideways
Relative strength	Stronger	Weaker
Electron density	Along internuclear axis	Above and below axis
Free rotation	Usually possible	Restricted

5. Bond Length, Bond Angle and Bond Strength

Bond length

Bond length is the average distance between the nuclei of two bonded atoms.

In general:

Single bond > double bond > triple bond

in terms of bond length.

Therefore, a C-C single bond is longer than a C=C double bond, which is longer than a C≡C triple bond.

Bond strength

Bond strength is related to the energy required to break a bond.

Generally:

Triple bond > double bond > single bond

in terms of bond strength.

Bond angle

Bond angle is the angle between two bonds around a central atom.

Examples:

sp → 180°
sp² → approximately 120°
sp³ → approximately 109.5°

6. Special Nature of Carbon

Carbon occupies a special position in organic chemistry.

Catenation

Catenation is the ability of an element to form bonds with other atoms of the same element.

Carbon shows extensive catenation because C-C bonds are sufficiently strong.

Carbon can form:

Long chains
Branched chains
Rings
Double bonds
Triple bonds

Tetravalency

Carbon has valency four and can form four covalent bonds.

Small atomic size

The small size of carbon allows strong overlap between its orbitals and orbitals of other atoms.

7. Polarity of Bonds

When two atoms have different electronegativities, the shared electrons are not distributed equally.

The atom with greater electronegativity attracts the bonding electrons more strongly.

Such a bond is called a polar covalent bond.

Example: C-Cl

Chlorine is more electronegative than carbon. Therefore:

Cδ+ — Clδ−

Carbon carries a partial positive charge while chlorine carries a partial negative charge.

Non-polar covalent bond

If the two bonded atoms have similar electronegativity, the electrons are shared more equally.

Example:

C — C

8. Dipole Moment

Dipole moment is a measure of the separation of positive and negative charges in a molecule.

μ = q × r

where:

μ = dipole moment
q = magnitude of charge
r = distance between the charges

The common unit is Debye (D).

Important point

Molecular dipole moment is the vector sum of individual bond dipoles.

Therefore, a molecule can contain polar bonds but still have zero net dipole moment if the bond dipoles cancel because of molecular symmetry.

Example: CO₂

O=C=O

Each C=O bond is polar, but the molecule is linear and the two bond dipoles act in opposite directions.

Therefore:

Net μ = 0

Example: H₂O

Water is bent, so its bond dipoles do not cancel. Therefore, water has a net dipole moment.

9. Resonance

Resonance is used when a single Lewis structure cannot adequately represent the actual distribution of electrons in a molecule or ion.

The different structures are called resonance structures or contributing structures.

The actual molecule is a resonance hybrid.

Important points

Atoms do not move between resonance structures.
Only electrons are redistributed.
The actual molecule is not rapidly changing between structures.
Resonance generally increases stability.
Equivalent resonance structures contribute equally.

Resonance structures are different drawings of the same electronic system. They should not be treated as separate molecules existing independently.

10. Resonance Examples

Carbonate ion

Carbonate ion can be represented by several equivalent structures in which the double bond appears at different oxygen atoms.

CO₃²⁻

The actual carbonate ion has equivalent C-O bonds rather than one permanently fixed C=O bond and two permanently fixed C-O single bonds.

Carboxylate ion

A carboxylate ion can be represented approximately as:

R-COO⁻ ↔ R-C(O⁻)=O

The negative charge is delocalized over the two oxygen atoms.

This delocalization increases the stability of the carboxylate ion.

Phenoxide ion

In phenoxide ion, the negative charge can be delocalized from oxygen into the aromatic ring.

This is one reason why phenol is considerably more acidic than a typical aliphatic alcohol.

Conditions for important resonance

Conjugated p orbitals should be available.
The structures should obey normal valence rules as far as possible.
Equivalent structures give especially strong resonance stabilization.
Charge separation generally makes a contributor less important unless other factors compensate.

11. Inductive Effect

The inductive effect is the permanent displacement of sigma electrons along a carbon chain due to an electronegative atom or group.

The effect is transmitted through sigma bonds.

It decreases rapidly as the distance from the substituent increases.

−I effect

Groups that withdraw electron density through sigma bonds show a negative inductive effect.

Examples include:

−NO₂, −CN, −COOH, −F, −Cl, −Br

+I effect

Alkyl groups generally release electron density through sigma bonds and show a positive inductive effect.

A common approximate order is:

tert-butyl > isopropyl > ethyl > methyl > H

Importance of inductive effect

It affects:

Acidity
Basicity
Stability of carbocations
Stability of carbanions
Dipole moments

12. Mesomeric Effect

The mesomeric effect is the permanent electron displacement through a conjugated pi-electron system.

It is also called the resonance effect in many contexts.

+M effect

Groups that donate electron density through resonance show a +M effect.

Common examples:

−OH, −OR, −NH₂

−M effect

Groups that withdraw electron density through resonance show a −M effect.

Examples:

−NO₂, −CHO, −COOH, −COOR, −CN

Inductive vs mesomeric effect

Inductive effect	Mesomeric effect
Transmitted through sigma bonds.	Involves conjugated pi systems.
Permanent.	Permanent.
Decreases rapidly with distance.	Requires suitable conjugation.
Related mainly to electronegativity.	Related to electron delocalization.

13. Electromeric Effect

The electromeric effect is a temporary complete transfer of the shared pi-electron pair of a multiple bond toward one of the bonded atoms when an attacking reagent approaches.

It is therefore observed only in compounds containing multiple bonds.

Example

Consider a carbonyl group:

R₂C=O

When a nucleophile approaches, the pi electrons of the C=O bond can shift toward oxygen:

R₂C=O + Nu⁻ → R₂C(Nu)-O⁻

The electron pair has shifted toward the more electronegative oxygen atom.

Important point

Unlike the inductive effect, the electromeric effect is temporary and occurs only in the presence of an attacking reagent.

14. Hyperconjugation

Hyperconjugation is the delocalization of electrons from a sigma bond, usually a C-H bond adjacent to a pi bond, carbocation or radical center.

It is sometimes described as "no-bond resonance".

Example: propene

CH₃-CH=CH₂

The C-H sigma bonds on the carbon adjacent to the double bond can interact with the neighboring pi system.

Hyperconjugation helps explain:

Stability of carbocations
Stability of alkenes
Electron-releasing nature of alkyl groups
Some effects on bond lengths and dipole moments

General carbocation stability

3° > 2° > 1° > CH₃⁺

Hyperconjugation and the +I effect both contribute to this general trend.

15. Bond Fission

Bond fission means breaking of a covalent bond.

There are two important types:

Homolytic fission
Heterolytic fission

Homolytic fission

In homolytic fission, each bonded atom takes one electron from the shared pair.

A:B → A· + ·B

Free radicals are produced.

Homolytic cleavage is commonly promoted by heat, light or radical conditions.

Heterolytic fission

In heterolytic fission, both bonding electrons go to one atom.

A:B → A⁺ + :B⁻

Ions are produced.

16. Electrophiles

An electrophile is an electron-deficient species that accepts an electron pair.

The word electrophile means electron-loving species.

Examples

H⁺, NO₂⁺, BF₃, AlCl₃, carbocations

Electrophiles attack electron-rich regions of organic molecules.

Example

CH₂=CH₂ + H⁺ → CH₃-CH₂⁺

The pi electrons of the alkene are donated to H⁺.

17. Nucleophiles

A nucleophile is an electron-rich species that donates an electron pair to an electron-deficient center.

Common nucleophiles include:

OH⁻, CN⁻, Cl⁻, Br⁻, NH₃, H₂O

A nucleophile attacks an electrophilic carbon atom in many organic reactions.

Example

CH₃Br + OH⁻ → CH₃OH + Br⁻

Here OH⁻ is the nucleophile and the carbon bonded to Br is the electrophilic center.

Electrophile	Nucleophile
Electron pair acceptor	Electron pair donor
Electron deficient	Electron rich
Examples: H⁺, NO₂⁺	Examples: OH⁻, CN⁻, NH₃

18. Free Radicals

A free radical is a species containing an unpaired electron.

It is commonly represented using a dot.

CH₃·

Free radicals are usually highly reactive.

Formation

Cl₂ → 2Cl·

Under suitable conditions such as ultraviolet light, the Cl-Cl bond can undergo homolytic cleavage.

General stability

3° radical > 2° radical > 1° radical > methyl radical

Hyperconjugation and inductive effects contribute to this order.

19. Carbocations

A carbocation is an organic ion in which carbon carries a positive charge.

CH₃⁺

The positively charged carbon is electron deficient.

Carbocations generally behave as electrophiles.

Geometry

A simple carbocation center is generally approximately trigonal planar and commonly described as sp² hybridized.

Stability

3° > 2° > 1° > CH₃⁺

Alkyl groups stabilize carbocations through the +I effect and hyperconjugation.

Resonance-stabilized carbocations

Carbocations adjacent to a double bond or aromatic ring can be strongly stabilized by resonance.

For example, an allylic carbocation:

CH₂=CH-CH₂⁺ ↔ ⁺CH₂-CH=CH₂

20. Carbanions

A carbanion is an organic species in which carbon carries a negative charge.

CH₃⁻

Carbanions are electron-rich and generally behave as nucleophiles or bases.

Simple carbanion stability

For simple alkyl carbanions, the trend is generally opposite to that of carbocations:

CH₃⁻ > 1° > 2° > 3°

Alkyl groups release electron density, which tends to destabilize a negative charge.

Electron-withdrawing groups can stabilize carbanions.

21. Stability of Reactive Intermediates

Species	General trend	Main reason
Carbocation	3° > 2° > 1° > methyl	+I effect and hyperconjugation
Carbanion	Methyl > 1° > 2° > 3°	Electron-releasing alkyl groups destabilize negative charge
Free radical	3° > 2° > 1° > methyl	Hyperconjugation and electron donation

Important exception

Resonance can be more important than simple alkyl substitution.

Therefore allylic and benzylic intermediates can be unusually stable because the unpaired electron or charge is delocalized.

22. Acidity of Organic Compounds

An acid is a species capable of donating a proton.

In organic chemistry, acidity often depends strongly on the stability of the conjugate base formed after proton removal.

HA ⇌ H⁺ + A⁻

The more stable A⁻ is, the easier it generally is for HA to lose H⁺.

Factors affecting acidity

Electronegativity
Atomic size
Resonance
Inductive effect
Hybridization
Solvation

Effect of hybridization

For carbon atoms bearing negative charge or hydrogen that is removed as a proton, greater s-character generally stabilizes electron density closer to the nucleus.

Therefore, for C-H acidity:

sp > sp² > sp³

in general acidity of the corresponding C-H bonds.

Effect of electron-withdrawing groups

Electron-withdrawing groups stabilize a nearby negative charge through the −I effect.

Therefore, they can increase acidity.

Example: chloroacetic acid

ClCH₂COOH

The chlorine atom withdraws electron density and helps stabilize the carboxylate ion.

Thus chloroacetic acid is more acidic than acetic acid.

23. Basicity of Organic Compounds

A base can accept a proton or donate an electron pair, depending on the definition being used.

In Lewis terms, a base is an electron-pair donor.

Example: ammonia

NH₃ + H⁺ → NH₄⁺

The lone pair on nitrogen is donated to H⁺.

Factors affecting basicity

Availability of the lone pair
Resonance
Inductive effect
Hybridization
Solvation
Steric effects

Resonance and basicity

If a nitrogen lone pair is delocalized by resonance, it becomes less available for accepting a proton.

This is one reason amides are much less basic than amines.

R-CONH₂

The nitrogen lone pair is delocalized toward the carbonyl group.

Inductive effect and basicity

Electron-releasing groups generally increase electron density and can increase basicity, while electron- withdrawing groups generally reduce it.

24. Steric Effect

Steric effect refers to changes in molecular behaviour caused by the physical crowding of atoms or groups in space.

Large groups can prevent a reagent from approaching a reaction center easily.

This can affect:

Reaction rate
Basicity
Nucleophilicity
Conformation
Stability

Example

A bulky group around a reaction center may physically block the approach of a nucleophile.

This is called steric hindrance.

25. Structural Features and Isomerism

Compounds can have the same molecular formula but different arrangements of atoms.

Such compounds are called isomers.

Structural isomerism

Structural isomers have the same molecular formula but different connectivity of atoms.

Chain isomerism

The carbon skeleton is different.

Example:

C₄H₁₀

can exist as n-butane and isobutane.

Position isomerism

The carbon skeleton is similar but the position of a functional group or multiple bond changes.

Functional isomerism

Compounds have the same molecular formula but different functional groups.

Example:

C₂H₆O

can represent ethanol and dimethyl ether.

Why structure matters

Different structures can have different:

Boiling points
Melting points
Solubilities
Acidities
Basicities
Reactivities

26. Solved Numericals

Numerical 1: Dipole moment

Two opposite charges of magnitude 1.0 × 10⁻¹⁹ C are separated by 2.0 × 10⁻¹⁰ m. Calculate the dipole moment in C m.

Given:

q = 1.0 × 10⁻¹⁹ C

r = 2.0 × 10⁻¹⁰ m

Formula:

μ = q × r

Substitution:

μ = (1.0 × 10⁻¹⁹)(2.0 × 10⁻¹⁰)

μ = 2.0 × 10⁻²⁹ C m

Answer: Dipole moment = 2.0 × 10⁻²⁹ C m.

Numerical 2: Dipole moment in Debye

Convert a dipole moment of 6.0 × 10⁻³⁰ C m into Debye.

Use:

1 D = 3.336 × 10⁻³⁰ C m

Given:

μ = 6.0 × 10⁻³⁰ C m

Formula:

μ(D) = μ(C m) / 3.336 × 10⁻³⁰

Substitution:

μ = 6.0 / 3.336

μ ≈ 1.80 D

Answer: Dipole moment ≈ 1.80 D.

Numerical 3: Percentage ionic character

Calculate the percentage ionic character of a bond having electronegativities 2.1 and 3.0 using the Pauling approximation:

% ionic character = [1 − e^{−0.25(Δχ)²}] × 100

Step 1: Find electronegativity difference

Δχ = 3.0 − 2.1 = 0.9

Step 2: Substitute

% ionic character = [1 − e^{−0.25(0.9)²}] × 100

First:

(0.9)² = 0.81

0.25 × 0.81 = 0.2025

% ionic character = [1 − e^−0.2025] × 100

Since:

e^−0.2025 ≈ 0.817

% ionic character ≈ (1 − 0.817) × 100

≈ 18.3%

Answer: Percentage ionic character ≈ 18.3%.

Numerical 4: Hybridization from steric number

Determine the hybridization of the central carbon atom in methane, CH₄.

Carbon forms four sigma bonds with four hydrogen atoms.

Number of electron domains = 4.

A steric number of 4 corresponds to:

sp³

Answer: Carbon in methane is sp³ hybridized.

Numerical 5: Hybridization of carbon in ethene

Determine the hybridization of each carbon atom in ethene.

H₂C=CH₂

Each carbon forms three sigma bonds:

Two C-H sigma bonds
One C-C sigma bond

Each carbon therefore uses three hybrid orbitals.

Hybridization = sp²

One unhybridized p orbital remains on each carbon and these overlap sideways to form the pi bond.

Answer: Both carbon atoms are sp² hybridized.

Numerical 6: Hybridization of carbon in ethyne

Determine the hybridization of carbon in ethyne.

HC≡CH

Each carbon forms two sigma bonds:

One C-H sigma bond
One C-C sigma bond

Therefore, two hybrid orbitals are required.

Hybridization = sp

Two unhybridized p orbitals remain and form two pi bonds.

Answer: Both carbon atoms are sp hybridized.

Numerical 7: Bond counting in ethene

How many sigma and pi bonds are present in ethene, C₂H₄?

H₂C=CH₂

There are four C-H bonds:

4 C-H = 4σ

The C=C double bond contains:

1σ + 1π

Therefore:

Total σ bonds = 4 + 1 = 5

Total π bonds = 1

Answer: Ethene contains 5 sigma bonds and 1 pi bond.

Numerical 8: Bond counting in ethyne

How many sigma and pi bonds are present in ethyne, C₂H₂?

HC≡CH

Two C-H bonds:

2σ

C≡C contains:

1σ + 2π

Therefore:

Total σ = 2 + 1 = 3

Total π = 2

Answer: Ethyne contains 3 sigma bonds and 2 pi bonds.

Numerical 9: Percentage s-character

Calculate the percentage s-character in sp, sp² and sp³ hybrid orbitals.

For sp:

s-character = 1/2 × 100 = 50%

For sp²:

s-character = 1/3 × 100 ≈ 33.3%

For sp³:

s-character = 1/4 × 100 = 25%

Answer:

sp = 50% s-character
sp² = 33.3% s-character
sp³ = 25% s-character

Greater s-character keeps electron density closer to the nucleus and is important when comparing the acidity of C-H bonds.

Numerical 10: Comparing acidity using hybridization

Arrange the following C-H bonds in increasing acidity: sp³ C-H, sp² C-H and sp C-H.

The conjugate base of an sp-hybridized carbon has greater s-character than that of an sp² or sp³ carbon.

More s-character means the negative charge can be held closer to the nucleus.

Therefore:

sp³ C-H < sp² C-H < sp C-H

Answer: sp C-H is the most acidic among the three.

Numerical 11: Percentage ionic character from dipole data

A bond has a measured dipole moment of 1.5 D and bond length of 1.0 Å. Calculate the approximate percentage ionic character.

For a completely ionic bond:

μ = e × r

Use:

1 e = 1.602 × 10⁻¹⁹ C

1 Å = 1.0 × 10⁻¹⁰ m

First calculate the dipole moment for 100% ionic character:

μ_ionic = (1.602 × 10⁻¹⁹)(1.0 × 10⁻¹⁰)

= 1.602 × 10⁻²⁹ C m

Convert to Debye:

μ_ionic = 1.602 × 10⁻²⁹ / 3.336 × 10⁻³⁰

≈ 4.80 D

Now:

% ionic character = (observed μ / ionic μ) × 100

= (1.5 / 4.80) × 100

≈ 31.25%

Answer: Percentage ionic character ≈ 31.3%.

27. Practice Questions

1. What is hybridization? Explain sp, sp² and sp³ hybridization.

Include orbital mixing, geometry, bond angle and examples.

2. Explain sigma and pi bonds with suitable examples.

3. Why is ethyne linear?

4. Explain resonance with an example of a carboxylate ion.

5. Explain +I and −I effects.

6. Differentiate between inductive and electromeric effects.

7. Explain hyperconjugation and its role in carbocation stability.

8. Distinguish between electrophiles and nucleophiles.

9. Arrange carbocations in order of stability and explain the reason.

10. Explain the factors affecting acidity of organic compounds.

11. Explain why amides are less basic than amines.

12. Calculate the number of sigma and pi bonds in benzene.

Answer: 12 sigma bonds and 3 pi bonds when counting the six C-H bonds and six C-C sigma bonds.

28. Common Mistakes

A double bond is not two sigma bonds. It contains one sigma and one pi bond.
A triple bond contains one sigma and two pi bonds.
Do not confuse resonance with rapid movement of molecules between structures.
In resonance, atoms generally remain in the same positions; electron distribution changes.
Inductive effect is transmitted through sigma bonds, while resonance requires conjugation.
The electromeric effect is temporary and associated with multiple bonds during attack by a reagent.
Electrophiles accept electron pairs.
Nucleophiles donate electron pairs.
Carbocations are electron deficient and generally electrophilic.
Carbanions are electron rich and generally nucleophilic or basic.
Greater s-character generally increases the acidity of C-H bonds.
Polar bonds do not automatically mean that the whole molecule has a non-zero dipole moment. Molecular geometry must also be considered.

29. TU Examination Preparation

Important 2-mark questions

Define hybridization.
What is a sigma bond?
What is a pi bond?
Define resonance.
What is inductive effect?
What is electromeric effect?
Define hyperconjugation.
What is an electrophile?
What is a nucleophile?
Define carbocation.
Define carbanion.
What is dipole moment?

Important 5-mark questions

Explain sp, sp² and sp³ hybridization with examples.
Differentiate between sigma and pi bonds.
Explain resonance and resonance stabilization.
Explain positive and negative inductive effects.
Explain mesomeric and electromeric effects.
Explain hyperconjugation with suitable examples.
Explain homolytic and heterolytic bond fission.
Explain electrophiles and nucleophiles.

Important 10-mark questions

Explain the structure and bonding in organic compounds with reference to hybridization, sigma bonds, pi bonds, bond length, bond angle and bond strength.
Discuss resonance, inductive effect, mesomeric effect, electromeric effect and hyperconjugation.
Explain the formation, structure and stability of carbocations, carbanions and free radicals.
Discuss the factors affecting acidity and basicity of organic compounds.
Explain polarity and dipole moment and discuss the factors affecting them.

Numericals to prepare

Dipole moment
Conversion of dipole moment into Debye
Percentage ionic character
Hybridization
Number of sigma and pi bonds
Percentage s-character
Acidity comparison based on hybridization

30. Quick Revision

Hybridization

sp → linear → 180°

sp² → trigonal planar → 120°

sp³ → tetrahedral → 109.5°

Bond composition

Single = 1σ

Double = 1σ + 1π

Triple = 1σ + 2π

Dipole moment

μ = q × r

Effects

Inductive → sigma bond

Mesomeric → conjugated pi system

Electromeric → temporary pi-electron shift

Hyperconjugation → sigma electron delocalization

Bond fission

Homolytic → radicals

Heterolytic → ions

Reactive species

Electrophile → electron-pair acceptor

Nucleophile → electron-pair donor

Carbocation → electron deficient

Carbanion → electron rich

Important stability orders

Carbocation: 3° > 2° > 1° > CH₃⁺

Simple carbanion: CH₃⁻ > 1° > 2° > 3°

Free radical: 3° > 2° > 1° > CH₃·

C-H acidity according to hybridization

sp > sp² > sp³

in general acidity of C-H bonds.

31. Quick Check

1. What is the hybridization of carbon in methane?

2. How many pi bonds are present in a triple bond?

3. Which species is an electron-pair acceptor?

4. Which effect is transmitted through sigma bonds?

5. Which carbocation is generally most stable?

6. Which carbon has the greatest s-character?

7. Homolytic bond cleavage produces: