Difficulty: Intermediate | Prerequisites: Bonding fundamentals, resonance and acid-base chemistry
This material ties the bonding and resonance concepts together into their most practical application: understanding how and why reactions happen. Polar bonds create sites of reactivity (electrophilic and nucleophilic centres). Curved arrows show exactly which electrons move and where they go. Molecular formulas and degrees of unsaturation tell you what kind of structure to expect before you even draw it. These skills form the backbone of every mechanism you will write for the rest of the course.
Polar covalent bonds arise from electronegativity differences and create the reactive sites in organic molecules. Reaction mechanisms use curved arrows to track electron movement step by step. Molecular formulas encode structural information: the degree of unsaturation tells you how many rings and pi bonds a molecule contains.
Polar covalent bond
A bond between two atoms with different electronegativities, where the bonding electrons are shared unequally. The more electronegative atom carries a partial negative charge (δ-), the less electronegative atom a partial positive charge (δ+). Think of it as a tug-of-war for electrons that one side is winning.
Reaction mechanism
A step-by-step account of bond-breaking and bond-forming events in a chemical reaction, shown using curved arrows. Each step shows which electrons move, from where, and to where.
Curved arrow (electron-pushing arrow)
An arrow drawn from an electron source (a lone pair or a bond) to an electron sink (an atom or a bond). The tail starts at the electrons; the head points to where they go. A full curved arrow moves two electrons. A fishhook (half-headed) arrow moves one electron (radical mechanisms).
Nucleophile
An electron-rich species that donates an electron pair to form a new bond. In simple terms, it is the "electron giver." Examples: OH minus, Cl minus, water, amines.
Electrophile
An electron-poor species that accepts an electron pair to form a new bond. In simple terms, it is the "electron taker." Examples: carbocations, carbonyl carbons, protons (H+).
Leaving group
An atom or group that departs with a bonding pair of electrons during a reaction. Good leaving groups are stable after they leave (weak bases). Examples: Cl minus, Br minus, water, tosylate.
Carbocation
A carbon atom bearing a positive formal charge, with only three bonds and an empty p orbital. It is sp2 hybridised and trigonal planar. Carbocations are strong electrophiles.
Degree of unsaturation (index of hydrogen deficiency, IHD)
The number of rings plus pi bonds in a molecule. Calculated from the molecular formula CnHm as: (2n + 2 - m) / 2. Each degree represents one ring or one pi bond (a double bond = 1, a triple bond = 2, a ring = 1).
When is a bond polar?
A covalent bond is polar when the two atoms have different electronegativities. The greater the difference, the more polar the bond. Organic chemistry's most common polar bonds:
C-O and C=O (oxygen is more electronegative)
C-N and C=N (nitrogen is more electronegative)
C-Cl, C-Br (halogens are more electronegative)
O-H, N-H (these are polar, with the hydrogen carrying δ+)
C-C and C-H bonds are essentially nonpolar.
Problem II from the exam: identifying bond types in a bioactive compound
The exam shows a complex molecule containing a cyclohexane ring, an OH group, a Cl, and an amide (C=O bonded to N). You are asked to find:
A. A polar covalent single bond
Any C-O, C-N, C-Cl, O-H, or N-H single bond qualifies. The C-Cl bond is a clear example: chlorine is more electronegative than carbon, creating a permanent dipole.
B. A polar covalent double bond
The C=O (carbonyl) bond is the textbook example. Oxygen pulls electron density from carbon, making the carbon electrophilic (δ+) and the oxygen nucleophilic (δ-).
C. A bond between two atoms with different hybridisations (excluding bonds to H)
Look for a bond connecting, say, an sp2 carbon to an sp3 carbon. The carbon of the C=O group is sp2; the carbon next to it in a saturated ring is sp3. The bond between them is a bond between differently hybridised atoms.
D. A heteroatom with sp3 hybridisation
The oxygen of the -OH group: it has two bonds (to C and to H) and two lone pairs, giving four electron groups and sp3 hybridisation.
E. An atom with trigonal planar geometry
The carbonyl carbon (C=O): it has three groups around it (the O, the N, and another C), all in a plane at 120°. This is sp2 hybridised with trigonal planar geometry. The nitrogen of the amide can also qualify if its lone pair is delocalised into the carbonyl, making it effectively sp2.
The extra credit problem: an SN1-type mechanism
The exam shows three steps of a substitution reaction and asks you to fill in curved arrows. This is a preview of mechanisms covered later in the course, but the arrow-pushing logic is what you already know.
Step 1: Ionisation (leaving group departure)
Starting material: a tertiary alkyl chloride (cyclohexane ring with a methyl group and a Cl).
The C-Cl bond breaks heterolytically: the bonding electrons go entirely to chlorine, which leaves as Cl minus. This generates a tertiary carbocation.
Curved arrow: from the C-Cl bond to the Cl atom. The arrow tail sits on the bond, the arrow head points to Cl.
Step 2: Nucleophilic attack by water
Water (a nucleophile with lone pairs) attacks the electrophilic carbocation.
Curved arrow: from a lone pair on the oxygen of water to the positively charged carbon. This forms a new C-O bond. The product is an oxonium ion (the oxygen now has three bonds and a positive charge).
Step 3: Deprotonation
Chloride ion (Cl minus, from Step 1) acts as a base and removes a proton from the oxonium ion.
Curved arrow: from a lone pair on Cl minus to the H on the positively charged oxygen. A second curved arrow goes from the O-H bond back to the oxygen (showing that the bonding electrons return to oxygen as the proton is removed).
The product is a tertiary alcohol plus HCl.
Key principles for drawing curved arrows
Arrows always flow from electron-rich to electron-poor: from a lone pair or a bond towards a positive centre or an electronegative atom.
Every arrow must start at a pair of electrons (a lone pair or a bond). Never start an arrow at a positive charge or at an atom without specifying which electrons are moving.
Use one arrow per electron-pair movement. If two things happen simultaneously (as in Step 3, where one bond forms while another breaks), draw both arrows.
The number of bonds and lone pairs must balance in each step. If a bond forms, a bond or lone pair must break somewhere to compensate.
Reading molecular formulas from line structures (Q2)
The exam shows a line structure and asks for its molecular formula. To determine it:
Count every carbon at each vertex and terminus of the line structure.
Add hydrogens to each carbon until it has four bonds.
Count any explicitly shown heteroatoms and their hydrogens.
For a structure that looks like a branched chain with one double bond:
Count the carbons carefully at each kink and endpoint.
A double bond means two fewer hydrogens than the saturated formula.
Constitutional isomers (Q4)
Constitutional isomers have the same molecular formula but different connectivity (different bonding sequences).
The exam asks: how many constitutional isomers does C5H14 have?
This is a trick. For five carbons, the maximum hydrogen count in a saturated acyclic alkane (CnH2n+2) is C5H12 (pentane). C5H14 has two more hydrogens than is possible for a structure made of only carbon and hydrogen. No valid structure exists. The answer depends on whether the exam intends this as a trick question (answer: 0 or the question is flawed) or whether it is a typo for C5H12 (which has 3 constitutional isomers: pentane, isopentane, neopentane).
Degree of unsaturation (IHD)
For a hydrocarbon CnHm:
\text{IHD} = \frac{2n + 2 - m}{2}IHD = 0: fully saturated, no rings or pi bonds
IHD = 1: one ring or one double bond
IHD = 2: could be two double bonds, one triple bond, two rings, or one ring + one double bond
IHD = 4: typical of a benzene ring (three double bonds + one ring)
For molecules containing nitrogen: each nitrogen adds 1 to the (2n + 2) count. For halogens: each halogen subtracts 1 from the hydrogen count. For oxygen: no change to the formula.
Applying IHD as a diagnostic tool
Before you draw a structure, calculate the IHD. It tells you how many rings and/or pi bonds to expect and prevents you from drawing something impossible (like a saturated structure when IHD = 4).
Students often draw curved arrows starting at a positive charge. Arrows must start at electrons (a lone pair or a bond). A positive charge tells you where electrons are going, not where they come from.
Students sometimes think that all bond-breaking is heterolytic (both electrons go to one atom). In radical reactions, bonds break homolytically (one electron to each atom). For this exam, though, the mechanisms shown are all ionic (heterolytic).
Students frequently forget to count implicit hydrogens in line structures, leading to wrong molecular formulas. Every vertex and line terminus is a carbon, and every carbon needs four bonds.
Students often confuse constitutional isomers with stereoisomers. Constitutional isomers differ in connectivity (which atoms are bonded to which). Stereoisomers have the same connectivity but differ in spatial arrangement.
⚠️ Curved-arrow mechanisms appear on every CHEM 225 exam from this point forward. The extra credit question is a preview; later exams will require full mechanism drawings for marks.
⚠️ Identifying polar bonds, hybridisation, and geometry in complex molecules (Problem II style) is a high-value question type. Practise on structures you have not seen before.
⚠️ Molecular formula determination from line structures is a quick-mark question. Losing points here is avoidable with careful carbon and hydrogen counting.
⚠️ The IHD formula is a diagnostic tool for spectroscopy and structure determination problems later in the course. Memorise it now.
⚠️ Watch for trick questions involving impossible molecular formulas (like C5H14 for a pure hydrocarbon). Checking the IHD first catches these immediately.
True or false: Curved arrows in mechanisms flow from electron-poor to electron-rich. (False, they flow from electron-rich to electron-poor.)
Fill in the blank: The degree of unsaturation for C6H6 (benzene) is ______. (4)
True or false: A C-Cl bond is polar because chlorine is more electronegative than carbon. (True)
Fill in the blank: In an SN1 mechanism, the leaving group departs ______ the nucleophile attacks. (before)
True or false: C5H14 is a valid molecular formula for a compound made only of carbon and hydrogen. (False)
Q: In the SN1-type mechanism shown on the exam, what type of species is the intermediate formed after the leaving group departs?
A: A tertiary carbocation. The carbon that lost the chloride now has only three bonds and an empty p orbital, giving it a positive formal charge.
Q: Why does water attack the carbocation in Step 2 rather than the other way around?
A: Water is electron-rich (it has lone pairs) and the carbocation is electron-poor (it has an empty orbital). Electrons flow from the nucleophile (water) to the electrophile (carbocation).
Q: Calculate the degree of unsaturation for C7H12.
A: IHD = (2 x 7 + 2 - 12) / 2 = (16 - 12) / 2 = 2. The molecule has two degrees of unsaturation, which could be two double bonds, one triple bond, one ring + one double bond, or two rings.
Q: A molecule has a carbonyl group (C=O) bonded to a nitrogen. Name one polar covalent single bond and one polar covalent double bond in this fragment.
A: Polar single bond: C-N (different electronegativities). Polar double bond: C=O (oxygen is much more electronegative than carbon).
Q: In a line structure, how do you determine the molecular formula?
A: Count every carbon (at each vertex and terminus), add hydrogens to each carbon until it has four bonds, then include any explicitly drawn heteroatoms and their hydrogens.
Curved-arrow mechanisms are the language of organic chemistry from this point on. Every named reaction (SN1, SN2, E1, E2, additions, eliminations) is described using the same arrow conventions introduced here. Polar bonds determine where nucleophiles and electrophiles sit in a molecule, which directly governs regioselectivity. Degrees of unsaturation will become essential when you tackle spectroscopy (IR, NMR, mass spec) and need to propose structures from molecular formulas.
polar covalent bond, bond polarity, electronegativity difference, reaction mechanism, curved arrows, electron pushing, arrow pushing, nucleophile, electrophile, leaving group, carbocation, SN1, substitution, ionisation, oxonium ion, deprotonation, molecular formula, line structure, constitutional isomers, degree of unsaturation, index of hydrogen deficiency, IHD, CHEM 225, organic chemistry II, UMich, tertiary alcohol, heterolytic cleavage