Meditaliano IMAT Prep
Lesson 3: Organic Chemistry I (Fundamentals & Reactions)
Part 1: The Magic of Carbon & Basics
Welcome to Organic Chemistry, the study of carbon-based compounds. This field is the absolute foundation of biochemistry, pharmacology, and life itself. But why Carbon? Out of all 118 elements, why is life built on element #6?
Why Carbon is Special
- Tetravalency: Carbon has 4 valence electrons. It forms exactly 4 strong covalent bonds, acting like a universal 4-way Lego block.
- Catenation: Carbon's superpower. It can bond to other carbon atoms infinitely to form long chains, branched structures, and rings without losing stability.
- Moderate Electronegativity: Carbon plays well with others. It shares electrons evenly with Hydrogen (non-polar) but can also bond with O, N, and Halogens to create reactive "polar" sites.
How We Draw Molecules
Chemists are lazy in a smart way. Writing every "C" and "H" is tedious. Enter Skeletal (Line-Angle) Structures:
- Every endpoint and corner (vertex) represents a Carbon atom.
- Hydrogens attached to Carbon are HIDDEN. We assume Carbon always makes 4 bonds, so any "missing" bonds are implicitly Hydrogens.
- Heteroatoms (O, N, Cl, etc.) and their attached Hydrogens must ALWAYS be drawn explicitly.
IMAT Exam Scope Clarification
Advanced substitution and elimination mechanisms ($S_N1$, $S_N2$, $E1$, $E2$) and complex carbonyl condensations (Aldol, Claisen) fall outside the standard IMAT syllabus. Do not waste time memorizing them. We will focus strictly on the structural principles, basic additions, aromaticity, and functional groups guaranteed to appear on your exam.
Challenge an IMAT Question!
Question 173 Official Paper: 2019 - Q42
worked solution & explanation
Concept Classifying chemical reactions.
Step 1 A double exchange (double replacement) reaction is represented by the general form: $\text{AB} + \text{CD} \rightarrow \text{AD} + \text{CB}$.
Step 2 In $\text{NiO} + 2\text{HNO}_3 \rightarrow \text{Ni(NO}_3)_2 + \text{H}_2\text{O}$, nickel and hydrogen exchange their anions (oxygen and nitrate).
Step 3 This represents an acid-base double exchange reaction, which fits Option A.
Part 2: Structure & Hybridization
To form 4 identical bonds, Carbon mathematically mixes its one s-orbital and three p-orbitals to create new, equivalent orbitals. This is called Hybridization.
2.1 Sigma ($\sigma$) vs. Pi ($\pi$) Bonds
| Bond Type | Orbital Overlap | Rotation | Strength & Reactivity |
|---|---|---|---|
| Sigma ($\sigma$) | Direct, head-on overlap along the internuclear axis. | Completely Free. Defines molecular conformation. | Strongest covalent bond. Highly unreactive. First bond made. |
| Pi ($\pi$) | Lateral, side-by-side overlap of unhybridized p-orbitals. | Restricted. Locks geometry, allowing Cis/Trans isomers. | Weaker than $\sigma$. Electron-rich and highly reactive (nucleophilic). |
2.2 Hybridization States of Carbon
Diagram 1: Carbon Hybridization Geometries
Carbon's Hybridization and Geometric Architecture: Visualizing orbital redistribution and the three distinct geometric architectures (sp³, sp², sp).
Index of Hydrogen Deficiency (IHD)
The IHD calculates the exact total number of rings and $\pi$ bonds in a molecule from its molecular formula.
- C: Carbons, H: Hydrogens.
- N (Nitrogen) adds 1 to the count.
- X (Halogens: F, Cl, Br, I) subtracts 1 from the count.
- O (Oxygen) and S (Sulfur) are completely ignored.
Example: Benzene ($C_6H_6$) $\rightarrow IHD = 6 + 1 - (6/2) = 4$.
(1 ring + 3 double bonds = 4).
Challenge an IMAT Question!
Question 135 Official Paper: 2020 - Q45
worked solution & explanation
Concept Hybridization of central atoms in molecules.
Step 1 In $\text{NH}_3$, nitrogen has 3 bonding pairs and 1 lone pair. Its steric number is 4, which corresponds to $sp^3$ hybridization.
Step 2 In $\text{HCN}$, carbon has 1 single bond and 1 triple bond. Steric number is 2, corresponding to $sp$ hybridization.
Step 3 In $\text{BCl}_3$, boron has 3 single bonds and 0 lone pairs. Steric number is 3, corresponding to $sp^2$ hybridization.
Challenge an IMAT Question!
Question 133 Official Paper: 2021 - Q40
worked solution & explanation
Concept Hybridization geometry and bond angles.
Step 1 $sp^3$ hybrid orbitals are formed by mixing one $s$ and three $p$ atomic orbitals.
Step 2 The four orbitals arrange themselves in a tetrahedral molecular geometry to minimize electron-pair repulsion.
Step 3 The ideal tetrahedral bond angle formed between these orbitals is $109.5^\circ$.
Part 3: IUPAC Nomenclature
The IUPAC system gives a systematic, unambiguous name to every organic molecule. For the IMAT, prioritizing functional groups is absolutely critical to getting the name right.
3.1 The Master Priority Table
When multiple groups are present, the one with the highest priority determines the suffix (ending) of the name, and the chain is numbered to give this group the lowest possible number.
| Priority | Functional Group | Formula | Suffix (High Priority) | Prefix (Low Priority) |
|---|---|---|---|---|
| 1 (Highest) | Carboxylic Acid | -COOH | -oic acid | carboxy- |
| 2 | Ester | -COOR | -oate | alkoxycarbonyl- |
| 3 | Amide | -CONH₂ | -amide | amido- |
| 4 | Nitrile | -C≡N | -nitrile | cyano- |
| 5 | Aldehyde | -CHO | -al | oxo- (formyl) |
| 6 | Ketone | -CO- | -one | oxo- |
| 7 | Alcohol | -OH | -ol | hydroxy- |
| 8 | Amine | -NH₂ | -amine | amino- |
| Lowest | Alkyl / Halogen / Ether | -R / -X / -OR | -ane | alkyl- / halo- / alkoxy- |
IUPAC Naming Master Algorithm
- Find the Longest Carbon Chain: It must contain the highest priority functional group and any double/triple bonds.
- Number the Chain: Start from the end closest to the highest priority functional group.
- Identify Substituents: Name branches (e.g., methyl, ethyl, bromo).
- Assemble the Name: List substituents alphabetically (ignore di-, tri-), use locant numbers, add the parent chain length (meth, eth, prop...), indicate double/triple bonds (-an-, -en-, -yn-), and end with the primary suffix.
Diagram 2: Anatomy of an IUPAC Name
Part 4: 3D Stereochemistry & Isomerism
Isomers share the same molecular formula but have distinctly different structural arrangements, profoundly impacting their physical properties, smell, and how they interact with enzymes and drugs in the human body.
4.1 Isomer Classification Tree
| Isomer Class | Definition | Examples |
|---|---|---|
| Constitutional (Structural) |
Different covalent connectivity. The atoms are attached in a completely different order. | n-Butane vs Isobutane. 1-propanol vs 2-propanol. Ethanol vs Dimethyl ether. |
| Conformational | Differ only by rotation around single ($\sigma$) bonds. They constantly flip and cannot be isolated. | Anti, Gauche, Eclipsed conformers. Chair flips in Cyclohexane. |
| Enantiomers | Stereoisomers that are perfect, non-superimposable mirror images. (Chiral molecules). | (R)-Lactic acid and (S)-Lactic acid. Rotate plane-polarized light in exactly opposite directions. |
| Diastereomers | Stereoisomers that are strictly NOT mirror images. Have completely different physical properties. | Cis/Trans geometric isomers. Molecules with multiple stereocenters where only some differ. |
Diagram 3: Constitutional Isomers ($C_4H_{10}$)
Drawn using skeletal (line-angle) structures. Every corner/end is a Carbon.
Constitutional Isomerism in Alkanes: Visualizing how distinct covalent connectivity in butane and pentane isomers alters molecular shape and physical properties.
4.2 Geometric Isomers (Cis-Trans & E/Z)
These occur when free rotation is impossible, typically due to a C=C double bond locking the structure in place, or a rigid cyclic ring.
Diagram 4: Geometric Isomers (2-Butene)
Restricted Rotation and Geometric Isomerism (Cis/Trans): Visualizing how the π bond locks geometry, forcing substituents into distinct cis or trans arrangements.
4.3 Enantiomers & Absolute Configuration (R/S)
An asymmetric (chiral) carbon is an $sp^3$ hybridized carbon bonded to exactly four completely different groups. It produces two non-superimposable mirror images called Enantiomers, just like your left and right hands.
Cahn-Ingold-Prelog (R/S) System Rules
- Assign Priorities (1 to 4): Based strictly on the Atomic Number (Z) of the attached atom. (e.g., -Br > -OH > -CH3 > -H).
- Orient the Molecule: Ensure the lowest priority group (#4, usually H) points away from you (dashed line).
- Trace the Path: Draw a curve from priority 1 $\rightarrow$ 2 $\rightarrow$ 3.
- Assign: Clockwise = (R) [Rectus/Right]. Counter-clockwise = (S) [Sinister/Left].
- Master Trick: If group #4 points towards you (solid wedge), trace normally but completely REVERSE your final answer! (R becomes S, S becomes R).
Diagram 5: Enantiomers of Lactic Acid
Assigning Absolute Configuration (R/S System): A step-by-step master class on the Cahn-Ingold-Prelog (CIP) system for assigning absolute configuration to a chiral center.
Challenge an IMAT Question!
Question 57 Official Paper: 2015 - Q46
worked solution & explanation
Concept Boiling points and branching in structural isomers.
Step 1 Both pentane and 2,2-dimethylpropane are structural isomers with the formula $\text{C}_5\text{H}_{12}$, meaning they share the same molecular mass.
Step 2 Pentane is a straight-chain molecule with a larger surface area, allowing for stronger van der Waals dispersion forces.
Step 3 2,2-dimethylpropane is highly branched and spherical, reducing its contact surface area, which leads to weaker dispersion forces and a lower boiling point.
Part 5: Alkanes & Radicals
Alkanes ($C_nH_{2n+2}$) are saturated, nonpolar hydrocarbons. They are notoriously unreactive (chemically inert) due to strong, stable C-C and C-H $\sigma$ bonds. They basically do two things: Combust (burn) and undergo Free-Radical Halogenation.
- Boiling Points: Increase significantly with chain length due to a larger surface area for London Dispersion Forces.
- Branching: Branched isomers are more spherical/compact. This reduces their surface area, lowering the boiling point compared to long, spaghetti-like straight-chain isomers.
Radicals are species with a single unpaired electron. They are highly reactive and electron-deficient. Alkyl groups stabilize them by donating electron density (hyperconjugation).
Diagram 6: Free-Radical Halogenation (Propagation Step)
Challenge an IMAT Question!
Question 168 Official Paper: 2024 - Q40
worked solution & explanation
Concept Conformational analysis and projections.
Step 1 A Newman projection is a visualization method used to view conformations of chemical bonds along the C-C bond axis.
Step 2 The front carbon is represented by a dot (vertex of three bonds), while the back carbon is represented by a circle.
Part 6: Alkenes, Alkynes & Addition
Alkenes have a double bond ($\sigma + \pi$). The $\pi$ bond is an exposed cloud of electrons above and below the plane. It acts as a powerful nucleophile (nucleus-lover, seeks positive charge), attacking electrophiles to undergo Electrophilic Addition. The $\pi$ bond breaks to form two new, stable $\sigma$ bonds.
"The rich get richer." In the addition of HX (like HCl) to an unsymmetrical alkene, the Hydrogen ($H^+$) adds exclusively to the carbon of the double bond that already possesses the greater number of hydrogen atoms.
Why does this happen? It's not magic, it's stability. The $\pi$ bond attacks $H^+$, creating an intermediate with a positive charge on Carbon (a Carbocation). The reaction pathway strictly follows the formation of the most stable carbocation (3° > 2° > 1°).
Diagram 7: Electrophilic Addition Mechanism (Markovnikov)
Electrophilic Addition and Markovnikov's Rule: Visualizing the reactive mechanism and carbocation intermediate stability that dictates product formation.
6.2 Key Addition Reactions to Know
| Reaction Type | Reagents | What Adds? | Regiochemistry/Stereo |
|---|---|---|---|
| Hydrohalogenation | HX (HCl, HBr) | H and X | Markovnikov |
| Hydration | $H_2O, H^+ (acid)$ | H and OH (Makes Alcohol) | Markovnikov |
| Halogenation | $Br_2$ or $Cl_2$ | X and X | Anti-Addition (Trans product). *Decolorizes Bromine water. |
| Hydrogenation | $H_2$ + Pt/Pd/Ni catalyst | H and H (Makes Alkane) | Syn-Addition (Cis product) |
Part 7: Aromaticity & Benzene
Benzene ($C_6H_6$) is incredibly special. Even though it looks like it has three double bonds, it is highly stable due to complete resonance delocalization of its $\pi$ electrons in a ring. Because of this extreme stability, Benzene does NOT undergo standard addition reactions (which would break the ring). Instead, it undergoes Electrophilic Aromatic Substitution (EAS), replacing an H with an electrophile to maintain its aromatic ring.
A molecule is considered strictly aromatic only if it meets ALL of these criteria:
- It must be Cyclic (a ring).
- It must be Completely Planar (flat, all atoms $sp^2$ hybridized).
- It must be Fully Conjugated (alternating single/double bonds around the entire ring).
- Has exactly $4n + 2$ $\pi$ electrons. (e.g., 2, 6, 10, 14...). Benzene has 6.
Diagram 8: Benzene Resonance
The true structure is a hybrid average (right).
Part 8: Functional Groups & Real-World Marvels
Functional groups are specific clusters of atoms that completely dictate the chemical reactions, physical properties, and biological role of an organic molecule. Adding a single -OH group turns toxic ethane gas into drinkable ethanol liquid!
8.1 Comprehensive Functional Group Map
| Group | Structure | Properties & Reactions | Real-World Examples |
|---|---|---|---|
| Alcohol | -OH | Polar, forms strong H-bonds (high boiling point). Can be oxidized to aldehydes/ketones/carboxylic acids. | 🍺 Ethanol (Beverages, hand sanitizer). ❄️ Menthol (Peppermint cooling effect). |
| Aldehyde | -CHO | Carbonyl (C=O) at the end of a chain. Easily oxidized to carboxylic acid. Positive Tollens' test (creates a silver mirror). | ☠️ Formaldehyde (Preservative). 🍦 Vanillin (Vanilla extract). 🍂 Cinnamaldehyde (Cinnamon flavor). |
| Ketone | -CO- | Internal carbonyl. Sandwiched between carbons. Cannot be easily oxidized further. Negative Tollens' test. | 💅 Acetone (Nail polish remover). 💪 Testosterone (Steroid hormone). |
| Carboxylic Acid | -COOH | Weak acid. Deprotonates to form a highly resonance-stabilized carboxylate ion ($-COO^-$). Sour taste. | 🥗 Acetic Acid (Vinegar). 🍋 Citric Acid (Citrus fruits). 🏃 Lactic Acid (Muscle fatigue). |
| Ester | -COOR | Formed by Esterification (Carboxylic Acid + Alcohol $\rightleftharpoons$ Ester + $H_2O$). Typically volatile with pleasant, fruity odors. Found in fats (triglycerides). | 🍌 Isoamyl acetate (Banana flavor). 🍍 Ethyl butyrate (Pineapple). 💊 Aspirin (Acetylsalicylic acid). |
| Amine | -NH₂ | Derivative of ammonia. Weak base (Nitrogen's lone pair accepts $H^+$). Acts as a nucleophile. Often smells fishy. | 🧠 Dopamine/Serotonin (Neurotransmitters). 🧟 Cadaverine (Smell of rotting flesh). |
| Amide | -CONH₂ | Formed by Condensation (Acid + Amine $\rightleftharpoons$ Amide + $H_2O$). Not basic! The lone pair is tied up in massive resonance with the carbonyl. Highly stable. | 🥩 Proteins (Linked by amide/peptide bonds). 🛡️ Kevlar (Bulletproof vests). 💊 Acetaminophen (Tylenol). |
Diagram 9: Oxidation Pathways of Alcohols
Challenge an IMAT Question!
Question 153 Official Paper: 2022 - Q47
worked solution & explanation
Concept Identification of functional groups.
Step 1 An alcohol contains a hydroxyl group ($\text{-OH}$) attached to a saturated carbon atom.
Step 2 Trifluoroethanol ($\text{CF}_3\text{CH}_2\text{OH}$) possesses the hydroxyl group and is an alcohol.
Step 3 The other options: Benzene is an aromatic hydrocarbon, cyclohexanone is a ketone, propanal is an aldehyde, and butene is an alkene.
Part 9: Elemental Analysis
Combustion analysis mathematically determines the Empirical Formula (the simplest integer ratio of atoms) of an unknown organic compound. The sample is completely burned in oxygen: all Carbon becomes $CO_2$, and all Hydrogen becomes $H_2O$.
Master Calculation Algorithm
- Moles of Carbon: Moles C = (Mass of $CO_2$ produced / 44.01 g/mol).
- Moles of Hydrogen: Moles H = (Mass of $H_2O$ produced / 18.02 g/mol) $\times$ 2 (because there are 2 H per $H_2O$).
- Find Oxygen: Calculate the mass of C and H found. Subtract this sum from the original sample mass. Convert the remaining mass to moles of O (divide by 16).
- Empirical Formula: Divide all mole values by the absolutely smallest mole value present to get the simplest integer ratio. (e.g. $C_1H_2O_1 = CH_2O$).
- Molecular Formula: Found by dividing the given total Molar Mass of the compound by the mass of the Empirical Formula. Multiply the empirical subscripts by this integer.
IMAT Organic Chemistry Exam
30 High-Yield Questions (Strictly within IMAT Syllabus)