- An **organic molecule** is one with at least one carbon atom covalently bonded to another carbon or hydrogen atom (i.e., at least one C-H or C to C bond)
- A **branched hydrocarbon** is one with at least one "side group" extending from the main hydrocarbon chain.
- A **functional group** is a group of atoms responsible for the characteristic properties of a molecule (e.g. C=C)
- A **homologous series** is a family of organic compounds with the same functional group but the hydrocarbon chain length changes by 1 $\ce{CH2}$ group.
The general formula for an **acyclic** hydrocarbon with no rings is as follows, where $n$ is the number of carbon atoms, $x$ is the number of double bonds, and $y$ is the number of triple bonds.
$$\ce{C_nH_{2n+2-2x-4y}}$$
### Representing organic compounds
A simple **molecular formula** is the least useful as it provides no information on structure and bonding.
$$\ce{C6H14}$$
A **complete structural diagram** shows all atoms by their chemical symbols and uses lines like a Lewis Dot diagram to represent bonds. VSEPR shapes do not need to be taken into account because these are 2D representations of the molecule.
A **condensed structural diagram** is a complete structural diagram but C-H bonds are aggregated into a formula.
$$\ce{CH3 - CH2 - CH2 - CH2 - CH2 - CH3}$$
A **structural formula** or **expanded molecular formula** is a condensed structural diagram but there are no bond lines. The bond organisation is inferred based on the number of hydrogens on each carbon. Carbon chain side groups (branches) are shown with parentheses.
$$\ce{CH3CH2CH2CH2CH2CH3}$$
A **condensed structural formula** is a structural formula but any consecutive repeated $\ce{CH2}$ groups are factored with parentheses.
$$\ce{CH3(CH2)_4CH3}$$
A **line diagram** or **skeletal structural formula** removes carbons and hydrogens and replaces all carbon-carbon bonds with lines, where the number of lines represents the type of bond. Each line is bent where a carbon atom would be, except for triple bonds as those are linear. Non-carbon groups such as $\ce{OH}$ can be shown in collapsed form.
- The numbers should be arranged in a way that the highest priority functional group in the chain is assigned the lowest number possible.
- Apply the **first branch rule** only if there is a tie: If there are side chains, the parent chain should be numbered such that the location of any side chains have the lowest number possible.
- If there is a tie, the location with the most branches wins.
- If there is a tie, the rest of the chain is compared in sequence applying the first branch rule.
- If there is a tie, the first location with the side chain group name that is alphabetically greater wins.
- If there is a tie, it doesn't matter which side is picked as the whole thing is symmetrical.
5. Arrange the name with each side group with their numbers in alphabetical order, discounting any prefixes due to duplicates, followed by the parent chain.
- Drop the ending vowel from the prefix if there is a double vowel unless it is "i".
- Separate numbers from words with dashes.
- Separate numbers from numbers with commas.
- Do not separate words from words.
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In hydrocarbons:
- Atoms with double or triple bonds share equal priority as the highest functional group.
- The main chain will be named as an alkane if there are only single bonds.
- If there is exactly one double or triple bond, it will be named as an alkene or alkyne with its position inserted between the prefix and ending.
- e.g., "pentane", "pent-2-ene"
- If there are multiple double or triple bonds, their numbers are also included, but an "a" is appended to the prefix and a Greek prefix added to the suffix.
- e.g., "penta-1,3-diene", "hexa-1,3,5-triyne"
- If there are both double and triple bonds, the "-ene" becomes "-en" and is always before "-yne".
- e.g., "pent-4-en-2-yne"
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Other **side chains** with equal priority as double or triple bonds _in side chains_ include:
- halogens, which have their "-ine" suffix replaced with "o" (e.g., "chloro")
Cyclic aliphatic hydrocarbons are named the same way as acyclic hydrocarbons except they have a "**cyclo-**" at the start of the name of their parent chain.
The initial double bond should be numbered such that the lowest number is assigned to both sides of the bond (numbers 1 and 2 should be to either side of the double bond). If there is more than one double bond, the ring should be numbered such that the lowest number is assigned to both.
Rings can be side chains, and are named accordingly (e.g., "cyclopropyl"). The "cyclo-" prefix is counted when sorting names alphabetically as it describes the group.
As benzene rings do not have double bonds, they are named according to the **first branch rule**.
### Isomers
**Structural isomers** are two chemicals that have the same chemical formulas but have different structural formulas, resulting in different chemical properties.
**Geometric** or **cis/trans isomers** are two chemicals have the same chemical formulas and atom arrangements but are positioned differently, thus having ambiguous names.
In order for this to occur, there must be two different atoms or groups of atoms bonded to each carbon atom in the double bond.
- **polarity**: a cis isomer may cause a molecule to be polar as opposed to its trans variant
- **packing efficiency**: a non-branching hydrocarbon chain will pack better than a branching one, and a continuously trans chain will pack better than a cis one
In reactions involving a benzene ring, the ring itself is **stable** and will not break apart because of the strength of delocalised pi bonds.
Therefore, only the hydrogens can be swapped out via **electrophilic substitution**, where an hydrogen atom is substituted with an electrophile. The concentration of electrons in the delocalised pi area attracts electrophiles to initiate the bond.
In the mechanism diagram below, $\ce{E+}$ represents the electrophile. Curly arrows are used to show the movement of electrons from the **delocalised area to the electrophile** and **hydrogen atom to the delocalised area**.
The **first step** (the change from the first to the second diagram) is the **slow step** due to the highest activation energy due to the requirement to break a bond.
In a **nitrating mixture**, benzene will react with positive nitronium ions at **~50°C** to form nitrobenzene, outlined in the reaction mechanism diagrams below.
The lone pair on the oxygen of the nitric acid attracts a hydrogen atom, which becomes an $\ce{H+}$ ion as sulfuric acid's oxygen takes its electrons. The hydrogen ion bonds to the nitric acid.
The oxygen-hydrogen group is conveniently able to form water by taking both electrons it was sharing with the nitrogen. The other single-bonded oxygen compensates with a dative covalent bond with the nitrogen to form the nitronium ion.
The second step is to **react with benzene** through electrophilic substitution, with electrons moving back from the dative oxygen-nitrogen bond back to the oxygen.
Because a sigma bond must be broken, alkanes are not very reactive. In the presence of light, alkanes will react with halogens in their standard state through halogenation, replacing one of their hydrogens. **Fluorine** is an exception that does not require light because it is highly reactive.
If the halogen is in excess and the reaction continues, more of the halogen (**not the hydrogen-halogen product**) will react with the alkane until all hydrogens have been substituted.
The order that hydrogens are substituted in is **random**. If there is more than one possibility, all of them are written as products, ignoring balancing.
1-bromoethane reacts with chlorine gas to form either 1,1-dibromoethane (40% chance) or 1,2-dibromoethane (60% chance) because each hydrogen is equally likely to be substituted, and there are 2 and 3 that would form them, respectively.
- A **free radical** is a species with a lone unpaired electron.
- **Homolytic fission** is the dissociation of a chemical bond in a neutral molecule where each product takes one electron, generating two free radicals.
- **Heterolytic fission** is the dissociation of a chemical bond in a neutral molecule where one product takes both electrons.
- A **carbocation** is a compound with a $\ce{C+}$ atom.
- The **primary** (1°), **secondary** (2°), and **tertiary** (3°) carbocations are carbocations bonded to one, two, and three other carbon atoms, respectively.
The presence of double/triple bonds make alkenes and alkynes more reactive and also allow the **addition** of species as pi bonds are easier to break. Addition always takes precedence over substitution when possible.
1. If the non-alkene/yne reactant does not have a dipole moment, the electrons concentrated in the double/triple bond of the alkene/yne induce a dipole by repelling the electrons closest to it.
2. The positive dipole (such as H in HBr) is attracted to the double bond, and **two electrons** in the bond are used to form a **dative** bond with the positive dipole.
3. No longer needing its old bond, the previously positive dipole loses **both electrons** in its old bond to the negative dipole.
4. The now positive carbon atom attracts the now negative ion.
5. The negative ion forms a **dative** bond with the positive carbon atom.
- If an **alkene is formed**, the same randomness of where the atoms attach applies, so it is possible that a cis/trans isomer is formed.
- If an **asymmetrical alkane** is formed, the same randomness of where the atoms attach applies after applying Markovnikov's rule, so it is possible that positional isomers are formed.
**Markovnikov's rule** states that in Soviet Russia, the rich get richer. Hydrogens preferentially bond to the carbon with the **most hydrogens** if there is one — otherwise it randomly chooses one available.
This is because carbocations with that are _more highly substituted_ (are bonded to more carbon atoms) are more stable, so they last longer and are more likely to form a bond with the negative dipole.
The preferred product is the **major product** while the other is the **minor product**. Some minor product will still be produced if the negative dipole is speedy enough, although it will be vastly outnumbered by the major product.
#### Halogenation
Unlike alkane substitution, addition halogenation is spontaneous.
This process is used to test for alkenes/alkynes in a solution. As bromine water is red-brown, if alkenes/alkynes are present, the water will be **decolourised** from red-brown to become more colourless.
Nucleophilic substitution replaces a group of atoms attached to a C with a nucleophile. Both processes involve the **leaving group** taking both electrons, becoming negative in the process, and forming a carbocation as the other product, which attracts and bonds with the nucleophile.
Effectively all reactions here involve the formation or stealing of dative covalent bonds.
This **two-step** reaction involves the heterolytic fission of the C-X bond to form a carbocation + halide ion (slow), followed by the nucleophile's lone pairs/negative charge attracting it to the carbocation.
The "1" refers to the order of the rate-limiting step being a **unimolecular** collision.
This **single-step** reaction has the nucleophile forming a bond with the central atom **opposite the leaving group** in a "back-side attack". The oncoming nucleophile repels the other groups, causing them to move away, effectively **reflecting** ("inverting") the remaining groups across the vertical axis.
The "2" refers to the order of the rate-limiting step being a **bimolecular** collision.
Dashes must be drawn for the transition state for bonds breaking/forming. In this case, drawing the front/back lines for the bottom two atoms may be ignored in favour of regular lines instead to avoid the ambiguity of forming bonds.
**Steric hindrance** is the effect of other parts of a molecule getting in the way to the central atom, preventing a reaction. If there is not enough space for a backside attack, S<sub>N</sub>2 cannot happen. Therefore, this makes 3° S<sub>N</sub>2 substitution not viable.
**Steric stress reduction** is the resistance of groups against being forced together. In a 3° carbocation, pushing the groups together for a backside attack increases steric stress. This encourages S<sub>N</sub>1 substitution **only for 3°** to maintain a tetrahedral geometry.
The **positive inductive effect** is the effect that causes more highly substituted carbons to be more stable. Electrons on neighbouring carbon atoms can move closer to the carbon ion, creating an electron-donating effect that slightly balances its charge, increasing its stability and thus window of opportunity for a **S<sub>N</sub>1** substitution.
The -ol suffix is a standard suffix following the same numbering rules as -en and -yne. As functional groups are ordered from lowest to highest priority in their name, similar to how a -yne can have an -en, an -ol can also have an -en and **-yn** before it.
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- Therefore, $\ce{CH3OH}$ is methanol, _not_ methol.
When choosing a new double bond to form in the alkene, it must bond to the carbon the OH group was attached to. In elimination, **Markovnikov's rule does not apply**.
In the presence of an oxidising agent that is **limited** and acid, **primary** alcohols will oxidise to form aldehydes, where a hydroxyl group becomes a carbonyl group and the hydrogen migrates to the carbon.
An aldehyde is named like an alcohol but has a higher naming priority, with a suffix of **-al**. As aldehydes must be at the end of a chain, numbering their position is not required.
Aldehydes will continue to react to ketones if the oxidising agent is not limited. To prevent this, the aldehyde is separated and removed from the mixture through distillation.
The mixture is heated to a temperature greater than the aldehyde's boiling point but less than the alcohol's, such that the gaseous aldehyde enters the condenser and is cooled by the water jacket.
An aldehyde can also be reduced in a process similar to **hydrogenation** to reverse the reaction.
$$\ce{aldehyde + H2 ->[\text{high temp, high pressure, Pt/Pd/Ni}] alcohol}$$
### Ketones
In the presence of an oxidising agent and acid, **secondary** alcohols will oxidise to form ketones, where the hydrogen plops off completely.
Ketones have equal priority to aldehydes and are named the same but with a suffix of **-one**. A position number _is_ required because ketones can be located anywhere on the chain.
Carboxylic acids have higher priority than aldehydes/ketones and are named the same but with a suffix of **-oic acid**. Similar to aldehydes, because the $\ce{COOH}$ can only exist on the end of a chain, position numbers are omitted.
If the ionising hydrogen is removed ($\ce{COOH -> COO-}$), a carboxylic acid can form a salt by reacting with a metal to form an **ionic compound**. Salts are named as an ionic compound would be, with the acid component resuffixed to **-oate**.
The **Lucas test** is used to in part determine the type of alcohol (primary/secondary/tertiary) through the **nucleophilic substitution** of OH with Cl. To perform this substitution, **anhydrous** zinc chloride and **concentrated** HCl must be present.
$$\ce{R-OH + HCl ->[ZnCl2] R-Cl + H2O}$$
This test is only valid on **small** alcohols because (<6carbons)aslongeronesareinsoluble.
The insoluble halogenoalkane becomes visible, making the solution **cloudy**. Because the reaction is an S<sub>N</sub>1 reaction:
Alternatively, **oxidising** alcohols to aldehydes/ketones through S<sub>N</sub>2 by reducing $\ce{Cr2O7^2-}$ (orange) to $\ce{Cr^3+}$ (green) will identify the alcohol.
When an alcohol and carboxylic acid react in sulfuric acid **and heat**, the only the $\ce{O}$ from the alcohol remains in the ester while that in the acid forms a water. The formed $\ce{COO}$ is known as the **ester linkage**.
Esters are named with the alcohol as the side group and the acid as its salt variant with a space in between. If the side chain looks like an alkane, its position number and -ane suffix can be dropped.
Amines are $\ce{NR3}$ derived from ammonia ($\ce{NH3}$), where R is either H or a carbon group. Similar to alcohols, they can be primary/secondary/tertiary depending on the number of carbon groups attached. The **main chain** is the longest carbon chain.
Amines have a priority between double/triple bonds and alcohols, and are named like alcohols but with a suffix of **-amine**.
If there are any side groups attached to the nitrogen, they are named as if they were side groups on the main chain with a **number of $N$**.
Amides are formed from a reaction between an amine and a carboxylic acid through dehydration synthesis, similar to the formation of an ester. The $\ce{N-C=O}$ link is known as the **amide link**.
Nitriles consist of a cyanide(s) attached at the end of a carbon chain.
$$\ce{R-C#N}$$
As they can only be placed at the end of a carbon chain, a positional number is not used. These have the highest priority of all organic compounds and use the suffix **-nitrile** and the prefix **cyano-**.
**Hydride reagents** include $\ce{LiAlH4}$ and $\ce{NaBH4}$, the former of which requires ether because it reacts violently with water. Always use $\ce{LiAlH4}$ unless specified otherwise.
**Aldehydes** can be reduced to **primary alcohols**.
$$\ce{aldehyde ->[LiAlH4, ether, then acid] 1^\circ alcohol}$$
**Amides** can be reduced to their **amines**, reacting twice such that the O pops off. The name is a simple `amide.replace("amide", "amine")`.
Retro-synthesis is basically a language of math but for chem, with products on the left and reactants on the right. The bottom right contains initial reactant(s) and the top left contains the product(s).
Polymer properties change based on the type of linkages, the presence of side chains, and the extent of crosslinking between other chains.
The **addition formation** of an **addition polymer** opens up pi bonds which are used to bond to other monomers. Monomers are continuously added until the process ends with hydrogen atoms capping the ends.
**Polymer notation** is the formula/condensed formula/structural diagram of the **repeating unit only** with crossed out brackets and the number of repetitions at the bottom right (or $n$ if unknown). Side groups should be clearly expressed as side groups. Polymers are named with the prefix **poly-** on the repeating unit.
The crosslinking between polymers depends on the side chains. If there are multiple double bonds in monomers, those can be used in different chains which can attach them together.
A **polyester** has monomers connected via an ester linkage on both ends. **Unlike addition polymers**, any carbons between the functional groups are included in the parent polymer chain.
The repeating unit should be **copy-pastable** — it should not end with oxygen on both ends. The link is broken where it would normally break — between the C-O of the ester linkage, such that the O goes to the side of the alcohol.
E/Z isomers are a generalised form of cis-trans isomers, where priority is determined by atomic number. If both sides with the higher atomic number are on the **same** side, the isomer is a Z-isomer (German: _ze zame zide_). E/Z isomers are placed at the beginning surrounded by parentheses.
If the atoms are of equal priority, the sum of atomic numbers that they are directly connected to are compared (double bonds count twice), repeating as necessary.
Optical isomers are mirrored across the y-axis with the same compounds put on the same bonds. **Four distinct groups** must be attached to the central carbon atom to have optical isomers.
In the data booklet, all amino acids are chiral except for glyine and proline.
An **optically active** species is one that can rotate the plane of polarised light. Please see [SL Physics 1#Polarisation](/sph3u7/#polarisation) for more information.
A species that rotates the plane clockwise is positive, while counter-clockwise is negative. Both enantiomers have the same magnitude of polarisation except for the direction. If there is a mixture of both enantiomers, the angle changes depending on the proportion of each isomer.
Enantiomers have the same physical properties except for the direction of polarised light. They also have mostly the same chemical properties except for chemical reactions with other enantiomers of different compounds.
**Alcohols** are able to form hydrogen bonds, so are soluble in water. Increasing the length of the main chain decreases solubility as the rest of the molecule is non-polar, but this can be compensated by adding more hydroxyls too.
Although the boiling point of an alcohol will always be higher than its corresponding alkane, the difference between the two will decrease as chain length increases as the proportion of force the alcohol provides decreases relative to the larger contributor in the LDF from the main chain.
Low mass **esters** smell good, and large mass esters are oily/waxy.
**Amines** smell bad and are all Bronsted-Lowry weak bases because they can accept protons and form dative bonds.
The solubility of compounds is directly related to their melting/boiling point — compounds that cannot hydrogen bond with themselves but can with water have an advantage.
