Chapter 4: Hydrocarbons (C6018015)

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Getting started

Functional groups are specific groups of atoms, which are responsible for notable characteristics, despite being a part of a larger molecule. Hydrocarbons, are compounds consisting of entirely hydrogen and carbon, and include:

• Alkanes, discussed
• Alkenes, discussed
• Alkynes ($R\equiv R'$), which are hydrocarbons containing at least one triple bond
• Aromatic hydrocarbons, discussed

Other notable functional groups that will not be covered in detail, include:

• Dihalide, which is the attachment of two halides. These can be either geminal, which means two functional groups are attached to the same atom; or vicinal, which means two functional groups are attached to adjacent carbon atoms. These are known as geminal dihalide (aka gem dihalide), which are two halides attached to the same carbon; and vicinal dihalide (aka vic dihalide), which are two halides attached to adjacent carbons. This can be memorized with the mnemonic, that a single carbon rich with halogens has “gems”
• Alkoxide ion ($RO^{-}$), which is an alcohol that has donated a proton (i.e. lost one of its hydrogens)
• Nitride ion ($N^{3-}$)
• Vinyl ($-CH=CH_{2}$)
• Nitro ($R-NO_{2}$) and nitroso ($R-NO$). These can be distinguished with the mnemonic, that the functional group with 2 O’s in the name has one O, and vice versa

 Frequently asked questions What are hydrocarbons?Compounds made of hydrogen and carbon. We'll be covering the alkane, alkene, alkyne, and aromatic hydrocarbon in detail.That's all the hydrocarbons there are?No. There are tons.

IUPAC nomenclature are international rules for naming organic compounds.

Arrangements can be classified as:

• Normal, where the carbon connecting the R-group is primary, meaning it is attached to a single carbon. These groups have a prefix “n-“. For example, n-butyl. These groups can be isomers, which will have the prefix “iso”. For example, if a butyl group’s carbon connecting the R-group is still attached to a single carbon, but the carbon connected to it is attached to two carbons; since butyl is no longer a straight chain, it is an isomer, and hence called isobutyl

[img]n-butyl.png[/img]

• Secondary, where the carbon connecting the R-group is attached to two carbons. These groups have a prefix “sec-
• Tertiary, which is where the carbon connecting the R-group is attached to three carbons. These groups have a prefix “tert-

The IUPAC nomenclature names organic compounds as follows:

[img]IUPAC-nomenclature-example.png[/img]

• The base is named after the longest carbon chain. In the example , this is 5 carbons, so pentane
• The carbons can then be numbered, such that the end carbon closest to a carbon with a functional group, is named carbon number one. If there is a tie, the end carbon with the next closest carbon with a functional group is chosen. However, a double or triple bond takes precedence, such that the carbons with the double or triple bond have the lowest numbers possible. Either way, the functional groups are then given the number of the carbon to which it’s attached to. In the example , the first functional group from an end is the methyl group on the 2nd carbon, so is named in this numbering order. The methyl group is attached to the 2nd carbon, and an ethyl group is attached to the 3rd carbon, so it is 3-ethyl-2-methyl-pentane
• If there are identical branches, the prefixes di-, tri-, etc. For example, if there are methyl groups on the 2 and 3 carbons of pentane, this would be 2,3-dimethylpentane
• The functional group names should then be ordered alphabetically, ignoring alpha and/or numeric prefixes

 Frequently asked questions What is the difference between normal, secondary, and tertiary?It describes what the carbon connecting the R-group, is attached to. In normal, it is attached to a single carbon. In secondary, it is attached to 2 carbons. In tertiary, it is attached to 3 carbons.How are organic compounds named?The base is obtained from the longest carbon chain. Carbons are then numbered starting on the end with a functional group closest to it. And functional groups are given the numbers of the carbon its attached to. If there is more than one functional group attached to a carbon, it is prefixed with di-, tri-, etc. All these functional groups are then ordered alphabetically [ignoring prefixes].

The order of acidity of hydrocarbons from weakest to strongest, include alkane, alkene, $\ce{H2}$, $\ce{NH3}$, alkyne, aldehyde, alcohol, water, carboxylic acid.

1 Alkanes

Alkanes ($R-R'$) are saturated hydrocarbons, in that they only contain single bonds. When alkane is a functional group, it is referred to as alkyl. Note that $R$ represents an unspecified generic group, and the dash afterwards is to indicate that it is distinct from the [former] generic group. Alkanes have the suffix “-ane”, alkenes have the suffix “-ene”, and alkynes have the suffix “-yne”. The prefixes indicating the number of carbons present include:

 Prefix Number of carbons Meth- 1 Eth- 2 Prop- 3 Prop- 4 Pent- 5 Hex- 6 Hept- 7 Oct- 8 Non- 9 Dec- 10

For example, the alkane with 5 carbons is pentane.

By increasing the number of carbons, both the boiling point and melting point increases.

“For example, methane $CH_{4}$ found in LPG gas, is a gas at room temperature.”

“In contrast, octane $C_{8}H_{18}$ found in petrol is a liquid at room temperature.”

“Again in contrast, hentriacontane $C_{31}H_{64}$, found in paraffin wax, is a solid at room temperature.”

 Frequently asked questions What are alkanese?Hydrocarbons which are saturated. This means they contain only single bonds.What's the difference between an alkane and alkyl?Alkyl is the "functional group" version of the alkane.Why do you use "R" in formulas?It's like using "x" in maths. It can be replaced with anything.How can you tell alkanes?They have the suffix "-ane".The 10 carbon prefixes. Do I need to know them all?They are relatively straightforward, so, yes.

Branching increases the melting point (solid to liquid), but decreases the boiling point (liquid to gas). This is because branching increases compactness, which means that the molecules are smaller, which therefore weakens the van der Waals forces, thereby lowering boiling point. However, this same compactness permits the solid to pack more easily, and therefore remain as a solid for longer, thus increasing the melting point.

Alkanes have a specific gravity of less than 1 [approximately 0.7], as oil floats on water, and is thus lighter than water. They are nonpolar, and so therefore insoluble/immiscible in water. If a polar group attaches to an alkane, it increases its solubility in water. The longer the alkane group is however, the more it will dominate over the polar group, thereby decreasing its solubility in water.

 Frequently asked questions What is branching?Where there are a lot of functional groups.How does this affect melting and boiling points?Branching increases melting point (solid to liquid). But decreases boiling point (liquid to gas).Why? Especially, why the difference?For boiling point: branching makes the molecules more compact, so the molecules are smaller. Decreased surface area weakens the van der Waals force. This makes it easier to convert from liquid to gas.For melting point: branching permits more compact packing as a solid, so it will want to remain a solid for longer. This increases melting point.What is specific gravity? What does 0.7 mean?Specific gravity is comparison to water. Water has a specific gravity of 1. Less than 1 is lighter than water (i.e. floats on it), and greater than 1 is heavier than water.Water and alkanes don't mix?No. Alkanes are fat-soluble, which don't mix with water. In general, hydrocarbons don't mix with water, unless it has a very large water-dominating functional group.

Cycloalkanes are alkanes that have ringed structures. Each carbon in an alkane can form 4 sigma bonds (single bonds, discussed ), meaning the bonds are $sp^3$ hybridized (meaning there is 4 sigma bonds, and no lone pairs). In these bonds, the predicted angle is the tetrahedral $109.5^{\circ}$. However, as the ring forces the angle to become smaller or larger, and even eclipse the bonds, this creates ring strain. However, cyclohexane $C_{6}H_{12}$ has no ring strain, because it takes on a 3D structure that can reduce ring strain.

Cyclohexane exists as several conformers, which are isomers with the same molecules, bonds, but have different orientation. For cyclohexane, these include:

[img]cyclohexane-conformers.png[/img]

• Chair, which exists at a lower energy than the boat, because [on the chair] the sides are bent towards the opposite sides of the ring. Even though the molecule exists as the different conformers, it prefers, and so therefore predominates as the lower energy chair conformation
• Boat, which exists at a higher energy than the chair, because [on the boat] the sides are bent towards the same side of the ring. Because the sides repel each other, there is higher energy in the boat
• Half-chair
• Twist, which is at a slightly lower energy than the boat

In the chair conformation, half of the carbon bonds are (1) equatorial, pointing out away from the ring; and the other half are (2) axial, pointing up or down, parallel to the axis of the ring. For carbon-hydrogen bonds, neither the equatorial nor axial are favored. For larger functional groups attached to a ring, equatorial has less steric hindrance (interference of large groups with another, discussed ), and is therefore favored.

Ring strain can be tested by the bomb calorimeter, which measures heat of combustion, which should be elevated due to high strain. This is because strain contributes to strain energy, per additional $CH_{2}$ unit, which is released upon combustion.

 Frequently asked questions What are cycloalkanes?Alkanes that have a ringed structure.Cycloalkanes have ring strain? What does that mean?The ring forces the angle to be indifferent (smaller or larger) from the predicted tetrahedral angle.How about cyclohexane?Unusually, it has no ring strain. This is because it takes on a 3D structure that can reduce ring strain.Cyclohexane exists as several conformers. What is a conformer?Conformers are isomers with the same molecules and bonds, but have different orientation.What are these conformers?The most important are the chair, and boat. Chair is lower energy, and boat is higher energy. This is because the sides of the boat are on the same side, so repel each other, so is higher energy.Which is the more popular conformer for cyclohexane to exist in?Evidently, the lower energy state. So the chair.

Alkanes are not very reactive [with ionic and other polar substances], but if energy is applied, react violently. Reactions include:

• Combustion is the reaction with oxygen, of which the activation energy is high. Once the reaction begins however, it is sufficiently exothermic (energy releasing) to be self-perpetuating. Combustion is $\ce{C_{n}H_{2n+2} + O_{2} -> H_{2}O + CO_{2}}$. Where there is incomplete combustion, free radicals are emitted
• (Free radical) halogenation is reaction with the halogens $F_{2}$, $Cl_{2}$, or $Br_{2}$. Fluorine is the most reactive halogen, followed by chlorine, then bromine. As a radical, bromine is the most selective, followed by chlorine and fluorine, which is the least selective. Selective means the preference to add/react with tertiary, whilst non-selective means ability to react with primary, secondary or tertiary. This is because the more selective substances [like bromine] have a more stable radical, it is less desperate to react, and so is more selective to choose the most favorable pathway. [Trending down the periodic table further] iodine $I_{2}$ does not react with alkanes. Halogenation is a chain reaction. This involves:
• Initiation, which is the formation of free radicals. In the presence of heat or UV light, the halogen is homolytically cleaved, meaning each atom receives one of the electrons [involved in the original bond]. As a result, there are two [free] radicals, which are atoms with an unpaired electron. [Note that radicals are not necessarily charged, and in fact, are usually electrically neutral.] However, they are highly unstable, and thus reactive, and therefore only exist in very low concentrations. For example, $\ce{Cl_{2} -> Cl\bullet + Cl\bullet}$
• Chain propagation, where the radical strips a hydrogen atom off an alkane, forming a hydrogen halide, and an alkyl radical. For example, $\ce{CH_{4}+Cl\bullet -> CH_{3}\bullet + HCl}$. Then, the alkyl radical ($CH_{3}\bullet$) can strip a halogen atom from a halogen molecule, forming another halogen radical. For example, $\ce{CH_{3}\bullet + Cl_{2} -> CH_{3}Cl + Cl\bullet}$. Alkyl radicals have analogous reactivity to carbocations (which rather than being radicals, have a positively charged carbon atom). Methyl alkyl radicals are the most reactive, followed by primary, secondary, then tertiary, so the most stable, tertiary, is most likely to form [statistically]. The net equation is thus $\ce{CH_{4}+Cl_{2} -> HCl + CH_{3}Cl}$. Note therefore, that the halogen radical $Cl\bullet$ is a catalyst, as it is neither a reactant nor product in the net reaction
• Chain termination, where the radicals are removed, by stabilization. Two free radicals can recombine, by pairing their lone electron

Because of the various products that halogenation creates, there is not a single product, but a statistical distribution of various products. Halogenation is an exothermic (energy releasing) reaction.

 Frequently asked questions Do alkanes react?They're not really reactive, unless energy is applied [where they will react violently].How do alkanes react?2 ways: combustion and halogenation.What is combustion?Reacting an alkane with oxygen. Energy must be applied, which is why combustion occurs at high temperature [as in burning of fossil fuels].What is halogenation?In the presence of high energy, a halogen is cleaved into 2 free radicals. The free radical then strips a hydrogen off an alkane, creating an alkyl radical. The radicals can then recombine, creating a stable molecule.Is there any one stable molecule that is produced from halogenation?No. It is a distribution of various products.

2 Alkenes

Alkenes ($R=R'$) are hydrocarbons containing at least one double bond. This double bond contains one [first] sigma bond, and one [subsequent] pi bond (discussed ). Remember from  that the pi bond occupies the space above and below the sigma bond, which is a large amount of space for the electron to cover. In order to counteract this, and therefore stabilize the bond, the carbon pulls negative charge from other bonds on the carbons. Thus, alkene is electron withdrawing [as contrasted with electron donating]. This explains why the most substituted alkene (covered ), is the most stable; because there is more opportunity to withdraw electrons.

Alkenes demonstrate analogous physical properties to alkanes, namely, that increasing molecular mass increases boiling point and density. Like alkanes, alkenes are less dense than water. Alkenes are insoluble in water, but slightly more soluble than alkanes. They are most soluble in nonpolar solvents.

 Frequently asked questions What is an alkene?Hydrogens containing at least 1 double bond.Why is the alkene, electron withdrawing?The double bond contains 1 pi bond. Because the sigma bond occupies the space between the 2 atoms, the pi bond is placed above and below this [sigma] bond. So to stabilize this bond, the carbon pulls negative charge from other bonds on the carbon. It wants negative charge.In terms of physical properties, how are alkenes distinct from alkenes?They're very similar. They are slightly more soluble in water than alkanes.

Alkenes are synthesized by elimination reaction, where a functional group is removed, usually to form a double bond (discussed ). This can occur either through the one-step E2 reaction, or the two-step E1 reaction. An example is:

• The E2 dehydrohalogenation, where a strong bulky base is added to an alkyl halide. The base is too bulky to act as a nucleophile (i.e. donate electron pair to an electrophile, and since it is an electron pair donator, is by definition a Lewis base, discussed ), but still, by the Bronsted-Lowry [base] definition, wants to accept a proton. Therefore, it rips a proton, and therefore a halogen, from the alkyl halide; thereby creating an alkene

[img]E2-dehydrohalogenation.png[/img]

• The E1 dehydration of alcohols, where a hot, concentrated acid is added to an alcohol, causing the alcohol to lose its hydroxyl group to become an alkene [water as a waste product], taking two steps. The acid is neither a reactant nor a product, so is a catalyst. During these [two] steps, the carbon skeleton rearranges, to form a more stable carbocation. Therefore, the statistically most common product, is the most stable product, which is the most substituted alkene [because of increased opportunity to withdraw electrons, from ], known as Zaitsev’s rule. Therefore, from the most stable to the least, is tertiary, secondary, primary, then methyl, which is analogous ordering to the alkyl groups (from )

[img]E1-dehydration-of-alcohols.png[/img]

 Frequently asked questions How are alkenes created?Using an elimination reaction (either E2 or E1), removing a functional group, to form a double bond.What's the difference between E2 and E1?E2 has 1-step, E1 has 2-steps.E2 involves adding a [bulky] base to an alkyl halide, which per a base (i.e. proton accepter), rips a proton [in the form of the alkyl], creating a double bond [i.e. alkene].E1 involves adding acid to an alcohol, with water coming out as a waste product.What is Zaitsev's rule?The most common product is the most stable product. This is the alkene most substituted [with functional groups].

There are also reactions where alkene is a reactant, including:

• Catalytic hydrogenation, where two hydrogens are added, one to each carbon in the double bond [of an alkene], via a syn addition. Syn addition is addition of the substituent [hydrogen] on the same side of the double bond. If the double bonded carbons are attached to two different functional groups, it can either produce threo (identical substituent on opposite sides in Fischer projection, discussed ) or erythro (identical substituent on same side in Fischer projection, discussed ) diastereomer (not a mirror image, discussed ) products. Catalytic hydrogenation uses a heterogeneous catalyst, which is a catalyst in a different phase from that of the reactants (cf. homogeneous catalyst, which is in the same phase from that of the reactants). Hydrogenation is an exothermic (energy releasing) reaction. Hydrogenation occurs analogously with the carbon-carbon triple bonded alkynes, reducing the number of triple bonds [to make a double bonded alkene]

[img]alkene-catalytic-hydrogenation.jpg[/img]

• Oxidation, such as in the reaction with high-energy ozone ($O_{3}$). Ozonolysis is where an alkene has its carbon-carbon double bond cleaved by ozone, forming two separate molecules, with carbonyl $C=O$ groups

[img]alkene-oxidation-with-ozone.png[/img]

Alkene can also be oxidized with other oxidizing agents, such as hot permanganate ions ($MnO_{4}^{-}$) [in acidic solution] can also cleave an alkene in two, adding a hydroxyl group to each carbon in the double bond. However, if an aldehyde ($R-CHO$ , discussed ) is one such product [of this split], permanganate will again cleave the double bond present, further oxidizing aldehyde to a carboxylic acid ($R-COOH$). It is important the permanganate ions are “hot”, because in the absence of heat, permanganate reacted with a base forms glycols, in a hydroxylation reaction. Glycols are a type of diol, which is a compound with two hydroxyl groups ($-OH$); but is distinct from other diols, because they are attached to adjacent carbons

[img]alkene-oxidation-with-permanganate-ions.jpg[/img]

• Electrophilic addition, where a double bond is replaced with a single bond. Note that an electrophile is electron loving, and so therefore positively, or partially positively [due to dipole] charged. Remembering from  that the alkene has a pi bond that is a large cloud of negative charge, electrophiles are attracted to this [electron cloud]. Electrophiles usually add by Markovnikov’s rule, which states that electrophiles (typically hydrogen atoms) will add to the carbon [of the carbon-carbon double bond] with the least alkyl substituents, which forms the most stable carbocation, remembering from  that the most substituted alkene is the most stable. For example, if hydrobromic acid ($HBr$) is added to an alkene, $\ce{H+}$ ions will act as an electrophile, attacking the carbon-carbon double bond, adding to the least substituted carbon, to form the most stable carbocation

Remember that whenever carbocations are formed, carbon skeleton rearrangement may occur, to form an even more stable carbocation. The alkyl shift [of carbon skeleton rearrangement] only occurs if the shifted alkyl group is adjacent to the positively charged carbon [of the carbocation]

In the presence of peroxides, $HBr$ can add anti-Markovnikov, where hydrogens add to the most substituted [rather than the least substituted] carbon. This occurs because peroxides generate free radicals, and so $HBr$ produces a $Br$ radical, which reacts with the double bond [before the hydrogen], on the least substituted carbon [for the same stability reasons hydrogen wanted to act there]. Therefore, hydrogen is left to add on the more substituted carbon. Note anti-Markovnikov only occurs with $HBr$, and not the other hydrogen halides

Other examples of electrophilic addition include:

• Hydration, which is the addition of water to an alkene. It is essentially the reverse process to dehydration of alcohol [used to form alkenes, discussed ]. Given dehydration of alcohol occurs in hot, concentrated acid; this means hydration of an alkene involves cold, dilute acid. The reverse process therefore [similarly] includes, the [intermediate] carbocation forming [and hence rearrangement may occur], adding of the hydroxyl group, thereby forming alcohol. This is Markovnikov addition

[img]alkene-hydration.png[/img]

• Oxymercuration[-demercuration], is a particular type of alkene hydration [forming alcohol], but doesn’t permit carbon skeleton rearrangement. Alkene is reacted with mercury ions, to attack the alkene double bond. Water then attacks on the other side of the double bond, in an anti addition, [which is the opposite of syn addition] meaning the substituent is added on the opposite side of the double bond

[img]alkene-oxymercuration-with-water.png[/img]

Instead of water, oxymercuration can also work [by reacting alkene] with alcohol, which like water, adds anti addition, adding an $-OR$ group instead of an $-OH$ group (as in water). Therefore, rather than forming an alcohol ($R-OH$), an ether ($R-OR'$) is formed

• Hydroboration, is a particular type of alkene hydration [forming alcohol], that is anti-Markovnikov. Borane $\ce{BH3}$ is the catalyst in the first reaction, and [hydrogen] peroxide $\ce{H2O2}$ is the catalyst in the second reaction. Remember that peroxides were implicated in $HBr$ for adding anti-Markovnikov too

[img]alkene-hydroboration.png[/img]

• [Electrophilic] halogenation, which is distinct from the free halogen halogenation of alkanes, because alkenes are more reactive than alkanes. This is so because the electrons of the [subsequent] pi bond are in a higher energy than the [initial] sigma bond [meaning double bonds are more reactive, as stated ]. As a result, alkanes required the generation of free radicals in an initiation step, to promote reaction with halogens (discussed ). Free radicals are not required in alkene halogenation, because of their greater reactivity. Alkene halogenation occurs anti addition, meaning one halogen is added on the opposite side of the double bond (discussed )

[img]alkene-halogenation.png[/img]

 Frequently asked questions Other than hydrobromic acid, can you use other electrophiles [as apart of electrophilic addition with alkenes]?Yes, you can add:Water (known as hydration; or where borane is used as a catalyst, hydroboration)Mercury ions (known as oxymercuration)Halogens (known as halogenation).
3 Aromatic hydrocarbons

Aromatic hydrocarbons (aka arene) are ringed structures of carbon atoms. An example benzene $\ce{C6H6}$, which is a flat, 6 carbon ring, with 1 hydrogen attached to each of the 6 carbons. It is often depicted either as a hexagon with a circle in the middle, or hexagon with three double bonds, though note benzene doesn’t actually have any double bonds:

[img]benzene.png[/img]

Instead, benzene has 6 partial [double] bonds, that are shorter than a [longer, thus weaker] single bond, but longer than a [shorter, thus stronger] double bond. This double bond character is represented by a dotted line. This is so because the electrons of the [subsequent] pi bonds are delocalized around the ring. Although pi bonds usually have higher energy [associated with lower stability] than a sigma bond [as they are further away from the nucleus than a sigma bond, discussed ], delocalization of these pi bond electrons permits lower orbital energies, thereby increasing stability.

 Frequently asked questions What are aromatic hydrocarbons?It's a ringed hydrocarbon. An example is benzene.Benzene's the hexagon with a circle in the middle?Yeah. It can also be drawn as a hexagon with 3 double bonds.So benzene has 3 double bonds?No, it actually doesn't have any double bonds. It has 6 partial double bonds, which are placed between the [lower energy] single and [higher energy] double bond [in terms of energy].Why is the energy of benzene's bond placed between a single and double bond?Benzene's pi bonds are delocalized. So whereas sigma is usually low energy, and pi is usually high energy, by delocalizing, there is increased stability [so lower energy].

This explains why benzene undergoes electrophilic [aromatic] substitution (discussed ), but not addition (replace a double bond with a single bond); as this would destroy its resonance.

Arene substitution patterns are the nomenclature by which substituents are named relative to another substituent, on an aromatic hydrocarbon. There are five such positions, which can be notated by naming the “R” position 1, and increasing on either side, and so forth, including:

• Ortho, found in positions 1 and 2, positioned next to each other
• Meta, found in positions 1 and 3
• Para, found in positions 1 and 4, positioned at opposite ends

[img]arene-substitution-patterns.png[/img]

This can be memorized with the mnemonic, that benzene likes to “ROMP”, which includes the R-group, ortho, meta and then para.

If the R substituent is electron withdrawing, it deactivates the ring, known as deactivating; and directs the next substituent to add at the meta position, known as meta directing. If the substituent is electron donating, it activates the ring, known as activating; and directs the next substituent to add at the ortho or para position, known as ortho-para directing.

This can be memorized with the mnemonic, that Uncle Ben (for benzene) was a couch potato. If he was given money (electron donating), Uncle Ben would start partying (become activated), and would want this person to sit either next to him (ortho) or opposite him (para) at dinner. For those who did not give Uncle Ben money (electron withdrawing), he would not be very happy about it (become deactivated), and would not want them to sit next to or opposite them (meta).

The exceptions are halogens, which are electron withdrawing, but ortho-para directing.

The mnemonic can then be modified, by Uncle Hal, who wouldn’t give Uncle Ben any money, but he liked Uncle Hal so he made an exception, always letting Uncle Hal sit next to or across from him.

 Frequently asked questions Why can't benzene undergo electrophilic addition?Removing one of its partial double bonds, would destroy what it means to be a benzene. It'd destroy its resonance.But benzene can undergo electrophilic substitution?Yes. And to notate this, we start by naming the "R" position as 1, and on either side, assigned the 2nd position as ortho, 3rd position as meta, and 4th position (i.e. opposite) as para.

Some rules for classification into electron donating and withdrawing groups:

• Reference is made to hydrogen, which is thus considered neither electron donating nor withdrawing)
• Groups that consistently donate electrons:
• Alkyl groups (based off alkane $\ce{CH3}$)
• Groups that consistently withdraw electrons:
• Halogens ($F, Cl$)
• Carbonyls ($CO$)
• Nitriles ($C\equiv N$)
• Groups that can donate or withdraw electrons:
• Nitrogens (with 3 bonds) donate electrons, but nitrogens (with 4 bonds) have a positive bond so therefore [by nature of charge] withdraw electrons
• Oxygens donate electrons through single bonds, and withdraw through double bonds
• Benzene is electron donating when attached to another benzene. In almost all other circumstances, benzene is electron withdrawing

Nucleophilic substitution can be either:

• $S_{N}1$, which like elimination E1, occurs in 2 steps. The number “1” represents the order of the rate law, namely, 1st order, meaning there is 1 reactant involved in the rate-determining step of the reaction. It takes place in 2 steps:

[img]nucleophilic-substitution-SN1.png[/img]

• Slow step, which is thus the rate-determining step. A tertiary alkyl halide is placed in the presence of a base ($\ce{OH-}$). Usually, if the base was strong enough, it rips a proton [and therefore a halogen] from the alkyl halide, thereby creating an alkene (discussed ), per dehydrohalogenation, and as it relates to a tertiary alkyl halide, in an E1 dehydrohalogenation. In fact, $S_{N}1$ and E1 often occur together to form mixed products. This however, does not occur as the base is not strong enough. Rather, a halide ion separates from the carbon backbone [now a carbocation]

When the carbocation is formed, the nucleophile (which donates an electron pair, discussed )  [namely, the halide] can add to either side, meaning if the alkyl halide is chiral (has handedness), it would generate a racemic mixture (mixture of enantiomers, which are mirror images that are not superimposable). Remember that since a carbocation is formed, carbon skeletal rearrangement is possible. Because the slow step is the rate determining step, the rate is solely dependent on the concentration of the reactant, which is alkyl halide
• Fast step [and therefore known as an “attack”], where the nucleophile [namely, the halide] reacts with the carbocation. Because the fast step requires a stable carbocation, it is uncommon that $S_{N}1$ occurs with secondary alkyl halides, and not at all with primary alkyl halides [because their carbocations are too unstable]. Polar solvents promote $S_{N}1$, as they help to stabilize the carbocation
• $S_{N}2$, which like elimination E2, occurs in 1 step. The number “2” [analogous to reasoning ] represents 2nd order, meaning there are 2 reactants involved in the rate-determining step of the reaction. The alkyl halide lacks the bulky alkyl [found in $S_{N}1$], so is therefore usually a primary or secondary alkyl halide, now permitting the nucleophile, a strong base, to attach to the alkyl halide, and strip off a halide ion in the same step. Even though a strong base is desired, if the base is too strong, the base is sterically hindered, and an E2 elimination (i.e. adding bulky base to alkyl halide, from ) may occur. In $S_{N}2$, there is Walden inversion, which is the inversion of a chiral center in a molecule, thereby converting from one enantiomer to another. This can be visualized as an umbrella turning inside-out in a strong wind

[img]nucleophilic-substitution-SN2.png[/img]

Because there is only 1 step, the rate of the reaction depends on both the concentration of the alkyl halide AND nucleophile. Unlike $S_{N}1$, polar solvents inhibit $S_{N}2$ since they stabilize the nucleophile. Note that as no carbocation is formed, there is no carbon skeleton rearrangement

Despite nucleophiles are by definition, Lewis bases, as they donate an electron pair, they are distinct, because whereas a base accepts a proton, nucleophiles bond to a carbon. Nucleophilicity is promoted by:

• High pH (more basic), as what makes a strong base also makes a strong nucleophile
• A less bulky base, as a bulkier base will experience more steric hindrance (interfere with one another, discussed ), and therefore slow the rate of reaction
• Polarizability, and thus negative charge, enhances the ability of the nucleophile to form bonds with carbon
• [Reduced] electronegativity, as by definition, it is how tightly held an electron pair is. Therefore, the ability to donate that pair is decreased. Reduced electronegativity therefore promotes nucleophilicity

# Assessment e-submission

(Formative assessments are not assessed for marks. Assessments are made on the unit level.

# (MED5118352)

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