MECH MONSTER 6 Flashcards

Question

which is a stronger nucleophile R-O^- R-OH

Answer 1

**S_N2 halogen exchange reaction** Alkyl fluorides are difficult to synthesize directly, and they are often made by treating alkyl chlorides or bromides with KF under conditions that use a crown ether to dissolve the fluoride salt in an aprotic solvent, which enhances the normally weak nucleophilicity of the fluoride ion.

Answer 2

**S_N2 halogen exchange reaction** Iodide is a good nucleophile, and many alkyl chlorides react with sodium iodide to give alkyl iodides.

Answer 3

**S_N2 halogen exchange reaction** Alkyl fluorides are difficult to synthesize directly, and they are often made by treating alkyl chlorides or bromides with KF under conditions that use a crown ether to dissolve the fluoride salt in an aprotic solvent, which enhances the normally weak nucleophilicity of the fluoride ion.

Answer 4

**S_N2 halogen exchange reaction** Iodide is a good nucleophile, and many alkyl chlorides react with sodium iodide to give alkyl iodides.

Answer 5

Nucleophilicity increases down the periodic table, following the increase in size and polarizability, and the decrease in electronegativity. * stronger* **I^- ** \> **Br^- ** \> **Cl^-** \> **F^-** *weaker* * stronger* **^-SeH** \> **^-SH** \> **^-OH** *weaker* * stronger* **(CH₃CH₂)₃P** \> ** (CH₃CH₂)₃N** *weaker*

Answer 6

**Alkoxide ion** is the strongest nucelophile (similar to hydroxide ion), a species with a negative charge is a stronger nucleophile than a similar neutral species. In particular, a base is a stronger nucleophile than its conjugate acid. **Carboxylate ion** is a moderate nucleophile, as carboxylic acids easily dissociate into a carboxylate anion and a positively charged hydrogen ion (proton), much more readily than alcohols do (into an alkoxide ion and a proton), because the carboxylate ion is stabilized by resonance. The negative charge that is left after deprotonation of the carboxyl group is delocalized between the two electronegativeoxygen atoms in a resonance structure. **Alcohol** (similar to water) is the weakest nucleophile of the group, as the conjugate acid of alkoxide.

Answer 7

**Basicity** is defined by the *equilibrium constant* for abstracting a proton. **Nucleophilicity** is defined by the *rate* of attack on an electrophilic carbon atom. In both cases, the nucleophile (or base) forms a new bond. If the new bond is to a pro- ton, it has reacted as a **base**; if the new bond is to carbon, it has reacted as a **nucleophile**.

Answer 8

Trends in size and polarisability reflect an atom’s ability to engage in partial bonding as it begins to attack an electrophilic carbon atom. As we go down a column in the periodic table, the atoms become larger, with more electrons at a greater distance from the nucleus. The electrons are more loosely held, and the atom is more polarisable: Its electrons can move more freely toward a positive charge, resulting in stronger bonding in the transition state. The increased mobility of its electrons enhances the atom’s ability to begin to form a bond at a relatively long distance.

Answer 9

Fluoride has tightly bound electrons that cannot begin to form a C—F bond until the atoms are close together. Iodide has more loosely bound outer electrons that begin bonding earlier in the reaction.

Answer 10

when bulky groups interfere with a reaction by virtue of their size

Answer 11

A **protic solvent** is one that has acidic protons, usually in the form of **O - H** or **N - H** groups.

Answer 12

Small anions are **solvated** _more strongly_ than large anions in a **protic solvent** because the solvent approaches a small anion more closely and _forms stronger hydrogen bonds_. When an anion reacts as a nucleophile, energy is required to “strip off” some of the solvent molecules, breaking some of the hydrogen bonds that stabilized the solvated anion. More energy is required to strip off solvent from a small, strongly solvated ion such as fluoride than from a large, diffuse, less strongly solvated ion like iodide.

Answer 13

solvents without **O - H** or **N - H** groups, most polar, ionic reagents are *insoluble* in simple aprotic solvents such as alkanes

Answer 14

In contrast with protic solvents, **aprotic solvents** (solvents *without* O - H or N - H groups) _enhance the nucleophilicity of anions_. An anion is *more* reactive in an aprotic solvent because it is not so strongly solvated. There are no hydrogen bonds to be broken when solvent must make way for the nucleophile to approach an electrophilic carbon atom.

Answer 15

**Polar aprotic solvents** have strong dipole moments to enhance solubility, yet they have no **O - H** or **N - H** groups to form hydrogen bonds with anions. Examples of useful polar aprotic solvents are acetonitrile, dimethylformamide, and acetone.

Answer 16

Fluoride ion, normally a poor nucleophile in hydroxylic (protic) solvents, can be a good nucleophile in an aprotic solvent. Although KF is not very soluble in acetonitrile, 18-crown-6 solvates the potassium ions, and the poorly solvated (and therefore nucleophilic) fluoride ion follows.

Answer 17

**Walden inversion** an asymmetric carbon atom will undergo inversion of configuration

Answer 18

A S_N1 reaction that takes place with the solvent acting as the nucleophile, its rate does not depend on the concentration of the nucleophile (usually methanol). The rate depends only on the concentration of the substrate.

Answer 19

formation of carbocation (rate limiting)

Answer 20

nucleophilic attack on the carbocation

Answer 21

loss of proton to solvent (only if the nucleophile is an uncharged molecule like water or an alcohol, the positively charged product must lose a proton to give the final uncharged product)

Answer 22

**Vinyl halides** and **aryl halides** generally do not undergo S_N1 or S_N2 reactions. An S_N1 reaction would require ionization to form a vinyl or aryl cation, either of which is less stable than most alkyl carbocations. An S_N2 reaction would require back-side attack by the nucleophile, which is made impossible by the repulsion of the electrons in the double bond or aromatic ring.

Answer 23

Ionization of an alkyl halide requires formation and separation of positive and negative charges, similar to what happens when sodium chloride dissolves in water. Therefore, S_N1 reactions require highly polar solvents that strongly solvate ions. Note that ionisation occurs much faster in highly polar solvents such as water and alcohols. Although most alkyl halides are *not* soluble in water, they often dissolve in highly polar mixtures of acetone and alcohols with water.

Answer 24

The S_N1 reaction is *not* stereospecific. In the S_N1 mechanism, the carbocation intermediate is sp2 hybridized and planar. A nucleophile can attack the carbocation from either face, because the carbocation is planar and achiral; and attack from both faces gives both enantiomers of the product. Such a process, giving both enantiomers of the product (whether or not the two enantiomers are produced in equal amounts), is called **racemization**. The product is either racemic or at least less optically pure than the starting material.

Answer 25

Primary halides and methyl halides rarely ionize to carbocations in solution. If a primary halide ionizes, it will likely ionize with **rearrangement**. The **methyl shift** occurs while bromide ion is leaving, so that only the more stable tertiary carbocation is formed.

Answer 26

**S_N1: Nucleophile strength is unimportant (usually weak).** **S_N2: Strong nucleophiles are required.** The nucleophile takes part in the slow step (the only step) of the SN2 reaction but not in the slow step of the SN1. Therefore, a strong nucleophile promotes the SN2 but not the SN1. Weak nucleophiles fail to promote the SN2 reaction; therefore, reactions with weak nucleophiles often go by the SN1 mechanism if the sub- strate is secondary or tertiary.

Answer 27

**S_N1 substrates: 3° \> 2° (1° and CH₃X are unlikely)** **S_N2 substrates: CH₃X \> 1° \> 2° (3° unstable) ** Most methyl halides and primary halides are poor substrates for S_N1 substitutions be- cause they cannot easily ionize to high-energy methyl and primary carbocations. They are relatively unhindered, however, so they make good S_N2 substrates. Tertiary halides are too hindered to undergo S_N2 displacement, but they can ionize to form tertiary carbocations. Tertiary halides undergo substitution exclusively through the S_N1 mechanism. Secondary halides can undergo substitution by either mechanism, depending on the conditions.

Answer 28

**S_N1: Good ionizing solvent required.** **S_N2: May go faster in a less polar solvent ** The slow step of the S_N1 reaction involves formation of two ions. Solvation of these ions is crucial to stabilizing them and lowering the activation energy for their formation. Very polar ionizing solvents such as water and alcohols are needed for the S_N1. The solvent may be heated to reflux (boiling) to provide the energy needed for ionization. Less charge separation is generated in the transition state of the S_N2 reaction. Strong solvation may weaken the strength of the nucleophile because of the energy needed to strip off the solvent molecules. Thus, the S_N2 reaction often goes faster in less polar solvents if the nucleophile will dissolve. Polar aprotic solvents may enhance the strength of weak nucleophiles.

Answer 29

**S_N1 rate = *k*_r[R-X]** **S_N2 rate = *k*_r[R-X][Nuc:^-] ** The rate of the S_N1 reaction is proportional to the concentration of the alkyl halide but not the concentration of the nucleophile. It follows a first-order rate equation. The rate of the S_N2 reaction is proportional to the concentrations of both the alkyl halide [R - X] and the nucleophile [Nuc:^-]. It follows a second-order rate equation.

Answer 30

**S_N1 stereochemistry: Mixture of retention and inversion; racemization.** **S_N2 stereochemistry: Complete inversion.** The S_N1 reaction involves a flat carbocation intermediate that can be attacked from either face. Therefore, the S_N1 usually gives a mixture of inversion and retention of configuration. The S_N2 reaction takes place through a back-side attack, which inverts the stereochemistry of the carbon atom. Complete inversion of configuration is the result.

Answer 31

**S_N1: Rearrangements are common.** **S_N2: Rearrangements are impossible.** The S_N1 reaction involves a carbocation intermediate. This intermediate can rearrange, usually by a hydride shift or an alkyl shift, to give a more stable carbocation. The S_N2 reaction takes place in one concerted step with no intermediates. No rearrangement is possible in the S_N2 reaction.

Answer 32

true, it is less electronegative

Answer 33

**nucleophile** is a species that donates a pair of electrons to form a new covalent bond

Answer 34

S_N1: a reluctant first-order substrate can be forced to ionize by adding some silver nitrate (one of the few soluble silver salts) to the reaction. Silver ion reacts with the halogen to form a silver halide (a highly exothermic reaction), generating the cation of the alkyl group.

Answer 35

An **elimination** involves the loss of two atoms or groups from the substrate, usually with formation of a π bond. Elimination reactions frequently accompany and compete with substitutions. By varying the reagents and conditions, we can often modify a reaction to favor substitution or to favor elimination.

Answer 36

The rate-limiting transition state involves a single molecule rather than a collision between two molecules. The slow step of an E1 reaction is the same as in the S_N1 reaction: *unimolecular ionization* to form a **carbocation**. In a fast second step, a base abstracts a proton from the carbon atom adjacent to the C⁺. The electrons that once formed the carbon–hydrogen bond now form a π bond between two carbon atoms.

Answer 37

*Unimolecular ionization* to give a **carbocation** (rate-limiting)

Answer 38

*Deprotonation* by a **weak base** (often the solvent) gives the alkene (fast).

Answer 39

The E1 reaction almost always competes with the S_N1 reaction. Whenever a carbocation is formed, it can undergo either substitution or elimination, and mixtures of products often result.

Answer 40

an **elimination** of hydrogen and a halogen atom

Answer 41

Ionization of the alkyl halide gives a carbocation intermediate, which loses a proton to give the alkene. Ethanol serves as a *base* in the **elimination**.

Answer 42

S_N1 results from nucleophilic attack on the carbocation. Ethanol serves as a *nucleophile* in the **substitution**.

Answer 43

Solvent (a weak nucleophile) attacks the rearranged (in a **hydride shift**, not shown) carbocation, **deprotonation** (not shown) gives the rearranged product.

Answer 44

The weakly basic solvent removes either adjacent proton on the rearranged (in a **hydride shift**, not shown) carbocation (fast).

Answer 45

1. React with its own leaving group to return to the **reactant**: R⁺ + :X^- -\> R - X 2. React with a nucleophile to form a substitution product (**S_N1**): R⁺ + Nuc:^- -\> R - Nuc 3. Lose a proton to form an elimination product (an alkene) (**E1**) (shown) 4. **Rearrange** to a more stable carbocation, then react further. The order of stability of carbocations is: resonance-stabilised, 3° \> 2° \> 1°.

Answer 46

In elimination reactions, the most substituted alkene usually predominates.

Answer 47

E1 and E2 elimination reactions that produce the most substituted alkene

Answer 48

In the general mechanism of the **E2** reaction, a strong *base* abstracts a proton on a carbon atom _adjacent_ to the one with the leaving group. As the *base* abstracts a proton, a _double bond forms_ and the leaving group leaves. Like the **S_N2** reaction, the **E2** is a **concerted** reaction in which bonds break and new bonds form at the same time, in a single step. (shown, methoxide reacts as a *base* rather than as a *nucleophile*).

Answer 49

The order of reactivity of alkyl halides toward **E2** **dehydrohalogenation** is found to be **3°** \> **2°** \> **1°**

Answer 50

The _concerted_ **E2** reaction takes place in a single step. A strong *base* abstracts a proton on a carbon _next to the leaving group_, and the leaving group leaves. The product is an *alkene*.

Answer 51

The **E2** is a **stereospecific** reaction, because different stereoisomers of the starting material react to give different stereoisomers of the product. This stereospecificity results from the _anti-coplanar transition state_ that is usually involved in the **E2**.

Answer 52

The **E2** is a **stereospecific** reaction, because different stereoisomers of the starting material react to give different stereoisomers of the product. This stereospecificity results from the _anti-coplanar transition state_ that is usually involved in the E2.

Answer 53

**E1: Base strength is unimportant (usually weak).** **E2: Requires strong bases.** The nature of the *base* is the single most important factor in determining whether an elimination will go by the E1 or E2 mechanism. If a _strong_ base is present, the rate of the bimolecular reaction will be _greater_ than the rate of ionization, and the **E2** reaction will predominate (perhaps accompanied by the **S_N2**). If no strong base is present, then a good solvent makes a unimolecular ionization likely. Subsequent loss of a proton to a _weak_ base (such as the solvent) leads to **elimination**. Under these conditions, the **E1** reaction usually predominates, usually accompanied by the **S_N1**.

Answer 54

**E1: Requires a good ionizing solvent.** **E2: Solvent polarity is not so important. ** The slow step of the E1 reaction is the formation of two ions. Like the S_N1, the **E1** reaction critically depends on _polar ionizing solvents_ such as water and the alcohols. In the **E2** reaction, the transition state spreads out the negative charge of the base over the entire molecule. There is no more need for solvation in the E2 transition state than in the reactants. The **E2** is therefore _less sensitive to the solvent_; in fact, some reagents are _stronger bases in less polar_ solvents.

Answer 55

**E1, E2: 3° \> 2° \> 1° (1° usually will not go E1) ** In the E1 reaction, the rate-limiting step is formation of a carbocation, and the reactivity order reflects the stability of carbocations. In the E2 reaction, the more substituted halides generally form more substituted, more stable alkenes.

Answer 56

**E1 rate = *k*_r[RX] E2 rate = *k*_r[RX][B:^-] ** The rate of the E1 reaction is proportional to the concentration of the alkyl halide [RX] but not to the concentration of the base. It follows a first-order rate equation. The rate of the E2 reaction is proportional to the concentrations of both the alkyl halide [RX] and the base [B:^-]. It follows a second-order rate equation.

Answer 57

**E1, E2: Usually Zaitsev orientation.** In most E1 and E2 eliminations with two or more possible products, the product with the most substituted double bond (the most stable product) predominates. This principle is called Zaitsev’s rule, and the most highly substituted product is called the Zaitsev product.

Answer 58

**E1: No particular geometry required for the slow step.** **E2: Coplanar arrangement (usually *anti*) required for the transition state.** The E1 reaction begins with an ionization to give a flat carbocation. No particular geometry is required for ionization. The E2 reaction takes place through a concerted mechanism that requires a coplanar arrangement of the bonds to the atoms being eliminated. The transition state is usually anti-coplanar, although it may be syn-coplanar in rigid systems.

Answer 59

**E1: Rearrangements are common.** **E2: No rearrangements. ** The E1 reaction involves a carbocation intermediate. This interme- diate can rearrange, usually by the shift of a hydride or an alkyl group, to give a more stable carbocation. The E2 reaction takes place in one step with no intermediates. No rearrangement is possible in the E2 reaction.