#4: Alcohols and Alkyl Halides Flashcards

1
Q

Functional Group

A

The atom of group in a molecule most responsible for the reaction the compound undergoes under a prescribed set of conditions.

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2
Q

Mechanism

A

How the structure of the reactant is transformed to that of the product.

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3
Q

Substitutive Nomenclature for Alkyl Halides

A

Halogens are treated as substituents on the alkane chain.

Name of the halogen is written as a halo (fluoro, chloro, bromo, and iodo).

Name of the parent alkane ends with an -ane.

The parent carbon chain is numbered so that the substituted carbon gets the lowest number.

When the parent carbon chain has both halogen and an alkyl substituent the two are considered equal ranked and the parent chain is numbered so that lower number is given to the substituent nearer the end of the chain.

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4
Q

Substitutive Nomenclature for Alcohols

A

Identify the longest alkyl chain that has the -OH group.

Number the alkyl chain so that the carbon containing the -OH gets the lowest number.

Name of the alkane chain ends with -ol.

Hydroxyl group (-OH) out ranks alkyl groups and halogens in determining the direction in which the carbon chain is numbered.

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5
Q

Classes of Alcohols and Alkyl Halides

A

Alcohols and alkyl halides are classified as primary, secondary, or tertiary according to the degree of substitution of the carbon that bears the functional group. Thus, primary alcohols and primary alkyl haldies are compounds of the type RCH2G (where G is the functional group), secondary alcohols and secondary alkyl halides are compounds of the type R2CHG, and tertiary alcohols and tertiary alkyl halides are compounds of the type R3CG.

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6
Q

Bonding in Alcohols and Alkyl Halides

A

The carbon thatbears the functional group is sp3-hybridized in alcohols and alkyl haldies. THe bond angles at carbon are approximately tetrahedeal, as is the C-O-H angle. A similar orbital bybridization model applies to alkyl halides, with the halogen connected to sp3-hybridized carbon by a sigma bond. Carbon-halogen bond distances in alkyl halides increase in the order C-F (140 pm) < C-Cl (179 pm) < C-Br (197 pm) < C-I (216 pm).

Carbon-oxygen and carbon-halogen bonds are polar covalent bonds, and carbon bears a partial positive charge in alcohols and in alkyl halides. Alcohols and alkyl halides are polar molecules. The dipole moments of methanol and chloromethane are very similar to each other and to water.

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7
Q

Van der Waals Forces

A

Van der Waals attractive forces between molecules are of three types.

1) Induced-dipole/induced-dipole forces
2) Dipole/induced-dipole forces
3) Dipole-dipole forces

Induced-dipole/induced-dipole forces are the only intermolecular attactive forces available to nonpolar molecules such as alkanes and are important in polar molecules as attractions.

The dipole-dipole attractive force is easiest to visualize. Two molecules of a polar substance experience a mututal attraction between the positively polaized region of one molecule and the negatively polarized region of the other.

As its name implies, the dipole/induced-dipole force combines features of both the other two. A polar region of one molecule alters the electron distribution in a nonpolar region of another in a direction that produces an attractive force between them.

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8
Q

Boiling Point

A

Van der Waals forces are very important in explaining boiling point of Alcohols and Alkyl Halides. Consider three compounds, alkane propane, alkyl halide fluoroethane, and the alcohol ethanol. Both of the polar compounds, ethanol and fluoroethane, have higher boiling points than the nonpolar one, propane. We attribute this to a combo of dipole/induced dipole and dipole-dipole attractive forces that are present in the liquid states of ethanol and fluoroethane, but absent in propane.

BPs

  • Propane (CH3CH2CH3): - 42 C
  • Fluoroethane (CH3CH2F): - 32 C
  • Ethanol (CH3CH2OH): 78 C

The most striking difference, however, is that despite the similarity in their dipole moments, ethanol has a much higher BP than fluoroethane. This suggests that the attractive forces in ethanol are unusually strong. They’re an example of a special type of dipole-dipole attraction called hydrogen bonding and involve, in this case, the positively polarized proton of the -OH group of one ethanol molecule with the negatively charged polarized oxygen of another. The oxygen of the -OH group of alcohols serve as a hydrogen bond acceptor, while the hydrogen attached to the oxygen serves as a hydrogen bond donor. Having both hydrogen bond acceptor and donor capability in the same molecule creates a strong network among ethanol molecules in the liquid phase.

Among alkyl halides, the boiling point increases with increasing size of the halogen, which means alkyl fluorides have lowest BPs on alkyl halides. But alcohols overall have higher BPs than alkyl halides. Both of them however do have higher BPs than alkanes.

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9
Q

Solubility in Water

A

Alkyl halides are all insoluble in water, but low-molecular weight alcohols are soluble in water in all proportions. Their ability to particupate in intermolecular hydrogen bonding not only affects the boiling point of alcohols, but also enhances their water solubility.

Higher alcohols become more “hydrocarbon-like” and less water soluble. 1-Octanol, for eample, dissolves to the extent of only 1 mL in 2000 mL of water. As the alkyl chain gets longer, the hydrophobic effect becomes more important, to the point that it, more than hydrogen bonding, governs the solubility of alcohols.

For alcohols…
1-4 Carbons = Water Soluble
5+ Carbons = Not water soluble

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10
Q

Preparation of Alkyl Halides

A

We’ll begin with the preparation of alkyl halides from alcohols by reaction with hydrogen halides.

R-OH + H-X –> R-X + H-OH

The order of reactivity of the hydrogen halides parallels their acidity: HI > HBr > HCl > > HF. Hydrogen iodide is used infrequently, however, and the reaction of alcohols with hydrogen flouride is not a useful method for the preparation of alkyl fluorides.

Among the various classes of alcohols, tertiary alcohols are observed to be the most reactive and primary alcohols the least reactive. Tertiary alcohols are converted to alkyl chlorides in high yield within minutes on reaction with hydrogen chloride at room temperature and below. Secondary and primary alcohols don’t react with HCl at rates fast enough to make the preparation of the corresponding alkyl chlorides a method of practical value. Therefore, the more reactive hydrogen halide HBr is used; even then, elevated temperatures are required to increase the rate of reaction.

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11
Q

Mechanism of the Reaction of Alcohols with Hydrogen Halides: Hammond’s Postulate

A

This reaction is a substitution. A halogen, usually chlorine or bromine, replaces a hydroxyl group as a substituent on carbon. Calling the reaction a substitution tells us the relationship between the organic reactant and product but does not reveal the mechanism. The mechanism is the step-by-step pathway of bond cleavage and bond formation that leads from reactants to products. In developing a mechanistic picture for a particular reaction, we combine some basic principles of chemical reactivity with experimental observations to deduce the most likely sequence of steps.

Consider the reaction of tert-butyl alcohol with hydrogen chloride:
(CH3)3COH + HCl –> (CH3)3CCl + H2O

The mechanism for this reaction is presented as a series of three equations. There is no way to truly prove this reaction is correct though. It can always be modified upon the findings of new data.

Each equation in the mechanism represents a single elementary step, which is one that involves only one transition state. A particular reaction might proceed by way of a single elementary step, in which it is described as a concerted reaction, or by a series of elementary steps.

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12
Q

Mechanism of the Reaction of Alcohols with Hydrogen Halides Step 1: Proton Transfer

A

ALcoholds resemble water in respect to their Bronsted acidity. They also resemble water in their Bronsted basicity. Just as proton transfer to a water molecule gives oxonium ion (H3O+), proton transfer to an alcohol gives an alkyloxonium ion (ROH2+).

(CH3)3C-O-H + H-Cl (CH3)3C-O-H2+ + Cl-

Furthermore, a strong acid such as HCl that ionizes completely when dissolved in water also ionizes completely when dissolved in an alcohol. Many important reactions of alcohols involve strong acids either as reactants or as catalysts. In all these reactions the first step is formation of an alkyloxonium ion by proton transfer from the acid to the alcohol.

For molecularity, the transfer of a proton from HCl to tert-butyl alcohol is bimolecular because two molecules [HCl and (CH3)3COH) undergo chemical change.

The tery-butyloxonium ion [(CH3)3C-O-H2+] formed in step 1 is an intermediate. It’s not one of the initial reactants, nor is it formed as one of the final products. Rather it’s formed in one elementary step, consumed in another, and lies ont he pathway from reactants to products.

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13
Q

Molecularity

A

The molecularity of an elementary step is given by the number of species that undergo a chemical change in that step.

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14
Q

Mechanism of the Reaction of Alcohols with Hydrogen Halides Step 1: Proton Transfer (Potential Energy Diagram)

A

Shown on page 150. These aspects worth noting.

  • The point of max potential energy encountered by the reactants as they proceed to products is called the transition state.
  • The difference in energy between the reactants and transition state is known as the energy of activation Eact.
  • Because this is an elementary step, it involves a single transition state.
  • This step is known to be exothermic, so the products are placed lower in energy than the reactants. It is exothermic because HCl is a stronger acid than the alkyloxonium ion.
  • Proton transfers from strong acids to water and alcohols rank among the most rapid chemical processes and occur almost as fast as the molecules collide with one another. Thus the height of the energy barrier, the Eact for proton transfer, must be quite low.

The concerted nature of proton transfer contributes to its rapid rate. The energy cost of breaking the H-Cl bond is partially offset by the energy offset by the energy released in forming the new bond between the transferred proton and the oxygen of the alcohol. Thus, the activation energy is far less than it would be for a hypothetical two-step process in which the H-Cl bond breaks first, followed by bond formation between H+ and the alcohol.

The species present at the transition state is not a stable structure and cannot be isolated or examined directly. In general, the bonds in transition states are partially rather than fully formed. Its structure is assumed to be one in which the proton being transferred is partially bonded to both chlorine and oxygen simultaneously, although not necessarily to the same extent.

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15
Q

Hammond’s Postulate

A

If two states are similar in energy, they are similar in structure.

One of its corollaries is that the structure of a transition state more closely resembles the immediately preceding or following state to which it is closer in energy.

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16
Q

Mechanism of the Reaction of Alcohols with Hydrogen Halides Step 2: Carbocation Formation

A

In the second step, the alkyloxonium ion dissociates to a molecule of water and a carbocation, an ion that contains a positively charged carbon.

(CH3)3C-O-H2 (CH3)3C+ + H2O

Only one species, tert-bytyloxonium ion, undergoes a chemical change in this step, therefore, the step is unimolecular.

Like tert-butyloxonium ion, tert-butyl cation is an intermediate along the reaction pathway. It is, however, a relatively unstable species and its formation by dissociation of the alkyloxonium ion is endothermic. Step 2 is the lowest step in the mechanism and has the highest Eact.

17
Q

Mechanism of the Reaction of Alcohols with Hydrogen Halides Step 2: Carbocation Formation (Potential Energy Diagram)

A

Can be seen of Page 151.

Because this step is endothermic, the products of it are placed higher in energy than the reactants.

The transition state is closer in energy to the carbocation, so, according to Hammond’s postulate, its structure more closely resembles the carbocation that it resembles the tert-butyloxonium ion. The transition state has considerable “carbocation character,” meaning that a significant degree of positive charge has developed at carbon.

There is ample evidence that carbocations are intermediates in some chemical reactions, but they are almost always too unstable to isolate. The simplest reason is that the positively charged carbon only has six electrons in valence shell-the octet rule is not satisfied.

With only six valence electrons, the positively charged carbon is sp2-hybridized. The unhybridized 2p orbital that remains on the positively charged carbon contains no electrons; its axis is perpendicular to the plane of the bonds connecting that carbon to the three methyl groups.

The positive charge on carbon and the vacant p orbital combine to make carbocations electrophilic.

18
Q

Electrophilic

A

Lewis Acids. Species that are electron loving. Positively charged.

19
Q

Mechanism of the Reaction of Alcohols with Hydrogen Halides Step 3: Reaction of tert-Butyl Cation with Chloride Ion

A

This is a Lewis Base/Acid reaction. The electrophile is tert-Butyl cation, nucleophile the chloride ion.

(CH3)3C+ + Cl- –> (CH3)3C-Cl

Step 3 is bimolecular because two species, the carbocation and chloride ion, react together.

20
Q

Nucleophilies

A

Lewis base. Species that are nucleus (proton) loving. Negatively charged with an unshared electron pair.

21
Q

Mechanism of the Reaction of Alcohols with Hydrogen Halides Step 3: Reaction of tert-Butyl Cation with Chloride Ion (Potential Energy Diagram)

A

The step is exothermic; it leads from the carbocation intermediate to the stable isolated products of the reaction.

The activation energy for this step is small, and bond formation between a positive ion and a negative ion occurs rapidly.

The transition state for this step involves partial bond formation between tert-butyl cation and chloride ion.

22
Q

Sn

A

Stands for subsitution nucleophilic. It’s followed by 1 or 2 according to whether the rate-determining step is unimolecular or bimolecular.

The reaction of tery-butyl alcohol with hydrogen chloride, for example, is said to follow an Sn1 mechanism because its slow step (dissociation of tert-butyloxonium ion) is unimolecular. Only the alkyloxonioum ion undergoes a chemical change in this step.

23
Q

Carbocation

A

The rate-determining step of tert-butyl alcohol with hydrogen chloride is formation of the carbocation (CH3)3C+. Convincing evidence tells us that carbocations can exist, but are unstable. When they’re in chemical reactions, it’s as reactive intermediates, formed slowly in one step and consumed rapidly in the next one.

Alkyl groups directly attached to the positively charged carbon stabilize a carbocation. Order of carbocation stability is methyl < primary < secondary < tertiary.

As carboncations go, CH3+ is particularly unstable, and its existence as an intermediate in chemical reactions has never been demonstrated. Primary carbocations, though more stable, are still too unstable to be involved as intermediates in chemical reactions. The threshold of stability is reached with secondary carbocations.

24
Q

Reaction of Methyl and Primary Alcohols with Hydrogen Halides: The Sn2 Mechanism

A

Unlike tertiary and seconday carbocations, methyl and primary carbocations are too high in energy to be intermediates in chemical reactions. However, methyl and primary alcohols are converted, albeit rather slowly, to alkyl halides on treatment with hydrogen halides. Therefore, they must follow a different mechanism, one that involves carbocation intermediates.

CH3(CH2)5CH2OH + HBr –> CH3(CH2)5CH2Br + H2O

The first step of this new mechanism is exactly the same as seen for the reaction of tery-butyl alcohol with hydrogen chloride-formation of an alkyloxonium ion by proton transfer from the hydrogen halide to the alcohol. Rapid, reversible Bronsted acid-base reaction.

CH3(CH2)5CH2-O-H + H-Br CH3(CH2)5CH2-H2O+ + Br-

The major difference is the second step. Heptyloxonium ion, instead of dissociating to an unstable primary carbocation, reacts differently. It’s attacked by bromide ion, which acts as a nucleophile. Bromide ion forms a bond to the primary carbon by “pushing off” a water molecule. This step is bimolecular because it involves both bromide and heptyloxonium ion. Step 2 is also rate-determining.

It’s important to note that although methyl and primary alcohols react with hydrogen halides by a mechanism that involves fewer steps than the corresponding reactions of secondary and tertiary alcohols, fewer steps don’t translate to faster reaction rates. Remember, the observed order of reactivity of alcohols with hydrogen halides is teritary > secondary > primary. Reaction rate is governed by the activate energy of the slowest step, regardless of how many steps there are.

25
Q

Halogenation of Alkanes

A

There is a completely different method for preparing alkyl halides, one that uses alkanes as reactants. It involves substitution of a halogen atom for one of the alkane’s hydrogens.

R-H + X2 –> R-X + H-X

The alkane is said to undergo fluorination, chlorination, bromination, or iodination according to whether X2 is F2, Cl2, Br2, or I2, respectively. The general term is halogenation. Chlorination and bromination are most widely used.

The reactivity of the halogens decreases in the order F2 > Cl2 > Br2 > I2. Fluorine is an extremely aggressive oxidizing agent, and its reaction with alkanes is strongly exothermic and difficult to control. Direct fluorination of alkanes requires special equipment and techniques, is not a reaction of general applicability, and will not be discussed further.

Chlorination of alkanes is less exothermic than fluorination, and bromination less exothermic than chlorination. Iodine is unique among halogens in that is reaction with alkanes is endothermic and alkyl iodides are never prepared by iodination of alkanes.

26
Q

Free Radicals

A

Species that contain unpaired electrons. The octet rule notwithstanding, not all compounds have all their electrons paired. Oxygen is the most familiar example of a compound with unpaired electrons; it has 2 of them. Compounds that have an odd number of electrons, such as nitrogen dioxide (NO2), must have at least one unpaired electron.

Simple alkyl radicals require special procedures for their isolation and study. We will encounter them here only as reactive intermediates, formed in one step of a reaction mechanism and consumed in the next. Alkyl radicals are classified as primary, secondary, or tertiary according to the number of carbon atoms directly attached to the carbon that bears the unpaired electron.

Experimental studies show that the planar sp3 model describes the bonding in alkyl radicals better than the pyramidal sp3 model.

27
Q

Stability of Free Radicals

A

Methyl < Primary < Secondary < Tertiary Radical

28
Q

Homolytic & Heterolytic Cleavage

A

Some of the evidence indicating that alkyl substituents stabilize free radicals comes from bond enthalpies. The strength of a bond is measured by the energy required to break it. A covalent bond can be broken in two ways.

In a homolytic cleavage, a bond between two atoms is broken so that each of them retains one of the electrons in the bond.

X:Y –> X. + .Y

In contrast, heterolytic cleavage one fragment retains both electrons.

X:Y –> X+ + :Y-

29
Q

Mechanism of Methane Chlorination

A

CH4 + Cl2 –> CH3Cl

The reaction is normally carried out in the gas phase at high temperature. The reaction itself is strongly exothermic, but energy must be put into the system to get it going. This energy goes into breaking the weakest bond in the system, which is the Cl-Cl bond with a bond dissociation enthalpy of 243 kJ/mol (58 kcal/mol). The step in which the Cl-Cl bond homolysis occurs is called the inituation step.

Each chlorine atom formed in the initiation step has seven valence electrons and is very reactive. Once formed, a chlorine atom abstracts a hydrogen atom from methane. Hydrogen chloride, one of the isolated products from the overall reaction, is formed in this step. A methyl radical is also formed, which then reacts with a molecule of Cl2 in step 3 giving chloromethane, the other product of the overall reaction, along with a chlorine atom. The chlorine atom then cycles back to step 2, and the process repeats.

Steps 2 and 3 are called the propagation steps of the reaction and, when added together, give the overall equation for the reaction. Because one initiation step can result in a great many propagation cycles, the overall process is called a free-radical chain reaction.

In practice, side reactions intervene to reduce the efficiency of the propagation steps. The chain sequence is interrupted whenever two odd-electron species combine to give an even-electron product. Reactions of this type are called chain-terminating steps.

Some commonly observed chain-terminating steps include:

  • Combination of a methyl radical with a chlorine atom.
  • Combination of two methyl radicals.
  • Combination of two chlorine atoms.

Termination steps are generally less likely to occur than propagation steps. Each of the termination steps requires two free radicals to encounter each other in a medium that contains far greater quantities of other materials with which they can react.

30
Q

Photochemical

A

Giving heat by shining light through it.