Ch 6 Alkyl Halides

Question 1

Which alkyl halides are less dense than water (1.00 g/mol)?

Answer 1

Alkyl fluorides and alkyl chlorides (those with just one chlorine atom) are less dense than water (1.00 g/mL).

Question 2

Which alkyl halides are more dense than water (1.00 g/mol)?

Answer 2

Alkyl chlorides with two or more chlorine atoms are denser than water, and all alkyl bromides and alkyl iodides are denser than water.

Question 3

Why is free-radical halogenation rarely an effective method for the synthesis of alkyl halides?

Answer 3

It usually produces mixtures of products because there are different kinds of hydrogen atoms that can be abstracted. Also, more than one halogen atom may react, giving multiple substitutions. For example, the chlorination of propane can give a messy mixture of products.

Question 4

allylic

Answer 4

A carbon atom next to a carbon–carbon double bond.

Question 5

How are allylic intermediates stabilised?

Answer 5

Resonance with the double bond, allowing the charge or radical to be delocalised.

Question 6

Compare the bond dissociation enthalpies between a resonance-stabilized primary allylic radical and a typical secondary radical.

Answer 6

Less energy (lower enthalpy) is required to form a resonance-stabilized primary allylic radical than a typical secondary radical.

Question 7

What is the first step of allylic bromination?

Answer 7

Initiation Step: Bromine absorbs light, causing formation of radicals.

Question 8

What is the second step of allylic bromination?

Answer 8

First Propagation Step: A bromine radical abstracts an allylic hydrogen.

Question 9

What is the third step of allylic bromination?

Answer 9

Second Propagation Step: Either radical carbon can react with bromine.

Question 10

allylic shift

Answer 10

In one of the products of allylic bromination (used as an example of any halogenation), the bromine atom appears in the same position where the hydrogen atom was abstracted. The other product results from reaction at the carbon atom that bears the radical in the second resonance form of the allylic radical. This second compound is said to be the product of an allylic shift.

Question 11

What are the three major trends concerning the strength of a nucleophile?

Answer 11

1) 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.
2) Nucleophilicity decreases from left to right in the periodic table, following the increase in electronegativity from left to right. The more electronegative elements have more tightly held nonbonding electrons that are less reactive toward forming new bonds.
3) Nucleophilicity increases down the periodic table, following the increase in size and polarizability, and the decrease in electronegativity.

Question 12

What effect does polarisability have on S_N2 reactions?

Answer 12

The increased mobility of a more poleraiseable atom’s electrons enhances the atom’s ability to begin to form a bond at a relatively long distance.

Question 13

Explain the effect of steric hindrance on basicity and nucleophilicity

Answer 13

Steric hindrance has little effect on basicity because basicity involves attack on an unhindered proton. When a nucleophile attacks a carbon atom, however, a bulky nucle- ophile cannot approach the carbon atom so easily. Most bases are also nucleophiles, capable of attacking either a proton or an electrophilic carbon atom. If we want a species to act as a base, we use a bulky reagent like tert-butoxide ion. If we want it to react as a nucleophile, we use a less hindered reagent, like ethoxide.

Question 14

protic solvent

Answer 14

A protic solvent is one that has acidic protons, usually in the form of O ¬ H or N ¬ H groups. These groups form hydrogen bonds to negatively charged nucleophiles. Protic solvents, especially alcohols, are convenient solvents for nucle- ophilic substitutions because the reagents (alkyl halides, nucleophiles, etc.) tend to be quite soluble.

Question 15

aprotic solvent

Answer 15

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 sol- vent 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.

Question 16

What is the disadvantage for using an aprotic solvent in a S_N2 reaction?

Answer 16

Most polar, ionic reagents are insoluble in simple aprotic solvents such as alkanes.

Question 17

polar aprotic solvent

Answer 17

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. We can add specific solvating reagents to enhance solubility without affecting the reactivity of the nucleophile. For example, the “crown ether” 18-crown-6 solvates potassium ions. Using the potassium salt of a nucleophile and solvating the potassium ions causes the nucle- ophilic anion to be dragged along into solution.

Question 18

Describe a good substrate for a S_N2 reaction

Answer 18

To be a good substrate for S_N2 attack by a nucleophile, a molecule must have an electrophilic carbon atom with a good leaving group, and that carbon atom must not be too sterically hindered for a nucleophile to attack.

Question 19

Explain what two purposes the leaving group serves and describe an ideal leaving group for a S_N2 reaction

Answer 19

A leaving group serves two purposes in the S_N2 reaction:

1) It polarizes the C - X bond, making the carbon atom electrophilic.
2) It leaves with the pair of electrons that once bonded it to the electrophilic carbon atom.

To fill these roles, a good leaving group should be:

1) electron withdrawing, to polarize the carbon atom
2) stable (not a strong base) once it has left
3) polarizable, to stabilize the transition state.

Question 20

Describe the S_N1 reaction

Answer 20

The S_N1 mechanism is a multistep process. The first step is a slow ionization to form a carbocation (Reactivity: 3° > 2° > 1°). The second step is a fast attack on the carbocation by a nucleophile. The carbocation is a strong electrophile; it reacts very fast with nucleophiles, including weak nucleophiles. The nucleophile in S_N1 reactions is usually weak, because a strong nucleophile would be more likely to attack the substrate and force some kind of second-order reaction. 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.

Question 21

Answer 21

free-radical halogenation

Question 22

Answer 22

free-radical halogenation