Haloalkanes are alkanes that contain one or more members of the halogen family. The halogens found in organic molecules are chlorine, bromine, fluorine, and iodine. Some texts refer to this class of compounds as halogenoalkanes or alkyl halides. This text will frequently use both haloalkane and alkyl halide, so it's important to remember that they are the same thing.
Note: The X in R-X represents a generic halogen atom.
Methods for preparation are found elsewhere in this text:
- Preparation from Alcohols (nucleophilic substitution)
- Preparation from Alkanes (radical substitution)
- Preparation from Alkenes (electrophilic addition)
- Preparation by halogen exchange.*It is generally used for preparing iodoalkanes.
- Preparation from silver salts of acids*
Haloalkanes are named by adding a prefix to the name of the alkane from which they are derived. The prefix denotes the particular halogen used.
F = Fluoro-
Cl = Chloro-
Br = Bromo-
I = Iodo-
If other substituents need to be named, all prefixes are still put in alphabetical order. When necessary, numbers identify substituent locations.
Example names of haloalkanes
R-X bond polarity: C—F > C—Cl > C—Br > C—I
|atom |||electronegativity |||difference from C (= 2.5) ||
The difference in electronegativity of the carbon-halogen bonds range from 1.5 in C-F to almost 0 in C-I. This means that the C-F bond is extremely polar, though not ionic, and the C-I bond is almost nonpolar.
Physical appearance: Haloalkanes are colourless when pure. However bromo and iodo alkanes develop colour when exposed to light. Many volatile halogen compounds have sweet smell.
Boiling point: Haloalkanes are generally liquids at room temperature. Haloalkanes generally have a boiling point that is higher than the alkane they are derived from. This is due to the increased molecular weight due to the large halogen atoms and the increased intermolecular forces due to the polar bonds, and the increasing polarizabilty of the halogen.
For the same alkyl group, the boiling point of haloalkanes decreases in the order RI > RBr > RCl > RF.This is due to the increase in van der Waals forces when the size and mass of the halogen atom increases.
For isomeric haloalkanes, the boiling point decrease with increase in branching. But boiling points of dihalobenzenes are nearly same; however the para-isomers have higher melting points as it fits into the crystal lattice better when compared to ortho- and meta-isomers.
Density: Haloalkanes are generally more dense than the alkane they are derived from and usually more dense than water. Density increases with the number of carbon and halogen atom. It also increases with the increase in mass of halogen atom.
Solubility: The haloalkanes are only very slightly soluble in water, but dissolves in organic solvents. This is because for dissolving haloalkanes in water the strong hydrogen bonds present in the latter has to be broken. When dissolved in organic (non polar) solvents, the intermolecular attractions are almost same as that being brocken.
Bond Length: C—F < C—Cl < C—Br < C—I
Larger atoms means larger bond lengths, as the orbitals on the halogen is larger the heavier the halogen is. In F, the orbitals used to make the bonds is 2s and 2p, in Cl, it's 3s and 3p, in Br, 4s and 4p, and in I, 5s and 5p. The larger the principal quantum number, the bigger the orbital. This is somewhat offset by the larger effective nuclear charge, but not enough to reverse the order.
Bond strength: C—F > C—Cl > C—Br > C—I
|bond||strength (kJ mol-1)|
The orbitals C uses to make bonds are 2s and 2p. The overlap integral is larger the closer the principal quantum number of the orbitals is, so the overlap is larger in the bonds to lighter halogens, making the bond formation energetically favorable.
Bond reactivity: C—F < C—Cl < C—Br < C—I
Stronger bonds are more difficult to break, making them less reactive. In addition, the reactivity can also be determined by the stability of the corresponding anion formed in solution. One of the many trends on the periodic table states that the largest atoms are located on the bottom right corner, implying that iodine is the largest and fluorine being the smallest. When fluorine leaves as fluoride (if it does) in the reaction, it is not so stable compared to iodide. Because there are no resonance forms and inductive stabilizing effects on these individual atoms, the atoms must utilize their own inherent abilities to stabilize themselves. Iodide has the greatest surface area out of these four elements, which gives it the ability to better distribute its negative charge that it has obtained. Fluorine, having the least surface area, is much more difficult to stabilize. This is the reason why iodine is the best leaving group out of the four halogens discussed.
Determination of Haloalkanes: A famous test used to determine if a compound is a haloalkane is the Beilstein test, in which the compound tested is burned in a loop of copper wire. The compound will burn green if it is a haloalkane. The numbers of fluorine, chlorine, bromine and iodine atoms present in each molecule can be determined using the sodium fusion reaction, in which the compound is subjected to the action of liquid sodium, an exceptionally strong reducing agent, which causes the formation of sodium halide salts. Qualitative analysis can be used to discover which halogens were present in the original compound; quantitative analysis is used to find the quantities.
Substitution reactions of haloalkanes
R-X bonds are very commonly used throughout organic chemistry because their polar bonds make them reasonably reactive. In a substitution reaction, the halogen (X) is replaced by another substituent (Y). The alkyl group (R) is not changed.
The ":" in a chemical equation represents a pair of unbound electrons.
A general substitution reaction
Y: + R—X → R—Y + X:
Substitutions involving haloalkanes involve a type of substition called Nucleophilic substitution, in which the substituent Y is a nucleophile. A nucleophile is an electron pair donor. The nucleophile replaces the halogen, an electrophile, which becomes a leaving group. The leaving group is an electron pair acceptor. Nuclephilic substition reactions are abbreviated as SN reactions.
"Nu" represents a generic nucleophile.
↑Jump back a section
General nucleophilic substitution reactions
Nu:- + R—X → R—Nu + X:-
Example: Suggest a reaction to produce the following molecule.
Any halogen could be used instead of Br
Nucleophilic substitution can occur in two different ways. SN2 involves a backside attack. SN1 involves a carbocation intermediate.
Comparison of SN1 and SN2 mechanism
SN2 - Configuration is inverted (i.e. R to S and vice-versa).
SN1 - Product is a mixture of inversion and retention of orientation because the carbocation can be attacked from either side. In theory the products formed are usually racemic due to the 50% change of attack from the planar conformation. Interestingly, the amount of the inverted product is often up to 20% greater than the amount of product with the original orientation. Saul Winstein has proposed that this discrepancy occurs through the leaving group forming an ion pair with the substrate, which temporarily shields the carbocation from attack on the side with the leaving group.
Rate of reaction:
SN2 - Rate depends on concentrations of both the haloalkane and the nucleophile. SN2 reactions are fast.
SN1 - Rate depends only on the concentration of the haloalkane. The carbocation forms much slower than it reacts with other molecules. This makes SN1 reactions slow.
Role of solvent:
SN2 - Polar aprotic solvents favored. Examples: Acetone, THF (an ether), dimethyl sulfoxide, n,n-dimethylformamide, hexamethylphosphoramide (HMPA).
Nonpolar solvents will also work, such as carbon tetrachloride (CCl4)
Protic solvents are the worst type for SN2 reactions because they "cage," or solvate, the nucleophile, making it much less reactive.
SN1 - Polar protic solvents favored. Examples: H2O, Formic acid, methanol.
Aprotic solvents will work also, but protic solvents are better because they will stabilize the leaving group, which is usually negatively charged, by solvating it. Nonpolar solvents are the worst solvent for SN1 reactions because they do nothing to stabilize the carbocation intermediate.
Role of nucleophile:
SN2 - Good nucleophiles favored
SN1 - Any nucleophile will work (since it has no effect on reaction rate)
3° carbon - most stable = SN1 favored
2° carbon - less stable = either could be favored
1° carbon - seldom forms = SN2 favored
CH3+ - never forms = SN2 favored
The reason why the tertiary carbocation is most favored is due to the inductive effect. In the carbocation intermediate, there is a resulting formal charge of +1 on the carbon that possessed the haloalkane. The positive charge will attract the electrons available. Because this is tertiary, meaning that adjacent carbon atoms and substituents are available, it will provide the most electron-density to stabilize this charge.
Predict whether the following reactions will undergo SN2 or SN1 and tell why.
1) SN2. Good nucleophile, polar solvent.
2) SN1. Tertiary carbon, polar solvent. Very slow reaction rate.
3) SN2. Primary carbon, good nucleophile, nonpolar solvent.
Grignard reagents are created by reacting magnesium metal with a haloalkane. The magnesium atom gets between the alkyl group and the halogen atom with the general reaction:
R-Br + Mg → R-Mg-Br
Gringard reagents are very reactive and thus provide a means of organic synthesis from haloalkanes. For example, adding water gives the alcohol R-OH. Basit R-X+Mg--->R-Mg-X For example X=Cl and R=CH3 CH3-CL+Mg--->CH3MgCl
With alcoholic potassium hydroxide, haloalkanes lose H-X and form the corresponding alkene. Very strong bases such as KNH2/NH3 convert vic-dihalides (haloalkanes with two halogen atoms on adjacent carbons) into alkynes.