# Organic Chemistry/Haloalkanes

**Haloalkanes** are otherwise simple alkanes that contain one or more
members of the halogen family. In practice, the halogens found in
organic molecules are chlorine (Cl), bromine (Br), fluorine (F), and
iodine (I). Some texts refer to this class of compounds as
**halogenoalkanes** or **alkyl halides**. This text (and the chemical
literature) will frequently use the terms **haloalkane** and **alkyl
halide** interchangeably.

*Note:* The X in R-X represents a generic halogen atom.

# Preparation

Methods for preparation are found elsewhere in this text:

-   Preparation from
    Alcohols
    (nucleophilic substitution)

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-   Preparation from /Alkanes/ (radical
    substitution)

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-   Preparation from
    Alkenes
    (electrophilic addition)
-   Preparation by halogen exchange.\*It is generally used for preparing
    iodoalkanes.
-   Preparation from silver salts of acids\*

# Properties

## Naming Haloalkanes

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

  ---------------------------------------------------------------------------------- --------------------------------------- ----------------------
                                                                                     **IUPAC name**                          **Common name**
  CH~3~---F                                                                          Fluoromethane                           Methyl fluoride
  CH~3~---Cl                                                                         Chloromethane                           Methyl chloride
  CH~3~---Br                                                                         Bromomethane                            Methyl bromide
  CH~3~---I                                                                          Iodomethane                             Methyliodide
  F---CH~2~---F                                                                      Difluoromethane                         Methylene fluoride
  Cl---CH~2~---Cl                                                                    Dichloromethane                         Methylene chloride
  F---CH~2~---Cl                                                                     Chlorofluoromethane                     
  CHBrClF                                                                            Bromochlorofluoromethane                
  HCCl~3~                                                                            Trichloromethane                        Chloroform
  CHX~3~                                                                             Haloforms (X=halogen)                   
  CCl~4~                                                                             Tetrachloromethane                      Carbon tetrachloride
  CH~3~CHCl~2~                                                                       1,1-Dichloroethane                      
  ![]((Dibromomethyl)cyclohexane.svg "(Dibromomethyl)cyclohexane.svg"){width="75"}   (Dibromomethyl)cyclohexane              
  ![](Dibromomethylcyclohexane2.png "Dibromomethylcyclohexane2.png")                 Equatorial (Dibromomethyl)cyclohexane   
  ![](1,6-Dichloro-2,3,5-trimethy.png "1,6-Dichloro-2,3,5-trimethy.png")             1,6-Dichloro-2,5-dimethylhexane         
  !50 px                             1,1-Dichloro-3-methylcyclobutane        
  ---------------------------------------------------------------------------------- --------------------------------------- ----------------------

## Physical properties

**R-X bond polarity:** C---F \> C---Cl \> C---Br \> C---I

  ------------- -------------------------- ----------------------------------
  **atom** \|   **electronegativity** \|   **difference from C** (= 2.5) \|
  F             4.0                        1.5
  Cl            3.0                        0.5
  Br            2.8                        0.3
  I             2.5                        0.0
  ------------- -------------------------- ----------------------------------

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 broken.

**Bond Length:** C---F \< C---Cl \< C---Br \< C---I

  ---------- -----------------
  **bond**   **length** (pm)
  C-F        138
  C-Cl       177
  C-Br       193
  C-I        214
  ---------- -----------------

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.

## Chemical properties

**Bond strength:** C---F \> C---Cl \> C---Br \> C---I

  ---------- ---------------------------
  **bond**   **strength** (kJ mol^-1^)
  C-F        484
  C-Cl       338
  C-Br       276
  C-I        238
  ---------- ---------------------------

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.

# Reactions

**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 S~N~ reactions.

  ------------------------------------------
  \"Nu\" represents a generic nucleophile.
  ------------------------------------------

+---------------------------------------------+
| ```{=html}                                  |
| <H2>                                        |
| ```                                         |
| General nucleophilic substitution reactions |
|                                             |
| ```{=html}                                  |
| </H2>                                       |
| ```                                         |
| Nu:^-^ + R---X → R---Nu + X:^-^\            |
| Nu: + R---X → R---Nu^+^ + X:^-^             |
+---------------------------------------------+

  ------------- ----------------- ---------------- -------------- -------------------
  **Reagent**   **Nucleophile**   **Name**         Product        Product name
  NaOH/KOH      :O^-^H            Hydroxide        R---OH         Alcohol
  NaOR\'        :O^-^R\'          Alkoxide         R---O---R\'    Ether
                :S^-^H            Hydrosulfide     R---SH         Thiol
  NH~3~         :NH~3~            Ammonia          R---NH~3~^+^   Alkylammonium ion
  KCN           :C^-^N            Cyanide          R---CN         Nitrile
  AgCN          Ag-CN:            Silver cyanide   R-NC           isonitrile
                :C^-^≡C---H       Acetylide        R-C≡C---H      Alkyne
  NaI           :I^-^             Iodide           R---I          Alkyl Iodide
  R\'^-^M^+^    :R\'^-^           Carbanion        R-R\'          Alkane
  KNO~2~        O=N---O           Nitrite          R--O---N=O     Alkyl nitrite
  AgNO~2~       Ag---Ö---N=O      Silver nitrite   R---NO~2~      Nitroalkane
  LiAlH~4~      H                 Hydrogen         RH             alkane
  R\'COOAg      R\'COO^-^         Alkanoate        R\'COOR        Ester
  ------------- ----------------- ---------------- -------------- -------------------

  : **Common Nucleophiles**

**Example:** Suggest a reaction to produce the following molecule.

![](1-ethoxy-3-methylbutane.gif "1-ethoxy-3-methylbutane.gif")

**Answer:**

![](1-bromo-3-methylbutane.gif "1-bromo-3-methylbutane.gif")`<span style="font-size:x-large;">`{=html}+`</span>`{=html}!Ethanolate{width="150"}

`<span style="font-size:x-large;">`{=html}OR`</span>`{=html}

![](3-methylbutan-1-olate.gif "3-methylbutan-1-olate.gif")`<span style="font-size:x-large;">`{=html}+`</span>`{=html}!Bromoethane{width="120"}

*Any halogen could be used instead of Br*

### Reaction mechanisms

Nucleophilic substitution can occur in two different ways. S~N~2
involves a **backside attack**. S~N~1 involves a **carbocation
intermediate**.

**S~N~2 mechanism** !Illustration of the Sn2 mechanism. First, the
electrons in the nucleophile attack the central carbon atom from the
side opposite the leaving group (in this case, a halogen). The electrons
forming the bond between the central carbon atom and the halogen move to
the halogen, causing the halogen to leave the
molecule.. The electrons forming the bond between the central carbon atom and the halogen move to the halogen, causing the halogen to leave the molecule.")

**S~N~1 mechanism** !Illustration of the Sn1 mechanism. First, in the
presence of a polar solvent, the C-X bond breaks, forming the
carbocation. This carbocation intermediate is highly reactive. In this
case, it reacts with water. Note that the water may attack from either
side.

### Comparison of S~N~1 and S~N~2 mechanism

**Stereochemistry:**\
S~N~2 - Configuration is inverted (i.e. R to S and vice-versa).\
S~N~1 - 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% chance 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:**\
S~N~2 - Rate depends on concentrations of both the haloalkane and the
nucleophile. S~N~2 reactions are fast.\
S~N~1 - Rate depends only on the concentration of the haloalkane. The
carbocation forms much slower than it reacts with other molecules. This
makes S~N~1 reactions slow.\
**Role of solvent:**\
S~N~2 - 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 (CCl~4~)\
Protic solvents are the worst type for S~N~2 reactions because they
\"cage,\" or solvate, the nucleophile, making it much less reactive.\
S~N~1 - Polar protic solvents favored. Examples: H~2~O, 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
S~N~1 reactions because they do nothing to stabilize the carbocation
intermediate.\
**Role of nucleophile:**\
S~N~2 - Good nucleophiles favored\
S~N~1 - Any nucleophile will work (since it has no effect on reaction
rate)\
**Carbocation stability:**\
3^°^ carbon - most stable = S~N~1 favored\
2^°^ carbon - less stable = either could be favored\
1^°^ carbon - seldom forms = S~N~2 favored\
CH~3~^+^ - never forms = S~N~2 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.

### Example

Predict whether the following reactions will undergo S~N~2 or S~N~1 and
tell why.

1:
![](haloalkanemechanismsproblem1.gif "haloalkanemechanismsproblem1.gif")

2:
![](haloalkanemechanismsproblem2.gif "haloalkanemechanismsproblem2.gif")

3:
![](haloalkanemechanismsproblem3.gif "haloalkanemechanismsproblem3.gif")

**Answers:**\
1) S~N~2. Good nucleophile, polar solvent.\
2) S~N~1. Tertiary carbon, polar solvent. Very slow reaction rate.\
3) S~N~2. Primary carbon, good nucleophile, nonpolar solvent.\

## Grignard reagents

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 as stated below:

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. Basic: R-X + Mg → R-Mg-X For example (X=Cl and R=CH3): CH3-Cl + Mg
→ CH3MgCl (methylmagnesiumchloride)

## Elimination reactions

With alcoholic potassium hydroxide, haloalkanes lose H-X and form the
corresponding alkene. Very strong bases such as KNH~2~/NH~3~ convert
vic-dihalides (haloalkanes with two halogen atoms on adjacent carbons)
into alkynes.


------------------------------------------------------------------------

\<\< Stereochemistry \|
Haloalkanes \| Alcohols \>\>

