Tertiary Halogenoalkanes: A Complete Guide
What is a Tertiary Halogenoalkane?
To understand a tertiary halogenoalkane, we first need to break down the name. A halogenoalkane (also called an alkyl halide) is an organic compound containing one or more halogen atoms (fluorine, chlorine, bromine, or iodine) bonded to a carbon atom in an alkane chain. The "tertiary" part describes the level of the carbon atom that holds the halogen.
Carbon atoms can be classified based on the number of other carbon atoms they are directly attached to:
- Primary (1°) carbon: Bonded to 1 other carbon.
- Secondary (2°) carbon: Bonded to 2 other carbons.
- Tertiary (3°) carbon: Bonded to 3 other carbons.
Therefore, a tertiary halogenoalkane has the general structure shown below, where the "R" groups represent alkyl groups (which are fragments of alkanes like -CH3, -C2H5, etc.).
Where $ X $ is a halogen ($ F, Cl, Br, I $) and $ R_1 $, $ R_2 $, and $ R_3 $ are alkyl groups, which can be the same or different.
A classic example is 2-bromo-2-methylpropane, formerly known as tert-butyl bromide. Its structure is $(CH_3)_3CBr$. The central carbon is bonded to three methyl groups ($-CH_3$) and one bromine atom, making it a perfect model of a tertiary halogenoalkane.
Naming and Identifying Tertiary Halogenoalkanes
Following IUPAC[1] naming rules is the best way to identify the class of a halogenoalkane. The number in the name often reveals the carbon's status.
Let's look at some examples with four-carbon chains:
| Name | Structural Formula | Carbon Type | Explanation |
|---|---|---|---|
| 1-bromobutane | $ CH_3CH_2CH_2CH_2Br $ | Primary | The Br is on a carbon bonded to only 1 other carbon. |
| 2-bromobutane | $ CH_3CH_2CHBrCH_3 $ | Secondary | The Br is on a carbon bonded to 2 other carbons. |
| 2-bromo-2-methylpropane | $(CH_3)_3CBr$ | Tertiary | The Br is on a carbon bonded to 3 other carbons. |
As you can see, the carbon holding the halogen in 2-bromo-2-methylpropane is a tertiary carbon, making the entire molecule a tertiary halogenoalkane.
How Tertiary Halogenoalkanes are Made
Tertiary halogenoalkanes can be produced through several methods. The most common one is the free radical substitution reaction of an alkane with a halogen. However, this method often gives a mixture of products. A more targeted approach is the addition of a hydrogen halide to an alkene.
Addition of HBr to an Alkene (Markovnikov's Rule): When an unsymmetrical alkene like propene reacts with HBr, the hydrogen atom attaches to the carbon of the double bond that already has more hydrogen atoms. The halogen ends up on the more substituted carbon, which can often result in a tertiary halogenoalkane if the alkene is branched.
Addition to 2-methylpropene: $ (CH_3)_2C=CH_2 + HBr \rightarrow (CH_3)_3CBr $
Here, 2-methylpropene reacts with hydrogen bromide to form the tertiary halogenoalkane 2-bromo-2-methylpropane.
Chemical Reactions: Why Structure Matters
The "tertiary" structure has a profound impact on how these molecules react. The carbon-halogen bond in a tertiary halogenoalkane is relatively weak because the three alkyl groups crowd the central carbon, making it easier for the bond to break. This is an example of steric hindrance[2] actually increasing reactivity in certain mechanisms.
The two most important reaction types for halogenoalkanes are nucleophilic substitution and elimination.
Nucleophilic Substitution Reactions
In this reaction, a nucleophile[3] (a species that loves positive charges) attacks the electron-deficient carbon attached to the halogen, replacing the halogen atom. Tertiary halogenoalkanes primarily undergo an $ S_N1 $ mechanism.
The $ S_N1 $ Mechanism (Substitution, Nucleophilic, Unimolecular):
- Step 1 (Slow): The carbon-halogen bond breaks to form a stable carbocation[4] intermediate. Tertiary carbocations are very stable because the three alkyl groups donate electron density to the positive carbon.
- Step 2 (Fast): The nucleophile rapidly attacks the carbocation to form the final substitution product.
For example, the hydrolysis of 2-bromo-2-methylpropane with aqueous sodium hydroxide is slow but follows the $ S_N1 $ path, producing 2-methylpropan-2-ol.
$ (CH_3)_3CBr + OH^- \rightarrow (CH_3)_3COH + Br^- $
(2-bromo-2-methylpropane + hydroxide ion → 2-methylpropan-2-ol + bromide ion)
Elimination Reactions
Under strong basic conditions (like hot ethanolic KOH), tertiary halogenoalkanes readily undergo elimination to form alkenes. In this reaction, a base removes a hydrogen atom from a carbon adjacent to the C-X carbon, leading to the loss of HX and the formation of a double bond. They favor the $ E1 $ mechanism, which also involves a carbocation intermediate.
$ (CH_3)_3CBr + OH^- \ (hot, ethanolic) \rightarrow (CH_3)_2C=CH_2 + H_2O + Br^- $
(2-bromo-2-methylpropane → 2-methylpropene)
Comparing Reactivity: Primary vs. Tertiary
The reactivity of halogenoalkanes depends heavily on their structure and the reaction mechanism. The table below provides a simplified comparison.
| Property / Reaction | Primary Halogenoalkane | Tertiary Halogenoalkane |
|---|---|---|
| Example | 1-bromobutane ($ CH_3CH_2CH_2CH_2Br $) | 2-bromo-2-methylpropane ($ (CH_3)_3CBr $) |
| Strength of C-X bond | Stronger | Weaker |
| Favored Substitution Mechanism | $ S_N2 $ | $ S_N1 $ |
| Rate of Hydrolysis with $ OH^- $ | Slower (for $ S_N2 $) | Faster (due to $ S_N1 $) |
| Favored Reaction with Strong Base | Substitution ($ S_N2 $) | Elimination ($ E1 $) |
A Practical Example: From Starting Material to Product
Let's follow a real-world scenario involving a tertiary halogenoalkane. Imagine a chemist wants to synthesize the alkene 2-methylpropene, a useful compound in making plastics and fuels.
- Step 1: Make the Tertiary Halogenoalkane. The chemist starts with 2-methylpropene, which might seem odd, but this is a common lab preparation. They react the alkene with hydrogen bromide (HBr). Following Markovnikov's rule, the H and Br add across the double bond to form the stable tertiary halogenoalkane, 2-bromo-2-methylpropane.
- Step 2: Convert it Back to the Alkene. Now, to get a purer sample of the original alkene, the chemist treats the 2-bromo-2-methylpropane with a hot, concentrated solution of potassium hydroxide in ethanol. This strong base promotes an elimination reaction ($ E1 $), removing HBr and regenerating 2-methylpropene.
This cycle demonstrates the interconversion between alkenes and halogenoalkanes and highlights the utility of tertiary halogenoalkanes as intermediates in organic synthesis.
Important Questions
Q1: Why are tertiary halogenoalkanes more reactive than primary ones in nucleophilic substitution?
Q2: What is the main product when a tertiary halogenoalkane reacts with a strong base?
Q3: How can I quickly identify if a halogenoalkane is tertiary from its name?
Footnote
[1] IUPAC: International Union of Pure and Applied Chemistry. This body establishes the standard rules for naming chemical compounds.
[2] Steric Hindrance: The slowing of a chemical reaction due to the physical bulk of groups within a molecule. In tertiary halogenoalkanes, the bulky groups hinder the $ S_N2 $ mechanism but favor the $ S_N1 $ mechanism by stabilizing the carbocation.
[3] Nucleophile: A chemical species that donates an electron pair to form a chemical bond. Examples include hydroxide ion ($ OH^- $), water ($ H_2O $), and ammonia ($ NH_3 $).
[4] Carbocation: An ion with a positively charged carbon atom. The stability of carbocations follows the order: tertiary > secondary > primary > methyl.