Chemistry | Organic Chemistry: Alcohols, Halogenoalkanes and Spectra

Welcome to Organic Chemistry: Alcohols, Halogenoalkanes and Spectra!

Hello future Chemist! This chapter is incredibly important because it bridges the gap between simple alkanes and more complex functional group chemistry. Halogenoalkanes and alcohols are versatile starting materials for synthesizing almost any other organic molecule.

Don't worry if mechanisms seem tricky at first. We will break down the key ideas using simple analogies. Focus on understanding "who attacks whom" and "why." Let's dive in!

Part 1: Halogenoalkanes (Haloalkanes)

1.1 Structure and Polarity

Halogenoalkanes are compounds where one or more hydrogen atoms in an alkane have been replaced by a halogen atom (\(X\): F, Cl, Br, or I).

The key feature is the Carbon-Halogen (C-X) bond. Halogens are significantly more electronegative than carbon.

• This causes the electrons in the bond to be pulled towards the halogen.
• The carbon atom becomes slightly positive (\(\delta+\)).
• The halogen atom becomes slightly negative (\(\delta-\)).

This creates a polar bond. The \(\delta+\) carbon is now highly vulnerable to attack by electron-rich species.

1.2 Reactions of Halogenoalkanes: Nucleophilic Substitution

Halogenoalkanes primarily undergo a reaction called Nucleophilic Substitution.

What is a Nucleophile?

A Nucleophile (from the Latin meaning 'nucleus loving') is an electron-pair donor. It seeks out the positively charged carbon atom (\(\delta+\)). Nucleophiles are often anions (like \(OH^-\)) or molecules with a lone pair (like \(H_2O\) or \(NH_3\)).

Analogy: Imagine the C-X bond is like a relationship. The C atom is vulnerable (\(\delta+\)). The Nucleophile (Nu) is the 'new partner' who swoops in, bonds with C, and kicks out the halogen (the leaving group).

Key Nucleophilic Substitution Reactions

In this mechanism, the nucleophile substitutes (replaces) the halogen atom:

1. Hydrolysis (Substitution by \(OH^-\)): Converting a halogenoalkane into an alcohol.

\(R-X + OH^- \rightarrow R-OH + X^-\)

2. Reaction with Ammonia (\(NH_3\)): Converting a halogenoalkane into an amine.
3. Reaction with Cyanide (\(CN^-\)): Increasing the carbon chain length by one atom (important synthesis step).

1.3 The Rate of Hydrolysis

The rate at which a halogenoalkane reacts depends primarily on the strength of the C-X bond.

Bond Strength Order (Strongest to Weakest):
C-F > C-Cl > C-Br > C-I

Reaction Rate Order (Fastest to Slowest):
Iodoalkanes > Bromoalkanes > Chloroalkanes > Fluoroalkanes

The weaker the bond, the easier it is for the nucleophile to break it and kick out the halogen, making the reaction faster.

Testing the Rate of Hydrolysis (Experimental)

We test the rate of hydrolysis using aqueous silver nitrate solution (\(AgNO_3(aq)\)) in the presence of water (which acts as the nucleophile) and ethanol (to help the reactants mix).

1. As the halogenoalkane reacts, halide ions (\(X^-\)) are released into the solution.
2. These halide ions react instantly with the silver ions to form a coloured silver halide precipitate.

\(Ag^+(aq) + X^-(aq) \rightarrow AgX(s)\)

Observation Summary:

• Chloroalkane (\(Cl^-\)): Slowest reaction, forms a white precipitate.
• Bromoalkane (\(Br^-\)): Medium speed, forms a cream precipitate.
• Iodoalkane (\(I^-\)): Fastest reaction, forms a yellow precipitate.

Part 2: Alcohols

2.1 Structure, Classification and Physical Properties

Alcohols contain the hydroxyl functional group (\(-OH\)). This group is responsible for their characteristic reactions.

Classification of Alcohols

The classification (Primary, Secondary, Tertiary) is critical because it dictates how the alcohol reacts, especially during oxidation.

We classify the alcohol based on the number of carbon atoms attached to the carbon atom carrying the \(-OH\) group:

1. Primary Alcohols (1°): The C atom attached to the OH group is attached to one other C atom. (e.g., Ethanol)
2. Secondary Alcohols (2°): The C atom attached to the OH group is attached to two other C atoms. (e.g., Propan-2-ol)
3. Tertiary Alcohols (3°): The C atom attached to the OH group is attached to three other C atoms. (e.g., 2-methylpropan-2-ol)

Prerequisite Concept: Hydrogen Bonding

Due to the highly polar O-H bond, alcohols can form hydrogen bonds with each other and with water.

• This gives alcohols significantly higher boiling points than alkanes of similar size.
• Smaller alcohols (up to 4 carbons) are fully miscible (soluble) in water because they can form strong H-bonds with \(H_2O\) molecules.

2.2 Reactions of Alcohols

A. Oxidation (The most important reaction)

Oxidation involves adding oxygen or removing hydrogen. The oxidizing agent commonly used is acidified potassium dichromate(VI) solution (\(K_2Cr_2O_7 / H^+\)).

The Colour Change: If oxidation occurs, the orange dichromate(VI) ions are reduced to green chromium(III) ions.

1. Primary Alcohols (1°):
   • Controlled Oxidation (Distillation): Forms an aldehyde. The aldehyde is volatile and boils off as soon as it forms, preventing further oxidation.
   • Vigorous Oxidation (Reflux): Forms a carboxylic acid. Heating under reflux ensures all the alcohol and the intermediate aldehyde are fully oxidized.

   Mnemonic: P is for Primary. If you reflux it, you get C (Carboxylic Acid).

2. Secondary Alcohols (2°):
   • Only oxidize to form a ketone. Ketones resist further oxidation under normal conditions. This reaction requires heating under reflux.
3. Tertiary Alcohols (3°):
   • Do not react with the oxidizing agent. The central carbon atom (the one bonded to the -OH) has no hydrogen atoms attached, which are required for the oxidation mechanism to proceed.

   This difference in reactivity is a perfect way to distinguish between the three classes in the lab!

B. Elimination (Dehydration)

Alcohols can undergo elimination (the opposite of hydration) to form an alkene and water.

Conditions: Alcohol is heated with a strong, concentrated acid catalyst, typically concentrated sulfuric acid (\(H_2SO_4\)) or phosphoric acid (\(H_3PO_4\)) at high temperatures (around \(170^\circ C\)).

The alcohol loses the \(-OH\) group and a hydrogen atom from an adjacent carbon atom, forming a double bond.

C. Substitution (Reaction with Halides)

Alcohols can be converted back into halogenoalkanes using suitable reagents, such as hydrogen halides (\(HBr\)) or phosphorus(V) chloride (\(PCl_5\)).

\(R-OH + HBr \rightarrow R-Br + H_2O\)

Part 3: Analytical Techniques (Spectroscopy)

Spectroscopic techniques allow chemists to determine the structure of unknown compounds. We focus on Mass Spectrometry (MS) and Infrared Spectroscopy (IR).

3.1 Mass Spectrometry (MS)

Mass spectrometry measures the mass-to-charge ratio (\(m/z\)) of ions. It helps determine the relative molecular mass (\(M_r\)) and structural fragments of a molecule.

A. Molecular Ion Peak (\([M]^+\))

• The peak corresponding to the highest \(m/z\) value (furthest right, ignoring isotope peaks) is the molecular ion peak.
• The \(m/z\) value of this peak is equal to the Relative Molecular Mass (\(M_r\)) of the molecule.

B. Fragmentation

The molecular ion often breaks up (fragments) into smaller pieces. These fragments also register as peaks, helping us determine the structure.

Specific Fragments to Identify:

• Alcohols: Often lose a water molecule (\(H_2O\)). This results in a peak at \(M_r - 18\).
• Halogenoalkanes (Bromine and Chlorine): Because bromine and chlorine have two common isotopes in significant abundance, they produce unique patterns:
  • Chlorine: \(\mathbf{^{35}Cl}\) and \(\mathbf{^{37}Cl}\) exist in a roughly 3:1 ratio. This means you will see the M peak and an M+2 peak with a height ratio of 3:1.
  • Bromine: \(\mathbf{^{79}Br}\) and \(\mathbf{^{81}Br}\) exist in a roughly 1:1 ratio. This means you will see the M peak and an M+2 peak with a height ratio of 1:1.

If you see M and M+2 peaks, you know immediately that Chlorine or Bromine is present!

3.2 Infrared (IR) Spectroscopy

IR spectroscopy identifies the functional groups present in a molecule. It works because chemical bonds absorb infrared radiation at specific frequencies, causing them to vibrate (stretch or bend).

Key IR Concepts

• Wavenumber: The unit used for IR frequency, typically measured in inverse centimeters (\(cm^{-1}\)).
• The Fingerprint Region: The region below \(1500\ cm^{-1}\). This area is complex and unique to every molecule, like a fingerprint. You are generally not expected to interpret this region.
• The Diagnostic Region: The region above \(1500\ cm^{-1}\), which shows the characteristic stretches of functional groups (the part you must learn!).

Crucial Absorption Peaks for Alcohols and Halogenoalkanes

You must memorize the characteristic absorption ranges for the following bonds:

Bond	Functional Group	Approximate Wavenumber Range (\(cm^{-1}\))	Appearance
\(\mathbf{O-H}\)	Alcohols	\(3200-3600\)	Strong and Very Broad (A 'hilly' peak)
\(\mathbf{C-H}\)	Alkanes / General	\(2850-3000\)	Medium/Strong and Sharp
\(\mathbf{C=O}\)	Aldehydes/Ketones/Acids (Oxidation Products)	\(1680-1750\)	Strong and Sharp (A 'dagger' peak)
\(\mathbf{C-X}\)	Halogenoalkanes	Varies, often below \(1000\)	(Less reliable diagnostic)

Did you know? The broadness of the O-H peak in alcohols is due to varying strengths of the hydrogen bonding network.

How to Use IR to Monitor Oxidation

IR is perfect for checking if an oxidation reaction (e.g., alcohol \(\rightarrow\) aldehyde) has completed:

1. Check the starting material (alcohol): It should show a strong, broad O-H peak around \(3300\ cm^{-1}\).
2. Check the product (aldehyde/ketone/acid):
   • The O-H peak should disappear (or become much smaller).
   • A new, very strong, sharp C=O peak should appear around \(1700\ cm^{-1}\).