Halogen Derivatives - Class 12 Chemistry | Chemca.in

Module 10 • Maharashtra HSC Board

Halogen Derivatives

Organic Chemistry: Mechanisms, Stereochemistry, Name Reactions & Solved PYQs

1. Classification of Halogen Derivatives

Halogen derivatives are compounds obtained by replacing one or more hydrogen atoms of an aliphatic or aromatic hydrocarbon by corresponding number of halogen atoms ($F, Cl, Br, I$).

A. On the basis of skeleton

Haloalkanes: Halogen attached to $sp^3$ hybridized carbon of an alkyl group. ($R-X$)
Haloalkenes: Halogen attached to $sp^2$ hybridized carbon of an alkene (Vinylic halides).
Haloalkynes: Halogen attached to $sp$ hybridized carbon of an alkyne.
Haloarenes: Halogen attached directly to the $sp^2$ hybridized carbon of an aromatic ring (Aryl halides).

B. On the basis of hybridization of Carbon

$sp^3$ C-X Bond

Alkyl Halides: Halogen attached to primary, secondary, or tertiary alkyl carbon.
Allylic Halides: Halogen attached to an $sp^3$ carbon next to a carbon-carbon double bond ($C=C-C-X$).
Benzylic Halides: Halogen attached to an $sp^3$ carbon next to an aromatic ring ($Ph-CH_2-X$).

$sp^2$ C-X Bond

Vinylic Halides: Halogen is bonded directly to one of the carbon atoms of a $C=C$ double bond ($CH_2=CH-X$).
Aryl Halides: Halogen is bonded directly to the aromatic ring.

2. Methods of Preparation

A. From Alcohols ($R-OH$)

The hydroxyl group of an alcohol is replaced by a halogen on reaction with halogen acids ($HX$), phosphorus halides ($PX_3, PX_5$), or thionyl chloride ($SOCl_2$).

Using $HCl$ (Lucas Reagent):
$R-OH + HCl \xrightarrow{\text{anhyd. } ZnCl_2} R-Cl + H_2O$
Using $SOCl_2$ (Darzen's Method): This is the best method for preparing alkyl chlorides because the byproducts ($SO_2$ and $HCl$) are gases and escape easily, leaving pure alkyl chloride.
$R-OH + SOCl_2 \xrightarrow{\text{Pyridine}} R-Cl + SO_2 \uparrow + HCl \uparrow$

B. From Hydrocarbons (Addition to Alkenes)

Addition of hydrogen halides ($HX$) to unsymmetrical alkenes follows Markovnikov's Rule: The negative part of the reagent ($X^-$) gets attached to the carbon atom containing fewer hydrogen atoms.

$CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3$ (2-Bromopropane, Major)

Peroxide Effect (Anti-Markovnikov's Rule / Kharasch Effect): In the presence of organic peroxides, addition of $HBr$ only to unsymmetrical alkenes yields the primary alkyl halide as the major product.
$CH_3-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3-CH_2-CH_2-Br$ (1-Bromopropane)

C. Halogen Exchange (Name Reactions)

Finkelstein Reaction

Used to prepare Alkyl Iodides. Alkyl chloride/bromide reacts with NaI in dry acetone.

$R-X + NaI \xrightarrow{\text{Dry Acetone}} R-I + NaX \downarrow$

($X = Cl, Br$. $NaX$ precipitates out).

Swarts Reaction

Used to prepare Alkyl Fluorides. Alkyl chloride/bromide is heated with metallic fluorides ($AgF, Hg_2F_2, CoF_3, SbF_3$).

$R-Br + AgF \rightarrow R-F + AgBr \downarrow$

3. Optical Isomerism

Stereochemistry plays a huge role in the reactions of halogen derivatives. Key terms to remember:

Chiral Carbon (Asymmetric Carbon): A carbon atom bonded to four different atoms or groups. Denoted by an asterisk ($C^*$). E.g., 2-chlorobutane.
Chirality: The property of a molecule being non-superimposable on its mirror image (like left and right hands). Such molecules are optically active.
Optical Activity: The ability of a chiral substance to rotate the plane of plane-polarized light.
- Dextrorotatory ($d$ or $+$): Rotates light to the right (clockwise).
- Laevorotatory ($l$ or $-$): Rotates light to the left (anti-clockwise).
Enantiomers: Stereoisomers that are non-superimposable mirror images of each other. They have identical physical properties but rotate plane-polarized light in opposite directions.
Racemic Mixture ($\pm$ or $dl$): An equimolar mixture of dextro and laevo enantiomers. It is optically inactive due to external compensation (rotations cancel out).

4. Nucleophilic Substitution Reactions ($S_N$)

When a stronger nucleophile ($Nu^-$) replaces a weaker nucleophile (halide ion, $X^-$) from an alkyl halide, it is called nucleophilic substitution.

$S_N2$ Mechanism (Substitution Nucleophilic Bimolecular)

Example: Alkaline hydrolysis of methyl bromide ($CH_3Br$) to form methanol.

Kinetics: It is a second-order reaction. $\text{Rate} = k [CH_3Br] [OH^-]$.
Mechanism: It is a single-step concerted process. Bond making and bond breaking occur simultaneously.
Transition State: Carbon is pentacoordinated (partially bonded to incoming $OH^-$ and departing $Br^-$). Highly unstable.
Stereochemistry: The nucleophile attacks from the backside (opposite to the leaving group) to avoid steric hindrance. This results in 100% Inversion of configuration (Walden Inversion).
Order of Reactivity: Primary ($1^\circ$) > Secondary ($2^\circ$) > Tertiary ($3^\circ$). Bulky groups hinder backside attack (steric hindrance).

$S_N1$ Mechanism (Substitution Nucleophilic Unimolecular)

Example: Alkaline hydrolysis of tert-butyl bromide to form tert-butyl alcohol.

Kinetics: It is a first-order reaction. $\text{Rate} = k [(CH_3)_3C-Br]$. It depends only on the concentration of the substrate.
Mechanism: It is a two-step process.
- Step 1 (Slow, Rate-determining): Heterolytic fission of C-Br bond to form a planar carbocation intermediate.
- Step 2 (Fast): Nucleophile ($OH^-$) attacks the carbocation.
Stereochemistry: Since the carbocation is planar ($sp^2$ hybridized), the nucleophile can attack from either front or back with equal probability. This results in Racemization (50% inversion, 50% retention).
Order of Reactivity: Tertiary ($3^\circ$) > Secondary ($2^\circ$) > Primary ($1^\circ$). Determined by the stability of the intermediate carbocation ($3^\circ$ carbocation is highly stable due to +I effect and hyperconjugation).

5. Elimination Reactions and Active Metals

Dehydrohalogenation ($\beta$-Elimination)

When an alkyl halide is heated with alcoholic KOH (or NaOH), it loses a hydrogen atom from the $\beta$-carbon and a halogen atom from the $\alpha$-carbon to form an alkene.

Saytzeff's Rule (Zaitsev's Rule)

In dehydrohalogenation reactions, if more than one alkene can be formed, the preferred product is the highly substituted alkene (the alkene with a greater number of alkyl groups attached to the double-bonded carbon atoms).

$CH_3-CH_2-CH(Br)-CH_3 \xrightarrow{\text{alc. KOH, } \Delta} CH_3-CH=CH-CH_3$
(2-Bromobutane $\rightarrow$ But-2-ene [Major 80%] + But-1-ene [Minor 20%])

Reaction with Active Metals

1. Formation of Grignard Reagent: Alkyl halides react with magnesium metal in dry ether to form alkyl magnesium halides ($R-Mg-X$).
$R-X + Mg \xrightarrow{\text{Dry Ether}} R-Mg-X$
Note: Grignard reagents are highly reactive and react with any source of proton (like water) to give hydrocarbons. Hence they must be prepared in anhydrous (dry) conditions.
2. Wurtz Reaction: Alkyl halides react with sodium metal in dry ether to form higher alkanes (symmetrical alkanes with double the carbon atoms).
$2 R-X + 2 Na \xrightarrow{\text{Dry Ether}} R-R + 2 NaX$

6. Reactions of Haloarenes (Aryl Halides)

Why are Haloarenes less reactive towards Nucleophilic Substitution?

Unlike alkyl halides, aryl halides do not readily undergo nucleophilic substitution under ordinary conditions due to:

Resonance Effect: The lone pairs on the halogen atom are in conjugation with the $\pi$-electrons of the benzene ring. This gives a partial double bond character to the C-X bond, making it difficult to break.
$sp^2$ Hybridization: The carbon attached to halogen is $sp^2$ hybridized, which is more electronegative and holds the electron pair of the C-X bond tightly.
Instability of Phenyl Cation: The phenyl cation formed (if $S_N1$ occurs) is highly unstable.

However, presence of electron-withdrawing groups (like $-NO_2$) at ortho and para positions greatly increases the reactivity towards nucleophilic substitution.

Electrophilic Substitution Reactions

Haloarenes undergo electrophilic substitution. The halogen atom is ortho, para-directing and mildly deactivating (due to its -I effect dominating over +R effect). Examples include halogenation, nitration, sulfonation, and Friedel-Crafts alkylation/acylation.

7. Polyhalogen Compounds and Environment

Compounds containing more than one halogen atom are called polyhalogen compounds. Some important ones are:

Dichloromethane ($CH_2Cl_2$): Used as a solvent, paint remover. Harmful to the central nervous system.
Chloroform ($CHCl_3$): Earlier used as an anesthetic. It oxidizes slowly in the presence of air and light to form a highly poisonous gas, phosgene ($COCl_2$). Hence, stored in dark-colored, completely filled bottles.
Iodoform ($CHI_3$): Used as an antiseptic due to the liberation of free iodine.
Freons (CFCs): Chlorofluorocarbons of methane and ethane. Used in refrigeration. They initiate radical chain reactions that cause ozone layer depletion in the stratosphere.
DDT (p,p'-Dichlorodiphenyltrichloroethane): A powerful insecticide. It is highly persistent in the environment and non-biodegradable, leading to biomagnification in the food chain. Its use is banned in many countries.

8. Solved Reasoning Questions

Question 1: Racemization in $S_N1$

Why does the $S_N1$ reaction of an optically active alkyl halide result in racemization?

Answer:

The $S_N1$ mechanism proceeds via the formation of a carbocation intermediate in the first slow step. A carbocation is $sp^2$ hybridized and has a planar geometry. Therefore, in the second fast step, the nucleophile can attack this planar carbocation from either the front side or the back side with equal probability. This results in the formation of a 1:1 mixture of two enantiomers (dextro and laevo), which is a racemic mixture. Thus, optical activity is lost (racemization).

Question 2: Preparation of Grignard Reagent

Why must Grignard reagents be prepared under strictly anhydrous conditions?

Answer:

Grignard reagents ($R-Mg-X$) are extremely reactive compounds. They act as strong bases and strong nucleophiles. If even a trace of moisture (water, $H-OH$) or any proton donor (like alcohols) is present, the Grignard reagent instantly reacts with it to form an alkane, destroying the reagent.

$R-Mg-X + H_2O \rightarrow R-H (\text{Alkane}) + Mg(OH)X$

Therefore, strictly anhydrous (dry) conditions (using dry ether) are essential.

9. Board PYQs with Complete Answers

Verified previous year questions from the Maharashtra State Board HSC Chemistry exams.

1 Mark Questions (VSA)

Q1. Write the name of the reaction used for the synthesis of alkyl fluorides. (Oct 2013, March 2022)

Answer: Swarts Reaction.

Q2. Define: Enantiomers. (March 2016, Oct 2020)

Answer: The stereoisomers which are non-superimposable mirror images of each other and rotate plane-polarized light in opposite directions are called enantiomers.

Q3. Give the formula of Freon-12. (March 2017)

Answer: Dichlorodifluoromethane ($CCl_2F_2$).

2 Mark Questions (SA-I)

Q4. State Saytzeff’s rule. Give one example. (March 2014, Oct 2019)

Answer:

Statement: In dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

Example: Dehydrohalogenation of 2-bromobutane with alcoholic KOH gives But-2-ene as the major product and But-1-ene as the minor product.

Q5. Explain the Wurtz reaction with a suitable example. (March 2015, March 2021)

Answer:

When an alkyl halide is treated with sodium metal in the presence of dry ether, it forms a higher alkane containing double the number of carbon atoms present in the alkyl halide. This is called Wurtz reaction.

$2CH_3Br + 2Na \xrightarrow{\text{Dry Ether}} CH_3-CH_3 (\text{Ethane}) + 2NaBr$

3 Mark Questions (SA-II)

Q6. Differentiate between $S_N1$ and $S_N2$ mechanisms. (Any three points). (Oct 2014, March 2019, March 2023)

Answer:

Point of Difference	$S_N1$ Mechanism	$S_N2$ Mechanism
Kinetics (Order)	First order (Rate depends only on substrate).	Second order (Rate depends on substrate and nucleophile).
Steps	It is a two-step process.	It is a single-step concerted process.
Stereochemistry	Leads to Racemization (retention + inversion).	Leads to 100% Inversion of configuration (Walden inversion).
Order of Reactivity	$3^\circ > 2^\circ > 1^\circ$ (Carbocation stability).	$1^\circ > 2^\circ > 3^\circ$ (Steric hindrance).

Q7. Write short notes on (a) Finkelstein reaction (b) Sandmeyer's reaction. (March 2018, Oct 2021)

Answer:

(a) Finkelstein Reaction:

It is a halogen exchange reaction used specifically for the preparation of alkyl iodides. Alkyl chlorides or bromides are heated with a solution of Sodium Iodide (NaI) in dry acetone.

$R-Cl + NaI \xrightarrow{\text{Dry Acetone}} R-I + NaCl\downarrow$

(b) Sandmeyer's Reaction:

It is used for the preparation of haloarenes (aryl halides). A primary aromatic amine (like aniline) is first treated with $NaNO_2$ and $HX$ at cold temperatures (0-5°C) to form a diazonium salt. This diazonium salt is then treated with Cuprous halide ($Cu_2X_2$) to yield the corresponding haloarene.

$Ph-N_2^+X^- + Cu_2Cl_2 \rightarrow Ph-Cl (\text{Chlorobenzene}) + N_2\uparrow$

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