Chapter 10: Halogen Derivatives
Highly Detailed Mark-wise Solutions for Board Exam Preparation
1 Mark Questions (Very Short Answer)
Answer: A carbon atom in a molecule which is bonded to four completely different atoms or groups of atoms is called a chiral carbon atom (or asymmetric carbon atom).
Answer: Swarts Reaction. (Heating alkyl chlorides/bromides with metallic fluorides like $AgF$, $Hg_2F_2$, or $CoF_3$).
Answer: An equimolar mixture of two enantiomers (dextrorotatory and laevorotatory forms of a compound) which is optically inactive due to external compensation is called a racemic mixture (denoted by $\pm$ or $dl$).
Answer: Thionyl chloride ($SOCl_2$). The reaction (Darzen's method) yields $SO_2$ and $HCl$ as by-products, which are gases and escape easily, leaving behind pure alkyl chloride.
2 Mark Questions (Short Answer-I)
Answer:
When an alkyl halide is treated with metallic sodium in the presence of dry ether, it undergoes a coupling reaction to form a symmetrical higher alkane containing double the number of carbon atoms present in the original alkyl halide. This is known as the Wurtz reaction.
$2 CH_3-Br + 2Na \xrightarrow{\text{Dry ether}} CH_3-CH_3 + 2NaBr$
(Methyl bromide) (Ethane)
Answer:
| Enantiomers | Diastereomers |
|---|---|
| 1. They are stereoisomers that are non-superimposable mirror images of each other. | 1. They are stereoisomers that are not mirror images of each other. |
| 2. They have identical physical properties (except the direction of rotation of plane-polarized light). | 2. They have different physical properties (melting point, boiling point, density). |
Answer:
Statement: When an unsymmetrical reagent (like $HX$) is added to an unsymmetrical alkene, the negative part of the reagent (e.g., $X^-$) gets attached to that carbon atom of the carbon-carbon double bond which carries a lesser number of hydrogen atoms.
Illustration:
Addition of HBr to Propene yields 2-Bromopropane as the major product.
$CH_3-CH=CH_2 + H-Br \rightarrow CH_3-CH(Br)-CH_3$
(Propene) (2-Bromopropane - Major)
3 Mark Questions (Short Answer-II / Mechanisms)
Answer:
| $S_N1$ Mechanism | $S_N2$ Mechanism |
|---|---|
| 1. Kinetics: First-order kinetics. Rate depends only on substrate concentration. ($\text{Rate} = k[RX]$). | 1. Kinetics: Second-order kinetics. Rate depends on both substrate and nucleophile. ($\text{Rate} = k[RX][Nu^-]$). |
| 2. Steps: It is a two-step process. | 2. Steps: It is a single-step concerted process. |
| 3. Intermediate: Proceeds via the formation of a stable carbocation intermediate. | 3. Intermediate: No intermediate is formed. It passes through a highly unstable transition state. |
| 4. Stereochemistry: Results in racemization (mixture of retention and inversion). | 4. Stereochemistry: Results in 100% complete inversion of configuration (Walden Inversion). |
| 5. Reactivity Order: $3^\circ > 2^\circ > 1^\circ > CH_3X$ (governed by carbocation stability). | 5. Reactivity Order: $CH_3X > 1^\circ > 2^\circ > 3^\circ$ (governed by steric hindrance). |
| 6. Solvent: Favored by polar protic solvents (like water, alcohols). | 6. Solvent: Favored by non-polar or polar aprotic solvents (like acetone). |
Answer:
Dehydrohalogenation:
The process in which a hydrogen atom and a halogen atom are eliminated from adjacent carbon atoms of an alkyl halide to form an alkene is called dehydrohalogenation (or $\beta$-elimination). It occurs when an alkyl halide is heated with alcoholic KOH.
Saytzeff's Rule:
When an unsymmetrical alkyl halide undergoes dehydrohalogenation to yield more than one alkene, the major product is the alkene that has the greater number of alkyl groups attached to the doubly bonded carbon atoms (i.e., the highly substituted alkene is the major product).
Illustration (2-bromobutane):
When 2-bromobutane is heated with alcoholic KOH, the $\beta$-hydrogen can be eliminated from $C_1$ or $C_3$.
Elimination from $C_3$:
$CH_3-CH_2-CH(Br)-CH_3 \xrightarrow{\text{alc. } KOH, \Delta} CH_3-CH=CH-CH_3$ (But-2-ene, Major 80%)
Elimination from $C_1$:
$CH_3-CH_2-CH(Br)-CH_3 \xrightarrow{\text{alc. } KOH, \Delta} CH_3-CH_2-CH=CH_2$ (But-1-ene, Minor 20%)
But-2-ene is the major product as it has two alkyl (methyl) groups attached to the $C=C$ bond, whereas But-1-ene has only one (ethyl) group.
4 Mark Questions (Long Answer / Combined)
Note: In the HSC Board Exam, 4-mark questions in Chemistry are frequently split into sub-questions (e.g., a 2-mark mechanism combined with a 2-mark reasoning question).
(b) Why is Grignard reagent prepared under strictly anhydrous conditions? [2 Marks] March 2019, March 2021
Answer (a): Reactivity of Haloarenes
Haloarenes (e.g., Chlorobenzene) are very unreactive towards nucleophiles due to the following reasons:
- Resonance Effect: The lone pair of electrons on the halogen atom is in conjugation with the $\pi$-electrons of the benzene ring. This gives the $C-X$ bond a partial double bond character. Since a double bond is shorter and stronger than a single bond, it is very difficult for a nucleophile to break it.
- Hybridization of Carbon: In haloarenes, the halogen is attached to an $sp^2$ hybridized carbon atom (which is more electronegative), whereas in haloalkanes it is attached to an $sp^3$ carbon. The $sp^2$ carbon holds the electron pair of the $C-X$ bond more tightly, preventing its cleavage.
Answer (b): Grignard Reagent Condition
Grignard reagents ($R-Mg-X$) are extremely highly reactive compounds and act as very strong bases (sources of carbanions, $R^-$). If even a trace amount of moisture (water) is present, the Grignard reagent will immediately react with the active hydrogen of water to form an alkane, thereby destroying the reagent.
$R-Mg-X + H_2O \rightarrow R-H \text{ (Alkane)} + Mg(OH)X$
To prevent this unwanted side reaction, it is absolutely necessary to prepare and store Grignard reagents under strictly anhydrous (dry) conditions using dry ether as a solvent.
(b) Define optical activity. [1 Mark] Oct 2013, March 2023
Answer (a): $S_N2$ Mechanism
Reaction:
$CH_3-Br + OH^- \rightarrow CH_3-OH + Br^-$
Kinetics:
The rate of the reaction depends on the concentration of both methyl bromide and the hydroxide ion. Hence, it is a second-order reaction.
$\text{Rate} \propto [CH_3Br][OH^-]$
Mechanism:
- The reaction proceeds in a single step. There is no intermediate formed.
- The nucleophile ($OH^-$) attacks the electrophilic carbon atom from the back side, exactly opposite to the leaving group ($Br^-$) to avoid steric hindrance and electronic repulsion.
- As the $OH^-$ starts forming a bond with carbon, the $C-Br$ bond simultaneously starts breaking. This leads to the formation of a highly unstable Transition State where the carbon atom is partially bonded to five atoms.
- Finally, the $Br^-$ ion completely departs, and the $C-OH$ bond is fully formed. Because the attack happens from the back, the configuration of the molecule flips inside out like an umbrella in a strong wind. This is called 100% Inversion of Configuration (Walden Inversion).
Answer (b): Theory
Optical Activity: The property of certain organic substances (having chiral molecules) to rotate the plane of plane-polarized light either to the right (clockwise) or to the left (anti-clockwise) when it is passed through their solutions is called optical activity.
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