Chapter 10: Halogen Derivatives
Exhaustive Revision Guide - 50 Subjective Questions with Detailed Solutions
Click on any question to reveal its answer.
1. What are halogen derivatives of hydrocarbons? How are they classified based on the hydrocarbon skeleton?
Halogen Derivatives: Compounds obtained by replacing one or more hydrogen atoms of an aliphatic or aromatic hydrocarbon by the corresponding number of halogen atoms ($F, Cl, Br, I$) are called halogen derivatives.
Classification based on hydrocarbon skeleton:
- Haloalkanes: Halogen is attached to an $sp^3$ hybridized carbon of an alkyl group.
- Haloalkenes: Halogen is attached to an $sp^2$ hybridized carbon of an alkene.
- Haloalkynes: Halogen is attached to an $sp$ hybridized carbon of an alkyne.
- Haloarenes: Halogen is directly attached to an $sp^2$ hybridized carbon of an aromatic ring.
2. Classify monohalogen derivatives (alkyl halides) based on the type of carbon atom.
Monohalogen derivatives are classified into three types based on the nature of the carbon atom to which the halogen is attached:
- Primary ($1^\circ$) Alkyl Halide: Halogen is attached to a primary carbon (a carbon attached to one or zero other carbons). Example: $CH_3-CH_2-Cl$ (Ethyl chloride).
- Secondary ($2^\circ$) Alkyl Halide: Halogen is attached to a secondary carbon (attached to two other carbons). Example: $CH_3-CH(Cl)-CH_3$ (Isopropyl chloride).
- Tertiary ($3^\circ$) Alkyl Halide: Halogen is attached to a tertiary carbon (attached to three other carbons). Example: $(CH_3)_3C-Cl$ (tert-Butyl chloride).
3. Define allylic and benzylic halides with one example each.
- Allylic Halides: The halogen atom is bonded to an $sp^3$ hybridized carbon atom which is directly attached to a carbon-carbon double bond ($C=C$).
Example: $CH_2=CH-CH_2-Cl$ (Allyl chloride). - Benzylic Halides: The halogen atom is bonded to an $sp^3$ hybridized carbon atom which is directly attached to an aromatic ring.
Example: $C_6H_5-CH_2-Cl$ (Benzyl chloride).
4. Define vinylic halides and aryl halides with one example each.
- Vinylic Halides: The halogen atom is bonded directly to an $sp^2$ hybridized carbon atom of an aliphatic carbon-carbon double bond.
Example: $CH_2=CH-Cl$ (Vinyl chloride). - Aryl Halides (Haloarenes): The halogen atom is bonded directly to an $sp^2$ hybridized carbon atom of an aromatic ring.
Example: $C_6H_5-Cl$ (Chlorobenzene).
5. Explain the nature of the $C-X$ bond in alkyl halides.
Halogen atoms (X) are more electronegative than carbon. Therefore, the shared electron pair in the $C-X$ covalent bond is pulled closer to the halogen atom.
This creates a partial negative charge ($\delta^-$) on the halogen atom and a partial positive charge ($\delta^+$) on the carbon atom: $C^{\delta^+} - X^{\delta^-}$.
Because of this polarity, alkyl halides are moderately polar molecules and are highly reactive towards nucleophilic substitution.
6. Why is Thionyl chloride ($SOCl_2$) the most preferred reagent for preparing alkyl chlorides from alcohols?
Reaction: $R-OH + SOCl_2 \xrightarrow{\text{Pyridine, } \Delta} R-Cl + SO_{2(g)} \uparrow + HCl_{(g)} \uparrow$
Thionyl chloride is preferred because the by-products formed, Sulfur dioxide ($SO_2$) and Hydrogen chloride ($HCl$), are both gases. They easily escape from the reaction mixture, leaving behind pure alkyl chloride without the need for complex separation processes.
7. What is Lucas Reagent? How does it react with primary, secondary, and tertiary alcohols?
Lucas Reagent: An equimolar mixture of concentrated Hydrochloric acid ($HCl$) and anhydrous Zinc chloride ($ZnCl_2$). It is used to distinguish between $1^\circ, 2^\circ, \text{ and } 3^\circ$ alcohols based on the formation of alkyl chloride (which appears as turbidity).
- $3^\circ$ alcohols: React immediately at room temperature to form turbidity.
- $2^\circ$ alcohols: React slowly, forming turbidity after 5 to 10 minutes.
- $1^\circ$ alcohols: Do not react at room temperature (require heating to show turbidity).
8. Write the chemical equations for the reaction of Ethanol with (a) $PCl_3$ and (b) $PCl_5$.
- (a) With $PCl_3$:
$3C_2H_5OH + PCl_3 \xrightarrow{\Delta} 3C_2H_5Cl + H_3PO_3$ (Phosphorous acid) - (b) With $PCl_5$:
$C_2H_5OH + PCl_5 \xrightarrow{\Delta} C_2H_5Cl + POCl_3 + HCl$ (Phosphoryl chloride)
9. State Markownikoff's rule. Give an example.
Markownikoff's Rule: When an unsymmetrical reagent (like $HX$) adds to an unsymmetrical alkene, the negative part of the reagent ($X^-$) gets attached to that carbon atom of the double bond which carries the lesser number of hydrogen atoms.
Example: Addition of $HBr$ to Propene.
$CH_3-CH=CH_2 + H-Br \rightarrow CH_3-CH(Br)-CH_3$ (2-Bromopropane, Major product)
10. State Anti-Markownikoff's rule (Kharasch effect / Peroxide effect). Give an example.
Anti-Markownikoff's Rule: When $HBr$ is added to an unsymmetrical alkene in the presence of an organic peroxide ($R-O-O-R$), the addition occurs opposite to Markownikoff's rule. The negative part ($Br^-$) goes to the carbon with more hydrogen atoms. (Note: This effect is only observed with $HBr$).
Example:
$CH_3-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3-CH_2-CH_2-Br$ (1-Bromopropane, Major product)
11. Write a short note on Finkelstein reaction.
Finkelstein Reaction: It is a halogen exchange reaction used specifically for the preparation of alkyl iodides. Alkyl chlorides or bromides are heated with Sodium iodide ($NaI$) in dry acetone.
$R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX \downarrow \quad (X = Cl, Br)$
The forward reaction is favored because $NaCl$ and $NaBr$ are insoluble in dry acetone and precipitate out, driving the equilibrium to the right (Le Chatelier's principle).
12. Write a short note on Swarts reaction.
Swarts Reaction: It is a halogen exchange reaction used specifically for the preparation of alkyl fluorides. Alkyl chlorides or bromides are heated in the presence of metallic fluorides such as $AgF, Hg_2F_2, CoF_2, \text{ or } SbF_3$.
$R-X + AgF \xrightarrow{\Delta} R-F + AgX \downarrow \quad (X = Cl, Br)$
Example: $CH_3-Br + AgF \rightarrow CH_3-F \text{ (Methyl fluoride)} + AgBr$
13. How are aryl halides prepared by Sandmeyer's reaction?
A primary aromatic amine (like aniline) is treated with nitrous acid ($NaNO_2 + HCl$) at cold temperatures (273-278 K) to form a diazonium salt (Benzenediazonium chloride).
When this freshly prepared diazonium salt is mixed with cuprous chloride ($Cu_2Cl_2$) or cuprous bromide ($Cu_2Br_2$), the diazonium group is replaced by $-Cl$ or $-Br$ respectively, forming the aryl halide. This is Sandmeyer's reaction.
$C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_{2(g)}$
14. Explain the trend in boiling points of alkyl halides having the same alkyl group but different halogens.
For the same alkyl group, the boiling point increases with the increase in the atomic mass and size of the halogen atom.
Order: $RI > RBr > RCl > RF$
Reason: As the size of the halogen increases, the magnitude of the Van der Waals dispersion forces increases due to higher polarizability. Stronger intermolecular forces require higher temperatures to boil.
15. What is the effect of branching on the boiling point of isomeric alkyl halides?
Among isomeric alkyl halides, the boiling point decreases with an increase in branching.
Reason: Branching makes the molecule more compact and spherical. This decreases the surface area of contact between molecules, which in turn weakens the intermolecular Van der Waals forces. Weaker forces result in a lower boiling point. (e.g., n-butyl chloride > isobutyl chloride > tert-butyl chloride).
16. Alkyl halides are polar but are insoluble in water. Explain why.
Although alkyl halides are polar, they cannot form hydrogen bonds with water molecules. Also, the energy released when new solute-solvent ($R-X \dots H_2O$) attractions are set up is less than the energy required to break the strong hydrogen bonds between water molecules. Therefore, they are virtually insoluble in water, but soluble in organic solvents.
17. What is Plane Polarized Light?
Ordinary light consists of electromagnetic waves vibrating in all possible planes perpendicular to the direction of propagation. When ordinary light is passed through a Nicol prism, the emergent light vibrates in only one single plane. This light is called Plane Polarized Light.
18. Define Optical Activity. What are dextro and laevo rotatory substances?
Optical Activity: The property of certain substances to rotate the plane of plane-polarized light when it is passed through their solutions.
- Dextrorotatory ($d$ or $+$): A substance that rotates the plane of polarized light to the right (clockwise).
- Laevorotatory ($l$ or $-$): A substance that rotates the plane of polarized light to the left (anticlockwise).
19. What is a Chiral (asymmetric) carbon atom? Give an example.
Chiral Carbon Atom (C*): A carbon atom that is bonded to four different atoms or groups of atoms. A molecule containing one chiral carbon is asymmetric and optically active.
Example: In 2-Chlorobutane ($CH_3-C^*H(Cl)-CH_2CH_3$), the 2nd carbon is attached to $-H, -Cl, -CH_3,$ and $-CH_2CH_3$. Since all four groups are different, it is a chiral carbon.
20. Define Enantiomers. Mention their key properties.
Enantiomers: The stereoisomers which are non-superimposable mirror images of each other are called enantiomers.
Properties: They have identical physical properties (melting point, boiling point, refractive index) and identical chemical properties (except towards chiral reagents). Their only difference is the direction in which they rotate plane-polarized light: one rotates it to the right ($d$), the other to the left ($l$) by exactly the same angle.
21. What is a Racemic Mixture (Racemate)? Why is it optically inactive?
Racemic Mixture: An equimolar mixture of the $d$-enantiomer and $l$-enantiomer of an optically active compound. It is represented as $(dl)$ or $(\pm)$.
Reason for inactivity: It is optically inactive due to external compensation. The clockwise rotation caused by the $d$-isomer is exactly cancelled by the equal and opposite anticlockwise rotation caused by the $l$-isomer.
22. What is a Nucleophilic Substitution Reaction? Give the general equation.
Nucleophilic Substitution: A reaction in which a stronger nucleophile (electron-rich species) replaces a weaker nucleophile (leaving group) from the substrate molecule.
General Equation: $R-X + Nu^- \rightarrow R-Nu + X^-$
(Where $Nu^-$ is the incoming nucleophile and $X^-$ is the leaving halide ion).
23. Explain the kinetics of the $S_N2$ mechanism.
$S_N2$ (Substitution Nucleophilic Bimolecular): Example: Hydrolysis of Methyl bromide with aqueous NaOH.
The rate of the reaction depends on the concentration of both the alkyl halide and the nucleophile. It follows second-order kinetics.
Rate $\propto [CH_3Br][OH^-] \implies \text{Rate} = k[CH_3Br][OH^-]$
Because the rate determining step (the only step) involves the collision of two molecules, its molecularity is 2 (bimolecular).
24. Describe the mechanism and stereochemistry of the $S_N2$ reaction.
Mechanism: It is a single-step concerted mechanism. The nucleophile ($OH^-$) attacks the electrophilic carbon from the backside (exactly $180^\circ$ opposite to the leaving group, $Br^-$). This forms a high-energy transition state where carbon is pentacoordinate (partially bonded to both $OH$ and $Br$). The $C-Br$ bond breaks and $C-OH$ bond forms simultaneously.
Stereochemistry: The backside attack causes the umbrella-like turning inside out of the other three groups attached to the carbon. This results in 100% Inversion of Configuration (Walden inversion).
25. Explain the kinetics of the $S_N1$ mechanism.
$S_N1$ (Substitution Nucleophilic Unimolecular): Example: Hydrolysis of tert-Butyl bromide with aqueous NaOH.
The rate of the reaction depends only on the concentration of the alkyl halide and is independent of the nucleophile concentration. It follows first-order kinetics.
Rate $\propto [(CH_3)_3C-Br] \implies \text{Rate} = k[(CH_3)_3C-Br]$
Since the rate-determining step involves only one molecule, its molecularity is 1 (unimolecular).
26. Describe the mechanism of the $S_N1$ reaction.
It is a two-step mechanism:
- Step 1 (Slow, Rate Determining Step): The alkyl halide slowly undergoes heterolytic cleavage to form a halide ion and a stable carbocation intermediate.
$(CH_3)_3C-Br \xrightarrow{\text{slow}} (CH_3)_3C^+ \text{ (carbocation)} + Br^-$ - Step 2 (Fast): The nucleophile ($OH^-$) rapidly attacks the planar carbocation to form the product.
$(CH_3)_3C^+ + OH^- \xrightarrow{\text{fast}} (CH_3)_3C-OH$
27. Explain the stereochemistry of the $S_N1$ reaction.
In step 1, a carbocation is formed which is $sp^2$ hybridized and planar. Therefore, in step 2, the nucleophile can attack the planar carbocation from either the front side or the back side with equal probability.
Attack from the front yields a product with retention of configuration (50%). Attack from the back yields a product with inversion of configuration (50%). Consequently, if the starting alkyl halide is optically active, the product is a Racemic mixture (Racemization occurs).
28. Discuss the order of reactivity of alkyl halides ($1^\circ, 2^\circ, 3^\circ$) towards $S_N1$ and $S_N2$ mechanisms.
- $S_N2$ Reactivity: $CH_3X > 1^\circ > 2^\circ > 3^\circ$. It is governed by steric hindrance. Bulky alkyl groups block the back-side attack of the nucleophile. Hence, $1^\circ$ halides are most reactive via $S_N2$.
- $S_N1$ Reactivity: $3^\circ > 2^\circ > 1^\circ > CH_3X$. It is governed by the stability of the carbocation formed in the first step. Tertiary carbocations are most stable due to inductive (+I) and hyperconjugation effects. Hence, $3^\circ$ halides prefer $S_N1$.
29. Distinguish between $S_N1$ and $S_N2$ mechanisms. (Any 4 points)
| Property | $S_N1$ Mechanism | $S_N2$ Mechanism |
|---|---|---|
| Kinetics/Order | First order (Rate = $k[RX]$) | Second order (Rate = $k[RX][Nu^-]$) |
| Steps | Two steps | Single step (concerted) |
| Intermediate | Carbocation intermediate formed. | No intermediate; Transition state formed. |
| Stereochemistry | Racemization (Retention + Inversion) | 100% Inversion of configuration |
| Reactivity | $3^\circ > 2^\circ > 1^\circ$ | $1^\circ > 2^\circ > 3^\circ$ |
30. How does the nature of the solvent affect the rate of $S_N1$ and $S_N2$ reactions?
- $S_N1$ reactions are favored by polar protic solvents (like water, alcohol). These solvents solvate and stabilize the formed carbocation and the halide ion through hydrogen bonding, lowering the activation energy for the first step.
- $S_N2$ reactions are favored by polar aprotic solvents (like Acetone, DMF, DMSO). They solvate the cation of the nucleophile but leave the nucleophile (anion) free and highly reactive for the backside attack.
31. What happens when ethyl bromide is boiled with aqueous KOH vs alcoholic KOH?
- Aqueous KOH (Nucleophilic Substitution): $OH^-$ acts as a nucleophile. It replaces $Br$ to form an alcohol.
$C_2H_5-Br + KOH_{(aq)} \xrightarrow{\Delta} C_2H_5-OH \text{ (Ethanol)} + KBr$ - Alcoholic KOH (Elimination/Dehydrohalogenation): $OH^-$ acts as a strong base in alcoholic medium. It removes $H^+$ and $Br^-$ to form an alkene.
$CH_3-CH_2-Br + KOH_{(alc)} \xrightarrow{\Delta} CH_2=CH_2 \text{ (Ethene)} + KBr + H_2O$
32. Write the reaction of alkyl halides with KCN and AgCN. Why do they yield different products?
The cyanide ion ($CN^-$) is an ambidentate nucleophile (can attack via C or N).
- $R-X + KCN \xrightarrow{\Delta} R-CN \text{ (Alkyl cyanide / Nitrile)} + KX$
Reason: $KCN$ is predominantly ionic and provides free $CN^-$ ions. Attack through carbon is favored as the $C-C$ bond is stronger than the $C-N$ bond. - $R-X + AgCN \xrightarrow{\Delta} R-NC \text{ (Alkyl isocyanide / Isonitrile)} + AgX$
Reason: $AgCN$ is largely covalent. The carbon atom is tied up with Silver, so the lone pair on Nitrogen is available for nucleophilic attack, resulting in isocyanides.
33. Write the reaction of alkyl halides with KNO2 and AgNO2.
The nitrite ion ($NO_2^-$) is also an ambidentate nucleophile.
- $R-X + KNO_2 \xrightarrow{\Delta} R-O-N=O \text{ (Alkyl nitrite)} + KX$
($KNO_2$ is ionic, Oxygen attacks). - $R-X + AgNO_2 \xrightarrow{\Delta} R-NO_2 \text{ (Nitroalkane)} + AgX$
($AgNO_2$ is covalent, Nitrogen attacks).
34. What is a Dehydrohalogenation ($\beta$-elimination) reaction? Give an example.
When an alkyl halide having at least one $\beta$-hydrogen is boiled with an alcoholic solution of a strong base (like KOH), the halogen atom from the $\alpha$-carbon and a hydrogen atom from the adjacent $\beta$-carbon are eliminated. This forms a double bond, resulting in an alkene.
Example: $CH_3-CH_2-Br + KOH_{(alc)} \xrightarrow{\Delta} CH_2=CH_2 + KBr + H_2O$
35. State Saytzeff's (Zaitsev's) Rule. Give an example.
Saytzeff's Rule: In dehydrohalogenation reactions, if more than one alkene can be formed, the highly substituted alkene (the one with the greater number of alkyl groups attached to the doubly bonded carbon atoms) is the major product because it is more stable.
Example: Elimination of 2-Bromobutane with alc. KOH yields two products:
Major: $CH_3-CH=CH-CH_3$ (But-2-ene, more substituted)
Minor: $CH_2=CH-CH_2-CH_3$ (But-1-ene, less substituted)
36. How is Grignard Reagent prepared? Write the chemical equation.
Alkyl halides react with active metals like Magnesium in the presence of pure and dry ether to form alkyl magnesium halides ($R-Mg-X$), which are known as Grignard Reagents.
$$R-X + Mg \xrightarrow{\text{dry ether}} R-Mg-X$$
Example: $CH_3-CH_2-Br + Mg \xrightarrow{\text{dry ether}} CH_3-CH_2-Mg-Br$ (Ethyl magnesium bromide)
37. Why is it strictly necessary to avoid even traces of moisture during the preparation of Grignard reagent?
Grignard reagents ($R-MgX$) are highly reactive compounds. The C-Mg bond is highly polar ($R^{\delta-} - Mg^{\delta+}X$). If even a trace of moisture ($H_2O$) is present, the strongly basic alkyl group ($R^-$) immediately abstracts a proton ($H^+$) from water to form an alkane, destroying the reagent.
$$R-Mg-X + H-OH \rightarrow R-H \text{ (Alkane)} + Mg(OH)X$$
38. Write a short note on Wurtz Reaction. Give an example.
Wurtz Reaction: Alkyl halides react with metallic sodium in the presence of dry ether to form higher alkanes containing double the number of carbon atoms present in the alkyl halide.
$$2R-X + 2Na \xrightarrow{\text{dry ether}} R-R \text{ (Alkane)} + 2NaX$$
Example: $2CH_3-Br + 2Na \xrightarrow{\text{dry ether}} CH_3-CH_3 \text{ (Ethane)} + 2NaBr$
39. Why are aryl halides extremely less reactive towards nucleophilic substitution reactions compared to alkyl halides? (Give two reasons).
- Resonance Effect: In haloarenes (e.g., chlorobenzene), the lone pairs of electrons on the halogen atom are in conjugation with the $\pi$-electrons of the benzene ring. This imparts a partial double bond character to the $C-X$ bond, making it shorter and much harder to break.
- $sp^2$ Hybridized Carbon: The $C-X$ bond in haloarenes is attached to an $sp^2$ hybridized carbon, which is more electronegative than the $sp^3$ carbon in alkyl halides. This holds the electron pair of the $C-X$ bond more tightly, making cleavage difficult.
40. Describe the Wurtz-Fittig reaction.
Wurtz-Fittig Reaction: A mixture of an alkyl halide and an aryl halide reacts with sodium metal in dry ether to form an alkylbenzene.
$C_6H_5-X \text{ (Aryl halide)} + 2Na + X-R \text{ (Alkyl halide)} \xrightarrow{\text{dry ether}} C_6H_5-R \text{ (Alkylbenzene)} + 2NaX$
Example: Chlorobenzene + Methyl chloride + 2Na $\rightarrow$ Toluene + 2NaCl.
41. Describe the Fittig reaction.
Fittig Reaction: When only aryl halides are treated with sodium metal in dry ether, two aryl groups join together to form a diaryl (or biaryl) compound.
$2C_6H_5-X + 2Na \xrightarrow{\text{dry ether}} C_6H_5-C_6H_5 \text{ (Diphenyl / Biphenyl)} + 2NaX$
42. In electrophilic substitution reactions of Chlorobenzene, the incoming electrophile is directed to ortho and para positions. Why?
The Chlorine atom attached to the benzene ring has a +R (resonance) effect. It donates its lone pair of electrons to the ring. This resonance increases the electron density specifically at the ortho and para positions relative to the halogen atom. Since electrophiles ($E^+$) seek electron-rich centers, they preferentially attack the ortho and para positions. Hence, halogens are ortho-para directing groups.
43. Write the chemical equation for the Nitration of Chlorobenzene. Identify major and minor products.
Chlorobenzene reacts with a nitrating mixture (conc. $HNO_3$ + conc. $H_2SO_4$) to introduce the nitro group ($-NO_2$).
$C_6H_5Cl + HNO_3 \xrightarrow{\text{conc. } H_2SO_4, \ \Delta}$
- 1-Chloro-4-nitrobenzene: (Para isomer) - Major product due to less steric hindrance.
- 1-Chloro-2-nitrobenzene: (Ortho isomer) - Minor product.
44. Write the chemical equation for Friedel-Crafts alkylation of Chlorobenzene.
Chlorobenzene reacts with an alkyl halide (e.g., Methyl chloride, $CH_3Cl$) in the presence of anhydrous Aluminum chloride ($AlCl_3$) as a Lewis acid catalyst.
$C_6H_5Cl + CH_3Cl \xrightarrow{\text{Anhyd. } AlCl_3}$
- 1-Chloro-4-methylbenzene (p-Chlorotoluene): Major product.
- 1-Chloro-2-methylbenzene (o-Chlorotoluene): Minor product.
45. How will you convert Propene to 1-Bromopropane?
This requires Anti-Markownikoff addition, so we add $HBr$ in the presence of an organic peroxide.
$$CH_3-CH=CH_2 \text{ (Propene)} + HBr \xrightarrow{\text{Peroxide}} CH_3-CH_2-CH_2-Br \text{ (1-Bromopropane)}$$
46. How will you convert Ethyl bromide to Ethene?
This requires a dehydrohalogenation (elimination) reaction. Boil ethyl bromide with alcoholic KOH.
$$CH_3-CH_2-Br + KOH_{(alc)} \xrightarrow{\Delta} CH_2=CH_2 \text{ (Ethene)} + KBr + H_2O$$
47. How will you convert Toluene to Benzyl chloride?
This is a side-chain free-radical halogenation. Treat toluene with chlorine gas in the presence of sunlight (UV) or heat.
$$C_6H_5-CH_3 + Cl_2 \xrightarrow{UV \text{ light or } \Delta} C_6H_5-CH_2Cl \text{ (Benzyl chloride)} + HCl$$
48. What are Freons? Mention their environmental impact.
Freons: The chlorofluorocarbon (CFC) compounds of methane and ethane are collectively known as freons. They are extremely stable, unreactive, non-toxic, non-corrosive, and easily liquefiable gases. (Example: Freon-12, $CCl_2F_2$). They are widely used as refrigerants in refrigerators and ACs.
Environmental Impact: In the stratosphere, Freons absorb UV radiation and break down to produce chlorine free radicals. These radicals catalyze the destruction of the ozone layer, leading to ozone depletion.
49. What is DDT? State its use and environmental effect.
DDT stands for 1,1,1-trichloro-2,2-bis(p-chlorophenyl)ethane. It is a powerful contact insecticide, originally used to combat malaria-carrying mosquitoes and typhus-carrying lice.
Environmental Effect: It is non-biodegradable and highly stable. It persists in the environment and accumulates in the fatty tissues of animals (biomagnification), posing severe toxicity risks to wildlife, especially birds (causes thinning of eggshells) and humans. Its use is now banned in many countries.
50. Write the chemical formula and use of Iodoform and BHC.
- Iodoform ($CHI_3$): It is a pale yellow solid with a characteristic unpleasant smell. It was earlier used as an antiseptic, especially for dressing wounds. Its antiseptic property is due to the liberation of free iodine, not iodoform itself.
- BHC (Benzene Hexachloride, $C_6H_6Cl_6$): Also known as Gammexane or Lindane. Its gamma isomer is a very powerful insecticide and pesticide used in agriculture.
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