Alkyl halides or halo alkanes are compounds in which a halogen is bonded to an alkyl group. They have the general formula RX (where R is alkyl group CnH2n+1) X is halogen atom.
Alkyl halides are classified as primary, secondary and tertiary alky halides depending on whether the halogen atom is attached to a primary, secondary or tertiary carbon atom respectively.
Aromatic halogen compounds or halo arenes are the halogen compounds which contain atleast one aromatic ring. Halogen derivatives of aromatic compounds are of two types.
- Aryl halides
|In this type of compounds, the halogen atom is directly linked to the carbon of benzene nucleus e.g.|
- Aralkyl halides
|In this type of compounds halogen is linked to the carbon atom of the side chain.|
NOMENCLATURE OF ALKYL & ARYL HALIDES
In the common system, aliphatic halogen derivatives are named as alkyl halides. The words
n⎯, sec⎯, tert⎯, iso⎯, neo⎯ & amyl are usually used in writing the common names. In IUPAC nomenclature they are named as halo alkanes.
|Formula||Common name||IUPAC name|
|CH3Cl||Methyl chloride||Chloro methane|
|CH3CH2CH2Cl||n – propyl chloride||1 – chloro propane|
|Isopropyl chloride||2 – chloro propane|
The dihalogen derivatives having same type of halogen atoms on the same carbon are known as geminal dihalides and are assigned common name alkylidene halides.
The dihalogen derivatives having 2 similar halogen atoms on adjacent carbon atoms are known as vicinal dihalides & are assigned common name alkylene halides.
Tri halo methanes are called haloforms in trivial system.
|Formula||Common name||IUPAC name|
|Ethylene chloride||1, 2 – dichloro ethane|
|Ethylidene chloride||1, 1 – dichloro ethane|
Haloarenes are named by prefixing the halogen and its position.
Illustration 1. Write structural formulae of 3 Bromo hexane.
Illustration 2. Write IUPAC name of following compounds.
Solution: (i) 1, 2 – dichloropropane
(i) Write IUPAC name for
(ii) Give structural formulae of
(a) Ethylene dibromide
(b) 2 – chloro – 3, 3 – dimethyl butane
(i) From Alkanes: Alkanes react with halogens in the presence of light to give alkyl halides.
Alkyl halides formed further react with halogen to give di, tri and tetra halogen compounds.
(ii) From alkenes: Alkenes add halogen acids to give halides. For example
Markonikoff’s rule: In the addition reactions of unsymmetrical alkenes the −ve part attaches to the carbon atom having lesser number of H-atoms. E.g.
In case of HBr if peroxide is added antimarkonikoff’s addition takes place which is also called Kharash effect or peroxide effect e.g.
(iii) From silver salt of carboxylic acids:
This reaction is called Borodine Hunsdiecker reaction.
(iv) Finkelstein reaction: Alkyl chlorides and bromides reacts with NaI in acetone to give alkyl iodides.
This reactions is possible because NaI is soluble in acetone but NaCl and NaBr are insoluble in acetone.
Illustration 3. Dry gaseous hydrohalogen acid and not their aqueous solutions are used to prepare alkyl halides from alkenes.
Solution: Dry gaseous halogen acids are better electrophiles than formed in water solutions. Also in aqueous solution H2O acting as nucleophile may produce alcohol.
- Boiling Points
The boiling points of haloalkanes are in the order RCl < RBr < RI. It is because with increase in size and mass of halogen atom the magnitude of Vander Waal’s forces of attraction increases.
Among isomeric alkyl halides, the boiling point decreases with increase in branching in alkyl group.
e.g the decreasing order of boiling point among the isomers of butane is
For same halogen, the boiling point increases with increase in molecular mass.
e.g. CH3Cl has lower boiling point than CH3CH2Cl
The boiling points of various halogen compounds increase with increase in number of halogen atoms.
For e.g. boiling point of CCl4 is more than boiling point of CHCl3 which is further more than CH2Cl2
- Boiling point
The boiling points of mono halogen derivatives of benzene follows the order:
Iodo > Bromo > Chloro
The boiling points of isomeric dihalo benzenes are nearly the same. However their melting points are quite different. The melting point of para isomer is generally 70 – 100 degrees more than the melting points of ortho & meta isomers.
B.P 453 K 446 K 448 K
M.P. 256 K 249 K 325 K
The higher melting point of p – isomer is due to its symmetry which leads to more close packing of its molecules in the crystal lattice.
Haloarenes are insoluble in water, acids or base but are soluble in organic solvents.
Haloarenes are insoluble in water because they can not form hydrogen bonds with water molecules.
They are all heavier than water. Their densities follow the order:
Iodo > Bromo > Chloro
Illustration 4. Explain why chloroform (CHCl3) is not soluble in water although it is polar.
Solution: Chloroform is insoluble is water inspite of its polar nature because its molecules can not form hydrogen bonds with water.
Illustration 5. Arrange the following in increasing order of density
CHCl3, CH2Cl2, CCl4, CH3Cl
Solution: CH3Cl < CH2Cl2 < CHCl3 < CCl4
Illustration 6. Melting and boiling points of alkyl halides are higher then their corresponding alkanes. Why?
Solution: The higher molecular mass and polar nature of alkyl halides is the reason of high melting and boiling points which are in the order
(a) p-dichlorobenzene is solid at room temperature whereas ‘o’ and
‘m’ – derivatives are liquids.
(b) Haloalkanes are used as solvents in industry are chloro compounds rather than bromo compounds.
(c) Ethyl chloride is a gas whereas ethyl iodide is a liquid at room temperature. Why?
(i) Arrange the following in order of increasing boiling points
bromobenzene, chlorobenzene, iodobenzene
(ii) Explain why haloarenes are insoluble in water but are soluble in benzene.
Haloalkanes are highly reactive class of aliphatic compounds. Their reactivity is due to the presence of polar carbon – halogen bond in their molecule. In general for a given alkyl group, the order of reactivity decreases as:
Iodides > Bromides > Chlorides
The explanation of above order is that reaction of alkyl halides involve cleavage of C – X bond. So higher the bond dissociation energy smaller reactivity & the bond dissociation energy decrease with increase in size of halogen atom.
The chemical reactions of halo alkanes are of 4 types:
- Nucleophilic substitution reactions
- Dehydro halogenation reactions
- Reactions with metals
- Reduction reactions
NUCLEOPHILIC SUBSTITUTION REACTION
|In halo alkanes, the halogen atom is attached to the carbon atom. As the halogen atom is more electronegative than carbon, the bond between carbon & halogen is polar in character.|
Due to the presence of partial positive charge on the carbon atom, the nucleophiles can attack on electron deficient carbon.
The order of reactivity of various alkyl halides towards nucleophilic substitution is in the order:
RI > RBr > RCl > RF
Nucleophilic substitution reactions in halides containing bond may take place through either of the two different mechanism – SN1& SN2.
SN1 mechanism (Unimolecular Nucleophilic Substitution)
In this type the rate of reaction is dependent only on the concentration of alkyl halide i.e.
Rate = k[RX]
In this step the alkyl halide slowly dissociates into halide ion & carbocation.
In the 2nd step carbocation at once combines with the nucleophile to form the final substituted product.
The order of reactivity of various alkyl halides through SN1 mechanism is
30 > 20 > 10
Allylic & benzylic halides show greatest reactivity through SN1 mechanism due to stability of allylic & benzylic carbocations.
SN2 Mechanism (Bimolecular Nucleophilic Substitution)
In this type the rate of reaction is dependent on the concentration of alkyl halide as well as nucleophile i.e.
Rate = K[RX] [Z−]
Primary alkyl halides react by SN2 mechanism via formation of transition state.
The order of reactivity of various alkyl halides through SN2 mechanism is
10 > 20 > 30
Factors affecting SN1 & SN2 Mechanism
The reaction mechanism, SN1 or SN2, followed by nucleophilic substitution depends upon a number of factors. These factors are
- Nature of alkyl halides: Primary alkyl halides react through SN2 & tertiary alkyl halides through SN1 mechanism.
- Nature of Nucleophile: Strong nuclophile favour SN2 mechanism whereas weak nucleophile favours SN1 mechanism.
- Concentration of Nucleophile: High concentration of nucleophile favours SN2 while low concentration favours SN1 mechanism.
- Nature of Solvent: Polar solvents favour SN1 mechanism.
Illustration 7. Why are alkyl halides very reactive?
Solution: Cδ+ − Xδ− bond in alkyl halides is polar and the negative charge on halogen atom is intensified by the positive inductive effect of alkyl groups.
Illustration 8. Amongst the following the most reactive alkyl halide is
(A) C2H5F (B) C2H5Cl
(C) C2H5Br (D) C2H5I
Illustration 9. Which of the following alkyl halides is hydrolysed by SN2 mechanism?
(A) C6H5CH2Br (B) CH3Br
(C) CH2 = CHCH2Br (D) (CH3)3CBr
Illustration 10. Which compound has a faster rate of reaction with HCl :
Solution: More stable the carbocation formed faster is the reaction.
Vinyl chloride does not give SN reaction but allyl chloride gives. Explain.
Some nucleophilic reactions are as follows:
- Replacement by hydroxyl group (formation of alcohols)
Haloalkanes on treatment with aqueous solution of KOH or moist silver oxide give alcohol.
- Replacement by Alkoxy (Formation of ethers) (Williamson’s synthesis)
Haloalkanes on treatment with alcoholic sodium or potassium hydroxide form ethers. This reaction is known as Williamson’s synthesis .
Exercise 5. What are the products of the following reactions?
- Replacement by cyano group
Haloalkanes on treatment with alcoholic KCN give alkyl nitriles or alkyl cyanides as major product.
When CH3—CH=CH–CH2Cl reacts with alcoholic KCN, a mixture of isomeric products is obtained. Explain.
- Replacement with Isocyanide Group (Formation of isocyanides)
On reaction with alcoholic silver cyanide solution, haloalkanes give alkyl carbylamines or alkyl isocyanides as the major product along with or small amount of alkyl cyanide.
RX + alc. AgCN ⎯⎯⎯→ R ⎯ NC + AgX
- Replacement by Amino Group (Formation of amines)
On heating haloalkanes with alcoholic ammonia solution in a sealed tube, halogen is replaced by ⎯NH2 group to form primary amine.
R ⎯ X + NH3 (alc.) ⎯⎯⎯→ R ⎯ NH2 + HX
In case haloalkanes is in excess, the other 2 hydrogen atoms of amino group are also replaced by alkyl groups leading to the formation of secondary & tertiary amines.
C2H5 ⎯ Br + HNHC2H5 ⎯⎯⎯→ (C2H5)2NH + HBr
- Replacement by Nitro group (Formation of nitro alkanes)
On treating ethanolic solution of haloalkanes with silver nitrite (Ag ⎯ O ⎯ N = O), nitro alkane is formed.
It is because the bond between Ag ⎯ O being covalent, the lone pair on nitrogen act as attacking site for nucleophilic substitution.
- Replacement by Nitrite group (Formation of alkyl nitrites)
On treatment of haloalkanes with potassium nitrite alkyl nitrite is formed.
R ⎯ X + KNO2 ⎯⎯⎯→ R ⎯ O ⎯ N = O + KX
- Replacement of halogens by Mercaptide (:SR Group)
On treating haloalkanes with sodium mercaptide, thio ethers are formed.
- Replacement by ⎯SH (Hydrosulphide) group (Formation of thiols or mercaptals)
On treating haloalkanes with or thioalcohols are formed.
- Replacement by Alkynyl Group (Formation of higher alknes)
On treating halo alkanes with sodium alkynide , higher alkynes are formed.
- Replacement by Carboxylate Group (Formation of esters)
Haloalkanes on treatment with silver salt of carboxylic acids in ethanol give esters.
- Replacement by Hydride Ion
Alkyl halides on reaction with lithium aluminium hydride in the presence of dry ether as solvent yield corresponding hydrocarbon.
Dehydrohalogenation Reactions or β – elimination Reactions
When halo alkanes are heated with alcoholic KOH, they undergo dehydrohalogenation to form alkanes. These reactions are called β – elimination because the hydrogen atom present at
β – position of halo alkanes is removed.
CH3 ⎯ CH2 ⎯ Br + KOH (alc.) ⎯⎯⎯→ H2C = CH2 + KBr + H2O
The reactivity of haloalkanes towards elimination reaction follows the order
Tertiary > secondary > Primary
This is because tertiary alkyl halides on dehydrohalogenation form most substituted alkenes which are more stable & are formed at faster rate.
Among various halides with same alkyl group the order of reactivity is
RI > RBr > RCl
In case the haloalkanes can eliminate hydrogen halide in 2 – different ways, the preferred alkane is the one which is maximum alkylated (most substituted).
Illustration 11. What is the function of anhyd. ZnCl2 in the reaction of alcohols with conc. HCl (or Lucas reagent)?
Solution: The function of anhydrous zinc chloride is to help in the cleavage of C – O bond. Being a lewis acid anhydrous ZnCl2 co-ordinates to the O-atoms of R – OH and thus weakens the C – O bond which then breaks to give carbocation. Moreover anhyd. ZnCl2 acts as a dehydrating agent and helps the reaction to go in the forward direction.
Reactions with Metals
- Reaction with sodium (Wurtz reaction)
Haloalkanes react with sodium in the presene of ether to form alkanes.
- Reaction with Magnesium
Haloalkanes react with magnesium in the presence of dry ether to form alkyl magnesium halide (Grignard reagents)
Grignard reagents are organometallic compounds, i.e. compounds having metal carbon bond. Grignard regents are highly reactive. They react with proton donors (acids) to give hydrocarbons.
- Reaction with Lithium
Haloalkanes react with lithium in the presence of dry ether to form alkyl lithium. These salts serve as strong bases.
- Reaction with lead sodium Alloy
Ethyl bromide react with lead sodium alloy in the presence of dry ether to form tetra ethyl lead (TEL).
Haloalkanes can be reduced to alkanes by any of following methods
- Reaction with H2/Ni
- Reaction with zinc copper couple
In the presence of alcohols, Zn – Cu couple reduces haloalkanes to alkanes.
Illustration 12. The order of reactivity of alkyl halides towards elimination reaction is :
Illustration 13. The end product “Z” in the following reaction, ethylamine is
(A) Methyl amine (B) Acetamide
(C) Ethylamine (D) Propylamine
Illustration 14. Why Grignard reagents should be prepared under anhydrous conditions?
Solution: Grignard reagents react with water & get decomposed (hydrolysed) hence they should be prepared under anhydrous conditions.
Illustration15. Predict the major product in each of following reaction.
Solution: Ph ⎯ C ≡ C ⎯ Ph
Illustration 16. Alkyl iodides darken on standing. Why?
Solution: Iodides being less stable, lose I2. The liberated iodine is absorbed by iodides which results in darkening of colour.
Illustration 17. Iodine reacts with alcohols to form alkyl iodide only in the presence of phosphorous. Explain why?
Solution: Phosphorus reacts with I2 to give PI3 which replaces OH group of alcohol to produce R − I.
(i) Arrange the following in order of increasing reactivity towards nucleophilic substitutions: – (CH3)2CHBr, (CH3)2CHI, (CH3)2CHCl.
(ii) Why neo-pentyl bromide undergoes nucleophilic substitution reactions very slowly?
Difference in Reactivity of C – X bond in Alkyl halides & Aryl halides
Aryl halides are much less reactive towards nucleophilic substitution reaction than haloalkanes. The less reactivity of aryl halides can be explained as follows:
- Withdrawal of Electrons by benzene & stabilization by resonance
In aryl halide, the electron pair of halogen atom is in conjugation with π electrons of benzene ring. Thus halobenzene is a resonance hybrid of following structures:
The contributing structures II, III & IV indicate that C ⎯ X bond has partial double bond characters.
As a resultant the C ⎯ X bond in halobenzene is shorter & hence stronger as compared to that in alkyl halides. Thus cleavage of C ⎯ X bond in halobenzene becomes difficult which makes it less reactive towards nucleophilic substitution.
- Different hybrid states of carbon atom
In haloalkanes, the carbon atom bearing halogen is sp3 hydridized while halogen bearing carbon atom is sp2 hybrized in haloarenes. sp2 hybrid orbital is smaller in size due to greater
s – character. As a result bond in haloarenes is smaller & is cleaved with difficulty.
- Polarity of C ⎯ X Bond
The C ⎯ X bond in halo alkanes is more polar than the C ⎯ X bond in haloarenes. Now greater the polarity of bond more is the reactivity.
Illustration18. Benzyl chloride is more reactive than chlorobenzene towards nuclephilic substitution. Explain.
Solution: In the molecule of benzyl chloride, the electron pairs on the chlorine atom are not in a position to conjugate with the π-electrons of the ring.
whereas electron pairs on the chlorine atom are in conjugation with the
π-electrons of the ring in the molecule of chlorobenzene. Consequently the
C-Cl bond in chlorobenzene acquires some double bond character which is not present in benzyl chloride. As a result, the reactivity of chlorobenzene towards nucleophilic substitution is much less than that of benzyl chloride.
Illustration 19. It is difficult to remove halogen from the aromatic ring by an attack of KOH. Why?
Solution: Only at 600 K and 300 K (very drastic conditions) Cl atoms of chlorobenzene is replaced by −OH on treatment with KOH. Haloarenes are less reactive due to resonance stabilization.
Reactions of Aryl Halides
Aryl halides being less reactive, can be made to react under drastic conditions.
Nucleophilic Substitution Reactions
- Replacement by Hydroxyl group
On heating chlorobenzene with an aq. solution of NaOH at 623 K under 300 atm pressure, sodium phenoxide is formed which on subsequent acidification produces phenol.
- Replacement by Cyano Group
When heated with anhydrous CuCN in the presence of pyridine or dimethyl formamide at 470 K, bromo benzene gives cyanobenzene.
- Replacement by Amino Group
Halogen atom of haloarenes is replaced by amino group by reacting it with aq. NH3; & in presence of catalyst, Cu2O
Reactions with metals
- Action with magnesium
Aryl bromides & iodides form Grignard’s reagents with Mg in dry ether.
- Reaction with Sodium
Aryl halides react with sodium in the presence of ether. During reaction two phenyl rings unite. The reaction is called Fittig reaction.
However aryl halides when treated with halo alkane & sodium in dry ether undergo Wurtz fitting reaction.
- Reaction with Lithium
Aryl halides react with lithium metal to form the corresponding organometallic compound.
These organometallic compounds behave like Grignard reagents.
- Reaction with Copper Powder
Iodo benzene when heated with copper powder in a sealed tube gives diphenyl
An aryl halide is reduced to parent hydrocarbon by the action of nickel aluminium alloy in the presence of an alkali.
The above transformation proceed through
(A) Electrophilic addition (B) Benzyne intermediate
(C) Activated nucleophilic substitution (D) Oxirane
The compound X is
(A) Phenol (B) Benzene
(C) o & p – chlorophenol (D) Benzol
Illustration 22. Which out of o – chloro nitrobenzene and 1 – chloro – 2, 4, 6 – tri nitrobenzene is more reactive towards nucleophilic substitution.
Solution: 1 – chloro – 2, 4, 6 – trinitrobenzene
Write equations for
(i) Borodine Hunsdiecker reaction
(ii) Finkelstein reaction
(iii) Baltz-Schimann reaction
RING SUBSTITUTION REACTIONS
An aryl halide undergoes electrophilic substitution reactions in benzene ring. The presence of halogen atom in the ring directs the incoming substituent to the ortho & para position.
Aryl halides are less reactive than benzene towards electrophilic substitution reactions.
Some ring substitution reactions of aryl halides are given below:
Halogentation takes place in the presence of iron or FeCl3 or anhydrous AlCl3 as a catalyst.
- Friedel Craft’s Alkylation
With alkyl halide in presence of anhydrous AlCl3, alkylation takes place for e.g.
- Friedel Craft’s Acylation
Acylation of haloarenes can be carried out with the reaction of acyl chlorides in the presence of anhydrous AlCl3. for example
Illustration 23. Complete the following giving structures of the principal organic products.
|Solution: (i)||⎯ CCl3 is meta – directing group.|
Illustration 24. How would you prepare 1 – iodopropane from?
(i) Propene (ii) 1 – propanol
Illustration 25. How is chlorobenzene obtained from?
(i) benzene and
(ii) benzene diazonium chloride
Solution: (i) Chlorobenzene is obtained by the direct chlorination of benzene in the presence of a halogen carrier.
(ii) Chlorobenzene is obtained by the treatment of benzene diazonium chloride with cuprous chloride (Cu2Cl2) and HCl. (Sandmeyer’s reaction).
Find X, Y & Z in the following sequence of reaction:
How may the two substances in each of the following pairs be distinguished from each other?
(i) Chlorobenzene and hexyl chloride.
(ii) CH2 = CH − CH2Br and CH3CH2CH2Br
(iii) p-Bromobenzyl chloride and chlorobenzyl bromide
(i) Identify X, Y & Z in the following reactions
(ii) How will you obtain 2, 6-dinitrophenol from chlorobenzene?
POLY HALOGEN COMPOUNDS
Carbon compounds containing more than one halogen atom are usually referred to as polyhalogen compounds. Some important polyhalogen compounds are:
Dichloromethane (Methylene Chloride)
It is prepared industrially by direct chlorination of methane. The mixture so obtained is separated by fractional distillation.
It is a colourless, sweet smelling, volatile liquid boiling point = 313K. Because of its low boiling point & low inflammability, it is an effective extraction solvent used in pharmaceutical & food industries
Trichloro Methane (Chloroform)
- From Methane:
Chloroform is manufactured by chlorination of methane in presence of light or catalyst.
CH4 + Cl2 ⎯⎯→ CH3Cl + HCl
CH3Cl + Cl2 ⎯⎯→ CH2Cl2 + HCl
CH2Cl2 + Cl2 ⎯⎯→ CHCl3 + HCl
CHCl3 + Cl2 ⎯⎯→ CCl4 + HCl
CHCl3 can be separated by fractional distillation.
- From Chloral Hydrate
- Laboratory Method
In this method chloroform obtained from ethanol or acetone by reaction with a paste of bleaching powder and water.
- From Carbon tetra chloride
By partial reduction of carbontetrachloride with iron filings & water.
CHCl3 is a colourless, oily liquid which has sweetish, sickly odour & taste. It is heavier than water.
- Action of Sun light & Air
In presence of sunlight chloroform is oxidized by air to a highly poisonous compound phosgene, COCl2.
When boiled with aqueous KOH, CHCl3 is hydrolysed to potassium formate.
Chloroform can be reduced with Zn & HCl.
Zn + 2HCl ⎯⎯→ ZnCl2 + 2[H]
CHCl3 + 2[H] ⎯⎯→ CH2Cl2 + HCl
- Reaction with acetone
Chloroform reacts with acetone in presence of a base such as KOH & forms the addition product, chloretone.
Chloretone is used as hypnotic (as sleep inducing) drug.
- Reaction with Nitric acid
Chloform reacts with conc. HNO3 on heating.
CHCl3 + HONO2 ⎯⎯⎯→ CCl3.NO2 + H2O
Chloropicrin is used as an insecticide & war gas.
- Reaction with primary amines (carbyl-amine reaction)
When chloroform is warmed with a primary amine aliphatic or aromatic) in the presence of alcoholic KOH an offensive smell of isocyanide or carbyl-amine is obtained.
Illustration 26. Which one of the following compounds when heated with KOH and a primary amine gives carbylamine test?
(A) CHCl3 (B) CH3Cl
(C) CH3OH (D) CH3CN
- Reaction with Silver Powder
On heating with silver powder, chloroform forms acetylene.
On further chlorination it gives CCl4
- Reimer – Tiemann reaction
CHCl3 reacts with phenol & KOH to form salicylaldehyde.
Uses of Chloroform
- As solvent in oils and varnishes
- In medicine
- As preservative
- As laboratory reagent
Illustration 27. When chloroform reacts with acetone, the product is
(A) ethylidene dichloride (B) mesitylene
(C) Chloretone (D) chloral
Illustration 28. H atom of chloroform is acidic. Why?
Solution: The three chlorine atoms attached on carbon atom make it partially positive. Due to negative inductive effect of Cl atom carbon attracts the shared pair of electrons of C and H bond more effectively to produce partial positive charge on H to behave as acid.
Illustration 29. Chloroform is stored in dark coloured bottles. Why?
Solution: To prevent oxidation of CHCl3 to COCl2 (poisonous) which occurs in the presence of sunlight.
Tri iodo Methane (Iodoform)
It is prepared by heating ethanol or acetone with sodium hydroxide & iodine or sodium carbonate & iodine in water.
Iodoform is a yellow coloured solid having m.p. = 382 K. It is insoluble in water but dissolves readily in organic solvents.
- It is used as antiseptic for dressing wounds.
- It is used to manufacture pharmaceuticals.
Tetrachloro methane (carbon tetra chloride)
CCl4 is manufactured by following methods:
(i) From methane: By chlorination of methane in the presence of sunlight.
(ii) From carbon disulphide: By reaction of chlorine with CS2 in presence of AlCl3 as catalyst.
CCl4 is colourless oily liquid with characteristic sickly smell. It is heavier than water & is insoluble in water but is soluble in organic solvents such as ether. CCl4 is used as solvent for organic reactions.
On reduction with moist iron filling, CCl4 gives chloroform.
On heating with alc. KOH, it undergoes hydrolysis & gives CO2 which dissolves in KOH to yield K2CO3.
CCl4 is stable to red heat but when the vapours come in contact with water, phosgene is liberated.
- As industrial solvent.
- for the manufacture of chloroform.
- As laboratory reagents.
- In dry cleaning.
- As a fire extinguisher.
p – Dichloro benzene
It is prepared by chlorination of benzene.
It is white volatile, solid m.p. = 325K, which readily sublimes.
- As insecticide
- As germicide
- As soil fumigant
- As deodrant & moth repellent.
Perfluoro carbon (PFC’s)
They are prepared by controlled fluorination of alkanes in vapour phase. The reaction mixture is diluted with nitrogen in order to control the reaction
Teflon is a common perfluoro carbon.
They are non – toxic, non-inflammable, non-corrosive & extremely stable compounds.
- As lubricants
- For surface coating
- As electrical insulator.
Benzene haxa chloride
These are popularly known as lindane & gammaxene. It is commercially prepared by the addition of chlorine to benzene in presence of ultraviolet light.
As a pesticide in agriculture.
p – p′ – Dichloro diphenyl tri chloro ethane (DDT)
It is manufactured by condensation of chloro benzene with trichloro acetaldehyde (chloral) in the presence of sulphuric acid.
It is a white powder insoluble in water but soluble in oils.
It is a powerful insecticide. However it is highly stable & is not easily decomposed in the environment. Therefore its long term effect could be potentially dangerous & its use is banned in many countries
Illustration 30. What does DDT stands for? What is its chemical name?
Solution: DDT stands for p, p′ – dichlorodiphenyl trichloro ethane. Its actual name is
2, 2 – bis (4 – chlorophenyl) – 1, 1, 1 – trichloro ethane.
(i) Why the hydrogen in CHCl3 is slightly acidic?
(ii) Pyrene can be used as a fire extinguisher, why?
ANSWERS TO EXERCISES
(i) (a) 1 – bromo 2, 2 – dimethyl propane
(b) 2 – bromo 2 – methyl propane
(c) 1, 3 – dichloro propane
(d) 3 – chloro toluene
(a) Amongst the three isomeric dichlorobenzens, the p-isomer has the highest melting point as this molecule is symmetric and affords a better fitting in crystal lattice. Thus it is solid.
(b) In alkyl chlorides the C-Cl bond is more polar than the C – Br bond in alkyl bromide, as chlorine is more electronegative than bromine. Due to greater polarity of molecules, alkyl chlorides are better solvents than alkyl bromides.
(c) Because ethyl iodide has greater molecular mass so greater van der Waal’s attraction hence ethyl iodide is liquid at room temperature.
(i) Iodobenzene > bromobenzene > chlorobenzene
(ii) Haloarenes are insoluble in water because of their incapability of formation of hydrogen bonds.
In vinyl chloride C—Cl bond is stable due to resonance (as in chlorobenzene).
Hence SN reaction in which Cl is replaced by nucleophile is not possible.
In allyl chloride, SN reaction is easier since allyl carbonium ion formed after removal of Cl– is stabilized by resonance.
(nucleophile) can’t attack 3° carbon having high electron – density hence elimination takes place.
Nucleophilic attack on methyl carbon is possible giving ether (Williamson synthesis).
It can undergo SN1 and SN2 reaction. By SN2 reaction only one product is formed. But by SN1 reaction, intermediate is carbonium ion.
Thus we get two isomeric products by SN1 reaction.
(i) (CH3)2CHI > (CH3)2CHBr > (CH3)2CHCl
(ii) Neo pentyl bromide being a primary halide reacts slowly through SN1 and being sterically hindered reacts slowly even through SN2.
(i) On heating with aqueous KOH solution C6H5Cl is not hydrolysed but C6H13Cl
(n hexyl chloride) is hydrolysed easily and resulting solution gives test for Cl− ion. After acidifying with dil. HNO3, the resulting solution forms a white precipitate with AgNO3 solution.
(ii) CH3 − CH2CH2Br will not discharge the pink colour of dilute alkaline KMnO4 (Baeyer’s reagent) whereas CH2 = CH − CH2Br will decolourise its colour.
(iii) On adding aq. solution of NaOH, dilute HNO3 and few drops of AgNO3 solution, p-bromobenzyl chloride give white precipitate of AgCl, while the other give yellow ppt.
(i) X = C6H5 ⎯ NH2
Y = C6H5 ⎯ N ≡ N+ ⎯ Cl−
Z = C6H5 ⎯ I
(i) H atom in CHCl3 is slightly acidic because of following two reasons
(a) Due to ⎯I effect of three Cl – atoms.
(b) The : CCl3− ion left after the removal of a proton from CHCl3 is stabilized by resonance in which Cl atom can expand its valence shell octet because of the presence of
d – orbitals.
(ii) Pyrene can be used as fire extinguisher because its vapours are non – inflammable.
Exercise 1: Arrange each set of compounds in order of increasing boiling point.
(i) Methyl bromide, methylene bromide, bromoform.
(ii) n – butyl chloride, iso – butyl chloride, tert – butyl chloride.
Exercise 2: Complete the following sequences of reactions by giving A, B & C
Exercise 3: Arrange the following compounds in decreasing order of SN1 reactions.
|(i)||CH3CH2CH2Cl, CH2 = CH CHClCH3, CH3CH2CHClCH3|
Exercise 4: Explain why:
Grignard reagents are prepared under anhydrous conditions.
Exercise 5: Write structural formula of the following
(i) Benzyl chloride
(ii) 4 – chloropent – 2 – ene
(iii) 1 – bromo – 2, 2 – dimethyl propane
(iv) 3 – chloro – 2, 4 – dimethyl pentane
Exercise 6: How many dichloro derivatives are possible for propane? Give their structures & IUPAC names.
Exercise 7: Give the structure & names of the chief organic products expected from the reaction of n – butyl bromide with each of the following reagents.
(i) Zn/H+ (ii) Mg/ether
Exercise 8: Why are haloarenes more stable than haloalkanes & undergo electrophilic substitution at ortho & para position?
Exercise 9: An alkyl chloride ‘X’ on reaction with magnesium in dry ether followed by treatment with ethanol give 2 – methyl butane. Write all the possible structures
Exercise 10: Why aryl halides are less reactive than benzene towards electrophilic substitution reactions?
ANSWERS TO MISCELLANEOUS EXERCISES
Exercise 1: (i) Methyl bromide < methylene bromide < bromoform
(ii) tert – butyl chloride < iso – butyl chloride < n – butyl chloride
|(ii)||A = CH3 − CH = CH2|
|Exercise 3: (i)||II > III > I|
Exercise 4: Grignard reagents react with water and get decomposed (hydrolysed) hence it should be prepared under anhydrous conditions.
|Exercise 6:||Dichloro derivatives of propane||IUPAC Name|
|CH3 ⎯ CH2 ⎯ CHCl2||1, 1 – dichloro propane|
|2, 2 – dichloro propane|
|1, 2 – dichloro propane|
|1, 3 – dichloro propane|
Exercise 8: Halo arenes are more stable than halo alkanes because C ⎯ X bond in former is shorter & stronger. Halo arenes undergo electrophilic substitutions at ortho & para position because halogen atom is slightly deactivating & o, p – directing.
In structure II, III & IV there is negative charge at o & p positions i.e. ortho and para position are electron rich due to resonance, so electrophilic substitution will occur at these sites.
|R ⎯ H is 2 methyl butane.
∴ Carbon skeleton is
The possible structures of alkyl chloride are
Exercise 10: This is due to the reason that halogen atom because of its electron with drawing inductive effect decreases electron density at the benzene ring.
Board Type Questions
Prob 1. Arrange each of following sets of compounds in order of increasing boiling points:Chloromoethane, 1 – chloropropane, isopropyl chloride, 1 – chlorobutane.
Sol. Since boiling point increase with increasing molecular mass due to greater magnitude of Vander Waal’s forces of attraction, the boiling points of these compounds increase in the order:
Prob 2. Give IUPAC name of:
Sol. (a) 2 – trichloromethyl – 1, 1, 1, 2, 3, 3, 3 – hepta chloropropane
(b) 1 – bromo – 1 – chloro – 1, 2, 2 – trifluoroethane
Prob 3. Explain why KCN reacts with R – I to give alkyl cyanide, while Silver cyanide forms isocyanide as a major product.
CN− is ambient nucleophile and can attack from either side and RCN is formed predominantly but AgCN being insoluble can attack only from N side hence RNC is major product.
Prob 4. How will you bring about following conversion?
Chlorobenzene to biphenyl
Prob 5. Identify the possible alkenes that would be formed on dehydrohalogenation of
1 – Chloropentene
IIT Level Questions
Prob 6. Draw structure of the major monohalo product in each of following reactions :
Prob 7. Predict the order of reactivity of the following compounds in reactions
Sol. The first compound is a alkyl halide while all others are alkyl halides. Since alkyl halides are more reactive than alkyl halides in reactions, therefore first compound is the least reactive. Further reactivity increases in the order: chloride < bromide < iodide. Thus the increasing order of reactivity in reactions is the same in which they are listed above.
Prob 8. Explain why, Allyl chloride is hydrolysed more readily than n – propyl chloride.
Sol. Allyl chloride readily undergoes ionization to produce resonance stabilized allyl carbocation. Since carbocations are reactive species, therefore allyl cation readily combines with ions to form allyl alcohol.
Prob 9. Compare the acidic behaviour of
Sol. is more acidic than It is because left after removal of proton from is stabilised by resonance due to presence of d – orbitals in Cl but left after the removal of proton from is not stabilised by resonance due to absence of d orbitals in F.
Prob 10. Which is a better nucleophile, a bromide ion or an iodide ion?
Sol. Iodide ion
Prob 11. Which alkyl halide has the highest density and why?
Sol. because of its smallest carbon content and heaviest halogen i.e. I.
Prob 12. A Grignard reagent [A] and a halo alkane [B] react together to give [C]. Compound [C] on heating with KOH yields a mixture of two geometrical isomers, [D] and [E] of which [D] predominates. [C] and [E] have same molecular formula and [C] gives 1 – bromo – 3 – phenyl propane on reaction with HBr in presence of a peroxide. Give structures of [A], [B] and [C] and configurations of [D] and [E] with reasons.
Sol. Since [C] gives 1 – bromo – 3 – phenyl propane on reaction with HBr in the presence of peroxide, it must be 3 – phenyl – 1 – propene.
[C] on heating with KOH undergo isomerisation to form more stable conjugated alkene 1 – phenyl – 1 – propene which can exist as geometrical isomer [D] and [E]. Since [D] is formed in more amount, it must be the trans – isomer.
Prob 13. Predict X, Y and Z in following reaction
Prob 14. An alkyl chloride (A) on reaction with Magnesium in dry ether followed by treatment with ethanol gave 2 – methyl butane. Write all the possible structures of (A).
Possible structures of alkyl chloride are:
Prob 15. Predict the major product in each of following reactions:
Prob 16. Identify A to C in following:
Prob 17. Give major product when the given compound is treated with sodium ethoxide.
Prob 18. 2-Bromo pentane when treated with alcoholic KOH yields a mixture of three alkenes A, B & C. Identify A, B & C which is predominant. (Assume reaction proceeds through mechanism).
Prob 19. Identify all the possible monochloro structural isomers that would be expected to form on free radical chlorination of .
Sol. In the given compound; there are four different types of halogen atoms as shown below:
The replacement of each give four monochloro derivatives:
Prob 20. Draw the structure of major mono halo product:
Prob 1. Of the following alkyl halides, one with the lowest boiling point is
(A) ethyl bromide (B) iso propyl bromide
(C) n – butyl bromide (D) methyl bromide
Prob 2. there X is
|(A)||C2H5 ⎯ O ⎯ N = O||(B)|
|(C)||C2H5 ⎯ N = O||(D)||C2H5 ⎯ N = N ⎯ C2H5|
Prob 3. . The compound X is
(A) phenol (B) benzene
(C) O & P chlorophenol (D) benzol
Prob 5. Identify Y in the following
(A) 1, 1, 2 – Trichloroethane (B) Acetaldehyde
(C) 1, 2 – Dichloroethane (D) chloroethylene
Prob 6. Benzene reacts with Cl2 in the presence of FeCl3 & in absence of sunlight to form
(A) benzyl chloride (B) benzal chloride
(C) chlorobenzene (D) benzenehexa chloride
Prob 7. Cl2 reacts with C2H6 in presence of light to form
(A) CHCl3 (B) CCl4
(C) C2H5Cl (D) CH3I
Prob 8. Chloroform is slowly oxidised by air in the presence of light & air to form
(A) formyl chloride (B) phosgene
(C) trichloro ethanol (D) formaldehyde
Prob 9. The order of reactivities of the following alkyl halides for SN2 reaction is:
(A) RF > RCl > RBr > RI (B) RF > RBr > RCl > RI
(C) RCl > RBr > RF > RI (D) RI > RBr > RCl > RF
Prob 10. Ethylene chloride and ethylidene chloride are :
(A) Chain isomers (B) Functional isomers
(C) Position isomers (D) Stereo isomers
Prob 11. Alkyl chloride is more reactive than vinyl chloride towards nucleophilic substitution reaction.
Prob 12. Benzene diazonium chloride reacts with HCl and Cu. The reaction is called Sandmyer’s reaction.
Prob 13. Dipole moment of CH3⎯F is more than CH3⎯Cl.
Prob 14. Chloroform reacts with silver powder to give acetylene.
Prob 15. The bond length of C⎯Cl bond in chlorobezene is less than in methyl chloride.
Fill in the Blanks
Prob 16. Carbyl amine reaction is used as a test for………………….amines.
Prob 17. Benzene reacts with chlorine in the presence of light to form………………………
Prob 18. CHCl3 is………………acidic than CH4.
Prob 19. When silver salt of ethanoic acid is treated with Br2/CCl4 it forms………………………
Sol. CH3 ⎯Br
Prob 20. CH2=CH⎯CH2Cl when treated with aqueous NaOH it follows…………….mechanism.
- Which alkyl halide has the highest density?
- Why chloroform is stored in dark bottles?
- Explain why methyl chloride is hydrolysed more readily than chlorobenzene.
- CCl4 does not dissolve in water. Why?
- Alkanes can not be easily converted to fluorides. Why?
- What are DDT & BHC? Give use of each.
- Chlorobenzene is less reactive towards chemical reactions. Why?
- Give IUPAC name of following compound
- Arrange the following in the order of increasing SN2 reactivity.
CH3Cl, CH3Br, CH3CH2Cl, (CH3)2CHCl
- 10. Alkyl halides do not produce a white ppt. with aq. AgNO3 why?
- 11. Complete the following by giving A, B, C & D.
- Why Benzyl chloride undergoes nucleophilic substitution much more easily than chlorobenzene.
- Explain why haloalkanes undergo nucleophilic substitution whereas haloarenes undergo electrophilic substitution reactions.
- How would you prepare 1 – iodo propane from?
(i) 1 – chloro propane
(ii) 2 – chloro propane
- Which of the following compounds will give positive iodoform test
(i) Propan – 1 – ol
(v) Pentan – 2 – one
- Give the product of bromination of toluene.
- Identify Z in the following series.
- Identify major & minor product in the following.
- Arrange the following compounds in each set in order of ease of dehydrohalogenation by concentrated alcoholic KOH.
, , ,
- Identify final product
- Write structures of all the isomeric alkyl bromides having the molecular formula C5H11Br.
- Explain why the boiling point of bromo ethane is greater than that of chloro ethane.
- Give the IUPAC name of (CH3)3CCH2CH2Cl.
- Write structure of the following: 1, 4 – dibromobut-2 – ene.
Level – II
- Which is faster in following pairs of halogen compound in SN2 reactions?
- Identify A to C in following series of reaction.
- Write structures of following compounds
(i) 2 – (2 – chlorophenyl) – 1 – iodoctane
(ii) 1 – bromo – 4 – sec.butyl – 2 – methyl benzene.
- Predict the order of reactivity of following compounds in dehydrohalogenation
CH3CH(Br)CH3, CH3CH2CH2Br, (CH3)2CH ⎯ CH2Br, (CH3)3C ⎯ CH2Br
- How will you bring about the following conversions?
(i) ethanol to but – 1 – yne
(ii) bromoethane to cis – hex – 3 – ene.
- Predict X, Y in the following sequence
- An organic compound C8H18 on monochlorination gives a single monochloride. Write the structure of hydrocarbon.
- Write the structure of possible major mono substituted products formed when Br+ attacks the following molecules. Justify your answer.
- Why the melting point of p-dichlorobenzene is greater than o-dichlorobenzene?
Level – I
- Toluene when heated with Br2/Fe, gives p – bromo toluene as the major product because the methyl group is
(A) p – directing (B) m – directing
(C) stabilising group (D) deactivating group
- Aryl halides are less reactive towards SN reactions as compared to alkyl halide due to
(A) formation of more stable carbocation
- B) resonance stabilization
(C) long carbon – halogen bond
(D) can not be predicted
- R ⎯ OH + HX ⎯⎯⎯→ R ⎯ X + H2O
In the above reaction, the reactivity of different alcohol is
(A) 30 > 10 > 20 (B) 30 > 20 > 1
(C) 10 < 20 > 30 (D) 20 < 10 < 30
- Which of the following does not occur during the formation of CHCl3 from C2H5OH & CaOCl2?
(A) hydrolysis (B) oxidation
(C) reduction (D) chlorination
- SN1 reaction of alkyl halides leads to
(A) retention of configuration
(B) inversion of configuration
(D) none of the above
- Of the following alkyl halides one with lowest boiling point is
(A) ethyl bromide (B) isopropyl bromide
(C) n – butyl bromide (D) methyl bromide
- Amongst the following the most reactive alkyl halide is
(A) C2H5F (B) C2H5Cl
(C) C2H5Br (D) C2H5I
- n – propyl bromide on treatment with ethanolic KOH produces
(A) propane (B) propene
(C) propyne (D) propanol
- Iodoform test is given by
(A) all types of alcohols
(B) all types of ketones
(C) all types of aldehydes
(D) only methtyl ketones & those alcohols which contain CH3CHOH group.
- When chloroform reacts with acetone the product is
(A) ethylidenedichloride (B) mesitylene
(C) chloretone (D) chloral
- Phenol gives Reimer tiemann reaction with
(A) CHCl3 (B) CCl4
(C) CHCl3 & CCl4 (D) C6H5CHCl
- Which of the following is tertiary alkyl halide?
(A) 2 – chloro – 2 – methyl butane (B) 1 – chloro propane
(C) 2 – chloro propane (D) cyclohexyl chloride
- Which of the following is least reactive for nucleophilic substitution?
(A) CH3 ⎯ CH2 ⎯ CH2 ⎯Br (B) CH2 = CH ⎯ Br
(C) C6H5 ⎯ CH2 ⎯ Br (D) C6H5 ⎯ Br
- Which one of following compounds on dehalogenation gives, 2, 3 – dimethyl – 2 – butene?
|(A)||(B)||CH3 ⎯ (CH2)4 ⎯ CHCl2|
|(C)||CH3 ⎯ CH2 ⎯ CHCl2||(D)||CH2Cl ⎯ (CH2)4 ⎯ CH2Cl|
- In the given reaction
, [X] will be
(A) CH4 (B) CH2Cl2
(C) CCl4 (D) CH3Cl
- Alkyl halide on heating with dry Ag2O gives
(A) ester (B) ether
(C) alcohol (D) alkane
- Alkyl halide is converted into alcohol by
(A) addition reaction (B) substitution reaction
(C) dehydrogenation (D) dehydrohalogenation
- When an alkyl halide reacts with an alkoxide the product is
(A) alkene (B) alkane
(C) ether (D) mixture of alkene & ether
- The conversion of 2, 3 – dibromobutane to 2-butene with Zn & C2H5OH is
(A) redox reaction
(B) α – elimination
(C) β – elimination
(D) both α – elimination and redox reaction
- The order of reactivity of following alkyl halides for SN1 reaction is
(A) RF > RCl > RBr > RI (B) RF > RBr > RCl > RI
(C) RCl > RBr > RF > RI (D) RI > RBr > RCl > RF
Level – II
- Identify ‘Z’ in following sequence of reactions
(A) (CH3)2CH ⎯CN (B) Br ⎯ CH = CH ⎯ CN
(C) CH2 = CH ⎯ CH2CN (D) CH2 = CH ⎯ CHBr ⎯ CN
- The well known insecticide gammaxene is one of the stereoisomers of hexachlorocyclohexane. The reagent useful for conversion of benzene into hexachloro cyclo hexane is
(A) HCl (B ) Cl2(AlCl3)
(C) Cl2(ZnCl2) (D) Cl2(hν)
- Which of the following pair gives DDT when treated with conc. H2SO4?
- Which of the following compounds will be most reactive for SN1 reactions?
- In the reaction,
, [X] will be
(A) NaCl (B) SOCl2
(C) Cl2 (D) KCl
- Iodoform test is not given by
(A) acetone (B) ethyl alcohol
(C) 1 – propanol (D) 2 – propanol
- Which of the following will not form iodoform with I2/OH−?
(A) ethanol (B) ethanal
(C) isopropyl alcohol (D) benzyl alcohol
- Compound [X] gives very unpleasant odour with CHCl3/alc. KOH. [X] is
(A) C6H5NHCH3 (B) C6H5⎯ CONH2
(C) C6H5NH2 (D) C6H5⎯NH⎯C2H5
- In chloroination of benzene with Cl2/FeCl3, the reactive species is
(C) (D) Cl2
- Which of the following compounds is used as tear gas?
(A) BHC (B) DDT
(C) chloropicrine (D) chloretone
- Which is highly selective?
(A) chlorination of hydrocarbons (B) bromination of hydrocarbons
(C) both are correct (D) none is correct
- Which one of the following is most rapidly hydrolysed by SN1mechanism?
|(A)||CH3CH = CHCl||(B)||ClCH2CH = CH2|
- Which of the following will give haloform test?
|(C)||(D)||all are correct|
- What is the major product of the following elimination reactions?
- RCl + AgCN ⎯⎯→ A B, A & B are
(A) RCN, RCH2NH2 (B) RNC, RNHCH3
(C) RCN, RNHCH3 (D) RNC, RCH2NH2
|18.||can undergo elimination to give major product as|
|(C)||(D)||None of these|
- CH3Br can be prepared by
(C) both are correct (D) none is correct
|(C)||(D)||None is correct|
ANSWERS TO ASSIGNMENT PROBLEMS
- Alkyl iodides.
- Chloroform is stored in dark bottles because in presence of sunlight CHCl3 is oxidised by air to produce a highly poisonous compound phosgene COCl2.
- Because C ⎯ Cl bond in chloro benzene is shorter & hence stronger as compared to that in alkyl halides.
- Because CCl4 is heavier than water. Also CCl4 is non polar.
- Because fluorination of hydrocarbons with pure F2 gas occurs explosively.
- DDT is 2, 2 – Bis (4 – chloro phenyl) – 1, 1, 1 – trichloro ethane. It is used as insecticide.
BHC is benzene hexa chloride. It is extensively used as pesticide.
- Because C ⎯ Cl bond is shorter & stronger.
- 2 – chloro 2 – phenyl butane.
- (CH3)2CHCl < CH3CH2Cl < CH3Cl < CH3Br
- Because C-Cl bond in alkyl halides is covalent & does not ionize in aqueous solution to produce chloride ions.
- Benzyl chloride undergoes nucleophilic substitution via resonance stabilized benzyl carbocation and hence undergoes substitution readily.
On the other hand C – Cl bond in chlorobenzene has partial double bond character due to resonance and hence is difficult to cleave.
- Haloalkanes are more polar than halo arenes consequently the carbon atom carrying the halogen in halo alkanes is more electron deficient than that in halo arenes. As a result halo alkanes undergo nucleophilic substitution more readily than haloarenes.
- Following compounds will give positive iodo form test:
Propanone and pentan – 2 – one
Level – I
- D < A < B < C
Reason: Conjugated system is more stable than non – conjugated system and dehydrohalogenation of D will form cyclo alkyne which is unstable.
|7. (i)||CH3CH2 ⎯ CH2CH2CH2Br
1 – bromo pentane
- This is because with the increase in the size & mass of the halogen (in bromo ethane) the magnitude of Vander Waal’s forces of attraction increases.
- BrH2C ⎯ HC = CH ⎯ CH2Br
Level – II
- First one is faster in both the cases because ease of backside attack decides which undergoes SN2 faster.
- (CH3)3CCH2Br > (CH3)2CH ⎯ CH2Br > CH3CH(Br)CH3 > CH3CH2CH2Br
- Since the hydrocarbon C8H18 on mono chloroniation gives a single mono chloride therefore, all the hydrogen atoms are equivalent. Therefore structure is
- In p-dichlorobenzene because of symmetrical structure better crystal fitting is possible as compared to o-dichlorobenzene. That’s why p-dichlorobenzene has greater melting point.
Level – I
- C 2. B 3. B
- C 5. C 6. D
- D 8. B 9. D
- C 11. C 12. A
- D 14. A 15. B
- B 17. B 18. D
- C 20. D
Level – II
- C 2. D 3. B
- D 5. B 6. C
- D 8. C 9. A
- C 11. B 12. A
- D 14. D 15. B
- B 17. C 18. A
- A 20. A