Haloalkanes and Haloarenes

Reasoning based questions

Q.1.The reactions of primary and secondary alcohols with HCl require the presence of a catalyst, ZnCl₂. With tertiary alcohols, the reaction is conducted by simply shaking with concentrated HCl at room temperature why?

Ans.It is due to order of reactivity of three classes of alcohols with HCl. Tertiary alcohols are most reactive hence do not need any catalyst. The order of reactivity of alcohols with a given haloacid is 3^o>2^o>1^o

Q.2.The preparation of aryl halides is not possible from phenol and HX why?

Ans.Because the carbon-oxygen bond in phenol has a partial double bond character and is difficult to break being stronger than a single bond.

Q.3.Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts like iron or iron(III) chloride but aryl iodide  and aryl fluoride is not possible by this method why?

Ans. Reactions with iodine are reversible in nature and require the presence of an oxidising agent (HNO₃, HIO₄) to oxidise the HI formed during iodination. Fluoro compounds are not prepared by this method due to high reactivity of fluorine.

Q.4.Boiling points of chlorides, bromides and iodides are considerably higher than those of the hydrocarbons of comparable molecular mass.

Ans.The chlorides, bromides and iodides are polar molecules and  thus having dipole-dipole interaction between their molecules .On other hand hydrocarbons are non-polar molecules hence they contain weak dispersion forces .Thus boiling points of chlorides, bromides and iodides are considerably higher than those of the hydrocarbons of comparable molecular mass.

Q.5.The boiling points of alkyl halides decrease in the order: RI> RBr> RCl> RF why.

Ans. As the mass of the molecules increases size increases and thus, the magnitude of van der Waal forces increases hence boiling point increases.

Q.6.Arrange n-butyl bromide,isobutyl bromide,sec-butyl bromide and tertiary butyl bromide in decreasing order of boiling point.

Ans.
 

As branching increases surface area decreases hence magnitude of intermolecular vander waal forces decreases thus boiling point decreases.The order is-

n-butyl bromide > sec-butyl bromide > isobutyl bromide > tertiary butyl bromide.

Q.7. Arrange o-dichlorobenzene,p-dichlorobenzene, m-dichlorobenzene in decreasing order of their boiling and melting point.

Ans. o-dichlorobenzene > p-dichlorobenzene > m-dichlorobenzene (B.P).

p-dichlorobenzene > o-dichlorobenzene >m-dichlorobenzene (M.P).

The p-isomer has higher melting as compared to their ortho and meta-isomers. It is due to symmetrical structure of para-isomers that fits in crystal lattice better as compared to ortho– and meta-isomers.

Q.8. The density of alkyl halide increases with increase in number of carbon atoms why?

Ans. As the number of carbon atoms increases, mass increases thus density increases.

Q.9.The haloalkanes are only very slightly soluble in water why?

Ans. Haloalkanes molecules have dipole-Dipole interaction between their molecules, but water molecules have hydrogen bonding .Since hydrogen bonding is stronger force hence very less energy is realesed when new attraction is setup between haloalkanes and water molecules.

Q.10. Arrange each set of compounds in order of increasing boiling points.

(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.

(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

Ans. (i) Chloromethane < Bromomethane < Dibromomethane < Bromoform

Above order is due to atomic mass which increases in same order.

(ii)  1-Chloropropane < Isopropyl chloride < 1-Chlorobutane.

1-Chlorobutane has highest molecular mass thus it has highest boiling point. Since Isopropyl chloride has more branching than 1-chloropropane hence former has lesser boiling than latter.

Ans. CN^– is ambident nucleophile hence reaction can occur through C or N. Since KCN is ionic compound hence nucleophile CN^– is available for bonding, since C-C bond is more stronger than C-N bond hence main product is cyanide. AgCN is covalent compound hence reaction occurs through N and main product is isocyanide.

Q.12.The order of reactivity towards S₂N decreases in following order why?

   Primary halide > Secondary halide > Tertiary halide

Ans.As the alkyl groups increases around carbon having leaving group it is getting very difficult to approach that carbon by nucleophile hence reactivity decreases in following order Primary halide > Secondary halide > Tertiary halide.

Q.13.S₁N reactions are generally carried out in polar protic solvents (like water, alcohol, acetic acid, etc.).

Ans. In S₁N mechanism 1^st step involves the C–X bond breaking for which the energy is obtained through solvation of halide ion with the proton of protic solvent.

Q.15.Why is S₁N reation 1^st order and unimolecular reaction ?

Ans. In S₁N mechanism 1^st step involves the C–X bond breaking which is slow process and thus rate of reaction depends only on concentration of alkyl halide thus reation is 1^st order and unimolecular.

Q.17. The order of reactivity towards S₁N decreases in following order why?

   Tertiary halide  > Secondary halide > Primary halide

Ans.In S₁N mechanism 1^st step involves the C–X bond breaking and an intermediate carbocation forms. Since the stability of carbocation decreases in following order

                                 3^o˃2^o˃1^o (+I effect decreases)

Hence reactivity of alkyl halide decreases in following order  Tertiary halide  > Secondary halide > Primary halide.

Q.18.Allylic and benzylic halides show high reactivity towards the S₁N reaction.

Ans. The carbocation thus formed as intermediate gets stabilised through resonance.

Q.19.For a given alkyl group, the reactivity of the halide, R-X, follows the same order in both the mechanisms R–I> R–Br>R–Cl>>R–F why?

Ans.As the size of halogen increases bond dissociation energyof C-X decreases thus rate of releasing of leaving group decreases in following order I^–˃Br^–˃Cl^–.Thus for a given alkyl group, the reactivity of the halide, R-X, follows the same order in both the mechanisms R–I> R–Br>R–Cl>>R–F.

Q.20.Grignard reagents should be prepared under anhydrous conditions?

R-MgX + H₂O   →  R-H + MgOH(X). Grignard reagents are highly reactive in protic solvent like water , alcohol etc.

Q.21.Aryl halides are extremely less reactive towards nucleophilic substitution reactions why?

Ans.

Due to partial double bond character of  C-Cl due to resonance , bond dissociation in haloarene is difficult than haloalkane and therefore, they are less reactive towards nucleophilic substitution reaction.

Q.22.Although chlorine is an electron withdrawing group, yet it is ortho,para   directing in electrophilic aromatic substitution reactions why?            

Ans

Cl on benzene ring produces two effects during electrophilic substitution one is inductive effect and other is resonance effect. By – I effect Cl decreases the stability of intermediate carbocation and by +R effect Cl increases the stability of intermediate carbocation. The inductive effect here is stronger than resonance and causes net electron deactivation and thus Cl is deactivating group.

Electron dcensity is high at ortho and para positions hence benzene ring donate electrons to the electrophile from these positions only hence Cl is ortho and para directing.

Q.23.Chloroform is stored in closed dark coloured bottles why?.

Ans,It is due to formation of poisonous gas phosgene.

2CHCl₃ +O₂  → 2COCl₂  + 2HCl( COCl₂ is phosgene)

Name reactions of Haloalkanes and Haloarenes

1.Wurtz reaction

R-X + 2Na + R-X →  R-R + 2NaX

2. Sandmeyer’s reaction

3. Finkelstein reaction

          R-X + NaI → R-I + NaX
    X=Cl, Br

4.Swarts reaction

H₃C-X +AgF →      H₃C-F + AgBr(X=Cl, Br)

5. Friedel-Crafts alkylation reaction

6. Friedel -Craft acylation reaction

7 . Wurtz –Fittig reaction

8.Fittig reaction

Other reactions

Preparation of haloalakane from alcohol        

1. R-OH  +  H-X      →       R-X  +  H₂O (ZnCl₂ is used as catalyst when 1^o and 2^o alcohols are reacted with HCl)

2. R-OH  +  NaBr + H₂SO₄        →       R-Br   + NaHSO₄ + H₂O

3. R-OH  +  PX₃     →       R-X   + H₃PO₃+ HCl (X=Cl,Br)

4. R-OH  +  PCl₅     →     R-Cl  +  R-X   + POCl₃ + HCl

5. R-OH  +   Red P/X₂    →   R-X   (X=Cl₂,I₂)

6.R-OH  +  SOCl₂     →   R-Cl   +SO₂  + HCl

Formation of haloalkane from alkane

1. CH₃CH₂CH₂CH₃    + Cl₂/UV light    → CH₃CH₂CH₂CH₂Cl + CH₃CH₂CHClCH₃

2. Addition of H-X to alkene

Addition takes place according to markonikov rule,But addition of HBr in presence of peroxide takes place antimarkonikov rule.

CH₂=CH₂ +HX     →  CH₃-CH₂X

3.Addition of halogen to alkene

CH₂=CH₂ +Br₂  /CCl₄  → CH₂Br  -CH₂Br

4.Formation of Grignard reagent

CH₃-CH₂-Br +  Mg   →     CH₃-CH₂MgBr (Grignard reagent)

Formation of iodobenzene

Formation of haloarene from benzene

Formation of phenol from chlorobenzene

Halogenation Of Chlorobenzene

Nitration of Chlorobenzene

Sulphonation of Chlorobenzene

Mechanisms

S_N1 Mechanism

1.S_N1 mechanism completes in two steps.

2.In first step an intermediate carbocation forms and in second step nucleophile attacks the carbocation and formation of  product can takes place.

3.Rate of reaction depends only on the first step and concentration of alkyl halide because it is slowest step thus S_N1 is first order unimolecular reaction.

S_N2 mechanism

1 S_N2 mechanism completes in one step.

2.A transition state forms where there is simultaneous attack of the nucleophile and displacement of the leaving group.

3. Rate of reaction depends  both on concentration of nucleophile and alkyl halide thus reaction is 2^nd order bimolecular reaction.

Sterochemistry involves during S_N2 and S_N1 mechanism

Q.1.What do you mean by optical activity of a particular compound?

Ans.Compound which  rotate plane polarized light in either a clockwise or anticlockwise direction is called optical active compound.

Q.2.Define enantiomers.

Ans.Enantiomers possess identical chemical structures (i.e. their atoms are the same and connected in the same order), but are mirror images and nonsuperimposable of one another.They are optically active compounds one is called dextro (+) and another is called laevo(-) compound. Dextro rotate the plane polarised light in clockwise(right) direction and laevo in anticlockwise(left) direction.

Q.3.Define racemic mixture.      

Ans. Mixture of equal amounts of dextro and laevo isomers results in no rotation called racemic mixture.racemic mixture is optically inactive.

Q.4.What is the criteria for optical activity?

Ans.A chiral compound is must be an optically active compound.

Q.5.What do you mean by chiral compound?

Ans. A compound which contain at least a chiral carbon(asymmetric carbonn) that is the carbon atom which is attached to four different types of atoms or four different groups of atoms  that cannot be superimposed on their own mirror image are said to be chiral compound.

Q.6.Define retention ,inversion and partial racemisation.

Ans. Retention :The configuration of the molecule is retained through the reaction, so the product molecule has the same configuration as the reactant that is dextro reactant  forms dextro product and laevo forms laevo.

Inversion: When the product’s configuration is opposite that of the substrate the process is called inversion.

Partial racemisation:The phenomena in which retention and inversion both takes place simultaneously called partial racemisation.

Q.7.Explain the sterochemistry involves during S_N2 and S_N1 mechanism.

Ans Experimental observation shows that all SN₂ reactions proceed with inversion of configuration; that is, the nucleophile will always attack from the backside in all SN₂ reactions. S_N1 involves both retention and inversion that is it undergo partial racemisation configuration.

Polyhalogen Compounds

Dichloromethane (Methylene chloride)   Trichloromethane (Chloroform)   Triiodo-methane (Iodoform) Tetrachloromethane (Carbon tetrachloride)   Freons   p,p-Dichlorodiphenyl trichloro ethane(DDT)

Cl2CH2 CHCl3   CHI3   CCl4   CCl2F2   CCl3CH(C6H5Cl)2

Uses of polyhalogen compounds

  Cl₂CH₂

• Dichloromethane is widely used as a solvent as a paint remover.
• As a propellant in aerosols
• As a process solvent in the manufacture of drugs.
• It is also used as a metal cleaning and finishing solvent

CHCl₃

• chloroform is employed as a solvent for fats, alkaloids, (Chloroform) the production of the freon refrigerant R-22

CHI₃

• It was used earlier as an antiseptic but the antiseptic properties are due to the liberation of free iodine and not due to iodoform itself

CCl₄

• It is produced in large quantities for use in the manufacture of refrigerants and propellants for aerosol cans.
• It is also used as feedstock in the synthesis of chlorofluorocarbons and other chemicals, pharmaceutical manufacturing.

CCl₂F₂

• These are usually produced for aerosol propellants, refrigeration and air conditioning purposes.

CCl₃CH(C₆H₅Cl)₂

• As insecticides