haloalkanes and haloarenes

haloalkanes/ alkyl halides

• the replacement of H atoms from a hydrocarbon, by halogens (F, Cl, Br, I) forms a haloalkane

• the X atom is attached to a sp3 hybridised C-atom

haloarenes

• the replacement of H atoms from a benzene ring, with a halogen (F, Cl, Br, I) forms a haloarene

• the X atom is attached to a sp2 hybridised C-atom

classification of haloalkanes/ arenes on the basis on no. of X atoms

• mono: has 1 X atom

• di: has 2 X atoms

• tri: has 3 X atoms

classification haloalkanes in terms of the C—X bond

• alkyl halides:

  • the X atom is bonded to an alkyl group R

  • they form a homologous series CnH2n+1X

    • primary: the C atom is attached to 1 alkyl group

    • secondary: the C atom is attached to 2 alkyl groups

    • tertiary: the C atom is attached to 3 alkyl groups

• allylic halides:

  • the X atom is bonded to a C atom which is bonded to another C atom which is double bonded to anotherr C atom

• benzyllic halides:

  • the X atom is bonded to a C atom which is bonded to benzene ring

classification of haloarenes in terms of the C—X bond

• vinylic halides:

  • in these, the X atom is bonded to the sp2 hybridised C atom (the one with a double bond

  • eg: halocycohexane

• aryl halides:

  • the X atom is directly bonded to an aromatic ring

  • eg: halobenzene

nomenclature for mono-substituted haloalkane

• common name: n-/ iso-/ sec-/ tert- + alkyl group name + X-ide

  • eg: n-propyl iodide

• IUPAC: position + Xo + alkane

  • eg: 1-iodo propane

how to use the prefixes

• n → straight chain

• iso → branch on second carbon

• sec → func. group on secondary carbon (a C atom attached to 2 other C atoms)

• tert → func. group on tertiary carbon (a C atom attached to 3 other C atoms)

• neo → highly branched C atom (C atom is attached to a C atom which is attached to 3 other C atoms)

nomenclature for disubstituted haloalkane

first, there are 2 types

• gem halides: the 2 X atoms are on the same C atom

• vic halides: the 2 X atoms are on adjacent C atoms

• common name:

  • gem halides → alkylidene halides

  • vic halides → alkylene dihalides

• IUPAC: dihaloalkanes

nomenclature for haloarenes

• mono-substituted: halobenzene for both common and iupac names

• disubstituted:

  • common name: o-/m-/p- + di + halobenzene

  • iupac: 1,2/1,3/1,4 + di + halobenzene

nature of the C—X bond

• electronegativity of X is more than C

• therefore, the shared pair electron is closer to the X atom

• this results in X atom bearing a partial -ve charge and C atom bearing a partial +ve charge

  • this implies C—X bond is polarised

polarity of C—X bond decreases as- F > Cl > Br > I (electronegativity decreases)

length of C—X bond decreases as- I > Br > Cl > F (atom size decreases → bond length decreases)

general formula of preparation of haloalkanes from alcohols

ROH + HX → RX + H2O

preparation of chloroalkanes from alcohols

• 1 and 2: andhyd. ZnCl2 acts as catalyst, because it acts as a lewis acid and helps in the cleavage of C—O bond

• 3: very reactive, can be done by shaking with HCl at room temp

preparation of bromoalkanes from alcohols

• method 1: by heating alcohol and HBr in presence of conc. H2SO4

• ONLY FOR 1* → method 2 (to generate HBr in situ): adding KBr/ NaBr, and conc. H2SO4 and heating it

preparation of iodoalkanes from alcohols

• method 1: just by heating an alcohol with HI (reflux)

• ONLY FOR 1* → method 2 (to generate HI in situ): add KI with H3PO4 and heat it

why cant 2* and 3* bromo/iodoalkanes be prepared from alcohols in presence of conc. H2SO4 as catalyst?

• its because the acid oxidises the compound into Br2/ I2

• as its highly dehydrating, it makes the alcohol undergo dehydration to form alkenes

order of reactivity of alcohols with haloacids

• 3* > 2* > 1* alcohols

• HI > HBr > HCl (due to order of bond dissociation energy which is the other way around)

preparation of chloroalkanes from alcohols by the action of phosphorus chloride

• PCl5: R—OH + PCl5 → R—Cl + POCl3 + HCl

• PCl3: R—OH + PCl3 → R—Cl + H3PO3

preparation of bromoalkanes/ iodoalkanes from alcohols by the action of PBr3/ PI3

it is done in situ by the reaction of phosphorus and Br/I

R—OH + P + Br2 → R—X (X = Br/ I)

preparation of chloroalkanes from alcohols by the action of thionyl chloride (SOCl2)

R—OH + SOCl2 →(pyridine) R—Cl + SO2 + HCl

• this is preferable as SO2 and HCl are escapable gases, giving us pure alkyl halide

why cant we use phenols to prepare haloarenes

this is because the C—O bond in phenols have a partial double bond character, and is difficult to break

preparation of haloalkanes by free radical halogenation of alkanes

alkane →(Cl2/ Br2 + UV light) n-alkylhalide + sec-alkylhalide

ease of substitution of H by free radical halogenation of alkanes

benzylic, allylic > 3* > 2* > 1* > vinylic, aryl

markovnikov and anti-markovnikov’s rule

• markovnikov’s rule: when a protic acid (H—X) is added to an unsymmetrical alkene, the X atom attaches itself to the most substituted C atom and the H atom attaches itself to the least substituted C atom

• anti-markovnikov’s rule: when a protic acid (H—X) is added to an unsymmetrical alkene in the presence of peroxide, the opposite happens; the X atom attached itself to the least substituted C atom and the H atom attaches itself to the most substituted C atom

preparation of haloalkanes from alkenes by addition of hydrogen halides

• for symmetrical alkenes: alkene + HX → haloalkane

• for unsymmetrical alkenes: alkene + HX → minor haloalkane + major haloalkane

  • from markovnikov’s rule, we know that the major haloalkane is the one where the X atom is attached to the most substituted C atom

  • anti-markovnikov’s rule: if we use peroxide as catalyst, the opposite happens and the minor product is now the major product

• reactivity: HI > HBr > HCl > HF

preparation of bromoalkanes from alkenes by addition of Br2

alkene + Br2 →(CCl4) vic-dibromide

• this results in the discharge of reddish-brown color of Br

  • this helps us detect the presence of double bond in a molecule

preparation of iodoalkanes from chloroalkanes/ bromoalkanes (finkelstein reaction)

R—X + NaI →(acetone + heat) R—I + NaX (X = Cl/ Br)

• the NaBr/ NaCl formed gets precipitated in the acetone, which facilitates the forward reaction (acc. to le-chatelier’s principle)

preparation of fluoroalkanes from chloroalkanes/ bromoalkanes (swarts reaction)

R—X + AgF → R—F + AgBr

• instead of AgF, we can also use any metallic fluoride like Hg2F2, CoF2 or SbF3

preparation of bromoalkanes from silver salts (borodine-hundsdiecker reaction)

silver salt (eg: CH3COOAg) + Br2 →(CCl4) bromoalkane + CO2 + AgBr

preparation of haloarenes by electrophilic substitution of aromatic hydrocarbons

• toulene + X2 →(Fe, dark) o-halotoulene (next to) + p-halotoulene (opp.)

• halobenzene + X2 →(FeCl3) o-dihalobenzene (minor) p-dihalobenzene (major)

X = Cl/ Br

NOTE: o and p isomers can be easily separated due to large difference in melting points

why cant we use electrophilic substitution to make iodo/fluoroarenes from aromatic hydrocarbons

• iodine: the reaction is reversible in nature and requires the presence of an oxidising agent (HNO3/ HIO4) to oxidise the HI formed during iodination

• fluorine: the reaction is very violent and cant be controlled

preparation of diazonium salts

aniline →(NaNO2 + HX, cold) diazonium halide (benzene ring with N2X)

preparation of bromo/ chloroarenes from diazonium salts

• diazonium salt →(Cu2X2) halobenzene + N2 (X = Br/ Cl) (sandmeyer’s reaction)

• diazonium chloride →(Cu, HCl) chlorobenzene + N2 (gattermann reaction)

preparation of iodoarenes from diazonium salts

diazonium salt (any) →(KI) iodobenzene + N22

raschig process

it is the preparation of chlorobenzene commerically

• benzene + HCl + ½ O2 →(CuCl2, heat) chlorobenzene + H2O

physical state of haloalkanes/arenes

• methyl chloride/bromide, ethyl chloride are gases at room temperature

• higher members are liquids or solids

• volatile halogen compounds have sweet smell

why is the b.p of haloalkanes/arenes higher than hydrocarbons of comparable molecular mass?

• halogen compounds are generally polar and have slightly higher m.m compared to parent hydrocarbon

• hence, the intermolecular forces (dipole-dipole and van der waals) are stronger

how various factors affect the b.p of halogen compounds

• as molecular mass increases, m.p and b.p increases

• for same alkyl group, the b.p decreases in the order RI > RBr > RCl > RF

  • this is because with increase in atom size, theres an increase in van der waals forces

• for same halogen atom, the b.p decreases with decrease in alkyl group size

• for isomeric alkyl halides, b.p decreases as branching increases

  • this is because with increase in branching, theres surface area decreases and hence the van der waals forces decrease

• more the no. of halogen atoms, more the b.p

• for isomeric dihalobenzenes, m.p of p > o > m

  • this is because the symmetry of p fits the crystal lattice better

density of haloalkanes and haloarenes

• fluoro/chloroalkanes are lighter than water

• bromo/iodo/polychloroalkanes are heavier than water

• more the no. of atoms/ atomic mass, more the density

• all haloarenes are heavier than water

  • their densities decrease from iodo to chloro

solubility of haloalkanes/haloarenes

• in water: it is low

  • its because a lot of energy is required to overcome the attraction between the haloalkane molecules and break the hydrogen bonds between water, and energy released is not enough for it

• in alcohol: it is high

  • its because the intermolecular attractions between haloalkane and alcohol are of same strength as the ones being broken

stability of haloalkanes/ haloarenes

• it decreases as the strength of C—X bond decreases

• RF > RCl > RBr > RI

dipole moment of haloalkanes/ haloarenes

• dipole moment decreases as electronegativity decreases

• BUT, fluorides have a lower dipole moment than chlorides because the very small size of F outweighs the effect of greater electronegativity

• RCl > RF > RBr > RI

nucleophilic substitution reaction

• when a nucleophile (electron rich species) stronger than the X ion reacts with a haloalkane (which has a partial +ve charge on the C atom), a substitution reaction takes place, which is called nucleophilic substitution reaction

• the halogen (called leaving group) leaves the ion

ambident nucleophiles

• groups like cyanides and nitrites possess 2 nucleophilic centres, and are called ambident nucleophiles

• cyanides and isocyanides:

  • when it links through C atom, it forms alkyl cyanides

  • when it links through N atom, it forms isocyanides

• nitroalkanes and alkyl nitrites:

• when it links through O atom, it forms alkyl nitrites

• when it links through N atom, it forms nitroalkanes

which X ions have the fastest and slowest rate of nucleophilic substitution reaction?

• I is the fastest as its the best leaving group

• F is the slowest as its the poorest leaving group

SN2 reaction

• when 2 molecules take part to determine the rate of reaction, its called SN2 eeaction

• it involves the reactant, transition state and product

mechanism of SN2 reaction

• the incoming nucleophile attacks the R—X molecule from behind and starts breaking the C—X bond and a new C—Nu bond is formed

• this process takes place simultaneously in a single step, and theres no intermediate formed

• it is done in the transition state, in which the C atom is simultaneously bonded to both X and Nu

• this state is unstable, and results in the formation of the product

why do simple halogen compounds react predominantly to SN2?

• SN2 reaction requires the backside approach of Nu to the C atom

• the presence of bulky alkyl groups blocks the approach due to steric hindrance

• hence, simple compounds are better

order of reactivity towards SN2

• CH3X > 1* > 2* > 3*

• non-polar solvents prefer SN2

SN1 reaction

• when only one molecule is involved in determining the rate of reaction, it is called SN1 reaction

• it involves the formation of carbocation intermediate

mechanism of SN1 reaction

• STEP 1: formation of carbocation intermediate

  • the formation takes place due to heterolytic cleavage of >C—X bond

  • its a slow and reversible process

  • it involves only one reactant, the halide. hence, it follows first order kinetics; energy for C—X bond breaking is obtained through solvation of halide ion with the proton of protic solvent

• STEP 2: attack of nucleophile on carbocation formed

  • the Nu attacks the carbocation and substitution reaction is completed

order of reactivity towards SN1

• 3* > 2* > 1* > CH3X

  • this is because more stable the carbocation, greater is the ease of formation and faster the rate. 3* forms the most stable carbocation

• allylic and benzylic halides also show high reactivity because the carbocation formed gets stabilised thru resonance

plane polarised light

• a beam of light consists of em waves, which vibrates in all planes in space

• when this light is passed through a nicol prism, it becomes plane polarised and only vibrates in one plane

optical activity

• optically active substances are those which rotate the light when placed in the path of plane polarised light

  • if it rotates to the right (clockwise): dextrorotatory/ d-form, +ve sign

  • if it rotates to the left (anti-clockwise): laevorotatoey/ l-form, -ve sign

• the d- and l-forms of a compound are called optical isomers

• the angle by which the light rotates can be measured by a polarimeter

chirality

• a molecule thats non-superimposable on its mirror image is called chiral

• they contain an asymmetric C atom (all 4 groups attached to the C are different)

• chiral molecules are optically active

• eg: butan-2-ol

• a molecule thats superimposable on its mirror image is called achiral

• they contain a symmetric C atom (all 4 groups arent same)

• achiral molecules are optically inactive

• eg: propan-2-ol

enantiomers

• they are the pair of optical isomers which are non-superimposable mirror images of each other

• they have identical physical and chemical properties, but their rotation of plane polarised light is opposite (d & l)

racemic mixtures & racemisation

• it is a mixture containing 2 enantiomers in equal proportions

  • there will be zero optical rotation, as they will be cancelled by each other

  • it is represented as dl

• the process of converting d and l form into dl form is called racemisation

retention

• when the spatial arrangement of bonds in a chiral molecules remains same before and after reaction, retention of configuration is said to occur

• however, optical rotation may be different

inversion

• if the spatial arrangement of a chiral molecule becomes opposite after the reaction, then its called inversion of configuration

what happens when a 50:50 mixture of retention and inversion is formed?

the process of racemisation takes place, and the product is optically inactive

stereochemical aspects of SN2

• in SN2, the nucleophile attacks the opposite side

• hence, it is always accompanied by inversion of configuration

• this is called walden inversion

• eg: 2-halobutane becomes butane-2-ol, where —OH takes opp. position of halogen

stereochemical aspects of SN1

• in SN1, if the haloalkane is optically active, we get a racemic mixture

• the intermediate carbocation is achiral (planar)

• hence, the —OH can attack from either side of the plane of carbocation, forming a mixture of 2 enantiomers

dehydrohalogenation (elimination reaction)

• when a haloalkane with \beta-H atom is heated with alc. KOH (or any other strong base), the H is eliminated from the \beta-C atom and X atom is lost from the \alpha-C atom. this results in the formation of an alkene

• it is also called \beta-elimination

haloalkane →(alc KOH) alkene + KX + H2O

saytzeff rule

• in dehydrohalogenation, there is a possibility of formation of more than 1 alkene as theres more than one \beta H atom present

• hence, according to the rule,

• the preferred product is the alkene which has the greater no. of alkyl groups attached to the doubly bonded C-atoms

what type of reaction a haloalkane will prefer depending on its nature

• primary haloalkanes will prefer SN2

• secondary haloalkanes will prefer SN2 or elimination (depending on strength of nucleophile/ base)

• tertiary haloakanes will prefer SN1 or elimination (depending on the stability of carbocation/ more substituted alkene)

what type of reaction a haloalkane would prefer depending on strength and size of nucleophile/ base

• bulkier nucleophile (like C2H5O-) will carry out elimination more easily than substitution

  • this is because it prefers to act as a base and abstract a proton rather than approach a carbon atom

• whereas nucleophiles like OH- will carry out substitution

reaction of haloalkanes with magnesium (formation of grignard reagent)

haloalkane + Mg →(dry ether) R—MgX

• here, the C—Mg bond is highly polar, and carbon pulling electrons from magnesium makes the latter acquire partial positive charge

why do we prepare grignard reagent in the presence of dry ether?

• this is because they are highly reactive and react with anything to form hydrocarbons

• so its necessary to avoid any traces of moisture

reaction of haloalkanes with sodium (wurtz reaction)

• it forms hydrocarbons containing double the no. of C-atoms present in the halide

2RX + 2Na →(dry ether) 2R + 2NAX

• its useful to prepare symmetrical alkanes with higher no. of C-atoms

• 3* halides dont undergo this

why does resonance effect result in less reactivity of haloarenes in nucleophilic sub. reactions?

• the lone pair of electrons in X atom are in conjugation with pi-electrons of the ring

• hence they get delocalised (draw the delocalisation)

• as a result, due to resonance, C—Cl bond acquires a partial double bond character

• hence, the bond is difficult to break, reducing the reactivity towards SN

how the sp2-hybridisation of carbon atom in C—X bond results in haloarenes not being reactive towards SN reactions

• the sp2 hybridised C-atom has greater s-character and is more electronegative

• it can hold the electron pair of C—X bond more tightly than the sp3 hybridised C-atom

• hen e, the C—X bond is shorter and more difficult to break

how unstability of phenyl cation and repulsive interaction results in haloarenes being less reactive towards SN reactions

1. in haloarenes, the phenyl cation formed due to self-ionisation cannot be stabilised by resonance

• hence, it SN1 cannot occur

2. it is unlikely for nucleophile to approach haloarenes due to repulsion

replacement of haloarenes by hydroxyl group

chlorobenzene + NaOH →(623K, 300atm, HCl) sodium phenoxide →(dil. HCl, -NaCl) phenol

how to activate the haloarenes towards nucleophilic sub. reaction

• the presence of —NO2, —CN at o/ p positions withdraws electron density from benzene ring and facilitates the attack of Nu. the carbanion formed is stabilised through resonance

• greater the no. of the groups at o/ p positions, more reactive the haloarene

—NO2 at para position during SN reaction of haloarenes

the negative charge that appears at o-position gets stabilised by the —NO2 group

(draw the structures)

—NO2 at ortho-position during SN reaction of haloarenes

the negative charge that appears at para-position gets stabilised by —NO2 group

(draw the structures)

—NO2 group at meta-position during SN reaction of haloarenes

• theres no structure in which the negative charge is on the C atom bearing the —NO2 group

• hence, it doesnt stabilise the -ve charge

• so theres no change in reactivity

(draw the structures)

why does electrophilic substitution in haloarenes occur at a slow rate?

• due to resonance, the electron density increases more at o and p positions than m

• hence, the reactions occur at o and p position

• the halogen has electron withdrawal tendency, due to which the electron density on the ring decreases (-I effect), and the ring gets deactivated

• hence, it occurs at a slower rate

halogenation

halobenzene + X2 →(anhyd. FeCl3) p-dihalobenzene (major) + o-dihalobenzene (minor)

nitration

halobenzene →(conc. HNO3+, conc. H2SO4) halo-p-nitrobenzene (major) + halo-o-nitrobenzene (minor)

sulphonation

halobenzene →(conc. H2SO4, heat) p-halobenzene sulphonic acid (major) + o-halobenzene sulphonic acid (minor)

sulphonic acid = SO3H