Grignard Reagent, Reduction & Alkane Theory_E

GRIGNARD REAGENT, REDUCTION & ALKANE

1.

GRIGNARD REAGENT

1.1

Organometallic compounds

Organometallic compounds are the organic compounds in which a metal atom is directly attached to carbon of organic moetiy through covalent bond or ionic bond. For example

C – M or

(Where C is a carbon atom of an organic molecule and M is a metal atom)

If the metal atom is attached to oxygen, nitrogen. sulphur, etc., then such an organic compound is not regarded as an organometallic compound. The following structural formula do not belong to the family of organometallic compounds.

RONa (Sodium alkoxide). CH₃COONa (Sodium acetate), CH₃COOAg(Silveracetate),RSK (Potassium mercaptide) RNHK

(N-Alkylpotassamide), (CH₃COO)₄Pb (Lead tetraacetate), etc.

Note : It should be noted that (CH₃)₄Si (Tetramethylsilane, TMS) is also not an organometallic compound because silicon is a nonmetal.

Most important examples of organometallic compounds are Grignard reagents. In Grignard reagent, the carbon and magnesium atoms are bonded with each other through polar covalent bond and magnesium atom is attached to halogen by ionic bond.

(Functional part of a Grignard reagent molecule)

In organometallic compounds, the metal atom can be bonded to carbon atom of a hydrocarbon parts (Saturated, unsaturated, aliphatic,

alicyclic or aromatic) or carbon atom of a heterocyclic part. Some examples are given below. Saturated Aliphatic Grignard reagent :- CH₃– Mg

I

(Methylmagnesium iodide)

Unsaturated Aliphatic Grignard reagent :

-(i) Alkenyl Grignard reagent : - CH₂= CH – CH₂– MgX (Allylmagnesium halide) (ii) Alkynyl Grignard reagent : - CH C – CH₂– MgX (Propargylmagnesium halide)

Alicyclic Grignard reagent : - (Cyclohexylmagnesium halide)

Aromatic Grignard reagent : (Phenylmagnesium halide)

Heterocyclic Aromatic Grignard reagent :- (Pyridine-4-magnesium halide) Heterocyclic Nonaromatic Grignard reagent :- (Piperidene-4-magnesium halide)

1.2

Preparation of Grignard Reagent

RX + Mg Ether pure and Dry_{}    _RMgX

Ether is used as a solvent because it is a Lewis base that donates its lone pair of electrons to electron-deficient magnesium atom, therefore providing stability to the Grignard reagent by completing the octet on magnesium atom.

 

1.3

Reactions of Grignard reagents

It has been found out by estimation that there is 35% ionic character in carbon-magnesium bond of Grignard reagent. Therefore, there is a tendency of forming carbanion by heterolysis of this polar coordinate bond as follows.



 +

The carbanion (a nucleophile) formed as shown above, attacks the positively, charged electrophilic centre of other compound. Therefore. It can be said that if a Grignard reagent is regarded as the substrate, then electrophile displaces MgX, i.e. electrophilic substitution (S_E) reaction takes place.

R – MgX  R – E (S_EProduct)

Grignard reagents form adducts by addition on the following types of pi bonds.

, , and C N , – N = O, eg. R – MgX +  Adduct | OMgX C R |  

On having same hydrocarbon radical, the order of reactivity of Grignard reagents will be as follows : RMg

I

> RMgBr > RMgCl

1.4

Synthetic importance of Grignard reagents

1.4.1 Synthesis of Alkanes

(i) With compounds having reactive hydrogen atom General reaction : R – MgX + H – Z  R – H + ZMgX CH₃– MgX + H – OH  CH₃– H + Mg(OH)X C₂H₅– MgX + H – OR  C₂H₅– H + Mg(OR)X C₃H₇– MgX + H – OC₆H₅  C₃H₇– H + Mg(OC₆H₅)X C₂H₅– MgX + H – NH₂  C₂H₅– H + Mg(NH₂)X CH₃– MgX + H – NHR  CH₃– H + Mg(NHR)X C₂H₅– MgX + H – NR₂  C₂H₅– H + Mg(NR₂)X CH₃– MgX + H – NHC₆H₅  CH₃– H + Mg(NHC₆H₅)X C₂H₅– MgX + H – SR  C₂H₅– H + Mg(SR)X CH₃– MgX + H – C N  CH₃– H + Mg(CN)X C₂H₅– MgX + H – C CH  C₂H₅– H + HC C – MgX (Ethynylmagnesium halide) CH₃– MgX + H – C CR  CH₃– H + R – C C – MgX (Alkynylmagnesium halide)

Methane gas is released on reacting methylmagnesium iodide with a compound containing reactive hydrogen atom. The reaction is used for estimation of reactive hydrogen atoms present in a molecule. This method is called Zerewitinoff method of estimation of reactive hydrogen atoms.

(ii) With alkyl halide (coupling) :

R – MgX + X – R  R – R + MgXX 1.4.2 Synthesis of alkenes R – MgX + X halide Allyl 2 2 CH CH CH     R – CH₂– CH = CH₂+ MgXX

1.4.3 Synthesis of higher alkynes Non-terminal alkynes R C  C – H RMgx_RH R – C  CMgX 2 MgX X R  _    _{R – C  C – R}

1.4.4 Synthesis of other organometallics

4R – MgCl + 2PbCl₂  R₄Pb + Pb + 4MgCl₂

Two important antiknocking compounds, tetraethyllead (T.E.L.) and tetramethyllead (T.M.L.) are manufactured by the above reaction.

2R – MgCl + HgCl₂  rcury Diallkylme MgCl 2 R Hg R   ₂ 2R – MgCl + CdCl₂  dmium Diallkylca MgCl 2 R Cd R   ₂ 4R – MgCl + SnCl₄  tin Tetraalkyl MgCl 4 Sn R₄  ₂ 1.4.5 Synthesis of Alcohols

There are following methods to obtain alcohols from Grignard reagent.

(i) From carbonyl compounds

R – MgX +  | OMgX C R |   _{ }HOH_ alcohol | OH C R |  

This is nucleophilic addition reaction (a) Primary or 1° Alcohols

Primary alcohols are formed on taking formaldehyde

R – MgX +  H | OMgX C R | H    HOH Alcohol 1 H | OH C R | H    (b) Secondary or 2° alcohols

(1) From RCHO Secondary alcohols are formed by the reaction of any aldehyde other than formaldehyde. R – MgX +  Alcohol 2 H | OH C R | R    

(2) From Formic Ester :

Secondary alcohols are obtained on hydrolysis of the product obtained by taking excess of Grignard reagent and adding formic ester to it.

O MgX || | OEt C R | H    O || C R | H  O H RMgX 2      OH | R C R | H  

(c) Tertiary or 3° alcohols

(1) Tertiary alcohols are formed by taking any ketone

R – MgX +      R | OMgX C R | R Alcohol 3 R | OH C R | R     

(2) Tertiary alcohols are also obtained on hydrolysis of the product obtained by taking excess of Grignard reagent and an ester of a higher homologue of formic acid.

O MgX || | OEt C R | R    O || C R | R  O H RMgX 2      OH | R C R | R  

Various alcohols can be prepared by changing R in the above synthesis. (iii) From Epoxides

  OMgX | CH CH R 2 2 C more two having Alcohol 1 OH | CH CH R _{2 2}   

(iv) From Oxygen

(a) Synthesis of alcohol

R – MgX + O = O  R – O – O – MgX

R – O – O – MgX + R – MgX  2R – O – MgX

R – O – MgX + HOH  R – O – H + Mg(OH)X

Primary, secondary and tertiary alcohols can be obtained by above reaction (b) Synthesis of phenols

Phenol is obtained on hydrolysis of the product obtained by reaction of arylmagnesium bromide with oxygen.

C₆H₅MgBr + O = O  C₆H₅O – OMgBr

C₆H₅O – OMgBr + C₆H₅MgBr  2C₆H₅– OMgBr C₆H₅– OMgBr + H₂O  C₆H₅– OH + Mg(OH)Br 1.4.6 Synthesis of Aldehydes

Corresponding aldehyde is obtained on hydrolysis of the product obtained by reacting of formic ester and Grignard reagent in equimolar ratio.

O MgX || | OEt C R | H    –EtO– O || C R | H  Tetrahedral intermediate

1.4.7 Synthesis of Ketones

(i) From Acid derivative

  –MgOG O || C R | R   Where G = -OEt, – Cl

The above reaction sequence can be simplified as follows for convenience. R – MgX + O || R C G    O || R C R   + G – MgX eg. R – MgX + O || R C EtO    O || R C R   + EtO – MgX eg. R – MgCl + Cl – CO – R  R – CO – R + MgCl₂

(ii) From Alkyl Cyanides

A ketimine is formed on hydrolysis of the adduct obtained by the reaction of Grignard reagent and an alkyl cyanide, which gives ketone on further hydrolysis.

RMgX + RCN  R₂C = NMgX  HOH e min Keti NH C R₂   HOH Ketone O C R₂ 

1.4.8 Synthesis of Carboxylic acids

A carboxylic acid is formed on hydrolysis of the adduct formed by passing carbon dioxide in the ethereal solution of a Grignard reagent.

O || O C MgX R    O || OMgX C R  X ) OH ( Mg HOH   O || OH C R 

1.4.9 Synthesis of Carboxylic acid esters

Esters are formed on reacting the ethylchloroformate with Grignard reagent.

RMgCl + ClCOOEt  RCOOEt + MgCl₂

1.4.10 Synthesis of Ethers

Higher ethers can be synthesised by reacting a lower chlorinated ether with Grignard reagent. R – MgCl + Cl – CH₂– O – R  R – CH₂– O – R + MgCl₂ CH₃MgCl + ether ethyl yl Chlorometh 5 2 2 O C H CH Cl    ether Diethyl 5 2 2 3 CH O C H CH    1.4.11 Synthesis of Mercaptans

Alkanethiols, i.e. mercaptan is formed on hydrolysis of the product obtained by adding sulphur to the ethereal solution of Grignard reagent.

2. REDUCTION

Introduction

Reduction covers both the addition of hydrogen (or deuterium) to a double bond and the replacement of an atom or group by hydrogen (or deuterium). In other words, reduction means hydrogenation or

hydrogenolysis.

Methods of Reduction :

2.1

Catalytic hydrogenation :

The catalysts used can be divided into two broad classes, both of which

mainly consist of transition metals and their compounds :

2.1.1 Heterogeneous catalysts : (catalysts insoluble in the reaction medium)

In heterogeneous catalytic hydrogenation catalysts are used in powdered form (raney nickel (Ni), Palladium on charcol (Pd/C), Platinum metal or its oxide). Substrate molecules are assumed to undergo homolysis into atoms which are chemisorbed at the surface of the catalyst. The substrate is also chemisorbed on the surface of the catalyst.

General Reaction C = C + H₂





Catalyst







) addition syn ( H H | | C C | |   

Hydrogenation of an alkene is an exothermic reaction (_{ ºH }–120 kJ mol–1_):

R –CH = CH – R + H₂ Ni R – CH₂– CH₂– R + heat

The process is exothermic, there is usually a high free energy of activation for uncatalyzed alkene hydrogenation, therefore, the uncatalyzed reaction does not take place at room temperature. Hydrogenation will take place readily at room temperature in the presence of a catalyst because the catalyst provides a new pathway for the reaction that involves lower free energy of activation .

Heterogeneous hydrogenation catalysts typically involve finely divided platinum, palladium, nickel, or rhodium deposition on the surface of powdered carbon ( charcoal). Hydrogenation actually takes place at the surface of the metal, where the liquid solution of the alkene comes in contact with hydrogen and the catalyst. Hydrogen gas is adsorbed into the surface of these metal catalysts and the catalyst weakens the H – H bond. In fact, if H₂and D₂are mixed in the presence of a Pt catalyst, the two isotopes quickly scramble to produce a random mixture of HD, H₂and D₂. (No scrambling occurs in the absence of the catalyst.) Hydrogenation is an example of heterogeneous catalysis, because the (solid) catalyst is in a different phase from the reactant solution. In contrast, homogeneous catalysis involves reactants and catalyst in the same phase, as in the acid-catalyzed dehydration of an alcohol.

Both hydrogen atoms usually add from the same side of the molecule. This mode of addition is called a syn addition.

The ease of reduction of various functional groups toward catalytic hydrogenation (Ni/Pd/Pt)

Lindlar’s catalys

t :

is a poisoned palladium catalyst, composed of powdered barium sulphate coated

with palladium, poisoned with quinoline. Nickel boride Ni₂B (P-2 catalyst) (made from sodium acetate and sodium borohidride) is an excellent alternative catalyst for the conversion of alkyne into alkene. (syn addition) The partial reduction of alkyne to alkene is heterogeneous hydrogenation with Lindlar's catalyst.

eg. CH₃– CH₂– C C – CH₃ B Ni or catalyst s ' Lindlar H 2 2    (syn addition)

Acid chloride reduced to aldehyde by using Pd /BaSO₄catalyst

      H2/PdBaSO4 CH 3– CHO

This reaction is called Rosenmund Reduction

e.g. CH₃–CC–CH₃

e.g. O₂N – CH = CH – CH₂– CH_{3 }H2/_Pd_ H

2N – CH2– CH2– CH2– CH3

2.2

Reduction by Dissolving Metals

The general mechanism of reduction by dissolving metals is based on the fact that the metal acts as a source of electrons first an electron adds to the substrate causing fission of a single bond into a free radical and an anion or it can adds to a double bond forming a resonance - stablized radical ion.

A – B

 – e _A

+

B–:

or

A–:

+

B

A = B

 – e   B – A :

or

:   B – A

2.2.1 Reduction by Na or Li /NH₃(Birch reduction)

Mechanism :

Reagents Na(or Li,K) + liq NH₃  Na+_{+ e¯ (solvated electron)}

NaNH₂+ (anti addition)

Typical example of reduction for aromatic system

:-    Na/NH3     Na/NH3 R – C C – R Na/NH3 (anti addition) e.g. _Na_/NH_3_

2.2.2 Reduction by Na/C₂H₅OH [Bouvealt - Blanc reduction]

Reduction of aldehydes ketones or esters by means of excess of Na/C₂H₅OH or n-butanol is called Bouvealt-Blanc reduction reduction.

Mechanism :    Na C2H5OH _{ }Na_ RCH₂OH_C2_H5OH_RCH 2– O– Na RCH2– O

·

C2H5OH  Na R CH = O + C2H5O– e.g. CH₃CHO CH₃CH₂OH Acetaldehyde Ethanol CH₃–COCH₃ Acetone propane-2-ol CH₃CH₂OH

Ethyl acetate Ethanol

2.3

Reduction by metal hydrides and alkoxides :

Certain complex metal hydrides, borane and aluminium isopropoxide, are reagents of choice for reduction of carbonyl group.

2.3.1 LiAlH₄(LAH) Lithium aluminium hydride : LAH is most common and versatile of this class of reagent. It is sensitive to protic solvent and therefore used in carefully prepared nonhydroxylic solvent generally ether is used as solvent, It may be regarded as derived from metal hydride.

LiH + AlH₃ 4 – H Al Li  Mechanism :   R2C O 4 R₂CHOH    O H3 _Al–_(OCHR 2)4  O R₂C  l HA (OCHR₂)₃

2.3.2 NaBH₄(Sodium borohydride) : It is more specific than LAH as a reducing agent. It is reduces only ketones and aldehydes to the corresponding alcohols without affecting other functional groups. It is effective even in protic solvent like alcohol. reduces acid chlorides but not and any other derivative of acid.

Mechanism :   R2C O 4 R₂CHOH    O H3 _B–_(OCHR 2)4  O R₂C  B H (OCHR₂)₃ eg. _LiAlH_4_ eg. NaBH4 eg.

2.3.3

DIBAL-H (Diisobutyl Aluminium Hydride) (ALANE) :

Most important alane is diisobutyl aluminium hydride. It runs parallel to LAH (Lithiom aluminium hydride) as a reducing agent but it is more selective.

eg. Ph–CH=CH—COOC₂H₅ O H THF LiAlH 2 4 _       Ph–CH₂–CH₂–CH₂OH + C₂H₅OH Ph–CH=CH—CH₂OH

By DIBAL at ordinary temperature esters are reduced to alcohols but it low temperature exters are reduced

to aldehyde.

eg. C₆H₅–C || O –OCH₃ C º 25 toluene DIBAL_    _C 6H5CH2OH C₆H₅CHO

LAH reduce RCN to amine but DIBAL is found to be reduce it to aldehyde. eg. CH₃–CN O H THF LiAlH 2 4 _       CH₃–CH₂–NH₂ CH₃–CHO

2.3.4

Reduction by isopropyl alcohol and aluminium isopropoxide. This is called the Meerwein-Pondorf-Verley reduction Mechanism :       (CH3)2CHOH eg. + +

2.4

Miscellaneous Reductions

2.4.1

Stephen’s Reductions

When reduction of compounds is carried out with acidified stannous chloride (SnCl₂/HCl) at room temperature, imine hydrochloride is obtained which on subsequent hydrolysis with boiling water gives aldehyde. This specific type of reduction of nitrile is called stephen’s reduction.

eg. CH₃CN O H ) ii ( HCl / SnCl ) i ( 2 2      CH₃CHO Acetaldehyde eg. O H ) ii ( HCl / SnCl ) i ( 2 2      Benzaldehyde

2.4.2

Clemmensen’s Reducation :

Used to get alkane from carbonyl compounds.

R – CHOZnHg/_conc.HCl RCH₃+ H₂O   C R R || O      ZnHg/conc.HCl RCH₂R + H₂O

eg. CH₃– CHO ZnHg/_conc.HClCH₃CH₃+ H₂O

eg. CH₃ C C₂H₅ || O



 + 4[H] ZnHg/_conc.HClCH₃CH₂C₂H₅+ H₂O

Clemmensen reduction is not used for compounds which have acid sensitive group.

2.4.3

Wolff-kishner reduction with NH₂NH₂/ KOH

Used to get alkane from carbonyl compounds

Mechanism : RCHO      NH2NH2/KOH RCH₃ RCO – R      NH2NH2/KOH RCH₂R

Wolff-kishner reduction is not used for compounds which have base sensitive groups.

2.4.4

By Red P & HI      RedP HI

R – CH

_{3 ;}      RedP HI

R – CH

₃      RedP HI

R – CH

_{3 ;}      RedP HI

R – CH

₂

– R

eg.      RedP HI CH₃– CH₃ eg. _  HI P d Re

CH

₃

– CH

₂

– CH

₃

(*) double bond can be reduced by LiAlH

₄

/ THF only in cinnamic system.

For eg. Ph–CH=CH—CHO

O H THF LiAlH 2 4 _       Ph–CH₂–CH₂–CH₂OH

3.

ALKANE

3.1

General method of preparation

3.1.1. By catalytic reduction of alkenes and alkynes R – C C – R Pd or Pt , Ni C 25 , H2_     R – CH₂– CH₂– R R – CH = CH – R Pd or Pt , Ni C 25 , H2_     R – CH₂– CH₂– R

Hydrogenation Addition of H₂to unsaturated bond. Hydrogenation is of two kinds

(a) Heterogeneous (b) Homogeneous

(a) Heterogeneous : It is two phase hydrogenation the catalyst is finely devided metal like Ni, Pt or Pd and a solution of alkene.

(b) Homogeneous : It is one phase hydrogenation both catalyst and alkenes are in solution. In this hydrogenation catalyst are organic complex of transition metal like Rh or

I

r.

Hydrogenation is exothermic, quantitative and during the hydrogenation, total heat evolved to hydrogenate one mole of unsaturated compound is called heat of hydrogenation. Heat of hydrogenation is the measurment of stability of isomeric alkenes.

stability of alkene _{Heatof hydrogenation} 1

Note : Hydrogenation of alkene or alkyne in presence of metal catalyst is syn addition

eg. + H₂ COOH CH Pd 3    (Meso)

eg. + H₂ COOH CH Pd 3    (Racemic mixture) 3.1.2. From alkyl halide

(i) By wurtz reaction

2R – X + 2Na dryether R – R + 2NaX R – X + R – X ) dry ( ether Na_      R – R, R – R, R – R

Mechanism  Two mechanisms are suggested (a) Ionic mechanism

2Na  2Na + ) 2 , 1 ( X R   + 2e¯     SN2 R – R   NaX

(b) Free radical mechanism

Na 

R – X +  R +



R + R  R – R

Note : The alkyl halide should be 1° or 2°, with 3° R – X S_N2 and free radical coupling is not possible due to steric hinderence so in that case elimination or disproportionation is possible.

I

n the ionic mechanism alkyl sodium gives strong base as well as nucleophile which gives S_N2

with R – X, ether should be dry otherwise if moisture is present than forms R – H instead of R – R with H₂O. e.g. CH₃– CH₂– Br _{Na } CH₃– CH₂– CH₂– CH₃

e.g. CH₃– Cl + CH₃– CH₂– Cl _{Na } CH₃– CH₃+ CH₃– CH₂– CH₃+ CH₃– CH₂– CH₂– CH₃ (ii) By reduction of alkyl halides

(A) with metal-acid

  1 X R R – H + HX Reducing agent Zn / HCl , Zn – Cu / H₂O or Zn – Cu + CH₃COOH Zn – Cu / C₂H₅OH, Na – Hg / HCl, Al – Hg / H₂O etc. Mechanism Metal 

(B) With metal hydrides

(a) TPH (Ph₃SnH) Triphenyltin hydride : It reduces 1°, 2° & 3° R – X

º 3 / º 2 / º 1 X – R Ph3SnH R – H (b) NaBH₄ :    3 & 2 X R NaBH4 R – H (c) LiAlH₄:    2 & 1 X R LiAlH4 R – H   3 X R LiAlH4 Alkene

3.1.3 From organometallic compound (i) By Grignard Reagent.

  ,2 3 1 X R _etherM g RMgX RMgX + compound containg H active all   R – H H O2 ROH NH3 R – C CH RSH R – COOH R – H + Mg (OH) X + Mg (OR) X + Mg (NH ) X2 + Mg (CCR ) + Mg (SR) X + Mg (OCOR) X RMgX R – H R – H R – H R – H R – H X

(ii) By corey house alkane synthesis

) 3 , 2 , 1 ( X R      2Li  CuX ) reagent s ' Gilman ( cuprate dialkyl Lithium CuLi R_{2 }__ _   ) 2 1 ( X R R – R Mechanism

R₂CuLi is the source ofR

R – R

R₂CuLi does not reacts with – NO₂, – CN, etc.

(iii) By Franklands reagent

Mechanism

3.1.4 From carboxylic acids:

-(i) By soda lime : Fatty acids are good source of hydrocarbon, continuous heating of sodium salt of carboxylic acid (R – COONa) with soda lime (NaOH – CaO) gives hydrocarbon, this process is known as decarboxyla-tion (e.g. replacement of – COOH group by – H) decarboxyladecarboxyla-tion also takes place on heating only, when compound is gem dicarboxylic acid or there is keto group or double bond on carbon.

O ||

OH C

R  NaOH  CaO_ R – H + Na₂CO₃

(ii) By Kolbe’s electrolysis

2RCOOK + 2HOH Electrolysis R – R + 2CO₂+ H₂+ 2KOH Mechanism R CO₂K R CO₂–_{+ K}+ At Anode: - R CO₂–  _R  2 CO + e– _(oxidation) (I) R CO₂ R + CO₂ (II)  R + R  R – R

If n is the number of carbon atoms in the salt of carboxylic acid, the alkane formed has 2(n–1) carbon atoms. e.g. 2CH₃– COOK + 2H₂O Electrolysis CH₃CH₃+ 2CO₂+ H₂+ 2KOH.

3.1.5 By Reduction

(i) By Clemmensen’s reduction : with Zn – Hg / conc. HCl R – CHOZnHg/_conc.HCl RCH₃+ H₂O   C R R || O      ZnHg/conc.HCl RCH₂R + H₂O

e.g. CH₃– CHO ZnHg/_conc.HClCH₃CH₃+ H₂O

5 2 3 C C H CH || O   + 4[H] ZnHg/_conc.HCl_CH 3CH2C2H5+ H2O

Clemmensen reduction is not used for compounds which have acid sensitive group. (ii) By Wolff-kishner reduction with NH₂NH₂/ KOH

RCHO      NH2NH2 /KOH RCH₃ RCO – R      NH2NH2/KOH RCH₂R

Wolff-kishner reduction is not used for compounds which have base sensitive groups.

eg. ZnHg/_conc.HCl

Clemmemsen’s reduction can be can be used to reduce the given compounds. (iii) By red P & H

I

Red P & H

I

is strong reducing agent

R – COOH RedPHI R – CH₃ O || Cl C R  Red PHI _{R – CH} 3 O || OEt C R  RedPHI R – CH₃ R – X RedPHI R – H R – OH RedPHI R – H

3.2

Physical Properties of Alkanes :

3.2.1 Physical state :

The first four members (C₁to C₄) are gases ; the next thirteen members, (C₅to C₁₇) are liquids while the higher members are waxy solids.

3.2.2. Boiling points :

The boiling points of n-alkanes increase regularly with the increase in the number of carbon atoms.

Among the isomeric alkanes, the branched chain isomers have relatively low boiling points as compared to their corresponding straight chain isomers. Greater the branching of the chain, lower is the boiling point. This is due to the fact that branching of the chain makes the molecule more compact and brings it close to a sphere, so the magnitude of vander wall forces decreases. 3.2.3. Melting Points

It is evident that the increase in melting point is relatively more in moving from an alkane having odd number of carbon atoms to the higher alkane with even no. of ‘C’ while it is relatively less in mov-ing from an alkane with even number of carbon atoms to the higher alkane.

3.2.4. Solubility

In keeping with the popular rule “like dissolves like” hydrocarbons are insoluble in polar solvent like water because they are predominantly non-polar in nature.

3.2.5. Density

The densities of alkanes increase with increasing molecular weight but become constant at about 0.8 g cm–3_{. This means that all alkanes are lighter than water.}

3.3

Chemical Reactions of Alkanes :

Characteristic reaction of alkanes are free radical substitution reaction, these reactions are generally chain reactions which are completed in three steps.

(i) chain initiation (ii) chain propagation. (iii) chain termination 3.3.1. Halogenation :

R – H + X₂ UV₂₅₀Light__₄₀₀ortemp_CR – X + HX

Mechanism of halogenation of alkane  (i) Chain initiation it is an endothermic step.

X₂ C 400 250 . temp or UV          2X•

(ii) Chain propagation

•

X + R – H  R• + HX

•

R + X – X  R – X + X•

(iii) Chain termination it is always exothermic

• X + X•  X₂ • R + R•  R – R • R + X•  R – X

Steps of halogenation, Value ofH for each step. (Kcal/mole)

F Cl Br

I

(i) X₂  2 • X + 38 + 58 +46 +36 (ii) • X + CH₄  ₃ • H C + HX – 32 + 1 + 16 + 33 (iii) C•H₃ + X₂  CH₃X + • X – 70 – 26 – 24 – 20 Reactivity of X₂: F₂> Cl₂> Br₂>

I

₂ Reactivity of H : 3°H > 2°H > 1° H

With F₂alkanes react so vigorously that even in the dark and at room temperature, reactant is diluted with an Inert gas. Iodination is reversible reaction, since H

I

formed as a by product and It is a strong reducing agent and reduces alkyl iodide back to alkane. Hence iodination can be done only in presence of strong oxidising agent like H

I

O₃, HNO₃or HgO.

R – H +

I

₂ R –

I

+ H

I

5H

I

+ H

I

O₃  3H₂O + 3

I

₂

2HI + HgO 

I

₂ + H₂O + Hg

2HI + 2HNO₃ 

I

₂ + 2H₂O + NO₂ Formation of alkyl free radical is rate determining step.

order of stability of Free radical • 3C Ph > Ph CH • 2 > 2 • H C Ph > ₂ • 2 CH CH CH   > (CH3)3  C > (CH₃)₂  CH > CH₃  2 CH > 3 • H C eg. CH₄  _ h Cl₂ HCl Cl CH₃   h Cl₂ HCl Cl CH_{2 2}   h Cl₂ HCl CHCl₃   h Cl₂ HCl CCl₄ 

When equimolar amount of methane and Cl₂is taken, a mixture of four possible products are formed, but when we take excess of CH₄then yield of CH₃Cl will be high.

Each photon of light cleaves one chlorine molecule to form two chlorine radicals, each chlorine atom starts a chain and on an average each chain contains 5000 repeatitions of the chain propagating cycle so about 10,000 molecules of CH₃Cl are formed by one photon of light.

In a chain reaction following reagents are involved

-(i) Initiators they initiate the chain reaction, Initiators are peroxide (R₂O₂), Perester's etc. R – O – O – R       or h • O R O O || || R C O O C R           or h O || O C R •  

(ii) Inhibitors A substance that slows down or stops the reaction is known as inhibitors For example O₂is a good inhibitor

•

R + O₂  ROO• + R•  R – O – O – R

all reactive alkyl free radicals are consumed so reaction stops for a period of time. Relative reactivity of halogen towards methane 

Order of reactivity is F₂> Cl₂> Br₂>

I

₂ which can be explained by the value ofH (enthalpy change)

Halogenation of higher alkane : (i) CH₃– CH₂– CH₃ C 25 , light Cl2        % 55 Cl | CH CH CH₃  ₃ + % 45 Cl CH CH CH₃ ₂ ₂ (ii) CH₃– CH₂– CH₂– CH₃ C 25 , light Cl2       % 72 Cl | CH CH CH CH3 2  3 + % 28 Cl CH CH CH CH₃ ₂ ₂ ₂ (iii) 3 3 3 CH | CH CH CH   C 25 , light Cl2       % 64 CH | Cl CH CH CH 3 2 3   + % 36 Cl | CH C CH | CH 3 3 3   (iv) CH₃– CH₂– CH₃ C 127 , heat Br₂        % 3 Br CH CH CH3 2 2 + % 97 Br | CH CH CH3  3

(v) 3 3 3 CH | CH CH CH   C 127 , heat Br₂        trace CH | Br CH CH CH 3 2 3  + % 99 over CH | Br C CH | CH 3 3 3  

Relative amounts of the various isomers differ remarkably depending upon the halogen used. From the above reaction, it is observed that chlorination gives mixture in which no isomer greatly dominates while, in bromi-nation gives a mixture in which one isomer dominates greatly (97% – 99%),

Factors affecting the relative yields

:-Factors determining the relative yields of the isomeric products.

(i) Probability factor : This factor is based on the number of each kind of H atom in the molecule. (ii) Reactivity of hydrogen : The order of reactivity is 3° > 2° > 1° the relative rate per hydrogen atom is found to be

(iii) Reactivity v/s selectivity principle The more reactive is halogen less selective it will be, so the more reactive chlorine free radical is less selective and more influenced by the probability factor and the less reactive

B

r

•

is more selective and less influenced by the probability factors.

Based on relative reactivity of different types of H, percentage of each in the product mixture can be calcu-lated. e.g. % yield A = 2 . 21 6 × 100 = 28.3 % ; % yield B = 2 . 21 2 . 15 × 100 = 71.%

3.3.2. Sulphonation : Lower alkanes are not easily sulphonated but hexane & higher members are sulphonated on heating with oleum (conc. H₂SO₄+ SO₃) at 400°C

C₆H₁₄+ H₂SO₄ 400C C₆H₁₃SO₃H + H₂O 3.3.3. Isomerisation : CH₃CH₂CH₂CH₃ C 300 HCl / AlCl . Anhyd 3         e tan Isobu 3 3 3 CH | CH CH CH   CH₃– (CH₂)₃– CH₃ AlCl₃₀₀3/_HCl_C e tan Neopen 3 3 3 3 CH | CH C CH | CH  

3.3.4. Alkylation : Isoalkanes add to isoalkenes in presence of conc. H₂SO₄or HF to give higher branched alkane e tan Isobu 3 3 3 CH | H C CH | CH   + Isobutene 3 3 2 CH | CH C C H   conc.H2SO4 e tan Isooc CH CH | | CH CH CH C CH | CH 3 3 3 2 3 3     3.3.5. Aromatisation :

Alkanes containing 6 or more number of C atoms are oxidised (dehydrogenate & cyclised) in presence of oxidising agent at higher temparature and form benzene or its derivatives.

CH₃– (CH₂)₄– CH₃ P High C 600 3 2 3 2O AlO Cr        + 4H₂ CH₃– (CH₂)₅– CH₃ P High C 600 3 2 3 2O AlO Cr        CH₃(CH₂)₆CH₃ P High C 600 3 2 3 2O AlO Cr        + 3 4 2 3 3 CH CH ) CH ( CH | CH    P High C 600 3 2 3 2O AlO Cr        3 2 3 2 3 3 CH CH CH ) CH ( CH | CH     P High C 600 3 2 3 2O AlO Cr        3.3.6 Pyrolysis / Cracking CH₃– CH₂– CH₃ CH₃CH₃+ CH₃CH = CH₂ + CH₂= CH₂+ CH₄+ H₂

Higher alkanes are heated in absence of air so these compounds break down into smaller alkanes which are better fuel. Mixture of products contains all lower alkanes, alkenes & hydrogens.

3.3.7. Combustion : C_nH_2n+2+ _  _ 2 1 n 3

O₂ combustion_  nCO₂+ (n + 1) H₂O (H_combustion= -ve) exothermic reaction C_xH_y+        4 y x _O 2  combustion xCO₂+ 2 y H₂O C₅H₁₂+ 8O₂ combustion_  5CO₂+ 6H₂O

Heat of combustion : Amount of heat liberated when 1 mole of hydrocarbon is completely burnt into CO₂& H₂O.

 H of combustion is a measurement of stability of alkane : Combustion is used as a measurment of stability.

More branched alkanes are more stable and have lower heat of combustion.

e.g. CH₃–

I

2 CH – CH2– CH3

II

H | CH C CH | CH 3 3 3   H_comb. :

I

>

II

Stability :

II

>

I

Points to remember forH of combustion

Homologes : Higher homologes have higher heat of combustion. CH₄< C₂H₆< C₃H₈

Isomers : Branched isomer has lower heat of combustion.More branched alkane has more no. of primary C – H bonds. (therefore it has more bond energy).

> >

Cyclo-alkanes : More strained ring has higher heat of combustion per – CH₂– unit.

> > > (H_comb.per ‘CH₂’ unit)

3.3.8 Octane Number : It is a scale of fuel efficiency when the fuel burns during combustion, more branched alkanes have lower knocking (cracking sound). so are better fuels. On commercial scale iso-octane has been alotted a rating i.e. octane no. of Isooctane is 100 & n-heptane is - ‘0’

If the octane no. of a fuel is 80, it means that the efficiency of the fuel is equivalent to the efficiency of mixture of 80% isoctane and 20% n-heptane.

MISCELLANEOUS SOLVED PROBLEMS

1. Which of the following formula represents Grignard reagent ?

(A) H₂NMgX (B) HC CMgX (C) R₂Mg (D) Mg(OH)Br

Ans. B

Sol. Grignard reagent is represented by RMgX. 2. Which of the following is not a Grignard reagent ?

(A) (CH₃)₂CHMgCl (B) CH₃COOMg

I

(C) C₆H₁₁MgBr (D) C₂H₅MgCl

Ans. B

Grignard reagent is represented by RMgX. Where R = Carbon.

3. What is the motive of adding iodine in small amount during preparation of methylmagnesium iodide ?

(A) It acts as a catalyst (B) It converts ether into alkyl iodide

(C) It decreases the violence of the reaction (D) It acts as an electrophilic

Ans. A

Sol. It acts as a catalyst

4. The reaction of an alkyl halide and magnesium is initially very slow, but later becomes very fast, because

(A) The reaction is autocatalytic (B) The Grignard reagent formed is ether-soluble

(C) The reaction is endothermic (D) The reaction is exothermic

Ans. D

R–X + MgR–MgX H = –ve

5. A Grignard reagent is formed on reacting magnesium with (A) Alkyl halide in the presence of dry ether

(B) Alkyl halide in the presence of phenol (C) Alkyl halide in the presence of alcohol (D) Alkyl halide in the presence of alcoholic ether

Ans. A

Sol. R–X + MgR–MgX (Grignard Reagent)

6. Zerewitinoff's method is used for

(A) the estimation of reactive hydrogen atom (B) the estimation of alkoxy group (C) the preparation of a higher ether (D) the preparation of a Grignard reagent

Ans. A

Sol. Z–H + R–MgXR–H + MgXZ

This is estimation of reactive hydrogen atom.

7. Primary alcohol cannot be obtained by the reaction of a Grignard reagent with

(A) The simplest alkanal (B) Dimethylene oxide

(C) Ethylene glycol (D) Oxygen

Ans. C

Sol. RMgX + HCHOR–CH₂OH

RMgX + (CH₂)₂OR–CH₂–CH₂–OH

R–MgX + (CH₂OH)₂R–H + MgXOCH₂CH₂OH

RMgX + O₂ROH

8. Primary, secondary and tertiary alcohols can be obtained by hydrolysis of the products formed on reaction of a Grignard reagent with

(A) Isobutyraldehyde (B) Butanone (C) Oxygen (D) Oxirane

Ans. C

Sol. R–MgX + O₂ROMgX _H₃O__ROH

9. Which of the following is formed on the reaction of one mole of acetyl chloride with one mole of methylmagnesium chloride ?

(A) 2-Butanol (B) Acetone (C) Isobutyl alcohol (D) t-Butyl alcohol

Ans. B

Sol. 

10. Identify the product - X

H₅C₆– C C – C₆H_{5 }Lindlar_'sCatalyst_{ X} Sol. 11. Identify X     H2/Pd _X Sol. X =

12. Identify the product x and y

+ D_{2 }(Ph_3P)_3RhCl_ X

+ D_{2 }(Ph_3P)_3RhCl_ Y

Sol. X = ; Y =

(Meso) (dl) Recemic mixture

13. Identify product ?     LiAlH4 P Sol. P = 14. Where x and y ?

15.

Write the products.

Sol.

(a)

(b)

16.

Sol. In all cases product is

17. Arrange the following alkenes in decreasing order of their stability

R₂C = CR₂, CH₂= CH₂, R₂C = CH₂, R – CH = CH – R (cis, trans), R₂C = CHR, RCH = CH₂ Sol. R₂C=CR₂> R₂C = CHR >R₂C = CH₂> trans R CH CH R   > cis CHR RCH  > R – CH = CH₂> CH₂= CH₂

18. What is the order of stability of following alkene

I

3 2 3 CH | CH CH CH CH    _, II 3 2 2 3 CH CH C CH | CH    _, III3 3 3 CH | CH CH C CH    Ans.

III > II > I

Sol. Stability of alkene Hyperconjugative Structure. 19. C₂H₅Cl _EtherNa  products

Write all possible product

Sol. CH₃– CH₂– Cl  major CH CH CH CH₃ ₂ ₂ ₃ CH₃– CH₂– Cl 2 3 CH CH        coupling CH₃– CH₂– CH₂– CH₃        disproportion CH₃– CH₃ + CH₂= CH₂

20. Wurtz reaction will not be observed in (A) (B) (C) 3 3 CH | Br CH CH   (D) none of these Ans. B Sol. _{ }Na_{ + H} 2 gas 21. CH₃– Br Li A  CuI B Y C if C is CH₃– CH₂– (CH₂)₅– CH₃, than what is Y. Ans. CH₃– (CH₂)₆– Br Sol. CH₃– Br Li CH₃Li  CuI (CH₃)₂CuLi _CH_3–_(CH_2)_5–_CH_2–_X_CH 3– CH2– (CH2)5– CH3 22.  A       / CaO NaOH B What are A and B

Sol. A is , B is

23. Explain why the chain initiating step in thermal chlorination of CH₄is

Cl₂  Cl • and not CH₄  ₃ • H C + H•

Sol. Because E_actof Cl₂is less than E_actof CH₄

24. Chlorination of CH₄involves following steps :

(i) Cl₂  2Cl • (ii) CH₄+ Cl •   C•H₃ + HCl (iii) ₃ • H C + Cl •   CH₃Cl

Which of the following is rate determining ?

(A) Step (i) (B) Step (ii) (C) Step (iii) (D) Step (ii) and (iii) both

Ans. B