//M2//QN1//SUB//DL0//EQ

What are haloalkane and haloarenes? Explain with examples.

//X

“The replacement of hydrogen atom(s) in an aliphatic or aromatic hydrocarbon by halogen atom(s) results in the formation of alkyl halide (haloalkane) and aryl halide (haloarene), respectively.”

Haloalkanes contain halogen atom(s) attached to the sp3 hybridised carbon atom of an alkyl group whereas haloarenes contain halogen atom(s) attached to sp2 hybridised carbon atom(s) of an aryl group.

//M2//QN2//SUB//DL0

Write down uses of halogen containing organic compounds.

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//M2//QN3//SUB//DL0//EQ

Explain classification of haloalkane and haloarene on the basis of number of halogen atoms.

//X

These may be classified as mono, di, or polyhalogen (tri-,tetra-, etc.) compounds depending on whether they contain one, two or more halogen atoms in their structures.

Examples:

//M4//QN4//SUB//DL0//EQ

Explain classification of monohalogen compounds on the basis of sp3 (C – X) bonds.

//X

These are the types:

(a) Alkyl halides or haloalkanes (R – X)

(b) Allylic halides

(c) Benzylic halides

(a) Alkyl halides or haloalkanes (R—X)

“In alkyl halides, the halogen atom is bonded to an alkyl group (R).”

They form a homologous series represented by CnH2n+1X.

They are further classified as primary, secondary or tertiary according to the nature of carbon to which halogen is attached.

If halogen is attached to a primary carbon atom in an alkyl halide, the alkyl halide is called primary alkyl halide or 1° alkyl halide.

Similarly, if halogen is attached to secondary or tertiary carbon atom, the alkyl halide is called secondary alkyl halide (2°) and tertiary (3°) alkyl halide, respectively.

(b) Allylic halides:

“These are compounds in which the halogen atom is bonded to an sp3-hybridised carbon atom adjacent to carbon-carbon double bond.”

(c) Benzylic halides:

“There are the compounds in which the halogen atom is bonded to an sp3-hybridised carbon atom attached to an aromatic ring.”

//M2//QN5//SUB//DL0//EQ

What are Vinylic halides and Aryl halides Explain with examples. OR Explain classification of halogen compound containing sp2 C – X bonds.

//X

(a) Vinylic halides: “These are compounds in which the halogen atom is bonded to a
sp2-hybridised carbon atom of a carbon-carbon double bond (C = C).”

(b) Aryl halides: “These are compounds in which the halogen atom is directly bonded to the sp2-hybridised carbon atom of an aromatic ring.”

//M2//QN6//SUB//DL0//EQ

Explain the Common and IUPAC nomenclature of haloalkane and haloarene compounds

with examples.

//X

The common names of alkyl halides are derived by naming the alkyl group followed by the name of halide.

In the IUPAC system of nomenclature, alkyl halides are named as halosubstituted hydrocarbons.

For mono halogen substituted derivatives of benzene, common and IUPAC names are the same.

For dihalogen derivatives, the prefixes o-, m-, p- are used in common system but in IUPAC system, the numerals 1,2; 1,3 and 1,4 are used.

Common name: n-Propyl bromide Isopropyl chloride Isobutyl chloride

IUPAC name: 1-Bromopropane 2-Chloropropane 1-Chloro-2-methylpropane

Common name: Bromobenzene m-Dibromobenzene sym-Tribromobenzene

IUPAC name: Bromobenzene 1,3-Dibromobenzene 1,3,5-Tribromobenzene

Common name: Neopentylchloride isopropylbromide

IUPAC name: 1-Chloro-2,2-dimethylpropane 2-Bromopropane

//M2//QN7//SUB//DL0//EQ

Explain geminal and vicinal dihalide with examples.

//X

When both the halogen atoms are present on the same carbon atom, it is known as geminal dihalide.

In common name system, gem-dihalides are named as alkylidene halides.

When halogen atoms are present on adjacent carbon atoms, it is known as vicinal dihalides.

In common name system, vic-dihalides are named as alkylene dihalides.

Common name: Ethylidene chloride (gem-dihalide) Ethylene dichloride (vic-dihalide)

IUPAC name: 1, 1-Dichloroethane 1, 2-Dichloroethane

Common and IUPAC Names of some Halide Compounds
Structure	Common Name	IUPAC Name
CH3CH2CH(Cl)CH3	sec-Butyl chloride	2-Chlorobutane
(CH3)3CCH2Br	neo-Pentyl bromide	1-Bromo-2,2-dimethylpropane
(CH3)3CBr	tert-Butyl bromide	2-Bromo-2-methylpropane
CH2 = CHCl	Vinyl chloride	Chloroethene
CH2 = CHCH2Br	Allyl bromide	3-Bromopropene
	o-Chlorotoluene	1-Chloro-2-methylbenzene OR 2-Chlorotoluene
	Benzyl chloride	Chlorophenylmethane
CH2Cl2	Methylene chloride	Dichloromethane
CHCl3	Chloroform	Trichloromethane
CHBr3	Bromoform	Tribromomethane
CCl4	Carbon tetrachloride	Tetrachloromethane
CH3CH2CH2F	n-Propyl fluoride	1-Fluoropropane

//M2//QN8//SUB//DL0//EQ

Write a note on nature of C-X bond.OR Explain polarity of C – X bond in Haloalkane.

//X

Halogen atoms are more electronegative than carbon, therefore, carbon-halogen bond of alkyl halide is polarised.

The carbon atom bears a partial positive charge whereas the halogen atom bears a partial negative charge.

As we go down the group in the periodic table, the size of halogen atom increases. Fluorine atom is the smallest and iodine atom is the largest.

Consequently the carbon-halogen bond length also increases from C–F to C–I.

Bond Length: C – F < C – Cl < C – Br < C – I

Bond enthalpy: C – F > C – Cl > C – Br > C – I

Dipole Moment: CH3 – Cl > CH3 – F > CH3

– Br > CH3 – I

//M4//QN9//SUB//DL0//EQ

Write down the preparation of alkyl halides from alcohols.

//X

The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides or thionyl chloride etc.

(i) Reaction with thionyl chloride (SOCl2):

Thionyl chloride is first preferred because in this reaction alkyl halide is formed along with gases SO2 and HCl. The two gaseous products are escapable, hence, the reaction gives pure alkyl halides.

R–OH + SOCl2 → R–Cl + SO2 + HCl

(ii) Reaction with halogen acids or Lucas Test:

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

Constant boiling with HBr (48%) is used for preparing alkyl bromide.

Good yields of R—I may be obtained by heating alcohols with sodium or potassium iodide in 95% orthophosphoric acid.

The order of reactivity of alcohols with a given haloacid is 3° > 2° > 1°.

R–OH + HCl R–Cl + H2O

(iii) Reaction with Phosphorus halides:

Phosphorus tribromide and triiodide are usually generated in situ (produced in the reaction mixture) by the reaction of red phosphorus with bromine and iodine respectively.

3R–OH + PX3 → 3R–X + H3PO3(X = Cl, Br)

R–OH + PCl5 → R–Cl + POCl3 + HCl

(iv) Reaction with NaBr and H2SO4:

R – OH + NaBr + H2SO4 → R–Br + NaHSO4 + H2O

(v) Reaction with halogen:

R–OH R–X

The above methods are not applicable for the preparation of aryl halides because the carbon-oxygen bond in phenols has a partial double bond character and is difficult to break being stronger than a single bond.

//M3//QN10//SUB//DL0//EQ

Explain the preparation of haloalkanes from hydrocarbon.

//X

(I) From alkanes by free radical halogenation :

Free radical chlorination or bromination of alkanes gives a complex mixture of isomeric mono- and polyhaloalkanes, which is difficult to separate as pure compounds. Consequently, the yield of any single compound is low.

Example:

(II) From alkenes:

(a) Addition of hydrogen halides:

An alkene is converted to corresponding alkyl halide by reaction with hydrogen chloride, hydrogen bromide or hydrogen iodide.

Propene yields two products, however only one predominates as per Markovnikov’s rule.

(b) Addition of halogens:

In the laboratory, addition of bromine in CCl4 to an alkene resulting in discharge of reddish brown colour of bromine constitutes an important method for the detection of double bond in a molecule.

The addition results in the synthesis of vic-dibromides, which are colourless.

//M2//QN11//SUB//DL0//EQ

Explain Finkelstein and Swarts reaction. OR Explain preparation of haloalkane by halogen exchange method.

//X

(1) Finkelstein reaction:

Alkyl iodides are often prepared by the reaction of alkyl chlorides/ bromides with NaI in dry acetone.

This reaction is known as Finkelstein reaction.

R–X + Nal R–I + NaX

X = Cl, Br

Example:

(i) CH3 - Cl + NaI → CH3 - I + NaCl

(ii) CH3 - CH2 - Br + NaI → CH3 - CH2 - I + NaBr

NaCl or NaBr thus formed is precipitated in dry acetone.

(2) Swarts reaction:

The synthesis of alkyl fluorides is best accomplished by heating an alkyl chloride/bromide in the presence of a metallic fluoride such as AgF, Hg2F2, CoF2 or SbF3.

The reaction is termed as Swarts reaction.

H3C - Br + AgF → H3C–F + AgBr

//M2//QN12//SUB//DL0

Explain preparation of haloarene compounds from hydrocarbons.

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//M2//QN13//SUB//DL0//EQ

Explain preparation of haloarene from amine compound. OR Preparation of hyloarene by Sandmayer reaction.

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When a primary aromatic amine, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed.

Mixing the solution of freshly prepared diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by –Cl or –Br.

This reaction is known as Sandmayer reaction.

Replacement of the diazonium group by iodine does not require the presence of cuprous halide and is done simply by shaking the diazonium salt with potassium iodide.

//M2//QN14//SUB//DL0

Write down physical state, colour and odour of halogen compound.

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//M3//QN15//SUB//DL0//EQ

Explain boiling point of alkyl halides.

//X

Molecules of organic halogen compounds are generally polar.

Due to greater polarity as well as higher molecular mass as compared to the parent hydrocarbon, the intermolecular forces of attraction (dipole-dipole and van der Waals) are stronger in the halogen derivatives.

That is why the boiling points of chlorides, bromides and iodides are considerably higher than those of the hydrocarbons of comparable molecular mass.

The pattern of variation of boiling points of different halides is depicted in Fig.

For the same alkyl group, the boiling points of alkyl halides decrease in the order: RI > RBr > RCl > RF.

This is because with the increase in size and mass of halogen atom, the magnitude of van der Waal forces increases.

The boiling points of isomeric haloalkanes decrease with increase in branching (1° > 2° > 3°).

For example, 2-bromo-2-methylpropane has the lowest boiling point among the three isomers.

//M2//QN16//SUB//DL0//EQ

Explain boiling point and melting point of dihalo-benzene.OR Give reason why melting point of p-Dichloro benzene is more than ortho and meta isomers.

//X

Boiling points of isomeric dihalobenzenes are nearly the same.

However, the para-isomers are high melting as compared to their ortho- and meta-isomers.

It is due to symmetry of para-isomers that fits in crystal lattice better as compared to ortho- and meta-isomers.

//M2//QN17//SUB//DL0

Explain density of haloalkane compound.

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//M3//QN18//SUB//DL0

Explain solubility of haloalkane.

//X

//M3//QN19//SUB//DL0//EQ

What is nucleophilic substitution reaction? Write down different products obtained from alkyl halides.

//X

Nucleophiles are electron rich species. Therefore, they attack at that part of the substrate molecule which is electron deficient.

The reaction in which a nucleophile replaces already existing nucleophile in a molecule is called nucleophilic substitution reaction.

Haloalkanes are substrate in these reactions.

In this type of reaction, a nucleophile reacts with haloalkane (the substrate) having a partial positive charge on the carbon atom bonded to halogen.

A substitution reaction takes place and halogen atom, called leaving group departs as halide ion. Since the substitution reaction is initiated by a nucleophile, it is called nucleophilic substitution reaction.

It is one of the most useful classes of organic reactions of alkyl halides in which halogen is bonded to sp3 hybridised carbon.

R – X + Nu– → R – Nu + X–

Reagent	Nucleophile (Nu–)	Substitution product R – Nu	Class of main product
NaOH (KOH)	HO–	ROH	Alcohol
H2O	H2O	ROH	Alcohol
NaOR'	R'O–	ROR'	Ether
Nal	I–	R-I	Alkyl iodide
NH3	NH3	RNH2	Primary amine
R'NH2	R'NH2	RNHR'	Sec. amine
R'R"NH	R'R"NH	RNR'R"	Tert. amine
KCN		RCN	Nitrile (cyanide
AgCN	Ag-CN	RNC	Isonitrile (isocyanide)
KNO2	O = N – O	R – O – N = O	Alkyl nitrite
AgNO2		R-NO2	Nitroalkane
R'COOAg	R'COO–	R'COOR	Ester
LiAlH4	H	RH	Hydrocarbon
R'–M+	R'	RR'	Alkane

//M0//QN20//SUB//DL0//EQ

What are ambident nucleophiles? Explain with an example.

//X

Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident nucleophiles have two sites through which they can attack.

Groups like cyanides and nitrites possess two nucleophilic centres are called ambident nucleophiles.

Cyanide group is a hybrid of two contributing structures and therefore, it can act as a nucleophile in two different ways .

The Linking through carbon atom resulting in alkyl cyanides and through nitrogen atom leading to isocyanides.

Similarly nitrite ion also represents an ambident nucleophile with two different points of linkage .

The linkage through oxygen results in alkyl nitrites while through nitrogen atom, it leads to nitroalkanes.

//M3//QN21//SUB//DL0//EQ

Explain bimolecular nucleophilic substitution (SN2) reaction with example.OR Explain Mechanism of following reaction : CH3Cl + : OH– → CH3OH + : Cl–

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In the year 1937, Edward Davies Hughes and Sir Christopher Ingold proposed a mechanism for an SN2 reaction.

The reaction between CH3Cl and hydroxide ion to yield methanol and chloride ion follows a second order kinetics.

The rate depends upon the concentration of both the reactants.

Rate =

The incoming nucleophile interacts with alkyl halide causing the carbon-halide bond to break and a new bond is formed between carbon and attacking nucleophile.

Here, it is C-O bond formed between C and -OH. These two processes take place simultaneously in a single step and no intermediate is formed.

As the reaction progresses and the bond between the incoming nucleophile and the carbon atom starts forming, the bond between carbon atom and leaving group weakens.

As this happens, the three C-H bonds of the substrate start moving away from the attacking nucleophile.

In transition state all the three C-H bonds are in the same plane and the attacking and leaving nucleophiles are partially attached to the carbon.

As the attacking nucleophile approaches closer to the carbon, C-H bonds still continue to move in the same direction till the attacking nucleophile attaches to carbon and leaving group leaves the carbon.

As a result configuration is inverted. This process is called inversion of configuration.

In the transition state, the carbon atom is simultaneously bonded to incoming nucleophile and the outgoing leaving group.

Such structures are unstable and cannot be isolated. Thus, in the transition state, carbon is simultaneously bonded to five atoms.

//M2//QN22//SUB//DL0//EQ

Explain reactivity order towards SN2 reaction of alkyl halides.

//X

Reactivity order of primary secondary and tertiary halides toward SN2 reaction is
CH3–X > 1° – halide > 2° – halide > 3° – halide.

The presence of bulky substituents on or near the carbon atom has a dramatic inhibiting effect.

Of the simple alkyl halides, methyl halides react most rapidly in SN2 reactions because there are only three small hydrogen atoms.

Tertiary halides are the least reactive because bulky groups hinder the approach of nucleophiles.

Thus, the order of reactivity followed is: Primary halide > Secondary halide > Tertiary halide.

//M3//QN23//SUB//DL0//EQ

Explain unimolecular nucleophilic substitution (SN1) reaction with mechanism.

//X

SN1 reactions are generally carried out in polar protic solvents (like water, alcohol, acetic acid etc.)

The reaction between tert- butyl bromide and hydroxide ion yields tert-butyl alcohol and follows the first order kinetics,

The rate of reaction depends upon the concentration of only one reactant, which is tert- butyl bromide.

Rate = K[(CH3)3 C Br]

(CH3)3CBr + –OH (CH3)3 COH + Br–

2-Bromo-2-methylpropane 2-methylpropan-2-ol

It occurs in two steps. In step I, the polarised
C–Br bond undergoes slow cleavage to produce a carbocation and a bromide ion.

Step I:

The carbocation formed is then attacked bynucleophile in step II to complete the substitution reaction.

Step II:

Step I is the slowest and reversible. It involves the C–Br bond breaking, for which the energy is obtained through solvation of halide ion with the proton of protic solvent.

Since the rate of reaction depends upon the slowest step, the rate of reaction depends only on the concentration of alkyl halide and not on the concentration of hydroxide ion.

Further, greater the stability of carbocation, greater will be its ease of formation from alkyl halide and faster will be the rate of reaction. In case of alkyl halides, 3o alkyl halides undergo SN1 reaction very fast because of the high stability of 3o carbocations.

Reactivity order for SN1 reaction

3° – halide > 2° – halide > 1° – halide > CH3 – X

//M2//QN24//SUB//DL0//EQ

Explain reactivity order of SN1 and SN2 reaction for alkyl halide, allylic halide and benzylic halide.

//X

The order of reactivity of alkyl halides towards SN1and SN2 reactions are as follows:

For the same reasons, allylic and benzylic halides show high reactivity towards the SN1 reaction. The carbocation thus formed gets stabilised through resonance as shown below:

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.

//M3//QN25//SUB//DL0//EQ

What is stereochemistry? Explain plane polarised light, optical activity and optical isomers.

//X

Stereochemistry is the branch of chemistry that involves "the study of the different spatial arrangements of atoms in molecules".

Plane polarised light (PPL) is obtained when ordinary light is passed through a nicol prism a light of single wavelength vibrating in a single plane is obtained this is known as plane polarised light.

Optical activity: “When plane polarised light is passed through a solution the light is rotated either in clockwise or in a anticlockwise direction by a certain angle. These types of compounds are known as optical active compound and such a property is known as optical activity.”

The angle by which the plane polarised light is rotated is measured by an instrument called polarimeter.

If the compound rotates the plane of plane polarised light to the right, i.e., in a clockwise direction, it is called dextrorotatory (Greek for right rotating) or the d-form and is indicated by placing a positive (+) sign before the degree of rotation.

If the light is rotated towards left (anticlockwise direction), the compound is said to be laevo-rotatory or the l-form and a negative (–) sign is placed before the degree of rotation.

Such (+) and (–) isomers of a compound are called optical isomers and the phenomenon is termed as optical isomerism.

//M3//QN26//SUB//DL0//EQ

Explain molecular asymmetry, chirality and enantiomers.

//X

The observation of Louis Pasteur (1848) that crystals of certain compounds exist in the form of mirror images laid the foundation of modern stereochemistry.

He demonstrated that aqueous solutions of both types of crystals showed optical rotation, equal in magnitude (for solution of equal concentration) but opposite in direction.

He believed that this difference in optical activity was associated with the three dimensional arrangements of atoms in the molecules (configurations) of two types of crystals.

Dutch scientist, J. Van’t Hoff and French scientist, C. LeBel in the same year (1874), independently argued that the spatial arrangement of four groups (valencies) around a central carbon is tetrahedral and if all the substituents attached to that carbon are different, the mirror image of the molecule is not superimposed (overlapped) on the molecule; such a carbon is called asymmetric carbon or stereocentre.

The resulting molecule would lack of symmetry and is referred to as asymmetric molecule.

Chirality: The symmetry and asymmetry are also observed in many day to day objects: a sphere, a cube, a cone, are all identical to their mirror images and can be superimposed.

However, many objects are non superimposable on their mirror images. For example, your left and right hand look similar but if you put your left hand on your right hand by moving them in the same plane, they do not coincide.

The objects which are non-superimposable on their mirror image (like a pair of hands) are said to be chiral and this property is known as chirality.

Chiral molecules are optically active, while the objects, which are, superimposable on their mirror images are called achiral. These molecules are optically inactive.

The above test of molecular chirality can be applied to organic molecules by constructing models and its mirror images or by drawing three dimensional structures and attempting to superimpose them in our minds.

Let us consider two simple molecules propan-2-ol and butan-2-ol and their mirror images.

As you can see very clearly, propan-2-ol (A)
does not contain an asymmetric carbon, as all the four groups attached to the tetrahedral carbon
are not different.

We rotate the mirror image (B) of the molecule by 180° (structure C) and try to overlap the structure (C) with the structure (A), these structures completely overlap. Thus propan-2-ol is an achiral molecule.

Butan-2-ol has four different groups attached to the tetrahedral carbon and as expected is chiral.

Some common examples of chiral molecules such as 2-chlorobutane, 2, 3-dihyroxypropanal, (OHC–CHOH–CH2OH), bromochloro-iodomethane (BrClCHI), 2-bromopropanoic acid (H3C–CHBr–COOH), etc.

The stereoisomers related to each other as non-superimposable mirror images are called enantiomers . A and B and D and E are enantiomers.

Enantiomers possess identical physical properties namely, melting point, boiling point, refractive index, etc.

They only differ with respect to the rotation of plane polarised light. If one of the enantiomer is dextro rotatory, the other will be laevo rotatory.

//M2//QN27//SUB//DL0

What is racemic mixture?

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//M2//QN28//SUB//DL0

What is retention? Explain with examples.

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//M2//QN29//SUB//DL0

Explain inversion, retention and racemisation.

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//M4//QN30//SUB//DL0//EQ

Explain stereochemistry of SN2 and SN1 reaction.

//X

In case of optically active alkyl halides, the product formed as a result of SN2 mechanism has the inverted configuration as compared to the reactant.

This is because the nucleophile attaches itself on the side opposite to the one where the halogen atom is present.

When (–)-2-bromooctane is allowed to react with sodium hydroxide, (+)-octan-2-ol is formed with the –OH group occupying the position opposite to what bromide had occupied.

Thus SN2 reactions of optically active halides are accompanied by inversion of configuration.

SN1 reaction:

In case of optically active alkyl halides, SN1 reactions are accompanied by racemisation.

Actually the carbocation formed in the slow step being sp2 hybridised is planar (achiral).

The attack of the nucleophile may be accomplished from either side of the plane of carbocation resulting in a mixture of products,

One having the same configuration (the –OH attaching on the same position as halide ion) and the other having opposite configuration (the –OH attaching on the side opposite to halide ion).

This may be illustrated by hydrolysis of optically active 2-bromobutane, which results in the formation of (±)-butan-2-ol.

//M3//QN31//SUB//DL0

Write down the difference between SN1 and SN2 reaction.

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//M3//QN32//SUB//DL0//EQ

Explain β – Elimination reaction with example. OR Explain Dehydrohalogenation reaction of Haloalkane. OR Explain Zaitsev or Saytzeff rule with example.

//X

When a haloalkane with β-hydrogen atom is heated with alcoholic solution of potassium hydroxide, there is elimination of hydrogen atom from β carbon and a halogen atom from the α-carbon atom.

As a result, an alkene is formed as a product. Since β-hydrogen atom is involved in elimination, it is often called β-elimination.

If there is possibility of formation of more than one alkene due to the availability of more than one
β-hydrogen atoms, usually one alkene is formed as the major product.

These form part of a pattern was first observed by Russian chemist, Alexander Zaitsev (also pronounced as Saytzeff).

A rule which can be summarised as in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.

Thus, 2-bromopentane gives pent-2-ene as the major product.

//M3//QN33//SUB//DL0//EQ

What is organo metallic compound? Explain Grignard reagent.

//X

Most organic chlorides, bromides and iodides react with certain metals to give compounds containing
carbon-metal bonds. Such compounds are known as organo-metallic compounds.

An important class of organo-metallic compounds discovered by Victor Grignard in 1900 is alkyl
magnesium halide, RMgX, referred as Grignard Reagents.

These reagents are obtained by the reaction of haloalkanes with magnesium metal in dry ether.

In the Grignard reagent, the carbon-magnesium bond is covalent but highly polar, with carbon pulling electrons from electropositive magnesium; the magnesium halogen bond is essentially ionic.

Grignard reagents are highly reactive and react with any source of proton to give hydrocarbons. Even water, alcohols, amines are sufficiently acidic to convert them to corresponding hydrocarbons.

RMgX + H2O → RH + Mg(OH)X

It is therefore necessary to avoid even traces of moisture from a Grignard reagent. That is why reaction is carried out in dry ether. On the other hand, this could be considered as one of the methods of converting halides to hydrocarbons.

//M3//QN34//SUB//DL0//EQ

Write a note on Wurtz reaction and its limitations.

//X

Alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide.

This reaction is known as Wurtz reaction.

2RX + 2Na R – R + 2NaX

Limitation:

(1) This method is suitable to produce an alkane only with even number of carbon.

(2) If alkyl halide is tertiary, the alkene is major product.

//M4//QN35//SUB//DL0//EQ

Why haloarenes are less reactive towards nucleophilic substitution reactions. Give reasons.

//X

Aryl halides are extremely less reactive towards nucleophilic substitution reactions due to the following reasons:

(i) Resonance effect: In haloarenes, the electron pairs on halogen atom are in conjugation with π-electrons of the ring and the following resonating structures are possible:

C-CI bond acquires a partial double bond character due to resonance.

As a result, the bond cleavage in haloarene is more difficult than haloalkane and therefore, they are less reactive towards nucleophilic substitution reaction.

(ii) Difference in hybridization of carbon atom in C-X bond: In haloalkane, the carbon atom attached to halogen is sp3 hybridised while in case of haloarene, the carbon atom attached to halogen is sp2 hybridised.

The sp2 hybridised carbon with a greater s-character is more electronegative and can hold the electron pair of C-X bond more tightly than sp3 hybridised carbon in haloalkane with less s-character.

Thus, C-CI bond length in haloalkane is 177 pm while in haloarenes is 169 pm.

Since it is difficult to break a shorter bond than a longer bond, therefore, haloarenes are less reactive than haloalkanes towards nucleophilic substitution reaction.

(iii) Instability of phenyl cation: In case of haloarenes, the phenyl cation formed as a result of self-ionization will not be stabilised by resonance and therefore, SN1 mechanism is ruled out.

(iv) Repulsion: Because of the possible repulsion, it is less likely for the electron rich in nucleophile to approach electron rich arenes.

//M4//QN36//SUB//DL0//EQ

Give the reaction of chlorobenzene with hydroxyl group.

//X

Chlorobenzene can be converted into phenol by heating in aqueous sodium hydroxide solution at a temperature of 623K and a pressure of 300 atmospheres. Because atomatic compounds undergo nucleoplhic reactions on High Temp and High presure.

The presence of an electron withdrawing group (-NO2) at ortho- and para- positions increases the reactivity of haloarenes.

The effect is pronounced when (-NO2) group is introduced at ortho- and para positions. However, no effect on reactivity of haloarenes is observed by the presence of electron withdrawing group at meta-position.

//M3//QN37//SUB//DL0//EQ

Why electron withdrawing groups such as – NO2 show their effect only at o and p – position and not at m – position.

//X

As shown, the presence of nitro group at ortho- and para-positions withdraws the electron density from the benzene ring and thus facilitates the attack of the nucleophile on haloarene.

The carbanion thus, formed is stabilised through resonance.

The negative charge appeared at ortho- and
para-positions with respect to the halogen substituent is stabilised by –NO2 group while in case of
meta-nitrobenzene, none of the resonating structures bear the negative charge on carbon atom bearing the –NO2 group.

Therefore, the presence of nitro group at meta- position does not stabilise the negative charge and no effect on reactivity is observed by the presence of –NO2 group at meta-position.

//M3//QN38//SUB//DL0//EQ

Why Haloarenes undergo electrophilic substitution reactions at o and p – position? Why it is less reactive than benzene? OR Although chlorine is an electron withdrawing group, yet it is ortho, para directing in electrophilic aromatic substitution reaction, Explain.

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Haloarenes undergo the usual electrophilic reactions of the benzene ring such as halogenation, nitration, sulphonation and Friedel-Crafts reactions.

Halogen atom besides being slightly deactivating is o, p- directing; therefore, further substitution occurs at ortho- and para- positions with respect to the halogen atom.

The o, p-directing influence of halogen atom can be easily understood if we consider the resonating structures of halobenzene as shown:

Due to resonance, the electron density increases more at ortho- and para-positions than at meta-positions.

Further, the halogen atom because of its -I effect has some tendency to withdraw electrons from the benzene ring.

As a result, the ring gets somewhat deactivated as compared to benzene and hence the electrophilic substitution reactions in haloarenes occur slowly and require more drastic conditions as compared to those in benzene.

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Explain Halogenation, Nitration, Sulphonation and Friedel crafts reaction of chlorobenzene.

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(i) Halogenation

(ii) Nitration

(iii) Sulphonation

(iv) Friedel-Crafts reaction

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Write down reaction of halogens with metals.

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(1) Fittig reaction:

2 mole of Aryl halides react with sodium in presence of dry ether to give diphenyl

(2) Wurtz-Fittig reaction:

A mixture of an alkyl halide and aryl halide gives an alkylarene when treated with sodium in dry ether is called Wurtz-Fittig reaction.

(3) Grignard reaction:

Aryl halide compounds react with Mg Metal in presence of dry ether & gives Aryl Magnesium halide.

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What are polyhalogen compounds? Write a note on dichloromethane (Methylene chloride)

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Write a note on trichloromethane (Chloroform).

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Uses : Chloroform CHCl3 is employed as a solvent for fats, alkaloids, iodine and other substances.

The major use of chloroform today is in the production of the freon refrigerant R-22.

It was once used as a general anaesthetic in surgery but has been replaced by less toxic, safer anaesthetics, such as ether.

Harmful effects:

As might be expected from its use as an anaesthetic, inhaling chloroform vapours depresses the central nervous system.

Breathing about 900 parts of chloroform per million parts of air (900 parts per million) for a short time can cause dizziness, fatigue and headache.

Chronic chloroform exposure may cause damage to the liver (where chloroform is metabolised to phosgene) and to the kidneys, and some people develop sores when the skin is immersed in chloroform.

Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carbonyl chloride, also known as phosgene.

It is therefore stored in closed dark coloured bottles completely filled so that air is kept out.

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Write a short note on triiodomethane (Iodoform).

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Write short note on tetra chloromethane (CCl4) carbon tetra chloride.

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Write note on freons.

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Write a note on DDT.

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DDT, the first chlorinated organic insecticides, was originally prepared in 1873, but it was not until 1939 that Paul Muller of Geigy Pharmaceuticals in Switzerland discovered the effectiveness of DDT as an insecticide.

Paul Muller was awarded the Nobel Prize in Medicine and Physiology in 1948 for this discovery.

The use of DDT increased enormously on a worldwide basis after world war II, primarily because of its effectiveness against the mosquito that spreads malaria and lice that carry typhus.

However, problems related to extensive use of DDT began to appear in the late 1940s.

Many species of insects developed resistance to DDT, and it was also discovered to have a high toxicity towards fish.

The chemical stability of DDT and its fat solubility compounded the problem.

DDT is not metabolised very rapidly by animals; instead, it is deposited and stored in the fatty tissues.

If ingestion continues at a steady rate, DDT builds up within the animal over time.

The use of DDT was banned in the United States in 1973, although it is still in use in some other parts of the world.

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Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

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Write down chemical equations to prepare following substances from 1-Chloropropane.

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(i)

(ii)

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Give conversion: Ethene into butane.

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State the reaction equation of ethyl chloride with the following compounds: (i) Aqueous KOH (ii) Alcoholic KOH

[June 2024] [2 Marks]

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