and Haloarenes X

ORGANO HALOGEN COMPOUNDS SP 3C-X: i.Alkyl halides: Halogen atom is bonded to an alkyl group. 3 ii. Allylic halides: SP C-X connected to carbon-carbon double bond.eg: CH2=CH-CH2X 3 iii. Benzylic halides: SP C-X connected to aromatic ring. Eg: C6H5-CH2-X SP 2C-X: 2 i.Vinyl halides: Halogen atom bonded to an SP hybridised carbon atom. Eg : CH2=CHX ii. Aryl halides: Halogen atom bonded to an SP2 hybridised carbon atom of an aromatic ring.

Eg: C6H5-X Same halogen atoms are present on the same carbon atom : Geminal halides

Eg: CH3-CHCl2. Same halogen atoms are present on the adjacent carbon atoms : Vicinal halides.

Eg : CH2Cl-CH2Cl. 2-Bromo-2-chloro-1,1,1-trifluoroethane was given the name Halothane. Which is widely used in hospitals as anaesthetics. C-F bond is very strong, hence it is very unreactive. C-Cl bond is more reactive than C-F bond. CFCs are responsible for serious damage to the stratospheric ozone layer which filters out most of the sun’s harmful UV radiation before it can reach us on earth.

In CFCs, the hemolytic fission of the C-Cl bond results Cl* radicals, which converts O3 to O2 and ClO* radicals. Hydrofluorocarbons, HFCs [Hydrofluoro alkanes,HFAs] replace CFCs.

Haloalkanes is boiled with aq. NaOH, then cooled and neutralized with dil.HNO 3. then AgNO3 solution is added. A white precipitate of AgCl indicates a chloroalkane, a pale yellow precipitate indicates a bromo alkane and a yellow precipitate indicates an iodo alkane. The C-X bond is polar. The positively charged carbon atom tends to be attacked by nucleophiles. The Ar-X bond is considerably less reactive than the R-X bond. Optically active isomers related to each other as object and mirror images are called enantiomers. Equimolal mixture of enantiomers are called racemic mixture. Conversion of enantiomers in to racemic mixture are called racemisation. In the presence of a strong base, haloalkanes can undergo elimination reactions, forming . According to Saytzeff rule,the neutral substrates in elimination reactions predominantly yield the most substituted , i.e., the one carrying the largest number of alkyl substituents.

Presence of electron withdrawing group (NO2) at ortho and para- positions increases the reactivity of haloarenes towards replacement reactions. None of the resonating structures bear the negative charge on carbon atom bearing electron withdrawing group at meta- positions. Hence no effect on reactivity.

Preparation of R-X : From hydrocarbons: Halogenations of alkanes.:

Addition of alkenes with hydrogen halides.:

R-CH=CH2 + H-Br →R-CH(Br)-CH3 Mechanism is in accordance with Markownikoff’s rule. From alcohols:

With HX and anhydrous ZnCl2

Anhydrous ZnCl2 and con. HCl is called Lucas reagent. With phosphorus halides

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Haloalkanes and Haloarenes X

Better yield of R-X is obtained by using PX3 or PX5 in comparison to HX, because (i).sec- and tert- alcohols tend to dehydrate in presence of HX, and (ii).the intermediate carbocation, can undergo rearrangement in presence of HX.

With SOCl2 (Darzen’s method)

By halide exchange reaction: R-Cl or R-Br is heated with con.NaI in acetone. [Finkelstein reaction]

R-X with mercurous fluoride,Hg2 F2 gives R-F [Swarts reaction]

Preparation of Ar-X: Direct halogenations of aromatic hydrocarbons.

From aromatic amines: Sandmeyer’s reaction: 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 t he replacement of the diazonium group by –Cl or –Br.

C6H5 N2X + CuCl/HCl  C6H5 Cl

+ CuBr/HBr  C6H5 Br + KI  C6H5 I From Phenol:

C6H5 OH + PCl5 → C6H5 Cl + HCl + POCl3

R-X: Reactions. 1. Nucleophilic Substitution Reactions: : R-X warm with RCOOAg gives esters , RCOOR : R-X heated with KCN in ethanol gives cyanides, RCN

: R-X heated under reflux (H2O) gives alcohols, ROH : R-X reflux with aq.NaOH gives alcohols, ROH

: Heat under reflux with con.NH3 (aq) gives amines, RNH2, R2NH, etc : R-X react with R’ONa (in ethanol) gives ethers, R’OR

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There are two mechanisms for nucleophilic substitution. (a) SN1 – substitution nucleophilic unimolecular: The SN1 mechanism involves two steps and an intermediate carbonium ion is formed. rate determining step consist of substrate only. Hence it follows first order kinetics. i.e. Rate, r = k [RX].

(b). SN2- substitution nucleophilic bimolecular: These reactions proceed in one step and is a second order reaction with r = k[RX] [Nu].

SN1 SN2 Follows first order kinetics Follows second order kinetics Nucleophile may attack the central carbon from the front Nucleophile may attack the central carbon from the back or back side. side. When the substrate is optically active, racemic mixture Inversion of configuration occurs. will be the product. Solvents of high polarity favour. Solvents of low polarity favour. Order of reactivity of alkyl halides is 30>20>10>methyl Order of reactivity of alkyl halides is 30<20<10

2. Elimination reactions When a 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.

This reaction takes place in accordance with Saytzeff rule and it states that, the neutral substrates in elimination reactions predominantly yield the most substituted alkene, i.e., the one carrying the largest number of alkyl substituents. Ease of dehydrohalogenation among halides: 3° > 2° > 1° 3. Reduction :

C2H5 –Br + H2 → C2H6 + HBr C2H5I + HI → C2H6 + I2 4. Reaction with Metals i. With Mg: R-X react with certain metals to give compounds containing carbon-metal bonds. Such compounds are known as organo- metallic compounds. Alkyl magnesium halide, RMgX, referred as Grignard Reagents, obtained by the reaction of haloalkanes with magnesium metal in dry ether.

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ii. with Na: Wurtz reaction. R-X react with Na in ether gives alkanes. This reaction is known as Wurtz reaction. R-X + 2Na + X-R → R-R + 2NaX iii. with Ar-X and Na : Wurtz- Fittig reaction. R-X and Ar-X react with Na gives alkyl substituted aromatic hydrocarbons. This reaction is known as Wurtz- Fittig reaction. R-X + 2Na + X-Ar →Ar-R + 2NaX Reactions of Haloarenes: 1. Nucleophilic substitution The Ar-X bond is considerably less reactive than the R-X bond ,due to 1. Resonance effect: C-X bond in Ar-X has some partial double bond character due to resonance.

2. Instability of phenyl cation: The phenyl cation formed as a result of self-ionisation will not be stabilised by resonance. 3. Difference in hybridisation of carbon atom in C—X bond: 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-chararcter. Thus, C—Cl bond length in haloalkane is greater than that in haloarene. 4. Because of the possible repulsion between the electron rich nucleophile and electron rich arenes.

However, aryl halides having electron withdrawing groups (like – NO2 , -SO3H, etc.) at ortho and para positions undergo nucleophilic substitution reaction easily.

Picric acid 2. Electrophilic substitution reactions: Halogens are deactivating but O, p-directing. Thus, chlorination, nitration, sulphonation and Friedel Craft’s reaction give a mixture of o- and P- chloro substituted derivatives. (i) Halogenation:

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(ii) Nitration:

(iii) Sulphonation:

(iv) Friedel-Crafts reaction:

3. Reaction with Metals: (i) Wurtz Fittig reaction;

(ii) Fitting reaction:

(iii) Ullmann reaction:

Polyhalogen Compounds: Carbon compounds containing more than one halogen atom are referred to as polyhalogen compounds. Dichloro- methane (Methylene chloride):

Dichloromethane (CH2Cl2 ) is widely used as a solvent, as a propellant in aerosols. Direct contact of dichloromethane in humans causes intense burning and mild redness of the skin.

Chloroform [Trichloromethane, CHCl3 ]:

CHCl3 is widely used in the production of freon refrigerant R-22. On nitration, it gives tear producing insecticide substance chloropicrin. On dehalogination, it gives acetylene. Oxidation of CHCl3 gives poisonous gas phosgene (carbonyl chloride). To avoid this oxidation CHCl3 will stored in dark brown bottles and filled to the brim. 1% ethanol is added to chloroform which converts harmful phosgene gas into diethyl carbonate.

2CHCl3 + O2 → 2COCl2 + 2HCl

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Iodoform (tri-iodornethane, CHl3 ):

Compounds containing either CH3 CO- or CH3 CH(OH) group form yellow colour iodoform with I2 and NaOH. Iodoform when comes in contact with organic matter, decomposes easily to free , an antiseptic. Due to its objectionable smell, it has been replaced by other formulations containing iodine.

Tetrachloromethane (Carbon Tetrachloride, CCl4 ): CCI4 is a colourless, non-inflammable, poisonous liquid, soluble in alcohol and ether. Carbon tetrachloride is used: 1. as a solvent for oils, fats, resins 2. in dry cleaning 3. as fire extinguisher under the name ‘pyrene’. Freons: The chlorofluorocarbon compounds of methane and ethane are collectively known as freons. These are usually produced for aerosol propellants, refrigeration and air conditioning purposes. Freon 12 (CCl2F2) is one of the most common freons in industrial use. Carbon tetra chloride when reacts with antimony trifluoride in the presence of SbCl5 as catalyst, dichlorofluromethane (freon) is obtained. DDT (p, p’-Dichlorodiphenyltrichloroethane): DDT is the first chlorinated organic insecticide. Its stability and fat solubility’is a great problem.

It is prepared from chloral and chlorobenzene in the presence of conc. H2SO4 ·

Q.1: Write the mechanism of the following reaction: The given reaction is:

- The given reaction is an SN2 reaction. In this reaction, CN acts as the nucleophile and attacks the carbon atom to which Br is attached. CN- ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C-atom.

Q.2: Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields (i) A single monochloride. (ii) Three isomeric monochlorides. (iii) Four isomeric monochlorides. (i) To have a single monochloride, there should be only one type of H-atom in the isomer of the alkane of the molecular formula C5H12. This is because, replacement of any H-atom leads to the formation of the same product. The isomer is neopentane.

Neopentane (ii) To have three isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain three different types of H-atoms. Therefore, the isomer is n-pentane. It can be observed that there are three types of H atoms labelled as a, b and c in n-pentane. c b a b c CH3-CH2-CH2-CH2-CH3 (iii) To have four isomeric monochlorides, the isomer of the alkane of the molecular formula C5H12 should contain four different types of H-atoms. Therefore, the isomer is 2-methylbutane. It can be observed that there are four types of H-atoms labelled as a, b, c, and d in 2-methylbutane.

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Q.3: How the following conversions can be carried out? (i) Propene to propan-1-ol (ii) Ethanol to but-1-yne (iii) 1-Bromopropane to 2-bromopropane (iv) Toluene to benzyl alcohol (v) Benzene to 4-bromonitrobenzene (vi) Benzyl alcohol to 2-phenylethanoic acid (vii) Ethanol to propanenitrile (viii) Aniline to chlorobenzene (ix) 2-Chlorobutane to 3, 4-dimethylhexane (x) 2-Methyl-1-propene to 2-chloro-2-methylpropane (xi) Ethyl chloride to propanoic acid (xii) But-1-ene to n-butyliodide (xiii) 2-Chloropropane to 1-propanol (xiv) Isopropyl alcohol to iodoform (xv) Chlorobenzene to p-nitrophenol (xvi) 2-Bromopropane to 1-bromopropane (xvii) Chloroethane to butane (xviii) Benzene to diphenyl (xix) tert-Butyl bromide to isobutyl bromide (xx) Aniline to phenylisocyanide

(i)

(ii)

(iii)

(iv)

(v)

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(vi)

(vii)

(viii)

(ix)

(x)

(xi)

(xii)

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(xiii)

(xiv)

(xv)

(xvi)

(xvii)

(xviii)

(xix)

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(xx)

Q.4: Explain why (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride? (ii) alkyl halides, though polar, are immiscible with water? (iii) Grignard reagents should be prepared under anhydrous conditions? (i)

In chlorobenzene, the Cl-atom is linked to a sp2 hybridized carbon atom. In cyclohexyl chloride, the Cl-atom is linked to a sp3 hybridized carbon atom. Now, sp2 hybridized carbon has more s-character than sp3 hybridized carbon atom. Therefore, the former is more electronegative than the latter. Therefore, the density of electrons of C-Cl bond near the Cl-atom is less in chlorobenzene than in cydohexyl chloride. Moreover, the +R effect of the benzene ring of chlorobenzene decreases the electron density of the C-Cl bond near the Cl-atom. As a result, the polarity of the C-Cl bond in chlorobenzene decreases. Hence, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. (ii) To be miscible with water, the solute-water force of attraction must be stronger than the solute-solute and water-water forces of attraction. Alkyl halides are polar molecules and so held together by dipole-dipole interactions. Similarly, strong H-bonds exist between the water molecules. The new force of attraction between the alkyl halides and water molecules is weaker than the alkyl halide-alkyl halide and water-water forces of attraction. Hence, alkyl halides (though polar) are immiscible with water. (iii) Grignard reagents are very reactive. In the presence of moisture, they react to give alkanes.

Therefore, Grignard reagents should be prepared under anhydrous conditions. Q.5: Give the IUPAC names of the following compounds: (i) CH3CH(Cl)CH(Br)CH3 (ii) CHF2CBrClF (iii) ClCH2C≡CCH2Br (iv) (CCl3)3CCl (v) CH3C(p-ClC6H4)2CH(Br)CH3 (vi) (CH3)3CCH=CClC6H4I-p (i) (v)

2-Bromo-3-chlorobutane (ii)

2-Bromo-3, 3-bis(4-chlorophenyl) butane (vi)

1-Bromo-1-chloro-1, 2, 2-trifluoroethane (iii)

1-Bromo-4-chlorobut-2-yne

(iv) 1-chloro-1-(4-iodophenyl)-3, 3-dimethylbut-1-ene

2-(Trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane

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Q.6: p-Dichlorobenzene has higher m.p. and lower solubility than those of o- and m-isomers. Discuss.

p-Dichlorobenzene is more symmetrical than o-and m-isomers. For this reason, it fits more closely than o-and m-isomers in the crystal lattice. Therefore, more energy is required to break the crystal lattice of p-dichlorobenzene. As a result, p-dichlorobenzene has a higher melting point and lower solubility than o-and m-isomers. Q.7: Write the equations for the preparation of 1-iodobutane from (i) 1-butanol (ii) 1-chlorobutane (iii) but-1-ene. (i)

(ii)

(iii)

Q.8: Arrange the compounds of each set in order of reactivity towards SN2 displacement: (i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane (ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2- methylbutane (iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane. (i)

An SN2 reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group is attached. When the nucleophile is sterically hindered, then the reactivity towards SN2 displacement decreases. Due to the presence of substituents, hindrance to the approaching nucleophile increases in the following order. 1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane 2 Hence, the increasing order of reactivity towards SN displacement is: 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane (ii)

Since steric hindrance in alkyl halides increases in the order of 1° < 2° < 3°, the increasing order of reactivity towards SN2 displacement is : 3° < 2° < 1°. 2 Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards SN displacement as: 2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane

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(iii)

The steric hindrance to the nucleophile in the SN2 mechanism increases with a decrease in the distance of the substituents from the atom containing the leaving group. Further, the steric hindrance increases with an increase in the number of substituents. Therefore, the increasing order of steric hindrances in the given compounds is as below: 1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane < 1-Bromo-2, 2-dimethylpropane Hence, the increasing order of reactivity of the given compounds towards SN2 displacement is: 1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1-Bromobutane Q.9: Arrange each set of compounds in order of increasing boiling points. (i) Bromomethane, Bromoform, Chloromethane, Dibromomethane. (ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane. (i)

For alkyl halides containing the same alkyl group, the boiling point increases with an increase in the atomic mass of the halogen atom. Since the atomic mass of Br is greater than that of Cl, the boiling point of bromomethane is higher than that of chloromethane. Further, for alkyl halides containing the same alkyl group, the boiling point increases with an increase in the number of halides. Therefore, the boiling point of Dibromomethane is higher than that of chloromethane and bromomethane, but lower than that of bromoform. Hence, the given set of compounds can be arranged in the order of their increasing boiling points as: Chloromethane < Bromomethane < Dibromomethane < Bromoform. (ii)

For alkyl halides containing the same halide, the boiling point increases with an increase in the size of the alkyl group. Thus, the boiling point of 1-chlorobutane is higher than that of isopropyl chloride and 1-chloropropane. Further, the boiling point decreases with an increase in branching in the chain. Thus, the boil ing point of isopropyl alcohol is lower than that of 1-chloropropane. Hence, the given set of compounds can be arranged in the increasing order of their boiling points as: Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane

Q. 10: Write the isomers of the compound having formula C4H9Br. There are four isomers of the compound having the formula C4H9Br. These isomers are given below. (a) (c)

1-Bromobutane (b) 1-Bromo-2-methylpropane (d)

2-Bromobutane

2-Bromo-2-methylpropane

- Q.11: Which compound in each of the following pairs will react faster in S N2 reaction with OH ? (i) CH3Br or CH3I (ii) (CH3)3CCl or CH3Cl (i) In the SN2 mechanism, the reactivity of halides for the same alkyl group increases in the order. This happens because as the size increases, the halide ion becomes a better leaving group. R-F << R-Cl < R-Br < R-I - Therefore, CH3I will react faster than CH3Br in SN2 reactions with OH .

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(ii)

(CH3)3CCl, the attack of the nucleophile at the carbon atom is hindered because of the presence of bulky substituents on that carbon atom bearing the leaving group. On the other hand, there are no bulky substituents on the carbon atom bearing the leaving group in CH Cl. Hence, CH Cl reacts The S 2 mechanism involves the attack of the nucleophile 3 3 N faster than (CH ) CCl in S 2 reaction with OH-. at the atom bearing the leaving group. But, in case of 3 3 N Q.12: What happens when: (i) n-butyl chloride is treated with alcoholic KOH, (ii) bromobenzene is treated with Mg in the presence of dry ether, (iii) chlorobenzene is subjected to hydrolysis, (iv) ethyl chloride is treated with aqueous KOH, (v) methyl bromide is treated with sodium in the presence of dry ether, (vi) methyl chloride is treated with KCN. (i) When n-butyl chloride is treated with alcoholic KOH, the formation of but-l-ene takes place. This reaction is a dehydrohalogenation reaction.

(ii) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide is formed.

(iii) Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.

(iv) When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol.

(v) When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. This reaction is known as the Wurtz reaction.

(vi) When methyl chloride is treated with KCN, it undergoes a substitution reaction to give methyl cyanide.

Q.13: A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon. A hydrocarbon with the molecular formula, C5H10 belongs to the group with a general molecular formula CnH2n. Therefore, it may either be an alkene or a cycloalkane. Since hydrocarbon does not react with chlorine in the dark, it cannot be an alkene. Thus, it should be a cycloalkane. Further, the hydrocarbon gives a single monochloro compound, C5H9Cl by reacting with chlorine in bright sunlight. Since a single monochloro compound is formed, the hydrocarbon must contain H-atoms that are all equivalent. Also, as all H-atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. Hence, the said compound is cyclopentane.

Cyclopentane (C5H10) The reactions involved in the question are:

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Q.14: Why is sulphuric acid not used during the reaction of alcohols with KI? In the presence of sulphuric acid (H2SO4), KI produces HI

Since H2SO 4 is an oxidizing agent, it oxidizes HI (produced in the reaction to I2).

As a result, the reaction between alcohol and HI to produce alkyl iodide cannot occur. Therefore, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidizing acid such as H3PO4 is used. Q.15: Give the uses of freon 12, DDT, carbon tetrachloride and iodoform. Uses of Freon - 12 Freon-12 (dichlorodifluoromethane, CF2Cl2) is commonly known as CFC. It is used as a refrigerant in refrigerators and air conditioners. It is also used in aerosol spray propellants such as body sprays, hair sprays, etc. However, it damages the ozone layer. Hence, its manufacture was banned in the United States and many other countries in 1994. Uses of DDT DDT (p, p-dichlorodiphenyltrichloroethane) is one of the best known insecticides. It is very effective against mosquitoes and lice. But due its harmful effects, it was banned in the United States in 1973. Uses of carbontetrachloride (CCl4) (i) It is used for manufacturing refrigerants and propellants for aerosol cans. (ii) It is used as feedstock in the synthesis of chlorofluorocarbons and other chemicals. (iii) It is used as a solvent in the manufacture of pharmaceutical products. (iv) Until the mid 1960’s, carbon tetrachloride was widely used as a cleaning fluid, a degreasing agent in industries, a spot reamer in homes, and a fire extinguisher. Uses of iodoform (CHI3) Iodoform was used earlier as an antiseptic, but now it has been replaced by other formulations-containing iodine-due to its objectionable smell. The antiseptic property of iodoform is only due to the liberation of free iodine when it comes in contact with the skin. Q.16: Primary alkyl halide C4H9Br (a) reacted with alcoholic KOH to give compound (b).Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), C 8H18 which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions. There are two primary alkyl halides having the formula, C4H9Br. They are n - bulyl bromide and isobutyl bromide.

Therefore, compound (a) is either n-butyl bromide or isobutyl bromide. Now, compound (a) reacts with Na metal to give compound (b) of molecular formula, C8H18, which is different from the compound formed when n−butyl bromide reacts with Na metal. Hence, compound (a) must be isobutyl bromide.

Thus, compound (d) is 2, 5-dimethylhexane. It is given that compound (a) reacts with alcoholic KOH to give compound (b). Hence, compound (b) is 2-methylpropene.

Also, compound (b) reacts with HBr to give compound (c) which is an isomer of (a). Hence, compound (c) is 2-bromo-2- methylpropane.

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Q.17: The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain. In an aqueous solution, KOH almost completely ionizes to give OH- ions. OH- ion is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.

On the other hand, an alcoholic solution of KOH contains alkoxide (RO -) ion, which is a strong base. Thus, it can abstract a hydrogen from the β-carbon of the alkyl chloride and form an alkene by eliminating a molecule of HCl.

OH- ion is a much weaker base than RO- ion. Also, OH- ion is highly solvated in an aqueous solution and as a result, the basic character of OH- ion decreases. Therefore, it cannot abstract a hydrogen from the β-carbon. Q.18: What are ambident nucleophiles? Explain with an example. Ambident nucleophiles are nucleophiles having two nucleophilic sites. Thus, ambident nucleophiles have two sites through which they can attack. For example, nitrite ion is an ambident nucleophile.

Nitrite ion can attack through oxygen resulting in the formation of alkyl nitrites. Also, it can attack through nitrogen resulting in the formation of nitroalkanes.

Q.19: How will you bring about the following conversions? (i) Ethanol to but-1-yne (ii) Ethane to bromoethene (iii) Propene to 1-nitropropane (iv) Toluene to benzyl alcohol (v) Propene to propyne (vi) Ethanol to ethyl fluoride (vii) Bromomethane to propanone (viii) But-1-ene to but-2-ene (ix) 1-Chlorobutane to n-octane (x) Benzene to biphenyl. (i)

(ii)

(iii)

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(iv) (vii)

(v)

(vi) (viii)

(ix)

(x)

Q.20: Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene: (i) 1-Bromo-1-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 2,2,3-Trimethyl-3-bromopentane. (i)

(ii)

1-bromo-1-methylcyclohexane In the given compound, all β-hydrogen atoms are equivalent. Thus, dehydrohalogenation of this compound gives only one alkene. In the given compound, there are two different sets of equivalent β-hydrogen atoms labelled as a and b. Thus, dehydrohalogenation of the compound yields two alkenes.

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Saytzeff’s rule implies that in dehydrohalogenation reactions, the alkene having a greater number of alkyl groups attached to a doubly bonded carbon atoms is preferably produced. Therefore, alkene (I) i.e., 2-methylbut-2-ene is the major product in this reaction. (iii)

2,2,3-Trimethyl-3-bromopentane In the given compound, there are two different sets of equivalent β-hydrogen atoms labelled as a and b. Thus, dehydrohalogenation of the compound yields two alkenes.

According to Saytzeff’s rule, in dehydrohalogenation reactions, the alkene having a greater number of alkyl groups attached t o the doubly bonded carbon atom is preferably formed. Hence, alkene (I) i.e., 3,4,4-trimethylpent-2-ene is the major product in this reaction. Q.21: Write structures of different dihalogen derivatives of propane. There are four different dihalogen derivatives of propane. The structures of these derivatives are shown below.

(i) 1, 1-Dibromopropane (ii) 2, 2-Dibromopropane

(iii) 1, 2-Dibromopropane (iv) 1, 3-Dibromopropane Q.22: Which one of the following has the highest dipole moment? (i) CH2Cl2 (ii) CHCl3 (iii) CCl4 (i)

Dichlormethane (CH2Cl2) μ = 1.60D (ii)

Chloroform (CHCl3) μ = 1.08D (iii)

Carbon tetrachloride (CCl4) μ = 0D CCl4 is a symmetrical molecule. Therefore, the dipole moments of all four C-Cl bonds cancel each other. Hence, its resultant dipole moment is zero. As shown in the above figure, in CHCl3, the resultant of dipole moments of two C-Cl bonds is opposed by the resultant of dipole moments of one C-H bond and one C-Cl bond. Since the resultant of one C-H bond and one C-Cl bond dipole moments is smaller than two C-Cl bonds, the opposition is to a small extent. As a result, CHCl3 has a small dipole moment of 1.08 D. On the other hand, in case of CH2Cl2, the resultant of the dipole moments of two C-Cl bonds is strengthened by the resultant of the dipole moments of two C-H bonds. As a result, CH2Cl2 has a higher dipole moment of 1.60 D than CHCl3 i.e., CH2Cl2 has the highest dipole moment. Hence, the given compounds can be arranged in the increasing order of their dipole moments as: CCl4 < CHCl3 < CH2Cl2

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Q.23: Draw the structures of major monohalo products in each of the following reactions:

(i) (ii)

(iii) (iv)

(v) (vi)

(i) (iv)

(ii) (v)

(vi)

(iii)

Q.24: Write the structure of the major organic product in each of the following reactions: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)

(ii)

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(viii)

Q.25: Which alkyl halide from the following pairs would you expect to react more rapidly by an S N2 mechanism? Explain your answer.

(i) (ii)

(iii)

(i)

2-bromobutane is a 2° alkylhalide whereas 1-bromobutane is a 1° alkyl halide. The approaching of nucleophile is more hindered in 2-bromobutane than in 1-bromobutane. Therefore, 1-bromobutane reacts more rapidly than 2-bromobutane by an SN2 mechanism. (ii)

2-Bromobutane is 2° alkylhalide whereas 2-bromo-2-methylpropane is 3° alkyl halide. Therefore, greater numbers of substituents are present in 3° alkyl halide than in 2° alkyl halide to hinder the approaching nucleophile. Hence, 2-bromobutane reacts more rapidly than 2-bromo-2-methylpropane by an SN2 mechanism. (iii)

Both the alkyl halides are primary. However, the substituent -CH3 is at a greater distance to the carbon atom linked to Br in 1- bromo-3-methylbutane than in 1-bromo-2-methylbutane. Therefore, the approaching nucleophile is less hindered in case of the former than in case of the latter. Hence, the former reacts faster than the latter by SN2 mechanism. Q.26: Write the structures of the following organic halogen compounds. (i) 2-Chloro-3-methylpentane (ii) p-Bromochlorobenzene (iii) 1-Chloro-4-ethylcyclohexane (iv) 2-(2-Chlorophenyl)-1-iodooctane (v) Perfluorobenzene (vi) 4-tert-Butyl-3-iodoheptane (vii) 1-Bromo-4-sec-butyl-2-methylbenzene (viii) 1,4-Dibromobut-2-ene (i) (iii)

2-Chloro-3-methylpentane (ii)

1-Chloro-4-ethylcyclohexane (iv)

p-Bromochlorobenzene

2-(2-Chlorophenyl)-1-iodooctane

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(v) (vii)

Perfluorobenzene (vi) 1-Bromo-4-sec-butyl-2-methylbenzene (viii)

1,4-Dibromobut-2-ene

4-Tert-Butyl-3-iodoheptane

Q.27: Out of C6H5CH2Cl and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH?

Hydrolysis by aqueous KOH proceeds through the formation of carbocation. If carbocation is stable, then the compound is easily hydrolyzed by aqueous KOH. Now, C6H5CH2Cl forms 1°-carbocation, while C6H5-CH(Cl)-C6H5 forms 2°-carbocation, which is more stable than 1°-carbocation. Hence, C6H5-CH(Cl)-C6H5 is hydrolyzed more easily than C6H5CH2Cl by aqueous KOH.

Q.28: In the following pairs of halogen compounds, which compound undergoes faster S N1 reaction? (i)

(ii)

(i)

SN1 reaction proceeds via the formation of carbocation. The alkyl halide (I) is 3° while (II) is 2°. Therefore, (I) forms 3° carbocation while (II) forms 2° carbocation. Greater the stability of the carbocation, faster is the rate of SN1 reaction. Since 3° carbocation is more stable than 2° carbocation. (I), i.e. 2-chloro-2-methylpropane, undergoes faster SN1 reaction than (II) i.e., 3- chloropentane. (ii)

The alkyl halide (I) is 2° while (II) is 1°. 2° carbocation is more stable than 1° carbocation. Therefore, (I), 2-chloroheptane, undergoes faster SN1 reaction than (II), 1-chlorohexane. Q.29: Identify A, B, C, D, E, R and R1 in the following:

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Since D of D2O gets attached to the carbon atom to which MgBr is attached, C is

Therefore, the compound R - Br is

When an alkyl halide is treated with Na in the presence of ether, a hydrocarbon containing double the number of carbon atoms as present in the original halide is obtained as product. This is known as Wurtz reaction. Therefore, the halide, R1-X, is

Therefore, compound D is

And, compound E is

Q.30: Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides: (i) (CH3)2CHCH(Cl)CH3 (ii) CH3CH2CH(CH3)CH(C2H5)Cl (iii) CH3CH2C(CH3)2CH2I (iv) (CH3)3CCH2CH(Br)C6H5 (v) CH3CH(CH3)CH(Br)CH3 (vi) CH3C(C2H5)2CH2Br (vii) CH3C(Cl)(C2H5)CH2CH3 (viii) CH3CH=C(Cl)CH2CH(CH3)2 (ix) CH3CH=CHC(Br)(CH3)2 (x) p-ClC6H4CH2CH(CH3)2 (xi) m-ClCH2C6H4CH2C(CH3)3 (xii) o-Br-C6H4CH(CH3)CH2CH3 (i) (ii)

2-Chloro-3-methylbutane (Secondary alkyl halide) 3-Chloro-4-methyhexane (Secondary alkyl halide)

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(iii) (viii)

3-Chloro-5-methylhex-2-ene (Vinyl halide) 1-Iodo-2, 2 -dimethylbutane (ix) (Primary alkyl halide) (iv)

4-Bromo-4-methylpent-2-ene (Allyl halide) 1-Bromo-3, 3-dimethyl-1-phenylbutane (x) (Secondary benzyl halide) (v)

2-Bromo-3-methylbutane (Secondary alkyl halide) (vi) 1-Chloro-4-(2-methylpropyl) benzene (Aryl halide) (xi)

1-Bromo-2-ethyl-2-methylbutane

(Primary alkyl halide) 1-Chloromethyl-3-(2, 2-dimethylpropyl) benzene (vii) (Primary benzyl halide) (xii)

3-Chloro-3-methylpentane 1-Bromo-2-(1-methylpropyl) benzene (Tertiary alkyl halide) (Aryl halide)

Q.31:Write structures of the following compounds: (i) 2-Chloro-3-methylpentane (ii) 1-Chloro-4-ethylcyclohexane (iii) 4-tert. Butyl-3-iodoheptane (iv) 1,4-Dibromobut-2-ene (v) 1-Bromo-4-sec. butyl-2-methylbenzene (i)

(iv)

2-Chloro-3-methyl pentane (ii) 1, 4-Dibromobut-2-ene (v)

1-Chloro-4-ethylcyclohexane (iii) 1-Bromo-4-sec-butyl-2-methylbenzene

4- tert-Butyl-3-iodoheptane

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1. The order of reactivity of following alcohols with halogen acids is

(i) (A) > (B) > (C) (ii) (C) > (B) > (A) (iii) (B) > (A) > (C) (iv) (A) > (C) > (B)

2. Which of the following alcohols will yield the corresponding alkyl chloride on reaction with concentrated HCl at room temperature?

(i) CH3 CH2 —CH2 —OH (ii) CH3CH2-CH(CH3)-OH (iii) CH3CH2-CH(CH3)-CH2 OH (iv) CH3CH2-C(CH3)2-OH 3. Identify the compound Y in the following reaction.

4. Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. The reaction is (i) Electrophilic elimination reaction (ii) Electrophilic substitution reaction (iii) Free radical addition reaction (iv) Nucleophilic substitution reaction

5. Which of the following is halogen exchange reaction? (i) R X + NaI → RI + NaX

6. Which reagent will you use for the following reaction?

CH3CH2CH2CH3 → CH3CH2CH2CH2Cl + CH3CH2CHClCH3 (i) Cl2 /UV light (ii) NaCl + H2SO4 (iii) Cl2 gas in dark

(iv) Cl2 gas in the presence of iron in dark 7. Arrange the following compounds in the increasing order of their densities.

(i) (a) < (b) < (c) < (d) (ii) (a) < (c) < (d) < (b) (iii) (d) < (c) < (b) < (a) (iv) (b) < (d) < (c) < (a)

8. Arrange the following compounds in increasing order of their boiling points.

(i) (b) < (a) < (c) (ii) (a) < (b) < (c) (iii) (c) < (a) < (b) (iv) (c) < (b) < (a)

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9. In which of the following molecules carbon atom marked with asterisk (*) is asymmetric?

(i) (a), (b), (c), (d) (ii) (a), (b), (c) (iii) (b), (c), (d) (iv) (a), (c), (d) 10. Which of the following structures is enantiomeric with the molecule (A) given below :

11. Which of the following is an example of vic-dihalide? (i) Dichloromethane (ii) 1,2-dichloroethane (iii) Ethylidene chloride (iv) Allyl chloride

12. The position of –Br in the compound in CH3CH==CHC(Br)(CH3)2 can be classified as __. (i) Allyl (ii) Aryl (iii) Vinyl (iv) Secondary

13. Chlorobenzene is formed by reaction of chlorine with benzene in the presence of AlCl3 . Which of the following species attacks the benzene ring in this reaction? – + – (i) Cl (ii) Cl (iii) AlCl3 (iv) [AlCl4] 14. Ethylidene chloride is a/an ____. (i) vic-dihalide (ii) gem-dihalide (iii) allylic halide (iv) vinylic halide 15. What is ‘A’ in the following reaction?

16. A primary alkyl halide would prefer to undergo ___. (i) S N1 reaction (ii) SN2 reaction (iii) α–Elimination (iv) Racemisation

17. Which of the following alkyl halides will undergo SN1 reaction most readily?

(i) (CH3)3 C—F (ii) (CH3)3 C—Cl (iii) (CH3)3 C—Br (iv) (CH3)3 C—I 18. Which is the correct IUPAC name for CH3-CH(C2H5)-CH2-Br ? (i) 1-Bromo-2-ethylpropane (ii) 1-Bromo-2-ethyl-2-methylethane (iii) 1-Bromo-2-methylbutane (iv) 2-Methyl-1-bromobutane

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19. What should be the correct IUPAC name for diethylbromomethane? (i) 1-Bromo-1,1-diethylmethane (ii) 3-Bromopentane (iii) 1-Bromo-1-ethylpropane (iv) 1-Bromopentane 20. The reaction of toluene with chlorine in the presence of iron and in the absence of light yields ___

(iv) Mixture of (ii) and (iii) 21. Chloromethane on treatment with excess of ammonia yields mainly (i) N, N-Dimethylmethanamine (ii) N–methylmethanamine (iii) Methanamine (iv) Mixture containing all these in equal proportion 22. Molecules whose mirror image is non superimposable over them are known as chiral. Which of the following molecules is chiral in nature? (i) 2-Bromobutane (ii) 1-Bromobutane (iii) 2-Bromopropane (iv) 2-Bromopropan-2-ol

23. Reaction of C6H5CH2Br with aqueous sodium hydroxide follows _. (i) SN1 mechanism (ii) SN2 mechanism (iii) Any of the above two depending upon the temperature of reaction (iv) Saytzeff rule 24. Which of the carbon atoms present in the molecule given below are asymmetric?

(i) a, b, c, d (ii) b, c (iii) a, d (iv) a, b, c 25. Which of the following compounds will give racemic mixture on nucleophilic substitution by OH – ion?

(i) (a) (ii) (a), (b), (c) (iii) (b), (c) (iv) (a), (c)

26. Which is the correct increasing order of boiling points of the following compounds? 1-Iodobutane, 1-Bromobutane, 1-Chlorobutane, Butane (i) Butane < 1-Chlorobutane < 1-Bromobutane < 1-Iodobutane (ii) 1-Iodobutane < 1-Bromobutane < 1-Chlorobutane < Butane (iii) Butane < 1-Iodobutane < 1-Bromobutane < 1-Chlorobutane (iv) Butane < 1-Chlorobutane < 1-Iodobutane < 1-Bromobutane

27. Which is the correct increasing order of boiling points of the following compounds? 1-Bromoethane, 1-Bromopropane, 1-Bromobutane, Bromobenzene (i) Bromobenzene < 1-Bromobutane < 1-Bromopropane < 1-Bromoethane (ii) Bromobenzene < 1-Bromoethane < 1-Bromopropane < 1-Bromobutane (iii) 1-Bromopropane < 1-Bromobutane < 1-Bromoethane < Bromobenzene (iv) 1-Bromoethane < 1-Bromopropane < 1-Bromobutane < Bromobenzene

1. 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. But why does preparation of aryl iodides requires presence of an oxidising agent? Iodination reactions are reversible in nature. To carry out the reaction in the forward direction, HI formed during iodination is removed by oxidation. HIO4 is used as an oxidising agent.

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2. Out of o-and p-dibromobenzene which one has higher melting point and why? p-Dibromobenzene has higher melting point than its o-isomer. It is due to symmetry of p-isomer which fits in crystal lattice better than the o-isomer.

3. Which of the compounds will react faster in SN1 reaction with the – OH ion?

CH3 — CH2 — Cl or C6H5 — CH2 — Cl

C6H5 — CH2 — Cl 4. Why iodoform has appreciable antiseptic property? Due to liberation of free iodine.

5. Which of the following compounds (a) and (b) will not react with a mixture of NaBr and H2SO4 . Explain why?

(a) CH3CH2CH2 OH (b), C—O bond is more stable in (b) because of resonance. 6. Classify the following compounds as primary, secondary and tertiary halides. (i) 1-Bromobut-2-ene (ii) 4-Bromopent-2-ene (iii) 2-Bromo-2-methylpropane (i) Primary (ii) Secondary (iii) Tertiary

7. Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on concentration of compound and KOH both. (i) Write down the structural formula of both compounds ‘A’ and ‘B’. (ii) Out of these two compounds, which one will be converted to the product with inverted configuration.

(i) Compound A : Compound B : (ii) Compound ‘B’

8. Write the structures and names of the compounds formed when compound ‘A’ with molecular formula, C7H8 is treated with Cl2 in the presence of FeCl3 .

9. Identify the products A and B formed in the following reaction :

(a) CH3 —CH2 —CH==CH—CH3 +HCl → A + B

10. Which of the following compounds will have the highest melting point and why?

II, due to symmetry of para-positions; it fits into crystal lattice better than other isomers.

11. Write down the structure and IUPAC name for neo-pentylbromide.

1-Bromo-2,2-dimethylpropane

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12. A hydrocarbon of molecular mass 72 g mol–1 gives a single monochloro derivative and two dichloro derivatives on photo chlorination. Give the structure of the hydrocarbon. –1 C5H12 , pentane has molecular mass 72 g mol , i.e. the isomer of pentane which yields single monochloro derivative should have all the 12 hydrogens equivalent.

The hydrocarbon is: Monochloro derivative:

Dichloro derivatives:

13. Name the alkene which will yield 1-chloro-1-methylcyclohexane by its reaction with HCl. Write the reactions involved.

14. Which of the following haloalkanes reacts with aqueous KOH most easily? Explain giving reason. (i) 1-Bromobutane (ii) 2-Bromobutane (iii) 2-Bromo-2-methylpropane (iv) 2-Chlorobutane (iii); The tertiary carbocation formed in the reaction is stable.

15. Why can aryl halides not be prepared by reaction of phenol with HCl in the presence of ZnCl2 ? C—O bond in phenols is more stable due to resonance effect and it has double bond character, hence breaking of this bond is difficult.

16. Which of the following compounds would undergo SN1 reaction faster and why?

(B) Undergoes SN1 reaction faster than (A) because in case of (B), the carbocation formed after the loss of Cl– is stabilised by resonance, whereas, no such stabilisation is possible in the carbocation obtained from (A).

17. Allyl chloride is hydrolysed more readily than n-propyl chloride. Why? Allyl chloride shows high reactivity as the carbocation formed by hydrolysis is stabilised by resonance while no such stabilisation of carbocation exists in the case of n-propyl chloride.

18. Why is it necessary to avoid even traces of moisture during the use of a ? Grignard reagents are highly reactive and react with water to give corresponding hydrocarbons.

RMgX + H2O → RH + Mg(OH)X 19. Predict the major product formed when HCl is added to isobutylene. Explain the mechanism involved.

The mechanism involved in this reaction is:

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20. How can you obtain iodoethane from ethanol when no other iodine containing reagent except NaI is available in the laboratory?

C2H5OH + HCl → C2H5Cl → C2H5I 21. Cyanide ion acts as an ambident nucleophile. From which end it acts as a stronger nucleophile in aqueous medium? Give reason for your answer. It acts as a stronger nucleophile from the carbon end because it will lead to the formation of C–C bond which is more stable than the C–N bond.

1. (ii) 2. (iv) 3. (i) 4. (ii) 5. (i) 6. (i) 7. (i) 8. (iii), boiling point of (a) 364 K. boiling point of (b) 375 K, boiling point of (c) 346 K 9. (ii) 10. (i), Hint : Make the models of all the molecules and superimpose (i) to (iv) molecules on molecule (A). 11. (ii) 12. (i) 13. (ii) 14. (ii) 15. (iii) 16. (ii) 17. (iv) 18. (iii) 19. (ii) 20. (iv) 21. (iii) 22. (i) + 1 23. (i), Hint : C6H5C H2 is stable cation so favours the progress of reaction by SN mechanism. 24. (ii) 25. (i) 26. (i) 27. (iv)

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