NCERT

Chemistry (Vol 2) Chapter 1 Haloalkanes and haloarenes

CBSE NCERT Solutions for Class 12 chemistry Chapter 1

Exercises

Q.1. Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:
CH32CHCHClCH3

Solution:

Substituents =-CH3 (Methyl) and -Cl (Chloro)

(Word root) Parent carbon chain =4 (but)
Primary suffix = ane
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
IUPAC name is 2-Chloro -3-methylbutane (secondary alkyl halide).

Q.2. Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:

CH3CH2CHCH3CHC2H5Cl

Solution:

Substituents =-CH3 (Methyl) and -Cl (Chloro)

(Word root) Parent carbon chain =6 (hex)
Primary Suffix = ane
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
IUPAC name is 3-chloro -4-methylhexane (secondary alkyl halide).

Q.3. Name the following halide according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

CH3CH2CCH32CH2I

Solution:

Substituents =-CH3 (Methyl) and -I (Iodo)

(Word root) Parent Carbon chain =4 (but)
Primary suffix = ane
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
IUPAC name is 1-Iodo-2,2-dimethylbutane (primary alkyl halide).

Q.4. Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:

CH33CCH2CHBrC6H5

Solution:

Substituents =-CH3 (Two methyl groups) and -Br (Bromo)

(Word root) Parent carbon chain =4 (but)
Primary Suffix =ane
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
1-Bromo-3, 3-dimethyl-1-phenylbutane (secondary benzyl halide)

Q.5. Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:

CH3CHCH3CHBrCH3

Solution:

Substituents =-CH3 (Methyl) and -Br (Bromo)

(Word root) Parent carbon chain = 4 (but)
Primary suffix = ane
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
2-Bromo-3-methylbutane (secondary alkyl halide)

Q.6. Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:

CH3CC2H52CH2Br

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NCERT Chemistry (Vol 2) Chapter 1 Haloalkanes and haloarenes
Solution:

Substituent =-C2H5 (ethyl)

Secondary prefix =-Br (Bromo)
(Word root) Parent carbon chain =4 (but)
Primary suffix = ane
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
1-Bromo-2-ethyl-2-methylbutane
(primary alkyl halide)

Q.7. Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:

CH3CClC2H5CH2CH3

Solution:

Substituents = -CH3 (Methyl) and -Cl (Chloro)

(Word root) Parent carbon chain =5 (pent)
Primary suffix = ane
IUPAC name = Secondary prefix + Primary prefix + Word root +Primary suffix + Secondary suffix
3-Chloro-3-methylpentane (Tertiary alkyl halide)

Q.8. Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:

CH3CH=CHCBrCH32

Solution:

Substitute =-CH3 (Methyl)

Secondary prefix = Br(Bromo)
(Word root) Parent carbon chain= 5 (pent)
Primary suffix = ene
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
4-Bromo-4-methylpent-2-ene
(Allyl halide)

Q.9. Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:

CH3CH=C(Cl)CH2CHCH32

Solution:

Substituent =-CH3 (Methyl)

Secondary prefix =-Cl (Chloro)
(Word root) Parent carbon chain =6(hex)
Primary suffix = ene
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
3-Chloro-5-methylhex-2-ene
(Vinyl halide)

Q.10. Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide: p-ClC6H4CH2CHCH32

Solution:

Substituent =CH32CHCH2- (Methylpropyl)

Secondary prefix =-Cl (Chloro)
(Word root) Parent carbon chain = benzene
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
1-Chloro-4-(2-methylpropyl)benzene
(Aryl halide)

Q.11. Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:

m-ClCH2C6H4CH2CCH33

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Solution:

Substituent = CH33CCH2- (Neopentyl) and -CH2Cl (Chloromethyl)

(Word root) Parent carbon chain = benzene
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
1-chloromethyl-3-(2, 2-dimethylpropyl)benzene (Primary benzyl halide)

Q.12. Name the following halide according to IUPAC system and classify it as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halide:

o-Br-C6H4CHCH3CH2CH3

Solution:

Substituent =

(methylpropyl)
Secondary prefix = -Br (Bromo)
(Word root) Parent carbon chain = benzene
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
1-Bromo-2-(1-methylpropyl)benzene
(Aryl halide)

Q.13. Predict all the alkenes that would be formed by dehydrohalogenation of 1-Bromo-1-methylcyclohexane with sodium ethoxide in ethanol and identify the major alkene.

Solution: 1-bromo-1-methylcyclohexane

In the given compound, all β-hydrogen atoms are equivalent. Thus, the dehydrohalogenation of this compound gives only one alkene.

Q.14. Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:
2-Chloro-2-methylbutane

Solution:

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.

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 i.e. more
stable alkene is formed as the major product
Therefore, alkene (I) i.e., 2-methylbut-2-ene is the major product in this reaction.

Q.15. Predict all the alkenes that would be formed by dehydrohalogenation of 2,2,3-Trimethyl-3-bromopentane with sodium ethoxide in ethanol and identify the major alkene:

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NCERT Chemistry (Vol 2) Chapter 1 Haloalkanes and haloarenes
Solution: 2,2,3-Trimethyl-3-bromopentane has two different sets of β-hydrogen atoms and hence, in principle, can give two alkenes (I and II). But according to Saytzeff's rule, more highly substituted
alkene (II), being more stable, is the major product.

Q.16. How will you bring about the following conversion?

Ethanol to but-1-yne

Solution: The conversion of Ethanol to But-1-yne is as follows:

CH3CH2OHEthanol→SOCl2, PyridineCH3CH2ClChloroethane+SO2+HCl

HC≡CHEthyne+NaNH2→Liq.NH3HC≡CNa-+Sodium acetylide CH3CH2-ClChloroethane+HC≡CNa-+→CH3CH2C≡CHBut-1-yne+NaCl In this conversion, ethanol reacts with SOCl2 in

the presence of pyridine through nucleophilic substitution reaction and gets convert into chloroethane. On further reaction of chloroethane with sodium acetylide, but-1-yne is formed.

Q.17. How will you bring about the following conversion?

Ethane to bromoethene

Solution: The bromination of ethane (Br2 in the presence of light at 520-670K) gives bromoethane which on further treatment with alcoholic KOH results in the alkene called ethene.

This again on bromination with Br2/CCl4 gives dibromide(vicinal bromide) called 1,2-dibromoethane which on treatment with alcoholic KOH gives bromoethene as the final product.

Q.18. How will you bring about the following conversion?

Propene to 1-nitropropane

Solution: The preparation of 1-nitropropane from propene is as follows:

In this conversion, first Anti-Markovnikoff's addition reaction takes place. Due to this, hydrogen of HBr attach to double bonded carbon, where less number of hydrogens are present. And
bromide is attach to that carbon, where more number of hydrogens are present. The reaction of AgNO2 with 1-bromo propane results in the formation of 1-nitropropane.

Q.19. How will you bring about the following conversion?

Toluene to benzyl alcohol

Solution: The reaction for the conversion of Toluene to benzyl alcohol is as follows:

In this conversion, toluene reacts with chlorine in the presence of UV-light or heat to form benzyl chloride. Benzyl chloride further react with a base like aqueous KOH and heating forms
benzyl alcohol.

Q.20. How will you bring about the following conversion?

Propene to propyne

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NCERT Chemistry (Vol 2) Chapter 1 Haloalkanes and haloarenes
Solution: The reaction that take place during conversion of propene to propyne is as follows:

CH3-CH=CH2→E+ additionBr2/CCl4CH3-CHBr-CH2Br→Liq.NH32NaNH2CH3-C≡CH

In this conversion, firstly electrophilic addition takes place in the presence of bromine to form 1,2-dibromo propane. 1,2-dibromo propane on further reaction with NaNH2 forms propyne.

Q.21. How will you bring about the following conversion?

Ethanol to ethyl fluoride

Solution: The conversion of Ethanol to ethyl fluoride is as follows:

In this conversion, firstly ethanol reacts with PCl5 to form chloroethane. Further, chloroethane reacts with AgF and forms ethyl fluoride. This is known as Swarts reaction.

Q.22. How will you bring about the following conversion?

Bromomethane to propanone

Solution: The conversion of Bromomethane to propanone is as follows:

In this conversion, firstly Bromomethane reacts with KCN according to SN2 reaction to form acetonitrile. Acetonitrile is further treated with Grignard reagent to form an adduct,
which on hydrolysis forms propanone.

Q.23. How will you bring about the following conversion?

But-1-ene to but-2-ene

Solution: The conversion of But-1-ene to But-2-ene is as follows:

In this conversion, firstly Markovnikov's addition takes place. Further, the elimination of HBr takes place in the presence of alcoholic KOH and finally we get But-2-ene.

Q.24. How will you bring about the following conversion?

1-Chlorobutane to n-octane

Solution: The conversion of 1-chlorobutane to n-octane in the presence of sodium metal and dry ether is called Wurtz reaction. Two equivalent Chlorobutane reacts to give n-octane as a final product
and sodium chloride as the by-product.

2CH3CH2CH2CH2-Cl1-Chlorobutane+2Na→-2NaClEtherCH3CH2CH2CH2CH2CH2CH2CH3n-octane

Q.25. How will you bring about the following conversion?

Benzene to biphenyl

Solution: The conversion of benzene to biphenyl using Na and dry ether is called Fittig reaction.

Firstly, the benzene is treated with Br2/FeBr3 results in the formation of bromobenzene. It is an electrophilic substitution reaction. Further, the bromobenzene is treated with sodium metal in
the presence of dry ether to form biphenyl as the final product and sodium bromide as the by-product.

Q.26. Explain why the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

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Solution:

sp2-hybrid carbon in chlorobenzene is more electronegative than a sp3-hybrid carbon in cyclohexylchloride, due to greater s-character. Thus, C atom of chlorobenzene has less tendency to
release electrons to Cl than carbon atom of cyclohexylchloride.

Also, in chlorobenzene, the lone pair of electron on the Cl atom is in resonance with a benzene ring. This gives partial double bond character to C-Cl bond and there is a decrease in the
magnitude of partial negative charge on Cl atom. However, in cyclohexyl chloride, there is a C-Cl single bond. The dipole moment is a product of charge and distance. Also C-Cl bond with
partial double bond character has lower bond length than with C-Cl single bond. Hence, chlorobenzene has lower dipole moment than cyclohexyl chloride.

Q.27. Explain why alkyl halides, though polar, are immiscible with water?

Solution: 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.

Q.28. Explain why Grignard reagents should be prepared under anhydrous conditions?

Solution: Grignard reagents are very reactive. In the presence of water, they react and give alkanes because alkyl group acts as a base and water has acidic hydrogen.

Q.29. Give the uses of freon 12, DDT, carbon tetrachloride, and iodoform.

Solution: 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.30. Write the structure of the major organic product in each of the following reaction:

Solution:

Alkyl iodides are often prepared by the reaction of alkyl chlorides/bromides with NaI in dry acetone. This reaction is known as the Finkelstein reaction. In the given equation the major
organic product is 1-Iodopropane.

This is one of the Halogen exchange reaction.

Q.31. Write the structure of the major organic product in each of the following reaction:

Solution:

In this reaction, dehyrdrohalogenation takes place which results in the formation of 2-methylpropene.

Q.32. Write the structure of the major organic product in each of the following reaction:

Solution:

In this reaction unimolecular nucleophilic substitution reaction takes place resulting in the formation of butan-2-ol.

Q.33. Write the structure of the major organic product in the following reaction:

CH3CH2Br+KCN→aq. ethanol

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Solution: CH3CH2BrBromoethane+KCN→aq.ethanolCH3CH2CNCyanoethane+KBr

In this reaction nucleophilic substitution reaction takes place where nucleophile attack on the carbon which is attached to leaving group. Here, bond breaking and bond making action occur
simultaneously.

Q.34. Write the structure of the major organic product in the following reaction:
C6H5ONa+C2H5Cl→

Solution: C6H5ONaSodium phenoxide+C2H5ClChloroethane→Williamson synthesisC6H5-O-C2H5Ethoxy benzene+NaCl

In this method, an alkyl halide(Ethyl chloride) is allowed to react with sodium phenoxide and produce Ethoxybenzene as a major product. This reaction is also known as Williamson
synthesis. The reaction involves SN2 attack of an alkoxide ion on primary alkyl halide.

Q.35. Write the structure of the major organic product in the following reaction:

Solution:

In this reaction, nucleophilic substitution mechanism takes place. The hydroxyl group of an alcohol is replaced by halogen on reaction with thionyl chloride. Thionyl chloride is 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.

Q.36. Write the structure of the major organic product in the following reaction:
CH3CH2CH=CH2+HBr→Peroxide

Solution: CH3CH2CH=CH2But-1-ene+HBr→PeroxideCH3CH2CH2CH2-Br1-Bromobutane
According to Anti-Markovnikov’s rule, hydrogen will add to that carbon which have less number of hydrogen atoms and Br- (Nu-) attach to that carbon which have more number
of hydrogen atoms.

Q.37. Write the structure of the major organic product in the following reaction:

Solution:

According to Markovnikov's rule H atom will add to that carbon which have more H atoms and Br- (Nucleophile) attack on that carbon which have less H atoms.

Q.38. Write the mechanism of the following reaction:

nBuBr+KCN →EtOH-H2O nBuCN

Solution: The given reaction is:

nBuBr+KCN →EtOH-H2O nBuCN, given reaction is a SN2 reaction. In this reaction CN- ion acts as the nucleophile and attack on the carbon atom at which Br leaving group is attached
and less hinderend CN- ion is an ambident nucleophile and can attack through both C and N. In this case, it attacks through the C atom.

Q.39. Arrange the compounds of each set in order of reactivity towards SN2 displacement:

2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane.

Solution: An SN2 reaction involves the approaching of the nucleophile to the carbon atom to which the leaving group (-Br) is attached. When the nucleophile is sterically hindered, then the reactivity
towards SN2 displacement decreases.

CH3CH2CH2CH2-Br1-Bromopentane1o CH3CH2CH2CHBrCH32-Bromopentane2o CH3CH2CBrCH322-Bromo-2-methylbutane3o

Due to the presence of substituents, hindrance to the approaching nucleophile decreases SN2 order 1°> 2°>3° (alkyl halide). 1-Bromopentane>2-bromopentane >2-Bromo-2-
methylbutane (nu- approach tendency).
Hence, the increasing order of reactivity towards SN2 displacement is: 2-Bromo-2-methylbutane <2-Bromopentane <1-Bromopentane.

Q.40. Arrange the compounds of each set in order of reactivity towards SN2 displacement:
1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2-methyl butane

Solution:

Rate of SN2 reaction is inversely proportional to steric hindrance. 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°, Hence, the given set of compounds can be arranged in the increasing order of their reactivity towards SN2 displacement as:
2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < l-Bromo-3-methylbutane

Q.41. Arrange the compounds of each set in order of reactivity towards SN2 displacement:

1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane

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NCERT Chemistry (Vol 2) Chapter 1 Haloalkanes and haloarenes
Solution: The structures of the given compounds are as follows:

The steric hindrance to the nucleophile in the SN2 reaction increases with a decrease in the distance of the substituent from the atom containing the leaving group. Further, the steric
hindrance increases with an increase in the number of the substituent. Therefore, the increasing order of steric hindrances in the given compounds is as follows:
1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane < 1-Bromo-2, 2-dimethylpropaneHence, 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.42. Out of C6H5CH2CI and C6H5CHClC6H5, which is more easily hydrolysed by aqueous KOH?

Solution:

Hydrolysis by aqueous KOH proceeds through SN1 and by the formation of carbocation. If carbocation is stable, then the compound is easily hydrolysed by aqueous KOH. (polar protic
solvent) Now, C6H5CH2Cl forms 1°-carbocation, while C6H5CHClC6H5 forms 2°-carbocation which is more stable than 1°-carbocation as positive charge is more delocalised.

Hence, C6H5CHCIC6H5 is hydrolysed more easily than C6H5CH2Cl by aqueous KOH.

Q.43. p-Dichlorobenzene has higher melting point and lower solubility than those of o- and m-isomers. Discuss.

Solution:

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. Since during melting or dissolution, the crystal
lattice breaks, therefore, more energy is required to melt or dissolve the p-isomer than the corresponding o-and m-isomers. As a result, p-dichlorobenzene has a higher melting point and
lower solubility than o- and m-isomers.

Q.44. How the following conversion can be carried out?

Propene to Propan-1-ol

Solution:

In this conversion first propene undergo anti-markovnikov addition due to which hydrogen of HBr attach to that carbon where less number of hydrogens are present and bromide attach to
that carbon where more number of carbon are present and convert into bromopropane then it will undergo with nucleophilic substitution reaction in the presence of aqueous KOH and heat
and finally convert into propan-1-ol.

Q.45. How the following conversion can be carried out?

Ethanol to but-1-yne

Solution: The conversion of Ethanol to but-1-yne is carried out in the following steps:

CH3CH2OHEthanol→SOCl2, PyridineCH3CH2ClChloroethane+SO2+HCl

HC≡CHEthyne+NaNH2→Liq.NH3HC≡C-Na+Sodium acetylide CH3CH2-ClChloroethane+HC≡C-Na+→CH3CH2C≡CHBut-1-yne+NaCl In the above reaction, ethanol is reacted with

thionyl chloride to form ethyl chloride. In the second reaction, ethyne is treated with sodamide to form sodium acetylide. This sodium acetylide is treated with chloroethane to form But-1-
yne.

Q.46. How the following conversion can be carried out?

1-Bromopropane to 2-bromopropane

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Solution:

In the given conversion, first dehydrohalogenation reaction takes place in the presence of alcoholic KOH then Markovnikov addition takes place.

Q.47. How the following conversions can be carried out?

2-Methyl-1-propene to 2-chloro-2-methylpropane

Solution:

In this conversion Markovnikov addition takes place as a result of which hydrogen attach to that carbon which has more number of hydrogens and chlorine attach to that carbon which
has less number of hydrogen.

Q.48. How the following conversion can be carried out?

Ethyl chloride to propanoic acid.

Solution: The conversion of ethyl chloride to propanoic acid is as follows:

CH3CH2ClEthylchloride+KCN→aqueous ethanolCH3CH2CNPropanenitrile+KClCH3CH2CNPropanenitrile→H+/H2OCH3CH2COOHPropanoic acid

In this conversion, firstly ethyl chloride reacts with KCN in aqueous ethanol (Nucleophilic substitution) and forms propanenitrile. This will undergo hydrolysis in acid medium and finally
convert into propanoic acid.

Q.49. How the following conversion can be carried out?

But-1-ene to n-butyliodide.

Solution: The conversion of But-1-ene to n-butyl iodide is as follows:

CH3-CH2-CH=CH2But-1-ene→Anti-Markovnikov additionHBr/PeroxideCH3-CH2-CH2-CH2-Br1-BromobutaneCH3-CH2-CH2-CH2-Br1-Bromobutane→Finkelstein reactionNaI, dry actoneCH

In this conversion, first but-1-ene reacts with HBr according to Anti-Markovnikov's addition. As a result of which hydrogen will attach to that carbon which has less hydrogen atoms and bromine will attach to
has more hydrogen atoms to form 1-Bromobutane. 1-Bromobutane reacts with sodium iodide in presence of acetone gives n-Butyliodide. This reaction is known as Finkelstein reaction.

Q.50. How the following conversion can be carried out?

2-Chloropropane to 1-propanol.

Solution: The conversion of 2-Chloropropane to 1-propanol is as follows:

In this conversion, firstly dehydrohalogenation takes place in the presence of alcoholic KOH to form propene. Then Anti-Markovnikov's addition takes place in the presence of HBr and
peroxide. The hydrogen from the hydrogen halide attaches to that carbon which have less number of hydrogen atoms. The bromine atom attaches to that carbon which have more number of
hydrogen atoms and forms 1-bromopropane.

1-bromopropane will react in aqueous KOH (nucleophilic substitution) and forms 1-propanol.

Q.51. How the following conversion can be carried out?

Isopropyl alcohol to iodoform

Solution: The conversion of isopropyl alcohol to iodoform is as follows:

In this conversion, firstly isopropyl alcohol undergoes oxidation to form acetone in presence of chromic acid which is a good oxidising agent. Further reaction of the ketone with NaOI,
results in the formation of iodoform.

Q.52. How the following conversion can be carried out?

Chlorobenzene to p-nitrophenol

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Solution: The conversion of Chlorobenzene to p-nitrophenol is as follows:

In this conversion, firstly chlorobenzene undergoes a nitration reaction to form p-chloronitrobenzene as the major product.

The p-chloronitrobenzene undergoes an aromatic nucleophilic substitution reaction in the presence of sodium hydroxide at high temperature and pressure and finally, p-nitrophenol is
obtained.

Q.53. How the following conversion can be carried out?

2-Bromopropane to 1-Bromopropane

Solution: The conversion of 2-bromopropane to 1-bromopropane is as follows:

In this conversion, 2-bromopropane undergo dehydrohalogenation in presence of alcoholic KOH and forms propene, which reacts with hydrogen bromide finally converts into 1-
bromopropane by Anti-Markovnikov addition.

Q.54. How the following conversion can be carried out?

Chloroethane to butane

Solution: The conversion of Chloroethane to butane is as follows:

2CH3-CH2-Cl →Wurtz reaction2Na/dry etherCH3-CH2-CH2-CH3 + 2NaCl

In this conversion, two moles of chloroethane react with sodium in presence of dry ether to form butane. This reaction is known as Wurtz reaction.

Q.55. How the following conversions can be carried out?

Benzene to diphenyl

Solution: The conversion of Benzene to diphenyl is as follows:

In this conversion, first, the benzene undergoes bromination and forms bromobenzene.

Further two moles of bromobenzene undergo Fittig reaction in presence of sodium metal and dry ether to form biphenyl.

Q.56. How the following conversion can be carried out?

Tert-butyl bromide to isobutyl bromide

Solution: The conversion of tert-butyl bromide to isobutyl bromide is as follows:

In this conversion, firstly tert-butyl bromide undergoes dehydrohalogenation in presence of alcoholic KOH resulting in the formation of 2-methyl propene, then Anti-Markovnikov addition
takes place in presence of hydrogen bromide to form isobutyl bromide.

Q.57. How the following conversions can be carried out?

Aniline to phenyl isocyanide

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Solution:

In carbylamine reaction, aniline converts into phenyl isocyanide. When aniline reacts with CHCl3 and KOH in the presence of heat, it forms phenyl isocyanide. In this reaction, the
electrophilic dichlorocarbene attacks the nucleophilic nitrogen of aniline.

Q.58. How the following conversions can be carried out?

Toluene to benzyl alcohol

Solution:

In this conversion first toluene reacts with chlorine in the presence of ultraviolet light and gets converted into benzyl chloride then it will react to aqueous KOH (nucleophilic substitution)
and convert into benzyl alcohol.

Q.59. How the following conversions can be carried out?

Benzene to 4-bromonitrobenzene

Solution:

In this conversion first benzene reacts with bromine (bromination) to form bromobenzene then it undergo nitration in the presence of conc. nitric acid/conc. sulphuric acid and finally it gets
converted into p-bromonitrobezene(4-bromonitrobenzene) as the major product.

Q.60. How the following conversions can be carried out?

Benzyl alcohol to 2-phenylethanoic acid

Solution:

In this conversion first benzyl alcohol reacts with PCl5 according to SN2 reaction then convert into benzyl chloride. It then reacts with KCN and aqueous ethanol according to SN1 reaction
and convert into benzyl cyanide then hydrolysis takes place and finally convert into 2-phenylethanoic acid.

Q.61. How the following conversions can be carried out?

Ethanol to propanenitrile

Solution: CH3 -CH2-OH→SN2red P/Br2CH3-CH2-Br→SN2KCN, Aq.ethanolCH3-CH2-CN

In this conversion first ethanol reacts with bromine in presence of red phosphorus, the compound formed is ethyl bromide then it will reacts with KCN and aqueous ethanol and finally
convert into propanenitrile.

Q.62. How the following conversion can be carried out?

Aniline to chlorobenzene

Solution:

In this conversion, first diazotisation takes place with aniline and aniline will convert into benzenediazonium chloride then it will react with Cu2Cl2 (Gattermann reaction) and forms
chlorobenzene.

Q.63. How the following conversions can be carried out?

2-Chlorobutane to 3, 4-dimethylhexane

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Solution: Wurtz’s reaction is an organic chemical coupling reaction in which alkyl halides reacts with sodium metal in presence of dry ether and form a higher alkane.

In this conversion Wurtz reaction takes place which results in the formation of 3,4-dimethylhexane.

Q.64. Give the IUPAC name of the following compound:

CH3CHClCHBrCH3

Solution:

Secondary prefix = -Br(Bromo),-Cl (Chloro)

(Word root) Parent Carbon chain = 4 (but)
Primary suffix = ane
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
2-Bromo-3-chlorobutane

Q.65. Give the IUPAC name of the following compound:

CHF2CBrClF

Solution:

Secondary prefix = -Br(Bromo), -Cl (Chloro), -F (Fluoro)

(Word root) Parent Carbon chain = 2(eth)
Primary suffix = ane
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
1-Bromo-1-chloro-1,2,2-trifluoroethane

Q.66. Give the IUPAC name of the following compound:

ClCH2C≡CCH2Br

Solution:

Secondary prefix = -Br(Bromo),-Cl (Chloro)

(Word root) Parent Carbon chain = 4 (but)
Primary suffix = yne
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
1-Bromo-4-chlorobut-2-yne

Q.67. Give the IUPAC name of the following compound:

CCl33CCl

Solution:

Secondary prefix =-Cl (Chloro) and -CCl3 (trichloromethyl)

(Word root) Parent Carbon chain = 3(prop)
Primary suffix = ane
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
2-(trichloromethyl)-1,1,1,2,3,3,3-heptachloropropane

Q.68. Give the IUPAC name of the following compound:

CH3Cp-ClC6H42CHBrCH3

Solution:

Secondary prefix = -Br (Bromo) and para-phCl (4-chlorophenyl)

(Word root) Parent Carbon chain = 4 (but)
Primary suffix = ane
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
2-Bromo-3,3-bis(4-chlorophenyl)butane

Q.69. Give the IUPAC name of the following compound:

CH33CCH=CClC6H4I-p

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Solution:

Secondary prefix = -Cl (Chloro) and para-phI (iodophenyl)

(Word root) Parent Carbon chain = 4 (but)
Primary suffix = ene
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
1-chloro-1-(4-iodophenyl)-3,3-dimethylbut-1-ene

Q.70. 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.

Solution: In an aqueous solution, KOH almost completely ionises to give OH- ion, which is a strong nucleophile, which leads the alkyl chloride to undergo a substitution reaction to form alcohol.
R-Cl +KOH(aq)→R-OH+KCl

On the other hand, an alcoholic solution of KOH contains alkoxide RO- ion as well, which is a nucleophile as well as strong base. Thus, it can abstract a hydrogen from the β-carbon of the
alkyl chloride favouring an elimination reaction and forms an alkene.

Thus, it depends on the condition as well, to favour an elimination and a substitution reaction.

Q.71. 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), C8H18 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.

Solution: The two primary alkyl halides having the formula, C4H9Br are n-butyl 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 (d) 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.

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. 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,

Q.72. What happens when n-butyl chloride is treated with alcoholic KOH.

Solution: When n-butyl chloride is treated with alcoholic KOH, the formation of but-1-ene takes place. This reaction is a dehydrohalogenation reaction.

Q.73. What happens when bromobenzene is treated with Mg in the presence of dry ether?

Solution: When Bromobenzene is treated with Mg in the presence of dry ether, Phenylmagnesium bromide (Grignard reagent) is formed.

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Q.74. What happens when chlorobenzene is subjected to hydrolysis?

Solution: Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated with sodium hydroxide solution at a temperature of 623 K and a
pressure of 300 atm to form phenol.

Q.75. What happens when ethyl chloride is treated with aqueous KOH?

Solution: When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol. In this reaction, nucleophilic substitution takes place.

Q.76. What happens when methyl bromide is treated with sodium in the presence of dry ether?

Solution: When methyl bromide is treated with sodium in the presence of dry ether, ethane is formed. This reaction is known as the Wurtz reaction. This is a type of coupling reaction, in which two
alkyl groups combine with each other.

Q.77. What happens when methyl chloride is treated with KCN?

Solution: When methyl chloride is treated with KCN, it undergoes a nucleophilic substitution reaction to give methyl cyanide.
CH3Cl+KCN→nucleophilic substitutionCH3CN+KCl

Q.78. Write the structure of the following organic halogen compound:

2-chloro-3-methylpentane

Solution:

Secondary prefix = -Cl (Chloro) and -CH3(methyl)

(Word root) Parent Carbon chain = 5(pent)
Primary suffix = ane
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
2-chloro-3-methylpentane

Q.79. Write the structure of the following organic halogen compound:

p-Bromochlorobenzene

Solution:

Secondary prefix = -Cl (Chloro), -Br (Bromo)

(Word root) Parent Carbon chain = 6(benzene)
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
p-Bromochlorobenzene

Q.80. Write the structure of the following organic halogen compound:

1-Chloro-4-ethylcyclohexane

Solution:

Secondary prefix = -Cl (Chloro) and -C2H5 (ethyl)

Primary prefix = Cyclo

(Word root) Parent Carbon chain = 6(benzene)

IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
1-Chloro-4-ethylcyclohexane

Q.81. Write the structure of the following organic halogen compound :

2-(2-Chlorophenyl)-1-iodooctane

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Solution:

Secondary prefix = -I (iodo) and Ph-Cl (Chlorophenyl)

(Word root) Parent Carbon chain = 8(oct)
IUPAC name = secondary prefix + primary prefix + word root + primary suffix + secondary suffix
2-(2- Chlorophenyl)-1-iodooctane.

Q.82. Write the structure of the following organic halogen compound:

2-Bromobutane

Solution:

Secondary prefix = -Br (Bromo)

(Word root) Parent Carbon chain = 4(but)
IUPAC name = secondary prefix + primary prefix + word root + primary suffix + secondary suffix
2-bromobutane.

Q.83. Write the structure of the following organic halogen compound :

4-tert-Butyl-3-iodoheptane

Solution:

Secondary prefix = -I (iodo) and -(CH3)3C (tert-Butyl)

(Word root) Parent Carbon chain = 7(hept)
IUPAC name = Secondary prefix + Primary prefix + Word root + Primary suffix + Secondary suffix
4-tert-Butyl-3-iodoheptane.

Q.84. Write the structure of the following organic halogen compound :

1-Bromo-4-sec-butyl-2-methylbenzene

Solution:

Secondary prefix = -Br (Bromo), (sec-butyl) and -CH3(methyl)

(Word root) Parent Carbon chain = 6(benzene)
IUPAC name = secondary prefix + primary prefix + word root + primary suffix + secondary suffix
1-Bromo-4-sec-butyl-2-methylbenzene.

Q.85. Write the structure of the following organic halogen compound :

1, 4-Dibromobut-2-ene

Solution:

Secondary prefix = -Br (Bromo)

(Word root) Parent Carbon chain = 4(but)
Primary suffix = ene
IUPAC name = secondary prefix + primary prefix + word root + primary suffix + secondary suffix
1, 4-Dibromobut-2-ene

Q.86. Which of the following has the highest dipole moment?

i CH2Cl2 ii CHCl3 iii CCl4

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Solution: Dipole moment = Charge ×Distance
μ=q×d
It is a vector quantity.

Dichloromethane CH2Cl2
μ=1.60 D
In CH2Cl2, the resultant of the dipole moments of two 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. CH2Cl2 has the highest dipole moment.

Chloroform CHCl3
μ=1.08 D
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 and CHCl3 molecule is asymmetrical. As a result, CHCl3 has a small dipole moment of
1.08 D.

Carbon tetrachloride CCl4

μ=0 D

CCl4 molecule is symmetrical. Therefore, the dipole moments of all four C-Cl bonds cancel each other. So, its resultant dipole moment is zero.
Hence, the given compounds can be arranged in the increasing order of their dipole moments as:
CCl4<CHCl3<CH2Cl2.

Q.87. A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.

Solution: A hydrocarbon with the molecular formula, C5H10 has the general molecular formula CnH2n. Therefore, it can be either be an alkene or a cycloalkane.
It is a cycloalkane because hydrocarbons do not react with chlorine in the dark; it cannot be an alkene.

Also, the hydrocarbon gives a single monochloro compound, C5H9Cl by reacting with chlorine in bright sunlight. A single monochloro compound means that the hydrocarbon must contain
hydrogen atoms that are all equivalent. And as all hydrogen atoms of a cycloalkane are equivalent, the hydrocarbon must be a cycloalkane. Hence, the compound is cyclopentane. The
reactions involved are:

Q.88. Write the isomers of the compound having formula C4H9Br.

Solution: There are four isomers of the compound having the formula C4H9Br. These isomers are given below.
(a) 1-Bromobutane

(b) 2-Bromobutane

(c) 1-Bromo-2-methylpropane

(d) 2-Bromo-2-methylpropane

Q.89. Write the equation for the preparation of 1-iodobutane from 1-butanol.

Solution: CH3-CH2-CH2-CH2-OH+HI →(ZnCl2)CH3-CH2-CH2-CH2-I+H2O

In presence of ZnCl2, alcohol reacts with HI and forms alkyl iodide and water. This is a substitution reaction.

Q.90. Write the equation for the preparation of 1-iodobutane from

1-chlorobutane

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Solution:

This reaction is known as Finkelstein reaction, where halogen exchange takes place.

Q.91. Write the equation for the preparation of 1-iodobutane from but-1-ene.

Solution:

An alkene is converted to corresponding alkyl halide by reaction with hydrogen bromide. Alkyl iodide are prepared by the reaction with of alkyl chlorides/bromides with NaI in dry acetone.

Q.92. What are ambident nucleophiles? Explain with an example.

Solution: Nucleophiles having two nucleophilic sites are ambident nucleophiles. Hence, ambident nucleophiles have two sites through which they can attack.

Nitrite ion is an example of an ambident nucleophile. It can either attack through oxygen resulting in the formation of alkyl nitrites or it can attack through nitrogen resulting in the formation
of nitroalkanes.

Q.93. Which compound in each of the following pairs will react faster in SN2 reaction with OH-?
CH3Br or CH3I

Solution: In SN2 mechanism, the reactivity of halides for the same alkyl group increases in the order R-F<<R-Cl<R-Br<R-I.
As the size of the ion increases, it becomes a better leaving group.
Therefore, CH3I will react faster than CH3Br in SN2 reaction with OH-.

Q.94. Which compound in each of the following pair will react faster in SN2 reaction with OH-?

(CH3)3CCl or CH3Cl

Solution: The SN2 mechanism involves the attack of the nucleophile at the atom bearing the leaving group. The order of reactivity of alkyl halides towards SN2 reaction, tertiary halide, secondary
halide primary halide and methyl halide. But, in the case of (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 CH3Cl. Hence, CH3Cl reacts faster than
(CH3)3CCl in SN2 reaction with OH-.

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Intext question

Q.1. Write the structure of the following compound:

2-chloro-3-methylpentane

Solution: In the given molecule, the root word is "pent". So it has five carbon atoms in the parent chain. Two substituents i.e., chlorine and methyl group present at second and third carbon. The suffix
is "-ane". So all carbon atoms bonded each other with single(sigma)bond. Hence, the structure is

Q.2. Write the structure of the following compound:

1-chloro-4-ethylcyclohexane

Solution: In the given molecule, the root word is "hex". So it has six carbon atoms in the parent chain. Two substituents i.e., chlorine Cl and ethyl-C2H5 group present at 1st and 4th carbon
respectively. The suffix is "-ane". So all carbon atoms bonded each other with single(sigma)bond. The word "cyclo" indicates all six carbon atoms form a ring or closed structure. Hence, the
structure is

Q.3. Write the structure of the following compound:

4-tert. Butyl-3-iodoheptane

Solution: In the given molecule, the root word is "hept". So it has seven carbon atoms in the parent chain. Two substituents(written in prefix part) i.e., iodo and tert. Butyl group present at 3rd and 4th
carbon. The suffix is "-ane". So all carbon atoms bonded each other with single(sigma)bond. Hence, the structure is

Q.4. Write the structure of the following compound:

1,4-Dibromobut-2-ene

Solution: In the given molecule, the root word is "But". So, it has four carbon atoms in the parent chain. Two bromine groups are attached at 1st and 4th carbon. So, the prefix "dibromo" is used. The
suffix is "-ene" which indicates the presence of a double bond which is located at second carbon. Hence, the structure is

Q.5. Write the structure of the following compound:

1-Bromo-4-sec. butyl-2-methylbenzene

Solution: 1-Bromo-4-sec. butyl-2-methylbenzene

Primary prefix =-Br (Bromo)

Substituents =-CH3 (methyl) and

CH3-CH|-CH2-CH3 (Secondary butyl)

Parent hydrocarbon = Benzene

Q.6. Why is sulphuric acid not used during the reaction of alcohols with KI?

Solution: In the presence of sulphuric acid H2SO4, KI produces HI.

2KI(s)Potassium iodide+H2SO4(l)Sulphuric acid→2KHSO4(s)Potassium hydrogen sulphate+2HI(g)Hydrogen iodide

Since H2SO4 is an oxidising agent, it oxidises HI (produced in the reaction to I2).

2HI(g)Hydrogen iodide+H2SO4(l)Sulphuric acid→SO2(g)Sulphur dioxide+I2(s)Iodine+H2O(l)Water

Thus, the reaction between alcohol and HI to form an alkyl iodide cannot occur. Hence, sulphuric acid is not used during the reaction of alcohols with KI. Instead, a non-oxidising acid such
as H3PO4 is used.

CH3OH(l)Methanol+KI(s)Potassium iodide+H3PO4(l)Phosphoric acid →∆ CH3-I(l)Iodomethane+KH2PO4(s)Potassium dihydrogen phosphate+H2O(l)Water

Q.7. Write structures of different dihalogen derivatives of propane.

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Solution: There are four different dihalogen derivatives of propane. The structures of these derivatives are shown below:

Note: Chemical compounds that have identical chemical formulae but differ in properties and the arrangement of atoms in the molecule are called isomers.

Q.8. Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields four isomeric monochlorides.

Solution: The isomer of the molecular formula C5H12 should contain four different types of H-atoms to have four isomeric monochlorides.
Thus, the isomer is 2-methylbutane. The four types of H-atoms are labelled as a, b, c and d.

Q.9. Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields a single monochloride.

Solution: Only one type of hydrogen atoms in the isomer of the alkane of the molecular formula C5H12 should be there to have a single monochloride. This is because the replacement of any
hydrogen atom leads to the formation of the same product. Thus, the isomer is neopentane.

Neopentane

Q.10. Among the isomeric alkanes of molecular formula C5H12, identify the one that on photochemical chlorination yields three isomeric monochlorides.

Solution: The isomer of the alkane of the molecular formula C5H12 should contain three different types of H-atoms to have three isomeric monochlorides.
Thus, the isomer is n-Pentane. The three types of H atoms are labelled as a, b and c.

Q.11. Draw the structure of major monohalo products in the following reaction:

Solution:

It is a substitution reaction.

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

Q.12. Draw the structure of major monohalo products in the following reaction:

Solution:

This reaction is associated with free radical mechanism and bromination does not take place on the benzene ring.

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Q.13. Draw the structure of major monohalo products in the following reaction:

Solution:

The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids.

Q.14. Draw the structure of major monohalo products in the following reaction:

Solution:

It is a type of addition reaction where Markovnikov product is a major product. Iodine will add to the carbon atom having lesser number of hydrogen atoms.

Q.15. Draw the structure of major monohalo products in the following reaction:
CH3CH2Br+NaI→

Solution: CH3CH2Br(l)Ethyl bromide + NaI(s)Sodium iodide→ CH3CH2I(s)Ethyl iodide + NaBr(s)Sodium bromide

Alkyl halide serves as leaving group at the reactive sp3 hybridised carbon. The iodide displaces the bromide leaving group, resulting in a substituted product (alkyl iodide) and sodium
bromide.

Q.16. Draw the structure of major monohalo products in the following reaction:

Solution:

This reaction proceeds through free radical mechanism and form allyl product.

Q.17. Arrange the set of compounds in order of increasing boiling points.

Bromomethane, Bromoform, Chloromethane, Dibromomethane.

Solution:

For alkyl halides having the same alkyl group, the boiling point increases with an increase in the atomic mass of the halogen atom. Since, the atomic mass of Bromine is greater than that of
Chlorine, the boiling point of bromomethane is higher than that of chloromethane due to Van der Waal forces.

Also, for alkyl halides having the same alkyl group, the boiling point increases with an increase in the number of halides. Therefore, the boiling point of dibromomethane (CH2Br2) is higher
than that of chloromethane (CH3Cl) and bromomethane (CH3Br), but lower than that of bromoform (CHBr3).

Q.18. Arrange the set of compounds in order of increasing boiling points.

1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

Solution:

For alkyl halides having the same alkyl group, the boiling point increases with an increase in the atomic mass of the halogen atom. Since all have chloride, so, all depends on alkyl chain.
Butane has one extra carbon than propane. So, 1-Chlorobutane has a higher boiling point than 1-Chloropropane. Also, the boiling point decreases as branching increases.

Boiling Point order: Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane

Q.19. Which alkyl halide from the following pair would you expect to react more rapidly by an SN2 mechanism? Explain your answer.

Solution: 2-bromobutane is a 2° alkyl halide, 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 SN2 mechanism.

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Q.20. Which alkyl halide from the following pair would you expect to react more rapidly by an SN2 mechanism? Explain your answer.

Solution: 2-bromobutane is 2°alkyl halide, whereas 2-bromo-2-methylpropane is 3°alkyl halide.

Thus, greater number of substituents are present in 3°alkyl halide than in 2°alkyl halide for hindering the approaching nucleophile.
Therefore, 2-bromobutane reacts more rapidly than 2-bromo-2-methylpropane by an SN2 mechanism.

Q.21. Which alkyl halide from the following pair would you expect to react more rapidly by an SN2 mechanism? Explain your answer.

Solution: 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-2methylbutane.
Therefore, the approaching nucleophile is less hindered in case of the former than in case of the latter. Hence, 1-Bromo-3-methylbutane reacts faster than 1-bromo-2-methylbutane by
SN2 mechanism.

Q.22. In the following pair of halogen compounds, which compound undergoes faster SN1 reaction?
(i)

and (ii)

Solution: 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.
3°carbocation is more stable due to 9 hyperconjugation and more +I effect of 3-CH3 groups while 2°carbocation formed has 4 hyperconjugation and 2-C2H5 (+I group) groups. So, it is less
stable.

Q.23. In the following pair of halogen compounds, which compound undergoes faster SN1 reaction?
(i)

and (ii)

Solution: The alkyl halide (i) is 2°while (ii) is 1°. 2° carbocation is more stable than 1°carbocation because in 2°carbocation

, 5 hyperconjugations are taking place and +I effect of two alkyl groups while in

, 2 hyperconjugations are taking place and +I effect of -C5H11 group. Therefore (i), 2-chloroheptane, undergoes faster SN1 reaction than (ii), 1-chlorohexane.

Q.24. Identify A, B, C, D, E, R and R1 in the following:

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Solution:

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

R'X on treatment with Mg gives D i.e.,

D on treatment with water gives E,

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