Haloalkanes and Haloareness Questions and Answers AP Inter 2nd Year Chemistry Chapter 6

Andhra Pradesh BIEAP AP Inter 2nd Year Chemistry Study Material Pdf 6th Lesson Haloalkanes and Haloareness Class 12 Textbook Exercise Questions and Answers.

AP Inter 2nd Year Chemistry 6th Lesson Haloalkanes and Haloareness Questions and Answers

Haloalkanes and Haloareness Class 12 Questions and Answers

I. Multiple Choice Questions

Question 1.
What is the general formula for haloalkanes? (X = halogen atom, n = 1, 2, 3 )
1) C_nH_2nX
2) C_nH_2n+1X
3) C_nH_2n-1X
4) C_nH_2n-3X
Answer:
General formula for haloalkanes is C_nH_2n+1 X

Question 2.
Which of the following reagents is used for replacement of – OH group in an alcohol by -Cl group?
1) SO₂Cl₂
2) Aqueous HCl
3) PCl₃, SOCl₂
4) NaCl
Answer:

Question 3.
Identify the compound Y in the following reaction sequence.

Answer:

Question 4.
A primary alkyl halide would prefer to undergo [ 2 ]
1) S_N1 reaction
2) S_N2 reaction
3) a-elimination
4) racemization
Answer:
S_N2 reactions order of priority in alkylhalides 1° > 2° > 3°.

Question 5.
Which of the following is the suitable medium for preparing Grignard reagents?
1) Dry acetone
2) Diy ether
3) Concentrated HCl
4) Alcoholic KOH
Answer:

So dry ether is the medium.

Question 6.
Compared to haloalkanes, the reactivity of haloarenes towards nucleophilic substitution reactions is
1) low
2) high
3) very high
4) equal
Answer:
Reactivity of haloarenes towards nucleophilic substitution reactions is low.

Question 7.
Chlorobenzene is formed by the reaction of chlorine with benzene in the presence of anhydrous AlCl₃. Which of the following species attacks the benzene ring in this reaction?
1) Cl^–
2) Cl⁺
3) AlCl₃
4) [AlCl₄]^–
Answer:
Cl⁺ is electrophile in the chlorination of benzene.

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

1) (b) < (a) < (c)
2) (a) < (b) < (c)
3) (c) < (a) < (b)
4) (c) < (b) < (a)
Answer:

Straight chain compounds have high boiling point compared to branched chains.

Question 9.
Which of the following alkyl halide will undergo SN1 reaction most readily?
1) (CH₃)₃C – F
2) (CH₃)₃C – Cl
3) (CH₃)₃C – Br
4)(CH₃)₃C – I
Answer:
The order of reactivity of alkyl halides in S_N1 reactions is Iodides > Bromides > Chlorides > Fluorides

Question 10.
Which reagent will you use for the following reaction?
CH₃CH₂CH₂CH₂₃ CH₃CH₂CH₂CH₂Cl + CH₃ CH₂CHCICH₃
1) Cl₂/UV light
2) NaCl + H₂SO₄
3) Cl₂ gas in dark
4) Cl₂ gas in the presence of iron in dark
Answer:
Cl₂/UV light

Question 11.
Which of the following represents the correct name of the compound, CH₃CHCl₂?
1) Ethylidene chloride
2) Ethylene dichloride
3) 1,2-Dichloroethane
4) Vic-dichloride
Answer:
Ethylidene chloride

Question 12.
In alkyl halides carbon-halogen bond is polarized due to.
1) greater electronegativity of halogen atom than carbon
2) greater electronegativity of carbon atom than halogen atom
3) partial positive charge on halogen atom and a partial negative charge on carbon
4) tendency of halogen atom to donate electrons
Answer:
greater electronegativity of halogen atom than carbon

Question 13.
Chlorobenzene is
1) less reactive than benzyl chloride
2) more reactive than ethyl bromide
3) nearly as reactive as methyl chloride
4) more reactive than isopropyl chloride
Answer:
less reactive than benzyl chloride

Question 14.
The set of compounds in which the reactivity of halogen atom in the ascending order is
1) Vinyl chloride, chloroethane, chlorobenzene
2) Vinyl chloride, chlorobenzene, chloroethane
3) Chloroethane, chlorobenzene, vinyl chloride
4) Chlorobenzene, vinyl chloride, chloroethane
Answer:
C₆H₅Cl, CH₂ = CH – Cl, CH₃ – CH₂ – Cl
Chlorobenzene, Vinyl chloride, Chloroethane

Question 15.
The Wurtz-Fitting reaction involves
1) two molecules of aryl halides
2) two molecules of alkyl halides
3) one molecule each of aryl halide and alkyl halide
4) one molecule each of aryl halide and phenol
Answer:
Mixture of alkyl and aryl halides

II. Fill in the Blanks

Question 1.
Our body produces an iodine containing hormone, thyroxine, the deficiency of which causes a disease called _________.
Answer:
Goiter disease

Question 2.
Alkyl iodides are often prepared by the reaction of alkyl chlorides/ bromides with Nal in dry acetone. This reaction is known as _________.
Answer:
Finkelstein reaction

Question 3.
Groups like cyanides and nitrites possess two nucleophilic centers and are called ______ .
Answer:
Ambident nucleophiles

Question 4.
The compounds which rotate the plane polarized light are called _______.
Answer:
optically active (or) chiral

Question 5.
The compounds which are obtained by the reaction of haloalkanes with magnesium metal in dry ether are called ________.

Answer:
Grignard reagents

III. One Word Answer Questions

Question 1.
Preservation of the spatial arrangement of bonds to an asymmetric center during a chemical reaction or transformation is called
Answer:
Retention of configuration

Question 2.
Diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by -Cl or – Br. This reaction is called
Answer:
Sandmeyer’s reaction

Question 3.
Synthesis of alkyl fluorides by heating an alkyl chloride/bromide in the presence of a metallic fluoride such as AgF is named as
Answer:
Swart’s reaction

Question 4.
If the compound rotates the plane of plane polarized light to the right, i.e., clockwise direction, it is called
Answer:
Dextro rotatory (d) (+)

Question 5.
The Slercoisomers which are related to each other as non-superimposable mirror images are called
Answer:
Enantiomers

IV. Very Short Answer Questions

Question 1.
Write the structures of the following compounds.
i) 2-chloro-3-methylpentane
ii) 1 – bromo- 4 – sec – butyl – 2 – methylbenzene
Answer:

1 -Bromo-4-sec-butyl-2-methyl benzene

Question 2.
Which one of the following has the highest dipole moment?
i) CH₂Cl₂
ii) CHCl₃
iii) CCl₄
Answer:
i) Dipole moment of CH₂Cl₂ has 1.62D
ii) Dipole moment of CHCl₃ has 1.15D
iii) Dipole moment of CCl₄ has 0 D
So CH₂Cl₂ has high dipole moment.

Question 3.
What are ambident nucleophiles?
Answer:
Ambident nucleophiles are the species having two reactive sites (nucleophilic centers).
Ex: CN^– \(\mathrm{NO}_2^{-}\) etc.

Question 4.
Write the isomers of the compound having molecular formula C₄H₉Br.
Answer:
Isomers of C₄H₉Br. are
i) CH₃ – CH₂ – CH₂ – CH₂ – Br
1 – Bromobutane
ii)
2 – Bromobutane

iii)
1 – Bromo 2 – Methylpropane

iv)
2-Bromo-2methyl propane

Question 5.
Which compound in each of the following pairs will react faster in SN² reaction with OH?
i) CH₃Br or CH₃I
ii) (CH₃)₃CCl or CH₃Cl
Answer:
i) In CH₃Br or CH₃I, CH₃I will react faster in S_N2 reaction.
ii) In (CH₃)₃CCl or CH₃Cl, CH₃Cl will react faster in S_N2 reaction.

Question 6.
Explain why the alkyl halides though polar are immiscible with water.
Answer:
Alkyl halides are polar compounds but they are immiscible with water because they can not form strong hydrogen bonds with water molecules. The energy required to break the hydrogen bonds between water molecules is not compensated by the weak attraction between water and alkyl halide molecules. Hence, they do not mix with water.

Question 7.
Between C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH?
Answer:
C₆H₅CHClC₆H₅ is more easily hydrolysed by aQuestion KOH. This is because it forms more stable carbocation, which is stabilised by resonance from two phenyl rings.

Question 8.
What is the stereochemical outcome of S_N1 and S_N2 reactions?
Answer:
S_N1 reactions produce racemic mixture. S_N2 reactions produce inversion product.

Question 9.
What type of isomerism is exhibited by o-, m-, and p – chlorobenzenes?
Answer:
Position isomerism exhibits by o-, m-, p- chloro benzenes are

Question 10.
What are enantiomers?
Answer:
Enantiomers are stereoisomers that are non-superimposable mirror images of each other.

V. Short Answer Questions

Question 1.
Give the IUPAC names of the following compounds.
i) CH₃CH(Cl) CH(I)CH₃
ii) ClCH₂CH = CHCH₂Br
iii) (CCl₃)₃CCl
iv) CH₃C(p-Cl-C₆H₄)₂CH(Br)CH₃
Answer:
i)
2 – Chloro, 3 – Iodo butane

ii) Cl CH₂– CH = CH – CH₂ Br
1 – Bromo, 4 – Chloro but 2 – ene

iii)
2 – (tri chloro methyl) -1, 1, 1, 2, 3, 3, 3 – hepta chloro propane

iv)
2 – Bromo 3, 3 – bis – (4 – chlorophenyl) butane.

Question 2.
Write the structures of the following organic halides.
i) 1 -Bromo-4-sec-butyl-2-methylbenzene
ii) 2-Chloro- 1 -phenylhutane
iii) p-bromochlorobenzene
iv) 4-t-butyl-3-Iodoheptane
Answer:

Question 3.
How will you carry out the following conversions?
i) Ethane to bromoethane
ii) Toluene to benzyl alcohol
Answer:
i) Ethane to bromoethane
Bromine replaces one hydrogen atom of ethane in the presence of sunlight/U.V light to form bromoethane.

ii) Toluene to benzyl alcohol:
Toluene undergoes chlorination in presence of U.V light to form benzyl chloride and then followed by hydrolysis with aqueous KOH gives Benzyl alcohol.

Question 4.
Explain why the dipole moment of chloro-benzene is lower than that of cyclohexyl-chloride.
Answer:

1. The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride due to resonance in chlorobenzene.
2. In chloro benzene the lone pair of chlorine takes part in resonance with the benzene ring, reducing the polarity of the C-Cl bond.

Question 5.
Explain the mechanism of S_N¹ with an example.
Answer:
S_N¹ reactions (Nucleophilic uni molecular substitution):
1) The nucleophilic substitution reactions in which rate depends on only concentration of substrate are called S_N¹ reactions.
2) These reactions mainly takes place in 3⁰ – alkyl halides.
3) The product formed in S_N¹ reactions is racemic mixture.

Mechanism :
It involves in 2 steps.
Step 1: It is slow step and carbocation is formed.

Step II: It is fast step and attack of nucleophile takes step.

Order of reactivity towards S_N1 reactions 3° > 2° > 1° – Alkyl halides

Question 6.
Predict the alkenes that would be formed in the following reactions and identify the major alkene.

ii) 2-Chloro-2-methylbutane
Answer:

Question 7.
Write the mechanism of the following reaction. n-Butylbromide + KCN
Answer:


n – Butyl bromide

1. Above reaction shows S_N2 mechanism.
2. It involves single step with a transition state.
3. KCN gives CN^– ion
4. CN^– attacks the carbon attached to Br from the back side.
5. Br^– leaves simultaneously.

Question 8.
Write the mechanism of dehydrohalogenation of 2-bromobutane.
Answer:
When 2-Bromobutane is heated with alcoholic KOH, hydrogen from (3-carbon and bromine from a-carbon are eliminated to form 2 – Butene.

1) OH ion removes a (β-hydrogen atom .
2) At the same time, Br^– leaves from the adjacent carbon.
3) Double bond is formed between the two carbon atoms.

Question 9.
Describe the stereo chemical effect on the hydrolysis of 2-bromobutane.
Answer:

1. When S – 2-Bromobutane undergoes hydrolysis, R-2-Butanol is formed through S_N2 mechanism.
2. OH^– attacks from the side opposite to Br atom, causing inversion of configuration. (Walden inversion)
   

Question 10.
Write the differences between S_N1 and S_N2 reactions.
Answer:

SN1 reaction	SN2 reaction
1) Unimolecular nucleophilic substitution reaction.	1) Bimolecular nucleophilic substitution reaction.
2) Takes place in two steps.	2) Takes place in one step.
3) Rate depends only on substrate concentration	3) Rate depends on substrate and nucleophile
4) It takes place in tertiary alkyl halides	4) It takes place in primary alkyl halides
5) Shows racemic mixture	5) Shows inversion product.
6) Carbocation is formed during mechanism	6) Transition state is observed during mechanism
7) It follows first order kinetics Ex: Hydrolysis of tert – butyl bromide.	7) It follows second order kinetics. Ex: Hydrolysis of methyl bromide.

 

VI. Long Answer Questions

Question 1.
Explain the mechanism of nucleophilic bimolecular substitution (S_N2) reaction with one example.
Answer:
S_N2 reactions (Nucleophilic substitution bimolecular reactions):
The nucleophilic substitution reactions in which rate of reaction depends on substrate and nucleophile are called S_N2 reactions.
S_N2 reactions take place in primary alkyl halides.
Ex: CH₃ – Br + aQuestion KOH → CH₃ – OH + KBr
Methyl Bromide Methanol
Mechanism :

• Mechanism involves in a single step.
• Nucleophile attacks from backside.
• Transition state is observed during mechanism.
• Inversion product is formed.
• The order of reactivity of S_N2 reactions is Methyl halides > 1° – halides > 2° – halides > 3° – halides
  KOH → K⁺ + OH^–
  

Question 2.
Define the following
i) Racemic mixture
ii)Retention of configuration
iii)Enantiomers
Answer:
i) Racemic mixture :
A mixture containing equal amounts of two enantiomers and is optically inactive, d (+) lactic acid + l(-) lactic acid ld, l (±) lactic acid (Racemic mixture)
ii) Retention of configuration:
Preservation of the spatial arrangement around a chiral carbon during a reaction.
Ex:
iii) Enantiomers :Enantiomers are the stereoisomers which are non-superimposable mirror images of each other.
Ex:

Question 3.
Explain the preparation and applications of Grignard reagents with suitable examples.
Answer:
Alkyl Magnesium halides are called Grignard reagents.
The General Formula is RMgX
where, R= Alkyl or aryl group
x= halogen(Cl,B,I)
Eg: Methyl Magnesium bromide(CH₃Mg Br)
Preparation:
It is prepared by reacting an alkyl or aryl halide with magnesium in dry ether.

Eg:

Applications:
i) Preparations of hydrocarbons:
Grignard reagents react with alcohol, water etc., to form hydrocarbons.
Eg:

ii) Preparation of Carboxylic acids:
Grignard reagents react with C0₂ followed by the hydrolysis to form carboxylic acids.

Question 4.
Explain how the following conversions are carried out.
i) propene to propanol
ii) ethanol to but-l-yne
iii) 1-broniopropane to 2-bromopropane
iv) aniline to chlorobenzene
Answer:
i) Propene to propanol:
Propene reacts with HBr in the presence of benzonyl peroxide to form 1-Bromo propane which on reaction
Propene which on reaction with aQuestion KOH to form propan 1-ol.

ii) Ethanol to but -1 – yne
Ethanol reacts with SOCl₂ to form ethyl chloride which on reaction with sodium acetylide to form but -1 yne.

iii) 1 – Bromo propane to 2 – Bromo propane
1 – Bromo propane on heating with alcoholic KOH gives propene which on reaction with HBr to form 2 – Bromo propane.

iv) Aniline to chlorobenzeneAniline reacts with NaNO₂ and 2HCl at o – 5° C to form benzene diazonium chloride which on reaction with Cu/HCl to form chloro benzene.

Haloalkanes and Haloarenes Class 12 NCERT Solutions

Question 1.
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
Answer:
i) 2-Chloro-3-methylbutane; 2° alkyl halide
ii) 3-Chloro-4-methylhexane; 2° alkyl halide
iii) 1-Iodo-2, 2-dimethylbutane; 1° alkyl halide
iv) 1-Bromo-3,3-dimethyl-l-phenylbutane; 2° benzylic halide
v) 2-Bromo-3-methylbutane; 2°alkyl halide
vi) 1-Bromo-2-ethyl-2-methylbutane; 1° alkyl halide
vii) 3-Chloro-3-methylpentane; 3° alkyl halide
viii) 3-Chloro-5-methylhex-2-ene; vinylic halideix)
ix) 4-Bromo-4-methylpent-2-ene; allylic halide
x) 1-Chloro-4-(2-methylpropyl) benzene; aryl halide
xi) 1-Chloromethyl-3-(2, 2-dimethylpropyl) benzene; 1° benzylic halide
xii) 1-Bromo-2-(l-methylpropyl) benzene; aryl halide

Question 2.
Give the IUPAC names of the following compounds:
i) CH₃CH(Cl)CH(Br)CH₃
ii) CHF₂CBrC/F
iii) ClCH₂C ≡ CCH₂Br
iv) (CCl₃)₃CCl
v) CH₃C(p-ClC₆H₄)₂CH(Br)CH₃
vi) (CH₃)₃CCH = CClC₆H₄I-p
Answer:
i) 2-Bromo-3-chlorobutane
ii) 1-Bromo-1-chloro-1, 2, 2-trifluoroethane
iii) 1-Bromo-4-chlorobut-2-yne
iv) 2-(TVichloromethyl)-1, 1, 1,2, 3, 3, 3- heptachloropropane
v) 2-Bromo-3, 3-Bis (4-chlorophenyl) butane
vi) 1-Chloro-1-(4-iodophenyl)-3,3-dimethylbut-1- ene

Question 3.
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) 2-Bromobutane
vi) 4-tert-Butyl-3-iodoheptane
vii) 1-Bromo-4-sec-butyl-2-methylbenzene
viii) l,4-Dibromobut-2-ene
Answer:

Question 4.
Which one of the following has the highest dipole moment?
Answer:
i) CH₂Cl₂
ii) CHCl₃
iii) CCl₄

Individual bond dipoles (C-Cl) cancel out each other due to symmetry. Therefore, CCl₄ has no overall dipole moment.

Net resultant of individual bond dipoles of CH₂Cl₂ is greater than that of CHCl₃.
Dipole moment of CHCl₃ = 3µ cos Φ; Φ ~ 75° ≃ 0.82 p
Dipole moment of CH₂Cl₂ = 2µ cos δ; δ ~ 55.75°≃ 1.12p
Therefore, CH₂Cl₂ has the highest dipole moment among the given compounds.

Question 5.
A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound, C₅H₉Cl in bright sunlight. Identify the hydrocarbon.
Answer:
i) The hydrocarbon with molecular formula C₅H₁₀ can be either a cycloalkane or an alkene.
ii) Since the hydrocarbon does not react with Cl₂ in the dark, it cannot be an alkene but must be a cycloalkane.
iii) As the cycloalkane reacts with Cl₂ in the presence of bright sunlight, to give a single monochloro compound. C₅H₉Cl, therefore all the ten hydrogen atoms of the cycloalkane
must be equivalent. Therefore, the cycloalkane is cyclopentane.

Question 6.
Write the isomers of the compound having formula C₄H₉Br.
Answer:
It has the following four isomers:

Question 7.
Write the equations for the preparation of 1-iodobutane from
(i) 1- butanol
(ii) 1-chlorobutane, and
(iii) but-l-ene.
Answer:

Question 8.
What are ambident nucleophiles? Explain with an example.
Answer:
The nucleophiles which can attack through two different sites are called ambident nucleophiles. For example, cyanide ion is a resonance hybrid of the following two structures:

It can attack through carbon to form cyanides and through N to form isocyanides or carbylamines.

Question 9.
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
Answer:
i) Since I^– ion is a better leaving group than Br^– ion, hence, CH₃I reacts faster than CH₃Br in S_N2 reaction with OH^– ion.
ii) On steric grounds, 1° alkyl halides are more reactive than tert-alkyl halides in S_N2 reactions. Hence, CH₃Cl will react at a faster rate than (CH₂)₂CCl in a S_N2 reaction with OH ion.

Question 10.
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-l-methylcyclohexane
ii) 2-Chloro-2-methylbutane
iii) 2, 2, 3-Trimethyl-3-bromopentane
Answer:
i) In 1-bromo-l-methylcyclohexane, the β-hydrogens on either side of the Br atom are equivalent, therefore, only 1-alkene is formed.

ii) All (β-hydrogens in 2-Chloro-2-methylbutane are not equivalent, hence on treatment with C₂H₅ONa/C₂H₅OH it gives two alkenes.

iii) 2, 2, 3-Trimethyl-3-bromopentane has two different sets of (β-hydrogen and therefore, in principle, can give two alkenes (I and II). But according to Saytzeff rule, more highly substituted alkene (II), being more stable is the major product.

Question 11.
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) Propeiie to propyne
vi) Ethanol to ethyl fluoride
vii) Bromomethane to propanone
viii) But-l-ene to but-2-ene
ix) 1 -Chlorobutane to n-octane
x) Benzene to biphenyl
Answer:

Question 12.
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?
Answer:
i) Due to greater s-character, sp² hybrid carbon atom is more electronegative than sp³ hybrid carbon atom. Therefore, the sp² hybrid carbon of C-Cl bond in chlorobenzene has less tendency to release electrons to Cl than sp³ hybrid carbon of cyclohexyl chloride. As a result, the magnitude of negative charge is less on Cl atom of chlorobenzene than in cyclohexyl chloride.

Further, due to resonance the C-Cl bond in chlorobenzene acquires partial double bond character whereas the bond in cyclohexyl chloride is a pure single bond. In other words, C-Cl bond in chlorobenzene is shorter than in cyclohexyl chloride.

C-Cl bond acqures partial double bond character.
As dipole moment is a product of charge and distance. Therefore, dipole moment of chlorobenzene is less than that of cyclohexyl chloride.

ii) Alkyl halides, though polar, are immiscible in water because they are unable to form hydrogen bonds with water molecules.

iii) Grignard reagents are very reactive. They react with moisture present in the apparatus or the starting materials to give hydrocarbons.

Grignard reagent
Hence, Grignard reagent must be prepared under anhydrous conditions.

Question 13.
Give the uses of Freon-12, DDT, carbon tetrachloride and iodoform.
Answer:
i) Freon – 12 (CCl₂F₂)

• Used as a refrigerant in refrigerators and air conditioners.
• Used in aerosol sprays (Propellant).
• Used in foam blowing points.

ii) DDT (Dichloro Diphenyl Trichloroethane)

• Used as an insecticide.
• Controls mosquitoes (malaria) and agricultural pests.

iii) Carbon Tetrachloride (CCl₄)

• Used as a slovent for fats, oils and grease.
• Used in dry cleaning.
• Used in fire extinguishers.

iv) Iodoform (CHI₃)

• Used as an antiseptic for wounds.
• Used in medicines and disinfectants.

Question 14.
Write the structure of the major organic product in each of the following reactions:

v) C₆H₅ONa + C₂H₅Cl →
vi) CH₃CH₂CH₂OH + SOCl₂ →

viii) CH₃CH = C(CH₃)₂ + HBr →
Answer:

Question 15.
Write the mechanism of the following reaction:

Answer:
The above reaction is an S_N2 reaction.

CN^– ion is an ambident nucleophile so it can attack the carbon atom of C-Br bond in n-BuBr through C or N. As C-C bond is stronger than C-N bond, therefore attack occurs through C to form n-BuCN.

Question 16.
Arrange the compounds of each set in order of reactivity towards S_N2 displacement:
i) 2-Bromo-2-methylbutane,
1 -Bromopentane, 2-Bromopentane.
ii) 1-Bromo-3-methylbutane, 2-Bromo-
2-methylbutane, 2-Bromo-
3-methylbutane.
iii) 1 -Bromobutane, l-Bromo-2,
2-dimethylpropane, 1-Bromo-
2-methylbutane, 1 -Bromo-
3-methylbutane.
Answer:
The reactivity in S_N2 reactions depends upon steric hindrance; more the steric hindrance, slower the reaction.

1. Due to steric reasons, the order of reactivity in S_N2 reactions follows the order: 1° > 2° >3°, therefore, order of reactivity of the given alkyl bromides is as follows:
   1-Bromopentane > 2-Bromopentane > 2-Bromo-2-methylpentane.
2. Due to steric reasons, the order of reactivity of alkyl halides in S_N2 reactions follows the order: 1° > 2° >3°, therefore, the order of reactivity of the given alkyl bromides is as follows: 1-Bromo-3-methylbutane > 2-Bromo-3-methylbutane > 2-Bromo-2-methylbutane.
3. As in case of 1° alkyl halides, steric hindrance increases in the order: n-alkyl halides, alkyl halide with a substituent at any position other than the β-position, one substituent at the β-position, two substituents at the β-position. Therefore, the reactivity decreases in the same order. Hence, the reactivity of the given alkyl bromides decreases in the order:
   1-Bromobutane > 1-Bromo-3-methylbutane > 1-Bromo-2-methylbutane > 1-Bromo-2, 2-dimethylpropane.

Question 17.
Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅ which is more easily hydrolysed by aqueous KOH?
Answer:
C₆H₅CH₂Cl is a l°aralkyl halide while C₆H₅CHClC₆H₅ is a 2° aralkyl halide.

In SN¹ reactions, the reactivity depends upon the stability of carbocations. Since the carbocation
(where the +ve charge is delocalised over two C₆H₅ rings) derived from C₆H₅ – CHClC₆H₅ is more stable than the carbocation, (where + ve charge is delocalised over one C₆H₅ ring) derived from C₆H₅CH₂Cl, therefore, C₆H₅CHClC₆H₅ gets hydrolysed more easily than C₆H₅CH₂Cl under SN¹ conditions.
Although, under SN² conditions, the reactivity depends on steric hindrance, therefore, under SN² conditions, C₆H₅CH₂Cl gets hydrolysed more easily than C₆H₅CHClC₆H₅.
Probability of reaction to occur by SN¹ is more than that by SN². Therefore C₆H₅CHClC₆H₅. is more easily hydrolysed by aqueous KOH as compared to C₆H₅CH₂Cl.

Question 18.
p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.
Answer:
The p-isomer being more symmetrical, fits closely in the crystal lattice and thus has stronger intermolecular forces of attraction than o- and m-isomers. As during melting, the crystal lattice breaks, therefore, a larger amount of energy is needed to melt the p-isomer than the corresponding o-and m-isomers. In other words, the melting point of the p-isomer is higher than the corresponding o- and m-isomers.

Question 19.
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-phenyl ethanoic acid
vii) Ethanol to propane nitrile
viii) Aniline to chlorobenzene
ix) 2-Chlorobutane to 3, 4-dimethyl hexane
x) 2-Methyl-1-propene to 2-chloro-2-methylpropane
xi) Ethyl chloride to propanoic acid
xii) But-1 -ene to n-butyl iodide
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 phenyl isocyanide
Answer:

Question 20.
The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in presence of alcoholic KOH, alkenes are major products. Explain.
Answer:
In aqueous solution, KOH is almost completely ionised to give OH^– ions which being a strong nucleophile brings about a substitution reaction on alkyl halides to form alcohols. In the aqueous solution, OH^–ions are highly hydrated. This reduces the basic character of OH^– ions which fail to abstract a hydrogen from the B-carbon of the alkyl chloride to form an alkene.

On the other hand, an alcoholic solution of KOH contains alkoxide (OR^–) ions which being a much stronger base than OH- ions preferentially eliminates a molecule of HCl from an alkyl chloride to form alkenes.

Question 21.
Primary alkyl halide C₄H₉Br (a) reacted with alcoholic KOH to given 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.
Answer:

Equations of the Reactions:

Question 22.
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
Answer:

Intext Questions

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

Question 2.
Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer:
H₂SO₄ cannot be used along with KI in the conversion of an alcohol to an alkyl iodide as it converts KI to corresponding HI and then oxidises it to I₂.

Question 3.
Write structures of different dihalogen derivatives of propane.
Answer:
i) ClCH₂CH₂CH₂Cl
ii) ClCH₂CHClCH₃
iii) Cl₂CHCH₂CH₃
iv) CH₃CCl₂CH₃

Question 4.
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
Answer:
i) All the hydrogen atoms are equivalent and replacement of any hydrogen will give the same product.

ii) The equivalent hydrogens are grouped as a, b and c. The replacement of equivalent hydrogens will give the same product.
C^aH₃C^bH₂C^cH₂C^bH₂C^aH₃ pentane

iii) Similarly the equivalent hydrogens are grouped as a, b, c and d. Thus, four isomeric products are possible.

Question 5.
Draw the structures of major monohalo products in each of the following reactions:

Answer:

Only alcoholic but not phenolic OH groups are replaced by Cl on heating with HCl.

Iodide ion is a strong nucleophile and hence it displaces bromide ion.

Question 6.
Arrange each set of compounds in the order of increasing boiling points.
i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.
ii) 1 -Chloropropane, Isopropyl chloride, 1-Chlorobutane.
Answer:
i) Chloromethane < Bromomethane < Dibromomethane < Bromoform.
Boiling point increases with increase in molecular mass,

ii) Isopropyl chloride < 1-Chloropropane < 1-Chlorobutane.
Isopropyl chloride being branched has lower b.p. than 1-Chloropropane.

Question 7.
Which alkyl halide from the following pairs would you expect to react more rapidly by an S_N2 mechanism? Explain your answer.

Answer:
i) CH₃CH₂CH₂CH₂Br
Being primary halide, there won’t be any steric hindrance.

ii)

Being a secondary halide, reacts faster than tertiary halide.

iii)

The presence of methyl group closer to the halide group will increase the steric hindrance and decrease the rate.

Question 8.
In the following pairs of halogen compounds, which compound undergoes faster S_N1 reaction?

Answer:
i)

2- Chloro-2-methylpropane tertiary halide reacts faster than secondary halide because of the greater stability of tert – carbocation.

ii)

2-Chloroheptane as the secondary carbocation is more stable than primary carbocation.

Question 9.
Identify A, B, C, D, E, R and R’ in the following:

Answer: