NCERT Solutions for Class 12 Science Chemistry Chapter 10 - Haloalkanes And Haloarenes

Answer:

The reactivity of alcohol with halogen acids proceeds through SN₁ mechanism via formation of intermediate carbocation. The order of stability of carbocation is of the order 3° > 2° > 1°. Therefore, the order of reactivity of alcohol with halogen acid is (C) > (B) > (A).

Hence, the correct answer is option (ii).

Answer:

The reaction of alcohol with concentrated HCl at room temperature proceeds through the SN₁ mechanism and thus, depends upon the stability of carbonation formed. Therefore, tertiary alcohol will be very reactive towards concentrated HCl to give alkyl chloride.


Hence, the correct answer is option (iv).

Answer:

Hence, the correct answer is option (i).

Answer:

Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. This reaction is an example of an electrophilic substitution reaction, where Cl⁺ acts as an electrophile.

Hence, the correct answer is option (ii).
 

Answer:

The chemical reaction in which one halide replaces another halogen from its compound is known as halogen exchange reaction.
For example,
$\mathrm{RX} +\mathrm{NaI} \underset{}{\to } \mathrm{RI} + \mathrm{NaX}$
The above reaction is know as Finkelstein reaction.

Hence, the correct answer is option (i).

Answer:

Free radical substitution reaction occurs in the presence of ultraviolet light or at high temperature
${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_3\hspace{0.17em}\hspace{0.17em}\stackrel{{\mathrm{Cl}}_2/\mathrm{Uv} \mathrm{light}}{\to }\hspace{0.17em}\hspace{0.17em}{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Cl} + {\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CHClCH}}_3$

Hence, the correct answer is option (i).

Answer:

The density of a chemical compound increase on increasing the molecular mass. Therefore the correct order of increasing densities for the given compounds is


Hence, the correct answer is option (i).

Answer:

The boiling point of compound decreases with increase in branching as on increasing branching the surface area decrease. Therefore, the correct increasing order of boiling point for the given compound is as follows:


Hence, the correct answer is option (iii).

Answer:

Asymmetric/ chiral carbon atom is one in which all the four different atoms/ groups satisfy four valencies of the carbon atom.
Thus, molecules (a), (b) and (c), all are asymmetric carbon, in the molecule (d) contains symmetrical carbon.

Hence, the correct answer is option (ii).

Answer:

The stereoisomers related to each other as non-superimposable mirror images are called enantiomers. Enantiomers possess identical physical properties. They only differ with respect to the rotation of plane polarised light. If one of the enantiomers is dextrorotatory, the other will be laevorotatory. The compound in option (i) is non-superimposable image of compound (A).

Hence, the correct answer is option (i).

Answer:

In vic-dihalides are those halides in which two halogen atoms are present on the adjacent carbon atoms. Therefore, 1, 2 - Dichloroethane is an example of vic-dihalides.


Hence, the correct answer is option (ii).

Answer:

In the given compound, halogen atom is bonded to sp³ hybridised carbon atom next to carbon - carbon double bond. Such groups are known as allylic groups and the carbon to which halogen is attached is known as the allylic carbon.



Hence, the correct answer is option (i).

Answer:

During halogenation of benzene, AlCl₃ activities the chlorine molecule which generate the Cl⁺ electrophile.


Hence, the correct answer is option (ii).
 

Answer:

In Ethylidene chloride both the chlorine atoms are present on the same carbon atom, thus, it is a gem-dihalide.

Hence, the correct answer is option (ii).

Answer:

According to Markovnikov's rule, the negative part of the reagent will get attached to the carbon atom containing lesser number of hydrogen. Thus, the reaction will be as follows:



Hence, the correct answer is option (iii).
 

Answer:

Primary alkyl halides undergo SN​₂ reactions because 1^∘ substrates have little steric hindrance to nucleophilic attack and 1^∘ carbocations are relatively unstable.

Answer:

For all the tertiary alkyl halide, carbon and iodine (C – I) bond will be the weakest due to large size of iodine. Thus it is a better learning group. So, (CH₃)₃ C – I will give S_N1 reaction readily.

Hence, the correct answer is option (iv).

Answer:

The correct IUPAC name of the given compound is 1-Bromo-2-methylbutane.



Hence, the correct answer is option (iii).

Answer:

The structure of diethylbromomethane.



IUPAC Name: 3-Bromopentane.

Hence, the correct answer is option (ii).

Answer:

Toluene on chlorination in the presence of lewis catalyst like Fe in the dark gives a mixture of ortho and para substituted product.



Hence, the correct answer is option (iv).

Answer:

The reaction of chloromethane with excess of ammonia yields Methanamine (CH₃NH₂) as a product.


Hence, the correct answer is option (iii).
 

Answer:

The structure of given compounds are as follows:


Since the carbon atom number 2 in 2-Bromobutane has four bonds with four different groups, it is chiral in nature.

Hence, the correct answer is option (i).

Answer:

During the reaction of C₆H₅CH₂Br with aqueous sodium hydroxide, the formation of a stable intermediate C₆H₅CH₂⁺ takes place. Therefore, it follows S_N1 mechanism.
 

Hence, the correct answer is option (i).

Answer:

In the given compound, carbons b and c are bonded to four different groups respectively, Therefore, these represent asymmetrical carbon atoms.

Hence, the correct answer is option (ii).

Answer:

Since the carbon atom-2 in  $$\begin{gathered} \stackrel{1}{\mathrm{C}}{\mathrm{H}}_3-\underset{|}{\stackrel{2}{\mathrm{C}}\mathrm{H}}-\mathrm{Br} \\ \underset{3}{\mathrm{C}}{\mathrm{H}}_2-\underset{4}{\mathrm{C}}{\mathrm{H}}_3 \end{gathered}$$ is chiral, it will give a racemic mixture on nucleophilic substitution by OH^– ion.

Hence, the correct answer is option (i).

Answer:

The -NO₂ group is an electron withdrawing group. The presence of an electron withdrawing group promotes nucleophilic substitution reaction. But the presence of an electron withdrawing group at meta position has very less effect on rate of nucleophilic substitution reaction. Therefore, the correct order is (a) < (c) < (b).

Hence, the correct answer is option (iii).

Answer:

An electron releasing group at ortho or para position decreases the rate of nucleophilic substitution. Methyl group (-CH₃) is an electron releasing group. Therefore, the correct order is (b) < (c) < (a).

Hence, the correct answer is option (iv).

Answer:

In case of aryl halides, the presence of electron withdrawing group increases the rate of nucleophilic substitution reaction. Also, more the number of electron withdrawing groups, more will be the reactivity. Thus, the correct order will be (a) < (b) < (c).

Hence, the correct answer is option (iv).

Answer:

The presence of an electron releasing group decreases the reactivity aryl halides towards nucleophilic substitution. It also depends upon the number of electron releasing groups. Lesser the number of electron releasing groups, more is the reactivity.
Thus, the correct order is (c) < (b) < (a).

Hence, the correct answer is option (iii).

Answer:

Boiling point increases with the increase of molecular mass of the compound. In the given substituted alkanes, the boiling point increases with the increase in molecular mass of the attached halogen atoms. Also, butane has no halogen atom. So it will have lower boiling point than other compounds followed by 1-chlorobutane, 1-Bromobutane and 1-Iodobutane respectively.

Hence, the correct answer is option (i).

Answer:

The boiling point of a compound is directly proportional to its molecular mass. Therefore, the correct increasing order of boiling point is as follows:
1–Bromoethane < 1–Bromopropane < 1–Bromo-butane < Bromobenzene

Hence, the correct answer is option (iv).

Answer:

Both OH^– and Cl^– are electron rich species as they are holding negative charge. Thus, they are nucleophiles. The above reaction follows (S_N2) mechanism, where the carbon of the alkyl halide is sp³ hybridised.
In the mechanism, breaking of C–X bond and formation of new bond (C–Nu) occur simultaneously through a transition state where the carbon is in sp² hybridisation state.

Hence, the correct answers are option (i) and (iii).

Answer:

In the given reaction, alkyl halide is primary in nature. Here, a transition state is observed in which one bond is broken and one bond is formed simultaneously in the same step. Thus, it follows S_N2 mechanism.
In the mechanism, the nucleophile attacks the carbon at 180° to the leaving group as a result of which reactants and products have opposite configurations.

Hence, the correct answers are options (i) and (ii).

Answer:

In the above reaction, intermediate (c) shows the transition state which is highly unstable. In this transition state, the carbon atom is sp² hybridised as it is partially bonded to two nucleophiles so it is highly unstable and is comparatively less stable than the reactant (b). In reactant (b), carbon atom is sp³ hybridised and is more stable than intermediate (c).

Hence, the correct answers are options (i) and (iv).

Answer:

The reactant involved in the above reaction is secondary alkyl halide. It follows S_N1 mechanism instead of S_N2 mechanism as the secondary alkyl halide contain bulky group. In S_N1 mechanism, a stable carbocation intermediate is formed which is further attacked by OH^– nucleophile.

Hence, the correct answers are options (i) and (iv).

Answer:

The above reaction follows S_N1 mechanism. In this, formation of carbocation is a slow step. So, the rate of reaction depends upon the concentration of only (b) and therefore, molecularity of reaction is one.

Hence, the correct answers are options (i) and (iii).

Answer:

In 2-bromopentane $\left({\text{CH}}_{\text{3}}\text{–}\underset{\text{Br}}{\underset{\left|\right.}{\text{CH}}}{\text{–CH}}_{\text{2}}{\text{–CH}}_{\text{2}}{\text{–CH}}_{\text{3}}\right)$ and trichloromethane $\left(\text{Cl}–\stackrel{\text{Cl}}{\stackrel{\left|\right.}{\underset{\underset{\text{Cl}}{\left|\right.}}{\text{C}}}}–\text{H}\right),$ halogen atoms are attached to the sp³ hybridised carbon atom of an alkyl group.

Hence, the correct answers are options (i) and (iv).

Answer:

(i) Ethylene chloride and ethylidene chloride on treatment with alcoholic KOH undergoes elimination reaction and forms ethyne.
$\begin{array}{l}{\text{Cl–CH}}_{\text{2}}{\text{–CH}}_{\text{2}}\text{–Cl}\underset{\left(\text{excess}\right)}{\overset{\text{Alc.KOH}}{\to }}\underset{\text{Ethyne}}{\text{CH}\equiv \text{CH}}\\ {\text{CH}}_3–{\text{CHCl}}_2\underset{\left(\text{excess}\right)}{\overset{\text{Alc.KOH}}{\to }}\text{CH}\equiv \text{CH}\end{array}$
(ii) Ethylene chloride and ethylidene chloride form different products on treatment with aq. NaOH.
$\text{Cl}–{\text{CH}}_2–{\text{CH}}_2–\text{Cl}\stackrel{\text{Aq.NaOH}}{\to }\underset{\text{Ethylene\hspace{0.17em}glycol}}{\text{HO}–{\text{CH}}_2–{\text{CH}}_2–\text{OH}}$
${\text{CH}}_3–{\text{CHCl}}_2\stackrel{\text{Aq.NaOH}}{\to }{\text{CH}}_3–\text{CH}{\left(\text{OH}\right)}_2\stackrel{–{\text{H}}_{\text{2}}\text{O}}{\to }\underset{\text{Ethanal}}{{\text{CH}}_{\text{3}}\text{CHO}}$
(iii) Ethylene chloride and ethylidene chloride form same products on reduction.
$\begin{array}{l}{\text{Cl–CH}}_{\text{2}}{\text{–CH}}_{\text{2}}\text{–Cl}\stackrel{\text{reduction}}{\to }\underset{\text{Ethane}}{{\text{H}}_{\text{3}}{\text{C–CH}}_{\text{3}}\text{+2HCl}}\\ {\text{CH}}_{\text{3}}{\text{–CHCl}}_{\text{2}}\stackrel{\text{reduction}}{\to }\underset{\text{Ethane}}{{\text{H}}_{\text{3}}{\text{C–CH}}_{\text{3}}}\text{+2HCl}\end{array}$
(iv) Ethylene chloride and ethylidene chloride are optically inactive.

Hence, the correct answers are options (i) and (iii).

 

Answer:

gem –dihalides are compounds that contains two halogen atoms which are bonded to the same carbon atom.
      
            (I)                                    (II)

                
            (III)                              (IV)
In figure (i) and (iii), two halogen atoms are present on the same carbon atom and therefore, are gem-dihalides.

Hence, the correct answers are options (i) and (iii).
 

Answer:

(i) ${\text{H}}_3\text{C}–\underset{{\text{CH}}_3}{\underset{\left|\right.}{\text{CH}}}–\text{Br}$

(ii) ${\text{H}}_3\text{C}–\underset{{\text{CH}}_3}{\underset{\left|\right.}{\stackrel{{\text{CH}}_3}{\stackrel{\left|\right.}{\text{C}}}}}–{\text{CH}}_2\text{Br}$

(iii) ${\text{H}}_{\text{3}}\text{C–}\underset{\underset{\text{Br}}{\left|\right.}}{\text{CH}}{\text{–CH}}_{\text{2}}{\text{–CH}}_{\text{3}}$

(iv) ${\text{H}}_3\text{C}–\underset{{\text{CH}}_3}{\underset{\left|\right.}{\stackrel{\text{Br}}{\stackrel{\left|\right.}{\text{C}}}}}–{\text{CH}}_2{\text{CH}}_{\text{3}}$  
In compound (i) and (iii), the α-carbon is bonded to two alkyl groups and therefore is secondary bromide.

Hence, the correct answers are options (i) and (iii).

Answer:

                                
                (i)                                          (ii)                              (iii)                                           (iv)
In the above structures, compound (i) and compound (iv), halogen atom is directly bonded to aromatic ring. Therefore, these compounds are aryl halides.

Hence, the correct answers are options (i) and (iv).

Answer:

Alcohol on treatment with HCl and ZnCl₂ forms alkyl halide.
$\text{ROH}+\text{HCl}\stackrel{{\text{ZnCl}}_2}{\to }\underset{\left(\text{Alkyl\hspace{0.17em}halide}\right)}{\text{RCl}+{\text{H}}_2\text{O}}$
Alcohol on treatment with red phosphorus and X₂ forms alkyl halide.
$\text{R–OH}\stackrel{{\text{Red\hspace{0.17em}P/Br}}_{\text{2}}}{\to }\underset{\left(\text{Alkyl\hspace{0.17em}halide}\right)}{\text{R–X}}$
Alcohol when treated with H₂SO₄ and KI, then H₂SO₄ oxidises KI to I₂. This does not produces HI. Therefore, alkyl iodide is not produced on treatment of alcohols with H₂SO₄ and KI.

Hence, the correct answers are options (i) and (ii).
 

Answer:

Alkyl fluorides are synthesised by alkyl chloride/bromide in the presence of CoF₂ and Hg₂F₂. Alkyl fluorides are formed when transition metal fluorides react with alkyl chloride/bromide. Alkyl metal fluoride like NaF and alkaline earth metal fluoride like CaF₂ do not react to form fluorides.

Hence, the correct answers are options (ii) and (iii).

Answer:

Iodination reactions are reversible in nature.

C₆H₆ + I₂ ​⇌ C₆H₅ I + HI
In the above reaction, hydrogen iodide (HI) is formed along with iodobenzene (C₆H₅I). The HI formed has to be removed from the reaction mixture in order to prevent the backward reaction. So, in order to do this, HI formed during the reaction is removed by using oxidising agent like HIO₃ or HNO₃. The reaction for the same is as follows:
5HI + HIO₃ → 3I₂ + 3H₂O
2HI + 2HNO₃ → I₂ + 2NO₂ + 2H₂O
​

Answer:

P-dibromobenzene has higher melting point than its o-isomer due to symmetry. Due to the presence of symmetry, p-isomerfits in the crystal lattice effectively than the o-isomer. Thus, p-dibromobenzene has higher melting point.

Answer:

S_N1 mechanism depends upon the stability of Carbocation that is formed as an intermediate during the mechanism.
C₆H₅-CH₂-Cl forms ${\mathrm{C}}_6{\mathrm{H}}_5\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2$ as an intermediate.
The carbocation is resonance stabalised and will react faster in S_N1 reaction.

CH₃-CH₂-Cl forms ${\mathrm{CH}}_3-\stackrel{+}{\mathrm{C}}{\mathrm{H}}_2$ carbocation. It is highly unstable and does not give S_N1 reaction with OH^– ion.

Answer:

Iodoform, on coming in contact with the skin, liberates I₂. The antiseptic property of iodine is due to the liberation of I₂.
$\underset{\mathrm{Iodoform}}{{\mathrm{CHI}}_3}\stackrel{\mathrm{Contact} \mathrm{with} \mathrm{skin}}{\to }\underset{\mathrm{Iodine}}{{\mathrm{I}}_2}$

Answer:

C–X bond in haloarenes and haloalkenes have some double bond character due to the presence of resonance. As a result of which C–X bond strengthens. Therefore, haloarenes and haloalkenes are less reactive than haloalkanes. 
The resonating structure of haloarenes and haloalkenes is as follows:

The stability of the compound increases as the number of resonating structure increases. Also, the reactivity of the compound decreases with the increase in the stability.
Since, more resonating structures are observed in haloarenes than haloalkenes, thus, haloarenes are less reactive than haloalkenes and haloalkenes.

Answer:

Lewis-acids are electron H deficient species. They are in charge of causing heterolytic fission in halogen molecules.
The role of Lewis acid is to generate an electrophile.
Aryl bromides and chlorides can be made from arenes by electrophilic substitution. This reaction is carried out by treating the arene with chlorine or bromine in the presence of iron (III) chloride in the absence of light.
Iron (III) chloride, i.e., FeCl₃ is a Lewis acid, that produces the electrophile needed to proceed the reaction.
The complex Cl + (FeCl₄] is formed when FeCl₃ and Cl₂ form a coordination compound.

Mechanism: Cl – Cl + FeCl₃ → FeCl₄ + Cl⁺

Answer:

When NaBr and H₂SO₄ are combined, Br² gas is produced. Molecule (b) will not react with Br² gas due to the stable molecule formed as a result of resonance stabilisation.
Due to resonance, partial double bond character arises in C – O bond of phenol and it becomes more stable than alcohal (CH₃CH₂CH₂OH).
Reaction - 2NaBr + 3H₂SO₄ → 2NaHSO₄ + SO₂ + Br₂ + 2H₂O

${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{OH} \stackrel{{\mathrm{Br}}_2}{\to } \mathrm{No} \mathrm{reaction}$

Answer:

The major product in the given reaction is (B) i.e., CH₃CHICH₃.
The mechanism of the reactions is ass follows →

Answer:

Holoalkanes are very slightly soluble in water because it takes energy to overcome the attractions between the holoalkane molecules as well as to break the hydrogen bonds between water molecules in order to dissolve a haloalkane in water.
When new attractions are formed between the haloalkane and the water molecules, however, less energy is released because these are not as strong as the original hydrogen bonds in water.

Answer:

The other resonance structures related to the given structure:-

The functional groups in these molecules are ortho-para. From the above resonating structures it is very clear that electron density is more at ortho and para positions.

Answer:

Given compounds are:-
(i) 1-Bromobut-2-ene
(ii) 4-Bromopent-2-ene
(iii) 2-Bromo-2-methylpropane
Classifying the above compounds as primary, secondary and tertiary halides:

• Primary Halides - Compounds in which halide ion is attached to the primary carbon are known as primary halides.
Here, 1-Bromobut-2-ene is a primary halide.

• Secondary Halides - Compounds in which the halide ion is attached to the secondary carbon
Here, 4-Bromopent-2-ene is a secondary halide.

• Tertiary Halides - Compounds that contains a halide ion attached to a tertiary carbon.
Here, 2-Bromo-2-methylpropane is a tertiary halide.

The structural formula of given compounds are:
(i)

(ii)

(iii)

Answer:

(i) The structural formula of both compounds 'A' and 'B'.

Answer:

O-chlorotoluene or P-chlorotoluene will be formed as the compound A with moleculare formulaC₇H₈ ios treated with Cl₂ in the presence of FeCl₃.

Answer:

In the given reaction, CH₃—CH₂—CH=CH—CH₃+ HCl → A + B, the following two products are possible -

Answer:

From the given compounds, compound II is the most symmetrical because it contains both CH₃ groups and Cl atoms at para positions to each other. As a result, it fits better in the crystal lattice than the other two isomers and has a higher melting point.

Answer:

The structure of neo-pentylbromide is as follows:-

The IUPAC name for neo-pentylbromide is → 1-bromo-2, 2-dimethylpropane.

Answer:

The hydrocarbon that has molecular mass of 72 g mol^–1 is the alkane C₅H₁₂ i.e., pentane.
Upon photo chlorination, it produces single monochloro derivative, so, all the hydrogen atoms must be equivalent. The structures of the compound and the derivatives are as follows-

Answer:

Here, two different alkenes such as methylene cyclohexane and 1-methylcyclohex-1-ene can be found which will yield 1-chloro-1-methylcyclohexane by its reaction with HCl.

The reactions with two possible alkenes are:

The reaction takes place according to Markownikoff’s rule i.e., negative part of the adding molecule will get attached to that carbon which has lesser number of hydrogen atom.

Answer:

From the given haloalkanes, (iii) 2-bromo-2-methylpropane will react with aqueous KOH most easily.
2-bromo-2-methylpropane is a tertiary alkyl halide and it will from a stable 3^o carbocation on ionisation. Then, OH⁻ will attack to form 2-methylpropan-2-ol.

Answer:

Due to resonance, a partial double bond character is observed between the C–O bond of phenol. This partial double bond character strengthens the bond. Hence, it becomes difficult to break the C–O bond of phenol. As a result aryl halides are not prepared by the reaction of phenol with HCl in the presence of ZnCl₂.

Answer:

Compound (B) will undergo S_N1 reaction faster than compound (A) because in case of compound (A) carbocation formed will be resonance stabilized. Where as in case of compound (A) there will be no resonance.

 

Answer:

Allyl chloride show high reactivity as the carbocation formed during the process is resonance stabilized whereas, carbocation formed during the hydrolysis of n-propyl chloride is not resonance stabilized.

Answer:

Grignard reagent is reading reacts with water to give corresponding hydrocarbons.
$\mathrm{RMgx}+{\mathrm{H}}_2\mathrm{O}\stackrel{}{\to }\mathrm{RH} +\mathrm{Mg}\left(\mathrm{OH}\right)\mathrm{x}$

Answer:

The energy required to break C–X bond during S_N1 mechanism is obtained through the energy released during the solvation of halide on with proton of polar solvent. Hence, S_N1 mechanism is favoured in polar solvent.

Answer:

The presence of double bond in a molecule can be detected by baeyer's reagent test. The molecules with a double or triple bond decolourise bromine water 

Answer:

Diphenyls like p, p-Dichlorodiphenyl trichloroethane (DDT) are highly chemically stable and fat soluble. As a result, its residue accumulates in the environment and its long-term effects is highly dangerous, due to these reasons, they are potential threat to the environment.
Diphenyl could be prepared by the Fittig reaction shown below.

Answer:

The the structure of DDT is as follows

IUPAC Name: 1, 1, 1-Trichloro-2, 2-bis(p-chlorophenyl)
The structure of henzenehexachloride

IUPAC Name: 1, 2, 3, 4, 5, 6-Hexachlorocyclohexane 

Answer:

Alkyl halides can undergo both nucleophilic as well as elimination reaction depending upon the exact choice of reagents and reaction condition. For elimination, usually a strong and bulky base at high temperature. On the other hand, a weaker and smaller bases and lower temperature gives substituted products. For example, ethyl bromide on heating with alcoholic KOH at about 473-523k undergoes elimination to give ethene, but in the presence of aqueous KOH, it generated ethanol as.
 

Answer:

The reaction to obtain monobromobenzene from aniline is an follows:

Answer:

The attachment of nucleophile on the carbon containing –Cl group makes the benzene ring, resonance stabilized. Also, –NO₂ group is electron with drawing in nature, thus, makes it easier for the nucleophile to attack the benzene ring. Hence, greater the number of –NO₂ groups attached to the ring, more will the ease with which the nucleophile will be attached.
Therefore, the order of reactivity is as follows:

Answer:

In case, the reaction of tert-Butyl bromide with NaOH, a tertiary carbocation is formed as an intermediate which is highly stable. Thus, S_Nl mechanism is preferred in this reaction.

But in the case of n-butylbromide, a primary carbocation is formed as an intermediate which is least stable. Therefore, it does not follow S_Nl mechanism. Also, due the less steric hinderance it proceeds through S_N2 mechanism.

​

Answer:

The reaction between isobutylene and HCl follows markonikov's rule. The reaction is as follows:

Mechanism:

Answer:

In the structure of haloarenes, a halogen is bonded to the sp² hybridised carbon of the benzene ring. So, C–X bond is polar. Also, the lone pair of electrons of halogens are involved in resonance with the benzene ring. Therefore, the C–X bond acquires a partial double bond character. The dipole moment of C–X bond in haloarenes is less than dipole moment of C–X bond in haloalkanes. As a result C–X bond in halorenes is less polar than the C–X bond in haloalkenes.

Answer:

The reaction to obtain iodoethane from ethanol is as follows:

 

Answer:

Carbon end of cyanide ion acts as a stronger nucleophile as it leads to C–C bond which is more stable than C–N bond.

Answer:

• Chloramphenicol is an antibiotic used in the treatment of typhoid fever.
• Thyroxine is a hormone which is secreted by thyroid gland. The dysregulation of thyroxine causes goiter.
• Chloroquine is used in the treatment of malaria.
• Chloroform is used as an anesthetic in some cases.

Answer:

• Racemic mixtures are those mixtures when any mixture has two enantiormers in equal proportions, then there will be zero optical rotation obsessed, When alkyl halide follows S_N1 mechanism then censiation is seen.
• In fire extinguishers Chlorobromocarbons are used.
• Bromination of akenes gives vicinal elihalides.
• The saytzeff rule is followed in the elimination of HX from alkylhalide.

Answer:

	Column I	Column II
(i)		(b) Alkyl halide
(ii)		(d) Allyl halide
(iii)		(a) Aryl halide
(iv)		(c) Vinyl halide

• In Alkyl halide, halogen atom is bonded to sp³ hybridised carbon atom, which perhaps bond with one, two or three alkyl group, i.e., CH₃ – CH (X) – CH₃​
• Allyl halides are those compounds in which halogen atom is bonded to sp³ hybridised carbon atom next to carbon- carbon double bond. That is why CH_3\ – = CH – CH₂​ – X.
• Aryl halides are those compounds where halogen atom is bonded to sp² hybridised carbon atom of an aromatic ring. That is why C₆H₅X.
• Vinyl halides are those compounded in which halogen atom is bonded to an sp² hybridised carbon atom of a carbon – Carbon double bond. That is why CH₂ = CH – X.

Answer:

​

	Column I	Column II
(i)		(b) Electrophilic aromatic substitution
(ii)		(d) Electrophilic addition
(iii)		(e) Nucleophilic substitution (SN1)
(iv)		(a) Nucleophilic aromatic substitution
(v)		(c) Saytzeff elimination

• In this reaction, substitution takes place as an electrophile Cl⁺ attacks on to the benzene ring.
• In this reaction, the addition of HBr takes place on to the doubly bonded carbon of propone in accordance with markonikov rule. Electrophilic addition also takes place.
• In this reaction, as the reactant is a secondary halide so it follnos S_N1 mechanism.
• In this reaction, the halogen atom is directly bonded to the aromatic ring. Hence, it is nucleophillic aromatic substitution as  -OH group has substituted halogen of given compounds.
• This is an elimination reaction, so it follows the saytzeff elimination rule.

Answer:

	Column I	Column II
(i)		(a) 4-Bromopent-2-ene
(ii)		(b) 4-Bromo-3-methylpent-2-ene
(iii)		(c) 1-Bromo-2-methylbut-2-ene
(iv)		(d) 1-Bromo-2-methylpent-2-ene

• The IUPAC name of (i) is 4-bromopent-2-ene.
• The UPAC name of (ii) is 4-bromo-3-methyl pent-2-ene.
• The IUPAC name of (iii) is 1-bromo-2-methylbut-2-ene.
• The IUPAC name of (iv) is 1-bromo-2-methylpent-2-ene.

Answer:

	Column I	Column II
(i)		(b) Wurtz fittig reaction
(ii)		(a) Fittig reaction
(iii)		(d) Sandmeyer reaction
(iv)		(c) Finkelstein reaction

• Wurtz-fittig reaction is formed when a mixture of an alkyl halide aryl halide forms alkylarene and it is treated with sodium in dry ether.
• Fittg reaction takes place when aryl halides forms analogous compounds when treated with sodium in dry ether in which two ayl groups are joined together.
• Sandmeyer's reaction takes place when diazonium valut is treated with cupross chloride or cuproxes bromide and it forms chlorobenzene or bromobenzene.
• Finkelstein reaction takes place when alkyl iodines are prepared by the reaction of alkyl chlorides with sodium iodine in dry acetone.

Answer:

Thionyl chloride is considered as best over PCl₃ and PCl₅ that is used for the preparation of alkyl chlorides from alcohols. It gives pure alkyle halide as the other two products (SO₂ + HCl). Both assertion and reason are wrong.

Hence, the correct answer is option (iv).

Answer:

For the same hydrocarbon part, the boiling point depends upon the atomic mass of the halogen atom. The higher the mass of the halogen atom, the higher will be the boiling point. So, we can say that the boiling point decreases with a decrease in the atomic mass of the halogen atom.

Hence, the correct answer is option (ii).

Answer:

KCN reacts with methyl chloride to produce a mixture of methyl cyanide and methyl isocyanide because of the stable C-C bond in methyl cyanide. The assertion is not true but reason is true.

Hence, the correct answer is option (iii).

Answer:

Alkyl halides reacts with sodium in dry ether to give hydrocarbon in the wurtz reaction.

$2{\left(C{H}_3\right)}_{3}CBr \stackrel{Naldry ether}{\to }$ $$\begin{gathered} C{H}_3-C{\left(C{H}_3\right)}_2-C{\left(C{H}_3\right)}_2-C{H}_3+2NaBr \\ 2,2,3,3- \mathrm{tetramethy} \mathrm{butane}. \end{gathered}$$

Thus, both assertion and reason is correct and the explanation is also correct.

Hence, the correct answer is option (i).
 

Answer:

The presence of a nitro group in the ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution because NO₂ group, which is an electron-withdrawing group, reduces the electron density over the benzene ring. Both assertion and reason are correct and reason is the correct explanation.

Hence, the correct answer is option (i).

Answer:

The halogen atom in mono haloarenes increases the electron density in the ortho and para positions. As a result, additional electrophilic substitution occurs at the ortho and para positions. As halogens are highly electronegative, they have strong –I effect. That is why they are deactivating groups. Thus, both assertion and reason are correct but the reason is not the correct explanation.

Hence, the correct answer is option (ii).

Answer:

Arenes react with iodine in the presence of an oxidizing agent, resulting in the formation of aryl iodine. To avoid a backward reaction, the oxidising agent tends to oxidize HI into I₂. Assertion is correct but reason is wrong.

Hence, the correct answer is option (iv).

Answer:

In chlorobenzene, replacing chlorine with - OH is more difficult than in chloroethane as the C – Cl bond in chlorobenzene has a partial double bond character due to resonance. Both assertion and reason are correct and reason in the correct explanation.

Hence, the correct answer is option (i).

Answer:

This reaction occurs via tne S_N2 mechanism, in which –OH ions attacks the halogen atom of 2-bromoctane at 180° and cause the configuration to invert. The assertion is correct but the reason is wrong.

Hence, the correct answer is option (iii).

Answer:

Chlorination of nitrobenzene results in the formation of m-nitrochlorobenzene because – NO₂ group deactivates the ring as it is meta directing. The assertion is wrong but the reason is a correct statement.

Hence, the correct answer is option (iii).

Answer:

The type of reaction an alkyl halide will undergo depends on the structure of alkyl halide.
In case of primary alkyl halides, S_N2 mechanism is followed as the nucleophile attacks from back to the halogen atom. A transition state is formed in which carbon is simultaneously bonded to incoming nucleophile and the outgoing leaving group and finally, halogen atom is released.

In case of tertiary alkyl halides, S_N1 mechanism is followed via the formation of tertiary carbocation.
If the reagent is a weak base the substitution occurs while if it is strong base then substitution elimination occurs.

Secondary alkyl halides can undergo both substitution or elimination both depending upon the temperature condition or solvent type.
 

Answer:

The chlorofluorocarbon compounds of methane and ethane are collectively known as freons. They deplete the ozone layer which increases human exposure to ultraviolet rays, leading to increased skin cancer, eye diseases and disorders, and possible disruption of the immune system.

The following steps should be undertaken to minimise the harmful effects of freons:

• Use a freon gas with zero ozone depletion potential (ODP) and low global warming potential (GWP). 
• Recover and dispose off freon gas correctly when maintaining, or decommissioning a system.
• Minimise the power consumption of refrigerator or AC system.
• Install new systems that are energy efficient.
• Enhance the insulation and install doors on commercial freezers and refrigerators.
• Change the temperature set-points for AC.

Answer:

Aryl halides have less reactivity towards nucleophilic substitution reactions than alkyl halides due to following reasons:
(i) The lone pairs of electrons on halogen atom in haloarenes are in resonance with the benzene ring which leads to gaining of partial double character in C – X bond. . As a result, the bond cleavage in haloarene is difficult than haloalkane and therefore, they are less reactive towards nucleophilic substitution reaction.

(ii) In haloarenes, the carbon atom attached to halogen is sp² hybridised. The sp² hybridised carbon with a greater s-character is more electronegative and can hold the electron pair of C—X bond more tightly than sp³ -hybridised carbon in haloalkane with less s-chararcter. This makes the C – X bond shorter in haloarenes than C – X bond in haloalkanes.
As it is difficult to break a shorter bond than a longer bond, hence, haloarenes are less reactive than haloalkanes.
(iii) Aryl halides are less reactive than alkyl halides due to the repulsion between the nucleophile and electron rich arenes.
(iv) In case of haloarenes, the phenyl cation formed as a result of self-ionisation will not be stabilised by resonance and therefore, S_N1 mechanism is ruled out.

Method to enhance the reactivity of aryl halides:
The presence of an electron withdrawing group –(NO₂) at ortho and para positions increases the reactivity of aryl halides. However, the presence of an electron withdrawing group in meta-position has no effect on the reactivity of haloarenes.