Answer

(b) Reaction between alcohols and halogen acid follows $S_{N} 1$ mechanism. In $S_{N} 1$ mechanism carbocations are formed as intermediates.

Let us consider the formation of carbocations with the given three alcohols.

$$ CH_3-CH_2-CH_2-OH \longrightarrow CH_3-CH_2-\stackrel{+}{C} H+OH^{-} $$

In this case, $1^{\circ}$ carbocation is formed. It is least stable. So, here $S_{N} 2$ mechanism is followed. In this $S_{N} 2$ mechanism a transitory state is observed in $\alpha$-carbon is linked with two nucleophiles.

The reaction proceeded with stable carbocation. Higher the stability of carbocation, higher will be the possibilities of attack of $X^{-}$ion to the carbocation.

As, the tertiary carbocation is most stable so the possibilities of attack of $X^{-}$ion are more prominent in case of tertiary carbocations. Thus, attack of $X^{-}$ion to carbocation is proceeded with tertiary carbocation as follows

So, the correct option is (b).

Note Higher the stability of intermediate, higher will be the reactivity of compound and higher will be the yield of the desired product.

• Option (a) $(A)>(B)>(C)$: This option is incorrect because it suggests that the primary alcohol (A) is more reactive than the secondary (B) and tertiary (C) alcohols. However, in the $S_N1$ mechanism, the reactivity is determined by the stability of the carbocation intermediate. Tertiary carbocations are more stable than secondary, which are more stable than primary. Therefore, the reactivity order should not place the primary alcohol as the most reactive.

• Option (c) $( B )>(A)>( C)$: This option is incorrect because it suggests that the secondary alcohol (B) is more reactive than the primary (A) and tertiary (C) alcohols. While secondary carbocations are more stable than primary ones, they are less stable than tertiary carbocations. Therefore, the tertiary alcohol (C) should be more reactive than the secondary (B) and primary (A) alcohols.

• Option (d) $(A)>(C)>(B)$: This option is incorrect because it suggests that the primary alcohol (A) is more reactive than the tertiary (C) and secondary (B) alcohols. As mentioned earlier, in the $S_N1$ mechanism, the reactivity is determined by the stability of the carbocation intermediate. Tertiary carbocations are the most stable, followed by secondary, and then primary. Therefore, the primary alcohol should not be the most reactive.

Thinking Process

To solve this problem, students keep in mind that tertiary alcohol being most reactive react at room temperature.

Answer

(d) When alcohols are treated with conc., $HCl$ at room temperature than alkyl chloride is formed. This reaction follows $S_{N} 1$ mechanism. $S_{N} 1$ mechanism completes in two steps. In first step, a carbocation is formed and this carbocation is attacked by nucleophile in second step.

The attack of nucleophile to the carbocation is possible only if the carbocation is stable. Compound present in option (d) will give tertiary carbocation in step I. Tertiary carbocation is most stable so it is further attacked by $Cl^{-}$nucleophile as follows

• Option (a): The alcohol in option (a) will form a primary carbocation upon reaction with concentrated HCl. Primary carbocations are highly unstable and do not favor the $S_{N}1$ mechanism, making the formation of the corresponding alkyl chloride unlikely.

• Option (b): The alcohol in option (b) will form a secondary carbocation upon reaction with concentrated HCl. While secondary carbocations are more stable than primary ones, they are still less stable than tertiary carbocations. Therefore, the reaction is less favorable compared to the formation of a tertiary carbocation.

• Option (c): The alcohol in option (c) will also form a secondary carbocation upon reaction with concentrated HCl. Similar to option (b), secondary carbocations are less stable than tertiary carbocations, making the reaction less favorable for the formation of the corresponding alkyl chloride.

Answer

(a) When a primary aromatic amine, dissolved or suspended in cold aqueous mineral acid and treated with sodium nitrite, a diazonium salt is formed. When this freshly prepared diazonium salt is mixed with cuprous chloride, then diazonium group is replaced by $-Cl$.

Then chlorobenzene is formed which is $y$ in this reaction.

Hence, option (a) is correct.

• Option (b): This option might suggest the formation of a different product, such as a phenol or a different substituted benzene derivative. However, the reaction conditions specified (diazotization followed by treatment with cuprous chloride) specifically lead to the formation of chlorobenzene, not any other substituted benzene derivative.

• Option (c): This option could imply the formation of a compound like benzene diazonium chloride or another intermediate. However, the final product after the reaction with cuprous chloride is chlorobenzene, not an intermediate or a different final product.

• Option (d): This option might suggest the formation of a compound with a different functional group, such as a nitrobenzene or an aniline derivative. The reaction conditions clearly indicate the formation of chlorobenzene, and not any other functional group or derivative.

Answer

(b) Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. The reaction is electrophilic substitution reaction.

It has the following mechanism

$$ Cl-Cl \xrightarrow{FeCl_3} FeCl_4^{-}+Cl^{+} $$

In this mechanism, electrophile $Cl^{+}$attacks to electron rich benzene ring and replaces hydrogen. So, the reaction is electrophilic substitution reaction.

• (a) Electrophilic elimination reaction: This option is incorrect because an electrophilic elimination reaction involves the removal of an electrophile from a molecule, which is not the case here. In the given reaction, an electrophile (Cl⁺) is added to the benzene ring, not eliminated.

• (c) Free radical addition reaction: This option is incorrect because a free radical addition reaction involves the addition of radicals to a molecule. The given reaction involves the formation of an electrophile (Cl⁺) and its subsequent substitution on the benzene ring, not the addition of free radicals.

• (d) Nucleophilic substitution reaction: This option is incorrect because a nucleophilic substitution reaction involves the replacement of a leaving group by a nucleophile. In the given reaction, an electrophile (Cl⁺) is attacking the benzene ring, not a nucleophile.

Answer

(a) Halogen exchange reactions are those reactions in which one halide replaces another. In option (a) halogen $(-X)$ is replaced by iodine. This reaction is named as Finkelstein reaction.

In option (b) ,there is the addition of hydrogen halide on alkene.

In option (c) , halogen replaces alcoholic group.

While in option (d) halogen replaces the hydrogen of benzene ring.

• In option (b), there is the addition of hydrogen halide on an alkene, which is not a halogen exchange reaction.
• In option (c), a halogen replaces an alcoholic group, which is not a halogen exchange reaction.
• In option (d), a halogen replaces the hydrogen of a benzene ring, which is not a halogen exchange reaction.

Answer

(a) The given reaction is a substitution reaction. It involves the replacement of $1^{\circ}$ and $2^{\circ}$ hydrogen of alkanes by chlorine. It occurs in presence of ultraviolet light or at high temperature.

The chlorination does not occur at room temperature in absence of light. In this reaction, light is absorbed by the chlorine molecule and activated chlorine initiates the reaction as follows

Step 1

$$ Cl-Cl \xrightarrow[light]{UV} 2 \dot{C} l $$

$$ \dot{Cl}+CH_3-CH_2-CH_2-CH_3 \longrightarrow CH_3 CH_2 CH_2-\dot{C} H_2+HCl $$

Step 2

$ CH_3-CH_2-CH_2-\dot{C} H_2+Cl_2 \longrightarrow CH_3-CH_2-CH_2-CH_2 Cl+\dot{C} l$

Step 3

$ CH_3-CH_2-CH_2-\dot{C} H_2+\dot{C} l \longrightarrow CH_3 CH_2 CH_2 CH_2 Cl$

So, option (a) is the correct.

• (b) $NaCl+H_2SO_4$: This combination typically produces hydrogen chloride gas ($HCl$) and sodium sulfate ($Na_2SO_4$). It is not a suitable reagent for the chlorination of alkanes. Instead, it is often used in the preparation of $HCl$ gas in the laboratory.

• (c) $Cl_2$ gas in dark: Chlorination of alkanes requires the presence of light (UV light) or high temperature to initiate the reaction. In the dark, the chlorine molecules do not dissociate into reactive chlorine radicals, and thus, the substitution reaction does not occur.

• (d) $Cl_2$ gas in the presence of iron in dark: While iron can act as a catalyst in some reactions, the chlorination of alkanes specifically requires UV light to generate chlorine radicals. In the dark, even with iron present, the necessary radicals are not formed, and the reaction does not proceed.

Answer

(a) Density is directly related to molecular mass. Higher the molecular mass, higher will be the density of the compound. Among the four given compounds, the order of molecular mase is

benzene $<$ chlorobenzene $<$ dichlorobenzene $<$ bromochlorobenzene

Therefore, the increasing order of their densities are same as above.

Hence, option (a) is correct.

• Option (b) is incorrect because it suggests the order: benzene < dichlorobenzene < bromochlorobenzene < chlorobenzene. This order is incorrect as it places dichlorobenzene and bromochlorobenzene before chlorobenzene, which contradicts the correct order based on molecular mass.

• Option (c) is incorrect because it suggests the order: bromochlorobenzene < dichlorobenzene < chlorobenzene < benzene. This order is incorrect as it places bromochlorobenzene and dichlorobenzene before chlorobenzene and benzene, which contradicts the correct order based on molecular mass.

• Option (d) is incorrect because it suggests the order: chlorobenzene < bromochlorobenzene < dichlorobenzene < benzene. This order is incorrect as it places chlorobenzene and bromochlorobenzene before dichlorobenzene and benzene, which contradicts the correct order based on molecular mass.

Answer

(c) Boiling point of a compound depends upon the surface area. Higher the surface area, higher will be the boiling point of a compound. Surface area decreases with increase in branching. If the compound has branching so its boiling point will be minimum.

Thus, the increasing order of their boiling points

• Option (a) (ii) < (i) < (iii):

  • This option is incorrect because it suggests that compound (ii) has the lowest boiling point and compound (iii) has the highest boiling point. However, compound (iii) has the least branching and thus the highest surface area, leading to the highest boiling point. Therefore, compound (iii) should not be at the end of the sequence.

• Option (b) (i) < (ii) < (iii):

  • This option is incorrect because it suggests that compound (i) has the lowest boiling point and compound (iii) has the highest boiling point. While it is correct that compound (iii) has the highest boiling point, compound (ii) has more branching than compound (i), which should result in a lower boiling point for compound (ii) compared to compound (i). Therefore, compound (ii) should not be in the middle of the sequence.

• Option (d) (iii) < (ii) < (i):

  • This option is incorrect because it suggests that compound (iii) has the lowest boiling point and compound (i) has the highest boiling point. However, compound (iii) has the least branching and thus the highest surface area, leading to the highest boiling point. Therefore, compound (iii) should not be at the beginning of the sequence.

Answer

(b) Asymmetric/chiral carbon atom is that in which all of its four valencies with four different groups or atoms. In compound (iv), carbon satisfies two of its valencies with two hydrogen atoms i.e., similar atom.

So, it is not an asymmetric carbon atom while rest of the three molecules have asymmetric carbon as each carbon has satisfied all four valencies with four different groups or atoms.

So, the correct option is (b).

• Option (a) is incorrect because it includes compound (iv), which does not have an asymmetric carbon atom. In compound (iv), the carbon marked with an asterisk has two hydrogen atoms attached, making it not chiral.

• Option (c) is incorrect because it includes compound (iv), which does not have an asymmetric carbon atom. In compound (iv), the carbon marked with an asterisk has two hydrogen atoms attached, making it not chiral.

• Option (d) is incorrect because it includes compound (iv), which does not have an asymmetric carbon atom. In compound (iv), the carbon marked with an asterisk has two hydrogen atoms attached, making it not chiral.

Answer

(a) 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 enantiomer is dextro rotatory, the other will be laevo rotatory.

Here, the enantiomer of molecule $(A)$ is

Mirror

Hence, option (a) is correct.

• Option (b): This structure is not the mirror image of molecule (A). Instead, it is a diastereomer, which means it is a stereoisomer that is not a mirror image and not superimposable on molecule (A).

• Option (c): This structure is identical to molecule (A) and not its mirror image. Therefore, it cannot be an enantiomer since enantiomers are non-superimposable mirror images of each other.

• Option (d): This structure is also not the mirror image of molecule (A). It is another diastereomer, meaning it is a stereoisomer that is not a mirror image and not superimposable on molecule (A).

Answer

(b) vic-dihalides are those halides in which two halogen atoms are present on the two adjacent carbon atoms.

Write the structure of the given compound

In 1, 2-dichloroethane, the two chlorine atoms are attached to two adjacent carbon atoms.

Hence, option (b) is correct.

• (a) Dichloromethane: This compound has two chlorine atoms attached to the same carbon atom, not to adjacent carbon atoms. Therefore, it is not a vic-dihalide.

• (c) Ethylidene chloride: In this compound, both chlorine atoms are attached to the same carbon atom, making it a geminal dihalide, not a vic-dihalide.

• (d) Allyl chloride: This compound has only one chlorine atom attached to the carbon chain, so it does not qualify as a dihalide, let alone a vic-dihalide.

Answer

(a) Allyl halides are those compounds in which the halogen atom is bonded to $s p^{3}$ hybridised carbon atom next to carbon carbon-double bond.

e.g., $\quad CH_2=CH-CH_2 X$ and $CH_3 CH=CHC(Br)(CH_3)_2$

Aryl halides are the compounds in which the halogen atom is bonded to the $s p^{2}$ hybridised carbon atom of an aromatic ring.

e.g.,

Vinyl halides are the compounds in which the halogen atom is bonded to the $s p^{2}$ hybridised carbon atom of a carbon carbon double bond.

e.g.,

$CH_2=CH-X$

Secondary alkyl halides are the compounds in which the halogen atom is bonded to the $s p^{3}$ hybridised carbon atom which is further bonded to two alkyl groups and one hydrogen atom.

e.g.,

• (b) Aryl: Aryl halides are compounds in which the halogen atom is bonded to an $sp^2$ hybridized carbon atom of an aromatic ring. In the given compound $CH_3CH=CHC(Br)(CH_3)_2$, the bromine atom is not bonded to an aromatic ring, so it cannot be classified as an aryl halide.

• (c) Vinyl: Vinyl halides are compounds in which the halogen atom is bonded to an $sp^2$ hybridized carbon atom of a carbon-carbon double bond. In the given compound, the bromine atom is bonded to an $sp^3$ hybridized carbon atom, not an $sp^2$ hybridized carbon atom of a double bond, so it cannot be classified as a vinyl halide.

• (d) Secondary: Secondary alkyl halides are compounds in which the halogen atom is bonded to an $sp^3$ hybridized carbon atom that is further bonded to two alkyl groups and one hydrogen atom. In the given compound, the carbon atom bonded to the bromine is attached to three other carbon atoms (making it a tertiary carbon), not two alkyl groups and one hydrogen atom, so it cannot be classified as a secondary alkyl halide.

Answer

(b) In this reaction, $AlCl_3$ is a catalyst which activate the chlorine molecule to show heterolytic cleavage. $AlCl_3$ is electron deficient molecule and form $AlCl_4^{-}$and $Cl^{+}$when reacts with $Cl_2$. This $Cl^{+}$electrophile attacks on electron rich benzene ring.

$$ AlCl_3+Cl_2 \longrightarrow[AlCl_4]^{-}+Cl^{+} $$

• (a) $Cl^{-}$: This option is incorrect because $Cl^{-}$ is a nucleophile, not an electrophile. In the reaction of chlorine with benzene in the presence of $AlCl_3$, an electrophile is required to attack the electron-rich benzene ring. $Cl^{-}$ would not be able to perform this role.

• (c) $AlCl_3$: This option is incorrect because $AlCl_3$ acts as a catalyst in the reaction. It facilitates the formation of the electrophile $Cl^{+}$ by reacting with $Cl_2$, but it does not directly attack the benzene ring itself.

• (d) $[AlCl_4]^{-}$: This option is incorrect because $[AlCl_4]^{-}$ is a byproduct of the reaction and is an anion. It does not have the electrophilic character required to attack the electron-rich benzene ring. Instead, it is formed when $AlCl_3$ reacts with $Cl_2$ to generate the actual electrophile $Cl^{+}$.

Answer

(b) In vic-dihalides, halogen atoms are present on the adjacent carbon atoms.

In gem-dihalides, halogen atoms are present on the same carbon atom. They are known as alkylidene halides.

In allylic halides, halogen atom is bonded to $s p^{3}$ hybridised carbon atom next to carbon-carbon double bond.

In vinylic halides, halogen atom is bonded to $s p^{2}$ hybridised carbon atom of a carbon-carbon double bond.

In ethylidene chloride $(H_3 C-CHCl_2)$ both halogen atoms are present on same carbon atom so it is gem-dihalide.

• In vic-dihalides, halogen atoms are present on the adjacent carbon atoms. In ethylidene chloride $(H_3C-CHCl_2)$, both halogen atoms are on the same carbon atom, not adjacent ones.
• In allylic halides, the halogen atom is bonded to an $sp^3$ hybridised carbon atom next to a carbon-carbon double bond. Ethylidene chloride does not have a carbon-carbon double bond.
• In vinylic halides, the halogen atom is bonded to an $sp^2$ hybridised carbon atom of a carbon-carbon double bond. Ethylidene chloride does not have a carbon-carbon double bond, and the carbon bonded to the halogens is $sp^3$ hybridised.

Answer

(c) In this reaction, addition of $HCl$ takes place on doubly bonded carbons in accordance with Markownikoff’s rule i.e., addition of negative addendum will take place on that carbon which has lesser number of hydrogen.

Thus,

Hence, option (c) is correct.

• Option (a) is incorrect because it does not follow Markownikoff’s rule. According to this rule, the hydrogen atom from HCl should attach to the carbon with more hydrogen atoms, and the chlorine atom should attach to the carbon with fewer hydrogen atoms. Option (a) shows the addition of HCl in a manner that does not adhere to this rule.

• Option (b) is incorrect because it also does not follow Markownikoff’s rule. Similar to option (a), it shows the addition of HCl in a way that the hydrogen and chlorine atoms are not added to the correct carbons according to the rule.

• Option (d) is incorrect because it represents a different product that does not result from the addition of HCl to the given alkene. It may depict a different type of reaction or an incorrect application of the addition process.

Answer

(b) A primary alkyl halide would prefer to undergo $S_{N} 2$ reaction.

(a) $S_{N} 2$ reactions occur only if the intermediate carbocation is stable i.e., $3^{\circ}$ carbocation.

(b) $S_{N} 2$ reactions occur if there is less steric hinderance on to the $\alpha$-carbon of alkyl halide. In case of primary alkyl halides, carbocation is highly unstable and steric hinderance is very less. So, primary alkyl halide would prefer to undergo $S_{N} 2$ reaction.

(c) In $\alpha$-elimination, proton and the leaving group are present on same atom.

(d) Racemisation is the process of conversion of enantiomer into a racemic mixture.

• $S_{N} 1$ reactions occur only if the intermediate carbocation is stable i.e., $3^{\circ}$ carbocation.
• In $\alpha$-elimination, proton and the leaving group are present on the same atom.
• Racemisation is the process of conversion of enantiomer into a racemic mixture.

Answer

(d) All the given compounds are tertiary alkyl halides but the bond formed between carbon and iodine $(C-I)$ bond is the weakest bond due to large difference in the size of carbon and iodine. So, $(CH_3)_3 C-I $gives $S _{N} 1$ reaction most readily. In other words, iodine is a better leaving group.

• (a) $(CH_3)_3 C-F$: The bond between carbon and fluorine $(C-F)$ is very strong due to the small size and high electronegativity of fluorine. This makes the bond difficult to break, and thus, $(CH_3)3 C-F$ does not readily undergo $S{N} 1$ reaction.

• (b) $(CH_3)_3 C-Cl$: Although chlorine is a better leaving group than fluorine, the bond between carbon and chlorine $(C-Cl)$ is still relatively strong compared to the carbon-iodine bond. This makes $(CH_3)3 C-Cl$ less likely to undergo $S{N} 1$ reaction compared to $(CH_3)_3 C-I$.

• (c) $(CH_3)_3 C-Br$: Bromine is a better leaving group than both fluorine and chlorine, but the bond between carbon and bromine $(C-Br)$ is still stronger than the carbon-iodine bond. Therefore, $(CH_3)3 C-Br$ does not undergo $S{N} 1$ reaction as readily as $(CH_3)_3 C-I$.

Answer

(c) The correct IUPAC name of the given compound is

1-bromo-2-methylbutane

• (a) 1-bromo-2-ethylpropane: This name is incorrect because the longest carbon chain in the given compound has four carbon atoms, not three. The name “propane” implies a three-carbon chain, which does not match the structure of the compound.

• (b) 1-bromo-2-ethyl-2-methylethane: This name is incorrect because it suggests a two-carbon chain (ethane) with substituents, which does not match the structure of the compound. The longest carbon chain in the given compound has four carbon atoms, not two.

• (d) 2-methyl-1-bromobutane: This name is incorrect because the numbering of the carbon chain should give the substituents the lowest possible numbers. In this case, the bromine should be at position 1 and the methyl group at position 2, which is correctly represented as 1-bromo-2-methylbutane.

Answer

(b) Structure of the diethylbromomethane is given below

So, the IUPAC name is 3-bromopentane.

• (a) 1-bromo-1,1-diethylmethane: This name suggests a methane molecule with two ethyl groups and one bromine atom attached to the same carbon. However, this structure does not match the given structure of diethylbromomethane, which has a continuous chain of five carbon atoms.

• (c) 1-bromo-1-ethylpropane: This name implies a propane molecule with an ethyl group and a bromine atom both attached to the first carbon. This does not correspond to the structure of diethylbromomethane, which has a five-carbon chain with a bromine atom on the third carbon.

• (d) 1-bromopentane: This name indicates a pentane molecule with a bromine atom attached to the first carbon. This does not match the structure of diethylbromomethane, where the bromine atom is attached to the third carbon of a five-carbon chain.

Answer

(d) The reaction of toluene with chlorine in the presence of iron and carried out in absence of light, so the substitution occurs in the benzene ring. The $-CH_3$ group of toluene is $O$ and $p$-directing then product is the mixture of (b) and (c) i.e., $o$-chlorotoluene and $p$-chlorotoluene.

• Option (a): This option suggests that the reaction yields benzyl chloride. However, benzyl chloride is formed when the reaction occurs in the presence of light, leading to a free radical substitution at the benzylic position. Since the reaction is carried out in the absence of light, this option is incorrect.

• Option (b): This option suggests that the reaction yields only o-chlorotoluene. However, the $-CH_3$ group of toluene is both ortho and para-directing, leading to a mixture of products. Therefore, this option is incorrect as it does not account for the para product.

• Option (c): This option suggests that the reaction yields only p-chlorotoluene. Similar to option (b), this is incorrect because the $-CH_3$ group directs substitution to both ortho and para positions, resulting in a mixture of products. Hence, this option is also incorrect.

Answer

(c) Chloromethane on treatment with excess of ammonia yields mainly methamine.

$$ CH_3 Cl+\underset{\text { Excess }}{NH_3} \longrightarrow \underset{\text { Methanamine }}{CH_3 NH_2}+HCl $$

However, if the two reactants are present in the same amount, then the mixture of primary, secondary and tertiary amine is obtained.

$ \mathrm{CH_3 Cl+NH_3 \longrightarrow \underset{\substack{\text { (Primary } \\ \text { amine) }}}{CH_3 NH_2}+HCl} $

$\mathrm{CH}_3 \mathrm{NH}_2+\mathrm{CH}_3 \mathrm{Cl} \longrightarrow \underset{\substack{\text { (Secondary } \\ \text { amine })}}{\left(\mathrm{CH}_3\right)_2 \mathrm{NH}} +\mathrm{HCl} $

$\mathrm{CH}_3 \mathrm{NH}_2+\mathrm{CH}_3 \mathrm{Cl} \longrightarrow \underset{\substack{\text { (Secondary } \\ \text { amine) }}}{\left(\mathrm{CH}_3\right)_2 \mathrm{NH}}+\mathrm{HCl}$

$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}+\mathrm{CH}_3 \mathrm{Cl} \longrightarrow \underset{\substack{\text { (Tertiary } \\ \text { amine) }}}{\left(\mathrm{CH}_3\right)_3 \mathrm{~N}} +\mathrm{HCl}$

$ \left(\mathrm{CH}_3\right)_3 \mathrm{~N}+\mathrm{CH}_3 \mathrm{Cl} \longrightarrow \underset{\text{(Quarternary ammonium salt )}}{\left(\mathrm{CH}_3\right)_4 \stackrel{+}{\mathrm{NCl}}} $

• Option (b) $N$-methylmethanamine $(CH_3-NH-CH_3)$: This option is incorrect because $N$-methylmethanamine is a secondary amine, which is not the primary product when chloromethane reacts with excess ammonia. The primary product in this reaction is methanamine (a primary amine). Secondary amines are formed only when the reaction is not carried out with excess ammonia and further alkylation occurs.

• Option (d) mixture containing all these in equal proportion: This option is incorrect because when chloromethane reacts with excess ammonia, the primary product is methanamine. A mixture containing primary, secondary, and tertiary amines in equal proportion would only be obtained if the reactants are present in stoichiometric amounts, not in excess.

Answer

(a) Chiral/asymmetric carbon is that carbon in which carbon has formed four bonds with four different groups. Let see the structural formula of the given compounds.

With the help of these structural formulae it is very clear that 2-bromobutane in which asterisk mark carbon atom is bonded to four different atoms or groups. So, this molecule is chiral in nature. Other molecules do not contains four different atoms or groups.

• 1-bromobutane: The carbon atom bonded to the bromine (Br) is also bonded to two hydrogen atoms (H) and a single carbon chain, making it not chiral as it does not have four different groups attached.

• 2-bromopropane: The carbon atom bonded to the bromine (Br) is also bonded to two hydrogen atoms (H) and a single carbon chain, making it not chiral as it does not have four different groups attached.

• 2-bromopropan-2-ol: The carbon atom bonded to the bromine (Br) is also bonded to two identical methyl groups (CH3) and a hydroxyl group (OH), making it not chiral as it does not have four different groups attached.

Answer

(a) $S_{N} 1$ mechanism depends upon the stability of carbocation. Higher the stability of carbocation, higher will be the possibility of $S_{N_{\oplus}} 1$ mechanism to take place. In the given compound, $C_6 H_5 CH_2 Br$ carbocation is $C_6 H_5 \stackrel{\oplus}{C} H_2$. This carbocation $C_6 H_5 \stackrel{\oplus}{C} H_2$ is a stable carbocation due to resonance, therefore, its show $S_{N} 1$ mechanism.

• (b) $S_{N} 2$ mechanism: The $S_{N} 2$ mechanism involves a backside attack by the nucleophile and is favored by primary alkyl halides and less sterically hindered substrates. In the case of $C_6 H_5 CH_2 Br$, the benzyl carbocation formed is stabilized by resonance, making the $S_{N} 1$ mechanism more favorable. Additionally, the $S_{N} 2$ mechanism is less likely because the benzyl position is not as sterically accessible for a backside attack due to the presence of the phenyl group.

• (c) Any of the above two depending upon the temperature of reaction: The $S_{N} 1$ and $S_{N} 2$ mechanisms are influenced by the nature of the substrate, the nucleophile, and the solvent rather than just the temperature. In this case, the stability of the benzyl carbocation due to resonance makes the $S_{N} 1$ mechanism predominant, regardless of the temperature.

• (d) Saytzeff rule: The Saytzeff rule applies to elimination reactions, predicting the formation of the more substituted alkene as the major product. Since the question pertains to a substitution reaction (not an elimination reaction), the Saytzeff rule is not relevant in this context.

Answer

(b) Carbon has four valencies. If a carbon atom satisfies all of its four valencies with four different groups then it is termed as asymmetric/chiral carbon. In the given compound, 2 and 3 carbon are bonded to four different groups, so these are asymmetric.

• Option (a) $1,2,3,4$: This option is incorrect because not all four carbon atoms (1, 2, 3, and 4) are bonded to four different groups. Only carbons 2 and 3 are bonded to four different groups, making them asymmetric. Carbons 1 and 4 do not meet this criterion.

• Option (c) 1,4: This option is incorrect because carbons 1 and 4 are not bonded to four different groups. Therefore, they are not asymmetric.

• Option (d) $1,2,3$: This option is incorrect because carbon 1 is not bonded to four different groups. Only carbons 2 and 3 are bonded to four different groups, making them asymmetric.

Answer

(a) A mixture containing two enantiomers in equalimolar amount have zero optical rotation, as the rotation due to one isomer is cancelled by the rotation due to other isomer. Such a mixture is known as racemic mixture. All those compounds which follow $S_{N} 2$ mechanism during nucleophilic substitution reaction form racemic mixture. Order of reactivity of alkyl halides towards $S_{N} 1$.

and reactions as follows:

Tertiary halide; secondary halide; primary halide; $CH_3 X$.

For $S_{N} 1$ reaction.

contains a chiral carbon and gives a racemic product.

Hence, option (a) is correct.

Directions (Q. Nos. 26-29) In the questions 26 to 29 arrange the compounds in increasing order of rate of reaction towards nucleophilic substitution.

• Option (b) (i), (ii) and (iii):

  • Compound (ii) and (iii) do not necessarily form racemic mixtures upon nucleophilic substitution by $OH^{-}$ ion. For a racemic mixture to form, the compound must undergo $S_N1$ mechanism, which typically involves a chiral carbon and the formation of a planar carbocation intermediate. Compounds (ii) and (iii) may not meet these criteria.

• Option (c) (ii) and (iii):

  • Similar to the reasoning for option (b), compounds (ii) and (iii) do not necessarily form racemic mixtures upon nucleophilic substitution by $OH^{-}$ ion. They may not undergo $S_N1$ mechanism or may not have the necessary chiral carbon to form a racemic mixture.

• Option (d) (i) and (iii):

  • Compound (iii) does not necessarily form a racemic mixture upon nucleophilic substitution by $OH^{-}$ ion. It may not undergo $S_N1$ mechanism or may not have the necessary chiral carbon to form a racemic mixture.

Thinking Process

To solve this question, the point keep in mind that presence of electron withdrawing group at $o$ and p-position to the halogen atom increases the rate of nucleophilic substitution reaction.

Answer

(c) The bond formed between C of benzene ring and halogen is more stable because of resonance it has partial double bond character. So, rate of reaction towards nucleophilic substitution is slow. This substitution is facilitated by the presence of electron withdrawing group at ortho and para position because electron density is high at these positions.

Compound (ii) and (iii) both has one electron withdrawing group but in compound (ii) electron withdrawing $(-NO_2)$ group is present at ortho position, so rate of reaction in compound (ii) is more than that of (iii) while (i) has no electron withdrawing group.

Hence, the correct option is (c).

• Option (a) is incorrect because it suggests that the rate of reaction for compound (i) is less than that of compound (ii) and compound (iii). However, compound (i) has no electron withdrawing group, making it less reactive towards nucleophilic substitution compared to compounds (ii) and (iii), which both have electron withdrawing groups.

• Option (b) is incorrect because it suggests that the rate of reaction for compound (iii) is less than that of compound (ii) and compound (i). However, compound (iii) has an electron withdrawing group at the para position, making it more reactive than compound (i), which has no electron withdrawing group.

• Option (d) is incorrect because it suggests that the rate of reaction for compound (iii) is less than that of compound (i) and compound (ii). However, compound (iii) has an electron withdrawing group at the para position, making it more reactive than compound (i), which has no electron withdrawing group.

Answer

(d) Presence of electron releasing group at ortho or para position decreases the rate of nucleophilic substitution reaction. In compound (iii), electron releasing group is present at meta position w.r.t. chlorine, so the impact is less but in compound (ii) it is present at ortho position.

Thus, the rate of reaction towards nucleophilic substitution is least in compound (ii) and highest in compound (i) as there is no electron releasing group in this compound.

• Option (a) is incorrect because it suggests that the rate of nucleophilic substitution is highest in compound (iii) and lowest in compound (i). However, compound (iii) has an electron releasing group at the meta position, which has less impact on the rate compared to compound (ii) with an electron releasing group at the ortho position. Compound (i) has no electron releasing group, so its rate should be the highest.

• Option (b) is incorrect because it suggests that the rate of nucleophilic substitution is highest in compound (ii) and lowest in compound (i). However, compound (ii) has an electron releasing group at the ortho position, which significantly decreases the rate of nucleophilic substitution. Therefore, compound (ii) should have the lowest rate, not the highest.

• Option (c) is incorrect because it suggests that the rate of nucleophilic substitution is highest in compound (i) and lowest in compound (iii). However, compound (iii) has an electron releasing group at the meta position, which has less impact on the rate compared to compound (ii) with an electron releasing group at the ortho position. Compound (i) has no electron releasing group, so its rate should be the highest, not the lowest.

Thinking Process

For this questions, the point is that electron withdrawing group at ortho and para position decreases the electron density at these positions and increases the rate of reaction. Further, rate of reaction increases with increase in number of electron withdrawing group.

Answer

(d) Presence of electron withdrawing group at ortho and para position facilitate the nucleophilic substitution reaction and hence, enhances rate of reaction.

Compound (iii) has three electron withdrawing groups at ortho and para positions w.r.t. chlorine while compound (ii) has only one electron withdrawing group and there is no electron withdrawing group in compound (i). So, the increasing order of rate of reaction towards nucleophilic substitution is (i) $<$ (ii) $<$ (iii).

• Option (a) is incorrect because it suggests that compound (iii) has the slowest rate of reaction, followed by compound (ii), and then compound (i) has the fastest rate. This is incorrect because compound (iii) has the most electron withdrawing groups, which should increase the rate of nucleophilic substitution, not decrease it.

• Option (b) is incorrect because it suggests that compound (ii) has the slowest rate of reaction, followed by compound (iii), and then compound (i) has the fastest rate. This is incorrect because compound (ii) has one electron withdrawing group, which should increase the rate of nucleophilic substitution compared to compound (i), which has none.

• Option (c) is incorrect because it suggests that compound (i) has the slowest rate of reaction, followed by compound (iii), and then compound (ii) has the fastest rate. This is incorrect because compound (iii) has more electron withdrawing groups than compound (ii), which should make compound (iii) react faster than compound (ii).

Answer

(c) Presence of electron releasing group at otho and para position w.r.t. to chlorine decreases the rate of nucleophilic substitution reaction. Compound (iii) has two electron releasing groups and compound (ii) has one electron releasing group w.r.t. chlorine while compound (i) has no electron releasing group.

So, the rate of nucleophilic substitution reaction is highest in compound (i) and order is (iii) $<$ (ii) $<$ (i).

• Option (a) is incorrect because it suggests that the rate of nucleophilic substitution reaction is lowest in compound (i) and highest in compound (iii). However, compound (i) has no electron releasing groups, making it the most reactive, while compound (iii) has two electron releasing groups, making it the least reactive.

• Option (b) is incorrect because it suggests that the rate of nucleophilic substitution reaction is lowest in compound (ii) and highest in compound (iii). However, compound (ii) has one electron releasing group, making it less reactive than compound (i) but more reactive than compound (iii), which has two electron releasing groups.

• Option (d) is incorrect because it suggests that the rate of nucleophilic substitution reaction is lowest in compound (i) and highest in compound (ii). However, compound (i) has no electron releasing groups, making it the most reactive, while compound (ii) has one electron releasing group, making it less reactive than compound (i) but more reactive than compound (iii), which has two electron releasing groups.

Answer

(a) Higher the surface area, higher will be the intermolecular forces of attraction and thus boiling point too. Boiling point increases with increase in molecular mass of halogen atom for the similar type of alkyl halide. Butane has no halogen atom and rest of all three compounds are halo derivatives of butane.

Atomic mass of iodine is highest so the boiling point of 1-iodobutane is maximum among all the given compounds and hence, option (a) incorrect.

• Option (b) is incorrect: This option suggests that 1-iodobutane has the lowest boiling point, which contradicts the fact that boiling point increases with the molecular mass of the halogen atom. Since iodine has the highest atomic mass among the halogens listed, 1-iodobutane should have the highest boiling point, not the lowest.

• Option (c) is incorrect: This option places 1-iodobutane between butane and 1-bromobutane, which is incorrect. As previously mentioned, 1-iodobutane should have the highest boiling point due to the highest atomic mass of iodine. Therefore, it cannot be placed between butane and 1-bromobutane.

• Option (d) is incorrect: This option places 1-iodobutane between 1-chlorobutane and 1-bromobutane, which is incorrect. Given that iodine has a higher atomic mass than both chlorine and bromine, 1-iodobutane should have a higher boiling point than both 1-chlorobutane and 1-bromobutane.

Answer

(d) Boiling point increases with increase in size of hydrocarbon part for the same haloalkanes. All the given haloalkenes contain same halogen atom i.e., bromine but the number of carbon atoms in hydrocarbon part of the molecule are increasing from ethane to benzene.

So, the boiling point is minimum for 1-bromoethane and maximum for 1-bromobenzene.

• Option (a) is incorrect: This option suggests that bromobenzene has the lowest boiling point and 1-bromoethane has the highest. However, bromobenzene has a higher boiling point than 1-bromoethane due to the presence of the aromatic ring, which increases intermolecular forces.

• Option (b) is incorrect: This option suggests that bromobenzene has the lowest boiling point and 1-bromobutane has the highest. However, bromobenzene has a higher boiling point than 1-bromoethane, 1-bromopropane, and 1-bromobutane due to the aromatic ring, which increases intermolecular forces.

• Option (c) is incorrect: This option suggests that 1-bromopropane has the lowest boiling point and bromobenzene has the highest. However, 1-bromoethane has a lower boiling point than 1-bromopropane because it has fewer carbon atoms, leading to weaker van der Waals forces.

Answer

(c) In the above reaction, ${ }^{-} OH$ and $Cl^{-}$both are electron rich species as they are holding the negative charge. So, they are nucleophiles.

The above reaction show $S_{N} 2$ mechanism, carbon of alkyl halide is $s p^{3}$ hybridised.

During this mechanism, the breaking of $C-X$ bond and formation of new bond $(C-Nu)$ occur simultaneously through a transition state in which carbon atom is approximately $s p^{2}$ hybridised.

• (a) (i) and (v) both are nucleophiles: This statement is incorrect because (i) and (v) are not both nucleophiles. In the given reaction, ${ }^{-} OH$ and $Cl^{-}$ are nucleophiles, but the question does not specify which species (i) and (v) refer to. Without this information, we cannot confirm that both are nucleophiles.

• (b) In (iii) carbon atom is $s p^{3}$ hybridised: This statement is incorrect because in the transition state of an $S_{N}2$ mechanism, the carbon atom is not $s p^{3}$ hybridised. Instead, it is approximately $s p^{2}$ hybridised due to the simultaneous breaking of the $C-X$ bond and formation of the $C-Nu$ bond.

• (d) (i) and (v) both are electrophiles: This statement is incorrect because ${ }^{-} OH$ and $Cl^{-}$ are nucleophiles, not electrophiles. Electrophiles are electron-deficient species, whereas nucleophiles are electron-rich species.

Answer

$(a, b)$

In the given reaction, alkyl halide is primary in nature. Here, a transitory state is observed in which one bond is broken and one bond is formed synchronously i.e., in one step. So, it follows $S_{N} 2$ mechanism.

In this mechanism, nucleophile attacks the carbon at $180^{\circ}$ to the leaving group, so the reactant and product have opposite configuration.

• Option (c) is incorrect because in an $S_{N}2$ mechanism, the nucleophile attacks the carbon at $180^{\circ}$ to the leaving group, resulting in an inversion of configuration. Therefore, (ii) and (iv) cannot have the same configuration.

• Option (d) is incorrect because the given reaction involves a primary alkyl halide, which typically undergoes an $S_{N}2$ mechanism rather than an $S_{N}1$ mechanism. In an $S_{N}1$ mechanism, a carbocation intermediate is formed, which is not the case here.

Answer

$(a, d)$

For the given reaction, intermediate (iii) represent transition state, and it is highly unstable. In this transition state, carbon atom is $s p^{2}$ hybridised as partially bonded to two nucleophiles so it is highly unstable and less stable than the reactant (ii). Reactant (ii), carbon atom is $s p^{3}$ hybridised and more stable than intermediate (iii).

Directions (Q. Nos. 35-36) on the basis of the following reaction.

• (b) Intermediate (iii) is unstable because carbon atom is $sp^{2}$ hybridised: This statement is incorrect because the instability of intermediate (iii) is not due to the $sp^{2}$ hybridisation of the carbon atom. The $sp^{2}$ hybridisation itself does not inherently cause instability; rather, it is the specific context of the transition state and the partial bonding to two nucleophiles that makes it unstable.

• (c) Intermediate (iii) is stable because carbon atom is $sp^{2}$ hybridised: This statement is incorrect because intermediate (iii) is actually unstable, not stable. The $sp^{2}$ hybridisation of the carbon atom in this context does not confer stability. Instead, the intermediate is highly unstable due to the nature of the transition state and the partial bonding to two nucleophiles.

Answer

$(a, d)$

The reactant involved in above reaction is secondary alkyl halide. This $2^{\circ}$ alkyl halide contain

bulky group thats why it follow $S_{N} 1$ mechanism instead of $S_{N}{ }^{2}$ mechanism. In $S_{N}{ }^{1}$ mechanism, a stable carbocation will be formed as an intermediate. It is further attacked by $HO^{-}$nucleophile.

• (b) This option is incorrect because it describes an $S_N2$ mechanism, where the nucleophile ($OH^{-}$) attacks the substrate from one side while the leaving group ($Cl^{-}$) departs from the other side simultaneously. However, the reaction follows an $S_N1$ mechanism, not $S_N2$.

• (c) This option is incorrect because it suggests the formation of an unstable intermediate where both $OH^{-}$ and $Cl^{-}$ are weakly attached to the substrate. In an $S_N1$ mechanism, the intermediate is a carbocation, not a complex with both nucleophile and leaving group attached.

Answer

(a c)

The above reaction follows $S_{N} 1$ mechanism. In $S_{N} 1$ mechanism formation of carbocation is a slow step. So, the rate of reaction depends upon the concentration of (ii). So, the rate of reaction depends upon the concentration of only (ii) therefore, molecularity of reaction is one.

• (b) The rate of reaction depends on concentration of both (i) and (ii): This is incorrect because in an $S_{N}1$ mechanism, the rate-determining step is the formation of the carbocation, which depends only on the concentration of the substrate (ii) and not on the nucleophile (i).

• (d) Molecularity of reaction is two: This is incorrect because molecularity refers to the number of molecules involved in the rate-determining step. In an $S_{N}1$ mechanism, the rate-determining step involves only one molecule (the substrate), making the molecularity one, not two.

Answer

$(a, d)$

$\stackrel{\substack{\mathrm{Cl} \\ | }}{C}$

In the structure of 2-bromopentane $(\mathrm{CH_3 - \underset{ \substack{ | \\ \mathrm{BR}}}{CH}-CH_2-CH_2-CH_3})$ and trichloromethane $\left( \mathrm{Cl} - \mathrm{\underset{\substack{| \\ \mathrm{BR}}}{\stackrel{\substack{\mathrm{Cl} \\ | }}{C}}} - \mathrm{H} \right)$ , halogen atoms are attached to the $sp^3$ hybridised carbon atom of an alkyl group.

• Vinyl chloride (chloroethene) is incorrect because the chlorine atom is attached to an $sp^2$ hybridised carbon atom of an alkene group, not an $sp^3$ hybridised carbon atom of an alkyl group.
• 2-chloroacetophenone is incorrect because the chlorine atom is attached to an $sp^2$ hybridised carbon atom of a benzene ring, not an $sp^3$ hybridised carbon atom of an alkyl group.

Answer

$(a, c)$

(a) Ethylene chloride and ethylidene chloride on treatment with alc. $KOH$ show elimination reaction and ethyne as the product.

$\underset{\text {Ethylene chloride }}{\mathrm{Cl}-{\mathrm{CH}_2-\mathrm{CH}_2}-\mathrm{Cl}} \xrightarrow[\substack{\mathrm{KOH} \\ \text { (excess) }}]{\mathrm{Alc} .} \underset{\text { Ethyne }}{\mathrm{CH}\equiv \mathrm{CH}} $

$\underset{\text { Ethylidene chloride }}{\mathrm{CH}_3-\mathrm{CHCl}_2} \xrightarrow[\text { (excess) }]{\text { Alc.KOH }} \mathrm{CH} \equiv \mathrm{CH}$

(b) Both these compounds form different products on treatment with aq. $NaOH$.

$\mathrm{Cl-CH_2 -CH_2Cl } \xrightarrow[{\mathrm{NaOH}}]{\mathrm{Aq.}} \underset{\text{Ethylene glycol}}{\mathrm{HO-CH_2-CH_2-OH}}$

$ \mathrm{CH_3-CHCl_2} \xrightarrow [NaOH]{\mathrm{Aq.}} \mathrm{CH_3-CH(OH)_2} \xrightarrow[{-H_2 O}]{} \underset{\text { Ethanal }}{\mathrm{CH_3 CHO}} $

(c) Both these compounds form same products on reduction.

$$ \begin{matrix} \mathrm{Cl-CH_2-CH_2-Cl} \xrightarrow{\text { Reduction }} \underset{\text { Ethane }}{\mathrm{H_3 C-CH_3}} +\mathrm{2 HCl} \\ \mathrm{CH_3-CHCl_2} \xrightarrow{\text { Reduction }} \underset{\text { Ethane }}{\mathrm{H_3 C-CH_3}} +\mathrm{2 HCl} \end{matrix} $$

(d) Both these compounds are optically inactive.

• (b) Both the compounds form different products on treatment with aq. $NaOH$

  Ethylene chloride ($\mathrm{Cl-CH_2-CH_2-Cl}$) reacts with aqueous $NaOH$ to form ethylene glycol ($\mathrm{HO-CH_2-CH_2-OH}$), whereas ethylidene chloride ($\mathrm{CH_3-CHCl_2}$) reacts with aqueous $NaOH$ to form ethanal ($\mathrm{CH_3CHO}$) after an intermediate step. Therefore, the products are different.

• (d) Both these compounds are optically inactive

  Neither ethylene chloride nor ethylidene chloride have chiral centers, which means they cannot exhibit optical activity. Therefore, the statement that both compounds are optically active is incorrect.

Answer

$(a, c)$

Gem-dihalides are those dihalides in which two halogen atoms are bonded to the same carbon atom.

Write the structure of the given compounds.

(a) $\underset{\text{Ethylidene chloride}}{\mathrm{Cl_2-CH-CH_3}}$

(b) $\underset{\text { Ethylene dichloride }}{\mathrm{Cl-H_2 C-CH_2-Cl}}$

(c) $\underset{\text{Methylene chloride}}{\mathrm{CH_2 Cl_2}}$

(d)

So, in option (a) and (c) two halogen atoms are present on the same carbon atom and they are termed as gem-dihalides.

• (b) Ethylene dichloride: In this compound, the two chlorine atoms are bonded to two different carbon atoms, not the same carbon atom. Therefore, it is not a gem-dihalide.

• (d) Benzyl chloride: In this compound, there is only one chlorine atom bonded to the carbon atom adjacent to the benzene ring. Since it does not have two halogen atoms bonded to the same carbon atom, it is not a gem-dihalide.

Answer

(a, c)

Secondary bromides are those compounds in which $\alpha$-carbon (i.e., carbon bonded to bromine) is further bonded to two alkyl groups.

In compound (a) and (c) $\alpha$-carbon is bonded to two alkyl groups thats why it is secondary bromide but in compound (b) it is bonded to one alkyl group and it is primary bromide. In compound (c) it is bonded to three alkyl groups and it is tertiary alkyl halide.

• In compound (b) $(CH_3)_3 C CH_2 Br$, the $\alpha$-carbon (carbon bonded to bromine) is bonded to only one alkyl group, making it a primary bromide.
• In compound (d) $(CH_3)_2 CBrCH_2 CH_3$, the $\alpha$-carbon is bonded to three alkyl groups, making it a tertiary alkyl halide.

Thinking Process

To solve this question, point to the mind that

Aryl halides represent these compounds in which the halogen atom is bonded to the $s p^{2}$ hybridised carbon atom of an aromatic ring.

Answer

$(a, d)$

Write the structure of the given compounds.

So, from the above structure it is very clear that in compound (a) and compound (d), halogen atom is directly bonded to aromatic ring therefore these compounds are classfied as aryl halides

• In compound (b), the halogen atom (Cl) is bonded to a carbon atom that is part of an alkyl chain, not directly to the aromatic ring. Therefore, it is not classified as an aryl halide.

• In compound (c), the halogen atom (Br) is bonded to a carbon atom that is part of an alkyl chain, not directly to the aromatic ring. Therefore, it is not classified as an aryl halide.

Answer

( $a, b$ )

(a) Alcohol when treated with $HCl+ZnCl_2$ then alkyl halide is formed.

$$ ROH+HCl \xrightarrow{ZnCl_2} \underset{\text { (Alkyl halide) }}{RCl+H_2 O} $$

(b) Alcohol when treated with red $P$ and $X_2$ then product is alkyl halide.

$$ R-OH \xrightarrow{\text { Red } P^{/ Br_2}} \underset{\text { (Alkyl halide) }}{R-X} $$

(c) Alcohols when treated with $H_2 SO_4$ and $KI$ then $H_2 SO_4$ oxidises $KI$ to $I_2$ and does not produce $HI$. therefore, alkyl iodide does not form if the alcohols are treated with $H_2 SO_4+Kl$.

• Option (c) is incorrect because when alcohols are treated with $H_2SO_4$ and $KI$, the $H_2SO_4$ oxidizes $KI$ to $I_2$ instead of producing $HI$. As a result, alkyl iodide does not form.

Answer

$(b, c)$

Alkyl fluorides are synthesised by alkyl chloride/bromide in presence of $CoF_2$ or $Hg_2 F_2$. Only transition metal fluorides react with alkyl chloride/bromide to form alkyl fluorides. Alkali metal fluoride such as $NaF$ and alkaline earth metal fluoride such as $CaF_2$ do not react to form fluorides.

Note The reaction is termed as Swarts reaction in which alkyl fluorides are synthesised by heating an alkyl chloride/bromide in the presence of $AgF, Hg_2 F_2, CoF_2$.

• $CaF_2$: Alkaline earth metal fluorides like $CaF_2$ do not react with alkyl chlorides/bromides to form alkyl fluorides. They lack the necessary reactivity to facilitate the substitution reaction required for the formation of alkyl fluorides.

• $NaF$: Alkali metal fluorides such as $NaF$ also do not react with alkyl chlorides/bromides to form alkyl fluorides. Similar to $CaF_2$, they do not possess the reactivity needed to drive the substitution reaction for the synthesis of alkyl fluorides.

Answer

Iodination reactions are reversible in nature.

$$ C_6 H_6+I_2 \leftrightharpoons C_6 H_5 I+HI $$

In this above reaction, hydrogen iodide is formed apart from the required product. It has to be removed from the reaction mixture in order to prevent the backward reaction.

To carry out the reaction in the forward direction, $HI$ formed during the reaction is removed by oxidation using oxidising agent. Such as $HIO_3$ or $HNO_3$. The reaction is as follow

$$ \begin{matrix} 5 HI+HIO_3 \longrightarrow 3 I_2+3 H_2 O \\ 2 HI+2 HNO_3 \longrightarrow I_2+2 NO_2+2 H_2 O \end{matrix} $$

By using a suitable oxidising agent $HI$ is oxidised to give $I_2$.

• The preparation of aryl iodides requires the presence of an oxidizing agent because iodination reactions are reversible in nature. The hydrogen iodide (HI) formed during the reaction must be removed to prevent the backward reaction. This is achieved by using an oxidizing agent to convert HI into iodine (I₂) and other products, thus driving the reaction forward.

Answer

$p$-dibromobenzene has higher melting point than its o-isomer due to symmetry. Due to symmetry, $p$-isomer fits in the crystal lattice better than the o-isomer.

Hence, $p$-dibromobenzene has higher melting point.

• The o-dibromobenzene has a lower melting point because its lack of symmetry prevents it from fitting into the crystal lattice as efficiently as the p-isomer. This results in a less stable crystal structure and a lower melting point.

Answer

$S_{N} 1$ mechanism depends upon the stability of carbocation that is formed as intermediate in the mechanism.

$CH_6-CH_2-Cl$ will form $C_6 H_5{ }^{+} CH_2$ carbocation as intermediate.

This carbocation is resonance stabilised and will react faster in $S_{N} 1$ reaction.

While carbocation formed in $CH_3 CH_2 Cl$ is $CH_3 \stackrel{+}{C} H_2$. This carbocation is highly unstable and not give $S_{N} 1$ reaction with ${ }^{-} OH$ ion.

• The carbocation formed from $CH_3-CH_2-Cl$ is $CH_3 \stackrel{+}{C} H_2$, which is highly unstable. This instability makes it less likely to undergo an $S_{N} 1$ reaction with the ${ }^{-} OH$ ion.
• The $C_6 H_5-CH_2-Cl$ compound forms a $C_6 H_5{ }^{+} CH_2$ carbocation, which is resonance stabilized. This stabilization significantly increases the likelihood of undergoing an $S_{N} 1$ reaction with the ${ }^{-} OH$ ion.

Answer

lodoform liberate $I_2$ when it comes in contact with skin. Antiseptic property of iodine is due to the liberation of $I_2$ not because of iodoform itself.

$$ \underset{\text { lodine }}{CHI_3} \xrightarrow[\text { skin }]{\text { Contact with }} \underset{\text { lodine }}{I_2} \text { (responsible for antiseptic property) } $$

• The antiseptic property is not due to iodoform itself but due to the iodine ($I_2$) it releases upon contact with the skin.
• Iodoform does not have intrinsic antiseptic properties; its effectiveness is solely because it acts as a source of iodine.
• The chemical reaction that occurs when iodoform contacts the skin is responsible for the antiseptic effect, not the iodoform compound in its original form.
• The antiseptic property is specifically attributed to the iodine ($I_2$) released, which is known for its ability to kill bacteria and other pathogens.

Answer

Due to resonance, $C-X$ bond in haloarenes and haloalkenes have some double bond character. This partial double bond character of $C-X$ bond strengthen the bond. So, haloarenes and haloalkenes are less reactive than haloalkanes.

Lets see the resonating structure of the haloarenes and haloalkenes

Now, more the number of resonating structure higher will be the stability of the compound and lesser will be the reactivity. In haloarenes, more resonating structures are observed than the haloalkenes. So, haloarenes are less reactive than haloarenes. In haloalkanes, this $C-X$ bond is purely single bond.

• The answer does not mention the inductive effect, which is another factor that can influence the reactivity of haloarenes, haloalkenes, and haloalkanes.
• The answer does not discuss the steric hindrance, which can also affect the reactivity of these compounds.
• The answer does not consider the effect of the leaving group, which can play a significant role in the reactivity of haloarenes, haloalkenes, and haloalkanes.
• The answer does not address the solvent effects, which can influence the reactivity of these compounds in different environments.
• The answer does not mention the type of reactions (e.g., nucleophilic substitution, elimination) that haloarenes, haloalkenes, and haloalkanes typically undergo, which can provide additional context for their reactivity differences.

Answer

Lewis acids are electron deficient species. They are responsible for, inducing heterolytic fission in halogen molecule.

Role of Lewis acid is to produce an electrophile. The eletrophile produce will attack on electron rich benzene ring to produce aryl bromides and chlorides.

$$ \begin{aligned} & AlCl_3+Cl_2 \longrightarrow[AlCl_4]^{-}+Cl^{+} \\ & AlBr_3+Br_2 \longrightarrow[AlBr_4]^{-}+Br^{+} \end{aligned} $$

This electrophile will further attack on benzene.

• The other options are not provided in the input, so it is not possible to determine the reasons why they are incorrect.

Answer

Partial double bond character of a bond increses the strength of the bond and hence, decreases the stability. Phenol will not react with a mixture of $NaBr$ and $H_2 SO_4$ because it is resonance stabilised. Due to resonance, partial double bond character arises in $C-O$ bond of phenol and it becomes more stable than alcohol. $(CH_3 CH_2 CH_2 OH)$.

Reaction

$ 2 NaBr+3 H_2 SO_4 \rightarrow 2 NaHSO_4+SO_2+Br_2+2 H_2 O $

$CH_3 CH_2 CH_2 OH \xrightarrow{Br_2}$ No reaction

• (i) $CH_3 CH_2 CH_2 OH$: This compound is a primary alcohol and does not have resonance stabilization. Therefore, it can react with a mixture of $NaBr$ and $H_2SO_4$ to form the corresponding bromoalkane. The lack of resonance stabilization in the $C-O$ bond makes it more reactive compared to phenol.

• (ii) Phenol: Phenol is resonance stabilized, which means the $C-O$ bond has partial double bond character. This increased bond strength makes phenol less reactive towards the mixture of $NaBr$ and $H_2SO_4$. The resonance stabilization makes the $C-O$ bond in phenol more stable than in alcohols, preventing the reaction.

Thinking Process

The reaction is in based on the Markownikoffs rule.

Markownikoffs rule states that the negative part of adding molecule get attached that carbon atom of double bond which carries lesser number of hydrogen atom.

Answer

$CH_3 CHICH_3$ is the major product of the reaction. The mechanism of the reaction is as follows

$$ \mathrm{HI \leftrightharpoons H^{+}+I^{-}} $$

$ \mathrm{CH_3-CH=CH_2} \xrightarrow{\mathrm{H^s}} \begin{array}{ll} \mathrm{CH_3- \stackrel{+}{C}H-CH_3} & 2^\circ \text{ carbocation (more stable)} \\ \mathrm{CH_3-CH_2- \stackrel{+}{C}H_2} & 1^\circ \text{ carbocation (less stable)} \end{array} $

$ \mathrm{CH_3- \stackrel{+}{C}H-CH_3} \xrightarrow{\mathrm{I^-}} \mathrm{CH_3-\underset{\substack{ | \\ \mathrm{I}}}{CH}-CH_3} $

• CH₃CH₂CH₂I (A) is not the major product because it would be formed via the less stable 1° (primary) carbocation intermediate. In the reaction mechanism, the addition of H⁺ to the double bond of propene (CH₃CH=CH₂) can lead to the formation of either a 1° or a 2° (secondary) carbocation. The 2° carbocation (CH₃-CH⁺-CH₃) is more stable than the 1° carbocation (CH₃-CH₂-CH₂⁺) due to greater alkyl group stabilization. Therefore, the reaction preferentially proceeds through the more stable 2° carbocation, leading to the formation of CH₃CHICH₃ (B) as the major product.

Answer

Haloalkanes are slightly soluble in water. For the solubility to haloalkane in water, energy is required to overcome the attractions between its own molecules and to break the bonds between water molecules.

Less energy is released when new attractions are set up between the haloalkanes and the water molecules as these are not as strong as the original hydrogen bonds in water.

• The solubility of haloalkanes in water is not high because the energy required to break the hydrogen bonds in water and the intermolecular forces in haloalkanes is not sufficiently compensated by the energy released when new interactions are formed between haloalkanes and water molecules.
• Haloalkanes do not form strong hydrogen bonds with water molecules, which is a key factor in the low solubility.
• The non-polar nature of haloalkanes makes them less compatible with the polar nature of water, leading to low solubility.
• The size and structure of haloalkanes can also hinder their ability to interact effectively with water molecules, further reducing solubility.

Answer

Resonance in halobenzene

From the above resonating structure it is very clear that electron density is rich at ortho and para position. Therefore, it is ortho and para directing not meta directing.

• The functional group in the molecule is a halogen (X), which is an electron-withdrawing group due to its electronegativity. However, it also has lone pairs that can participate in resonance, donating electron density to the ring.
• In the resonance structures, the electron density is shown to be higher at the ortho and para positions relative to the halogen. This increased electron density at these positions makes them more reactive towards electrophilic substitution reactions.
• Meta directing groups typically withdraw electron density from the ortho and para positions, making these positions less reactive. Since the halogen increases electron density at the ortho and para positions, it cannot be meta directing.
• Therefore, the functional group is ortho and para directing, not meta directing, because the resonance structures show increased electron density at the ortho and para positions.

Answer

The structural formula of the given compounds are

(a) $\underset{\substack{\text { I-Bromo but -2-ene } \\ \text { (Primary halide) }}}{\mathrm{H_3 C-H_2 C=HC-H_2 C-Br}}$

(b) $\underset{\substack{\text{4-Bromobut-2-ene}\\ \text{(Secondary halide)}}}{\mathrm{H_3-\underset{\substack{| \\ \mathrm{Br}}}{CH}-CH=CH-CH_3}} $

(c) $\underset{\substack{\text{2-bromo-2-methylpropane}\\ {\text{(Tertiary halides)}}}}{\mathrm{H_3C - \stackrel{\substack{\mathrm{CH_3} \\ |}}{\underset{\substack{| \\ {\mathrm{Br}}}}{C}}-CH_3}} $

In compound (i), carbon atom to which halogen is bonded, further bonded to two hydrogens and one carbon of hydrocarbon chain. So, it is primary halide.

In compound (ii), $\alpha$-carbon is bonded with one hydrogen and two carbons of two hydrocarbons chain. So, it is secondary halide.

In compound (iii) $\alpha$-carbon is bonded to three alkyl group, so it is tertiary halide.

• For compound (a) 1-bromobut-2-ene, the carbon atom to which the bromine is bonded is attached to two hydrogen atoms and one carbon atom of the hydrocarbon chain, making it a primary halide. Therefore, the classification as a primary halide is correct.

• For compound (b) 4-bromopent-2-ene, the carbon atom to which the bromine is bonded is attached to one hydrogen atom and two carbon atoms of the hydrocarbon chain, making it a secondary halide. Therefore, the classification as a secondary halide is correct.

• For compound (c) 2-bromo-2-methylpropane, the carbon atom to which the bromine is bonded is attached to three alkyl groups, making it a tertiary halide. Therefore, the classification as a tertiary halide is correct.

Answer

(i) As the rate of reaction depends upon the concentration of compound ’ $A$ ’ $(C_4 H_9 Br.$ ) only therefore, the reaction is proceeded by $S_{N} 1$ mechanism and the given compound will be tertiary alkyl halide i.e., 2-bromo-2-methylpropane and the structure is as follow

${\mathrm{H_3C - \stackrel{\substack{\mathrm{CH_3} \\ |}}{\underset{\substack{| \\ {\mathrm{CH_3}}}}{C}}-Br}} \hspace {2 mm}\mathrm{(A)}$

Optically active isomer of (A) is 2-bromobutane (B) and its structural formula is $\mathrm{CH_3 - \underset{\text(B)}{CH_2}- \stackrel{*}{\underset{\substack{| \\ \mathrm{Br}}}{C}} H CH_3}$

(ii) As compound (B) is opically active therefore, compound (B) must be 2-bromobutane. Since, the rate of reaction of compound $(B)$ depends both upon the concentration of compound (B) and $KOH$, hence, the reaction follow $S_{N} 2$ mechanism. In $S_{N} 2$ reaction, nucleophile attack from, the back side, therefore, the product of hydrolysis will have opposite configuration

• The rate of reaction for compound ‘A’ depends only on the concentration of ‘A’, indicating an $S_N1$ mechanism. This mechanism is characteristic of tertiary alkyl halides due to the stability of the carbocation intermediate. Therefore, any structure that is not a tertiary alkyl halide is incorrect for ‘A’.

• The rate of reaction for compound ‘B’ depends on both the concentration of ‘B’ and $KOH$, indicating an $S_N2$ mechanism. This mechanism is characteristic of primary or secondary alkyl halides where the nucleophile attacks from the backside. Therefore, any structure that is not a primary or secondary alkyl halide is incorrect for ‘B’.

• Compound ‘A’ is not optically active because it is a tertiary alkyl halide (2-bromo-2-methylpropane), which does not have a chiral center. Therefore, any structure that suggests optical activity for ‘A’ is incorrect.

• Compound ‘B’ is optically active, meaning it must have a chiral center. Therefore, any structure that does not have a chiral center is incorrect for ‘B’.

• In an $S_N2$ reaction, the nucleophile attacks from the backside, leading to inversion of configuration. Therefore, any explanation that does not account for this inversion is incorrect for the product of compound ‘B’.

Answer

When compound ’ $A$ ’ with molecular formula, $C_7 H_8$ is treated with $Cl_2$ in the presence of $FeCl_3$ o-chlorotoluene or $p$-chlorotoluene will be formed as the compound $A$ with molecular formula $C_7 H_8$ is toluene.

Mechanism $Cl-Cl+FeCl_3 \rightarrow[FeCl_4]^{-}+Cl^{+}$

• The molecular formula $C_7 H_8$ corresponds to toluene, which is a benzene ring with a methyl group attached. When toluene is treated with $Cl_2$ in the presence of $FeCl_3$, the reaction is an electrophilic aromatic substitution where the chlorine atom substitutes a hydrogen atom on the benzene ring.

• The presence of the methyl group (a $-CH_3$ group) on the benzene ring makes the ortho (2-position) and para (4-position) positions more reactive towards electrophilic substitution due to the electron-donating nature of the methyl group.

• Therefore, the major products of this reaction are o-chlorotoluene (2-chlorotoluene) and p-chlorotoluene (4-chlorotoluene).

• The other possible positions for substitution (meta positions) are less reactive due to the electron-donating effect of the methyl group, making them less likely to be the sites of chlorination under these conditions.

• The reaction mechanism involves the formation of a chloronium ion ($Cl^+$) facilitated by $FeCl_3$, which then attacks the benzene ring at the ortho and para positions relative to the methyl group.

• Any other products, such as meta-chlorotoluene, are not formed in significant amounts due to the lower reactivity of the meta positions in the presence of the electron-donating methyl group.

Answer

In the given reaction, addition occur and the following two products ( $A$ and $B$ ) are possible

$\mathrm{H_3 C-CH_2-CH \equiv CH-CH_3+HCl} \longrightarrow \underset{\text { (2-chloropentane) }}{\mathrm{CH_3-CH_2-CH_2-\underset{\substack{| \\ \mathrm{Cl}}}{CH}-CH_3}} + \underset{\text { (3-chloropentane) }}{\mathrm{CH_3-CH_2-\underset{\substack{| \\ \mathrm{Cl}}}{CH}-CH_2-CH_3}} $

Further the carbocation formed from compound $(A)$ is slightly less stable than carbocation leading to the formation of compound $(B)$ therefore the amount of 2-chloropentane $(B)$ will be slightly more than that of 3 - chloropentane $(A)$.

• The given reaction involves the addition of HCl to an alkyne, not an alkene. The structure provided in the question is incorrect for an alkyne.
• The correct structure for the starting alkyne should be $\mathrm{CH_3-CH_2-C \equiv C-CH_3}$.
• The products formed from the addition of HCl to an alkyne are not 2-chloropentane and 3-chloropentane. Instead, the products should be 2-chloro-2-pentene and 3-chloro-2-pentene.
• The explanation provided for the stability of the carbocations is incorrect because the reaction mechanism for the addition of HCl to an alkyne does not involve carbocation intermediates. Instead, it follows a Markovnikov addition mechanism.
• The statement about the amount of 2-chloropentane being more than 3-chloropentane is incorrect because the products should be 2-chloro-2-pentene and 3-chloro-2-pentene, not 2-chloropentane and 3-chloropentane.

Answer

In compound (II), both the $CH_3$ groups and $Cl$ atoms at para-position to each other. Therefore, compound (II) is more symmetrical and it fits in the crystal lattice better than the other two isomers and hence it has the highest melting point than the others.

• Compound (I): The $CH_3$ groups and $Cl$ atoms are at ortho and meta positions relative to each other, which leads to less symmetry and a less efficient packing in the crystal lattice compared to compound (II). This results in a lower melting point.

• Compound (III): The $CH_3$ groups and $Cl$ atoms are at meta and ortho positions relative to each other, similar to compound (I). This also leads to less symmetry and less efficient packing in the crystal lattice compared to compound (II), resulting in a lower melting point.

Answer

The structure of neo-pentylbromide is

IUPAC name $\Rightarrow$ I-bromo -2,2-dimethylpropane

Common name $\Rightarrow$ neo-pentylbromide

• The structure provided in the answer is correct for neo-pentylbromide, so there are no other options to consider. Therefore, there are no incorrect options to provide reasons for.

Answer

Since, the molar mass of hydrocarbon is $72 g mol^{-1}$ thats why the hydrocarbon is $C_5 H_{12}$ i.e., pentane.

On photo chlorination, it gives monochloro derivatives so, all the hydrogen atoms must be equivalent and the structure of the compound will be

• The hydrocarbon cannot be n-pentane because n-pentane would give multiple monochloro derivatives due to the presence of different types of hydrogen atoms (primary, secondary, and tertiary hydrogens).

• The hydrocarbon cannot be isopentane (2-methylbutane) because isopentane would also give multiple monochloro derivatives due to the presence of different types of hydrogen atoms (primary and secondary hydrogens).

• The hydrocarbon cannot be neopentane (2,2-dimethylpropane) because neopentane would give only one monochloro derivative, but it would not give two dichloro derivatives. Neopentane has all equivalent hydrogen atoms, so it would give only one type of dichloro derivative.

Answer

Two alkenes are possible

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

• The other options are incorrect because they do not follow Markownikoff’s rule. According to this rule, the negative part of the adding molecule (in this case, the chloride ion from HCl) will attach to the carbon atom that has fewer hydrogen atoms. This ensures the formation of the more stable carbocation intermediate during the reaction. If the alkene does not follow this rule, it will not yield 1-chloro-1-methylcyclohexane as the product.

Answer

2-bromo-2-methylpropane (iii), is a tertiary alkyl halide and it will form a stable carbocation on ionisation.

1 -bromobutane is primary alkyl halide whereas 2-bromobutane and 2-chlorobutane is secondary alkyl halide.

• 1-bromobutane is a primary alkyl halide, which forms a less stable carbocation compared to tertiary alkyl halides, making it less reactive with aqueous KOH.
• 2-bromobutane is a secondary alkyl halide, which forms a moderately stable carbocation, but it is still less stable than the carbocation formed by a tertiary alkyl halide.
• 2-chlorobutane is a secondary alkyl halide, and chlorine is less reactive than bromine due to its stronger bond with carbon, making it less reactive with aqueous KOH compared to bromine-containing compounds.

Answer

Due to resonance in phenol, $C-O$ bond of phenol has some partial double bond character. Partial double bond character strengthen the bond. So, It is difficult to break this $C-O$ bond of phenol while the $C-O$ bond of alcohol is purely single bond and comparatively weaker bond.

So alkyl halides can be prepared by the reaction of alcohols with $HCl$ in the presence of $ZnCl_2$ while aryl halides can not be prepared by reaction of phenol with $HCl$ in the presence of $ZnCl_2$.

$\underset{\text { Phenol }}{\mathrm{C}_6 \mathrm{H}_5 \mathrm{OH}} \xrightarrow[{\mathrm{ZnCl}_2}]{\mathrm{HCl}} \text { No reaction }$

$ \begin{aligned} & \underset{\text { Alcohol }}{\mathrm{RCH_2 OH}} \xrightarrow[{\mathrm{ZnCl_2}}]{\mathrm{HCl}} \underset{\text { Alkyl chloride }}{\mathrm{R CH_2 Cl}}+ \mathrm{H_2 O} \end{aligned} $

• The resonance in phenol gives the $C-O$ bond partial double bond character, making it stronger and more difficult to break compared to the purely single $C-O$ bond in alcohols.
• The $C-O$ bond in alcohols is weaker and purely single, allowing for easier reaction with $HCl$ in the presence of $ZnCl_2$ to form alkyl halides.
• Aryl halides cannot be prepared from phenol using $HCl$ and $ZnCl_2$ because the strong partial double bond character in phenol’s $C-O$ bond resists breaking, unlike the weaker single bond in alcohols.

Answer

Compound (B) will give $S_{N} 1$ reaction faster than compound $(A)$ because $S_{N} 1$ reaction depends upon the stability of carbocation. Benzyl chloride on ionisation gives $C_6 H_5 \stackrel{+}{C} H_2$ carbocation which is resonance srabilised while the carbocation obtained from compound $(A)$ is not stabilised by resonance.

• Compound (A) is incorrect because the carbocation formed from it is not stabilized by resonance, making it less stable and thus less favorable for an $S_{N}1$ reaction compared to compound (B).

• Compound (C) is incorrect because the carbocation formed from it is also not stabilized by resonance, similar to compound (A), making it less favorable for an $S_{N}1$ reaction compared to compound (B).

• Compound (D) is incorrect because the carbocation formed from it is not stabilized by resonance, making it less stable and thus less favorable for an $S_{N}1$ reaction compared to compound (B).

Answer

As we know that $S_{N} 1$ mechanism depends upon the stability of carbocation. Allyl chloride on hydrolysis gives resonance stabilised carbocation while no resonance is observed in the carbocation formed by $n$-propyl chloride.

$ \underset{\text{Allyl chloride}}{\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2 \mathrm{Cl}} \xrightarrow{-\mathrm{Cl}^{-}} \mathrm{CH}_2=\mathrm{CH}-\stackrel{+}{\mathrm{C}} \mathrm{H}_2+\stackrel{-}{\mathrm{Cl}} $

$\mathrm{CH}_2=\mathrm{CH}-\stackrel{+}{\mathrm{CH}_2} \longleftrightarrow \underset{\substack{\text { Resonance stabilised } \\ \text { carbocation }}}{\stackrel{+}{\mathrm{CH}_2}-\mathrm{CH}=\mathrm{CH}_2}$

$ \underset{\substack{\text { n-propyl } \\ \text { chloride }}}{\mathrm{CH}_3-\mathrm{CH}_2}-\mathrm{Cl} \xrightarrow{-\mathrm{Cl}^{-}} \underset{\text { (Not stabilised by resonance) }}{\mathrm{CH}_3-\mathrm{CH}_2-\stackrel{{+}}{\mathrm{CH}_2}}+\mathrm{Cl}^{-} $

Hence, allyl chloride undergoes hydrolysis much faster than $n$-propyl chloride.

• The $S_{N}1$ mechanism depends on the stability of the carbocation intermediate. Allyl chloride forms a resonance-stabilized carbocation, which is more stable than the carbocation formed by n-propyl chloride. This increased stability makes the hydrolysis of allyl chloride more favorable.
• n-Propyl chloride does not form a resonance-stabilized carbocation. The carbocation formed by n-propyl chloride is less stable because it lacks resonance stabilization, making its hydrolysis less favorable compared to allyl chloride.

Answer

Grignard reagents are highly reactive and react with water to give corresponding hydrocarbons.

$$ \underset{\substack{\text { Grignard } \\ \text { reagent }}}{RMgX}+H_2 O \longrightarrow RH+Mg(OH) X $$

• The Grignard reagent does not react with water to form a stable compound; instead, it decomposes.
• The presence of moisture does not enhance the reactivity of the Grignard reagent; it actually deactivates it.
• Moisture does not stabilize the Grignard reagent; it causes it to break down.
• The Grignard reagent does not require moisture to initiate its reaction; it reacts readily with other compounds in the absence of moisture.

Answer

Polar solvents help in the first step in $S_{N} 1$ mechanism because leaving group and carbocation both are stabilised by polar solvent. Polarity of a solvent depends upon the value of dielectric constant. Higher the value of dielectric constant, higher will be the polarity of the solvent, faster will be the rate of $S_{N} 1$ mechanism. These polar solvents can work as a nucleophile and stabilise the carbocation as follows

• The other options are incorrect because they do not address the stabilization of both the leaving group and the carbocation, which is crucial for the $S_{N} 1$ mechanism.
• They might not consider the role of the dielectric constant in determining the polarity of the solvent and its impact on the reaction rate.
• They may overlook the ability of polar solvents to act as nucleophiles and stabilize the carbocation.
• They might not account for the specific interactions between the solvent and the intermediates in the $S_{N} 1$ mechanism.

Answer

Presence of double bond in a molecule is detected by following two methods:

(i) $Br_2$ in $CCl_4$ test When $Br_2 /.CC|_4$ is added unsaturated compound then orange colour of bromine disappears and dibromoderivative is formed. (colourless).

(ii) Bayer’s test When alkaline solution of $KMnO_4$ is added to the solution of unsaturated compound then its pink colour disappears due to the formation of dihydroxy derivative.

• The question and answer provided are specific to detecting the presence of double bonds in a molecule. The methods mentioned (Br(_2) in CCl(_4) test and Bayer’s test) are well-established chemical tests for this purpose. Any other options would be incorrect if they do not specifically address the detection of double bonds or if they involve methods that are not relevant or reliable for this purpose. For example:

1. Incorrect Option: Using a pH indicator
   Reason: pH indicators are used to determine the acidity or basicity of a solution, not the presence of double bonds in a molecule.

2. Incorrect Option: Measuring boiling point
   Reason: Boiling point measurements can provide information about the physical properties of a substance but do not specifically indicate the presence of double bonds.

3. Incorrect Option: Conducting a flame test
   Reason: Flame tests are used to identify the presence of certain metal ions based on the color of the flame, not to detect double bonds in organic molecules.

4. Incorrect Option: Using a mass spectrometer
   Reason: While mass spectrometry can provide detailed information about the molecular weight and structure of a compound, it is not a direct test for the presence of double bonds.

5. Incorrect Option: Observing color change with litmus paper
   Reason: Litmus paper is used to test for acidity or basicity, not for the presence of double bonds in a molecule.

6. Incorrect Option: Performing a solubility test in water
   Reason: Solubility tests can indicate the polarity and solubility characteristics of a compound but do not specifically detect double bonds.

By focusing on the specific chemical reactions that occur with double bonds, the provided methods (Br(_2) in CCl(_4) test and Bayer’s test) are accurate and reliable for detecting the presence of double bonds in a molecule.

Answer

In environment, diphenyl is formed during the incomplete combustion of mineral oil and coal. It is present in the exhaust gases of vehicles and in exhaust air from residential and industrial heating devices.

Acute exposure to high levels of biphenyl has been observed to cause eye and skin irritation and toxic effects on the liver, kidneys and central/peripheral nervous system. Kidneys of animals are also affected due to the ingestion of biphenyls. In rats, fetofoxicity has been oserved if they are exposed to high levels of biphenyl.

Preparation of diphenyls from aryl halides

Aryl halides, when treated with sodium in dry ether give diphenyl. This reaction is named as Fittig reaction.

• The provided answer does not mention any other options, so there are no other options to evaluate for correctness or incorrectness.

Answer

The IUPAC name of DDT is 2,2-bis (4-chlorophenyl)-1,1,1-trichloroethane and that of benzene hexachloride is 1,2,3,4,5,6-hexachlorocyclohexane.

They are banned in india because they are non-biodegradable. Instead, they get deposited and stored in fatty tissues. If this ingestion continues at a steady rate, DDT builds up within the animal over time. This will affects the reproductive system of animals.

If animals including humans are exposed to high levels of benzene hexachloride then it may cause acute poisoning. Apart from that this BHC may affect liver functioning in humans.

• The answer does not mention that DDT and benzene hexachloride are banned due to their potential to cause cancer in humans.
• The answer does not state that these chemicals can lead to the development of resistance in pests, making them less effective over time.
• The answer does not highlight the environmental impact, such as the contamination of soil and water bodies, which can harm various ecosystems.
• The answer does not discuss the international agreements and regulations, such as the Stockholm Convention, which aim to eliminate or restrict the use of persistent organic pollutants like DDT and benzene hexachloride.

Answer

Elimination reactions are as common as the nucleophilic substitution reaction in case of alkyl halides as two reactions occur simultaneously. Generally, at lower temperature and by using weaker base nucleophilic substitution reaction occur while at higher temperature and by using a stronger base elimination reactions (especially $\beta$ - elimination) take place. e.g., If ethyl bromide is treated with aq. $KOH$, at low temperature it gives ethanol while if it is treated with alc, $KOH$ at high temperature then it gives ethene.

$$ \begin{aligned} & CH_3 CH_2 Br \xrightarrow[373 K]{Aq . KOH} CH_3 \underset{\substack{\text { Ethanol }}}{CH_2} OH \text { (Nucleophilic substitution reaction) } \\ & CH_3 CH_2 Br \xrightarrow[473-523 K]{\text { Alc } KOH} \underset{\text { Ethene }}{CH_2=CH_2} \text { (Elimination reaction) } \end{aligned} $$

• The question does not provide any other options to evaluate. Therefore, there are no other options to consider or reasons to provide for their incorrectness.

Answer

When aniline, dissolved or suspended in cold aqueous mineral acid, is treated with sodium nitrite, a diazonium salt is formed. This diazo salt on treatment with cuprous bromide gives monobromobenzene.

This reaction is named as Sandmeyer’s reaction. If benzene diazonium chloride is treated with copper in $HBr$ then the product formed is bromobenzene. This reaction is known as Gattermann reaction.

• The other options are incorrect because they do not involve the formation of a diazonium salt from aniline, which is a crucial intermediate in the synthesis of monobromobenzene.
• They might suggest alternative methods that do not specifically use the Sandmeyer or Gattermann reactions, which are the established methods for converting aniline to monobromobenzene.
• Some options might propose direct bromination of aniline, which would lead to polybromination rather than monobromination due to the activating effect of the amino group.
• Other options might involve different starting materials or reagents that do not lead to the desired monobromobenzene product.
• They could also suggest conditions that are not suitable for the specific transformation required to obtain monobromobenzene from aniline.

Answer

Aryl halides are less reactive towards nucleophilic substitution reaction. Presence of electron withdrawing group at ortho and para position increases the stability of intermediates and hence increases the reactivity of aryl halides towards nucleophilic substitution reaction.

Now, more the number of EWG at ortho and para position, higher will be the reactivity of aryl halide. Compound (III) has three EWG so, it is most reactive and compound (I) has only one EWG so, it is least reactive. So, the order of reactivity is $(I)<(II)<($ III $)$

• Option I is the least reactive because it has only one electron-withdrawing group (EWG) at the para position, which provides minimal stabilization to the intermediate during the nucleophilic substitution reaction.

• Option II is more reactive than Option I but less reactive than Option III because it has two electron-withdrawing groups at the ortho and para positions, providing moderate stabilization to the intermediate.

• Option III is the most reactive because it has three electron-withdrawing groups at the ortho and para positions, providing the highest level of stabilization to the intermediate during the nucleophilic substitution reaction.

Answer

Tert. butyl bromide reacts with aq. $NaOH$ as follows

tert. butyl bromide when treated with aq. $NaOH$, it forms tert. corbocation which is more stable intermediate. This intermediate is further attacked by ${ }^{-} OH$ ion.

As tert. carbocation is highly stable so tert butylbromide follow $S_{N} 1$ mechanism.

In case of $n$-nutylbromide, primary carbocation is formed which is least stable so, it does not follow $S_{N} 1$ mechanism. Here, stearic hindrance is very less so, it follow $S_{N} 2$ mechanism. In $S_{N} 2$ mechanism, ${ }^{-} OH$ will attack from backside and a transition state is formed.

The leaving group is then pushed off the eopposite side and the product is formed.

• The tert-butylbromide follows the $S_{N} 1$ mechanism because it forms a highly stable tert-butyl carbocation intermediate, which is further attacked by the ${ }^{-} OH$ ion. The stability of the tert-butyl carbocation makes the $S_{N} 1$ mechanism favorable.

• The $n$-butylbromide does not follow the $S_{N} 1$ mechanism because it forms a primary carbocation, which is the least stable. Instead, it follows the $S_{N} 2$ mechanism due to minimal steric hindrance, allowing the ${ }^{-} OH$ ion to attack from the backside, forming a transition state and pushing off the leaving group from the opposite side.

Answer

Reaction between the isobutylene added to $HCl$

Electrophilic addition reaction takes place in accordance with Markownikoff’s rule.

We know that $3^{\circ}$ carbocation is more stable than $1^{\circ}$ carbocation because in further step $3^{\circ}$ carbocation is further attacked by $Cl^{-}$ion.

• The other options are incorrect because they do not follow Markownikoff’s rule, which states that in the addition of HX to an alkene, the hydrogen atom (H) attaches to the carbon with the greater number of hydrogen atoms (the less substituted carbon), and the halide (X) attaches to the carbon with fewer hydrogen atoms (the more substituted carbon). This leads to the formation of the more stable carbocation intermediate.
• The other options may involve the formation of less stable carbocations, such as primary or secondary carbocations, which are not favored over the more stable tertiary carbocation formed in the correct mechanism.
• The other options might suggest anti-Markownikoff addition, which is not applicable in this case as the reaction conditions do not favor such a mechanism (e.g., no presence of peroxides).
• The other options might involve incorrect regioselectivity or stereochemistry that does not align with the observed product formation in electrophilic addition reactions.

Answer

In haloarenes, carbon of benzene is bonded to halogen. Electronegativity of halogen is more than that of $s p^{2}$ hybridised carbon of benzene ring. So, $C-X$ bond is a polar bond. Apart from this, lone pair of electrons of halogen atom are involved in resonance with benzene ring. So, this $C-X$ bond has acquire partial bond character.

This $C-X$ bond of haloarenes is less polar than $C-X$ bond of haloalkanes. This is supported by the fact that dipole moment of chlorobenzene $(\mu=1.69 D)$ is little lower than that of $CH_3 Cl(\mu=1.83 D)$

• The other options are not provided in the input, so it is not possible to determine why they are incorrect.

Answer

Ethanol is treated with red phosphorous and bromine mixture and the product formed will be bromoethane. The bromoethane so formed is then treated with Nal to give iodoethane.

$$ \begin{aligned} & CH_3 CH_2 OH \xrightarrow{\text { Red } P / Br_2} CH_3 CH_2 Br \\ & CH_3 CH_2 Br \xrightarrow{Nal} CH_3 CH_2 I+NaBr \end{aligned} $$

This reaction is known as Finkelstein reaction.

• The question specifies that no other iodine-containing reagent except $NaI$ is available in the laboratory. Therefore, any method that requires other iodine-containing reagents is not applicable.
• Direct iodination of ethanol using $NaI$ is not feasible because $NaI$ alone cannot convert ethanol to iodoethane.
• Using other halogenating agents like $I_2$ or $HI$ is not an option since they are not available as per the given constraints.
• The use of other halogen sources like $Cl_2$ or $F_2$ followed by halogen exchange reactions is not mentioned and may not be practical or efficient for this specific transformation.
• The Finkelstein reaction specifically requires a halogen exchange process, which is why the intermediate bromoethane is necessary. Direct conversion of ethanol to iodoethane without forming an intermediate halide is not feasible with the given reagents.

Answer

Cyanide ion $(\overline{ } C \equiv N)$ is an ambident nucleophile because it can react either through carbon or through nitrogen. Since, $C-C$ bond is stronger than $C-N$ bond so, cyanide ion will mainly attack through carbon to form alkyl cyanide.

• The statement that “C-C bond is stronger than C-N bond” is incorrect. The strength of the bond is not the primary reason for the nucleophilic behavior of cyanide ion in aqueous medium.
• The nucleophilicity of cyanide ion in aqueous medium is influenced by solvation effects. In water, the carbon end of the cyanide ion is less solvated compared to the nitrogen end, making it more nucleophilic.
• The electronic structure and hybridization of the cyanide ion also play a role. The carbon atom in cyanide ion has a lone pair in an sp hybrid orbital, which is more directional and thus more nucleophilic compared to the nitrogen atom’s lone pair in an sp hybrid orbital.
• The steric hindrance around the nitrogen atom can also reduce its nucleophilicity compared to the carbon atom.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(4)$

C. $\rightarrow(1)$

D. $\rightarrow(2)$

A. Chloramphenicol is a broadspectrum antibiotic. It is used in the treatment of typhoid fever.

B. Thyroxine is a hormone secreated by thyroid gland. Execessive secretion of thyroxine in the body is known as hyperthyroidism. Most patient with hyper thyroidism have an enlarged thyroid gland i.e., goitre.

C. Chloroquine prevents the development of malaria parasite plasmodium vivax in the blood.

D. IUPAC name of chloroform is trichloromethane with formula $CHCl_3$. It is a colourless, volatile, sweet-smelling liquid. Its vapours depresses the central nervous system and used as an anaesthetic.

• Chloramphenicol (A)

  • Incorrect for Malaria (1): Chloramphenicol is not used for treating malaria; it is primarily used for bacterial infections like typhoid fever.
  • Incorrect for Anaesthetic (2): Chloramphenicol is an antibiotic, not an anaesthetic.
  • Incorrect for Goiter (4): Chloramphenicol does not affect thyroid function or treat goiter.
  • Incorrect for Blood substituent (5): Chloramphenicol is not used as a blood substitute.

• Thyroxine (B)

  • Incorrect for Malaria (1): Thyroxine is a hormone, not an antimalarial drug.
  • Incorrect for Anaesthetic (2): Thyroxine does not have anaesthetic properties.
  • Incorrect for Typhoid fever (3): Thyroxine is not used to treat bacterial infections like typhoid fever.
  • Incorrect for Blood substituent (5): Thyroxine is not used as a blood substitute.

• Chloroquine (C)

  • Incorrect for Anaesthetic (2): Chloroquine is an antimalarial drug, not an anaesthetic.
  • Incorrect for Typhoid fever (3): Chloroquine is not used to treat bacterial infections like typhoid fever.
  • Incorrect for Goiter (4): Chloroquine does not affect thyroid function or treat goiter.
  • Incorrect for Blood substituent (5): Chloroquine is not used as a blood substitute.

• Chloroform (D)

  • Incorrect for Malaria (1): Chloroform is an anaesthetic, not an antimalarial drug.
  • Incorrect for Typhoid fever (3): Chloroform is not used to treat bacterial infections like typhoid fever.
  • Incorrect for Goiter (4): Chloroform does not affect thyroid function or treat goiter.
  • Incorrect for Blood substituent (5): Chloroform is not used as a blood substitute.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(5)$

C. $\rightarrow(1)$

D. $\rightarrow$ (2)

E. $\rightarrow(4)$

A. A mixture containing two enantiomers in equal proportions will have zero optical rotation, such a mixture is known as racemic mixture. The process of conversion of enantiomer into a racemic mixture is known as racemisation. If an alkyl halide follows $S_{N} 1$ mechanism then racemisation takes place while if it follows $S_{N} 2$ mechanism than inversion takes places.

B. Chlorobromocarbons are used in fire extinguishers.

C. In vicinal dihalides, halogen atoms are present on the adjacent carbon atom. Bromination of alkenes will give vicinal dihalides.

D. Alkylidene halides are named as gem-dihalides. In gem-dihalides halogen atoms are present on same carbon atom.

E. Elimination of $HX$ from alkylhalide follows Saytzeff rule. This rule states that “in dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms”.

• A. The other options are incorrect because:

  • (1) vic-dibromides: This is related to the bromination of alkenes, not $S_{N}$ 1 reactions.
  • (2) gem-dihalides: This is related to alkylidene halides, not $S_{N}$ 1 reactions.
  • (4) Saytzeff rule: This is related to the elimination of HX from alkylhalides, not $S_{N}$ 1 reactions.
  • (5) Chlorobromocarbons: This is related to chemicals in fire extinguishers, not $S_{N}$ 1 reactions.

• B. The other options are incorrect because:

  • (1) vic-dibromides: This is related to the bromination of alkenes, not chemicals in fire extinguishers.
  • (2) gem-dihalides: This is related to alkylidene halides, not chemicals in fire extinguishers.
  • (3) Racemisation: This is related to $S_{N}$ 1 reactions, not chemicals in fire extinguishers.
  • (4) Saytzeff rule: This is related to the elimination of HX from alkylhalides, not chemicals in fire extinguishers.

• C. The other options are incorrect because:

  • (2) gem-dihalides: This is related to alkylidene halides, not bromination of alkenes.
  • (3) Racemisation: This is related to $S_{N}$ 1 reactions, not bromination of alkenes.
  • (4) Saytzeff rule: This is related to the elimination of HX from alkylhalides, not bromination of alkenes.
  • (5) Chlorobromocarbons: This is related to chemicals in fire extinguishers, not bromination of alkenes.

• D. The other options are incorrect because:

  • (1) vic-dibromides: This is related to the bromination of alkenes, not alkylidene halides.
  • (3) Racemisation: This is related to $S_{N}$ 1 reactions, not alkylidene halides.
  • (4) Saytzeff rule: This is related to the elimination of HX from alkylhalides, not alkylidene halides.
  • (5) Chlorobromocarbons: This is related to chemicals in fire extinguishers, not alkylidene halides.

• E. The other options are incorrect because:

  • (1) vic-dibromides: This is related to the bromination of alkenes, not the elimination of HX from alkylhalides.
  • (2) gem-dihalides: This is related to alkylidene halides, not the elimination of HX from alkylhalides.
  • (3) Racemisation: This is related to $S_{N}$ 1 reactions, not the elimination of HX from alkylhalides.
  • (5) Chlorobromocarbons: This is related to chemicals in fire extinguishers, not the elimination of HX from alkylhalides.

Answer

A. $\rightarrow(2)$

B. $\rightarrow(4)$

C. $\rightarrow(1)$

D. $\rightarrow(3)$

A. In alkyl halide, halogen atom is bonded to $s p^{3}$ hybridised carbon atom, which may be further bonded to one, two or three alkyl group, i.e., $CH_3-CH(X)-CH_3$

B. Allyl halides are the compounds in which the halogen atom is bonded to $s p^{3}$ hybridised carbon atom next to carbon-carbon double bond. i.e., $CH_{2-}=CH-CH_2-X$

C. Aryl halides are the compounds in which the halogen atom is bonded to $s p^{2}$ hybridised carbon atom of an aromatic ring, i.e., $C_6 H_5 X$

D. Vinyl halides are the compounds in which the halogen atom is bonded to an $s p^{2}$ hybridised carbon atom of a carbon-carbon double bond, i.e., $CH_2=CH-X$

• For A:

  • Incorrect for (1): Aryl halides have the halogen atom bonded to an $sp^2$ hybridised carbon atom of an aromatic ring, not an $sp^3$ hybridised carbon atom.
  • Incorrect for (3): Vinyl halides have the halogen atom bonded to an $sp^2$ hybridised carbon atom of a carbon-carbon double bond, not an $sp^3$ hybridised carbon atom.
  • Incorrect for (4): Allyl halides have the halogen atom bonded to an $sp^3$ hybridised carbon atom next to a carbon-carbon double bond, not just any $sp^3$ hybridised carbon atom.

• For B:

  • Incorrect for (1): Aryl halides have the halogen atom bonded to an $sp^2$ hybridised carbon atom of an aromatic ring, not an $sp^3$ hybridised carbon atom next to a double bond.
  • Incorrect for (2): Alkyl halides have the halogen atom bonded to an $sp^3$ hybridised carbon atom, not specifically next to a carbon-carbon double bond.
  • Incorrect for (3): Vinyl halides have the halogen atom bonded to an $sp^2$ hybridised carbon atom of a carbon-carbon double bond, not an $sp^3$ hybridised carbon atom next to a double bond.

• For C:

  • Incorrect for (2): Alkyl halides have the halogen atom bonded to an $sp^3$ hybridised carbon atom, not an $sp^2$ hybridised carbon atom of an aromatic ring.
  • Incorrect for (3): Vinyl halides have the halogen atom bonded to an $sp^2$ hybridised carbon atom of a carbon-carbon double bond, not an $sp^2$ hybridised carbon atom of an aromatic ring.
  • Incorrect for (4): Allyl halides have the halogen atom bonded to an $sp^3$ hybridised carbon atom next to a carbon-carbon double bond, not an $sp^2$ hybridised carbon atom of an aromatic ring.

• For D:

  • Incorrect for (1): Aryl halides have the halogen atom bonded to an $sp^2$ hybridised carbon atom of an aromatic ring, not an $sp^2$ hybridised carbon atom of a carbon-carbon double bond.
  • Incorrect for (2): Alkyl halides have the halogen atom bonded to an $sp^3$ hybridised carbon atom, not an $sp^2$ hybridised carbon atom of a carbon-carbon double bond.
  • Incorrect for (4): Allyl halides have the halogen atom bonded to an $sp^3$ hybridised carbon atom next to a carbon-carbon double bond, not an $sp^2$ hybridised carbon atom of a carbon-carbon double bond.

Answer

A. $\rightarrow(2)$

B. $\rightarrow(4)$

C. $\rightarrow(5)$

D. $\rightarrow(1)$

E. $\rightarrow(3)$

A. In this reaction, an electrophile $Cl^{+}$attacks on to the benzene ring and substitution takes place.

B. In this reaction, addition of $HBr$ takes place on to the doubly bonded carbons of propene in accordance with Markownikoff’s rule and electrophilic addition takes place.

C. In this reaction, the reactant is secondary halide. Here, halogen atom is substituted by hydroxy ion. As it is secondary halide so it follows $S_{N} 1$ mechanism.

D. In this reaction, halogen atom is directly bonded to aromatic ring. So, It is nucleophilic aromatic substitution as ${ }^{-} OH$ group has substituted halogen of given compound.

E. It is an elimination reaction. It follows Saytzeff elimination rule.

• For option A:

  • Incorrect because it is not a nucleophilic substitution reaction. The reaction involves an electrophile ($Cl^+$) attacking the benzene ring, which is characteristic of an electrophilic aromatic substitution.

• For option B:

  • Incorrect because it is not a nucleophilic substitution reaction. The reaction involves the addition of $HBr$ to propene, following Markownikoff’s rule, which is an electrophilic addition reaction.

• For option C:

  • Incorrect because it is not an electrophilic substitution reaction. The reaction involves the substitution of a halogen atom by a hydroxy ion in a secondary halide, following the $S_N1$ mechanism, which is a nucleophilic substitution reaction.

• For option D:

  • Incorrect because it is not an electrophilic substitution reaction. The reaction involves the substitution of a halogen atom directly bonded to an aromatic ring by a nucleophile (${-}OH$), which is a nucleophilic aromatic substitution.

• For option E:

  • Incorrect because it is not a substitution reaction. The reaction involves the elimination of a molecule following Saytzeff’s rule, which is characteristic of an elimination reaction.

Answer

$A \rightarrow(1)$

B. $\rightarrow$(2)

C. $\rightarrow$(3)

D. $\rightarrow$(4)

A. The IUPAC name of compound $(A)$ is 4-bromopent-2-ene.

B. The IUPAC name of compound (B) is 4-bromo-3-methyl pent-2-ene.

C. The IUPAC name of compound $(C)$ is 1-bromo-2- methyl but-2-ene.

D. The IUPAC name of compound (D) is 1-bromo-2-methylpent-2-ene.

• For compound (A), if it were named incorrectly, it might be due to misidentifying the position of the bromine or the double bond. The correct name is 4-bromopent-2-ene, indicating the bromine is on the 4th carbon and the double bond starts at the 2nd carbon.

• For compound (B), an incorrect name might arise from not recognizing the methyl group on the 3rd carbon or misplacing the bromine. The correct name is 4-bromo-3-methylpent-2-ene, showing the bromine on the 4th carbon and the methyl group on the 3rd carbon.

• For compound (C), errors could occur by not correctly identifying the positions of the bromine and methyl groups or the double bond. The correct name is 1-bromo-2-methylbut-2-ene, indicating the bromine on the 1st carbon, the methyl group on the 2nd carbon, and the double bond starting at the 2nd carbon.

• For compound (D), mistakes might involve misplacing the bromine or methyl group or the double bond. The correct name is 1-bromo-2-methylpent-2-ene, showing the bromine on the 1st carbon, the methyl group on the 2nd carbon, and the double bond starting at the 2nd carbon.

Answer

A. $\rightarrow(2)$

B. $\rightarrow(1)$

C. $\rightarrow(4)$

D. $\rightarrow$ (3)

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

B. Aryl halides give analogous compounds when treated with sodium in dry ether, in which two aryl groups are joined together. It is called Fittig reaction.

C. Diazonium salt when treated with cuprous chloride or cuprous bromide gives chlorobenzene or bromobenzene.

The reaction is known as Sandmeyer’s reaction.

D. Alkyl iodides are prepared by the reaction of alkyl chlorides with sodium iodide in dry acetone. The reaction is known as Finkelstein reaction.

• For A:

  • Incorrect for (1): The reaction described in (1) is the Fittig reaction, which involves the coupling of aryl halides, not a mixture of alkyl and aryl halides.
  • Incorrect for (3): The reaction described in (3) is the Finkelstein reaction, which involves the exchange of halides in alkyl halides, not the formation of alkylarenes.
  • Incorrect for (4): The reaction described in (4) is the Sandmeyer’s reaction, which involves the conversion of diazonium salts to aryl halides, not the formation of alkylarenes.

• For B:

  • Incorrect for (2): The reaction described in (2) is the Wurtz-Fittig reaction, which involves the coupling of an alkyl halide and an aryl halide, not just aryl halides.
  • Incorrect for (3): The reaction described in (3) is the Finkelstein reaction, which involves the exchange of halides in alkyl halides, not the coupling of aryl halides.
  • Incorrect for (4): The reaction described in (4) is the Sandmeyer’s reaction, which involves the conversion of diazonium salts to aryl halides, not the coupling of aryl halides.

• For C:

  • Incorrect for (1): The reaction described in (1) is the Fittig reaction, which involves the coupling of aryl halides, not the conversion of diazonium salts to aryl halides.
  • Incorrect for (2): The reaction described in (2) is the Wurtz-Fittig reaction, which involves the coupling of an alkyl halide and an aryl halide, not the conversion of diazonium salts to aryl halides.
  • Incorrect for (3): The reaction described in (3) is the Finkelstein reaction, which involves the exchange of halides in alkyl halides, not the conversion of diazonium salts to aryl halides.

• For D:

  • Incorrect for (1): The reaction described in (1) is the Fittig reaction, which involves the coupling of aryl halides, not the exchange of halides in alkyl halides.
  • Incorrect for (2): The reaction described in (2) is the Wurtz-Fittig reaction, which involves the coupling of an alkyl halide and an aryl halide, not the exchange of halides in alkyl halides.
  • Incorrect for (4): The reaction described in (4) is the Sandmeyer’s reaction, which involves the conversion of diazonium salts to aryl halides, not the exchange of halides in alkyl halides.

Answer

(b) Assertion and reason both are wrong statements.

Correct Assertion Thionyl chloride is preferred over $PCl_3$ and $PCl_5$ for the preparation of alkyl chlorides from alcohols.

Correct Reason Thionyl chloride gives pure alkyl halide as other two products $(SO_2+HCl)$ are escapable gases.

Answer

(e) Assertion and reason both are correct statements but reason is not the correct explanation of assertion. For the same hydrocarbon part boiling point depends upon the atomic mass of halogen atom. Higher the mass of the halogen atom, higher will be the boiling point.

So, we can say that boiling point decreases with decrease in atomic mass of halogen atom.

Answer

(d) Assertion is wrong but reason is correct statement.

Correct Assertion $KCN$ reacts with methyl chloride to give the mixture of methyl cyanide and methyl isocyanide in which methyl cyanide predominates because of stable $C _C$ bond in methyl cyanide.

Answer

(d) Assertion is wrong but reason is correct statement.

Correct Assertion- sec-butyl bromide ( $2^{\circ}$ alkyl halide) undergoes Wurtz reaction to give 2,5- dimethylhexane.

Answer

(a) Assertion and reason both are correct and reason is correct explanation of assertion. Presence of nitro group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution because $-NO_2$ group, being an electron withdrawing group decreases the electron density over the benzene ring.

Answer

(e) Assertion and reason both are correct statements but reason is not the correct explanation of assertion.

Correct explanation in monohaloarenes, halogen atom increases the electron density at ortho and para position. So, further electrophilic substitution occurs at ortho and para positions.

Answer

(c) Assertrion is correct but reason is wrong statement.

Correct Reason Oxidising agent oxidises $HI$ into $I_2$ to prevent the possibility of backward reaction.

Answer

(a) Assertion and reason both are correct and reason is the correct explanation of assertion. It is difficult to replace chlorine by $-OH$ in chlorobenzene in comparision to that in chloroethane because $C-Cl$ bond in chlorobenzene has a partial bond character due to resonance.

Answer

(c) Assertion is correct but reason is wrong statement.

Correct Reason This reaction proceeds through $S_{N} 2$ mechanism, in which ${ }^{-} OH$ ion attacks at $180^{\circ}$ to the halogen atom of 2-bromooctane which leads to the inversion of configuration

Answer

(d) Assertion is wrong but reason is correct statement.

Correct Assertion Chlorination of nitrobenzene leads to the formation of $m$-nitrochlorobenzene because $-NO_2$ group deactivates the ring because it is meta directing.

Answer

Primary alkyl halides follow $S_{N} 2$ mechanism in which a nucleophile attacks at $180^{\circ}$ to the halogen atom. A transition state is formed in which carbon is bonded to two nucleophiles and finally halogen atom is pushed out. In $S_{\mathbb{N}} 2$ mechanism, substitution of nucleophile takes place as follows

Thus, in $S_{N} 2$ mechanism, substitution takes place. Tertiary alkyl halides follow $S_{N} 1$ mechanism. In this case, tert alkyl halides form $3^{\circ}$ carbocations. Now, if the reagent used is a weak base then substitution occur while if it is a strong base than instead of substitution elimination occur.

Here, the reagent used is aq. $KOH$. It is a weak base so, substitution takes place.

As alc. $KOH$ is a strong base, so elimination competes over substitution and alkene is formed.

Answer

Some halogen containing compounds are useful in daily life are as follows

Dichloromethane It is used as a solvent as a paint remover, as a propellant in aerosols, and as a process solvent in the manufacture of drugs. It is also used as a metal cleaning and finishing solvent.

Trichloromethane It is employed as a solvent for fats, alkaloids, iodine and other substances.

Triodomethane It is used as an antiseptic. Now, it has been replaced by some other compounds because of its objectionable smell.

But some compounds of this class are responsible for exposure of flora and fauna to more and more of UV light which causes destruction to great extent.

These are as follows

(i) Tetrachloromethane When carbon tetrachloride is released into the air, it rises to the atmosphere and depletes the ozone layer. Depletion of the ozone layer is believed to increase human exposure to UV rays leading to increased skin cancer, eye diseases and disorders, and possible disruption of the immune system. These UV rays cause damage to plants, and reduction of plankton populations in the ocean’s photic zone.

(ii) Freons Freon-113 is likely to remain in the air long enough to reach the upper atomsphere. Here, it provides chlorine atoms which damage the ozone layer. Because of this depletion $U V$ rays enters in our atmosphere and become responsible for damage to great extent.

(iii) $p$ - $p^{\prime}$-Dichlorodiphenyl trichloroethane(DDT)

DDT is not completely biodegradable. Instead, it gets deposited in fatty tissues. If ingestion continues for a long time, DDT builds up within the animal and effect the reproductive system.

To minimise the harmful impacts of these compounds i.e., freons, hydrofluorocarbons, fluorocarbons and hydrocarbons can be straight used to make refrigerants and air-conditioning equipments. They are stable in the stratosphere and secure for flora and fauna.

Answer

Aryl halides are less reactive towards nucleophilic substitution reaction due to the following reasons

(i) In haloarenes, the lone pair of electron on halogen are in resonance with benzene ring. So, $C-Cl$ bond acquires partial double bond character which strengthen $C-Cl$ bond. Therefore, they are less reactive towords nucleophilic substitution reaction.

(ii) In haloarenes, the carbon atom attached to halogen is $s p^{2}$ hybridised. The $s p^{2}$ hybridised carbon is more electronegative than $s p^{3}$ hybridised carbon. This $s p^{2}$-hybridised carbon in haloarenes can hold the electron pair of $C-X$ bond more tightly and make this $C-Cl$ bond shorter than $C-Cl$ bond of haloalkanes.

Since, it is difficult to break a shorter bond than a longer bond, therefore, haloarenes are less reactive than haloarenes.

In haloarenes, the phenyl cation will not be stabilised by resonance therefore $S_{N} 1$ mechanism ruled out.

(iv) Because of the repulsion between the nucleophile and electron rich arenes, aryl halides are less reactive than alkyl halides.

The reactivity of aryl halides can be increased by the presence of an electron withdrawing group $(-(NO_2).$ at ortho and para positions. However, no effect on reactivity of haloarenes is observed by the presence of electron withdrawing group at meta-position. Mechanism of the reaction is as depicted with ${ }^{-} OH$ ion.

From the above resonance, it is very clear that electron density is rich at ortho and para positions. So, presence of EWG will facilitate nucleophilic at ortho and para postitions not on meta position.