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.

$$ CH_3-CH_2-CH_2-OH \longrightarrow CH_3-CH_2-CH_2^{+}+OH^{-} $$

In this case, $1^{\circ}$ carbocation is formed. It is least stable. So, here $S_{N} 1$ mechanism is not favourable.

The reaction proceeds with a $2^{\circ}$ carbocation. So this reaction will be faster in comparison to previous reaction which involves $1^{\circ}$ carbocation.

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

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.

Answer

(d) When alcohols are treated with conc., $HCl$ at room temperature then 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.

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.

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 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 electrophillic aromatic substitutuion, 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.

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

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).

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

Hence, option (a) is correct.

Options (b), (c) and (d) are the same molecules and are similar to the given molecule.

Answer

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

The structure of the given compound are given below:

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 a gem-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.

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.,

$CH_3 CH_2-CH_2CH(Br)(CH_3)$

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.

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

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.

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.

Answer

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

(a) $S_{N} 1$ 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.

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.

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.

Answer

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

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.

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.

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 $S_{N} 1$ mechanism during nucleophilic substitution reaction form racemic mixture.

The compound given in option (2) will form a tertiary carbocation following the $S_{N} 1$ mechanism. So, the product will be racemic mixture.

Answer

(c) The bond formed between carbon 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 have 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).

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.

Answer

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.

(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).

Answer

(c) Presence of electron releasing group at ortho 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).

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) correct.

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.

Answer

Options (a) and (c) are correct because:

(a) Both $OH^{-}$ and $Cl^-$ are electron rich as they carry a negative charge. So, they will act as nucleophile

(c) In transition state of $S_N2$ mechanism, the carbon atom is $sp^2$ hybridised due to its planar structure. At this point, carbon almost acquires ‘pentavalency’ with three full bonds and two ‘partial’ bonds, and a planar complex is formed, which is $sp^2$ hybridised.

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.

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.

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.

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.

Answer

$(a, d)$

In the structure of 2-bromopentane and trichloromethane , 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.

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) $\underset{\text{Ethylidene chloride}}{\mathrm{CH_3-CHCl_2}}$

(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.

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 (d) 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.

Answer

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

$(a, d)$

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.

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

(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 $H_2 SO_4+Kl$.

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 $AgF, Hg_2 F_2, CoF_2$.

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

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

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.

Answer

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

• 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.

Let’s 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.

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.

Answer

A mixture of $NaBr$ and $H_2SO_4$ gives $Br_2$ gas as a product. Molecule (ii) will not react with $Br_2$ gas because of the stable molecule that is formed due to resonance stabilization.

Answer

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.

$CH_3 CH(I)CH_3$ 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{ | \ \normalsize{\mathrm{I}}}}{C}{H}-CH_3} $

Answer

• 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.

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{CH_3-\underset{\substack{| \\ \ \small {\mathrm{Br}}}}{C}{H}-CH=CH-CH_3}} $

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

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

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

(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

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:

Answer

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

$\mathrm{H_3 C-CH_2-CH = CH-CH_3 +H Cl} \longrightarrow \underset{\text { (2-chloropentane) }}{\mathrm{CH_3-CH_2-CH_2-\underset{\substack{ }}{CH(Cl)}-CH_3}} + \underset{\text { (3-chloropentane) }}{\mathrm{CH_3-CH_2-\underset{}{CH(Cl)}-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 $(A)$.

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.

Answer

The structure of neo-pentylbromide is

To name the compound I-bromo-2,2-dimethylpropane, follow these steps:

1. Identify the longest carbon chain: The longest continuous chain of carbon atoms in the compound is propane, which consists of three carbon atoms.

2. Number the carbon atoms in the main chain: Number the carbon atoms in the main chain starting from the end nearest to the substituent (bromine in this case). This ensures that the substituent gets the lowest possible number. For propane, you can number the carbons as 1, 2, and 3.

3. Identify and name the substituents:

   • The bromine atom is a substituent and is named as “bromo”.
   • There are two methyl groups attached to the second carbon atom of the propane chain. Each methyl group is named as “methyl”.

4. Assign numbers to the substituents:

   • The bromine atom is attached to the first carbon atom, so it is given the number 1.
   • The two methyl groups are both attached to the second carbon atom, so they are given the number 2.

5. Combine the substituents with the main chain name:

   • List the substituents in alphabetical order, regardless of their position numbers. In this case, “bromo” comes before “methyl”.
   • Use prefixes to indicate the number of identical substituents (e.g., “di-” for two methyl groups).
   • Place the position numbers before the substituent names, separated by commas if necessary.

Putting it all together, the name of the compound is: 1-bromo-2,2-dimethylpropane

This name indicates that there is a bromine atom on the first carbon and two methyl groups on the second carbon of the propane chain.

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

Common name $\Rightarrow$ neo-pentylbromide

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

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.

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.

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} $

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.

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.

Answer

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

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

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:

Answer

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

(i) $Br_2$ in $CCl_4$ test When $Br_2 /CCl_4$ 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.

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 observed 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.

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.

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.

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

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.

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 $(I)<(II)<(III )$

Answer

Tert-butylbromide reacts with aq. $NaOH$ as follows:

tert-butylbromide when treated with aq. $NaOH$, it forms tert-carbocation 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 on the opposite side and the product is formed.

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.

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 $CH_3 Cl(\mu=1.83 D)$

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.

• 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 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.

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.

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”.

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., $CH_2=CH-X$

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.

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.

1. Identify the longest carbon chain containing the double bond:

   • The longest chain has 5 carbon atoms, so the base name is “pent”.

2. Number the carbon chain:

   • Number the chain from the end nearest the double bond to give the double bond the lowest possible number. The double bond starts at carbon 2, so we use “pent-2-ene”.

3. Identify and number the substituents:

   • There is a bromine (Br) substituent on carbon 4.

4. Combine the substituents with the base name:

   • Place the substituent name and its position number before the base name. The substituent is “bromo” and it is on carbon 4, so we get “4-bromo”.

5. Combine everything together:

   • The final IUPAC name is “4-bromopent-2-ene”.

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

1. Identify the longest carbon chain containing the double bond:

   • The longest chain that includes the double bond has 5 carbon atoms, so the base name is “pentene.”

2. Number the carbon chain:

   • Number the chain from the end nearest the double bond to give the double bond the lowest possible number. The double bond starts at carbon 2, so the base name is “pent-2-ene.”

3. Identify and number the substituents:

   • There is a bromine (Br) substituent on carbon 4.
   • There is a methyl (-CH₃) substituent on carbon 3.

4. Combine the substituents with the base name:

   • List the substituents in alphabetical order with their corresponding numbers.
   • The substituents are “4-bromo” and “3-methyl.”

5. Construct the full name:

   • Combine the substituents and the base name, ensuring the substituents are listed in alphabetical order and the position of the double bond is indicated.

The final IUPAC name is: 4-bromo-3-methylpent-2-ene.

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

1. Identify the longest carbon chain containing the double bond:

   • The longest chain containing the double bond has 4 carbon atoms, which corresponds to “butene”.

2. Number the carbon atoms in the chain:

   • Number the chain from the end nearest the double bond to give the double bond the lowest possible number. In this case, the double bond starts at carbon 2, so we number the chain from left to right: 1, 2, 3, 4.

3. Identify and number the substituents:

   • There is a bromine (Br) substituent on carbon 1.
   • There is a methyl (-CH₃) substituent on carbon 2.

4. Combine the substituents with the base name:

   • The base name is “but-2-ene” because the double bond starts at carbon 2.
   • The substituents are “1-bromo” and “2-methyl”.

5. Assemble the name in alphabetical order of the substituents:

   • “1-bromo” comes before “2-methyl” alphabetically.

Putting it all together, the IUPAC name is “1-bromo-2-methylbut-2-ene”.

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

1. Identify the longest carbon chain containing the double bond:

   • The longest chain that includes the double bond has 5 carbon atoms, so the base name is “pentene”.

2. Number the carbon chain:

   • Number the chain from the end nearest the double bond to give the double bond the lowest possible number. The double bond starts at carbon 2, so it is “pent-2-ene”.

3. Identify and number the substituents:

   • There is a bromine (Br) substituent on carbon 1, so it is “1-bromo”.
   • There is a methyl (-CH₃) substituent on carbon 2, so it is “2-methyl”.

4. Combine the substituents with the base name:

   • Combine the substituents in alphabetical order with the base name, ensuring to include the position numbers.

Putting it all together, the IUPAC name is: 1-bromo-2-methylpent-2-enes

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 a halogen exchange reaction and is known as Finkelstein reaction.

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.

$$ \mathrm{R}-\mathrm{OH}+\mathrm{SOCl}_2 \longrightarrow \mathrm{R}-\mathrm{Cl}+\mathrm{SO}_2+\mathrm{HCl} $$

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 $KCN$ is ionic in nature.

Answer

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

Correct Assertion- tert-butyl bromide undergoes elimination when treated with sodium in dry ether.

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 Comparison 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.

Triiodomethane: 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) $p$ - $p^{\prime}$-Dichlorodiphenyltrichloroethane(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.

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

(v) 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 positions not on meta position.