Answer

(a) The structure

While writing IUPAC name, alkyl groups are written in alphabetical priority, thus lower locant 3 is assigned to ethyl.

Note Prefix di, tri, tetra are not included in alphabetical order.

• (b) 4,4-dimethyl-3-ethylheptane: The name does not follow the alphabetical order rule for substituents. “Ethyl” should be given a lower locant number than “dimethyl” because “ethyl” comes before “dimethyl” alphabetically.

• (c) 5-ethyl-4,4-dimethylheptane: The numbering of the carbon chain is incorrect. The substituents should be given the lowest possible locant numbers. In this case, numbering from the other end of the chain would give the ethyl group a lower number (3 instead of 5).

• (d) 4,4-bis(methyl)-3-ethylheptane: The use of “bis(methyl)” is incorrect. The correct term is “dimethyl” for two methyl groups. Additionally, the alphabetical order rule is not followed, as “ethyl” should be given a lower locant number than “dimethyl”.

Answer

(d) When more than one functional group lie in the main chain, nomenclature is done according to that functional group which has higher priority.

Carboxylic acid $(-\mathrm{COOH})$ has more priority than ketone $(>\mathrm{C}=\mathrm{O})$

• (a) 1-hydroxypentane-1,4-dione: This name incorrectly suggests the presence of a hydroxyl group (-OH) and two ketone groups (diones) at positions 1 and 4. However, the compound actually contains a carboxylic acid group (-COOH) and a ketone group, not two ketone groups.

• (b) 1,4-dioxopentanol: This name implies the presence of two ketone groups (diones) at positions 1 and 4 and an alcohol group (-OH) at an unspecified position. The compound actually has a carboxylic acid group (-COOH) and a ketone group, not an alcohol group.

• (c) 1-carboxybutan-3-one: This name suggests a carboxylic acid group (-COOH) at position 1 and a ketone group at position 3 on a butane chain. However, the correct structure has a carboxylic acid group at one end and a ketone group at position 4 on a pentane chain, not a butane chain.

Answer

(b) For tri or higher substituted benzene derivatives, the compounds are named by identifying substituent, positions on the ring by following the lowest locant rule.

Substituent of the base compound is assigned number 1 and then the direction of numbering is chosen such that the next substituent gets the lowest number. The substituents appear in the name in alphabetical order.

• (a) 1-chloro-2-nitro-4-methylbenzene: This option is incorrect because the numbering does not follow the lowest locant rule. According to the lowest locant rule, the substituents should be numbered in such a way that the sum of the locants is minimized. In this case, the numbering should start from the chlorine substituent, and the correct numbering would be 1-chloro, 2-nitro, and 4-methyl, which does not match the given name.

• (c) 2-chloro-1-nitro-5-methylbenzene: This option is incorrect because the substituents are not listed in alphabetical order. According to IUPAC nomenclature rules, substituents should be listed in alphabetical order regardless of their position on the ring. The correct name should list “chloro” before “methyl” and “nitro”.

• (d) $m$-nitro-p-chlorotoluene: This option is incorrect because it uses common names (meta and para) instead of the IUPAC systematic naming convention. IUPAC names should use numerical locants to indicate the positions of the substituents on the benzene ring. Additionally, the substituents should be listed in alphabetical order.

Answer

(c) Electronegativity of carbon atom depends on their state of hybridisation. More $s$-character more the electronegativity.

$ s p^{3}<s p^{2}<s p \\ $

$25 $%s $ 33$% s $50 $% s

Thus, $s p$-carbon has the highest electronegativity, i.e., option (c)

$\left(CH_{3}-CH_{2} -C \equiv{ }^{*} CH\right)$ is correct.

• Option (a): The carbon marked with an asterisk is in an sp³ hybridized state, which has the least s-character (25%) and therefore the lowest electronegativity among the given options.

• Option (b): The carbon marked with an asterisk is in an sp² hybridized state, which has more s-character (33%) than sp³ but less than sp. Therefore, it is not the most electronegative.

• Option (d): The carbon marked with an asterisk is in an sp² hybridized state, similar to option (b), with 33% s-character. It is more electronegative than sp³ but less than sp, making it not the most electronegative.

Answer

(c) Two or more compounds having the same molecular formula but different functional groups are called functional isomers.

Functional isomer of alcohol is ether.

Functional isomer of aldehyde is ketone.

Functional isomer of cyanide is isocyanide

However, alkyl halides do not show functional isomerism. Hence, option (c) is correct.

• Alcohols can exhibit functional group isomerism with ethers, as they have the same molecular formula but different functional groups.
• Aldehydes can exhibit functional group isomerism with ketones, as they have the same molecular formula but different functional groups.
• Cyanides can exhibit functional group isomerism with isocyanides, as they have the same molecular formula but different functional groups.

Answer

(d) As we know, essential oils are insoluble in water and have high vapour-pressure at $373 \mathrm{~K}$ but are miscible with water-vapour in vapour phase, it means these are steam volatile. Hence, steam distillation technique is used for the extraction of essential oils.

• (a) Distillation: This method is typically used for separating components based on differences in their boiling points. However, since essential oils are steam volatile and not soluble in water, regular distillation would not be effective in separating them from the flowers.

• (b) Crystallisation: This technique is used to purify solid compounds based on their solubility differences in a solvent. Essential oils are liquid and do not crystallize, making crystallisation an unsuitable method for their extraction.

• (c) Distillation under reduced pressure: This method is used to distill compounds with high boiling points at lower temperatures by reducing the pressure. While it can be used for heat-sensitive compounds, it is not specifically suited for separating steam volatile compounds that are immiscible with water, like essential oils. Steam distillation is more appropriate for this purpose.

Answer

(d) Thin layer chromatography (TLC) is an another type of adsorption which involves separation of substances of a mixture over a thin layer of an adsorbent coated on a glass plate.

A thin layer of an adsorbent is spread over a glass plate and glass plate is placed in an eluant. As eluant rises, components of the mixture move up along with the eluant to different distances depending on their degree of adsorption and separation takes place. Therefore, this TLC technique will give best results in identifying the different types of ink used at different places in the documents.

• Column chromatography: While column chromatography is effective for separating mixtures, it is more suited for larger scale separations and requires more sample preparation. It is not as efficient or practical for analyzing small samples like ink spots on documents.

• Solvent extraction: Solvent extraction is a technique used to separate compounds based on their solubility in two different immiscible liquids. It is not suitable for analyzing inks on documents because it does not provide the necessary resolution to distinguish between different inks.

• Distillation: Distillation is a method used to separate components based on differences in their boiling points. It is not applicable for analyzing inks on documents as it is designed for separating liquid mixtures and not for analyzing solid samples or thin layers of substances.

Answer

(b) Partition chromatography is based on continuous differential partioning of components of a mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography.

• Adsorption: Paper chromatography primarily relies on the partitioning of substances between the stationary phase (water trapped in the fibers of the paper) and the mobile phase (solvent). Adsorption chromatography, on the other hand, involves the adhesion of substances onto the surface of the stationary phase, which is not the main mechanism in paper chromatography.

• Solubility: While solubility plays a role in the movement of substances in paper chromatography, it is not the primary principle. The main principle is the differential partitioning between the stationary and mobile phases, rather than just the solubility of substances in the solvent.

• Volatility: Volatility refers to the tendency of a substance to vaporize. Paper chromatography does not rely on the volatility of substances; instead, it depends on the partitioning of substances between the stationary and mobile phases. Volatility is more relevant in techniques like gas chromatography.

Answer

(a) Stability of the given cations can be understood by the following structures

Hence, the stability of carbocation decreases

$$ |>|>|| \mid $$

• Option (b) II > III > I: This option is incorrect because cation III has an additional -OCH₃ group attached to the carbon chain, which provides a +M (mesomeric) effect, stabilizing the cation more than cation I. However, cation II has the -OCH₃ group directly attached to the positively charged carbon, providing a stronger +M effect and making it more stable than both I and III. Therefore, II should be more stable than III, but III should not be more stable than I.

• Option (c) III > I > II: This option is incorrect because cation II, with the -OCH₃ group directly attached to the positively charged carbon, is the most stabilized due to the strong +M effect. Cation III, although it has an -OCH₃ group, it is further away from the positively charged carbon, providing less stabilization compared to cation II. Therefore, II should be more stable than both I and III, making this order incorrect.

• Option (d) I > II > III: This option is incorrect because cation II, with the -OCH₃ group directly attached to the positively charged carbon, is more stabilized due to the strong +M effect compared to cation I, which only has alkyl groups providing hyperconjugation and inductive effects. Therefore, II should be more stable than I, making this order incorrect.

Answer

(b)

• (a) 2-ethyl-3-methylpentane: This name is incorrect because the longest carbon chain in the given structure is six carbons long, not five. Therefore, the base name should be “hexane” rather than “pentane.”

• (c) 2-sec-butylbutane: This name is incorrect because the longest carbon chain in the given structure is six carbons long, not four. Therefore, the base name should be “hexane” rather than “butane.” Additionally, the substituent “sec-butyl” is not present in the structure.

• (d) 2, 3-dimethylbutane: This name is incorrect because the longest carbon chain in the given structure is six carbons long, not four. Therefore, the base name should be “hexane” rather than “butane.”

Answer

(a) Electronegativity of $\mathrm{Cl}, \mathrm{Br}, \mathrm{C}$ and $\mathrm{Mg}$ follows the order $\mathrm{Cl}>\mathrm{Br}>\mathrm{C}>\mathrm{Mg}$

$$ \begin{aligned} & ^* CH_3 \rightarrow CH_2 \rightarrow Cl \quad \quad \quad (-I-effect)\\ & ^* CH_3 \leftarrow CH_2 \leftarrow Mg^+ Cl^- \\ & ^* CH_3 \rightarrow CH_2 \rightarrow Br \quad \quad \quad (-I-effect)\\ & ^* CH_3 CH_2 CH_3 \quad \quad \quad (+I-effect)\\ & \text -/effect \quad of \quad Cl>Br . \end{aligned} $$

Hence, $CH_{3} \quad CH_{2} \quad Cl$ has the greatest positive charge.

• (b) $ ^{*} CH_{3}-CH_{2}-Mg^{+} Cl^{-}$: In this compound, the carbon marked with an asterisk is bonded to a magnesium atom, which is less electronegative than carbon. This results in the carbon having a partial negative charge rather than a positive charge because the electron density is pulled towards the carbon from the magnesium.

• (c) $ ^{*} CH_{3}-CH_{2}-Br$: Although bromine is more electronegative than carbon, it is less electronegative than chlorine. Therefore, the inductive effect (-I effect) of bromine is weaker than that of chlorine, resulting in a smaller positive charge on the carbon marked with an asterisk compared to the chlorine-containing compound.

• (d) $ ^{*} CH_{3}-CH_{2}-CH_{3}$: In this compound, the carbon marked with an asterisk is bonded to another carbon atom, which has a similar electronegativity. There is no significant inductive effect causing a positive charge on the carbon. In fact, the inductive effect from the alkyl groups is slightly electron-donating (+I effect), which would reduce any positive charge on the carbon.

Answer

(d) In all the given carbocations, the negative charge is dispersed which stabilises these carbocations. Here, the negative charge is dispersed by two factors, i.e., $+R$-effect of the carboxylate ion (conjugation) and $I$-effect of the halogens.

These effects are shown below in the carbocations

(a)

(b)

(c)

As it is clearly evident from the above structures, that $+R$-effect is common in all the four structures, therefore, overall dispersal of negative charge depends upon the number of halogen atoms and electronegativity. Since, $\mathrm{F}$ has the highest electronegativity and two F-atoms are present in option (d), thus, dispersal of negative charge is maximum in option (d).

Note In above structure (a), methyl group $\left(\mathrm{CH}_{3}\right)$ increases the density on $\mathrm{C}$-atom.

• Option (a): The presence of a methyl group ($\mathrm{CH}_{3}$) increases the electron density on the carbon atom, which reduces the dispersal of the negative charge and thus decreases the stability of the carboxylate ion.

• Option (b): This structure has only one fluorine atom, which provides some dispersal of the negative charge due to its high electronegativity, but it is less effective compared to option (d) which has two fluorine atoms.

• Option (c): This structure has a chlorine atom, which is less electronegative than fluorine. Therefore, the dispersal of the negative charge is less effective compared to the structure with fluorine atoms.

Answer

(c) When electrophile attacks $CH_{3}-CH=CH_{2}$ delocalisation of electrons can take place, in two possible ways

As $2^{\circ}$ carbocation is more stable than $1^{\circ}$ carbocation thus first addition is more feasible.

Note Stability of carbocations is the basis of Markownikoffs rule.

• (a) $2^{\circ}$ carbanion: This option is incorrect because the intermediate formed in the first step of an electrophilic addition reaction is a carbocation, not a carbanion. A carbanion would result from the addition of a nucleophile, not an electrophile.

• (b) $1^{\circ}$ carbocation: This option is incorrect because, although a carbocation is formed, the $2^{\circ}$ carbocation is more stable than the $1^{\circ}$ carbocation. The stability of carbocations follows the order $3^{\circ} > 2^{\circ} > 1^{\circ}$ due to hyperconjugation and inductive effects.

• (d) $1^{\circ}$ carbanion: This option is incorrect for the same reason as option (a). The intermediate formed in the first step of an electrophilic addition reaction is a carbocation, not a carbanion.

Answer

(b) Arrow denotes the direction of movement of electrons

Since, $\mathrm{Br}$ is more electronegative than carbon, hence heterolytic fission occurs in such a way that $\mathrm{CH}_{3}$ gets the positive charge and $\mathrm{Br}$ gets the negative charge. Thus, option (b) is correct.

• Option (a): This option shows the bond breaking in such a way that both electrons from the bond go to the carbon atom, resulting in a negatively charged carbon (carbanion) and a positively charged bromine (bromonium ion). This is incorrect because bromine is more electronegative than carbon and would attract the electrons towards itself, not the other way around.

• Option (c): This option depicts the bond breaking in a homolytic manner, where each atom gets one electron from the bond, resulting in the formation of two radicals. This is incorrect because the question specifically asks for heterolytic fission, not homolytic fission.

• Option (d): This option shows the bond breaking in such a way that both electrons from the bond go to the bromine atom, resulting in a positively charged carbon (carbocation) and a negatively charged bromine (bromide ion). However, the direction of the arrow is incorrect as it should point towards the bromine atom to indicate the movement of electrons towards the more electronegative atom.

Answer

(b)

Step I $\pi$-bonds creates an electron cloud, Electrophile $\left(\mathrm{H}^{+} \mathrm{C}\right)$ from $\mathrm{H}-\mathrm{Cl}$ attacks the electron cloud, delocalising the electrons. And, a carbocation is formed.

Step II The chloride anion attacks the carbocation.

• Option (a) is incorrect because it does not account for the formation of a carbocation intermediate, which is a crucial step in the mechanism of the addition of HCl to an alkene.
• Option (c) is incorrect because it suggests a different mechanism that does not involve the initial attack of the proton (H⁺) on the π-electron cloud of the alkene, which is essential for the formation of the carbocation intermediate.
• Option (d) is incorrect because it implies a mechanism that bypasses the formation of the carbocation intermediate, which is a necessary step for the subsequent attack by the chloride anion (Cl⁻).

Answer

$(a, d)$

Hybridisation of carbon atoms in different compounds is shown below

(a) $HC \equiv C-C \equiv CH$

$s p \quad s p \quad s p \quad s p$

(b) $CH_{3}-C \equiv C-CH_{3}$

$\begin{array}{llll}s p^{3} & s p & s p & s p^{3}\end{array}$

(c) $CH_{2}=C=CH_{2}$

$s p^{2} \quad s p \quad s p^{2}$

(d) $CH_{2}=CH-CH=CH_{2}$

$\begin{array}{lll}s p^{2} & s p^{2} \quad s p^{2} \quad s p^{2}\end{array}$

In options (a) and (d), all carbon atoms are in same hybridisation state i.e., in $s p$ and $s p^{2}$ hybridisation respectively.

• In option (b) $CH_3-C \equiv C-CH_3$, the carbon atoms are in different hybridisation states: $sp^3$ for the $CH_3$ groups and $sp$ for the carbon atoms involved in the triple bond.
• In option (c) $CH_2=C=CH_2$, the carbon atoms are in different hybridisation states: $sp^2$ for the $CH_2$ groups and $sp$ for the central carbon atom involved in the double bonds.

Answer

$(a, c, d)$

Hence, only option (b) has same spatial arrangement of group/atom as in (A), i.e.,

clocwise, while in tootof the option (a) , (c) and (d) it is different in anti-clockwise.

• Option (a): The spatial arrangement of groups/atoms in option (a) is different from that in structure ‘A’ because the sequence of the remaining groups, after bringing $\mathrm{H}$ below the plane of the paper, is anti-clockwise instead of clockwise.

• Option (c): The spatial arrangement of groups/atoms in option (c) is different from that in structure ‘A’ because the sequence of the remaining groups, after bringing $\mathrm{H}$ below the plane of the paper, is anti-clockwise instead of clockwise.

• Option (d): The spatial arrangement of groups/atoms in option (d) is different from that in structure ‘A’ because the sequence of the remaining groups, after bringing $\mathrm{H}$ below the plane of the paper, is anti-clockwise instead of clockwise.

Answer

$(b, c)$

All $AlCl_{3}, SO_{3}$ (Lewis acids), $NO_{2}^{+}, CH_{3}^{+}, CH_{3}-\stackrel{+}{C}=O$ are electron deficient species. Hence, these are electrophiles.

Direction (Q. Nos. 19-20) Consider the following four compounds.

• Option (a): $BF_3, NH_3, H_2O$

  • $NH_3$ (ammonia) and $H_2O$ (water) are not electrophiles. They are nucleophiles because they have lone pairs of electrons and can donate electrons.

• Option (d): $C_2H_5^-, \dot{C}_2H_5, C_2H_5^+$

  • $C_2H_5^-$ (ethyl anion) is not an electrophile; it is a nucleophile because it has an extra electron.
  • $\dot{C}_2H_5$ (ethyl radical) is not a typical electrophile; it is a neutral species with an unpaired electron and can act as a radical.

Answer

(b) When two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton, they are called position isomers.

• Option (a) I and II: These compounds do not differ in the position of the substituent atom or functional group on the carbon skeleton. They have the same functional group in the same position.

• Option (c) II and IV: These compounds do not differ in the position of the substituent atom or functional group on the carbon skeleton. They have the same functional group in the same position.

• Option (d) III and IV: These compounds do not differ in the position of the substituent atom or functional group on the carbon skeleton. They have the same functional group in the same position.

Answer

$(a, c)$

Two or more compounds having the same molecular formula but different functional groups are called functional isomers.

I. Aldehyde

II. Ketone

III. Ketone

IV. Aldehyde

Here, II and III; I and IV are not functional group isomers. Thus, option (a) and (c) are correct.

• Option (b) is incorrect because II and IV are functional group isomers. II is a ketone and IV is an aldehyde, and they have the same molecular formula but different functional groups.
• Option (d) is incorrect because I and II are functional group isomers. I is an aldehyde and II is a ketone, and they have the same molecular formula but different functional groups.

Answer

$(a, c)$

Nucleophile (nucleus-loving) is a chemical species that donates an electron pair to an electrophile (electron-loving). Hence, a nucleophile should have either a negative charge or an electron pair to donate. Thus, option (a) and (c) are correct.

• Option (b) is incorrect because a nucleophile is characterized by its ability to donate electrons, and a positive charge would indicate a deficiency of electrons, making it an electrophile rather than a nucleophile.

• Option (d) is incorrect because an electron-deficient species would be an electrophile, which seeks electrons, rather than a nucleophile, which donates electrons.

Answer

$(a, b)$

Hyperconjugation is the delocalisation of sigma electron also known as sigma-pi conjugation. Presence of $\alpha-\mathrm{H}$ with respect to double bond, triple bond or carbon containing positive charge (in carbonium ion) or unpaired electron (in free radical) is a condition for hyperconjugation.

(I) $ CH_3-CH_2-CH_2-CH_2-\stackrel{\stackrel{\large \text{O}}{\text{||}}}{C}- H $

(II) $ CH_3-CH_2-CH_2-\stackrel{\stackrel{\large \text{O}}{\text{||}}}{C}- CH_3 $

(III) $ CH_3-CH_2-\underset{\underset{\large \text{O}}{\text{||}}}{C}- CH_2-CH_3 $

(IV) $ CH_3-\underset{\underset{\large CH_3}{|}}{CH}- CH_2-\underset{\underset{\large \text{O}}{\text{||}}}{C}-H $

• (c) r-electrons of carbon-carbon bond: This option is incorrect because hyperconjugation specifically involves the delocalization of sigma (σ) electrons, not pi (π) electrons. The delocalization of π-electrons is a different phenomenon known as resonance.

• (d) lone pair of electrons: This option is incorrect because hyperconjugation involves the delocalization of sigma (σ) electrons from C-H bonds, not lone pairs of electrons. The delocalization of lone pairs is typically associated with resonance or other types of conjugation, not hyperconjugation.

Answer

Metamerism arises due to different alkyl chains on either side of the functional group in the molecule. In the given structures $\mathrm{V}$ and $\mathrm{VI}$ or $\mathrm{VI}$ and $\mathrm{VII}$ form a pair of metamers because they differ in carbon atoms on the either side of the functional group, i.e., O-atom.

Answer

Two or more compounds having the same molecular formula but different functional groups are called functional isomers.

In the given structure I, II, III, IV represent alcohols as functional group, whereas V, VI, VII are ethers.

Hence, I and V, I and VI, I and VII, II and V, II and VI, II and VII, III and V, III and VI, III and VII, IV and $\mathrm{V}, \mathrm{IV}$ and $\mathrm{VI}, \mathrm{IV}$ and $\mathrm{VII}$ all are functional group isomers.

Answer

When two or more compounds differ in position of substituent atom or functional group on the carbon skeleton, they are position isomers. In the given structures, I and II; III and IV, and $\mathrm{VI}$ and $\mathrm{VII}$ are position isomers.

(I) $ \underset{\text{1-butanol}}{CH_3-CH_2-CH_2-CH_2-OH} $ and II. $ \underset{\text{2-butanol}}{CH_3-CH_2-\underset{\underset{\large OH}{|}}{CH}-CH_3} $

III $ \underset{\text{2-methyl propanol-2}}{CH_3-\underset{\underset{\large CH_3}{|}}{\stackrel{\stackrel{\large CH_3}{|}}{C}}-CH_3} $

IV $ \underset{2-methyl propanol-1}{CH_3-\underset{\underset{\large OH}{|}}{CH}-CH_2-CH_3} $

VI. $CH_{3}-O\underset{\text { Methoxy propane }}{-CH_{2}-CH_{2}-CH_{3}}$

and

VII $ \underset{\substack{\text{Methoxy isopropane } \\ \text{or methoxy 1-methyl ethane }}}{CH_3-O-\underset{\underset{\large CH_3}{|}}{CH}-CH_3} $

Answer

When two or more compounds have similar molecular formula but different skeletons, these are referred to as chain isomer.

In the following structure

Answer

On adding dilute $H_{2} SO_{4}$ for testing halogens in an organic compound with $AgNO_{3}$, white precipitate of $Ag_{2} SO_{4}$ is formed. This will interfere with the test of chlorine and this $Ag_{2} SO_{4}$ may be mistaken for white precipitate of chlorine as $AgCl$. Hence, dilute $HNO_{3}$ is used instead of dilute $H_{2} SO_{4}$.

Answer

The given structure is of allene $\left(C_{3} H_{4}\right)$

$$ H_{2} \stackrel{1}{C}=\stackrel{2}{C}=\stackrel{3}{C} H_{2} $$

In allene, carbon atoms 1 and 3 are $s p^{2}$-hybridised as each one of them is joined by a double bond. And, carbon atom 2 is sp-hybridised as it has two double bonds at each of its side. Therefore, the two $\pi$-bonds are perpendicular to each other, in allene, as shown below.

$H_{a}$ and $H_{b}$ lie in the plane of paper while $H_{c}$ and $H_{d}$ lie in a plane perpendicular to the plane of the paper. Hence, the allene molecule as a whole is non-planar.

Answer

Electronegativity of carbon atom, also depends on the hybridisation of the carbon atom. Since, s-electrons are more strongly attracted by the nucleus than $p$-electrons, thus, electronegativity increases with increase in s-character of the hybridised orbital i.e.,

$ \xrightarrow[\large sp^3<sp^2,sp>]{\large \text{25 \% 33.3 \% 50 \% }} $ Hybridisation

$ \xrightarrow[\substack{\text{Increasing order} \\ \text{of electronegativity} }]{\substack{\text{Increasing order} \\ \text{of \% s-character}}} $

Thus, $s p$-hybridised carbon is the most electronegative carbon.

Answer

Carbon (2.5) is more electronegative than magnesium (1.2) therefore, $\mathrm{Mg}$ acquires a partial positive charge while carbon attached to it acquires a partial negative charge.

$$ \begin{array}{llllll} CH_{3}-CH_{2}-CH_{2}-\stackrel{-\delta}{C} H_{2}- \stackrel{+\delta}{Mg}-x \end{array} $$

Answer

The two isomers which differ in the position of the functional group on the carbon skeleton are called position isomers and this phenomenon as position isomerism.

Thus, (A) and (B) may be regarded as position isomers and further they cannot be regarded as metamers since metamers are those isomers which have different number of carbon atoms on either side of the functional group.

But here, the number of carbon atoms on either side of sulphur atom (functional group) is the same, i.e., 1 and 3 .

(A)

(B)

Answer

Among the following given compounds, according to IUPAC, the longest carbon chain having maximum number of functional group is being selected.

Thus, carbon-chain containing 4-carbon atoms and which also includes both functional group will be selected. While the other three $\mathrm{C}$-chains are incorrect since none of them contains both the functional groups.

Answer

DNA and RNA have nitrogen in the heterocyclic rings. Nitrogen present in rings, azo and nitro groups cannot be converted into $(NH_4)_2 SO_4$. That’s why Kjeldahl method cannot be used for the estimation of nitrogen present in DNA and RNA.

Answer

If a compound decomposes at its boiling point but is steam volatile, water-insoluble and stable at low pressure, steam distillation can be used for its purification. This technique is applied to separate substances which are steam volatile and immiscible with water.

Direction (Q. Nos. 35-38) On the basis of information given below

“Stability of carbocations depends upon the electron releasing inductive effect of groups adjacent to positively charged atom involvement of neighbouring groups in hyperconjugation and resonance.”

Answer

The given carbocation has two resonance structures.

Structure (II) is more stable because both the carbon atoms and the oxygen atom have an octet of electrons.

Answer

Carbocation (A) is more stable than carbocation (B). Carbocation. (A) is more planar and hence is stabilised by resonance while carbocation (B) is non-planar and hence it does not undergo resonance. Further, double bond is more stable within the ring in comparison to outside the ring.

Answer

In triphenylmethyl cation, due to resonance, the positive charge can move at both the $o-$ and $p$-position of each benzene ring. This is illustrated below

Since, there are three benzene rings, hence, there are, in all, nine resonance structures. Thus, triphenylmethyl cation is highly stable due to these nine resonance structures.

Answer

2-methyl butane has four different sets of equivalent $\mathrm{H}$-atoms.

$ CH_3-\underset{\underset{\large CH_3}{|}}{CH}- CH_2-CH_3 $

Removal of 1-1-1 atom from any of these equivalent sets gives four different carbocations.

(III) $\left(3^{\circ}\right)\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad\quad$ (IV) $\left(1^{\circ}\right)$

Stability of carbocation decreases in the order $3^{\circ}>2^{\circ}>1^{\circ}$. So, III $\left(3^{\circ}\right.$ carbocation $)$ is most stable followed by II $\left(2^{\circ}\right.$ carbocation). Out of I and IV (both are $1^{\circ}$ carbocation) I has a $CH_{3}$ group at $\beta$-carbon while II has a $CH_{3}$ group at $\alpha$-carbon. As $H$-effect decreases with distance, hence IV is more stable than I. Therefore, the overall stability of these four carbocations increases in the order.

$$ \mathrm{I}<\mathrm{IV}<\mathrm{II}<\mathrm{III} $$

Answer

If the organic compound contains both $\mathrm{N}$ and $\mathrm{S}$, then while fusion it may for form either a mixture of sodium cyanide $(\mathrm{NaCN})$ and sodium sulphide $\left(\mathrm{Na}_{2} \mathrm{~S}\right)$ or sodium thiocyanate (NaSCN) depending on the amount of Na metal used.

If Less sodium metal is used, only NaSCN is obtained.

This then gives red colour on reacting with $\mathrm{Fe}^{3+}$ ions (produced by oxidation of $\mathrm{Fe}^{2+}$ ions while preparing Lassaigne’s extract) due to the formation of ferric thiocyanate.

$$ Fe^{2+} \xrightarrow{\text{Aerial oxidation}} Fe^{3+} $$

$$ Fe^{3+}+3NaSCN \longrightarrow \underset{\substack{\text{Ferric thiocyanate}\\ \text{(red)}}}{Fe(SCN)_3}+3Na^{+} $$

In case, excess of sodium metal is used, the initally formed sodium thiocyanate decomposes as follows:

$$ \underset{thiocyanate}{\underset{Sodium}{NaSCN}} +2Na \xrightarrow{\triangle} \underset{cyanide}{\underset{Sodium}{NaCN}} + \underset{sulphide}{\underset{Sodium}{Na_2 S}} $$

This $\mathrm{NaCN}$ then reacts with $\mathrm{FeSO}_{4}, \mathrm{Fe}^{3+}$ ions and $\mathrm{NaCN}$, it gives prussian blue colour due to the formation of ferric ferrocyanide or iron (III) hexacyanoferrate (II).

$$ 2NaCN + FeSO_4 \longrightarrow Na_2SO_4 + Fe(CN)_2 $$

$$ Fe(CN)_2 + 4NaCN \longrightarrow \underset{ferrate (II.)}{\underset{Sodium hexacyano^-}{Fe_4[Fe(CN)_6]}} $$

$$ 3Na_4[Fe(CN)_6] + 4Fe^{3+} \longrightarrow \underset{f(II) (prussian blue)}{\underset{Iron (II) hexacyanoferrate}{Fe_4[Fe(CN)_6]_3}} + 12Na^+ $$

On the basis of above results, it is clear that Ramesh used less sodium and hence NaSCN formed in the Lassaigne’s extract which gave red colouration due to $Fe(SCN)_3$ formation while Manish and Rajni used excess sodium and hence NaCN formed in the Lassaigne’s extract which gave prussian blue colour of $Fe_4[Fe(CN)_6]$.

Answer

3-ethyl-4-methylhept-5-en-2-one (C-atoms of the longest possible chain are numbered in such a way that the functional group, $>\mathrm{C}=\mathrm{O}$, gets the lowest possible locant)

3-nitrocyclohex-1-en (C-atoms of the ring are numbered in such a manner that double bond gets the lowest possible locant followed by the $-\mathrm{NO}_{2}$ group)

Answer

(a) $\stackrel{7}{C} H_{3}-\stackrel{6}{C} H_{2}-\stackrel{5}{C} H_{2}-\stackrel{4}{C} H_{2}-\stackrel{3}{C} H_{2}-\stackrel{2}{C} H_{2}-\stackrel{1}{C} H_{2}-Br$

1-bromoheptane

5-bromoheptanoic acid

Answer

(a)

(b)

(c)

Answer

(a) $\stackrel{+}{\mathrm{C}} \mathrm{H}_{3}$ is the most stable species because the replacement of $\mathrm{H}$ by $\mathrm{Br}$ increases positive charge ( $-I$-effect) on carbon atom and destabilises the species and, more the number of Br-atoms, less stable is the species.

(b) $\stackrel{\ominus}{C} Cl_{3}$ is is the most stable species because on replacing $\mathrm{H}$ by $\mathrm{Cl}$, negative charge on carbon is dispersed due to $-I$-effect of $\mathrm{Cl}$ and thus get reduced and species is stabilised. Further, more the number of $\mathrm{Cl}$ atoms, more is the dispersal of the negative charge and hence more stable is the species.

Answer

Difference between inductive effect and resonance effect is as follows

Answer

(a) $\mathrm{CH}_{3} \mathrm{OH}$ As it lacks $\pi$-electrons hence it will not exist as resonance hybrid.

(b) $\mathrm{R}-\mathrm{CONH}_{2}$ Due to the presence of $n$-electrons on $\mathrm{N}$ and $\pi$-electrons on $\mathrm{C}=\mathrm{O}$ bond, hence amide can be represented as a resonance hybrid of the following three resonating structures.

(c) $CH_{3} CH=CHCH_{2} NH_{2}$ As the lone pair of electrons on the $N$-atom is not conjugated with the $\pi$-electrons of the double bond, thus, resonance is not possible and hence no resonance hybrid will exist.

Answer

Three highly electronegative oxygen atoms are attached to sulphur atom in $SO_{3}$ which makes sulphur atom electron-deficient. Further, due to resonance, sulphur acquires positive charge. Both these factors, make $SO_{3}$ an electrophile.

Answer

The structure having more covalent bonds in a resonating structure, has more stability. Further, there is charge separation in structure (II) and the terminal carbon has only a sextet of electrons in (II). These two factors makes structure (II) less stable.

Hence, I > II in terms of stability.

Answer

The difference in boiling point of two liquids is more than $20^{\circ} \mathrm{C}$. Hence, simple distillation can be used and since at the boiling point of low boiling liquid, the vapours would consist entirely of only low boiling liquid without any contamination of vapours of high boiling liquid and vice-versa. Thus, both the liquids can be distilled without any decomposition.

Answer

Resonating structures of $(A)$ and $(B)$ are as follows

Structure (II) is less stable than structure (I) because later carries separation of positive and negative charges. Therefore, contribution of structure (II) is less than that of (I) towards the resonance hybrid of compound (A), i.e., $CH_{3} COOH$. On contrary, structure (III) and (IV) are of equal energy and hence contribute equally towards the resonance hybrid of compound $(B)$. Therefore, structure $(B)$ is more stable than structure $(A)$ i.e., $CH_{3} COO^{\ominus}$.

Answer

A. $\rightarrow$ (5)

B. $\rightarrow$ (4)

C. $\rightarrow(1)$

D. $\rightarrow(2)$

E. $\rightarrow$ (3)

Answer

A. $\rightarrow$ (3)

B. $\rightarrow$ (6)

C. $\rightarrow$ (2)

D. $\rightarrow(1)$

E. $\rightarrow(4) \quad$

F. $\rightarrow(5)$

Answer

A. $\rightarrow$ (3)

B. $\rightarrow(5)$

C. $\rightarrow$ (1)

D. $\rightarrow(2)$

E. $\rightarrow$ (4)

Answer

A. $\rightarrow$ (1) $\quad$

B. $\rightarrow$ (1) $\quad$

C. $\rightarrow$ (2)

	Column I	Column II	Explanation
A.	Free radical	Trigonal planar	Free radicals are formed by homolytic fission e.g… $\stackrel{\circ}{\mathrm{C}} \mathrm{H}_{3}$ hybridisation $s p^{2}$
B.	Carbocation	Trigonal planar
C.	Carbanion	Pyramidal	Formed by heterolytic fission when carbon is attached to more electropositive atom e.g., $\mathrm{CH}_{3}^{-}$hybridisation $s p^{3}$

Answer

A. $\rightarrow(1,2)$

B. $\rightarrow(2)$

C. $\rightarrow(2)$

D. $\rightarrow(3,4)$

Assertion and Reason

In the following questions a statement of Assertion (A) followed by a statement of Reason $(\mathrm{R})$ is given. Choose the correct option out of the choices given below in each question.

Answer

(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$.

Simple distillation can be used to separate a mixture of two liquids which do not react and have boiling point difference of more than $20^{\circ} \mathrm{C}$. Hence, a mixture of propan-1-ol and propanone can be separated.

Answer

(d) A is not correct but $\mathrm{R}$ is correct.

Resonance hybrids are always more stable than any of the canonical structures would be, if they existed. The delocalisation of the electrons lowers the orbital energies, imparting stability. The gain in stability of the resonance hybrid over the most stable of the canonical structure is called resonance energy.

A canonical structure that is lower in energy makes a relating greater contribution to resonance hybrid.

Thus, the correct assertion will be energy of resonance hybrid is equal to the sum of energies of all canonical forms in proportion of their contribution towards the resonance hybrid.

Answer

(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$.

When two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton then it is position isomerism. Double bond is a functional group whose position varies.

Answer

(d) A is not correct but $\mathrm{R}$ is correct.

Hybridisation can be determined by counting $\sigma$ - bond

$$ \begin{gathered} 3 \sigma \quad 2 \sigma \quad 3 \sigma \\ H_{2} C=C=CH_{2} \\ 3 \sigma-s p^{2} \text { hybridisation } \\ 2 \sigma-\text { sphybridisation } \end{gathered} $$

Correct assertion In $H_{2} C=C=CH_{2}$, the central carbon is sp-hybridised whereas the terminal carbons are $s p^{2}$-hybridised.

Answer

(c) Both $A$ and $R$ are correct but $R$ is not the correct explanation of $A$.

$S$ present in an organic compound can be estimated quantitatively by Carius method. In this method, a known weight of organic compound is heated with fuming $HNO_{3}$, $S$ present in it gets converted into $H_{2} SO_{4}$. On adding $BaCl_{2}, H_{2} SO_{4}$ gets precipitated as $BaSO_{4}$ which may be of light yellow or white in colour.

If light yellow colour is obtained, it means some impurities are present. It is then filtered, washed, purified and then dried and finally pure $\mathrm{BaSO}_{4}$ of white colour is obtained.

Answer

(a) Both $\mathrm{A}$ and $\mathrm{R}$ are correct and $\mathrm{R}$ is the correct explanation of $\mathrm{A}$.

In paper chromatography, a chromatography paper is used. It contains water in it,which acts as the stationary phase. A strip of chromatography paper spotted at the base with ink is suspended in a suitable solvent. Solvent acts as the mobile phase.

The solvent rises up the paper by capillary action and flows over the spot. The paper selectively retains different components according to their differing partition in two phases.

Hence, components of ink will migrate at different rates and are separated.

Answer

Hybridisation is mixing of atomic orbitals to form new hybrid orbitals. The new orbital have the same total electron capacity as the old ones. The properties and energies of the new hybridised orbitals are an average of the unhybridised orbitals.

Hybridisation can be found by counting number of $\sigma-$ bonds around the carbon atom.

$$ \begin{aligned} & 3 \sigma=s p^{2} \text {-hybridisation } \\ & 2 \sigma=s p \text {-hybridisation } \end{aligned} $$

In allene, carbon atoms 1 and 3 are $s p^{2}$-hybridised as each one of them is joined by a double bond. And, carbon atom 2 is $s p$-hybridised as it has two double bonds at each of its side. Therefore, the two $\pi$-bonds are perpendicular to each other, in allene, as shown below.

$H_{a}$ and $H_{b}$ lie in the plane of paper while $H_{c}$ and $H_{d}$ lie in a plane perpendicular to the plane of the paper. Hence, the allene molecule as a whole is non-planar.

Answer

Benzoic acid can be purified by hot water because of following characteristics

(i) Benzoic acid is more soluble in hot water and less soluble in cold water.

(ii) Impurities present in benzoic acid are either insoluble in water or more soluble in water to such an extent that they remain in solution as the mother liquor upon crystallisation.

Answer

If the difference in boiling points of two liquids is not much, fractional distillation is used to separate them. In this technique, fractionating column is fitted over the mouth of the round bottom flask.

When vapours of a liquid mixture are passed through a fractionating column, the vapours of the low boiling liquid $(A)$ will move up while those of the high boiling liquid will condense and fall back into the flask. Therefore, liquid (A) with low boiling point will distill first.

Answer

The boiling points are in the order of $A>C>B$.

Liquid $A$ can be separated from rest of the mixture of liquid $B$ and $C$ by simple distillation $B$ and $C$ can be separated by fractional distillation.

Due to the fact that boiling point of $A$ is much higher than those of liquids $B$ and $C$. This can be done by using apparatus as shown in figure (I). As the boiling points of liquid $(B)$ and $(C)$ are quite close but much lower than that of $A$, hence, mixture of liquids $(B)$ and $(C)$ will distill together leaving behind liquid $(A)$.

On further heating liquid $(A)$ will distill over. Now, place the mixtures of liquid $(B)$ and $(C)$ in a flask fitted with fractionating column as illustrated in figure (II). On fractional distillation, liquid $(B)$ will distill over first and then liquid $(C)$ as former possess lower boiling point than that of later.

Answer

If the difference in boiling points of two liquids is not much, fractional distillation is used. The techniques is, vapors of liquid mixture are passed through a fractionating column before condensation, fitted over the mouth of the round bottom flask.

Vapours of the liquid with higher boiling point condense before the vapours of the liquid with lower boiling point. The vapours rising up in the fractionating column become rich in more volatile component. Fractionating column provides many surfaces for heat exchange between ascending vapours and descending condensed liquid. The vapours become richer in low boiling component.

One of technological application of fractional distillation is to separate different fraction of crude oil in petroleum industry into various fractions like gasoline, kerosene oil, diesel oil, lubricating oil, etc.

Another application is the separation of acetone and methanol from pyroligneous acid obtained by destructive distillation of wood.

Answer

In steam distillation, the distillating mixture consists of steam and the vapour of organic substance. In steam distillation, the liquid boils when the sum of the vapour pressure of the organic substance $\left(p_{1}\right)$ and that of steam $\left(p_{2}\right)$ becomes equal to the atmospheric pressure $(p)$ at the temperature of distillation.

$$ p=p_{1}+p_{2} \text { or } p_{1}=p-p_{2} $$

Since, the vapour pressure of the organic substance is lower than $p$, it vaporises below its normal boiling point without decomposition e.g., aniline which normally boils at $457 \mathrm{~K}$ can be distilled at $371.5 \mathrm{~K}$ by this process.