NCERT Exemplar Questions

Hint

Hint: The longest carbon chain contains seven carbon atoms.

(a) While writing IUPAC name, the alkyl groups are written in alphabetical order. Thus lower locant 3 is assigned to ethyl. Prefix, di, tri, and tetra are not included in alphabetical order.

The longest carbon chain contains seven carbon atoms. Thus parent chain is heptane. Numbering of the chain is done in such a way that ethyl group gets lowest position. The compound with numbering is as follows:

The IUPAC name of the compound is 3-Ethyl-4,4-dimethylhexane. Hence option one is the correct answer. 4,4-Dimethyl-3-ethylheptane is wrong because alphabetically ethyl will come first than methyl.
5-Ethyl-4,4-dimethylheptane is wrong because lowest sum rule does not follow while numbering the chain.
4,4-Bis(methyl)-3-ethylheptane is wrong because bis is not used.

Alternate:

While writing the name of the substituents in alphabetical order, the prefixes such as di, tri, tetra, penta, hexa, etc. are not considered. Therefore, as “e” of ethyl comes before ” m ” of dimethyl thus, lower locant 3 assigned to ethyl group and is written first.
Hence, the correct IUPAC name is 3-Ethyl-4,4-dimethylbutane.

Hint

(d) Hint: Acid group gets higher priority than ketone group

In the given molecule, the acid group gets higher priority than ketone. So, while numbering the chain acid group carbon gets the lowest priority than ketone group carbon. The longest carbon chain contains five carbon atoms. Numbering the chain is as follows:

For parent chain pent is used, for ketone group, “oxo” is used and for an acid group, “oic” is used as a suffix. The IUPAC name of the compound is 4-Oxopentanoic acid.

Note: 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 the keto group \((>\mathrm{C}=0)\).

Hint

(b) Hint: Numbering should be done according to the lowest locant rule
For tri or higher substituted benzene derivatives, the compounds are named by identifying substituent, positions on the ring by following the lowest locant rule.
The 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. Numbering the compound is as follows:

The IUPAC name of the compound is 1-Chloro-4-methyl-2-nitrobenzene.

Hint

(c) Electronegativity increases as the state of hybridization changes from \(\mathrm{sp}^3\) to \(\mathrm{sp}^2\) and \(\mathrm{sp}^2\) to \(\mathrm{sp}\). Thus, sp hybridized carbon has the highest electronegativity.

Hint: Percentage of s character directly proportional to electronegativity

The electronegativity of carbon atoms depends on their state of hybridization. More s-character more is the electronegativity.
\(
\begin{aligned}
& \mathrm{sp}^3<\mathrm{sp}^2<\mathrm{sp} \\
& 25 \% \mathrm{~s} 33 \% \text { s } 50 \% \text { s }
\end{aligned}
\)
\(
\text { Thus, sp-carbon has the highest electronegativity, i.e., option (c) }\left(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{C} \equiv{ }^* \mathrm{CH}\right) \text { is correct. }
\)

Hint

(c) Alkyl halides do not show functional isomerism. Alcohols and ethers, aldehydes and ketones, cyanides and isocyanides are functional isomers.

Explanation: 

Hint: Halogen containing compound didnot show functional group isomerism
Functional isomerism occurs when substances have the same molecular formula but different functional groups.
Functional isomer of alcohol is ether.
Functional isomer of aldehyde is ketone.
Funtional isomer of aldehyde is ketone.
Functional isomer of cyanide is isocyanide
However, alkyl halides do not show functional isomerism. Hence, option (c) is correct.

Hint

(d) Essential oils are insoluble in water, soluble in steam and have high vapour pressure. Therefore, they can be separated by steam distillation.

Explanation: 

Hint: Essential oil is steam volatile
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.

Hint

(d) Thin layer chromatography (TLC) is used to separate the pigments present in ink.

Explanation:

Hint: Layer of adsorbent is spread over a glass plate
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.

Hint

(b) In paper chromatography, separation of the components of a mixture depends upon their partitioning between water held in the stationary phase (i.e. adsorbent paper) and the liquid present in the mobile phase.

Explanation:

Hint: Substances are distributed or partitioned between liquid phases.

Partition chromatography is based on the continuous differential partitioning of components of a mixture between stationary and mobile phases. Paper chromatography is a type of partition chromatography.

Hint

(a) Hint: Electron donating group stabilizes the carbocation.
The stability of given cations can be understood by the following structures.

The second carbocation is stabilized by the mesomeric effect of oxygen. The first carbocation is stabilized by the inductive effect of two methyl groups. The third carbocation is destabilized by the -l effect of the oxygen atom.
Hence, the decreasing order of stability of carbocation is II > I > III.

Hint

(b) Hint: The parent Chain contains six carbon atoms.
The given compound contains six carbon atoms chain as a parent chain. Numbering should be done in such a way that the two methyl groups’ positions must be the lowest. The structure of the molecule with numbering is as follows:

The parent chain name is hexane and it contains two methyl groups at 3rd and 4th positions. The IUPAC name of the compound is 3,4-Dimethylhexane.

Hint

(a) In \(\stackrel{*}{\mathrm{C}} \mathrm{H}_3-\mathrm{CH}_2-\mathrm{Cl}\), the carbon marked with asterisk has the greatest positive charge due to high electronegativity of \(\mathrm{Cl}\).

Explanation:

Hint: consider inductive effect of substituents

Electronegativity of \(\mathrm{Cl}, \mathrm{Br}, \mathrm{C}\) and \(\mathrm{Mg}\) follows the order \(\mathrm{Cl}>\mathrm{Br}>\mathrm{C}>\mathrm{Mg}\)
\(* \mathrm{CH}_3 \rightarrow \mathrm{CH}_2 \rightarrow \mathrm{Cl}(-I-\) effect \()\)
\(* \mathrm{CH}_3 \leftarrow \mathrm{CH}_2 \leftarrow \mathrm{Mg}^{+} \mathrm{Cl}^{-}\)
\(* \mathrm{CH}_3 \rightarrow \mathrm{CH}_2 \rightarrow \mathrm{Br}(-I-\) effect \()\)
\(* \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_3\) (+l-effect)
– I effect of \(\mathrm{Cl}>\mathrm{Br}\)

Hence, \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Cl}\) has the greatest positive charge.

Hint

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

(a) No electron withdrawing groups are attached at neighbouring position of carboxylate ion. So, extra stabilisation of carboxylate ion doesn’t occur due to absence of electron withdrawing groups.

(b) Chlorine atom is attached at neighbouring position of carboxylate ion, which can act as electron withdrawing group due to it’s -I effect.
So, extra stabilisation of carboxylate ion occurs due to presence of electron withdrawing group. Hence, it is more stable than option (a).

(c) Fluorine atom is attached at neighbouring position of carboxylate ion, which can act as electron withdrawing group due to it’s -I effect.
Since, fluorine is more electronegative than chlorine. So, it will exert stronger -I effect than chlorine. Hence, it will have more stabilisation of carboxylate ion than option (b).

(d) 2 Fluorine atoms are attached at neighbouring position of carboxylate ion, which can act as electron withdrawing groups due to their -I effect. Since, two fluorine atoms will exert stronger -I effect than single fluorine. Hence, it will have most stabilisation of carboxylate ion.
So, correct answer is option (d)

Explanation:

Hint: Resonance effect and inductive effect both working
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., \(+\mathrm{R}\) effect of the carboxylate ion (conjugation) and -I-effect of the halogens (fluorine).

These effect are shown below in the carbocations

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 \(\mathrm{F}\)-atoms are present in option (d), thus, dispersal of negative charge is maximum in option (d).

Hint

(c) When the electrophile attacks \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2\), delocalisation of electrons can take place in two possible ways

As \(2^{\circ}\) carbocation is more stable than \(1^{\circ}\) carbocation, the first addition is more feasible

Explanation:

Hint: Stability of carbocation
When electrophile attacks \(\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2\) delocalisation of electrons can take place, in two possible ways

Carbocation \(\mathrm{X}\) is \(2^{\circ}\) and carbocation \(\mathrm{Y}\) is \(1^{\circ}\). As \(2^{\circ}\) carbocation is more stable than \(1^{\circ}\) carbocation. Thus, option third correct.

Hint

(ii) Hint: More electronegative element gets negative charge and less electronegative element gets positive charge.
In heterolytic cleavage, unequal distribution of bonding electrons take place. More electronegative element takes bonding electrons and and acquire negative charge while less electronegative element acquire positive charge.
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 Br gets the negative charge. Thus, option (ii) is correct.

Hint

(b) Since double bond is a source of electrons and the charge flows from source of more electron density, therefore, \(\mathrm{n}\) electrons of the double bond attack the proton.

Hint

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

Hybridisation of carbon atoms in different compounds is shown below
\(
\text { (a) } \mathrm{H}-\stackrel{s p}{\mathrm{C}} \equiv \stackrel{s p}{\mathrm{C}}-\stackrel{s p}{\mathrm{C}} \equiv \stackrel{s p}{\mathrm{C}}-\mathrm{H}
\)
\(
\text { (b) } \stackrel{s p^3}{\mathrm{CH}}_3-\stackrel{s p}{\mathrm{C}} \equiv \stackrel{s p}{\mathrm{C}}-\stackrel{s p^3}{\mathrm{CH}}_3
\)
\(
\text { (c) } \stackrel{s p^2}{\mathrm{CH}}_2=\stackrel{s p}{\mathrm{C}}=\stackrel{s p^2}{\mathrm{CH}}_2
\)
\(
\text { (d) } \stackrel{s p^2}{\mathrm{CH}}_2=\stackrel{s p^2}{\mathrm{CH}}-\stackrel{s p^2}{\mathrm{CH}}=\stackrel{s p^2}{\mathrm{CH}}_2
\)

Hint

(a) (i, iii, iv) Different groups are present towards and away from the observer.

Hint

(b, c)
In (a), \(\mathrm{NH}_3\) and \(\mathrm{H}_2 \mathrm{O}\) are nucleophiles but \(\mathrm{BF}_3\) is an electrophile.
In (d), \(\mathrm{C}_2 \mathrm{H}_5^{-}\) is a nucleophile but \(\dot{\mathrm{C}}_2 \mathrm{H}_5\) and \(\mathrm{C}_2 \mathrm{H}_5^{+}\)are electrophiles.

Note: All \(\mathrm{AlCl}_3, \mathrm{SO}_3\) (Lewis acids), \(\mathrm{NO}_2^{+} . \mathrm{CH}_3^{+}, \mathrm{CH}_3-\mathrm{C}^{+}=\mathrm{O}\) are electron deficient species. Hence, these are electrophiles.

Hint

(b) II and III are position isomers as they differ in the position of -C-group.

Explanation:

Hint: Position of Functional group is different

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

Here, the position of the ketone functional group is varied and the molecular formula is the same. Thus, II and III are position isomers.

Hint

(a, c) (a) II and III have the same functional group.
(c) I and IV have the same functional group.

Explanation:

Two or more compounds having the same molecular forumla but different functional groups are called functional isomers.
I. Aldehyde
II. Ketone
III. Ketone
IV. Aldehyde

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

Hint

(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, options (a) and (c) are correct.

Hint

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

Hint

(a) Hint: Purification method
\(
\text { (i) } \rightarrow \text { (e) } \quad \text { (ii) } \rightarrow \text { (d) } \quad \text { (iii) } \rightarrow \text { (a) } \quad \text { (iv) } \rightarrow \text { (b) } \quad \text { (v) } \rightarrow \text { (c) }
\)

Hint

(a) Hint: Electrophile is an electron-deficient species

(i) \(\rightarrow\) (c) (ii) \(\rightarrow\) (f) (iii) \(\rightarrow\) (b) (iv) \(\rightarrow\) (a) (v) \(\rightarrow\) (d) (vi) \(\rightarrow\) (e)

Hint

(b)

(i) \(\rightarrow\) (c) (ii) \(\rightarrow\) (e) (iii) \(\rightarrow\) (a) (iv) \(\rightarrow\) (b) (v) \(\rightarrow\) (d)

Explanation:
(i) Used for \(\mathrm{N}\) containing compounds.
(ii) Nitrogen converts to ammonium sulphate.
(iii) Compound is heated in the presence of \(\mathrm{AgNO}_3\).
(iv) Adsorbent used is silica gel.
(v) Free radicals are formed by homolytic fission.

Hint

(c)

(i) \(\rightarrow\) (a) (ii) \(\rightarrow\) (a) (iii) \(\rightarrow\) (b)

\(
\begin{array}{|c|c|c|c|}
\hline & \text { Column I } & \text { Column II } & \text { Explanation } \\
\hline \text { A. } & \text { Free radical } & \text { Trigonal planar } & \begin{array}{l}
\text { Free radicals are formed by homolytic } \\
\text { fission e.g., } \dot{\mathrm{C}} \mathrm{H}_3 \text { hybridisation } s p^2
\end{array} \\
\hline \text { B. } & \text { Carbocation } & \text { Trigonal planar } & \begin{array}{l}
\text { Formed by heterolytic fission when carbon } \\
\text { is attached to a more electronegative atom } \\
\text { e.g., } \stackrel{+}{\mathrm{C}} \mathrm{H}_3 \text { hybridisation } s p^2
\end{array} \\
\hline \text { C. } & \text { Carbanion } & \text { Pyramidal } & \begin{array}{l}
\text { Formed by heterolytic fission when carbon } \\
\text { is attached to more electropositive atom } \\
\text { e.g., } \mathrm{CH}_3^{-} \text {hybridisation } s p^3
\end{array} \\
\hline
\end{array}
\)

Hint

(d)

(i) \(\rightarrow\) (a), (b), (d), (ii) \(\rightarrow\) (b), (iii) \(\rightarrow\) (b) , (iv) \(\rightarrow\) (c), (d)

The above carbocation is stabilised because of dispersal of positive charge via resonance. -I effect (i.e., electron withdrawing effect) of oxygen destabilize the carbocation.
Also, positively charged carbon ( carbocation) have three a-hydrogens ( \(\mathrm{C}-\mathrm{H}\) bonds next to \(\mathrm{C}^{+}\)). So, it can also stabilized via hyperconjugation. Hence, the correct match is:
(i) \(\rightarrow\) (a), (b), (d).

(ii) (b), Here, carbocation is destabilized by the negative \((-I)\) inductive effect of three fluorine atoms. It increases the positive charge density on carbocation center.

(iii) (b), Here, carbanion is destabilised due to the positive \((+I)\) inductive effect of three methyl groups. As methyl being an electron releasing group, increases the electron density on carbanion.

(iv) (c,d), Since positively charged carbon is directly attached with two other carbons, it is a secondary carbocation.
Here, secondary carbocation is stabilized by hyperconjugation via neighbouring
\(\mathrm{C}-\mathrm{H} \text { bonds (involving a-hydrogens ). }\)

Hint

(a) Hint: Liquids with a difference of more than \(20^{\circ} \mathrm{C}\) in their boiling points can be separated by simple distillation.
Simple distillation can be used to separate a mixture of two liquids that do not react with each other and have a boiling point difference of more than \(20^{\circ} \mathrm{C}\). Hence, a mixture of propan-1-ol and propanone can be separated out. Hence, both assertion and reason are true and the reason is the correct explanation of assertion.

Hint

(d)

Hint: energy of the resonance hybrid is equal to the sum of energies of all canonical forms in a proportion to their contribution towards the resonance hybrid.

Resonance hybrids are always more stable than any of the canonical structures would be, if they existed. The delocalization of the electrons lowers the orbital energies, imparting stability. The gain in the 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 the energy of the resonance hybrid is equal to the sum of energies of all canonical forms in a proportion to their contribution towards the resonance hybrid. Hence, the Assertion is false but the reason is true.

Hint

(a) Hint: The position of the double bond is different
When two or more compounds differ in the position of substituent atom or functional group on the carbon skeleton then it is position isomerism. A double bond is a functional group whose position varies.
The structure is as follows:
\(
\stackrel{5}{\mathrm{C}} \mathrm{H}_3-\stackrel{4}{\mathrm{C}} \mathrm{H}_2-\stackrel{3}{\mathrm{C}} \mathrm{H}_2-\stackrel{2}{\mathrm{H}}=\stackrel{1}{\mathrm{C}} \mathrm{H}_2
\)
\(
\quad \quad \quad \quad \quad \text { Pent-1-ene }
\)
\(
\stackrel{5}{\mathrm{C}} \mathrm{H}_3-\stackrel{4}{\mathrm{C}} \mathrm{H}_2-\mathrm{H} \stackrel{3}{\mathrm{C}}==\stackrel{2}{\mathrm{C}} \mathrm{H}-\stackrel{1}{\mathrm{C}} \mathrm{H}_3
\)
\(
\quad \quad \quad \quad \quad \text { Pent-2-ene }
\)
Hence, both assertion and reason are true and the reason is the correct explanation of assertion.

Hint

(d) Hint: second carbon is sp hybridized
Hybridization can be determined by counting \(\sigma\) – bond
\(
\stackrel{3 \sigma}{\mathrm{H}_2} \mathrm{C}=\stackrel{2 \sigma}{\mathrm{C}}=\stackrel{3 \sigma}{\mathrm{C}} \mathrm{{H}_2}
\)
\(3 \sigma-\mathrm{sp}^2\) hybridisation
\(2 \sigma-\mathrm{sp}\) hybridisation
In \(\mathrm{H}_2 \mathrm{C}=\mathrm{C}=\mathrm{CH}_2\), the central carbon is sp-hybridised whereas the terminal carbons are \(\mathrm{sp}^2\)-hybridised. Thus, the Assertion is false but the reason is true.

Hint

(c) Hint: Sulphur present in an organic compound can be estimated quantitatively by the Carius method
Sulphur present in an organic compound can be estimated quantitatively by the Carius method. In this method, a known weight of the organic compound is heated with fuming \(\mathrm{HNO}_3\), \(\mathrm{S}\) present in it gets converted into \(\mathrm{H}_2 \mathrm{SO}_4\). On adding \(\mathrm{BaCl}_2\), \(\mathrm{H}_2 \mathrm{SO}_4\) gets precipitated as \(\mathrm{BaSO}_4\) 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 the white colour is obtained.
Thus, both assertion and reason are true and the reason is the correct explanation of assertion.

Hint

(a) Hint: Chromatography
In paper chromatography, a chromatography paper is used. It contains water, which acts as the stationary phase. A strip of chromatography paper spotted at the base with ink is suspended in a suitable solvent. The 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. Thus, both assertion and reason are true and the the reason is the correct explanation of assertion.