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NCERT Solutions for Class 11 Chemistry Chapter 8 - Organic Chemistry

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Class 11 Chemistry Chapter 8 NCERT Solutions FREE PDF Download

Class 11 Chemistry Chapter Organic chemistry NCERT solutions Chapter 8 allows diving into the fascinating chemical bonding world, exploring molecular interactions and atomic structures. Access your FREE PDF download now, empowering your journey towards excellence in Chemistry. With step-by-step explanations and insightful guidance, conquer the challenges of chemical bonding effortlessly. Elevate your learning experience and master ch 8 chemistry class 11 NCERT solutions with our invaluable resource.


Download the FREE PDF of Class 11 NCERT Solutions for Chapter 7 Redox reactions prepared by master teachers. Check out the Revised class 11 chemistry syllabus and get started with Vedantu to embark on a journey of academic excellence!


Quick Insights of Class 11 Chemistry Chapter 8 Organic Chemistry NCERT Solutions

  • Class 11 Chemistry Chapter 8 NCERT solutions will give you insights about the General Introduction: methods of purification, qualitative and quantitative analysis, classification, and IUPAC nomenclature of organic compounds.

  • The section will give you crisp learnings on Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance, and hyperconjugation.

  • The understanding related to topics like the importance of Homolytic and heterolytic fission of a covalent bond: free radicals, carbocations, carbanions, electrophiles and nucleophiles, and types of organic reactions.

  • Using these NCERT solutions in class 11 chemistry chapter 8 can help students analyse their level of preparation and understanding of concepts.

  • Class 11 chemistry chapter Organic chemistry NCERT solutions topics are included according to the revised academic year 2024-25 syllabus.

  • Ch 8 Chemistry Class 11 NCERT Solutions Provide links to the other study materials, such as class notes, important concepts, and exemplar solutions.

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Access Chapter 8 -Organic Chemistry Class 11 NCERT Solutions

1. What are hybridization states of each carbon atom in the following compounds?

${\text{C}}{{\text{H}}_{\text{2}}}{\text{  =  C  =  O, C}}{{\text{H}}_{\text{3}}}{\text{CH  =  C}}{{\text{H}}_{\text{2}}},{\text{ }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CO, C}}{{\text{H}}_{\text{2}}}{\text{  =  CHCN, }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}$

Ans: 

  1. $\mathop {\text{C}}\limits^1 {{\text{H}}_{\text{2}}}{\text{  =  }}\mathop {\text{C}}\limits^2 {\text{  =  O}}$

In the given compound, carbon-1 is bonded to two hydrogen atoms and one carbon atom. So, according to the VSEPR theory, the steric number is 3 which corresponds to  $s{p^2}$ hybridization. Carbon-2 is bonded to one carbon atom and one oxygen atom. So, the steric number is 2 which corresponds to $sp$ hybridization.

  1. $\mathop {\text{C}}\limits^1 {{\text{H}}_{\text{3}}}\mathop {\text{C}}\limits^2 {\text{H  =  }}\mathop {\text{C}}\limits^3 {{\text{H}}_{\text{2}}}$

In the given compound, carbon-1 is bonded to three hydrogen atoms and one carbon atom. So, according to the VSEPR theory, the steric number is 4 which corresponds to  $s{p^3}$ hybridization. Carbon-2 is bonded to two carbon atoms and one hydrogen atom. So, the steric number is 3 which corresponds to $s{p^2}$ hybridization.

  1. $\mathop {\text{C}}\limits^1 {{\text{H}}_{\text{3}}}\mathop {\text{C}}\limits^2 {\text{O}}\mathop {\text{C}}\limits^3 {{\text{H}}_{\text{3}}}$

In the given compound, carbon-1 and carbon-3 are bonded to three hydrogen atoms and one carbon atom. So, according to the VSEPR theory, the steric number is 4 which corresponds to  $s{p^3}$ hybridization. Carbon-2 is bonded to two carbon atoms and one oxygen atom. So, the steric number is 3 which corresponds to $s{p^2}$ hybridization. 

  1. $\mathop {\text{C}}\limits^1 {{\text{H}}_{\text{2}}}{\text{  =  }}\mathop {\text{C}}\limits^2 {\text{H}}\mathop {\text{C}}\limits^3 {\text{N}}$

In the given compound, carbon-1 is bonded to two hydrogen atoms and one carbon atom. So, according to the VSEPR theory, the steric number is 3 which corresponds to  $s{p^2}$ hybridization. Carbon-2 is bonded to two carbon atoms and one hydrogen atom. So, the steric number is 3 which corresponds to  $s{p^2}$ hybridization. Carbon-3 is bonded to one carbon atom and one nitrogen atom. So, the steric number is 2 which corresponds to $sp$ hybridization.

  1. ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}$

In the given compound, all carbon atoms are bonded to two carbon atoms and one hydrogen atom. So, according to the VSEPR theory, the steric number is 3 which corresponds to  $s{p^2}$ hybridization.


2. Indicate the $\sigma {\text{ and }}\pi $ bonds in the following molecules: 

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}{\text{, }}{{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}{\text{, C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}{\text{, C}}{{\text{H}}_{\text{2}}}{\text{  =  C  =  C}}{{\text{H}}_{\text{2}}}{\text{, C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{O}}_{\text{2}}}{\text{, HCONHC}}{{\text{H}}_{\text{3}}}$

Ans: The structure of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}$ is shown below.

(image will be uploaded soon)

Single bond contains only one sigma bond and double bond contains one sigma and one pi bond. In this structure, nine single bonds and three double bonds are present. So, there are $12{\text{ }}\sigma {\text{ and 3 }}\pi $  bonds present.

The structure of ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{12}}}}$ is shown below.

(image will be uploaded soon)

In this structure, eighteen single bonds are present. So, there are only $18{\text{ }}\sigma {\text{ }}$ bonds present.

The structure of ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$ is shown below.

(image will be uploaded soon)

In this structure, four single bonds are present. So, there are only $4{\text{ }}\sigma {\text{ }}$ bonds present.

The structure of ${\text{C}}{{\text{H}}_{\text{2}}}{\text{  =  C  =  C}}{{\text{H}}_{\text{2}}}$ is shown below.

(image will be uploaded soon)

In this structure, four single bonds and two double bonds are present. So, there are $6{\text{ }}\sigma {\text{ and 2 }}\pi $ bonds present.

The structure of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{N}}{{\text{O}}_{\text{2}}}$ is shown below.

(image will be uploaded soon)

In this structure, five single bonds and one double bond are present. So, there are $6{\text{ }}\sigma {\text{ and 1 }}\pi $ bonds present.

The structure of ${\text{HCONHC}}{{\text{H}}_{\text{3}}}$ is shown below.

(image will be uploaded soon)

In this structure, seven single bonds and one double bond are present. So, there are $8{\text{ }}\sigma {\text{ and 1 }}\pi $ bonds present.


3. Write bond line formulas for: 

Isopropyl alcohol, ${\text{2,3 }} - {\text{ Dimethylbutanal, Heptan }} - {\text{ 4 }} - {\text{ one}}$

Ans: The bond line formulas for the given compounds are shown below.

  1. Isopropyl Alcohol: 

(image will be uploaded soon)

  1. ${\text{2,3 }} - {\text{ Dimethylbutanal}}$

(image will be uploaded soon)

  1. ${\text{Heptan }} - {\text{ 4 }} - {\text{ one}}$

(image will be uploaded soon)


4. Give the IUPAC names of the following compounds:

(image will be uploaded soon )

(f) ${\text{C}}{{\text{l}}_{\text{2}}}{\text{CHC}}{{\text{H}}_{\text{2}}}{\text{OH}}$

Ans: (a)

(image will be uploaded soon)

In this structure, the longest chain of carbon contains 3 carbon atoms and one phenyl group is present. So, the IUPAC name is 1-phenyl propane. 

(b) 

(image will be uploaded soon)

In this structure, the longest chain of carbon contains four carbon atoms. Also, one methyl and one cyanide groups are attached with carbon-2 and carbon-1 respectively. So, the IUPAC name is 2-methyl-1-cyanobutane.

(c) 

(image will be uploaded soon)

In this structure, the longest chain of carbon contains 7 carbon atoms and 2 methyl are positioned at carbon-2 and carbon-5. So, the IUPAC name of the compound is 2, 5-dimethyl heptane.

(d) 

(image will be uploaded soon)

In this structure, the longest chain of carbon contains 7 carbon atoms. Also, 1 chloro and 1 bromo groups are present at carbon-3. So, the IUPAC name of the compound is 3-bromo-3-chloroheptane.

(e) 

(image will be uploaded soon)

In this structure, the longest chain of carbon contains 3 carbon atoms. Also, 1 aldehyde and 1 chloro functional groups are present at carbon-1 and carbon-3 respectively. So, the IUPAC name of the compound is 3-chloropropanal.

(f) 

(image will be uploaded soon)

In this structure, the longest chain of carbon contains 2 carbon atoms. Also, 1 alcohol and 2 chloro functional groups are present at carbon-1 and carbon-2 respectively. So, the IUPAC name of the compound is 2,2-dichloro-1-ethanol.


5. Which of the following represents the correct IUPAC name for the compounds concerned? 

  1. ${\text{2,2 }} - {\text{ Dimethylpentane or 2}} - {\text{Dimethylpentane}}$

  2. ${\text{2,4,7 }} - {\text{ Trimethyloctane or 2,5,7 }} - {\text{ Trimethyloctane}}$

  3. ${\text{2 }} - {\text{ Chloro }} - {\text{ 4 }} - {\text{ methylpentane or 4 }} - {\text{ Chloro }} - {\text{ 2 }} - {\text{ methylpentane}}$

  4. ${\text{But }} - {\text{ 3 }} - {\text{ yn }} - {\text{ 1 }} - {\text{ ol or But }} - {\text{ 4 }} - {\text{ ol }} - {\text{ 1 }} - {\text{ yne}}$

Ans:

  1. In the IUPAC nomenclature, the prefix Di- denotes that the parent chain has two identical substituent groups. As there are two methyl groups in the C–2 of the parent chain of the given chemical compound, the correct IPUAC name is ${\text{2,2 }} - {\text{ Dimethylpentane}}{\text{.}}$

  2. The locant number 2, 4, 7 is less than the locant number 2, 5, 7. As a result, the given compound's IUPAC nomenclature is ${\text{2,4,7 }} - {\text{ Trimethyloctane}}$

  3. If the functional groups are present at the parent chain's equivalent position, the lower number is assigned to the one that appears first in the name in alphabetical order. As a result, the proper IUPAC nomenclature for the given compound is ${\text{2 }} - {\text{ Chloro }} - {\text{ 4 }} - {\text{ methylpentane}}{\text{.}}$

  4. The given molecule contains two functional groups: alcoholic and alkyne. The alcoholic group is the principal functional group. As a result, the parent chain will be terminated with ol. The alkyne group can be found in the parent chain's C–3. As a result, ${\text{But }} - {\text{ 3 }} - {\text{ yn }} - {\text{ 1 }} - {\text{ ol}}$ is the correct IUPAC nomenclature for the given compound.


6. Draw formulas for the first five members of each homologous series beginning with the following compounds. 

$\left( {\text{a}} \right){\text{ H -- COOH }}$

$  \left( {\text{b}} \right){\text{ C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{3}}}{\text{ }} $

$  \left( {\text{c}} \right){\text{ H -- CH =  C}}{{\text{H}}_{\text{2}}} $

Ans: The following are the first 5 members of each homologous series, starting with the given compounds: 

$  {\left( {\text{a}} \right){\text{ H -- COOH: Methanoic acid}}} $

$  {{\text{C}}{{\text{H}}_{\text{3}}}{\text{ -- COOH : Ethanoic acid}}} $

$  {{\text{C}}{{\text{H}}_{\text{3}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- COOH : Propanoic acid}}} $ 

$  {{\text{C}}{{\text{H}}_{\text{3}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- COOH : Butanoic acid}}} $

$  {{\text{C}}{{\text{H}}_{\text{3}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- COOH : Pentanoic acid}}} $

${{\text{}}\left( {\text{b}} \right){\text{ C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{3}}}{\text{ : Propanone}}} $  ${{\text{C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{ : Butanone}}} $

${{\text{C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{ : Pentan - 2 - one}}} $  ${{\text{C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{ : Hexan - 2 - one}}} $

${{\text{C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{ : Heptan - 2 - one}}} $

 ${\left( {\text{c}} \right){\text{ H -- CH  =  C}}{{\text{H}}_{\text{2}}}{\text{ : Ethene}}}$

$  {{\text{C}}{{\text{H}}_{\text{3}}}{\text{ -- CH  =  C}}{{\text{H}}_{\text{2}}}{\text{ : Propene}}} $ 

$  {{\text{C}}{{\text{H}}_{\text{3}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- CH  =  C}}{{\text{H}}_{\text{2}}}{\text{ : 1 - Butene}}} $

$  {{\text{C}}{{\text{H}}_{\text{3}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- CH  =  C}}{{\text{H}}_{\text{2}}}{\text{ : 1 - Pentene}}} $

$  {{\text{C}}{{\text{H}}_{\text{3}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- CH  =  C}}{{\text{H}}_{\text{2}}}{\text{ : 1 - Hexene}}} $


7. Give condensed and bond line structural formulas and identify the functional group(s) present, if any, for: 

 ${\left( {\text{a}} \right){\text{ }}2,2,4{\text{ }} - {\text{ Trimethylpentane}}} $

 ${\left( {\text{b}} \right){\text{ }}2{\text{ }} - {\text{ Hydroxy }} - {\text{ }}1,2,3{\text{ }} - {\text{ propanetricarboxylic acid}}} $

 $ \left( {\text{c}} \right){\text{ Hexanedial}} $

Ans: 

  1. $2,2,4{\text{ }} - {\text{ Trimethylpentane}}$

Condensed formula: ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHC}}{{\text{H}}_{\text{2}}}{\text{C }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{3}}}$

Bond Line Formula:

(image will be uploaded soon)

  1. $2{\text{ }} - {\text{ Hydroxy }} - {\text{ }}1,2,3{\text{ }} - {\text{ propanetricarboxylic acid}}$

Condensed Formula: $\left( {{\text{COOH}}} \right){\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}\left( {{\text{OH}}} \right)\left( {{\text{COOH}}} \right){\text{C}}{{\text{H}}_{\text{2}}}\left( {{\text{COOH}}} \right)$

Bond Line Formula:

(image will be uploaded soon)

In the given compound, the functional groups present are carboxylic acid $\left( { - {\text{COOH}}} \right)$ and alcoholic group $\left( { - {\text{OH}}} \right).$

  1. Hexanedial 

Condensed Formula: $\left( {{\text{CHO}}} \right){\left( {{\text{C}}{{\text{H}}_{\text{2}}}} \right)_{\text{4}}}\left( {{\text{CHO}}} \right)$

Bond Line Formula:

(image will be uploaded soon)

In the given compound, the functional group present is aldehyde $\left( { - {\text{CHO}}} \right).$


8. Identify the functional groups in the following compounds.

(a) 

(image will be  uploaded soon)

(b) 

(image will be  uploaded soon)

(c) 

(image will be  uploaded soon)

Ans: 

  1. In the given compound, aldehyde $\left( {--{\text{CHO}}} \right),$ hydroxyl $\left( {--{\text{OH}}} \right)$ and methoxy $\left( {--{\text{OMe}}} \right)$ functional groups are present.

  2. In the given compound, amino $\left( {--{\text{N}}{{\text{H}}_{\text{2}}}} \right),$ ketone $\left( {{\text{C }} = {\text{ O}}} \right),$ diethylamine $\left( {{\text{N}}{{\left( {{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{5}}}} \right)}_{\text{2}}}} \right)$ functional groups are present.

  3. In the given compound, nitro group $\left( { - {\text{N}}{{\text{O}}_{\text{2}}}} \right)$ and ${\text{C }} = {\text{ C}}$ double bond are present.


9. Which of the two: ${{\text{O}}_{\text{2}}}{\text{NC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{O --}}$ or ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{O --}}$ is expected to be more stable and why?

Ans: The nitro group is an electron-withdrawing group. As a result, it exhibits the -I effect. This group reduces the compound's negative charge by attracting electrons towards it and stabilizing it. The ethyl group, on the other hand, is an electron-releasing group. As a result, the ethyl group exhibits the +I effect. This raises the compound's negative charge, making it unstable. As a result, ${{\text{O}}_{\text{2}}}{\text{NC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{O --}}$ is projected to be more stable than ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{O --}}{\text{.}}$


10. Explain why alkyl groups act as electron donors when attached to a $\pi $ system.

Ans: When an alkyl group is linked to a $\pi $ system, the process of hyperconjugation causes it to act as an electron-donor group. Let us use propene as an example to better comprehend this topic.

(image will be uploaded soon)

The sigma electrons of an alkyl group's C-H bond are delocalised during hyperconjugation. This group is directly connected to an unsaturated system's atom. The delocalisation happens as a result of a partial overlap of a $s{p^3}$ sigma bond orbital with an empty p orbital of the pi bond of a nearby carbon atom's bond.

(image will be uploaded soon)


11. Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation. 

$\left( {\text{a}} \right){\text{ }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH }} $

$  \left( {\text{b}} \right){\text{ }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{O}}_{\text{2}}}{\text{ }}$

$  \left( {\text{c}} \right){\text{ C}}{{\text{H}}_{\text{3}}}{\text{CH  =  CHCHO }} $

$  \left( {\text{d}} \right){\text{ }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{ -- CHO }} $

$  \left( {\text{e}} \right){\text{ }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{ -- }}\mathop {\text{C}}\limits^ +  {{\text{H}}_{\text{2}}}{\text{ }} $

$  \left( {\text{f}} \right){\text{ C}}{{\text{H}}_{\text{3}}}{\text{CH  =  CH}}\mathop {\text{C}}\limits^ +  {{\text{H}}_{\text{2}}} $

Ans: 

  1. The resonating structures for ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{OH}}$ are shown below.

(image will be uploaded soon)

  1. The resonating structures for ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{N}}{{\text{O}}_{\text{2}}}$ are shown below.

(image will be uploaded soon)

  1. The resonating structures for ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CH  =  CHCHO}}$ are shown below.

(image will be uploaded soon)

  1. The resonating structures of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{ -- CHO}}$ are shown below.

(image will be uploaded soon)

  1. The resonating structures of ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{ -- }}\mathop {\text{C}}\limits^ +  {{\text{H}}_{\text{2}}}$ are shown below.

(image will be uploaded soon)

  1. The resonating structures of ${\text{C}}{{\text{H}}_{\text{3}}}{\text{CH  =  CH}}\mathop {\text{C}}\limits^ +  {{\text{H}}_{\text{2}}}$ are shown below.

(image will be uploaded soon)


12. What are electrophiles and nucleophiles? Explain with examples.

Ans: An electrophile is a reagent that removes a pair of electrons. In other words, an electrophile $\left( {{{\text{E}}^ + }} \right)$ is a reagent that seeks electrons. Electrophiles are electron-deficient and receiver of an electron pair.

Electrophiles include carbocations and $\left( {{\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}^ + } \right)$ neutral compounds with functional groups such as the carbonyl group.

A reagent that produces an electron pair is known as a nucleophile. A nucleus-seeking reagent is referred to as a nucleophile (Nu:).

For instance, ${\text{O}}{{\text{H}}^ - },{\text{ N}}{{\text{C}}^ - },$ carbanions $\left( {{{\text{R}}_3}{{\text{C}}^ - }} \right),$ and so on.

Because of the presence of a lone pair, neutral molecules such as water and ammonia also serve as nucleophiles.


13. Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:  

$\left( {\text{a}} \right){\text{ C}}{{\text{H}}_{\text{3}}}{\text{COOH  +  }}{\mathbf{O}}{{\mathbf{H}}^{\mathbf{ - }}}{\text{ }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - }{\text{  +  }}{{\text{H}}_{\text{2}}}{\text{O}} $

$  \left( {\text{b}} \right){\text{ C}}{{\text{H}}_{\text{3}}}{\text{COC}}{{\text{H}}_{\text{3}}}{\text{  +  }}{\mathbf{C}}{{\mathbf{N}}^{\mathbf{ - }}}{\text{ }} \to {\text{ }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{C}}\left( {{\text{CN}}} \right)\left( {{\text{OH}}} \right) $

$  \left( {\text{c}} \right){\text{ }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}{\text{  +  }}{\mathbf{C}}{{\mathbf{H}}_{\mathbf{3}}}\mathop {\mathbf{C}}\limits^{\mathbf{ + }} {\mathbf{O }} \to {\text{ }}{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{5}}}{\text{COC}}{{\text{H}}_{\text{3}}} $

Ans: Electrophiles are entities that are electron-deficient and can accept an electron pair. Nucleophiles, on the other hand, are electron-rich entities that can transfer their electrons.

  1. In this equation, ${\text{O}}{{\text{H}}^ - }$ serves as a nucleophile because it is an electron-rich species.

  2. In this equation, ${\text{C}}{{\text{N}}^ - }$serves as a nucleophile because it is an electron-rich species.

  3. In this equation, ${\text{C}}{{\text{H}}_{\text{3}}}\mathop {\text{C}}\limits^{\text{ + }} {\text{O}}$ serves as an electrophile because it is an electron-deficient species.


14. Classify the following reactions in one of the reaction types studied in this unit. 

$\left( {\text{a}} \right){\text{ C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Br  +  H}}{{\text{S}}^{\text{ - }}}{\text{ }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{HS  +  B}}{{\text{r}}^{\text{ - }}} $

$  \left( {\text{b}} \right){\text{ }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{C  =  C}}{{\text{H}}_{\text{2}}}{\text{  +  HCl }} \to {\text{ }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{ClC  -  C}}{{\text{H}}_{\text{3}}}$

$  \left( {\text{c}} \right){\text{ C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{Br  +  H}}{{\text{O}}^{\text{ - }}}{\mathbf{ }} \to {\text{ C}}{{\text{H}}_{\text{2}}}{\text{  =  C}}{{\text{H}}_{\text{2}}}{\text{  +  }}{{\text{H}}_{\text{2}}}{\text{O  +  B}}{{\text{r}}^{\text{ - }}} $

$  \left( {\text{d}} \right){\text{ }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{3}}}{\text{C  -  C}}{{\text{H}}_{\text{2}}}{\text{OH  +  HBr}}{\mathbf{ }} \to {\text{ }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CBrC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}}{\text{  +  }}{{\text{H}}_{\text{2}}}{\text{O }}$

Ans: 

  1. The given reaction is a type of a substitution reaction because the bromine group in bromoethane is substituted by thionyl group.

  2. The given reaction is a type of an addition reaction because two reactant molecules react to produce a single product.

  3. The given reaction is a type of an elimination reaction because hydrogen and bromine are eliminated by bromoethane in order to produce ethene.

  4. Substitution takes place in this reaction, followed by atom and atom group rearrangement.


15. What is the relationship between the members of the following pairs of structures? Are they structural or geometrical isomers or resonance contributors?

(a) 

(image will be uploaded soon)

(b)

 (image will be uploaded soon)

(c)

  (image will be uploaded soon)

Ans: 

  1. Compounds with the same chemical formula but different structures are referred to as structural isomers. The mentioned compounds have the same molecular formula, but the position of the functional group differs (ketone group).

The ketone group is at C-3 of the parent chain (hexane chain) in structure I, and at C-2 of the parent chain in structure II (hexane chain). As a result, the given pair of compounds are structural isomers.

  1. Geometrical isomers are compounds that have the same chemical formula, composition, and series of covalent bonds but have different relative positions of their atoms in space.

The relative positions of Deuterium (D) and hydrogen (H) in space differ in structures I and II. As a result, the given pairs are geometrical isomers.

  1. The structures shown are either canonical or contributing structures. They are fictitious and do not represent any actual molecule. As a result, the given pair represents resonance structures known as resonance isomers.


16. For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion.

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Ans: 

  1. The bond cleavage can be illustrated using curved arrows to show the electron flow of the given reaction as

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As one of the shared pairs in a covalent bond travels with the connected atom, it is an example of homolytic cleavage. A free radical is produced as a reaction intermediate.

  1. The bond cleavage can be illustrated using curved arrows to show the electron flow of the given reaction as

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It is an example of heterolytic cleavage because the bond breaks in such a way that the shared pair of electrons stay with the carbon of propanone. The reactive intermediate created is carbanion.

  1. The bond cleavage can be illustrated using curved arrows to show the electron flow of the given reaction as

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It is an example of heterolytic cleavage because the bond breaks in such a way that the bromine ion retains the shared pair of electrons. A carbocation is produced as a reaction intermediate.

  1. The bond cleavage can be illustrated using curved arrows to show the electron flow of the given reaction as

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The bond breaks in such a way that the shared pair of electrons stays with one of the fragments, resulting in a heterolytic cleavage. A carbocation is created as an intermediate.


17. Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? 

$  {\left( {\text{a}} \right){\text{ C}}{{\text{l}}_{\text{3}}}{\text{CCOOH  >  C}}{{\text{l}}_{\text{2}}}{\text{CHCOOH  >  ClC}}{{\text{H}}_{\text{2}}}{\text{COOH}}} $

 $\left( {\text{b}} \right){\text{ C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOH  >  }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHCOOH  >  }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{3}}}{\text{C}}{\text{.COOH}}$

Ans: Inductive effect: It refers to the permanent shift of sigma electrons along a saturated chain whenever an electron withdrawing or electron donating group is present. The inductive effect could be either a $ + {\text{ }}I$ effect or $ - {\text{ }}I$ effect. When an atom or group attracts electrons more strongly than hydrogen, it is said to have the $ - {\text{ }}I$ effect. Example, 

${\text{F}} -  \leftarrow {\text{C}}{{\text{H}}_{\text{2}}} -  \leftarrow {\text{C}}{{\text{H}}_{\text{2}}} -  \leftarrow {\text{C}}{{\text{H}}_{\text{2}}} -  \leftarrow {\text{C}}{{\text{H}}_{\text{3}}}$

When an atom or set of atoms attracts electrons less strongly than hydrogen, this is referred to as the $ + {\text{ }}I$ effect. For instance, ${\text{C}}{{\text{H}}_{\text{3}}} -  \to {\text{C}}{{\text{H}}_{\text{2}}} -  \to {\text{Cl}}$

Electromeric effect: In the presence of an attacking agent, it involves the complete transfer of the shared pair of pi electrons to either of the two atoms joined by multiple bonds. For example,

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Electrometric effects might be $ + {\text{ }}E$ or $ - {\text{ }}E.$ The $ + {\text{ }}E$ effect occurs when electrons are transferred to the assaulting reagent. Whereas when electrons are moved away from the attacking reagent, this is referred to as the $ - {\text{ }}E$ effect.

$\left( {\text{a}} \right){\text{ C}}{{\text{l}}_{\text{3}}}{\text{CCOOH  >  C}}{{\text{l}}_{\text{2}}}{\text{CHCOOH  >  ClC}}{{\text{H}}_{\text{2}}}{\text{COOH}}$ 

On the basis of the inductive effect $\left( { - {\text{ }}I} \right)$, the order of acidity can be explained. The $ - {\text{ }}I$ effect increases as the number of chlorine atoms increases. The acid strength increases in proportion to the increase in $ - {\text{ }}I$ effect.

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$\left( {\text{b}} \right){\text{ C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{COOH  >  }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHCOOH  >  }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{3}}}{\text{C}}{\text{.COOH}}$

The inductive effect ($ + {\text{ }}I$ effect) can be used to explain the order of acidity. The $ + {\text{ }}I$ effect increases in proportion to the quantity of alkyl groups. The acid strength increases in proportion to the increase in $ + {\text{ }}I$ effect.

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18. Give a brief description of the principles of the following techniques taking an example in each case. (a) Crystallisation (b) Distillation (c) Chromatography

Ans: 

  1. Crystallisation: It is one of the most widely used methods for purifying solid chemical molecules.

Principle: The differential in the solubilities of the compound and the impurities in a solvent. The impure substance dissolves in the solvent, which is only sparingly soluble at ambient temperature but completely soluble at higher temperatures. The solution is concentrated until it is virtually saturated. When the solution is cooled, the pure compound crystallises and is extracted by filtration. 

Pure aspirin, for example, is created by recrystallising crude aspirin. 2 - 4 g of crude aspirin is dissolved in approximately 20 mL of ethyl alcohol. To ensure complete decomposition, the solution is heated.

  1. Distillation: This approach is used to separate volatile liquids from nonvolatile contaminants or a mixture of those liquids with a large enough difference in boiling points.

Principle: It works on the principle that liquids with different boiling points vapourize at different temperatures. The vapours are then cooled, and the liquids that form are separated. 

For example: Distillation can be used to separate a combination of chloroform and aniline. The mixture is placed in a condenser-equipped round bottom flask. After that, it is heated. Because chloroform is more volatile, it vaporizes first and enters the condenser. The vapours condense in the condenser, and chloroform trickles down. Aniline is left in the round bottom flask.

  1. Chromatography: It is one of the most effective technologies for organic chemical separation and purification.

Principle: The differential in movement of separate components of a mixture through the stationary phase under the influence of the mobile phase is the basis for the principle.

Example: Chromatography, can separate a combination of red and blue ink. On the chromatogram, a drop of the combination is inserted. The component of the ink that is less adsorbed on the chromatogram travels with the mobile phase, whereas the component that is more adsorbed remains stationary.


19. Describe the method, which can be used to separate two compounds with different solubilities in a solvent${\text{S}}{\text{.}}$

Ans: The process of fractional crystallisation is used to separate two compounds with varying solubilities in a solvent ${\text{S}}{\text{.}}$ The fractional crystallisation process is divided into four phases.

  1. Solution’s Preparation: The powdered combination is placed in a flask, and the solvent is slowly and simultaneously added and agitated. The solvent is added until the solute is completely dissolved in it. This saturated solution is then boiled.

  2. Solution’s Filtration: In a China dish, the hot saturated solution is filtered through filter paper.

  3. Fractional Crystallisation: Allow the solution in a China dish to cool. To begin with, the less soluble molecule crystallizes, while the more soluble compound remains in solution. During the separation of these crystals from the liquor, the latter is concentrated once more. The hot solution is allowed to cool, resulting in the formation of crystals of the more soluble chemical.

  4. Isolation and Drying: Filtration is used to separate the crystals from the mother liquor. The crystals are finally dried.


20. What is the difference between distillation, distillation under reduced pressure and steam distillation?

Ans: The differences between distillation, reduced pressure distillation, and steam distillation are shown below.

Distillation: It is used to purify chemicals that are related with non-volatile impurities or liquids that do not disintegrate when heated. In other words, distillation is used to separate volatile liquids from nonvolatile contaminants or a mixture of such liquids with significant boiling point difference. This process separates a mixture of fuel and kerosene.

Reduced Pressure Distillation: This procedure is used to purify a liquid that decomposes when heated. Under reduced pressure, the liquid will boil at a lower temperature than its boiling point and so will not decompose.

This Process is Used to Purify Glycerol. At a temperature of ${\text{593 K,}}$ it boils with decomposition. It boils at ${\text{453 K}}$ without breakdown at a lower pressure.

Steam Distillation: It is used to purify an organic molecule that is steam volatile and water immiscible. The compound is heated by passing steam, and the steam is condensed to water. After some time, the water-liquid mixture begins to boil and flows through the condenser. A separating funnel is then used to separate this condensed mixture of water and liquid.

A solution of water and aniline can be separated by this method.


21. Discuss the chemistry of Lassaigne’s test.

Ans: Lassaigne's Test: This test detects the presence of nitrogen, sulfur, halogens, and phosphorus in an organic compound. An organic compound has these elements in covalent form. By fusing the chemical with sodium metal, these are transformed into the ionic form.

$  {{\text{Na  +  C  +  N }} \to {\text{ NaCN}}} $

$  {{\text{Na  +  S  +  C  +  N }} \to {\text{ NaSCN}}} $

$  {{\text{2Na  +  S }} \to {\text{ N}}{{\text{a}}_{\text{2}}}{\text{S}}} $

$  {{\text{Na  +  X }} \to {\text{ NaX}}} $

The sodium cyanide, sulphide, and halide are removed from the fused material by boiling it in distilled water. Lassaigne's extract is the name given to the extract obtained in this manner. The presence of nitrogen, sulphur, halogens, and phosphorus is then determined in this Lassaigne's extract.

  1. Chemistry of Nitrogen Test: The sodium fusion extract is heated with iron (II) sulphate and then acidified with sulphuric acid in the Lassaigne's test for nitrogen in an organic molecule. Sodium cyanide first interacts with iron (II) sulphate to generate sodium hexacyanoferrate in this method (II). The iron (II) is then oxidized by sulphuric acid to generate iron (III) hexacyanoferrate (II), which is Prussian blue. The reactions involved are shown below.

$ {{\text{6C}}{{\text{N}}^{{\text{ - }}}}{\text{ +  F}}{{\text{e}}^{{\text{2 + }}}}{\text{}} \to {\text{ }}{{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]}^{{\text{4 - }}}}} $

$  {{\text{3}}{{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]}^{{\text{4 - }}}}{\text{  +  4F}}{{\text{e}}^{{\text{3 + }}}}{\text{}} \to {\text{ F}}{{\text{e}}_{\text{4}}}{{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]}_{\text{3}}}{\text{}}} $

  1. Test for Sulphur: 

$\left( {\text{i}} \right){\text{ Lassaignes's extract  +  Lead acetate}} \to {\text{ Black precipitate}}$

Chemistry of the test: The sodium fusion extract is acidified with acetic acid and then lead acetate is added to it in the Lassaigne's test for sulphur in an organic compound. The presence of sulphur in the combination is shown by the precipitation of lead sulphide, which is black in color.

${{\text{S}}^{{\text{2 - }}}}{\text{  +  P}}{{\text{b}}^{{\text{2 + }}}}{\text{ }} \to {\text{ PbS}}$

$\left( {{\text{ii}}} \right){\text{ Lassaignes's extract  +  Sodium nitroprusside }} \to {\text{ Violet colour}}$

Chemistry of the Test: Sodium nitroprusside is used to treat the sodium fusion extract. The presence of sulphur in the compound is also indicated by the presence of violet color.

${{\text{S}}^{{\text{2 - }}}}{\text{ +  [Fe}}{\left( {{\text{CN}}} \right)_{\text{5}}}{\text{NO}}{{\text{]}}^{{\text{2 - }}}}{\text{}} \to {\text{ [Fe}}{\left( {{\text{CN}}} \right)_{\text{5}}}{\text{NOS}}{{\text{]}}^{{\text{ - 4}}}}$ 

If both nitrogen and sulphur are present in an organic molecule, ${\text{NaSCN}}$ is formed instead of ${\text{NaCN}}{\text{.}}$

${\text{Na  +  C  +  N  +  S }} \to {\text{ NaSCN}}$

This ${\text{NaSCN}}$ (sodium thiocyanate) color is blood red. The absence of free cyanide ions prevents the formation of prussian color.

  1. Test for Halogen: The sodium fusion extract is acidified with nitric acid and then treated with silver nitrate in the Lassaigne's test for halogens in an organic molecule. If the organic compound contains both nitrogen and sulphur, the Lassaigne's extract is heated to remove the nitrogen and sulphur, which would otherwise interfere with the halogen test.


22. Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahl’s method.

Ans: In the Dumas method, a specified amount of a nitrogen-containing organic molecule is burned strongly with an excess of copper oxide in a carbon dioxide atmosphere to yield free nitrogen as well as carbon dioxide and water.

The process can also yield traces of nitrogen oxides, which can be converted to dinitrogen by passing the gaseous mixture over a heated copper gauge. The dinitrogen generated is collected over a potassium hydroxide aqueous solution. At room temperature and atmospheric pressure, the volume of nitrogen generated is then measured.

Kjeldahl's method, on the other hand, involves heating a known quantity of a nitrogen-containing organic molecule with concentrated sulphuric acid. The compound's nitrogen is quantitatively transformed into ammonium sulphate. It is then distilled with a high concentration of sodium hydroxide. The ammonia produced by this method is injected into a known volume of ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{.}}$The chemical equations involved are shown below.

Volumetric analysis (titration against a standard alkali) is used to measure the quantity of acid that is left unused, and the amount of ammonia produced can be calculated. As a result, the percentage of nitrogen in the compound can be calculated. This approach is inapplicable to compounds with nitrogen in a ring structure, and it is also inapplicable to compounds with nitro and azo groups.


23. Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.

Ans: Estimation of Halogens: The Carius method is used to calculate halogens. In this approach, a known amount of organic chemical is heated with fuming nitric acid in the presence of silver nitrate in a Carius tube, which is placed in a furnace. The carbon and hydrogen in the compound are oxidized to generate ${\text{C}}{{\text{O}}_{\text{2}}}{\text{ and }}{{\text{H}}_{\text{2}}}{\text{O,}}$ respectively, while the halogen in the compound is transformed into the form of ${\text{AgX}}$ . After that, the ${\text{AgX}}$ is filtered, washed, dried, and weighed.

Estimation of Sulphur: In this approach, a known amount of organic substance is heated in a hard glass tube called the Carius tube with either fuming nitric acid or sodium peroxide. The compound's sulphur is oxidized to generate sulphuric acid. The precipitation of barium sulphate occurs when an excess of barium chloride is added to it. After that, the precipitate is filtered, washed, dried, and weighed.

Estimation of Phosphorus: A known amount of organic substance is heated with fuming nitric acid in this procedure. The compound's phosphorus is oxidized to generate phosphoric acid. Phosphorus can be precipitated as ammonium phosphomolybdate by adding ammonia and ammonium molybdate to the solution. Phosphorus can also be measured by precipitating it as ${\text{MgN}}{{\text{H}}_{\text{4}}}{\text{P}}{{\text{O}}_{\text{4}}}$ and then on igniting it yeild ${\text{M}}{{\text{g}}_{\text{2}}}{{\text{P}}_{\text{2}}}{{\text{O}}_{\text{7}}}{\text{.}}$


24. Explain the principle of paper chromatography. 

Ans: Paper chromatography employs the use of chromatography paper. This paper includes trapped water, which behaves as the stationary phase. The solution of the mixture is spotted on the base of this chromatography paper. The paper strip is then hung in a suitable solvent, which serves as the mobile phase. By capillary action, this solvent moves up the chromatography paper and flows over the spot during the procedure. Different component spots travel to different heights with mobile phase. The resulting paper is known as a chromatogram.


25. Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?

Ans: The Lassaigne's extract is first heated with weak nitric acid before being tested for the presence of halogens. This is done to convert ${\text{NaCN to HCN and N}}{{\text{a}}_{\text{2}}}{\text{S to }}{{\text{H}}_{\text{2}}}{\text{S}}$ and then expel these gases. If there is any nitrogen or sulphur present in the form of ${\text{NaCN or N}}{{\text{a}}_{\text{2}}}{\text{S,}}$ it is eliminated. The chemical equations involved are shown below.

$  {\text{N}}{{\text{a}}}{\text{CN  +  HN}}{{\text{O}}_{\text{3}}}{\text{ }}$ $\to {\text{ }}{{\text{H}}}{\text{CN  +  NaN}}{{\text{O}}_{\text{3}}} $

$  {\text{N}}{{\text{a}}_{\text{2}}}{\text{S  +  2HN}}{{\text{O}}_{\text{3}}}{\text{ }} \to {\text{ }}{{\text{H}}_{\text{2}}}{\text{S  +  2NaN}}{{\text{O}}_{\text{3}}} $


26. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens. 

Ans: Organic chemicals are covalently bonded to nitrogen, sulfur, and halogens. They must first be changed to ionic form in order to be detected. This is accomplished by combining the organic chemical with sodium metal. This is known as the "Lassaigne's test." The chemical equations used in the test are as follows:

  ${{\text{Na  +  C  +  N }} \to {\text{ NaCN}}} $

$  {{\text{Na  +  S  +  C  +  N }} \to {\text{ NaSCN}}} $

$  {{\text{2Na  +  S }} \to {\text{ N}}{{\text{a}}_{\text{2}}}{\text{S}}} $

$  {{\text{Na  +  X }} \to {\text{ NaX}}} $

Here, X is halogen like chlorine, bromine, iodine.

Organic compounds contain carbon, nitrogen, sulfur, and halogen.


27. Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor. 

Ans: A combination of camphor and calcium sulphate is separated by sublimation. The sublimable chemical converts from solid to vapour without going through the liquid state throughout this procedure. Camphor is a sublimable molecule, whereas calcium sulphate is a non-sublimable solid. As a result, when heated, camphor will sublime while calcium sulphate will remain.


28. Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation? 

Ans: In the process of steam distillation, the organic liquid begins to boil when the total of the organic liquid’s vapour pressure $\left( {{p_1}} \right)$ and the water’s vapour pressure $\left( {{p_2}} \right)$ equals atmospheric pressure $\left( p \right),{\text{ i}}{\text{.e}}.{\text{ }}p{\text{ }} = {\text{ }}{p_1}{\text{ }} + {\text{ }}{p_2}.$ Therefore,

${p_1}{\text{  <  }}{p_2}$

 Hence, organic liquid will vapourize at a lower temperature than its boiling point.


29. Will ${\text{CC}}{{\text{l}}_{\text{4}}}$ give white precipitate of ${\text{AgCl}}$ on heating it with silver nitrate? Give reason for your answer.

Ans: When ${\text{CC}}{{\text{l}}_{\text{4}}}$ heated with silver nitrate, it does not produce the white precipitate of ${\text{AgCl}}{\text{.}}$ This is because the chlorine atoms in ${\text{CC}}{{\text{l}}_{\text{4}}}$ are covalently connected to carbon. It must be present in ionic form in order to obtain the precipitate, which requires preparing the Lassaigne's extract of ${\text{CC}}{{\text{l}}_{\text{4}}}.$


30. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?

Ans: Carbon dioxide is an acidic substance, whereas potassium hydroxide is a strong basic. As a result, when carbon dioxide combines with potassium hydroxide, potassium carbonate and water are formed.

${\text{2KOH + C}}{{\text{O}}_{\text{2}}}{\text{}} \to {\text{}}{{\text{K}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}{\text{ +  }}{{\text{H}}_{\text{2}}}{\text{O}}$

As a result, the mass of the ${\text{KOH}}$containing U-tube increases. This rise represents the amount of ${\text{C}}{{\text{O}}_{\text{2}}}$ created. The proportion of carbon in the organic compound can be determined based on its mass.


31. Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?

Ans: Due to the common ion effect, the addition of sulphuric acid would precipitate lead sulphate, but the addition of acetic acid would assure complete precipitation of sulphur in the form of lead sulphate. As a result, acetic acid must be used to acidify sodium extract before testing for sulphur using the lead acetate test.


32. An organic compound contains $69{\text{ }}\% $ carbon and $4.8{\text{ }}\% $ hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when ${\text{0}}{\text{.20 g}}$of this substance is subjected to complete combustion.

Ans: The percentage of carbon in organic compound is $69{\text{ }}\% .$

It means ${\text{100 g}}$ of organic compound contains ${\text{69 g}}$ of carbon. So, the mass of carbon contained in ${\text{0}}{\text{.20 g}}$ of organic compound is calculated as

$  {\text{Mass of C  =  }}\dfrac{{69{\text{ }} \times {\text{ }}0.20}}{{100}}{\text{  }} $

   $= {\text{ }}0.138{\text{ g of C}} $

The molecular mass of carbon dioxide is ${\text{44 g}}/{\text{mol}}{\text{.}}$ Therefore, ${\text{12 g}}$of carbon is contained in ${\text{44 g}}$ of ${\text{C}}{{\text{O}}_{\text{2}}}.$

Therefore, the mass of ${\text{C}}{{\text{O}}_{\text{2}}}$ contained in ${\text{0}}{\text{.138 g}}$ of carbon is calculated as 

$ {\text{Mass of C}}{{\text{O}}_{\text{2}}}{\text{  =  }}\dfrac{{44{\text{ }} \times {\text{ }}0.138}}{{12}}{\text{  }} $

  $ = {\text{ }}0.506{\text{ g}} $

Thus, 0.506 g of CO2 will be produced on complete combustion of 0.2 g of organic compound.

The percentage of hydrogen in organic compound is $4.8{\text{ % }}{\text{.}}$ It means ${\text{100 g}}$ of organic compound contains ${\text{4}}{\text{.8 g}}$ of hydrogen.

Therefore, the mass of hydrogen contained in ${\text{0}}{\text{.20 g}}$ of organic compound is calculated as

$ {\text{Mass of H  =  }}\dfrac{{4.8{\text{ }} \times {\text{ }}0.20}}{{100}}{\text{  }} $

  $ = {\text{ }}0.0096{\text{ g of H}} $

It is known that the molecular mass of water is ${\text{18 g}}/{\text{mol}}{\text{.}}$

Thus, ${\text{2 g}}$ of hydrogen is contained in ${\text{18 g}}$ of water. Therefore, the mass of water contained in ${\text{0}}{\text{.0096 g}}$ of hydrogen is calculated as

 ${\text{Mass of }}{{\text{H}}_{\text{2}}}{\text{O  =  }}\dfrac{{18{\text{ }} \times {\text{ }}0.0096}}{2}{\text{  }}$ 

$   = {\text{ }}0.0864{\text{ g}} $ 


33. A sample of ${\text{0}}{\text{.50 g}}$ of an organic compound was treated according to Kjeldahl’s method. The ammonia evolved was absorbed in ${\text{50 ml of 0}}{\text{.5 M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{.}}$ The residual acid required ${\text{60 mL of 0}}{\text{.5 M}}$ solution of ${\text{NaOH}}$ for neutralisation. Find the percentage composition of nitrogen in the compound.

Ans: Given: The total mass of organic compound is ${\text{0}}{\text{.50 g}}{\text{.}}$

${\text{60 mL of 0}}{\text{.5 M}}$ solution of ${\text{NaOH}}$ was required for neutralisation by residual acid.

$  {\text{60 mL of 0}}{\text{.5 M NaOH solution  =  }}\dfrac{{60}}{2}{\text{ mL of 0}}{\text{.5M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{}} $

$  {\text{ }}\dfrac{{60}}{2}{\text{ mL of 0}}{\text{.5M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{  =  30 mL of 0}}{\text{.5 M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} $

Therefore, acid consumed in absorption of evolved ammonia is calculated as 

${\text{ = }}\left( {{\text{50 - 30}}} \right){\text{ mL }} $

$  {\text{ =  20 mL}} $

Again, ${\text{20 mL of 0}}{\text{.5 M }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ =  40 mL of 0}}{\text{.5 M N}}{{\text{H}}_{\text{3}}}$

Also, since ${\text{1000 mL of 1 M N}}{{\text{H}}_{\text{3}}}$contains ${\text{14 g}}$ of nitrogen, the mass of nitrogen in ${\text{40 mL of 0}}{\text{.5 M N}}{{\text{H}}_{\text{3}}}$ is calculated as

${\text{Mass of N  =  }}\dfrac{{14{\text{ }} \times {\text{ }}40{\text{ }} \times {\text{ }}0.5}}{{1000}}{\text{ }} $

$   = {\text{ }}0.28{\text{ g}} $

Therefore, the percentage of nitrogen in ${\text{0}}{\text{.50 g}}$ of organic compound is calculated as

${\text{Percentage of nitrogen  =  }}\dfrac{{0.28}}{{0.5}}{\text{ }} \times {\text{ 100% }} $

 $ {\text{ =  56 % }}$ 

 

34. ${\text{0}}{\text{.3780 g}}$ of an organic chloro compound gave ${\text{0}}{\text{.5740 g}}$ of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

Ans: Given: The mass of organic compound is ${\text{0}}{\text{.3780 g}}$ and the mass of silver chloride formed is ${\text{0}}{\text{.5740 g}}{\text{.}}$

The molar mass of ${\text{AgCl}}$ is $143.32{\text{ g}}/{\text{mol}}$ and the molar mass of chlorine is $35.5{\text{ g}}/{\text{mol}}{\text{.}}$

One mole of silver chloride contains one mole of chlorine. So, the mass of chlorine in ${\text{0}}{\text{.5740 g}}$ of silver chloride is calculated as

$ {\text{Mass of AgCl  =  }}\dfrac{{{\text{35}}{\text{.5 }} \times {\text{ 0}}{\text{.5740}}}}{{143.32}}{\text{ }}$ 

$  {\text{ =  0}}{\text{.1421 g}}$

Thus, the percentage of chlorine is calculated as

${\text{Percentage of chlorine  =  }}\dfrac{{0.1421}}{{0.3780}}{\text{ }} \times {\text{ 100% }}$

  ${\text{ =  37}}{\text{.59 % }} $

Hence, the percentage of chlorine is ${\text{37}}{\text{.59 % }}{\text{.}}$


35. In the estimation of sulphur by Carius method, $0.468{\text{ g}}$ of an organic sulphur compound afforded ${\text{0}}{\text{.668 g}}$ of barium sulphate. Find out the percentage of sulphur in the given compound.

Ans: Given: The total mass of organic compound is $0.468{\text{ g}}$ and the mass of barium sulphate formed is ${\text{0}}{\text{.668 g}}{\text{.}}$

The molar mass of ${\text{BaS}}{{\text{O}}_{\text{4}}}$ is $233{\text{ g}}/{\text{mol}}$ and the molar mass of sulphur is $32{\text{ g}}/{\text{mol}}{\text{.}}$

So, 

${\text{1 mol of BaS}}{{\text{O}}_{\text{4}}}{\text{  =  233 g of BaS}}{{\text{O}}_{\text{4}}}{\text{ }} $

$  {\text{ =  32 g of sulphur}} $

Thus, 

$0.668{\text{ g of BaS}}{{\text{O}}_{\text{4}}}{\text{  =  }}\dfrac{{{\text{32 }} \times {\text{ 0}}{\text{.668}}}}{{233}}{\text{ g of sulphur}} $

$  {\text{ =  0}}{\text{.0917 g of sulphur}}$

Therefore, the percentage of sulphur is calculated as

${\text{Percentage of sulphur  =  }}\dfrac{{0.0917}}{{0.468}}{\text{ }} \times {\text{ 100% }} $

 $ {\text{ =  19}}{\text{.59 % }} $

Hence, the percentage of sulphur is ${\text{19}}{\text{.59 % }}{\text{.}}$


36. In the organic compound ${\text{C}}{{\text{H}}_{\text{2}}}{\text{  =  CH -- C}}{{\text{H}}_{\text{2}}}{\text{ -- C}}{{\text{H}}_{\text{2}}}{\text{ -- C }} \equiv {\text{ CH,}}$the pair of hydridised orbitals involved in the formation of: ${\text{C2 -- C3}}$ bond is: 

$\left( {\text{a}} \right){\text{ }}sp{\text{ }}--{\text{ }}s{p^2}{\text{ }} $

  $\left( {\text{b}} \right){\text{ }}sp{\text{ }}--{\text{ }}s{p^3}{\text{ }} $

 $ \left( {\text{c}} \right){\text{ }}s{p^2}{\text{ }}--{\text{ }}s{p^3}{\text{ }} $

$  \left( {\text{d}} \right){\text{ }}s{p^3}{\text{ }}--{\text{ }}s{p^3} $ 

Ans: The given organic compound is shown below.

$\mathop {\text{C}}\limits^1 {{\text{H}}_{\text{2}}}{\text{  =  }}\mathop {\text{C}}\limits^2 {\text{H -- }}\mathop {\text{C}}\limits^3 {{\text{H}}_{\text{2}}}{\text{ -- }}\mathop {\text{C}}\limits^4 {{\text{H}}_{\text{2}}}{\text{ -- }}\mathop {\text{C}}\limits^5 {\text{ }} \equiv {\text{ }}\mathop {\text{C}}\limits^6 {\text{H}}$

The hybridization of carbon atoms numbered as 1, 2, 3, 4, 5 and 6 are $sp,{\text{ }}sp,{\text{ }}s{p^3}{\text{ }},{\text{ }}s{p^3}{\text{ }},{\text{ }}s{p^2}{\text{ }},{\text{ and }}s{p^2}$ respectively. Therefore, the pair of hybridized orbitals involved in the formation of ${\text{C2 -- C3}}$ bond is $sp{\text{ }}--{\text{ }}s{p^3}.$


37. In the Lassaigne’s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of: 

$ \left( {\text{a}} \right){\text{ N}}{{\text{a}}_{\text{4}}}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]{\text{ }} $

  $\left( {\text{b}} \right){\text{ F}}{{\text{e}}_{\text{4}}}{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]_{\text{3}}}{\text{ }} $

 $ \left( {\text{c}} \right){\text{ F}}{{\text{e}}_{\text{2}}}\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]{\text{ }} $

  $\left( {\text{d}} \right){\text{ F}}{{\text{e}}_{\text{3}}}{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]_{\text{4}}} $ 

Ans: The sodium fusion extract is heated with iron (II) sulphate and then acidified with sulphuric acid in the Lassaigne's test for nitrogen in an organic molecule. Sodium cyanide initially interacts with iron (II) sulphate to generate sodium hexacyanoferrate (II). The iron (II) is then oxidized by sulphuric acid to generate iron (III) hexacyanoferrate (II), which is Prussian blue. The chemical reactions involved are shown below. 

$6{\text{C}}{{\text{N}}^ - }{\text{ }} + {\text{ F}}{{\text{e}}^{2 + }}{\text{ }} \to {\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{4 - }} $

  $3{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]^{4 - }}{\text{ }} + {\text{ 4F}}{{\text{e}}^{3 + }}{\text{ }}\xrightarrow{{x.{{\text{H}}_{\text{2}}}{\text{O}}}}{\text{ }}\mathop {{\text{F}}{{\text{e}}_4}{{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]}_3}.x{{\text{H}}_{\text{2}}}{\text{O}}}\limits_{{\text{Prussian Blue}}}  $ 

Therefore, the Prussian blue color is formed due to the formation of complex ${\text{F}}{{\text{e}}_4}{\left[ {{\text{Fe}}{{\left( {{\text{CN}}} \right)}_{\text{6}}}} \right]_3}.$


38. Which of the following carbocation is most stable? 

$\left( {\text{a}} \right){\text{ }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_3}{\text{C}}{\text{.}}\mathop {\text{C}}\limits^{\text{ + }} {{\text{H}}_{\text{2}}} $

  $\left( {\text{b}} \right){\text{ }}{\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_3}\mathop {\text{C}}\limits^{\text{ + }}  $

 $ \left( {\text{c}} \right){\text{ C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_2}\mathop {\text{C}}\limits^{\text{ + }} {{\text{H}}_{\text{2}}} $

  $\left( {\text{d}} \right){\text{ C}}{{\text{H}}_{\text{3}}}\mathop {\text{C}}\limits^ +  {\text{HC}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{H}}_{\text{3}}} $ 

Ans: ${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_3}\mathop {\text{C}}\limits^{\text{ + }} $ is a tertiary carbocation. Because of the electron-releasing effect of three methyl groups, a tertiary carbocation is the most stable carbocation. Three methyl groups increase the + I effect, which stabilizes the positive charge on the carbocation. The stability of tertiary carbocations is also due to hyperconjugation and resonance energy.


39. The best and latest technique for isolation, purification and separation of organic compounds is: 

  1. Crystallization 

  2. Distillation 

  3. Sublimation 

  4. Chromatography

Ans: Chromatography is the most effective and up-to-date method of separating and purifying organic molecules. Initially, it was employed to separate a combination of colored substances.


40. The reaction: ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{I  +  KO}}{{\text{H}}_{\left( {aq} \right)}}{\text{ }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH  +  KI}}$ is classified as: 

  1. Electrophilic substitution 

  2. Nucleophilic substitution 

  3. Elimination 

  4. Addition

Ans: The reaction is given as:

${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{I  +  KO}}{{\text{H}}_{\left( {{\text{aq}}} \right)}}{\text{ }} \to {\text{ C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH  +  KI}}$

It shows a nucleophilic substitution reaction. The hydroxyl group of potassium hydroxide with a lone pair serves as a nucleophile, substituting iodide ion in ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{I}}$ to produce ethanol.


Class 11 Chemistry Chapter 8 Quick Overview of Topics

Chemistry class 11 Chapter 8 NCERT Solutions - Quick Overview of Detailed Structure of Topics and Subtopics Covered.


Topic

Subtopics

Introduction to Organic Molecules

Definition of organic molecules, the distinction between organic and inorganic compounds

Classification of Organic Compounds

Classification is based on functional groups, such as alkanes, alkenes, alkynes, alcohols, carboxylic acids, esters, etc.

Isomerism

Structural isomerism, stereoisomerism, geometric isomerism, optical isomerism, examples and explanations

Nomenclature of Organic Compounds

IUPAC nomenclature rules for naming organic compounds, naming alkanes, alkenes, alkynes, and other functional groups

Introduction to Hydrocarbons

Definition of hydrocarbons, types of hydrocarbons (alkanes, alkenes, alkynes), properties and examples

Structure and Bonding in Organic Molecules

Covalent bonding in organic molecules, hybridisation, sigma and pi bonds, structural formulas, molecular formulas

Chemical Properties of Organic Compounds

Combustion, substitution, addition, and elimination reactions in organic compounds



Some Basic Important Concepts for Organic Chemistry Class 11

Class 11 NCERT solutions help the students to go through the concepts easily. Here, find the Important concepts of  Chapter 8 -Organic Chemistry -Some Basic Principles and Techniques to crack your exams.


  1. Nomenclature:

    • IUPAC (International Union of Pure and Applied Chemistry) rules are used to name organic compounds systematically based on their structure and functional groups.

    • Prefixes, suffixes, and numbering systems are employed to denote substituents, parent chains, and functional groups.

  2. Reactivity of Organic Compounds:

    • Organic compounds exhibit diverse reactivity patterns based on functional groups and structural features.

    • Common reactions include addition, elimination, substitution, oxidation, and reduction reactions.

  3. Isomerism:

    • Isomers are compounds with the same molecular formula but different structural arrangements or spatial orientations.


Benefits of Referring to Vedantu’s NCERT Solutions for Class 12 Chemistry Chapter 8 -Organic Chemistry 

The Vedantu’s Class 11 NCERT Solutions of Chapter 8 -Organic Chemistry, provided here in PDFs, offer various benefits, including:


  • Detailed explanations and step-by-step solutions for all topics in Organic Chemistry.

  • Solutions curated by experienced educators to ensure accuracy and clarity.

  • Solutions to a variety of problems to strengthen analytical and problem-solving abilities.

  • Step-by-step solutions for numerical problems and reaction mechanisms.

  • class 11 Chemistry Chapter 8 NCERT solutions will give you insights about the General Introduction: methods of purification, qualitative and quantitative analysis, classification, and IUPAC nomenclature of organic compounds.

  • The section will give you crisp learnings on Electronic displacements in a covalent bond: inductive effect, electromeric effect, resonance, and hyperconjugation.


Important Study Materials for Class 11 Chemistry Chapter 8 NCERT solutions

Students can access extra study materials on Organic Chemistry -Some Basic Principles and Techniques. These resources are available for download and offer additional support for your studies.




Conclusion

Class 11 chemistry chapter organic chemistry NCERT solutions provide a comprehensive and reliable resource for students studying this chapter. These solutions cover the essential concepts and topics related to organic chemistry, including nomenclature, isomerism, and reactions of alkanes, alkenes, and alkynes. By utilising these solutions, students can enhance their understanding, clarify their doubts, and improve their problem-solving skills. The step-by-step explanations and detailed answers offered in the NCERT Solutions ensure that students grasp the concepts effectively. Overall, the NCERT organic chemistry class 11 solutions are a valuable tool for students to excel in their exams and gain a strong foundation in organic chemistry.


NCERT Solutions Class 11 Chemistry | Chapter-wise Links 

Access Vedantu’s chapter-wise NCERT Chemistry Class 11 Solutions PDFs below for all other chapters.




NCERT Solutions Class 11 Chemistry - Related Links

FAQs on NCERT Solutions for Class 11 Chemistry Chapter 8 - Organic Chemistry

1. What is Isomerism and What are its Types?

If two or more molecules have the same molecular formula but different structural formulas along with different physical and chemical characteristics, then this phenomenon is termed isomerism. Isomerism is of two types:

  • Structural Isomerism - When two molecules have the same molecular formula, but their structural formula differs concerning the arrangement of atoms. Those molecules are said to exhibit structural isomerism.

  • Stereoisomerism - When atoms or groups in space are arranged differently, then that phenomenon is called stereoisomerism. Molecules that are stereoisomers of another molecule have the same structural formula, but the arrangement of atoms in space is different.

  • Stereoisomerism is further divided into two types:

  • Geometrical isomerism

  • Optical isomerism

2. What Happens in an Organic Reaction?

The fundamental concept of organic reaction is based on the fission of covalent bonds. A covalent bond undergoes fission in two ways:

  • Homolytic Fission - This is also referred to as homolytic fission where each of the atoms acquires one of the bonding electrons.

  • Heterolytic Fission - When the bond is broken in such fission, one of the atoms acquire both of the bonding electrons.

The process of fission gives rise to reaction intermediaries that are short-lived fragments. Some of the important reaction intermediaries are carbanions, carbenes, carbonium ions, and carbon-free radicals. Students can also Download organic chemistry class 11 NCERT solutions .

3. What are the topics in Chapter 8, Organic Chemistry of Class 11 Chemistry?

Organic chemistry class 11 NCERT solutions that deal with only carbon compounds and where other elements will be limited. The topics  of organic chemistry are as follows;

  1. Tetravalency of carbon.

  2. Homologous Series

  3. Organic Compounds and Their Classification

  4. Isomerism - Structural Isomerism and Stereoisomerism (with Geometrical Isomerism and Optical Isomerism)

  5. Attacking Reagents - Electrophile and Nucleophiles

  6. Fundamental concepts.

  7. Methods of purification of organic compounds.

  8. Quantitative analysis.

4. What are the four main organic compounds found in the body, as discussed in Chapter 8, Organic Chemistry of Class 11 Chemistry?

There are four major organic compounds found in the body. They are:

  • Carbohydrates- Carbohydrates are the organic compounds that contain carbon, hydrogen and oxygen atoms in the ratio of 1:2:1.

  • Lipids or Fats- lipids consist of fats, oils and waxes. These help in storing energy from the structural cells which serve as insulation in organisms.

  • Proteins- These are responsible for certain reactions in organisms.

  • Nucleic Acid - Two types of nucleic acids are deoxyribonucleic acid(DNA) and ribonucleic acid(RNA). DNA is responsible for the genetic code of the organism.  The production of proteins is done by RNA.

5. What is the importance of Chapter 12, Organic Chemistry of Class 11 Chemistry?

Class 11 chemistry chapter organic chemistry NCERT solutions is very important, not only for board exams in Class 12 but also because a lot of concepts are included in the syllabus for competitive exams such as NEET and JEE. It will also help you understand future concepts in Chemistry. The conceptual understanding of this ch 8 chemistry class 11 NCERT solutions will be more advanced, and unless the basic concepts are clear, it will not be possible to get a grasp of the subject. Students who want to continue the science stream need to properly grasp the concepts.

6. Define isomerism and the types according to Class 11 Chemistry ch 8 NCERT Solutions.

Isomerism is defined as the two molecules having the same molecular formula but different structural formulas. This will have different physical and chemical characteristics. The types are as below.

  • Structural Isomerism - In structural Isomerism, the two molecules have the same molecular formula but the structural formula may be different according to the arrangements of the atoms.

  • Stereo Isomerism - In this Isomerism, the atoms and the groups in the space are arranged differently. Molecules in this will have the same structural formula with different atoms arrangement.

Students Can also download the class 11 chemistry ch 8 NCERT solutions.

7. Define Homolytic Fission and Heterolytic Fission according to Class 11 Chemistry Chapter Organic Chemistry NCERT solutions.

The fission of the covalent bonds is the main concept of the organic reaction. Fission is of two types.

  • Homolytic Fission - In this type of fission, each atom gets the bonding electrons.

  • Heterolytic Fission- In this type when the bond breaks then the atom will get both the bonding electrons.

The class 11 chemistry chapter 8 NCERT solutions is very well explained in NCERT solutions from Vedantu in a very simplified way. The study material is available to download for free from Vedantu’s website (vedantu.com).

8. What is the additional reaction in organic chemistry?

In organic chemistry, the additional reaction involves the addition of one or more atoms or groups of atoms to a molecule, resulting in the formation of a single product. Download the Organic Chemistry Class 11 NCERT Solutions PDF for good practices to crack your exams.

9. What are the conditions for an additional reaction?

For an additional reaction to occur, unsaturated organic compounds like alkenes or alkynes must be present, along with a suitable reagent such as a halogen or hydrogen halide, and often a catalyst or appropriate temperature and pressure. Download the Organic Chemistry Class 11 NCERT Solutions PDF for good practices to crack your exams.

10. What is the hardest topic in class 11 organic chemistry NCERT solutions?

The hardest topic in chemistry Class 11 is subjective, but some students find organic chemistry challenging due to its extensive nomenclature, diverse reactions, and conceptual intricacies.

11. What is another name for the addition reaction?

Another name for the addition reaction in organic chemistry is addition polymerisation, where monomers join together to form a polymer by adding to double or triple bonds. Download class 11 organic chemistry NCERT solutions to get better expertise in the topic.

12. Which is the toughest chapter in organic chemistry?

The toughest chapter in organic chemistry varies among students. Still, some may find chapters dealing with reaction mechanisms, such as electrophilic addition or nucleophilic substitution, to be particularly challenging due to their conceptual depth and complexity.