{"id":866520,"date":"2025-08-13T18:05:59","date_gmt":"2025-08-13T12:35:59","guid":{"rendered":"https:\/\/leverageedu.com\/discover\/?p=866520"},"modified":"2025-08-13T18:05:59","modified_gmt":"2025-08-13T12:35:59","slug":"ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry","status":"publish","type":"post","link":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/","title":{"rendered":"NCERT Solutions Class 11 Chemistry (Part-2) Chapter 8: Organic Chemistry (Free PDF)"},"content":{"rendered":"\n

In this unit, you were introduced to the basic concepts of organic chemistry, such as the tetravalence of carbon, shapes of organic compounds, structural representations, classification based on structure and functional groups, types of organic reactions, purification methods, and qualitative and quantitative analysis of organic compounds. Below, we have provided the NCERT Solutions for Class 11 Chemistry (Part II), Chapter 8: Organic Chemistry \u2013 Some Basic Principles and Techniques, which will help you revise these concepts effectively.<\/p>\n\n\n\n\n\n\n

Explore Notes of Class 11 Chemistry<\/strong><\/p>\n\n\n\n

Chapter 1<\/a><\/strong><\/td>Chapter 2<\/a><\/strong><\/td>Chapter 3<\/a><\/strong><\/td>Chapter 4<\/strong><\/a><\/td>Chapter 5<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

NCERT Solutions Class 11 Chemistry (Part-2) Chapter 8: Organic Chemistry<\/h2>\n\n\n\n

Below, we have provided you with the exercises mentioned in the NCERT Solutions for Class 11 Chemistry (Part II), Chapter 8: Organic Chemistry \u2013 Some Basic Principles and Techniques<\/p>\n\n\n\n

Exercises<\/h2>\n\n\n\n
    \n
  1. What are the hybridisation states of each carbon atom in the following compounds?
    CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6<\/li>\n\n\n\n
  2. Indicate the \u03c3 and \u03c0 bonds in the following molecules:
    C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3<\/li>\n\n\n\n
  3. Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.<\/li>\n\n\n\n
  4. Which of the following represents the correct IUPAC name for the compounds concerned?
    (a) 2,2-Dimethylpentane or 2-Dimethylpentane
    (b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane
    (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane
    (d) But-3-yn-1-ol or But-4-ol-1-yne<\/li>\n\n\n\n
  5. Draw formulas for the first five members of each homologous series beginning with the following compounds:
    (a) H\u2013COOH (b) CH3COCH3 (c) H\u2013CH=CH2<\/li>\n\n\n\n
  6. Give condensed and bond-line structural formulas and identify the functional group(s) present, if any, for:
    (a) 2,2,4-Trimethylpentane
    (b) 2-Hydroxy-1,2,3-propane-tricarboxylic acid
    (c) Hexanedial<\/li>\n\n\n\n
  7. Identify the functional groups in the following compounds:
    (a) (b) (c)<\/li>\n\n\n\n
  8. Which of the two: O2NCH2CH2O\u2013 or CH3CH2O\u2013 is expected to be more stable, and why?<\/li>\n\n\n\n
  9. Explain why alkyl groups act as electron donors when attached to a \u03c0 system.<\/li>\n\n\n\n
  10. Draw the resonance structures for the following compounds. Show the electron shift using curved-arrow notation:
    (a) C6H5OH (b) C6H5NO2 (c) CH3CH=CHCHO (d) C6H5\u2013CHO (e) C6H5\u2013C+H2 (f) CH3CH=CHC+H2<\/li>\n\n\n\n
  11. What are electrophiles and nucleophiles? Explain with examples.<\/li>\n\n\n\n
  12. Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
    (a) CH3COOH + HO\u2013 \u2192 CH3COO\u2013 + H2O
    (b) CH3COCH3 + C\u2013N \u2192 (CH3)2C(CN)(OH)
    (c) C6H6 + CH3C+O \u2192 C6H5COCH3<\/li>\n\n\n\n
  13. Classify the following reactions in one of the reaction types studied in this unit:
    (a) CH3CH2Br + HS\u2013 \u2192 CH3CH2SH + Br\u2013
    (b) (CH3)2C = CH2 + HCl \u2192 (CH3)2CCl \u2013 CH3
    (c) CH3CH2Br + HO\u2013 \u2192 CH2 = CH2 + H2O + Br\u2013
    (d) (CH3)3C\u2013CH2OH + HBr \u2192 (CH3)2CBrCH2CH2CH3 + H2O<\/li>\n\n\n\n
  14. Explain the terms Inductive and electromagnetic effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids?
    (a) Cl3CCOOH > Cl2CHCOOH > ClCH2COOH
    (b) CH3CH2COOH > (CH3)2CHCOOH > (CH3)3CCOOH<\/li>\n\n\n\n
  15. Give a brief description of the principles of the following techniques, taking an example in each case:
    (a) Crystallisation (b) Distillation (c) Chromatography<\/li>\n\n\n\n
  16. Describe the method that can be used to separate two compounds with different solubilities in a solvent S.<\/li>\n\n\n\n
  17. What is the difference between distillation, distillation under reduced pressure, a nd steam distillation?<\/li>\n\n\n\n
  18. Discuss the chemistry of Lassaigne\u2019s test.<\/li>\n\n\n\n
  19. Differentiate between the principles of estimation of nitrogen in an organic compound by:
    (i) Dumas method and (ii) Kjeldahl\u2019s method.<\/li>\n\n\n\n
  20. Discuss the principle of estimation of halogens, sulphur, and phosphorus present in an organic compound.<\/li>\n\n\n\n
  21. Explain the principle of paper chromatography.<\/li>\n\n\n\n
  22. Why is nitric acid added to the sodium extract before adding silver nitrate for testing halogens?<\/li>\n\n\n\n
  23. Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulfur, and halogens.<\/li>\n\n\n\n
  24. Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.<\/li>\n\n\n\n
  25. Explain why an organic liquid vaporises at a temperature below its boiling point in its steam distillation.<\/li>\n\n\n\n
  26. Will CCl4 give a white precipitate of AgCl on heating it with silver nitrate? Give a reason for your answer.<\/li>\n\n\n\n
  27. Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?<\/li>\n\n\n\n
  28. Why is it necessary to use acetic acid and not sulphuric acid for acidification of the sodium extract for testing sulphur by the lead acetate test?<\/li>\n\n\n\n
  29. An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.<\/li>\n\n\n\n
  30. A sample of 0.50 g of an organic compound was treated according to Kjeldahl\u2019s method. The ammonia evolved was absorbed in 50 mL of 0.5 M H2SO4. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.<\/li>\n\n\n\n
  31. 0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.<\/li>\n\n\n\n
  32. In the estimation of sulphur by the Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.<\/li>\n\n\n\n
  33. In the organic compound CH2 = CH \u2013 CH2 \u2013 CH2 \u2013 C \u2261 CH, the pair of hybridised orbitals involved in the formation of the C2 \u2013 C3 bond is:
    (a) sp \u2013 sp2 (b) sp \u2013 sp3 (c) sp2 \u2013 sp3 (d) sp3 \u2013 sp3<\/li>\n\n\n\n
  34. In Lassaigne\u2019s test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:
    (a) Na4[Fe(CN)6] (b) Fe4[Fe(CN)6]3 (c) Fe2[Fe(CN)6] (d) Fe3[Fe(CN)6]4<\/li>\n\n\n\n
  35. Which of the following carbocations is most stable?
    (a) (CH3)3C.C+H2 (b) (CH3)3C+ (c) CH3CH2C+H2 (d) CH3C+HCH2CH3<\/li>\n\n\n\n
  36. The best and latest technique for isolation, purification, and separation of organic compounds is:
    (a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography<\/li>\n\n\n\n
  37. The reaction:
    CH3CH2I + KOH(aq) \u2192 CH3CH2OH + KI
    It is classified as:
    (a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition<\/li>\n<\/ol>\n\n\n\n

    Also Read: <\/strong>NCERT Notes Class 11 Political Science Indian Constitution at Work Chapter 6: Judiciary (Free PDF)<\/strong><\/a><\/p>\n\n\n\n

    Solutions<\/h2>\n\n\n\n
      \n
    1. For CH\u2082=C=O the terminal CH\u2082 carbon is sp\u00b2 and the central carbon (connected to CH\u2082 and O) is sp; for CH\u2083CH=CH\u2082 the CH\u2083 carbon is sp\u00b3, the middle =CH is sp\u00b2 and the terminal =CH\u2082 is sp\u00b2; for (CH\u2083)\u2082CO the two methyl carbons are sp\u00b3 and the carbonyl carbon is sp\u00b2; for CH\u2082=CHCN the vinyl CH\u2082 is sp\u00b2, the vinyl CH is sp\u00b2 and the nitrile carbon is sp; for C\u2086H\u2086 all six ring carbons are sp\u00b2.<\/li>\n\n\n\n
    2. Benzene (C\u2086H\u2086) has 12 \u03c3 bonds (six C\u2013C and six C\u2013H) and a delocalised 6\u03c0 system usually represented as three \u03c0 bonds; saturated C\u2086H\u2081\u2082 (cyclohexane assumed) has only \u03c3 bonds (6 C\u2013C + 12 C\u2013H = 18 \u03c3, 0 \u03c0); CH\u2082Cl\u2082 has four single bonds (4 \u03c3, 0 \u03c0); CH\u2082=C=CH\u2082 (allene) has six \u03c3 bonds (two C\u2013C and four C\u2013H) and two \u03c0 bonds (two orthogonal \u03c0 systems); CH\u2083NO\u2082 has six \u03c3 bonds (three C\u2013H, one C\u2013N, two N\u2013O) and a delocalised \u03c0 system in the nitro group (commonly treated as one delocalised \u03c0 set across N\u2013O); HCONHCH\u2083 (N-methylformamide) has one C=O \u03c0 and the remaining bonds are \u03c3, with C\u2013N having partial \u03c0 character by resonance.<\/li>\n\n\n\n
    3. Isopropyl alcohol: condensed formula (CH\u2083)\u2082CHOH and bond-line is the central carbon bonded to OH and two methyl groups; 2,3-Dimethylbutanal: condensed CH\u2083\u2013CH(CH\u2083)\u2013CH(CH\u2083)\u2013CHO and bond-line is a four-carbon chain with \u2013CHO at one end and methyl substituents at C2 and C3; Heptan-4-one: condensed CH\u2083\u2013CH\u2082\u2013CH\u2082\u2013CO\u2013CH\u2082\u2013CH\u2082\u2013CH\u2083 and bond-line is a seven-carbon chain with a C=O at carbon 4.<\/li>\n\n\n\n
    4. The correct IUPAC forms are: 2,2-Dimethylpentane (not \u201c2-Dimethylpentane\u201d); 2,4,7-Trimethyloctane (the lowest set of locants); 2-Chloro-4-methylpentane (substituents listed alphabetically, showing the 2-chloro, 4-methyl locants); but-3-yn-1-ol (correct locant placement and suffixing).<\/li>\n\n\n\n
    5. First five members starting from H\u2013COOH are: HCOOH (methanoic acid), CH\u2083COOH (ethanoic acid), C\u2082H\u2085COOH (propanoic acid), C\u2083H\u2087COOH (butanoic acid), C\u2084H\u2089COOH (pentanoic acid). First five members starting from CH\u2083COCH\u2083 (ketone at C2) are CH\u2083COCH\u2083 (propanone), CH\u2083CH\u2082COCH\u2083 (butan-2-one), CH\u2083CH\u2082CH\u2082COCH\u2083 (pentan-2-one), hexan-2-one, heptan-2-one. First five members starting from H\u2013CH=CH\u2082 are CH\u2082=CH\u2082 (ethene), CH\u2083CH=CH\u2082 (propene), CH\u2083CH\u2082CH=CH\u2082 (but-1-ene), pent-1-ene, hex-1-ene.<\/li>\n\n\n\n
    6. 2,2,4-Trimethylpentane condensed formula is CH\u2083\u2013C(CH\u2083)\u2082\u2013CH\u2082\u2013CH(CH\u2083)\u2013CH\u2083, bond-line is the pentane backbone with methyls at C2 and C4, and it has no functional group (an alkane). 2-Hydroxy-1,2,3-propane-tricarboxylic acid (citric acid) can be written HO\u2013CH(COOH)\u2013CH(COOH)\u2013CH\u2082\u2013COOH, bond-line shows three carboxyl groups and one hydroxyl; functional groups are three \u2013COOH and one \u2013OH. Hexanedial is OHC\u2013(CH\u2082)\u2084\u2013CHO (hexane-1,6-dial), bond-line shows aldehyde groups at both ends, and the functional groups are two \u2013CHO (aldehydes).<\/li>\n\n\n\n
    7. O\u2082NCH\u2082CH\u2082O\u207b is more stable than CH\u2083CH\u2082O\u207b because the nitro group is strongly electron withdrawing (inductive and resonance effects) and stabilises negative charge, whereas an ethyl group is electron donating and destabilises an adjacent negative charge.<\/li>\n\n\n\n
    8. Alkyl groups act as electron donors to a \u03c0 system mainly by the inductive (+I) effect through \u03c3 bonds and by hyperconjugation, where C\u2013H (or C\u2013C) \u03c3 orbitals overlap with adjacent \u03c0 orbitals or empty p orbitals and delocalise electron density into the \u03c0 system, thereby increasing electron density on the \u03c0 system and stabilising adjacent positive charge.<\/li>\n\n\n\n
    9. Phenol (C\u2086H\u2085OH) resonates by donation of an oxygen lone pair into the ring, producing resonance forms with negative charge at ortho and para positions and positive charge on oxygen (lone pair \u2192 C\u2013O \u03c0, then \u03c0 shifts arouthe nd ring). Nitrobenzene (C\u2086H\u2085NO\u2082) resonates by withdrawing electron density from the ring toward the nitro group, giving ring-positive forms and a negative charge on nitro oxygens (ring \u03c0 \u2192 N, N=O \u03c0 \u2192 O). CH\u2083CH=CHCHO (an \u03b1,\u03b2-unsaturated aldehyde) resonates by conjugation of C=C and C=O: one form has C=O and C=C, another has C=O \u03c0 electrons on O (O\u207b) with the \u03c0 electrons of C=C shifted to place positive character at the \u03b2 carbon. Benzaldehyde (C\u2086H\u2085\u2013CHO) resonates with ring \u2192 carbonyl electron donation, producing O\u207b and positive character delocalised on the ring at ortho\/para. The benzylic carbocation (C\u2086H\u2085\u2013C\u207aH\u2082) is stabilised by resonance into the aromatic ring with positive charge delocalised to ortho\/para positions. An allylic carbocation CH\u2083CH=CHCH\u2082 is stabilised by resonance with the positive charge delocalised across the allylic system (\u03c0 electrons shift to place the + on different carbons). (If you want the curved-arrow diagrams, I can draw them for any lettered case.)<\/li>\n\n\n\n
    10. Electrophiles are electron-poor species that accept an electron pair (Lewis acids), for example, H\u207a, NO\u2082\u207a, R\u2013C\u207a; nucleophiles are electron-rich species that donate an electron pair (Lewis bases), for example,e OH\u207b, CN\u207b, NH\u2083.<\/li>\n\n\n\n
    11. In CH\u2083COOH + HO\u207b \u2192 CH\u2083COO\u207b + H\u2082O, the reagent HO\u207b is a nucleophile (base); in CH\u2083COCH\u2083 + C\u207bN \u2192 (CH\u2083)\u2082C(CN)(OH), the cyanide ion C\u207bN is a nucleophile; in C\u2086H\u2086 + CH\u2083C\u207aO \u2192 C\u2086H\u2085COCH,\u2083 the acyl-type CH\u2083C\u207aO species is an electrophile.<\/li>\n\n\n\n
    12. CH\u2083CH\u2082Br + HS\u207b \u2192 CH\u2083CH\u2082SH + Br\u207b is a nucleophilic substitution (S_N2); (CH\u2083)\u2082C=CH\u2082 + HCl \u2192 (CH\u2083)\u2082CCl\u2013CH\u2083 is electrophilic addition to an alkene (Markovnikov addition); CH\u2083CH\u2082Br + HO\u207b \u2192 CH\u2082=CH\u2082 + H\u2082O + Br\u207b is an elimination (E2); (CH\u2083)\u2083C\u2013CH\u2082OH + HBr \u2192 (CH\u2083)\u2082CBrCH\u2082CH\u2082CH\u2083 + H\u2082O indicates nucleophilic substitution via a carbocation with rearrangement (S_N1 with rearrangement).<\/li>\n\n\n\n
    13. The inductive effect is a permanent electron displacement along \u03c3 bonds due to electronegativity differences (it falls off with distance); the electromeric effect is a temporary, reagent-induced transfer of \u03c0 electrons during a reaction (instantaneous). The acidity order Cl\u2083CCOOH > Cl\u2082CHCOOH > ClCH\u2082COOH is explained by the inductive (\u2212I) effect of chlorine atoms stabilising the conjugate base, and the order CH\u2083CH\u2082COOH > (CH\u2083)\u2082CHCOOH > (CH\u2083)\u2083CCOOH is due to +I and hyperconjugation of alkyl groups destabilising the conjugate base (more alkyl substitution \u2192 less acidic).<\/li>\n\n\n\n
    14. Crystallisation (recrystallisation) separates and purifies solids based on different solubilities in a solvent, using a hot solvent to dissolve the compound, then cooling to give pure crystals (example: purifying benzoic acid). Distillation separates liquids by differences in boiling point; fractional distillation is used when bps are close (example: separating miscible liquids of different bps). Chromatography separates components by differential partitioning\/adsorption between a stationary phase and a mobile phase (example: column chromatography to separate plant pigments).<\/li>\n\n\n\n
    15. To separate two compounds with different solubilities in solvent S, dissolve the mixture in S where one compound (A) is soluble and the other (B) is insoluble, then filter off B and evaporate S to recover A; if needed, repeat or use extraction with immiscible solvents to partition components.<\/li>\n\n\n\n
    16. Distillation heats a mixture to boil and condense the more volatile component at atmospheric pressure; distillation under reduced pressure (vacuum distillation) lowers the external pressure so components boil at lower temperatures (useful for high-boiling or thermally sensitive compounds); steam distillation co-distils volatile organic compounds with water so they vaporise at a temperature below their normal boiling point because the total vapour pressure of water plus organic reaches atmospheric pressure at a lower temperature.<\/li>\n\n\n\n
    17. Lassaigne\u2019s test involves fusing an organic sample with metallic sodium to convert covalently bound nitrogen, sulfur and halogens into ionic forms (NaCN, Na\u2082S, NaX) which are extracted with water and tested: nitrogen yields Prussian blue with Fe\/acid (indicates CN\u207b \u2192 Fe\u2084[Fe(CN)\u2086]\u2083), sulfur gives PbS (black) or BaSO\u2084 after oxidation, and halogens give AgX precipitates on acidification and addition of AgNO\u2083.<\/li>\n\n\n\n
    18. The Dumas method estimates nitrogen by combustion of the sample to produce N\u2082 gas which is measured (instrumental; detects total N), while Kjeldahl\u2019s method digests the sample with concentrated acid to convert organic N to NH\u2084\u207a, distils the liberated NH\u2083 into standard acid and titrates to determine N (wet chemical method; some forms of nitrogen may need modification to be measured).<\/li>\n\n\n\n
    19. Estimation of halogens, sulfur, and phosphorus is typically done by oxidative decomposition (Carius method or similar), converting halogens to halide ions, which are precipitated and weighed as AgX, sulfur to sulfate precipitated as BaSO\u2084 and weighed, and phosphorus to orthophosphate, which can be precipitated or determined colorimetrically as phosphomolybdate or gravimetrically after suitable conversion.<\/li>\n\n\n\n
    20. Paper chromatography separates components by partition between the mobile solvent and the stationary water film on cellulose paper; components move with the solvent front to different extents according to their relative solubilities and affinities, giving distinct Rf values.<\/li>\n\n\n\n
    21. Nitric acid is added to the sodium fusion extract before adding AgNO\u2083 to acidify the extract (convert NaX to HX) and to oxidise interfering sulfides or thiocyanates; acidification prevents precipitation of Ag\u2082O by OH\u207b and avoids interfering side reactions, giving a clean AgX precipitate for halogen detection.<\/li>\n\n\n\n
    22. Fusion with metallic sodium is done because many heteroatoms (N, S, halogens) in organic compounds are covalently bound and not present as ionic species; sodium fusion cleaves bonds and converts these elements into water-soluble ionic forms (NaCN, Na\u2082S, NaX) that can be extracted and detected by specific qualitative tests.<\/li>\n\n\n\n
    23. A suitable separation technique for calcium sulphate and camphor is sublimation (preferably under reduced pressure): camphor sublimes and can be collected as vapour condensate while CaSO\u2084 remains as nonvolatile residue.<\/li>\n\n\n\n
    24. An organic liquid vaporises below its normal boiling point in steam distillation because the vapour pressure of the mixture equals atmospheric pressure when the sum of the partial pressures of water and the organic equals atmospheric pressure, so the organic does not need to reach its pure-component boiling point to enter the vapour phase.<\/li>\n\n\n\n
    25. Carbon tetrachloride (CCl\u2084) will not give a white AgCl precipitate when heated with aqueous AgNO\u2083 because CCl\u2084 is nonpolar and covalent and does not release free Cl\u207b ions into aqueous solution under such conditions; AgCl forms only if chloride ions are present in the aqueous phase.
      <\/li>\n\n\n\n
    26. Potassium hydroxide is used to absorb CO\u2082 evolved during carbon estimation because KOH chemically fixes CO\u2082 as carbonate\/bicarbonate (preventing its escape), so quantitative collection or measurement is possible.<\/li>\n\n\n\n
    27. Acetic acid (a mild, non-oxidising acid) is used rather than sulphuric acid to acidify the sodium extract before lead acetate test for sulfur because strong oxidising or strongly acidic conditions (H\u2082SO\u2084) can oxidise sulfide or produce interfering insoluble salts (e.g., PbSO\u2084) and lead to false results; acetic acid acidifies without such interference, allowing formation\/detection of PbS where appropriate.<\/li>\n\n\n\n
    28. For a 0.20 g sample that is 69% C and 4.8% H (remainder O): mass of carbon = 0.20 \u00d7 0.69 = 0.1380 g, which gives CO\u2082 mass = 0.1380 \u00d7 (44\/12) = 0.5060 g; mass of hydrogen = 0.20 \u00d7 0.048 = 0.00960 g, which gives H\u2082O mass = 0.00960 \u00d7 (18\/2) = 0.08640 g.<\/li>\n\n\n\n
    29. In the Kjeldahl problem: initial H\u2082SO\u2084 moles = 0.050 L \u00d7 0.5 M = 0.02500 mol; NaOH used = 0.060 L \u00d7 0.5 M = 0.03000 mol which neutralises 0.01500 mol H\u2082SO\u2084 (because 2 NaOH per H\u2082SO\u2084), so H\u2082SO\u2084 consumed by NH\u2083 = 0.02500 \u2212 0.01500 = 0.01000 mol; that neutralises 0.02000 mol NH\u2083 (2 NH\u2083 per H\u2082SO\u2084), so moles N = 0.02000 and mass N = 0.02000 \u00d7 14.007 \u2248 0.28014 g; percentage N = (0.28014 \/ 0.50) \u00d7 100 \u2248 56.03%.<\/li>\n\n\n\n
    30. From the Carius chlorination: 0.5740 g AgCl corresponds to moles 0.5740 \/ 143.3212 \u2248 0.0040050 mol Cl, mass Cl = 0.0040050 \u00d7 35.453 \u2248 0.14199 g, percentage Cl = (0.14199 \/ 0.3780) \u00d7 100 \u2248 37.56%.<\/li>\n\n\n\n
    31. From the Carius sulfur estimation: 0.668 g BaSO\u2084 corresponds to moles 0.668 \/ 233.383 \u2248 0.0028623 mol, mass S = 0.0028623 \u00d7 32.06 \u2248 0.09176 g, percentage S = (0.09176 \/ 0.468) \u00d7 100 \u2248 19.61%.<\/li>\n\n\n\n
    32. In CH\u2082=CH\u2013CH\u2082\u2013CH\u2082\u2013C\u2261CH, the C2\u2013C3 bond pairs an sp\u00b2 orbital (from the = carbon) with an sp\u00b3 orbital (from the CH\u2082), so the correct answer is sp\u00b2\u2013sp\u00b3 (option c).<\/li>\n\n\n\n
    33. In Lassaigne\u2019s test, Prussian blue corresponds to Fe\u2084[Fe(CN)\u2086]\u2083 (option b).<\/li>\n\n\n\n
    34. The most stable carbocation among the choices is the tertiary carbocation (CH\u2083)\u2083C\u207a (option b).<\/li>\n\n\n\n
    35. The best and latest technique for isolation, purification, and separation of organic compounds is chromatography (option d).<\/li>\n\n\n\n
    36. The reaction CH\u2083CH\u2082I + KOH(aq) \u2192 CH\u2083CH\u2082OH + KI is classified as nucleophilic substitution (option b).<\/li>\n<\/ol>\n\n\n\n

      Also Read<\/strong>: NCERT Notes Class 11 Geography Fundamentals of Geography Chapter 6 Landforms and their Evolution (Free PDF<\/a>)<\/p>\n\n\n\n

      NCERT Solutions Class 11 Chemistry (Part-2) Chapter 8: Organic Chemistry<\/h2>\n\n\n\n
      Download PDF of NCERT Solutions Class 11 Chemistry (Part-II) Chapter-8: Organic Chemistry- Some Basic Principles and Techniques<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

      Download the Solutions of Other Chapters of Class 11 Chemistry<\/strong><\/p>\n\n\n\n

      Chapter 1<\/strong><\/a><\/td>Chapter 2<\/strong><\/a><\/td>Chapter 3<\/strong><\/a><\/td>Chapter 4<\/strong><\/a><\/td>Chapter 5<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

      Related Reads<\/strong><\/p>\n\n\n\n

      Theories of Learning in Psychology: Definition, Types, and Concepts Involved<\/strong><\/a><\/td>How to Prepare for UPSC Optional Psychology? Tips and Tricks<\/strong><\/a><\/td><\/tr>
      NCERT Notes Class 11 Geography Fundamentals of Physical Geography Chapter 14: Biodiversity and Conservation (Free PDF)<\/strong><\/a><\/td>NCERT Notes Class 11 Geography Fundamentals of Physical Geography Chapter 13: Movements of Ocean Water (Free PDF)<\/strong><\/a><\/td><\/tr>
      NCERT Notes Class 11 Geography Fundamentals of Physical Geography Chapter 12: Water (Oceans) (Free PDF)<\/strong><\/a><\/td>NCERT Notes Class 11 Geography Fundamentals of Physical Geography Chapter 11: World Climate and Climate Change (Free PDF)<\/strong><\/a><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

      For more topics, follow LeverageEdu NCERT Study Material<\/strong><\/a> today! <\/p>\n","protected":false},"excerpt":{"rendered":"In this unit, you were introduced to the basic concepts of organic chemistry, such as the tetravalence of…\n","protected":false},"author":133,"featured_media":866561,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"editor_notices":[],"footnotes":""},"categories":[477,389],"tags":[],"class_list":{"0":"post-866520","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-ncert-study-material","8":"category-school-education"},"yoast_head":"\nNCERT Solutions Class 11 Chemistry (Part-2) Chapter 8: Organic Chemistry (Free PDF) - Leverage Edu Discover<\/title>\n<meta name=\"description\" content=\"Download free NCERT Solutions for Class 11 Chemistry (Part-II) Chapter-8: Organic Chemistry in PDF format. Get accurate, exam-ready answers and understand key concepts.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions Class 11 Chemistry (Part-2) Chapter 8: Organic Chemistry (Free PDF)\" \/>\n<meta property=\"og:description\" content=\"Download free NCERT Solutions for Class 11 Chemistry (Part-II) Chapter-8: Organic Chemistry in PDF format. Get accurate, exam-ready answers and understand key concepts.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/\" \/>\n<meta property=\"og:site_name\" content=\"Leverage Edu Discover\" \/>\n<meta property=\"article:published_time\" content=\"2025-08-13T12:35:59+00:00\" \/>\n<meta property=\"og:image\" content=\"https:\/\/blogassets.leverageedu.com\/media\/uploads\/sites\/9\/2025\/08\/21082029\/NCERT-Solutions-Class-11-Chemistry-Part-II-Chapter-8-Organic-Chemistry-Free-PDF.webp\" \/>\n\t<meta property=\"og:image:width\" content=\"1024\" \/>\n\t<meta property=\"og:image:height\" content=\"640\" \/>\n\t<meta property=\"og:image:type\" content=\"image\/webp\" \/>\n<meta name=\"author\" content=\"Devanshu Srivastava\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:label1\" content=\"Written by\" \/>\n\t<meta name=\"twitter:data1\" content=\"Devanshu Srivastava\" \/>\n\t<meta name=\"twitter:label2\" content=\"Est. reading time\" \/>\n\t<meta name=\"twitter:data2\" content=\"15 minutes\" \/>\n<!-- \/ Yoast SEO Premium plugin. -->","yoast_head_json":{"title":"NCERT Solutions Class 11 Chemistry (Part-2) Chapter 8: Organic Chemistry (Free PDF) - Leverage Edu Discover","description":"Download free NCERT Solutions for Class 11 Chemistry (Part-II) Chapter-8: Organic Chemistry in PDF format. Get accurate, exam-ready answers and understand key concepts.","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/","og_locale":"en_US","og_type":"article","og_title":"NCERT Solutions Class 11 Chemistry (Part-2) Chapter 8: Organic Chemistry (Free PDF)","og_description":"Download free NCERT Solutions for Class 11 Chemistry (Part-II) Chapter-8: Organic Chemistry in PDF format. Get accurate, exam-ready answers and understand key concepts.","og_url":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/","og_site_name":"Leverage Edu Discover","article_published_time":"2025-08-13T12:35:59+00:00","og_image":[{"width":1024,"height":640,"url":"https:\/\/blogassets.leverageedu.com\/media\/uploads\/sites\/9\/2025\/08\/21082029\/NCERT-Solutions-Class-11-Chemistry-Part-II-Chapter-8-Organic-Chemistry-Free-PDF.webp","type":"image\/webp"}],"author":"Devanshu Srivastava","twitter_card":"summary_large_image","twitter_misc":{"Written by":"Devanshu Srivastava","Est. reading time":"15 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/#article","isPartOf":{"@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/"},"author":{"name":"Devanshu Srivastava","@id":"https:\/\/leverageedu.com\/discover\/#\/schema\/person\/4100048e2da2b688952b61b843288395"},"headline":"NCERT Solutions Class 11 Chemistry (Part-2) Chapter 8: Organic Chemistry (Free PDF)","datePublished":"2025-08-13T12:35:59+00:00","mainEntityOfPage":{"@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/"},"wordCount":3103,"commentCount":0,"image":{"@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/#primaryimage"},"thumbnailUrl":"https:\/\/blogassets.leverageedu.com\/media\/uploads\/sites\/9\/2025\/08\/21082029\/NCERT-Solutions-Class-11-Chemistry-Part-II-Chapter-8-Organic-Chemistry-Free-PDF.webp","articleSection":["NCERT Study Material","School Education"],"inLanguage":"en-US","potentialAction":[{"@type":"CommentAction","name":"Comment","target":["https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/#respond"]}]},{"@type":"WebPage","@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/","url":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/","name":"NCERT Solutions Class 11 Chemistry (Part-2) Chapter 8: Organic Chemistry (Free PDF) - Leverage Edu Discover","isPartOf":{"@id":"https:\/\/leverageedu.com\/discover\/#website"},"primaryImageOfPage":{"@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/#primaryimage"},"image":{"@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/#primaryimage"},"thumbnailUrl":"https:\/\/blogassets.leverageedu.com\/media\/uploads\/sites\/9\/2025\/08\/21082029\/NCERT-Solutions-Class-11-Chemistry-Part-II-Chapter-8-Organic-Chemistry-Free-PDF.webp","datePublished":"2025-08-13T12:35:59+00:00","author":{"@id":"https:\/\/leverageedu.com\/discover\/#\/schema\/person\/4100048e2da2b688952b61b843288395"},"description":"Download free NCERT Solutions for Class 11 Chemistry (Part-II) Chapter-8: Organic Chemistry in PDF format. Get accurate, exam-ready answers and understand key concepts.","breadcrumb":{"@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/"]}]},{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/#primaryimage","url":"https:\/\/blogassets.leverageedu.com\/media\/uploads\/sites\/9\/2025\/08\/21082029\/NCERT-Solutions-Class-11-Chemistry-Part-II-Chapter-8-Organic-Chemistry-Free-PDF.webp","contentUrl":"https:\/\/blogassets.leverageedu.com\/media\/uploads\/sites\/9\/2025\/08\/21082029\/NCERT-Solutions-Class-11-Chemistry-Part-II-Chapter-8-Organic-Chemistry-Free-PDF.webp","width":1024,"height":640,"caption":"NCERT Solutions Class 11 Chemistry (Part-II) Chapter 8 Organic Chemistry (Free PDF)"},{"@type":"BreadcrumbList","@id":"https:\/\/leverageedu.com\/discover\/school-education\/ncert-solutions-class-11-chemistry-part-ii-chapter-8-organic-chemistry\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/leverageedu.com\/discover\/"},{"@type":"ListItem","position":2,"name":"NCERT Solutions Class 11 Chemistry (Part-2) Chapter 8: Organic Chemistry (Free PDF)"}]},{"@type":"WebSite","@id":"https:\/\/leverageedu.com\/discover\/#website","url":"https:\/\/leverageedu.com\/discover\/","name":"Leverage Edu Discover","description":"","potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/leverageedu.com\/discover\/?s={search_term_string}"},"query-input":{"@type":"PropertyValueSpecification","valueRequired":true,"valueName":"search_term_string"}}],"inLanguage":"en-US"},{"@type":"Person","@id":"https:\/\/leverageedu.com\/discover\/#\/schema\/person\/4100048e2da2b688952b61b843288395","name":"Devanshu Srivastava","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/secure.gravatar.com\/avatar\/e726853ff3f67fc16c6b1ea001b66527a82338787be9ae414260eab48d358048?s=96&d=mm&r=g","url":"https:\/\/secure.gravatar.com\/avatar\/e726853ff3f67fc16c6b1ea001b66527a82338787be9ae414260eab48d358048?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/e726853ff3f67fc16c6b1ea001b66527a82338787be9ae414260eab48d358048?s=96&d=mm&r=g","caption":"Devanshu Srivastava"},"description":"Devanshu Srivastava is a Content Strategist and Writer with over 4 years of experience in the education technology space, and 1.5 years specifically in the study abroad and content writing domains. His expertise lies in simplifying complex information into engaging, student-friendly content. Passionate about global education, Devanshu crafts stories that empower students to make informed decisions about their academic and career journeys.","sameAs":["https:\/\/www.linkedin.com\/in\/devsrivastava3"],"url":"https:\/\/leverageedu.com\/discover\/author\/devanshusrivastava\/"}]}},"acf":[],"_links":{"self":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts\/866520","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/users\/133"}],"replies":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/comments?post=866520"}],"version-history":[{"count":0,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/posts\/866520\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/media\/866561"}],"wp:attachment":[{"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/media?parent=866520"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/categories?post=866520"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/leverageedu.com\/discover\/wp-json\/wp\/v2\/tags?post=866520"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}