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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 – Some Basic Principles and Techniques, which will help you revise these concepts effectively.
Contents
Explore Notes of Class 11 Chemistry
NCERT Solutions Class 11 Chemistry (Part-2) Chapter 8: Organic Chemistry
Below, we have provided you with the exercises mentioned in the NCERT Solutions for Class 11 Chemistry (Part II), Chapter 8: Organic Chemistry – Some Basic Principles and Techniques
Exercises
- What are the hybridisation states of each carbon atom in the following compounds?
CH2=C=O, CH3CH=CH2, (CH3)2CO, CH2=CHCN, C6H6 - Indicate the σ and π bonds in the following molecules:
C6H6, C6H12, CH2Cl2, CH2=C=CH2, CH3NO2, HCONHCH3 - Write bond line formulas for: Isopropyl alcohol, 2,3-Dimethylbutanal, Heptan-4-one.
- 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 - Draw formulas for the first five members of each homologous series beginning with the following compounds:
(a) H–COOH (b) CH3COCH3 (c) H–CH=CH2 - 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 - Identify the functional groups in the following compounds:
(a) (b) (c) - Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable, and why?
- Explain why alkyl groups act as electron donors when attached to a π system.
- 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–CHO (e) C6H5–C+H2 (f) CH3CH=CHC+H2 - What are electrophiles and nucleophiles? Explain with examples.
- Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles:
(a) CH3COOH + HO– → CH3COO– + H2O
(b) CH3COCH3 + C–N → (CH3)2C(CN)(OH)
(c) C6H6 + CH3C+O → C6H5COCH3 - Classify the following reactions in one of the reaction types studied in this unit:
(a) CH3CH2Br + HS– → CH3CH2SH + Br–
(b) (CH3)2C = CH2 + HCl → (CH3)2CCl – CH3
(c) CH3CH2Br + HO– → CH2 = CH2 + H2O + Br–
(d) (CH3)3C–CH2OH + HBr → (CH3)2CBrCH2CH2CH3 + H2O - 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 - Give a brief description of the principles of the following techniques, taking an example in each case:
(a) Crystallisation (b) Distillation (c) Chromatography - Describe the method that can be used to separate two compounds with different solubilities in a solvent S.
- What is the difference between distillation, distillation under reduced pressure, a nd steam distillation?
- Discuss the chemistry of Lassaigne’s test.
- Differentiate between the principles of estimation of nitrogen in an organic compound by:
(i) Dumas method and (ii) Kjeldahl’s method. - Discuss the principle of estimation of halogens, sulphur, and phosphorus present in an organic compound.
- Explain the principle of paper chromatography.
- Why is nitric acid added to the sodium extract before adding silver nitrate for testing halogens?
- Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulfur, and halogens.
- Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.
- Explain why an organic liquid vaporises at a temperature below its boiling point in its steam distillation.
- Will CCl4 give a white precipitate of AgCl on heating it with silver nitrate? Give a reason for your answer.
- Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?
- 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?
- 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.
- A sample of 0.50 g of an organic compound was treated according to Kjeldahl’s 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.
- 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.
- 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.
- In the organic compound CH2 = CH – CH2 – CH2 – C ≡ CH, the pair of hybridised orbitals involved in the formation of the C2 – C3 bond is:
(a) sp – sp2 (b) sp – sp3 (c) sp2 – sp3 (d) sp3 – sp3 - In Lassaigne’s 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 - Which of the following carbocations is most stable?
(a) (CH3)3C.C+H2 (b) (CH3)3C+ (c) CH3CH2C+H2 (d) CH3C+HCH2CH3 - The best and latest technique for isolation, purification, and separation of organic compounds is:
(a) Crystallisation (b) Distillation (c) Sublimation (d) Chromatography - The reaction:
CH3CH2I + KOH(aq) → CH3CH2OH + KI
It is classified as:
(a) electrophilic substitution (b) nucleophilic substitution (c) elimination (d) addition
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Solutions
- For CH₂=C=O the terminal CH₂ carbon is sp² and the central carbon (connected to CH₂ and O) is sp; for CH₃CH=CH₂ the CH₃ carbon is sp³, the middle =CH is sp² and the terminal =CH₂ is sp²; for (CH₃)₂CO the two methyl carbons are sp³ and the carbonyl carbon is sp²; for CH₂=CHCN the vinyl CH₂ is sp², the vinyl CH is sp² and the nitrile carbon is sp; for C₆H₆ all six ring carbons are sp².
- Benzene (C₆H₆) has 12 σ bonds (six C–C and six C–H) and a delocalised 6π system usually represented as three π bonds; saturated C₆H₁₂ (cyclohexane assumed) has only σ bonds (6 C–C + 12 C–H = 18 σ, 0 π); CH₂Cl₂ has four single bonds (4 σ, 0 π); CH₂=C=CH₂ (allene) has six σ bonds (two C–C and four C–H) and two π bonds (two orthogonal π systems); CH₃NO₂ has six σ bonds (three C–H, one C–N, two N–O) and a delocalised π system in the nitro group (commonly treated as one delocalised π set across N–O); HCONHCH₃ (N-methylformamide) has one C=O π and the remaining bonds are σ, with C–N having partial π character by resonance.
- Isopropyl alcohol: condensed formula (CH₃)₂CHOH and bond-line is the central carbon bonded to OH and two methyl groups; 2,3-Dimethylbutanal: condensed CH₃–CH(CH₃)–CH(CH₃)–CHO and bond-line is a four-carbon chain with –CHO at one end and methyl substituents at C2 and C3; Heptan-4-one: condensed CH₃–CH₂–CH₂–CO–CH₂–CH₂–CH₃ and bond-line is a seven-carbon chain with a C=O at carbon 4.
- The correct IUPAC forms are: 2,2-Dimethylpentane (not “2-Dimethylpentane”); 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).
- First five members starting from H–COOH are: HCOOH (methanoic acid), CH₃COOH (ethanoic acid), C₂H₅COOH (propanoic acid), C₃H₇COOH (butanoic acid), C₄H₉COOH (pentanoic acid). First five members starting from CH₃COCH₃ (ketone at C2) are CH₃COCH₃ (propanone), CH₃CH₂COCH₃ (butan-2-one), CH₃CH₂CH₂COCH₃ (pentan-2-one), hexan-2-one, heptan-2-one. First five members starting from H–CH=CH₂ are CH₂=CH₂ (ethene), CH₃CH=CH₂ (propene), CH₃CH₂CH=CH₂ (but-1-ene), pent-1-ene, hex-1-ene.
- 2,2,4-Trimethylpentane condensed formula is CH₃–C(CH₃)₂–CH₂–CH(CH₃)–CH₃, 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–CH(COOH)–CH(COOH)–CH₂–COOH, bond-line shows three carboxyl groups and one hydroxyl; functional groups are three –COOH and one –OH. Hexanedial is OHC–(CH₂)₄–CHO (hexane-1,6-dial), bond-line shows aldehyde groups at both ends, and the functional groups are two –CHO (aldehydes).
- O₂NCH₂CH₂O⁻ is more stable than CH₃CH₂O⁻ 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.
- Alkyl groups act as electron donors to a π system mainly by the inductive (+I) effect through σ bonds and by hyperconjugation, where C–H (or C–C) σ orbitals overlap with adjacent π orbitals or empty p orbitals and delocalise electron density into the π system, thereby increasing electron density on the π system and stabilising adjacent positive charge.
- Phenol (C₆H₅OH) 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 → C–O π, then π shifts arouthe nd ring). Nitrobenzene (C₆H₅NO₂) resonates by withdrawing electron density from the ring toward the nitro group, giving ring-positive forms and a negative charge on nitro oxygens (ring π → N, N=O π → O). CH₃CH=CHCHO (an α,β-unsaturated aldehyde) resonates by conjugation of C=C and C=O: one form has C=O and C=C, another has C=O π electrons on O (O⁻) with the π electrons of C=C shifted to place positive character at the β carbon. Benzaldehyde (C₆H₅–CHO) resonates with ring → carbonyl electron donation, producing O⁻ and positive character delocalised on the ring at ortho/para. The benzylic carbocation (C₆H₅–C⁺H₂) is stabilised by resonance into the aromatic ring with positive charge delocalised to ortho/para positions. An allylic carbocation CH₃CH=CHCH₂ is stabilised by resonance with the positive charge delocalised across the allylic system (π electrons shift to place the + on different carbons). (If you want the curved-arrow diagrams, I can draw them for any lettered case.)
- Electrophiles are electron-poor species that accept an electron pair (Lewis acids), for example, H⁺, NO₂⁺, R–C⁺; nucleophiles are electron-rich species that donate an electron pair (Lewis bases), for example,e OH⁻, CN⁻, NH₃.
- In CH₃COOH + HO⁻ → CH₃COO⁻ + H₂O, the reagent HO⁻ is a nucleophile (base); in CH₃COCH₃ + C⁻N → (CH₃)₂C(CN)(OH), the cyanide ion C⁻N is a nucleophile; in C₆H₆ + CH₃C⁺O → C₆H₅COCH,₃ the acyl-type CH₃C⁺O species is an electrophile.
- CH₃CH₂Br + HS⁻ → CH₃CH₂SH + Br⁻ is a nucleophilic substitution (S_N2); (CH₃)₂C=CH₂ + HCl → (CH₃)₂CCl–CH₃ is electrophilic addition to an alkene (Markovnikov addition); CH₃CH₂Br + HO⁻ → CH₂=CH₂ + H₂O + Br⁻ is an elimination (E2); (CH₃)₃C–CH₂OH + HBr → (CH₃)₂CBrCH₂CH₂CH₃ + H₂O indicates nucleophilic substitution via a carbocation with rearrangement (S_N1 with rearrangement).
- The inductive effect is a permanent electron displacement along σ bonds due to electronegativity differences (it falls off with distance); the electromeric effect is a temporary, reagent-induced transfer of π electrons during a reaction (instantaneous). The acidity order Cl₃CCOOH > Cl₂CHCOOH > ClCH₂COOH is explained by the inductive (−I) effect of chlorine atoms stabilising the conjugate base, and the order CH₃CH₂COOH > (CH₃)₂CHCOOH > (CH₃)₃CCOOH is due to +I and hyperconjugation of alkyl groups destabilising the conjugate base (more alkyl substitution → less acidic).
- 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).
- 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.
- 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.
- Lassaigne’s test involves fusing an organic sample with metallic sodium to convert covalently bound nitrogen, sulfur and halogens into ionic forms (NaCN, Na₂S, NaX) which are extracted with water and tested: nitrogen yields Prussian blue with Fe/acid (indicates CN⁻ → Fe₄[Fe(CN)₆]₃), sulfur gives PbS (black) or BaSO₄ after oxidation, and halogens give AgX precipitates on acidification and addition of AgNO₃.
- The Dumas method estimates nitrogen by combustion of the sample to produce N₂ gas which is measured (instrumental; detects total N), while Kjeldahl’s method digests the sample with concentrated acid to convert organic N to NH₄⁺, distils the liberated NH₃ into standard acid and titrates to determine N (wet chemical method; some forms of nitrogen may need modification to be measured).
- 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 BaSO₄ and weighed, and phosphorus to orthophosphate, which can be precipitated or determined colorimetrically as phosphomolybdate or gravimetrically after suitable conversion.
- 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.
- Nitric acid is added to the sodium fusion extract before adding AgNO₃ to acidify the extract (convert NaX to HX) and to oxidise interfering sulfides or thiocyanates; acidification prevents precipitation of Ag₂O by OH⁻ and avoids interfering side reactions, giving a clean AgX precipitate for halogen detection.
- 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, Na₂S, NaX) that can be extracted and detected by specific qualitative tests.
- 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 CaSO₄ remains as nonvolatile residue.
- 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.
- Carbon tetrachloride (CCl₄) will not give a white AgCl precipitate when heated with aqueous AgNO₃ because CCl₄ is nonpolar and covalent and does not release free Cl⁻ ions into aqueous solution under such conditions; AgCl forms only if chloride ions are present in the aqueous phase.
- Potassium hydroxide is used to absorb CO₂ evolved during carbon estimation because KOH chemically fixes CO₂ as carbonate/bicarbonate (preventing its escape), so quantitative collection or measurement is possible.
- 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 (H₂SO₄) can oxidise sulfide or produce interfering insoluble salts (e.g., PbSO₄) and lead to false results; acetic acid acidifies without such interference, allowing formation/detection of PbS where appropriate.
- For a 0.20 g sample that is 69% C and 4.8% H (remainder O): mass of carbon = 0.20 × 0.69 = 0.1380 g, which gives CO₂ mass = 0.1380 × (44/12) = 0.5060 g; mass of hydrogen = 0.20 × 0.048 = 0.00960 g, which gives H₂O mass = 0.00960 × (18/2) = 0.08640 g.
- In the Kjeldahl problem: initial H₂SO₄ moles = 0.050 L × 0.5 M = 0.02500 mol; NaOH used = 0.060 L × 0.5 M = 0.03000 mol which neutralises 0.01500 mol H₂SO₄ (because 2 NaOH per H₂SO₄), so H₂SO₄ consumed by NH₃ = 0.02500 − 0.01500 = 0.01000 mol; that neutralises 0.02000 mol NH₃ (2 NH₃ per H₂SO₄), so moles N = 0.02000 and mass N = 0.02000 × 14.007 ≈ 0.28014 g; percentage N = (0.28014 / 0.50) × 100 ≈ 56.03%.
- From the Carius chlorination: 0.5740 g AgCl corresponds to moles 0.5740 / 143.3212 ≈ 0.0040050 mol Cl, mass Cl = 0.0040050 × 35.453 ≈ 0.14199 g, percentage Cl = (0.14199 / 0.3780) × 100 ≈ 37.56%.
- From the Carius sulfur estimation: 0.668 g BaSO₄ corresponds to moles 0.668 / 233.383 ≈ 0.0028623 mol, mass S = 0.0028623 × 32.06 ≈ 0.09176 g, percentage S = (0.09176 / 0.468) × 100 ≈ 19.61%.
- In CH₂=CH–CH₂–CH₂–C≡CH, the C2–C3 bond pairs an sp² orbital (from the = carbon) with an sp³ orbital (from the CH₂), so the correct answer is sp²–sp³ (option c).
- In Lassaigne’s test, Prussian blue corresponds to Fe₄[Fe(CN)₆]₃ (option b).
- The most stable carbocation among the choices is the tertiary carbocation (CH₃)₃C⁺ (option b).
- The best and latest technique for isolation, purification, and separation of organic compounds is chromatography (option d).
- The reaction CH₃CH₂I + KOH(aq) → CH₃CH₂OH + KI is classified as nucleophilic substitution (option b).
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