Organic Chemistry Basics Class 11 Chemistry NCERT Solutions

Step-by-step solution

1. 1A carbon with two σ bonds and no lone pair, as at a carbonyl carbon, is sp hybridised; a carbon with three σ bonds is sp² hybridised; and a saturated carbon with four σ bonds is sp³ hybridised.
2. 2In CH₂=C=O, the terminal CH₂ carbon is sp² and the central carbon is sp.
3. 3In CH₃CH=CH₂, the methyl carbon is sp³ and both alkene carbons are sp².
4. 4In (CH₃)₂CO, the carbonyl carbon is sp² and the two methyl carbons are sp³.
5. 5In CH₂=CH–CN, the two alkene carbons are sp² and the nitrile carbon is sp; in benzene all six carbons are sp².

Step-by-step solution

1. 1Every atom-to-atom connection contains one σ bond; a multiple bond adds one or more π bonds.
2. 2Benzene has a six-membered carbon ring with six C–C σ bonds, six C–H σ bonds and three delocalised π bonds.
3. 3Cyclohexane has six C–C σ bonds and twelve C–H σ bonds, with no π bond.
4. 4CH₂Cl₂ has two C–H and two C–Cl σ bonds and no π bond; CH₂=C=CH₂ has two C–C σ bonds, four C–H σ bonds and two π bonds.
5. 5In one nitro resonance contributor, CH₃NO₂ has three C–H σ bonds, one C–N σ bond, one N–O σ bond and one N=O π bond. HCONHCH₃ has four C–H σ bonds, one N–H σ bond, two C–N σ bonds, one C–O σ bond and one C=O π bond.

Step-by-step solution

1. 1Isopropyl alcohol has a three-carbon chain with the hydroxyl group on the middle carbon: (CH₃)₂CH–OH.
2. 22,3-Dimethylbutanal has the aldehyde carbon in the parent chain and methyl groups on C-2 and C-3: OHC–CH(CH₃)–CH(CH₃)–CH₃.
3. 3Heptan-4-one has seven carbons with the carbonyl group at C-4: CH₃CH₂CH₂–CO–CH₂CH₂CH₃.
4. 4A bond-line drawing leaves the carbon skeleton and attached heteroatoms visible; the condensed formulae above identify every vertex and substituent unambiguously.

Step-by-step solution

1. 1Choose the parent that contains the principal functional group, number it from the end nearest that group, and list substituents in alphabetical order.
2. 2For the benzene compound, benzene is the preferred parent because it contains six ring carbons, giving propylbenzene; 1-phenylpropane is an older, non-preferred form.
3. 3The nitrile carbon is C-1, the chain has five carbons and the methyl substituent is at C-3, giving 3-methylpentanenitrile.
4. 4The alkane names are 2,5-dimethylheptane and 3-bromo-3-chloroheptane; in the latter, bromo is alphabetised before chloro.
5. 5The aldehyde has three carbons and chlorine at C-3, giving 3-chloropropanal. In Cl₂CHCH₂OH, the alcohol has priority, both chlorines are at C-2, and the name is 2,2-dichloroethan-1-ol.

Step-by-step solution

1. 1The multiplier di- or tri- is retained when identical substituents are present.
2. 2Compare locant sets at the first point of difference: 2,4,7 is lower than 2,5,7.
3. 3When the two numbering directions give the same set, give the lower locant to the substituent cited first alphabetically; chloro precedes methyl.
4. 4The alcohol is the principal functional group, so the suffix is -ol and receives the lowest locant; the triple bond is then described as but-3-yn-1-ol.

Step-by-step solution

1. 1Members of a homologous series differ by CH₂ and retain the same functional group.
2. 2The acid series is methanoic, ethanoic, propanoic, butanoic and pentanoic acid: HCOOH, CH₃COOH, CH₃CH₂COOH, CH₃CH₂CH₂COOH and CH₃CH₂CH₂CH₂COOH.
3. 3The ketone series is propanone, butanone, pentan-2-one, hexan-2-one and heptan-2-one.
4. 4The alkene series is ethene, propene, but-1-ene, pent-1-ene and hex-1-ene, with the double bond at the end of the written chain.

Step-by-step solution

1. 1For 2,2,4-trimethylpentane, select the five-carbon chain CH₃–C–CH₂–CH–CH₃; put two methyl groups on C-2 and one on C-4.
2. 2Its condensed formula is (CH₃)₃C–CH₂–CH(CH₃)–CH₃, not a formula that places all three methyl groups on the terminal quaternary carbon.
3. 3Citric acid has three carboxyl groups and one alcohol group: HOOC–CH₂–C(OH)(COOH)–CH₂–COOH.
4. 4Hexane-1,6-dial is OHC–(CH₂)₄–CHO and therefore contains two aldehyde groups.

Step-by-step solution

1. 1An aldehyde is recognised by –CHO, an alcohol by –OH, an ether by –O–, and an alkene by a carbon–carbon double bond.
2. 2The second structure contains a primary amino group (–NH₂), an ester linkage (–O–CO–), and a tertiary amine whose nitrogen has three carbon attachments and no N–H bond.
3. 3A nitro group is –NO₂; it is not the same as a nitrate ester or a nitrile group.
4. 4Report every group present rather than naming only the group used to name the compound.

Step-by-step solution

1. 1A nitro group withdraws electron density through the –I effect and pulls it toward the nitro group.
2. 2In O₂N–CH₂–CH₂–O⁻, this withdrawal reduces and stabilises the negative charge on the terminal oxygen.
3. 3The ethyl group in CH₃CH₂–O⁻ has a +I effect and increases electron density near the negatively charged oxygen, destabilising the anion.

Step-by-step solution

1. 1An alkyl group directly attached to a π-bonded carbon contains σ(C–H) or σ(C–C) bonds on the adjacent saturated carbon.
2. 2A filled σ orbital can overlap partially with the neighbouring empty p orbital involved in the π system.
3. 3This overlap delocalises electron density into the π system; the contributors are often called no-bond resonance structures because a C–H bond is temporarily represented as broken.
4. 4The resulting electron donation is called hyperconjugation and stabilises the conjugated or unsaturated system.

Step-by-step solution

1. 1Resonance changes only electron placement; atoms are not rearranged and all contributors must have the same sigma framework.
2. 2For phenol, a lone pair on oxygen can form a π bond to the ring, placing positive charge on oxygen and negative charge at the ortho and para ring positions in separate contributors.
3. 3For nitrobenzene, the two N–O bonds are represented by equivalent charge-separated contributors, and the ring can also delocalise into the nitro group at ortho and para positions.
4. 4For CH₃CH=CHCHO, the C=C and C=O bonds are conjugated; π electrons can shift toward the carbonyl, giving contributors with negative charge on oxygen and positive charge at the remote conjugated carbon.
5. 5Benzaldehyde has ring-to-carbonyl conjugation, so charge-separated contributors place negative charge on oxygen and positive charge at ring positions. In the benzyl carbocation, the positive charge is delocalised to the ortho and para ring positions; in the allylic carbocation, it is shared between the two terminal allylic carbons.

Step-by-step solution

1. 1An electrophile is electron-deficient and accepts an electron pair; a nucleophile is electron-rich and donates an electron pair.
2. 2Positive ions such as H⁺, carbocations and acylium ions are electrophiles, as are neutral molecules with an electron-deficient atom such as BF₃.
3. 3Anions such as OH⁻, CN⁻ and carbanions are nucleophiles because they possess available electron density.
4. 4Neutral molecules with a lone pair, including NH₃ and H₂O, can also act as nucleophiles.

Step-by-step solution

1. 1OH⁻ has a negative charge and a lone pair, so it donates an electron pair to the proton of acetic acid.
2. 2CN⁻ is also an electron-rich anion; its carbon end attacks the electrophilic carbonyl carbon of propanone.
3. 3CH₃CO⁺ is the acylium ion. Its positively charged carbonyl carbon is electron-deficient and accepts a pair from the benzene π electrons.

Step-by-step solution

1. 1In (a), the –SH group replaces Br without changing the carbon skeleton, so the reaction is nucleophilic substitution.
2. 2In (b), H and Cl add across the C=C bond and the two reactants form one product, so it is an electrophilic addition reaction.
3. 3In (c), a β-hydrogen and Br are removed from adjacent positions to form a π bond, so it is elimination.
4. 4In (d), the neopentyl alcohol framework rearranges while –OH is replaced by Br; the carbocation rearrangement precedes substitution, giving 2-bromo-2-methylbutane.

Step-by-step solution

1. 1Pair (a) contains hexan-2-one and hexan-3-one: the molecular formula is the same but the position of the ketone group changes, so the compounds are structural (positional) isomers.
2. 2Pair (b) has the same constitution and bond sequence, but D and H occupy different relative positions across the same double-bond framework; these are geometrical isomers.
3. 3Pair (c) has identical atom connectivity and differs only in electron placement, so the drawings are canonical resonance contributors, not separate molecules.

Step-by-step solution

1. 1Homolysis divides the shared pair equally, giving two radicals; the first cleavage is therefore homolysis and produces a free radical.
2. 2If both electrons remain on carbon, the carbon fragment is negatively charged, so the second cleavage is heterolysis and produces a carbanion.
3. 3If both electrons remain on bromine, the carbon fragment is positively charged, so the third cleavage is heterolysis and produces a carbocation.
4. 4The fourth displayed cleavage is also heterolysis; retention of the electron pair by the carbon fragment gives a carbocation.

Step-by-step solution

1. 1The inductive effect is permanent polarisation of σ electrons through a saturated chain by an electron-withdrawing –I group or electron-releasing +I group.
2. 2The electromeric effect is a temporary shift of π electrons in a multiple bond caused by an attacking reagent; it can be +E or –E according to the direction of displacement.
3. 3Each chlorine exerts –I, and the effect becomes stronger with more chlorine atoms. The conjugate base is stabilised and acidity increases in the order in (a).
4. 4More alkyl groups exert a stronger +I effect, which destabilises the carboxylate conjugate base. Acidity therefore decreases from propanoic to isobutyric to pivalic acid in (b).

Step-by-step solution

1. 1Crystallisation separates a solid from impurities when its solubility changes greatly with temperature. Crude aspirin, for example, is dissolved in a minimum of hot ethanol and crystallised on cooling.
2. 2Distillation separates volatile liquids from non-volatile material, or liquids with sufficiently different boiling points, by vaporising and condensing them. Chloroform can be distilled from aniline because their boiling points differ greatly.
3. 3Chromatography separates components by their different adsorption or partition between a stationary phase and a mobile phase. A mixture of red and blue ink can be separated on paper because the components travel differently.

Step-by-step solution

1. 1Add S slowly to the powdered mixture with stirring until the mixture is just dissolved, then heat the saturated solution so that any insoluble impurity can be removed by hot filtration.
2. 2Cool the clear hot solution. The less soluble compound reaches saturation first and crystallises; filter off those crystals before the more soluble compound crystallises.
3. 3Concentrate the mother liquor and cool it again. The more soluble compound now crystallises in a later fraction.
4. 4Filter and dry each fraction separately, repeating the concentration and cooling if necessary for better separation.

Step-by-step solution

1. 1Ordinary distillation separates a volatile liquid from non-volatile impurities or separates liquids with sufficiently different boiling points when both are stable on heating; petrol and kerosene are an example.
2. 2Reduced-pressure or vacuum distillation lowers the boiling point by reducing external pressure. It purifies heat-sensitive liquids that decompose at their normal boiling point; glycerol can be distilled under reduced pressure without decomposition.
3. 3Steam distillation is used for an organic liquid that is steam-volatile and immiscible with water. The mixed vapours are condensed together and the organic layer is separated with a separating funnel; water and aniline are an example.

Step-by-step solution

1. 1Sodium fusion converts covalently bound elements into ionic sodium salts: NaCN from nitrogen, Na₂S from sulphur and NaX from a halogen, where X is Cl, Br or I. The fused mass is boiled in water to make the sodium extract.
2. 2For nitrogen, CN⁻ combines with Fe²⁺ to form [Fe(CN)₆]⁴⁻. Oxidation of part of the Fe²⁺ to Fe³⁺ gives hydrated Prussian blue, Fe₄[Fe(CN)₆]₃·xH₂O.
3. 3For sulphur, acetic acid releases H₂S from sulphide; lead acetate gives black PbS, and sodium nitroprusside gives a violet colour. If both N and S are present, NaSCN can form instead of free CN⁻, giving a blood-red complex with Fe³⁺.
4. 4For halogens, X⁻ reacts with Ag⁺ to form insoluble AgX after the extract is acidified with dilute HNO₃ and interfering CN⁻ and S²⁻ are expelled. AgCl is white, AgBr is pale yellow and AgI is yellow.

Step-by-step solution

1. 1In the Dumas method, a known mass is heated with excess CuO in a carbon dioxide atmosphere. Carbon and hydrogen form CO₂ and H₂O, while nitrogen is converted to N₂; nitrogen oxides, if formed, are reduced over heated copper.
2. 2The N₂ is collected over aqueous KOH to remove CO₂ and its volume is corrected to the original temperature and pressure. From the volume, the mass and percentage of nitrogen are calculated.
3. 3In the Kjeldahl method, the compound is digested with concentrated H₂SO₄ so that nitrogen becomes ammonium salt. Excess NaOH releases NH₃, which is absorbed in a known excess of standard H₂SO₄.
4. 4The unconsumed acid is back-titrated with standard NaOH. Kjeldahl is unsuitable for nitrogen in aromatic rings and for many nitro or azo compounds, whereas the Dumas principle is more broadly applicable.

Step-by-step solution

1. 1For halogen estimation, the compound is heated with fuming HNO₃ and AgNO₃ in a Carius tube. Carbon and hydrogen oxidise to CO₂ and H₂O, while X becomes AgX. The filtered, washed and dried AgX is weighed.
2. 2If m is the sample mass and m₁ is the AgX mass, the halogen percentage is 100 × (atomic mass of X)m₁/(molar mass of AgX × m).
3. 3For sulphur, oxidation gives H₂SO₄ and addition of BaCl₂ gives BaSO₄. Since 233 g of BaSO₄ contains 32 g of sulphur, %S = 100 × 32m₁/(233m).
4. 4For phosphorus, oxidation gives H₃PO₄. It can be precipitated as ammonium phosphomolybdate of molar mass 1877, giving %P = 100 × 31m₁/(1877m), or as MgNH₄PO₄ followed by ignition to Mg₂P₂O₇, giving %P = 100 × 62m₁/(222m).

Step-by-step solution

1. 1Spot a solution of the mixture near the base of chromatography paper. Water held by the paper acts as the stationary phase and the selected solvent acts as the mobile phase.
2. 2The solvent rises by capillary action and carries the components to different heights because they partition differently between the water in the paper and the moving solvent.
3. 3The developed paper is the chromatogram. For a component, Rf = distance travelled by the component / distance travelled by the solvent front.

Step-by-step solution

1. 1If the original compound contained nitrogen or sulphur, the extract may contain CN⁻ and S²⁻, which can react with Ag⁺ or otherwise interfere with the halide precipitate.
2. 2Dilute HNO₃ acidifies CN⁻ and S²⁻ to volatile HCN and H₂S; boiling expels both gases.
3. 3After the interfering species have been removed, X⁻ can react cleanly with Ag⁺ to give AgX: Ag⁺ + X⁻ → AgX.

Step-by-step solution

1. 1Nitrogen, sulphur and halogens are covalently bonded in the organic compound and cannot be tested directly as free ions.
2. 2Fusion with sodium breaks the covalent framework and converts the elements into water-soluble ionic sodium salts: NaCN, NaSCN or Na₂S, and NaX.
3. 3Boiling the fused mass in water extracts these ions, so conventional qualitative tests for CN⁻, S²⁻ and X⁻ become possible.

Step-by-step solution

1. 1Camphor is sublimable: on heating it changes directly from solid to vapour and later condenses to crystals.
2. 2Calcium sulphate is non-sublimable under these conditions and remains in the apparatus.
3. 3Collect the condensed camphor on a cold surface, leaving calcium sulphate behind.

Step-by-step solution

1. 1The organic liquid and water are immiscible, so each contributes its own vapour pressure to the vapour mixture.
2. 2The mixture boils when the sum of the partial vapour pressures equals atmospheric pressure: p_atm = p_organic + p_water.
3. 3The organic component therefore contributes part of the total pressure and reaches the vapour phase at a temperature below its pure-liquid boiling point.

Step-by-step solution

1. 1The chlorine atoms in CCl₄ are covalently bonded to carbon and are not present as free chloride ions.
2. 2AgCl precipitation requires Cl⁻ + Ag⁺ → AgCl, so the covalent chlorine must first be converted to an ionic halide.
3. 3A Lassaigne sodium fusion would produce NaCl in the extract, after which AgNO₃ could give the white AgCl precipitate.

Step-by-step solution

1. 1Carbon dioxide is acidic and reacts quantitatively with the strong base KOH: 2KOH + CO₂ → K₂CO₃ + H₂O.
2. 2The absorbing U-tube gains the mass of the retained CO₂ while the carbon dioxide is removed from the gas stream.
3. 3The increase in mass of the KOH tube gives the mass of CO₂, from which the mass and percentage of carbon are calculated.

Step-by-step solution

1. 1The extract contains S²⁻. Mild acidification with acetic acid releases H₂S without introducing sulfate into the solution.
2. 2Lead acetate then reacts with H₂S to form the diagnostic black precipitate PbS: Pb²⁺ + S²⁻ → PbS.
3. 3Sulphuric acid would introduce SO₄²⁻, which can also form an insoluble white PbSO₄ precipitate and mask the black PbS result.

Step-by-step solution

1. 1Mass of carbon in the sample: 0.20 × 69.0/100 = 0.138 g.
2. 2In 44 g of CO₂, 12 g is carbon, so m(CO₂) = 0.138 × 44/12 = 0.506 g.
3. 3Mass of hydrogen in the sample: 0.20 × 4.8/100 = 0.0096 g.
4. 4In 18 g of H₂O, 2 g is hydrogen, so m(H₂O) = 0.0096 × 18/2 = 0.0864 g.

Step-by-step solution

1. 1Initial H₂SO₄ = 0.050 L × 0.5 mol L⁻¹ = 0.025 mol.
2. 2Residual H₂SO₄ = (0.060 L × 0.5 mol L⁻¹)/2 = 0.015 mol, because one mole of H₂SO₄ neutralises two moles of NaOH.
3. 3H₂SO₄ consumed by NH₃ = 0.025 − 0.015 = 0.010 mol; 2NH₃ + H₂SO₄ → (NH₄)₂SO₄, so n(NH₃) = 0.020 mol.
4. 4Mass of nitrogen = 0.020 × 14 = 0.28 g; percentage N = 0.28/0.50 × 100 = 56%.

Step-by-step solution

1. 1One mole of AgCl contains one mole of Cl, so m(Cl) = m(AgCl) × 35.5/143.32.
2. 2m(Cl) = 0.5740 × 35.5/143.32 = 0.1421 g approximately.
3. 3%Cl = 0.1421/0.3780 × 100 = 37.59%.

Step-by-step solution

1. 1One mole of BaSO₄ has mass 233 g and contains 32 g of sulphur.
2. 2m(S) = 0.668 × 32/233 = 0.0917 g approximately.
3. 3%S = 0.0917/0.468 × 100 = 19.59% (about 19.6%).

Step-by-step solution

1. 1Number the chain from the terminal alkyne carbon: C₁–C₂≡C₃–C₄–C₅=C₆.
2. 2C₁ and C₂ of the triple bond are sp hybridised; C₃ and C₄ are sp³; C₅ and C₆ are sp².
3. 3The C₂–C₃ single bond therefore joins an sp carbon to an sp³ carbon, giving sp–sp³ overlap.

Step-by-step solution

1. 1CN⁻ first forms hexacyanoferrate(II): 6CN⁻ + Fe²⁺ → [Fe(CN)₆]⁴⁻.
2. 2Acid and heat oxidise part of Fe²⁺ to Fe³⁺, which combines with hexacyanoferrate(II) to form hydrated Prussian blue.
3. 3The characteristic formula is Fe₄[Fe(CN)₆]₃·xH₂O, corresponding to option (b).

Step-by-step solution

1. 1A positive carbon is stabilised by electron donation from adjacent alkyl groups through the +I effect and hyperconjugation.
2. 2(a) and (c) are primary carbocations, (d) is secondary, and (b) is tertiary.
3. 3The tertiary carbocation has three methyl groups supplying hyperconjugative C–H bonds and the strongest +I effect, so it is the most stable.

Step-by-step solution

1. 1Crystallisation, distillation and sublimation depend on particular physical properties and work well for selected mixtures.
2. 2Chromatography uses differential adsorption or partition between a stationary and a mobile phase and can separate a broad range of organic mixtures, including compounds with similar physical properties.
3. 3Its general applicability and ability to resolve mixtures into separate components make it the best choice among the options.

Step-by-step solution

1. 1Aqueous KOH supplies OH⁻, which has a lone pair and acts as a nucleophile.
2. 2OH⁻ attacks the carbon bonded to iodine and the C–I bond breaks, replacing I with OH to form ethanol.
3. 3Because one group is replaced without a net change in the degree of unsaturation, the reaction is nucleophilic substitution rather than elimination or addition.