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The Vault
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The Vault
CBSE Class 11 Chemistry · 10 questions · 24 marks
Carbon makes millions of compounds from a handful of elements, so organic chemistry needs a system rather than a memory list. Here you learn the naming rules that turn any structure into a unique IUPAC name, the electronic effects — inductive, resonance and hyperconjugation — that explain why one molecule reacts and its neighbour does not, and the laboratory methods used to purify an unknown and work out its formula.
The IUPAC name of the compound CH₃−CH₂−CH(CH₃)−CH(CH₃)−CH₃ is:
Answer
2,3-dimethylpentane is correct. The longest continuous chain runs through five carbon atoms, so the parent is pentane, and there are two methyl branches. Numbering from the right-hand end gives the methyls positions 2 and 3, a locant set of (2,3). Numbering from the left gives (3,4). The rule of lowest locants picks (2,3), which rules out 3,4-dimethylpentane. The third option is wrong because it fails to select the longest chain — an ethyl group is being named as a substituent when it should be part of the parent chain. The fourth is wrong because the molecule contains only seven carbons in total and just five in the parent chain, so hexane is impossible.
The correct order of decreasing stability of the carbocations (CH₃)₃C⁺, (CH₃)₂CH⁺, CH₃CH₂⁺ and CH₃⁺ is:
Answer
(CH₃)₃C⁺ > (CH₃)₂CH⁺ > CH₃CH₂⁺ > CH₃⁺ is the correct order. A carbocation carries a positive charge on an sp² carbon with an empty p orbital, so anything that pushes electron density towards it lowers the energy. Alkyl groups do this in two ways: through the +I inductive effect, and more importantly through hyperconjugation, where the electrons of an adjacent C−H sigma bond overlap with the empty p orbital. The tertiary cation has nine alpha hydrogens available for hyperconjugation, the secondary has six, the ethyl cation three and the methyl cation none. More alpha hydrogens means more delocalisation and greater stability, which produces exactly the order given. This order explains why tertiary halides ionise fastest and why rearrangements in organic reactions almost always run towards the more substituted cation.
Two liquids in a mixture have boiling points of 78 °C and 82 °C. The most suitable technique to separate them is:
Answer
Fractional distillation is the most suitable technique. Simple distillation works only when the boiling points differ by roughly 25 °C or more, because the vapour must be almost pure in the more volatile component in a single step. A gap of only 4 °C would give vapour rich in both liquids and no useful separation. A fractionating column packed with beads or glass rings provides a large surface on which the rising vapour repeatedly condenses and re-evaporates. Each of these cycles is effectively another distillation, so after many of them the vapour reaching the top is essentially pure in the lower-boiling component. Steam distillation is for water-insoluble substances that are steam volatile and would decompose at their normal boiling point, and crystallisation applies to solids, so neither fits a pair of close-boiling liquids.
In the sodium fusion extract of an organic compound, the addition of freshly prepared iron(II) sulphate followed by acidification with sulphuric acid gives a Prussian blue colour. This shows the presence of:
Answer
Nitrogen is present. Fusion with sodium converts any nitrogen in the compound into sodium cyanide. The cyanide ion then reacts with the added Fe²⁺ to form the hexacyanoferrate(II) complex, and on acidification some Fe²⁺ is oxidised to Fe³⁺, which combines with that complex to give ferriferrocyanide — the deep Prussian blue solid that gives the test its name. Sulphur would instead give a violet colour with sodium nitroprusside, or a black precipitate with lead acetate, because it is converted to sulphide. A halogen is detected by acidifying with nitric acid and adding silver nitrate, which gives a white, pale yellow or yellow precipitate depending on the halide. The blue colour is specific to nitrogen.
Assertion (A): All six carbon–carbon bonds in benzene have the same length, 139 pm, which lies between a normal single and a normal double bond. Reason (R): Benzene is a resonance hybrid of two equivalent Kekulé structures, so its π electrons are delocalised over all six carbon atoms.
Answer
Both A and R are true and R is the correct explanation of A. A single C−C bond is about 154 pm and a C=C bond about 134 pm. Experiment shows every bond in benzene to be identical at about 139 pm, neatly between the two, so the assertion is a correct statement of fact. The reason supplies the explanation. If benzene were a fixed structure with alternating single and double bonds, three bonds would be long and three short, and the ring would not be a regular hexagon. Because the two Kekulé structures are equivalent, the true molecule is a hybrid in which each of the six p orbitals contributes to one delocalised π cloud above and below the ring plane, giving every bond an identical bond order of about 1.5. This delocalisation also lowers the energy by the resonance energy of roughly 150 kJ/mol, which is why benzene prefers substitution to addition.
Distinguish between the inductive effect and the resonance effect, giving one example of each.
Answer
The inductive effect operates through sigma bonds. An atom or group more electronegative than carbon pulls the shared sigma electrons towards itself, creating a small permanent dipole that is relayed weakly along the chain and dies away within about three carbons. In 3-chloropropanoic acid the chlorine shows a −I effect that drains electron density along the chain, and this is why the substituted acid is stronger than propanoic acid itself. The resonance effect operates through π bonds and lone pairs, and involves the actual delocalisation of electrons over several atoms rather than mere polarisation. It requires a conjugated system, and unlike the inductive effect it is not weakened by distance along that conjugated path. In the carboxylate ion the negative charge is shared equally between both oxygen atoms, which makes the two C−O bonds identical and stabilises the ion.
On complete combustion, 0.220 g of an organic compound containing only carbon, hydrogen and oxygen gave 0.440 g of carbon dioxide and 0.180 g of water. Determine its empirical formula, and its molecular formula if its molar mass is 88 g/mol. (C = 12, H = 1, O = 16)
Answer
Mass of carbon = 0.440 × (12 ÷ 44) = 0.120 g. Mass of hydrogen = 0.180 × (2 ÷ 18) = 0.020 g. Mass of oxygen = 0.220 − 0.120 − 0.020 = 0.080 g (oxygen is always found by difference, never from the CO₂ or H₂O, since those also contain oxygen from the air). Moles: C = 0.120 ÷ 12 = 0.0100; H = 0.020 ÷ 1 = 0.020; O = 0.080 ÷ 16 = 0.0050. Divide by the smallest, 0.0050: C = 2, H = 4, O = 1. Empirical formula = C₂H₄O, with empirical formula mass = 24 + 4 + 16 = 44. n = molar mass ÷ empirical formula mass = 88 ÷ 44 = 2. Molecular formula = C₄H₈O₂.
(i) State the four main rules used to select and number the parent chain in IUPAC nomenclature. (ii) Write the IUPAC name of each of the following: (a) CH₃−CH(CH₃)−CH₂−CH₂−OH, (b) CH₃−CH₂−CO−CH₂−CH₃, (c) CH₂=CH−CH₂−CH₂−CH₃, (d) CH₃−CH(Cl)−COOH. (iii) Draw and name two chain isomers and one position isomer of C₅H₁₂ or C₅H₁₁OH as appropriate. (iv) Explain why 2-ethylbutane is not an acceptable IUPAC name. (v) Explain what a functional group is and why it, rather than the carbon skeleton, decides the reactions of a compound.
Answer
(i) First, choose as the parent the longest continuous carbon chain that contains the principal functional group, even if it has to bend on the page. Second, number the chain from the end that gives the principal functional group the lowest possible locant; if there is no functional group, give the multiple bond the lowest locant, and only then consider substituents. Third, cite substituents in alphabetical order, ignoring the multiplying prefixes di, tri and tetra when alphabetising but counting prefixes such as iso and cyclo. Fourth, when a choice still remains, apply the rule of lowest locants to the whole set of substituent numbers, comparing term by term. (ii) (a) The −OH group is the principal group, so the chain must include it and numbering starts from the carbon nearer to it. Four carbons in the chain with a methyl branch on carbon 3 gives 3-methylbutan-1-ol. (b) A carbonyl group flanked by two carbons is a ketone; the chain has five carbons with the C=O at the middle carbon, giving pentan-3-one. (c) A five-carbon chain with a double bond starting at carbon 1 gives pent-1-ene. (d) The −COOH group always takes carbon 1, and the chlorine sits on carbon 2 of a three-carbon acid, giving 2-chloropropanoic acid. (iii) For C₅H₁₂ the three chain isomers are: pentane, a straight chain of five carbons; 2-methylbutane, a four-carbon chain carrying one methyl on carbon 2; and 2,2-dimethylpropane, a three-carbon chain with two methyls on the middle carbon. These differ only in how the skeleton is branched, so they are chain isomers. For a position isomer, take the five-carbon alcohols pentan-1-ol and pentan-2-ol, which have the identical skeleton and the identical −OH group but differ only in where that group is attached. (iv) The name 2-ethylbutane implies a four-carbon parent with an ethyl group on carbon 2. Writing the structure out, CH₃−CH(C₂H₅)−CH₂−CH₃, and tracing the longest continuous chain reveals five carbons running from a methyl of the ethyl group through to the far end, not four. The correct parent is therefore pentane and the true name is 3-methylpentane. The error is a failure of rule one — the longest chain was not selected. (v) A functional group is the atom or group of atoms that gives a molecule its characteristic set of reactions, such as −OH, −CHO, −COOH, −NH₂ or a carbon–carbon double bond. The hydrocarbon skeleton is made of strong, non-polar C−C and C−H sigma bonds that are attacked only under vigorous conditions. The functional group, by contrast, contains polar bonds, lone pairs or π electrons, so it provides the electron-rich or electron-poor site that a reagent can attack. This is why every member of a homologous series behaves in essentially the same way chemically while its physical properties change gradually with chain length.
(i) Explain the difference between homolytic and heterolytic fission, showing the arrow notation used for each. (ii) Define electrophile and nucleophile and give two examples of each. (iii) Arrange the free radicals CH₃•, CH₃CH₂•, (CH₃)₂CH• and (CH₃)₃C• in increasing order of stability and justify the order. (iv) Explain, with a stepwise mechanism, how a carbocation intermediate is formed and then captured when 2-methylpropan-2-ol reacts with concentrated hydrochloric acid. (v) Why is the resonance hybrid always lower in energy than any single canonical structure?
Answer
(i) In homolytic fission the covalent bond breaks symmetrically and each fragment keeps one of the two shared electrons, producing two neutral free radicals: X−Y → X• + Y•. It is shown with half-headed (fish-hook) arrows, each representing the movement of a single electron, and it is favoured by ultraviolet light, high temperature and non-polar solvents. In heterolytic fission the bond breaks unsymmetrically and the more electronegative fragment takes both electrons, producing a cation and an anion: X−Y → X⁺ + Y⁻. It is shown with a full curved arrow representing an electron pair, and it is favoured by polar solvents that can solvate and stabilise the resulting ions. (ii) An electrophile is an electron-deficient species that accepts an electron pair, so it attacks electron-rich sites. Examples: the nitronium ion NO₂⁺ and the carbocation (CH₃)₃C⁺; neutral electron-poor species such as BF₃ and AlCl₃ also qualify. A nucleophile is an electron-rich species with a lone pair or a π bond that it donates to an electron-poor centre. Examples: the hydroxide ion OH⁻ and ammonia NH₃; the cyanide ion CN⁻ and water are others. (iii) Increasing stability: CH₃• < CH₃CH₂• < (CH₃)₂CH• < (CH₃)₃C•. A free radical has an unpaired electron in a p orbital on a roughly planar sp² carbon, so it is electron-deficient. Alkyl groups stabilise it by hyperconjugation, in which electrons of the adjacent C−H sigma bonds are delocalised into the half-filled orbital, and by their +I effect. The methyl radical has no alpha hydrogen, the ethyl has three, the isopropyl six and the tertiary butyl nine, so stability rises in exactly that order. The same reasoning gives the parallel order for carbocations. (iv) Step 1 — protonation. The oxygen of the alcohol carries lone pairs and acts as a nucleophile towards H⁺ from the acid. A curved arrow runs from an oxygen lone pair to the proton, converting the −OH into a protonated alcohol, −OH₂⁺. This is essential because OH⁻ is a poor leaving group whereas neutral water is an excellent one. Step 2 — ionisation. The C−O bond breaks heterolytically, the curved arrow running from the C−O bond to the oxygen. Water departs and a tertiary carbocation (CH₃)₃C⁺ is left behind. This slow step controls the rate, and it happens readily here because the tertiary cation is well stabilised by hyperconjugation from nine alpha hydrogens. Step 3 — capture. The chloride ion, a nucleophile, attacks the empty p orbital of the planar carbocation, the curved arrow running from a lone pair on Cl⁻ to the positive carbon. The product is 2-chloro-2-methylpropane, and the overall change is a nucleophilic substitution proceeding through a carbocation. (v) Delocalisation always lowers energy because the electrons are spread over a larger volume and are no longer confined between two nuclei; a bigger box means a lower kinetic energy for the electrons, and the charge is shared by more atoms so electron–electron repulsion is reduced. The canonical structures are only imperfect drawings, none of which exists in isolation, and the real hybrid lies below the most stable of them by an amount called the resonance energy. The more equivalent and comparably stable canonical structures can be written, the greater this stabilisation, which is why benzene and the carboxylate ion are so much more stable than their individual drawings suggest.
A student isolates a crude white organic solid from a reaction. She dissolves it in the minimum volume of hot ethanol, filters the hot solution and lets it cool slowly, obtaining shining crystals. She then checks their purity by thin layer chromatography, in which the spot travels 3.6 cm while the solvent front travels 6.0 cm. Finally, she sends a 0.80 g sample for Kjeldahl analysis; the ammonia liberated is absorbed in 50.0 mL of 0.500 M H₂SO₄, and the unreacted acid needs 60.0 mL of 0.500 M NaOH for neutralisation. Answer: (a) Explain the purpose of each step of the crystallisation. (b) Calculate the R_f value and state what it indicates. (c) Calculate the percentage of nitrogen in the sample. (d) State one type of nitrogen compound for which Kjeldahl's method fails, and give a reason. (N = 14)
Answer
(a) The solid is dissolved in the minimum volume of hot solvent so that the solution is saturated at the higher temperature; using excess solvent would keep the compound dissolved on cooling and the yield would collapse. Ethanol is chosen because the compound is much more soluble in it hot than cold, while the impurities behave differently. Filtering while hot removes insoluble impurities such as dust or unreacted solid before any crystallisation begins, and it must be done hot or the product itself will crystallise in the filter paper. Slow cooling is deliberate: crystals that grow slowly build an ordered lattice that excludes foreign molecules, whereas rapid cooling traps impurities inside and gives a fine, less pure powder. The soluble impurities stay behind in the mother liquor. (b) R_f = (distance moved by the spot) ÷ (distance moved by the solvent front) = 3.6 ÷ 6.0 = 0.60. A single spot with a definite R_f indicates a single component, so the crystallisation appears to have worked; two or more spots would have shown the sample was still a mixture. The R_f value is characteristic of a compound only for a fixed adsorbent and solvent system at a fixed temperature, so it is used for comparison against a known reference run on the same plate, not as an absolute identity. (c) Moles of H₂SO₄ taken = 0.500 × 0.0500 = 0.0250 mol. Moles of NaOH used on the excess acid = 0.500 × 0.0600 = 0.0300 mol. Since 2 mol of NaOH neutralise 1 mol of H₂SO₄, the excess acid = 0.0300 ÷ 2 = 0.0150 mol. Acid actually neutralised by ammonia = 0.0250 − 0.0150 = 0.0100 mol. Each mole of H₂SO₄ reacts with 2 mol of NH₃, so moles of NH₃ = 2 × 0.0100 = 0.0200 mol, and each NH₃ carries one nitrogen atom. Mass of nitrogen = 0.0200 × 14 = 0.280 g. Percentage of nitrogen = (0.280 ÷ 0.80) × 100 = 35.0%. (d) The method fails for nitrogen present in a ring, as in pyridine, and for nitro and azo compounds. Kjeldahl's digestion with hot concentrated sulphuric acid must convert every nitrogen atom quantitatively into ammonium sulphate. Nitrogen locked inside a stable aromatic ring is not attacked under these conditions, and nitrogen already in a high oxidation state, as in a nitro or azo group, is not reduced to ammonia by the digestion mixture. In both cases some nitrogen escapes conversion, less ammonia is liberated than it should be, and the result comes out too low. Such compounds are analysed by Duma's method instead, which burns the sample with copper(II) oxide and measures the nitrogen gas released.
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