NCERT Solutions · Class 11 Chemistry Organic Chemistry – Some Basic Principles and Techniques

Sodium fusion extract (Lassaigne's extract) turns any sulphur in the compound into sodium sulphide, and the test for that sulphide only works if the precipitate you get back is unambiguously lead sulphide — sulphuric acid ruins that by making its own precipitate.Where the sulphide comes from. When the organic compound is fused with sodium metal, any sulphur present is converted to sodium sulphide, which dissolves in water along with sodium cyanide and sodium halide (if present) to give the sodium fusion extract:\[\text{Na} + \text{C} + \text{N} + \text{S} \;\xrightarrow{\text{fusion}}\; \text{NaCN}, \; \text{Na}_2\text{S}, \; \text{NaX} \]The test itself. To confirm sulphur, the extract is acidified and treated with lead acetate solution, \(\displaystyle (\text{CH}_3\text{COO})_2\text{Pb} \) — "lead acetate" is lead(II) acetate, \(\displaystyle \text{Pb}^{2+} \) paired with two acetate ions. If sulphide is present, it forms a black precipitate of lead sulphide, \(\displaystyle \text{PbS} \):\[\text{Na}_2\text{S} + (\text{CH}_3\text{COO})_2\text{Pb} \;\longrightarrow\; \text{PbS}\downarrow \;(\text{black}) + 2\,\text{CH}_3\text{COONa} \]Check the balance: $\displaystyle 2$ Na on the left (in \(\displaystyle \text{Na}_2\text{S} \)) match $\displaystyle 2$ Na on the right (in \(\displaystyle 2\,\text{CH}_3\text{COONa} \)); $\displaystyle 1$ Pb matches $\displaystyle 1$ Pb; $\displaystyle 1$ S matches the $\displaystyle 1$ S locked in PbS; the two acetate groups that started on lead end up on the two sodium atoms. Nothing needs any other coefficient.Why acetic acid, and not sulphuric acid, is used to acidify. Acetic acid, \(\displaystyle \text{CH}_3\text{COOH} \), is a weak acid that does not react with lead acetate at all — it simply supplies the acidic medium needed for the test without introducing any new precipitating anion. Sulphuric acid, \(\displaystyle \text{H}_2\text{SO}_4 \), does the opposite: it reacts with the lead acetate itself, because sulphate ions form an insoluble salt with \(\displaystyle \text{Pb}^{2+} \):\[\text{H}_2\text{SO}_4 + (\text{CH}_3\text{COO})_2\text{Pb} \;\longrightarrow\; \text{PbSO}_4\downarrow \;(\text{white, insoluble}) + 2\,\text{CH}_3\text{COOH} \]Balance check: $\displaystyle 2$ H on the left end up as the $\displaystyle 2$ H's carried by the two \(\displaystyle \text{CH}_3\text{COOH} \) molecules on the right; the one sulphate group goes onto lead; $\displaystyle 1$ Pb matches $\displaystyle 1$ Pb. This is balanced with no extra coefficients needed elsewhere.The aside people miss: this is not just "sulphuric acid also happens to react" — that reaction produces a precipitate (\(\displaystyle \text{PbSO}_4\), white) every single time you add sulphuric acid to lead acetate, whether or not the compound being tested contains sulphur. That precipitate would appear over the black \(\displaystyle \text{PbS} \) (if any sulphide were present) and mask it, or appear by itself and be mistaken for a positive result even when there is no sulphur in the sample at all — a false positive. Acetic acid gives no such precipitate with lead acetate, so any precipitate that does form when the extract is added must be \(\displaystyle \text{PbS} \), and only \(\displaystyle \text{PbS} \), making the test reliable.Answer: Acetic acid is used because it does not react with lead acetate and so cannot form any precipitate of its own; sulphuric acid, in contrast, reacts with lead acetate to give a white precipitate of lead sulphate \(\displaystyle (\text{PbSO}_4) \), which would mask the black lead sulphide \(\displaystyle (\text{PbS}) \) precipitate that is meant to indicate sulphur, or falsely mimic a positive test even in its absence — so sulphuric acid would make the test unreliable.

Every carbon atom in the sample turns into one molecule of \(\displaystyle CO_2 \), and every two hydrogen atoms turn into one molecule of \(\displaystyle H_2O \) — combustion analysis is just carrying the percentage composition through mole ratios into product masses.Step $\displaystyle 1$: Work out what fraction of the sample is carbon and what fraction is hydrogen.The compound is $\displaystyle 69$% carbon and $\displaystyle 4.8$% hydrogen by mass, so the rest is oxygen: \[\%O = 100 - 69 - 4.8 = 26.2\% \] Oxygen doesn't matter for this question — combustion adds its own \(\displaystyle O_2 \), so only the carbon and hydrogen masses carry through to the products.Step $\displaystyle 2$: Find the actual mass of C and H burned, in the $\displaystyle 0.20$ g sample.A common slip here is using $\displaystyle 69$ g and $\displaystyle 4.8$ g (the amounts in $\displaystyle 100$ g of compound) instead of scaling down to the $\displaystyle 0.20$ g actually burned. The percentages apply to this sample, so: \[m(C) = 69\% \times 0.20\ g = 0.69 \times 0.20 = 0.138\ g \] \[m(H) = 4.8\% \times 0.20\ g = 0.048 \times 0.20 = 0.0096\ g \]Step $\displaystyle 3$: Convert each mass to moles of atoms, using molar mass = mass ÷ (mass per mole).Molar mass of C is \(\displaystyle 12\ g/mol \) and of H is \(\displaystyle 1\ g/mol \): \[n(C) = \frac{0.138\ g}{12\ g/mol} = 0.0115\ mol \] \[n(H\ atoms) = \frac{0.0096\ g}{1\ g/mol} = 0.0096\ mol \]Step $\displaystyle 4$: Turn moles of atoms into moles of product, using the balanced combustion equations.Carbon burns one atom to one molecule of carbon dioxide — already balanced as written, one C on each side: \[C + O_2 \rightarrow CO_2 \] So \(\displaystyle 1 \) mol C gives \(\displaystyle 1 \) mol \(\displaystyle CO_2 \): \[n(CO_2) = n(C) = 0.0115\ mol \]Hydrogen burns two atoms to one molecule of water. Writing it as a balanced equation makes the $\displaystyle 2$:$\displaystyle 1$ ratio explicit — $\displaystyle 4$ H atoms and one \(\displaystyle O_2 \) ($\displaystyle 2$ O atoms) on the left match $\displaystyle 2$ molecules of \(\displaystyle H_2O \) ($\displaystyle 4$ H, $\displaystyle 2$ O) on the right: \[4H + O_2 \rightarrow 2H_2O \] This is the step people get wrong: it is tempting to treat H and \(\displaystyle H_2O \) as $\displaystyle 1$:$\displaystyle 1$, the way C and \(\displaystyle CO_2 \) are. But every water molecule needs two hydrogen atoms, so: \[n(H_2O) = \frac{n(H\ atoms)}{2} = \frac{0.0096}{2} = 0.0048\ mol \]Step $\displaystyle 5$: Convert moles of product back to mass, using mass = moles × molar mass.Molar mass of \(\displaystyle CO_2 = 12 + 2(16) = 44\ g/mol \). Molar mass of \(\displaystyle H_2O = 2(1) + 16 = 18\ g/mol \). \[m(CO_2) = 0.0115\ mol \times 44\ g/mol = 0.506\ g \] \[m(H_2O) = 0.0048\ mol \times 18\ g/mol = 0.0864\ g \]Answer: Combustion of the $\displaystyle 0.20$ g sample produces $\displaystyle 0.506$ g of carbon dioxide and $\displaystyle 0.0864$ g of water.

In Kjeldahl's method, the nitrogen escapes as ammonia and gets trapped by a KNOWN excess of acid — what you actually measure is not the ammonia itself, but how much of that acid the ammonia used up.The ammonia liberated from the compound is passed into a measured volume of sulphuric acid — more acid than is needed, so all the ammonia is absorbed: \[2NH_3 + H_2SO_4 \rightarrow (NH_4)_2SO_4 \] (ammonia + sulphuric acid → ammonium sulphate). This is already balanced: sulphuric acid has two acidic hydrogens, so it takes two molecules of ammonia to use them both up.Whatever acid is left over (the "residual acid") is then found by titrating it against a standard sodium hydroxide solution: \[H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O \] (sulphuric acid + sodium hydroxide → sodium sulphate + water). Again balanced two-for-two, for the same reason.The step everyone trips on here: sulphuric acid is diprotic, so its molarity is NOT its normality. You cannot just subtract "volume of acid minus volume of base" when the acid is diprotic and the base is monoprotic — you have to convert both to milliequivalents (meq) first.Total acid taken, before any of it touches the ammonia: \[\text{mmol } H_2SO_4 = M \times V = 0.5\ \text{mol L}^{-1} \times 50\ \text{mL} = 25\ \text{mmol} \] Each \(\displaystyle H_2SO_4 \) molecule supplies $\displaystyle 2$ acidic \(\displaystyle H^+ \), so its normality is \(\displaystyle 2 \times 0.5 = 1.0\ \text{N} \), and \[\text{total meq } H_2SO_4 = 1.0\ \text{N} \times 50\ \text{mL} = 50\ \text{meq} \]Acid left over after the ammonia had its share, found from the NaOH titration: \[\text{mmol NaOH used} = 0.5\ \text{mol L}^{-1} \times 60\ \text{mL} = 30\ \text{mmol} \] NaOH is monoprotic, so its normality equals its molarity, and \(\displaystyle 30\ \text{mmol NaOH} = 30\ \text{meq NaOH} \). At the point of neutralisation, meq of acid reacting = meq of base added, so the residual (unreacted) acid was also \[30\ \text{meq } H_2SO_4 \]Acid actually consumed by the ammonia is what's left after removing the residual part from the total: \[50\ \text{meq (total)} - 30\ \text{meq (residual)} = 20\ \text{meq } H_2SO_4 \text{ reacted with } NH_3 \]One equivalent of acid always neutralises exactly one equivalent of base — that equivalence is what turns "acid consumed" into "ammonia present."\(\displaystyle NH_3 \) is a monoacidic base (it accepts a single \(\displaystyle H^+ \) to become \(\displaystyle NH_4^+ \)), so its milliequivalents equal its millimoles: \[n(NH_3) = 20\ \text{meq} = 20\ \text{mmol} \] Every \(\displaystyle NH_3 \) molecule carries exactly one nitrogen atom, so the moles of nitrogen equal the moles of ammonia: \[n(N) = 0.020\ \text{mol}, \qquad \text{mass of } N = 0.020\ \text{mol} \times 14\ \text{g mol}^{-1} = 0.28\ \text{g} \]Percentage composition is always (mass of the element ÷ mass of the WHOLE ORIGINAL SAMPLE) × $\displaystyle 100$ — here the $\displaystyle 0.50$ g of organic compound you started with, not the mass of acid or ammonia.Using the formula \(\displaystyle \%N = \dfrac{\text{mass of N}}{\text{mass of compound}} \times 100 \), where "mass of compound" is the weighed sample \(\displaystyle W \): \[\%N = \frac{0.28\ \text{g}}{0.50\ \text{g}} \times 100 = 56\% \]As a check, the same result follows from the compact Kjeldahl relation \(\displaystyle \%N = \dfrac{1.4 \times \text{meq}(NH_3)}{W(\text{g})} \), where the constant \(\displaystyle 1.4 = \dfrac{14}{10} \) folds together the atomic mass of nitrogen ($\displaystyle 14$) and the mmol-to-percentage conversion: \[\%N = \frac{1.4 \times 20}{0.50} = 56\% \]Both routes agree, confirming the residual-acid bookkeeping was done correctly.Answer: The compound contains $\displaystyle 56$% nitrogen by mass.

Carius chlorine estimation converts every atom of chlorine in the sample into silver chloride, so you scale by the fraction of AgCl's mass that is chlorine, then compare that chlorine mass to the mass of compound you started with.Step $\displaystyle 1$ — Find what fraction of AgCl's mass is chlorine.Silver chloride is \(\displaystyle \text{AgCl} \), one silver atom bonded to one chlorine atom. Its molar mass is\[M(\text{AgCl}) = 108 \ (\text{Ag}) + 35.5 \ (\text{Cl}) = 143.5\ \text{g mol}^{-1} \]Since every mole of AgCl carries exactly one mole of Cl, the mass fraction of chlorine locked inside AgCl is\[\frac{\text{mass of Cl}}{\text{mass of AgCl}} = \frac{35.5}{143.5} \]This is the step people skip past: you are not weighing chlorine directly — you are weighing AgCl and then asking "how much of that weight is the chlorine part?"Step $\displaystyle 2$ — Convert the AgCl mass obtained into a mass of chlorine.The Carius estimation gave $\displaystyle 0.5740$ g of AgCl from the sample. Using the fraction from Step $\displaystyle 1$,\[\text{mass of Cl} = 0.5740\ \text{g} \times \frac{35.5}{143.5} \]Work the arithmetic in one line:\[0.5740 \times 35.5 = 20.377 \] \[\text{mass of Cl} = \frac{20.377}{143.5} = 0.1420\ \text{g} \]Step $\displaystyle 3$ — Turn that into a percentage of the original compound.Percentage composition always means: (mass of the element) divided by (mass of the whole sample you weighed out), not by the mass of AgCl and not by any intermediate quantity. Here the sample taken was $\displaystyle 0.3780$ g of the organic chloro compound.\[\% \text{Cl} = \frac{\text{mass of Cl}}{\text{mass of compound}} \times 100 = \frac{0.1420}{0.3780} \times 100 \]\[\% \text{Cl} = \frac{14.20}{3.780} = 37.566\ldots \approx 37.57\% \]So, packing both steps into a single working formula (the one worth remembering for every Carius chlorine problem):\[\% \text{Cl} = \frac{35.5}{143.5} \times \frac{\text{mass of AgCl}}{\text{mass of compound}} \times 100 \]Answer: The compound contains $\displaystyle 37.57$% chlorine by mass.

In the Carius method, all the sulphur in the compound ends up locked inside a fixed amount of \(\displaystyle \text{BaSO}_4\), so you find the mass of sulphur by scaling down the mass of \(\displaystyle \text{BaSO}_4\) by the fraction of its molar mass that is sulphur.What actually happens in the tube. The organic sulphur compound is heated with fuming nitric acid in a sealed Carius tube. The nitric acid oxidises every sulphur atom in the compound to sulphate ion, \(\displaystyle \text{SO}_4^{2-}\). This sulphate is then precipitated completely as barium sulphate (a white solid) by adding excess barium chloride solution to the solution:\[\text{BaCl}_2 + \text{H}_2\text{SO}_4 \rightarrow \text{BaSO}_4\downarrow + 2\text{HCl} \]Check the balance: $\displaystyle 1$ barium atom and $\displaystyle 2$ chlorine atoms on each side, $\displaystyle 2$ hydrogen atoms on each side, and $\displaystyle 1$ sulphur with $\displaystyle 4$ oxygens (as the sulphate group) carried straight through onto \(\displaystyle \text{BaSO}_4\) — nothing needs a coefficient beyond the $\displaystyle 2$ already in front of HCl. Every sulphur atom that started in the organic compound is now sitting inside one formula unit of \(\displaystyle \text{BaSO}_4\), so weighing the barium sulphate precipitate tells you exactly how much sulphur was present.Setting up the mass relationship. One mole of \(\displaystyle \text{BaSO}_4\) (molar mass \(\displaystyle 137 + 32 + 4(16) = 233\ \text{g mol}^{-1}\), using \(\displaystyle \text{Ba} = 137\), \(\displaystyle \text{S} = 32\), \(\displaystyle \text{O} = 16\)) contains exactly one mole of S (atomic mass \(\displaystyle 32\ \text{g mol}^{-1}\)). So the mass of sulphur is:\[\text{mass of S} = \dfrac{32}{233} \times \text{mass of } \text{BaSO}_4 \]This is the step people skip past — the mass of the precipitate you weigh is not the mass of sulphur; you must scale it down by the ratio (atomic mass of S) / (molar mass of \(\displaystyle \text{BaSO}_4\)), because most of that precipitate's mass is barium and oxygen, not sulphur.Percentage of sulphur in the original compound is then this mass of S expressed as a percentage of the mass of compound taken:\[\%\ \text{S} = \dfrac{32}{233} \times \dfrac{\text{mass of BaSO}_4}{\text{mass of compound}} \times 100 \]Here "mass of compound" is the $\displaystyle 0.468$ g of organic substance actually weighed out for the Carius tube — not any other mass in the experiment, since percentage composition is always mass of the element per unit mass of the original sample.Substituting the numbers. Mass of \(\displaystyle \text{BaSO}_4\) formed = $\displaystyle 0.668$ g, mass of compound taken = $\displaystyle 0.468$ g:\[\%\ \text{S} = \dfrac{32 \times 0.668 \times 100}{233 \times 0.468} \]Working the numerator and denominator separately:\[32 \times 0.668 \times 100 = 2137.6 \] \[233 \times 0.468 = 109.044 \]So:\[\%\ \text{S} = \dfrac{2137.6}{109.044} = 19.60 \]Answer: The organic compound contains $\displaystyle 19.60$% sulphur by mass.

The hybridisation of a carbon atom is decided by how many σ (sigma) bonds surround it — a double bond counts as one σ + one π, a triple bond as one σ + two π — not by which functional group it happens to sit next to.Number the six carbons of the chain so each one can be pointed to individually:\[\underset{C_1}{CH_2} = \underset{C_2}{CH} - \underset{C_3}{CH_2} - \underset{C_4}{CH_2} - \underset{C_5}{C} \equiv \underset{C_6}{CH} \]Now go carbon by carbon and count only the σ bonds around each one (that count is what fixes the hybridisation): A step people get wrong here: it is tempting to think \(\displaystyle C_3\) and \(\displaystyle C_4\) must be "half-hybridised" because they sit between a double bond and a triple bond. They do not — neither carbon is itself part of a multiple bond, so each one is counted purely by its own four single-bond neighbours and comes out as an ordinary sp³ –\(\displaystyle \mathrm{CH_{2}}\)– carbon, exactly as in an alkane.The question asks about the \(\displaystyle C_2-C_3\) bond. This bond is a plain single bond that joins the last carbon of the double bond, \(\displaystyle C_2\) (sp²), to the first carbon of the –\(\displaystyle \mathrm{CH_{2}}\)–\(\displaystyle \mathrm{CH_{2}}\)– bridge, \(\displaystyle C_3\) (sp³). So it is formed by the end-on overlap of an sp² orbital on \(\displaystyle C_2\) with an sp³ orbital on \(\displaystyle C_3\) — an sp²–sp³ σ bond.Answer: (c) sp²–sp³

Prussian blue is not the sodium salt that forms first — it appears only once that sodium salt meets iron in the +$\displaystyle 3$ state, giving the compound \(\displaystyle \text{Fe}_4[\text{Fe(CN)}_6]_3\). Track the iron and the cyanide through each step of the test and the answer falls out of the bookkeeping.Step $\displaystyle 1$ — sodium fusion makes cyanide, not any iron compound yet. When the nitrogen-containing organic compound is fused with sodium metal, its carbon and nitrogen atoms combine with sodium to give sodium cyanide in the fusion extract: \[\text{Na} + \text{C} + \text{N} \longrightarrow \text{NaCN} \] There is no iron anywhere in this step — the extract is just a solution containing \(\displaystyle \text{Na}^+\) and \(\displaystyle \text{CN}^-\) (plus \(\displaystyle \text{NaOH}\), \(\displaystyle \text{Na}_2\text{S}\), etc., from other elements). Iron is added only in the next step, so option (a), which already has iron in it, cannot be what forms here.Step $\displaystyle 2$ — boiling with FeSO₄ builds the ferrocyanide ion. The alkaline extract is boiled with freshly prepared iron(II) sulphate, \(\displaystyle \text{FeSO}_4\) (iron here is \(\displaystyle \text{Fe}^{2+}\), the ferrous state). Excess \(\displaystyle \text{CN}^-\) wraps around each \(\displaystyle \text{Fe}^{2+}\) to give the complex hexacyanidoferrate(II) ion, isolated as sodium ferrocyanide: \[\text{FeSO}_4 + 6\,\text{NaCN} \longrightarrow \text{Na}_4[\text{Fe(CN)}_6] + \text{Na}_2\text{SO}_4 \] Check the balance: iron, \(\displaystyle 1=1\); sodium, \(\displaystyle 6\) on the left equals \(\displaystyle 4+2\) on the right; \(\displaystyle \text{CN}\), \(\displaystyle 6=6\); sulphate, \(\displaystyle 1=1\). In \(\displaystyle \text{Na}_4[\text{Fe(CN)}_6]\) the iron is still \(\displaystyle \text{Fe}^{2+}\) — this is option (a). It is pale/colourless in solution, not blue, because the blue colour needs a second, higher oxidation state of iron to react with it. This is the step people stop at and wrongly pick (a): forming \(\displaystyle \text{Na}_4[\text{Fe(CN)}_6]\) is necessary for the test, but it is an intermediate, not the coloured product asked for.Step $\displaystyle 3$ — some iron is already Fe³⁺. \(\displaystyle \text{FeSO}_4\) solution is never perfectly pure \(\displaystyle \text{Fe}^{2+}\); air oxidises part of it to iron(III) sulphate on standing and boiling, so the mixture always carries some \(\displaystyle \text{Fe}^{3+}\) alongside the \(\displaystyle \text{Fe}^{2+}\) that became ferrocyanide.Step $\displaystyle 4$ — acidifying brings Fe³⁺ and ferrocyanide together, and that reaction is Prussian blue. On cooling and acidifying with dilute sulphuric acid, any iron hydroxide precipitate dissolves back into solution as \(\displaystyle \text{Fe}^{3+}\), and these ferric ions combine with the sodium ferrocyanide made in Step $\displaystyle 2$ to throw down an intensely coloured, insoluble precipitate: \[3\,\text{Na}_4[\text{Fe(CN)}_6] + 2\,\text{Fe}_2(\text{SO}_4)_3 \longrightarrow \text{Fe}_4[\text{Fe(CN)}_6]_3 + 6\,\text{Na}_2\text{SO}_4 \] Balance check: sodium, \(\displaystyle 12=12\); iron, \(\displaystyle 3\) (inside the three ferrocyanide units) \(\displaystyle +\,4\) (from the ferric sulphate) \(\displaystyle =7\) on the left, and on the right \(\displaystyle 4\) (outside) \(\displaystyle +\,3\) (inside) \(\displaystyle =7\); \(\displaystyle \text{CN}\), \(\displaystyle 18=18\); sulphate, \(\displaystyle 6=6\).The product, \(\displaystyle \text{Fe}_4[\text{Fe(CN)}_6]_3\), in words is ferric ferrocyanide (iron(III) hexacyanidoferrate(II)) — this is exactly Prussian blue. Its charges make sense with real oxidation states: the four outer iron atoms are \(\displaystyle \text{Fe}^{3+}\) (\(\displaystyle 4\times(+3)=+12\)), and each \(\displaystyle [\text{Fe(CN)}_6]\) unit inside carries the inner iron as \(\displaystyle \text{Fe}^{2+}\) with a \(\displaystyle 4-\) charge (three units give \(\displaystyle 3\times(-4)=-12\)), so the compound is neutral — a genuine, stable substance.Aside — why the other options fail on this same charge check. Option (c), \(\displaystyle \text{Fe}_2[\text{Fe(CN)}_6]\), would need the outer iron to also be \(\displaystyle \text{Fe}^{2+}\) (ferrous), giving \(\displaystyle 2\times(+2)=+4\) against the ferrocyanide's \(\displaystyle -4\) — this is a real compound (ferrous ferrocyanide, pale/white, not blue), but it needs Fe²⁺ everywhere, and this test always oxidises some iron to Fe³⁺, so it isn't what forms in the Prussian-blue step. Option (d), \(\displaystyle \text{Fe}_3[\text{Fe(CN)}_6]_4\), does not even balance charge for any simple oxidation state: \(\displaystyle 3\times(+3)=+9\) against \(\displaystyle 4\times(-4)=-16\) — the numbers don't match, so this is not a real product of the reaction at all.Answer: (b) \(\displaystyle \text{Fe}_4[\text{Fe(CN)}_6]_3\) — ferric ferrocyanide, the Prussian blue compound.

Chromatography is the technique that works when every other separation method runs out of things it can tell apart.Look at what each option actually depends on, and where that dependence breaks down.The step people undervalue here: chromatography does not need the components to differ in a single fixed property like boiling point or solubility-at-one-temperature. Its separating power comes from a continuous, repeated partitioning between two phases (adsorption/desorption or dissolution/re-dissolution happening over and over as the mixture moves), so even compounds that are extremely close in behaviour — geometrical isomers, closely related compounds in a natural extract, trace impurities — get pulled apart given enough length of stationary phase (a longer column, a bigger sheet) to repeat the partitioning enough times.This is also why it is the "best and latest" of the four: Answer: (d) Chromatography.

A nucleophile is an electron-rich species that attacks an electron-poor (electrophilic) carbon — so you decide "nucleophilic vs electrophilic" by asking which atom is short of electrons, not which atom is attacking.First, check that the equation you're given is even balanced, since a wrong count of atoms is where a lot of mistakes start silently:\[\text{CH}_3\text{CH}_2\text{I} + \text{KOH} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{KI} \]Here \(\displaystyle \text{CH}_3\text{CH}_2\text{I}\) is ethyl iodide (iodoethane), \(\displaystyle \text{KOH}\) is potassium hydroxide, \(\displaystyle \text{CH}_3\text{CH}_2\text{OH}\) is ethanol, and \(\displaystyle \text{KI}\) is potassium iodide. Counting atoms on each side: left side has $\displaystyle 2$ C, $\displaystyle 6$ H ($\displaystyle 5$ on the ethyl group + $\displaystyle 1$ on OH), $\displaystyle 1$ I, $\displaystyle 1$ K, $\displaystyle 1$ O; right side has $\displaystyle 2$ C, $\displaystyle 6$ H ($\displaystyle 5$ on the ethyl group + $\displaystyle 1$ on OH), $\displaystyle 1$ O, $\displaystyle 1$ K, $\displaystyle 1$ I. Every atom matches with a coefficient of $\displaystyle 1$ in front of each species, so the equation is already balanced as written — no multiplying needed.Now look at what is actually happening at the carbon that carries the iodine, since that carbon is where the reaction takes place.Step $\displaystyle 1$ — find the electrophilic centre. Iodine is more electronegative than carbon, so it pulls the bonding electron pair of the \(\displaystyle \text{C–I}\) bond toward itself. That leaves the carbon attached to I carrying a partial positive charge (\(\displaystyle \delta^{+}\)), while iodine carries a partial negative charge (\(\displaystyle \delta^{-}\)). A carbon that is short of electron density like this is an electrophilic carbon — it is the site that something electron-rich will be drawn to.Step $\displaystyle 2$ — identify the attacking species. The reagent supplying the reactive species here is aqueous \(\displaystyle \text{KOH}\), which is fully ionic in water and furnishes hydroxide ions, \(\displaystyle \text{OH}^{-}\). The oxygen of \(\displaystyle \text{OH}^{-}\) carries lone pairs of electrons and a full negative charge, so it is strongly electron-rich. A species that seeks out an electron-poor centre and donates its own electron pair to form a new bond is, by definition, a nucleophile.Step $\displaystyle 3$ — the mechanism in words. The oxygen of \(\displaystyle \text{OH}^{-}\) uses one of its lone pairs to attack the \(\displaystyle \delta^{+}\) carbon of ethyl iodide from the side opposite the C–I bond. As the new \(\displaystyle \text{C–O}\) bond forms, the \(\displaystyle \text{C–I}\) bond breaks heterolytically, with both electrons of that bond leaving on iodine. This kicks iodine out as iodide ion, \(\displaystyle \text{I}^{-}\), which then pairs with \(\displaystyle \text{K}^{+}\) to give the by-product potassium iodide (KI). The carbon ends up bonded to \(\displaystyle \text{OH}\) instead of \(\displaystyle \text{I}\), giving the product ethanol, \(\displaystyle \text{CH}_3\text{CH}_2\text{OH}\).The step people mix up: it is tempting to call this "electrophilic" because you might think of the halogen-bearing carbon as the reactive part being replaced. But the classification of a substitution reaction is named after the incoming attacking species, not the carbon skeleton. Here the species doing the attacking — hydroxide — is electron-rich (a nucleophile), and it is substituting for a leaving group (iodide) at an electron-poor carbon. That combination is a nucleophilic substitution. An electrophilic substitution would instead need an electron-poor (electrophile) species attacking an electron-rich centre, such as the benzene ring in nitration — the opposite situation from what happens here.So option (a), electrophilic substitution, does not describe this reaction; option (b), nucleophilic substitution, does — \(\displaystyle \text{OH}^{-}\) is the nucleophile that displaces \(\displaystyle \text{I}^{-}\) from the electrophilic carbon of ethyl iodide to give ethanol and potassium iodide.**Answer: (b) nucleophilic substitution — hydroxide ion (\(\displaystyle \text{OH}^{-}\)), an electron-rich nucleophile, attacks the electron-poor carbon of \(\displaystyle \text{CH}_3\text{CH}_2\text{I}\) and displaces iodide ion, converting ethyl iodide into ethanol (\(\displaystyle \text{CH}_3\text{CH}_2\text{OH}\)) with KI as the by-product; the balanced equation is \(\displaystyle \text{CH}_3\text{CH}_2\text{I} + \text{KOH} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{KI}\).