NEB Class 12 · Past paper
The complete NEB Class 12 2073 exam paper for Chemistry, all 33 questions with solved model answers.
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Ammonia molecule has a trigonal pyramidal geometry even though nitrogen of ammonia gets sp3 hybridization. Give reason.
In $\ce{NH3}$ the nitrogen atom undergoes sp3 hybridization, giving four sp3 orbitals directed towards the corners of a tetrahedron. Three of these orbitals form $\ce{N-H}$ bonds while the fourth holds a lone pair. Because one corner is ...
Why is crystalline oxalic acid regarded as a good substance for the preparation of a primary standard solution?
Crystalline oxalic acid ($\ce{H2C2O4.2H2O}$) is a good primary standard because it: - is available in a pure, definite crystalline form with an exactly known formula and equivalent weight, - is non-hygroscopic and stable in air (does not...
Define i) Degree of ionisation ii) Ostwald's dilution law.
i) Degree of ionisation ($\alpha$): the fraction of the total number of molecules of an electrolyte that dissociate into ions in solution.
ii) Ostwald's dilution law: for a weak binary electrolyte, the dissociation constant is $$K = \frac{\alpha^2 C}{1-\alpha}$$ where $C$ is the concentration and $\alpha$ the degree of ionisation. For very weak electrolytes ($\alpha \ll 1$), $K \approx \alpha^2 C$, so $\alpha = \sqrt{K/C}$: the degree of ionisation increases on dilution.
Predict whether the following reaction will occur spontaneously or not, and why. Pb^2+ + 2Ag -> 2Ag^+ + Pb; given E(Ag+/Ag) = 0.80 V and E(Pb2+/Pb) = -0.13 V.
For the given reaction, $\ce{Ag}$ is oxidised (anode) and $\ce{Pb^2+}$ is reduced (cathode). $$ \begin{aligned} E^\circ{cell} &= E^\circ{cathode} - E^\circ{anode} \ &= E^\circ{Pb^{2+}/Pb} - E^\circ{Ag^+/Ag} \ &= (-0.13) - (0.80) \ &= ...
What is meant by enthalpy of reaction? If the standard enthalpy of formation of ammonia is -46 kJ/mol, what is the enthalpy change of the reaction N2(g) + 3H2(g) -> 2NH3(g)?
Enthalpy of reaction: the heat change (at constant pressure) when the number of moles of reactants shown in the balanced equation react completely to form products.
For
$$ \ce{N2(g) + 3H2(g) -> 2NH3(g)} $$
, two moles of $\ce{NH3}$ are formed, so
$$ \begin{aligned} \Delta H &= 2\times\Delta H_f(\ce{NH3}) \ &= 2\times(-46) \ &= -92\ \text{kJ} \end{aligned} $$
The reaction is exothermic, $\Delta H = -92$ kJ per mole of reaction (per 2 mol $\ce{NH3}$).
Under what conditions is the reaction expected to occur i) spontaneously ii) non-spontaneously, if dH and dS are both negative?
When $\Delta H$ and $\Delta S$ are both negative, $\Delta G = \Delta H - T\Delta S$; the term $-T\Delta S$ becomes positive. i) Spontaneous: at low temperature, the negative $\Delta H$ outweighs the positive $-T\Delta S$ term, so
What are the essential conditions for the effective collision of reacting species?
For a collision between reactant molecules to be effective (to lead to product), two conditions must be met:
Only collisions satisfying both conditions form the activated complex and give products.
Starting from phenol, how would you obtain cyclohexane?
First reduce phenol to benzene with zinc dust, then hydrogenate benzene to cyclohexane: $$ \begin{aligned} \ce{C6H5OH + Zn - C6H6 + ZnO} \ \ce{C6H6 + 3H2 -[Ni][\Delta] C6H12} \end{aligned} $$ The product $\ce{C6H12}$ is cyclohexane. (Al...
Identify A and B in the sequence and give their IUPAC names: CH3-CH=CH2 -(HBr)-> A -(Na/dry ether)-> B.
A: propene adds HBr by Markovnikov's rule (Br on the more substituted carbon): $$\ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3}$$ A $= \ce{CH3CHBrCH3}$, IUPAC name 2-bromopropane.
B: Wurtz coupling of two molecules of A with sodium: $$\ce{2CH3-CHBr-CH3 + 2Na ->[dry ether] (CH3)2CH-CH(CH3)2 + 2NaBr}$$ B $= \ce{(CH3)2CH-CH(CH3)2}$, IUPAC name 2,3-dimethylbutane.
How is phenol obtained from i) benzene diazonium chloride ii) chlorobenzene?
(i) From benzene diazonium chloride: warm its aqueous solution (hydrolysis): $$\ce{C6H5N2Cl + H2O -[\Delta] C6H5OH + N2(g) + HCl}$$ (ii) From chlorobenzene (Dow process): heat with $\ce{NaOH}$ at high temperature and pressure, then acidi...
Write down an isomeric ether of isopropyl alcohol. What happens when the isomeric ether is heated with excess HI?
Isopropyl alcohol is $\ce{(CH3)2CHOH}$ ($\ce{C3H8O}$). Its isomeric ether is methoxyethane, $\ce{CH3-O-C2H5}$ (same formula, different functional group). With excess HI (heated) the ether is cleaved completely; each part becomes an iodid...
Write an example of i) Cannizzaro's reaction ii) Aldol condensation.
i) Cannizzaro's reaction (aldehyde without alpha-H, conc. alkali, self redox): $$\ce{2HCHO + NaOH - CH3OH + HCOONa}$$ ii) Aldol condensation (aldehyde with alpha-H, dilute base, giving a beta-hydroxy aldehyde): $$\ce{2CH3CHO -[dil. NaOH]...
Why is chloroacetic acid a stronger acid than acetic acid?
In chloroacetic acid, $\ce{ClCH2COOH}$, the chlorine atom is strongly electron withdrawing (negative inductive effect, $-I$). It pulls electron density away from the $\ce{-COOH}$ group, which: - makes it easier to release the $\ce{H+}$, ...
What happens when nitrobenzene is i) subjected to electrolytic reduction ii) treated with Zn/NaOH?
(i) Electrolytic reduction (strongly acidic medium, at cathode): nitrobenzene is reduced via phenylhydroxylamine, which rearranges to p-aminophenol: $$\ce{C6H5NO2 -[electrolytic][H+] HO-C6H4-NH2}$$ (ii) With Zn/NaOH (alkaline reduction):...
Give a suitable chemical test to distinguish ethanamine from N-methyl methanamine.
Ethanamine ($\ce{C2H5NH2}$) is a primary amine and N-methylmethanamine ($\ce{CH3-NH-CH3}$) is a secondary amine, so use the carbylamine (isocyanide) test: Warm each with chloroform and alcoholic $\ce{KOH}$. $$\ce{C2H5NH2 + CHCl3 + 3KOH -...
Distinguish between homopolymer and co-polymer with an example of each.
Homopolymer: a polymer made by the repeated joining of only one kind of monomer, e.g. polythene (from ethene, $\ce{CH2=CH2}$) or PVC (from vinyl chloride). Copolymer: a polymer made from two or more different monomers, e.g. Buna-S (from ...
Write down the structure of i) aspirin ii) paracetamol, and mention one use of each.
i) Aspirin (acetylsalicylic acid): a benzene ring bearing $\ce{-COOH}$ and, ortho to it, $\ce{-OCOCH3}$; condensed as $\ce{C6H4(OCOCH3)COOH}$. Use: analgesic and antipyretic (relieves pain and fever); also an anti-blood-clotting agent. i...
Illustrate the formation of a peptide bond with an example.
A peptide bond ($\ce{-CO-NH-}$) forms when the carboxyl group of one amino acid condenses with the amino group of another, eliminating a water molecule. Example (two glycine molecules): $$\ce{H2N-CH2-COOH + H2N-CH2-COOH - H2N-CH2-CO-NH-C...
What are non-reducing sugars? Write an example of it.
Non-reducing sugars are carbohydrates that cannot reduce Tollen's reagent or Fehling's solution, because their reducing groups (the anomeric $\ce{-OH}$/free carbonyl of both monosaccharide units) are locked in the glycosidic linkage.
Example: sucrose, $\ce{C12H22O11}$ (in which the reducing groups of glucose and fructose are joined together, so none is free).
Write the action of heat on blue vitriol.
Blue vitriol is $\ce{CuSO4.5H2O}$. On heating it loses water of crystallisation in steps and finally, on strong heating, the anhydrous salt decomposes to black copper oxide: $$ \begin{aligned} \ce{CuSO4.5H2O -[100^{\circ}C] CuSO4.H2O + 4...
Give the reactions for the extraction of metallic zinc from zinc blende.
Zinc blende ($\ce{ZnS}$) is concentrated by froth flotation, then: 1. Roasting to the oxide: $$\ce{2ZnS + 3O2 - 2ZnO + 2SO2(g)}$$ 2. Reduction of the oxide with coke (carbon) in a fire-clay retort at about 1400 degrees; zinc distils over...
Write down the chemical reactions that occur in the zone of reduction of the blast furnace during the extraction of iron.
In the zone of reduction (around 400 to 900 degrees) carbon monoxide reduces the iron oxides stepwise to metallic iron: $$ \begin{aligned} \ce{3Fe2O3 + CO - 2Fe3O4 + CO2} \ \ce{Fe3O4 + CO - 3FeO + CO2} \ \ce{FeO + CO - Fe + CO2} \end{a...
State Faraday's laws of electrolysis. Establish the relationship between electrochemical equivalent and chemical equivalent. 0.197 gm of copper is deposited by a current of 0.2 A in 50 minutes. Calculate its electrochemical equivalent.
Faraday's first law: the mass of a substance deposited/liberated at an electrode is directly proportional to the quantity of electricity passed, $m = z,I,t$. Faraday's second law: when the same quantity of electricity is passed through...
Define redox titration. 10 gm of NaOH was added to 200 cc of N/2 (f = 1.5) H2SO4. The volume was diluted to two litres. Predict whether the dilute solution is acidic, basic or neutral and also calculate the resulting molarity of the dilute solution.
Redox titration: a titration based on an oxidation-reduction reaction between the titrant and the analyte (e.g. $\ce{KMnO4}$ vs $\ce{FeSO4}$), the end point being detected by a change in oxidation state (often self-indicating).
Calculation:
$$ \begin{aligned} &= 0.5\times1.5 \ &= 0.75 \end{aligned} $$
N; in 200 cc, eq
$$ \begin{aligned} &= 0.75\times0.200 \ &= 0.15 \end{aligned} $$
eq of acid.
Base (0.25 eq) exceeds acid (0.15 eq), so the solution is basic.
Excess
$$ \begin{aligned} \ce{OH-} &= 0.25 - 0.15 \ &= 0.10 \end{aligned} $$
eq (mol) left after neutralization.
Diluted to 2 L, the resulting concentration of excess alkali:
$$ \begin{aligned} \text{molarity} &= \frac{0.10}{2} \ &= 0.05\ \text{M}\ (\text{NaOH}) \end{aligned} $$
Write down the chemistry of calomel.
Calomel is mercurous chloride, $\ce{Hg2Cl2}$, a white insoluble powder.
Preparation: $$ \begin{aligned} \ce{HgCl2 + Hg -> Hg2Cl2} \ \ce{Hg2(NO3)2 + 2NaCl -> Hg2Cl2(s) + 2NaNO3} \end{aligned} $$
Action of ammonia (test): it turns black due to disproportionation into finely divided mercury (black) and a white mercuric amido chloride: $$\ce{Hg2Cl2 + 2NH3 -> Hg(NH2)Cl + Hg + NH4Cl}$$ The black colour of free mercury confirms a mercurous salt.
Uses: in the calomel reference electrode and, formerly, as a purgative in medicine.
Define rate law. The reaction P + Q -> Z is first order with respect to P and zero order with respect to Q. Fill in the blanks: Expt I ([P]=0.1, [Q]=0.1, rate=2x10^-2), Expt II ([P]=?, [Q]=0.2, rate=4x10^-2), Expt III ([P]=0.4, [Q]=0.4, rate=?), Expt IV ([P]=?, [Q]=0.2, rate=2x10^-2).
Rate law: the experimentally determined equation relating reaction rate to the molar concentrations of the reactants raised to appropriate powers. Here, since the reaction is first order in P and zero order in Q:
$$\text{rate} = k[P]$$
Find k from Expt I:
$$ \begin{aligned} k &= \dfrac{2\times10^{-2}}{0.1} \ &= 0.2\ \text{min}^{-1} \end{aligned} $$
Fill the blanks:
$$ \begin{aligned} [P] &= \dfrac{\text{rate}}{k} \ &= \dfrac{4\times10^{-2}}{0.2} \ &= 0.2 \end{aligned} $$
M.
$$ \begin{aligned} &= k[P] \ &= 0.2\times0.4 \ &= 8\times10^{-2} \end{aligned} $$
M min$^{-1}$ (Q has no effect).
$$ \begin{aligned} [P] &= \dfrac{2\times10^{-2}}{0.2} \ &= 0.1 \end{aligned} $$
M.
Describe the laboratory method of preparation of ethoxyethane. What happens when ethoxyethane is exposed to air?
Laboratory preparation: ethanol is heated with excess concentrated $\ce{H2SO4}$ at 140 degrees; ethyl hydrogen sulphate forms first, then reacts with more ethanol (continuous etherification): $$ \begin{aligned} \ce{C2H5OH + H2SO4 - C2H5H...
Give chemical reactions for the preparation of ethanal from i) 1,1-dibromoethane ii) ethyne iii) ethanoyl chloride. How is ethanal converted into propan-2-ol?
Ethanal from 1,1-dibromoethane by hydrolysis of the gem-dihalide with aqueous alkali: $$\ce{CH3CHBr2 + 2NaOH(aq) - CH3CHO + 2NaBr + H2O}$$ Ethanal from ethyne by hydration (Kucherov reaction): $$\ce{C2H2 + H2O -[HgSO4/H2SO4] CH3CHO}$$ Et...
An aliphatic compound A reacts with SOCl2 to give B. B on dehydrohalogenation yields C. C on ozonolysis gives a mixture of ethanal and methanal. If A is an alcohol which responds to the iodoform test, identify A, B and C. What product would you expect when compound B is heated with H2/Ni?
C on ozonolysis gives ethanal + methanal, so C = propene, $\ce{CH3-CH=CH2}$ (since $$ \ce{CH3CH=CH2 -[O3] CH3CHO + HCHO} $$ ). - C comes from B by dehydrohalogenation, so B = 2-chloropropane, $\ce{CH3CHClCH3}$. - B comes from the alcohol...
How is pure and dry aniline prepared in the laboratory? Identify the major products A, B, C and D in the sequence: A -(CuCl/HCl)-> B -(CH3Br, dry ether)-> C -(oxidation)-> D. Compound D undergoes Clemmensen reduction to give toluene.
Preparation of aniline: reduce nitrobenzene with tin and conc. HCl, liberate with $\ce{NaOH}$ and purify by steam distillation: $$\ce{C6H5NO2 + 6[H] -[Sn/HCl] C6H5NH2 + 2H2O}$$ The aniline is dried over $\ce{KOH}$ and distilled. Reaction...
How would you distinguish propan-2-ol from 2-methyl propan-2-ol by Victor Meyer's method? Write suitable methods for the conversion of i) Chloroform into dimethylamine ii) Ethanamine into methanamine.
Victor Meyer's method: convert each alcohol to the alkyl iodide (red P + $\ce{I2}$), then to the nitroalkane with $\ce{AgNO2}$, and treat with $\ce{HNO2}$ followed by $\ce{NaOH}$. Propan-2-ol (secondary) forms a secondary nitro compound that gives a blue colour with $\ce{NaOH}$, while 2-methylpropan-2-ol (tertiary) has no alpha-hydrogen on the nitro carbon and so gives no colour (it stays colourless).
(i) Chloroform into dimethylamine. First reduce chloroform to methyl chloride:
$$\ce{CHCl3 + 6[H] ->[Zn/HCl] CH3Cl + 2HCl}$$
then treat methyl chloride with ammonia, from which the secondary amine is separated:
$$\ce{2CH3Cl + NH3 -> (CH3)2NH + 2HCl}$$
(ii) Ethanamine into methanamine goes one carbon shorter. First convert the amine to ethanol with nitrous acid:
$$\ce{C2H5NH2 ->[HNO2] C2H5OH}$$
oxidise the ethanol to ethanoic acid:
$$\ce{C2H5OH + [O] -> CH3COOH}$$
convert the acid to its amide:
$$\ce{CH3COOH ->[NH3][\Delta] CH3CONH2}$$
and finally apply Hofmann bromamide degradation to lose one carbon and give methanamine:
$$\ce{CH3CONH2 ->[Br2/KOH] CH3NH2}$$
State the solubility product constant. What is the proper condition for precipitation of a salt from its solution? Explain the application of the solubility product principle and the common ion effect. What is the minimum volume of water required to dissolve 1 gm of calcium sulphate at 298 K? [Ksp of CaSO4 = 9.1x10^-6]
Solubility product ($K_{sp}$): for a sparingly soluble salt in its saturated solution, the product of the molar concentrations of its ions, each raised to its stoichiometric coefficient. For
$$ \ce{CaSO4 <=> Ca^2+ + SO4^2-} $$
, $K_{sp}=[\ce{Ca^2+}][\ce{SO4^2-}]$.
Condition for precipitation: a precipitate forms only when the ionic product exceeds the solubility product, i.e. $Q > K_{sp}$ (if $Q = K_{sp}$ the solution is just saturated; if $Q < K_{sp}$ no precipitate).
Application: by adding a common ion (common ion effect) the ionic product is raised above $K_{sp}$, so the salt precipitates; this controls selective precipitation in qualitative analysis (e.g. passing $\ce{H2S}$ with dilute $\ce{HCl}$ precipitates only sulphides of very low $K_{sp}$).
Calculation: let solubility $= s$ mol/L. Then $K_{sp}=s^2$:
$$ \begin{aligned} s &= \sqrt{9.1\times10^{-6}} \ &= 3.02\times10^{-3}\ \text{mol/L} \end{aligned} $$
Molar mass of $\ce{CaSO4}=136$, so solubility
$$ \begin{aligned} &= 3.02\times10^{-3}\times136 \ &= 0.41\ \text{g/L} \end{aligned} $$
Minimum volume of water to dissolve 1 g:
$$ \begin{aligned} V &= \frac{1}{0.41} \ &= 2.44\ \text{L} \end{aligned} $$
Write short notes on any two: i) Extraction of blister copper from copper pyrites ii) Rusting of iron iii) Hess's law of constant heat summation and its applications iv) Laboratory preparation of formic acid.
ii) Rusting of iron: this is an electrochemical corrosion that requires both water and oxygen. Iron acts as the anode, where it is oxidised:
$$\ce{Fe -> Fe^2+ + 2e-}$$
At the cathode, oxygen is reduced:
$$\ce{O2 + 2H2O + 4e- -> 4OH-}$$
The $\ce{Fe^2+}$ is then oxidised further to hydrated ferric oxide:
$$ \begin{aligned} \ce{4Fe(OH)2 + O2 + 2H2O -> 4Fe(OH)3} \ \ce{4Fe(OH)3 -> 2Fe2O3.3H2O} \end{aligned} $$
Rust ($\ce{Fe2O3.xH2O}$) is a brown flaky solid. It is prevented by painting, galvanising or cathodic protection.
iv) Laboratory preparation of formic acid (methanoic acid): heat oxalic acid with glycerol at about 100 to 110 degrees, where the glycerol acts through glyceryl monoxalate:
$$\ce{(COOH)2 ->[glycerol][\Delta] HCOOH + CO2(g)}$$
The methanoic acid that distils over is collected. On stronger heating with glycerol it can decompose further, so the temperature is carefully controlled.
In the nitrogen atom undergoes sp3 hybridization, giving four sp3 orbitals directed towards the corners of a tetrahedron. Three of these orbitals form bonds while the fourth holds a lone pair. Because one corner is ...
Crystalline oxalic acid () is a good primary standard because it: - is available in a pure, definite crystalline form with an exactly known formula and equivalent weight, - is non-hygroscopic and stable in air (does not...
i) Degree of ionisation (): the fraction of the total number of molecules of an electrolyte that dissociate into ions in solution.
ii) Ostwald's dilution law: for a weak binary electrolyte, the dissociation constant is where is the concentration and the degree of ionisation. For very weak electrolytes (), , so : the degree of ionisation increases on dilution.
For the given reaction, is oxidised (anode) and is reduced (cathode). $$ \begin{aligned} E^\circ{cell} &= E^\circ{cathode} - E^\circ{anode} \ &= E^\circ{Pb^{2+}/Pb} - E^\circ{Ag^+/Ag} \ &= (-0.13) - (0.80) \ &= ...
Enthalpy of reaction: the heat change (at constant pressure) when the number of moles of reactants shown in the balanced equation react completely to form products.
For
, two moles of are formed, so
The reaction is exothermic, kJ per mole of reaction (per 2 mol ).
When and are both negative, ; the term becomes positive. i) Spontaneous: at low temperature, the negative outweighs the positive term, so
First reduce phenol to benzene with zinc dust, then hydrogenate benzene to cyclohexane: The product is cyclohexane. (Al...
A: propene adds HBr by Markovnikov's rule (Br on the more substituted carbon): A , IUPAC name 2-bromopropane.
B: Wurtz coupling of two molecules of A with sodium: B , IUPAC name 2,3-dimethylbutane.
(i) From benzene diazonium chloride: warm its aqueous solution (hydrolysis): (ii) From chlorobenzene (Dow process): heat with at high temperature and pressure, then acidi...
Isopropyl alcohol is (). Its isomeric ether is methoxyethane, (same formula, different functional group). With excess HI (heated) the ether is cleaved completely; each part becomes an iodid...
i) Cannizzaro's reaction (aldehyde without alpha-H, conc. alkali, self redox): ii) Aldol condensation (aldehyde with alpha-H, dilute base, giving a beta-hydroxy aldehyde): $$\ce{2CH3CHO -[dil. NaOH]...
In chloroacetic acid, , the chlorine atom is strongly electron withdrawing (negative inductive effect, ). It pulls electron density away from the group, which: - makes it easier to release the , ...
(i) Electrolytic reduction (strongly acidic medium, at cathode): nitrobenzene is reduced via phenylhydroxylamine, which rearranges to p-aminophenol: (ii) With Zn/NaOH (alkaline reduction):...
Ethanamine () is a primary amine and N-methylmethanamine () is a secondary amine, so use the carbylamine (isocyanide) test: Warm each with chloroform and alcoholic . $$\ce{C2H5NH2 + CHCl3 + 3KOH -...
Homopolymer: a polymer made by the repeated joining of only one kind of monomer, e.g. polythene (from ethene, ) or PVC (from vinyl chloride). Copolymer: a polymer made from two or more different monomers, e.g. Buna-S (from ...
i) Aspirin (acetylsalicylic acid): a benzene ring bearing and, ortho to it, ; condensed as . Use: analgesic and antipyretic (relieves pain and fever); also an anti-blood-clotting agent. i...
A peptide bond () forms when the carboxyl group of one amino acid condenses with the amino group of another, eliminating a water molecule. Example (two glycine molecules): $$\ce{H2N-CH2-COOH + H2N-CH2-COOH - H2N-CH2-CO-NH-C...
Non-reducing sugars are carbohydrates that cannot reduce Tollen's reagent or Fehling's solution, because their reducing groups (the anomeric /free carbonyl of both monosaccharide units) are locked in the glycosidic linkage.
Example: sucrose, (in which the reducing groups of glucose and fructose are joined together, so none is free).
Blue vitriol is . On heating it loses water of crystallisation in steps and finally, on strong heating, the anhydrous salt decomposes to black copper oxide: $$ \begin{aligned} \ce{CuSO4.5H2O -[100^{\circ}C] CuSO4.H2O + 4...
Zinc blende () is concentrated by froth flotation, then: 1. Roasting to the oxide: 2. Reduction of the oxide with coke (carbon) in a fire-clay retort at about 1400 degrees; zinc distils over...
Faraday's first law: the mass of a substance deposited/liberated at an electrode is directly proportional to the quantity of electricity passed, . Faraday's second law: when the same quantity of electricity is passed through...
Redox titration: a titration based on an oxidation-reduction reaction between the titrant and the analyte (e.g. vs ), the end point being detected by a change in oxidation state (often self-indicating).
Calculation:
N; in 200 cc, eq
eq of acid.
Base (0.25 eq) exceeds acid (0.15 eq), so the solution is basic.
Excess
eq (mol) left after neutralization.
Diluted to 2 L, the resulting concentration of excess alkali:
Calomel is mercurous chloride, , a white insoluble powder.
Preparation:
Action of ammonia (test): it turns black due to disproportionation into finely divided mercury (black) and a white mercuric amido chloride: The black colour of free mercury confirms a mercurous salt.
Uses: in the calomel reference electrode and, formerly, as a purgative in medicine.
Rate law: the experimentally determined equation relating reaction rate to the molar concentrations of the reactants raised to appropriate powers. Here, since the reaction is first order in P and zero order in Q:
Find k from Expt I:
Fill the blanks:
M.
M min (Q has no effect).
M.
Laboratory preparation: ethanol is heated with excess concentrated at 140 degrees; ethyl hydrogen sulphate forms first, then reacts with more ethanol (continuous etherification): $$ \begin{aligned} \ce{C2H5OH + H2SO4 - C2H5H...
Ethanal from 1,1-dibromoethane by hydrolysis of the gem-dihalide with aqueous alkali: Ethanal from ethyne by hydration (Kucherov reaction): Et...
C on ozonolysis gives ethanal + methanal, so C = propene, (since ). - C comes from B by dehydrohalogenation, so B = 2-chloropropane, . - B comes from the alcohol...
Preparation of aniline: reduce nitrobenzene with tin and conc. HCl, liberate with and purify by steam distillation: The aniline is dried over and distilled. Reaction...
Victor Meyer's method: convert each alcohol to the alkyl iodide (red P + ), then to the nitroalkane with , and treat with followed by . Propan-2-ol (secondary) forms a secondary nitro compound that gives a blue colour with , while 2-methylpropan-2-ol (tertiary) has no alpha-hydrogen on the nitro carbon and so gives no colour (it stays colourless).
(i) Chloroform into dimethylamine. First reduce chloroform to methyl chloride:
then treat methyl chloride with ammonia, from which the secondary amine is separated:
(ii) Ethanamine into methanamine goes one carbon shorter. First convert the amine to ethanol with nitrous acid:
oxidise the ethanol to ethanoic acid:
convert the acid to its amide:
and finally apply Hofmann bromamide degradation to lose one carbon and give methanamine:
Solubility product (): for a sparingly soluble salt in its saturated solution, the product of the molar concentrations of its ions, each raised to its stoichiometric coefficient. For
, .
Condition for precipitation: a precipitate forms only when the ionic product exceeds the solubility product, i.e. (if the solution is just saturated; if no precipitate).
Application: by adding a common ion (common ion effect) the ionic product is raised above , so the salt precipitates; this controls selective precipitation in qualitative analysis (e.g. passing with dilute precipitates only sulphides of very low ).
Calculation: let solubility mol/L. Then :
Molar mass of , so solubility
Minimum volume of water to dissolve 1 g:
ii) Rusting of iron: this is an electrochemical corrosion that requires both water and oxygen. Iron acts as the anode, where it is oxidised:
At the cathode, oxygen is reduced:
The is then oxidised further to hydrated ferric oxide:
Rust () is a brown flaky solid. It is prevented by painting, galvanising or cathodic protection.
iv) Laboratory preparation of formic acid (methanoic acid): heat oxalic acid with glycerol at about 100 to 110 degrees, where the glycerol acts through glyceryl monoxalate:
The methanoic acid that distils over is collected. On stronger heating with glycerol it can decompose further, so the temperature is carefully controlled.