NEB Class 12 ยท Exam intelligence
From 6 NEB Class 12 past papers: the chapters that keep coming back and their most important questions, each with a solved model answer. No guarantees; study the whole syllabus.
From the most-tested chapters first, each with a solved model answer.
Describe the laboratory method of preparation of pure and dry nitrobenzene. Identify A, B, C and D in the following reaction sequence: A --(NaOH+CaO)--> B --(CH3Cl/heat)--> C --(CeO2/H+)--> D. The compound B can be prepared by heating phenol with zinc-dust.
Laboratory preparation of pure and dry nitrobenzene.
Nitrobenzene is prepared by nitrating benzene with a mixture of concentrated nitric acid and concentrated sulphuric acid, called the nitrating mixture. The sulphuric acid is there to generate the attacking electrophile, the nitronium ion:
$$ \begin{aligned} \ce{HNO3 + 2H2SO4 -> NO2+ + H3O+ + 2HSO4-} \ \ce{C6H6 + HNO3 ->[conc. H2SO4][328-333 K] C6H5NO2 + H2O} \end{aligned} $$
Procedure. Concentrated sulphuric acid is added slowly to concentrated nitric acid in a round bottomed flask, with cooling, and benzene is then added in small portions from a dropping funnel, the flask being shaken constantly. The temperature is held between $328\ \text{K}$ and $333\ \text{K}$ ($55$ to $60^\circ\text{C}$) using a water bath, and the flask is fitted with a reflux condenser and a thermometer. The temperature must not be allowed to rise above about $333\ \text{K}$, or a second nitro group enters and m-dinitrobenzene is formed. The mixture is refluxed for about half an hour and then poured into a large volume of cold water, when nitrobenzene separates as a heavy pale yellow oil.
Purification. The crude oil is washed in a separating funnel, first with water, then with dilute sodium carbonate solution to remove the acids adhering to it, and again with water. It is then dried over anhydrous calcium chloride, and finally distilled, the fraction boiling at $484\ \text{K}$ ($211^\circ\text{C}$) being collected as pure nitrobenzene. It is a pale yellow oily liquid with the smell of bitter almonds, and it is poisonous.
The reaction sequence.
The clue settles the middle of the chain: heating phenol with zinc dust reduces it, removing the oxygen, so
$$\ce{C6H5OH + Zn ->[\Delta] C6H6 + ZnO}$$
and therefore $\ce{B}$ is benzene, $\ce{C6H6}$.
$\ce{B}$ is obtained from $\ce{A}$ by heating with soda lime, which is the decarboxylation of the salt of an aromatic acid, so $\ce{A}$ is sodium benzoate, $\ce{C6H5COONa}$:
$$\ce{C6H5COONa + NaOH ->[CaO][\Delta] C6H6 + Na2CO3}$$
Benzene with methyl chloride on warming undergoes Friedel Crafts alkylation (anhydrous aluminium chloride being the catalyst), so $\ce{C}$ is toluene, $\ce{C6H5CH3}$:
$$\ce{C6H6 + CH3Cl ->[anhyd. AlCl3][\Delta] C6H5CH3 + HCl}$$
Toluene treated with the cerium(IV) oxidant in acid undergoes oxidation of the side chain only, which stops at the aldehyde, so $\ce{D}$ is benzaldehyde, $\ce{C6H5CHO}$:
$$\ce{C6H5CH3 ->[CeO2/H+] C6H5CHO}$$
(This is the same conversion that the Etard reaction brings about with chromyl chloride, $\ce{CrO2Cl2}$, followed by hydrolysis. A stronger oxidising agent such as acidified $\ce{KMnO4}$ would have carried the side chain on to benzoic acid.)
In summary: $\ce{A}$ = sodium benzoate, $\ce{B}$ = benzene, $\ce{C}$ = toluene, $\ce{D}$ = benzaldehyde.
a) How is propanone prepared from i) 2,2-dichloropropane ii) isopropyl alcohol iii) propyne? Give suitable reactions for the conversion of ethanoic acid into i) methane ii) methyl ethanoate.
b) Write the structural formula of primary, secondary and tertiary amines from C3H9N. How would you apply Hoffmann's method to separate them?
a) Propanone from: i) 2,2-dichloropropane (hydrolysis of gem-dihalide):
$$ \ce{CH3CCl2CH3 + 2NaOH(aq) -> CH3COCH3 + 2NaCl + H2O} $$
ii) isopropyl alcohol (oxidation):
$$ \ce{CH3CH(OH)CH3 ->[K2Cr2O7/H2SO4] CH3COCH3} $$
iii) propyne (Markovnikov hydration):
$$ \ce{CH3C#CH + H2O ->[HgSO4/H2SO4] CH3COCH3} $$
Ethanoic acid conversions: i) to methane (decarboxylation with soda-lime):
$$ \ce{CH3COONa + NaOH ->[CaO][\Delta] CH4 + Na2CO3} $$
ii) to methyl ethanoate (esterification):
$$ \ce{CH3COOH + CH3OH <=>[conc. H2SO4] CH3COOCH3 + H2O} $$
b) Amines of $\ce{C3H9N}$:
Hoffmann's separation: treat the mixture with diethyl oxalate; the primary amine gives a solid oxamide, the secondary gives a liquid oxamic ester, the tertiary does not react. Distil off the tertiary amine first, then hydrolyse the residue with $\ce{NaOH}$ to regenerate the primary amine (from the solid) and the secondary amine (from the liquid), and separate them by distillation.
State Ostwald's dilution law. What is the limitation of this law? Define the terms i) ionic product of water ii) common ion effect iii) degree of ionisation iv) pH value. What will be the resultant pH when 200 ml of aqueous HCl (pH = 2) is mixed with 300 ml of an aqueous solution of NaOH (pH = 12)?
Ostwald's dilution law: for a weak electrolyte, $K = \dfrac{\alpha^2 C}{1-\alpha}$ (for very weak electrolytes $\alpha=\sqrt{K/C}$), so the degree of ionisation increases on dilution. Limitation: it holds only for weak electrolytes; it f...
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:
$$ \begin{aligned} V &= \frac{1}{0.41} \ &= 2.44\ \text{L} \end{aligned} $$
What is meant by i) common ion effect ii) solubility product constant (Ksp)? Explain the common ion effect and the solubility product principle in qualitative salt analysis. What will be the resulting pH of a solution prepared by mixing 200 ml of aqueous HCl (pH = 2) with 300 ml of aqueous NaOH (pH = 12)?
i) Common ion effect: the suppression of the ionization of a weak electrolyte when a strong electrolyte having a common ion is added (e.g. adding $\ce{NH4Cl}$ to $\ce{NH4OH}$ lowers $[\ce{OH-}]$).
ii) Solubility product ($K_{sp}$): for a sparingly soluble salt, the product of the molar concentrations of its ions in a saturated solution, each raised to its stoichiometric power.
In qualitative analysis: the common ion effect and $K_{sp}$ control group precipitation. For group II, $\ce{H2S}$ is passed in the presence of dilute $\ce{HCl}$; the common $\ce{H+}$ suppresses ionization of $\ce{H2S}$, keeping $[\ce{S^2-}]$ low so that only the very insoluble sulphides (small $K_{sp}$) precipitate. For group III, $\ce{NH4Cl}$ (common $\ce{NH4+}$) lowers $[\ce{OH-}]$ from $\ce{NH4OH}$ so only hydroxides with the smallest $K_{sp}$ (Fe, Al, Cr) precipitate.
pH calculation:
$\ce{H+}$ from HCl: $[\ce{H+}]=10^{-2}$ M, moles
$$ \begin{aligned} &= 10^{-2}\times0.200 \ &= 2\times10^{-3} \end{aligned} $$
$\ce{OH-}$ from NaOH: $[\ce{OH-}]=10^{-2}$ M (pH 12 means pOH 2), moles
$$ \begin{aligned} &= 10^{-2}\times0.300 \ &= 3\times10^{-3} \end{aligned} $$
Excess
$$ \begin{aligned} \ce{OH-} &= 3\times10^{-3}-2\times10^{-3} \ &= 1\times10^{-3} \end{aligned} $$
mol in total 500 mL:
$$ \begin{aligned} [\ce{OH-}] &= \frac{10^{-3}}{0.5} \ &= 2\times10^{-3}\ \text{M},\quad pOH \ &= 2.70,\quad pH \ &= 14-2.70 \ &= 11.30 \end{aligned} $$
a) Explain the Cannizzaro reaction and Perkin condensation. What happens when propanone is treated with 2,4-dinitrophenylhydrazine?
b) In the sequence A -(NH3)-> B -(Br2/aq.KOH)-> C -(HNO2)-> D -(NaOH/I2)-> E, compound E produces ethyne when heated with silver powder. Identify A, B, C, D, E.
a) Cannizzaro reaction: an aldehyde with no alpha-hydrogen, with conc. alkali, undergoes self oxidation-reduction: $$\ce{2HCHO + NaOH - CH3OH + HCOONa}$$ Perkin condensation: an aromatic aldehyde condenses with an acid anhydride in the p...
Define molality of a solution. Calculate the molality of one litre of 93% H2SO4 solution (weight by volume). The density of the solution is 1.84 g/mL.
Molality (m): the number of moles of solute dissolved per kilogram of solvent. $$m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}}$$ Calculation (basis: 1 litre = 1000 mL of solution): - 93% w/v means 930 g $\ce{H2SO4}$ in 1...
What is meant by i) ionic product of water ii) solubility product constant (Ksp)? The solubility product of BaSO4 is 1x10^-10. Will a precipitate form if equal volumes of 2x10^-3 M BaCl2 and 2x10^-4 M Na2SO4 solutions are mixed?
(i) Ionic product of water ($Kw$): the product of the molar concentrations of $\ce{H+}$ and $\ce{OH-}$ ions in water, $Kw=[\ce{H+}][\ce{OH-}] = 10^{-14}$ at 25 degrees. (ii) Solubility product ($K{sp}$): for a sparingly soluble salt, the...
What is a standard solution? What volume of deci-normal HCl is required to neutralize 25 ml of NaOH containing 8 g NaOH in one litre of solution?
Standard solution: a solution whose concentration (normality/molarity) is exactly and accurately known.
Calculation: normality of the $\ce{NaOH}$ solution (equivalent weight of $\ce{NaOH}=40$):
$$ \begin{aligned} N_{NaOH} &= \frac{8}{40\times 1} \ &= 0.2\ \text{N} \end{aligned} $$
Deci-normal $\ce{HCl}$ is $0.1$ N. Using $N_1V_1 = N_2V_2$ for neutralization:
$$ \begin{aligned} 0.1 \times V_{HCl} &= 0.2 \times 25 ; \ &\Rightarrow ; V_{HCl} = \frac{0.2\times 25}{0.1} \ &= 50\ \text{mL} \end{aligned} $$
Distinguish between end point and equivalence point of a reaction. What volume of water should be evaporated from 2 liters semi-normal solution of Na2CO3 to make it exactly 2N ?
End point and equivalence point.
The equivalence point is the stage in a titration at which the two reactants have been mixed in exactly equivalent amounts, that is when the number of gram equivalents of the acid added equals the number of gram equivalents of the base present. It is a theoretical, exact point fixed by the stoichiometry of the reaction, and it does not depend on the indicator.
The end point is the stage at which the indicator changes colour and the titration is stopped. It is what the experimenter actually observes.
The two are not identical: the end point is the experimental estimate of the equivalence point, and it coincides with it only if the indicator has been properly chosen, so that its range of colour change falls within the sharp change of pH at the equivalence point. A badly chosen indicator gives an end point measurably before or after the equivalence point, and that difference is the indicator error.
Numerical. A semi-normal solution is one of normality $0.5\ \text{N}$, so
$$ \begin{aligned} N_1 &= 0.5\ \text{N} \ \qquad V_1 &= 2\ \text{L} \ &= 2000\ \text{mL} \ \qquad N_2 &= 2\ \text{N} \end{aligned} $$
Evaporating water removes solvent only, so the amount of solute is unchanged and the number of milliequivalents before and after is the same:
$$ \begin{aligned} N_1V_1 &= N_2V_2 \ 0.5\times2000 &= 2\times V_2 \ V_2 &= \frac{1000}{2} \ &= 500\ \text{mL} \end{aligned} $$
The solution must therefore be concentrated from $2000\ \text{mL}$ down to $500\ \text{mL}$, so the volume of water to be evaporated is
$$2000-500=\mathbf{1500\ mL=1.5\ L}$$
What is meant by : i) common ion effect. ii) ionic product of water. The solubility product of BaSO4 is 2x10^-10. Will precipitate occur or not if equal volume of 2x10^-8 M BaCl2 and 2x10^-3 M Na2SO4 are mixed ?
(i) Common ion effect. The common ion effect is the suppression of the ionisation of a weak electrolyte by adding to its solution a strong electrolyte that has an ion in common with it. By Le Chatelier's principle the added ion shifts the ionisation equilibrium backwards, so the degree of ionisation of the weak electrolyte falls. Adding ammonium chloride to ammonium hydroxide, for instance, supplies $\ce{NH4+}$ and pushes the equilibrium
$$\ce{NH4OH <=> NH4+ + OH-}$$
to the left, lowering $[\ce{OH-}]$. The effect is used in qualitative analysis to control the concentration of the precipitating ion.
(ii) Ionic product of water. Water is a very weak electrolyte and ionises slightly as
$$ \ce{H2O <=> H+ + OH-} $$
. The product of the molar concentrations of the hydrogen and hydroxyl ions in water, or in any aqueous solution, at a given temperature is a constant called the ionic product of water:
$$ \begin{aligned} K_w &= [\ce{H+}][\ce{OH-}] \ &= 1\times10^{-14}\ \text{mol}^2\text{L}^{-2}\ \text{at }25^\circ\text{C} \end{aligned} $$
It increases with temperature, and in pure water $[\ce{H+}]=[\ce{OH-}]=10^{-7}\ \text{M}$, giving pH $=7$.
Numerical. Equal volumes are mixed, so the total volume is doubled and each concentration is halved:
$$ \begin{aligned} [\ce{Ba^{2+}}] &= \frac{2\times10^{-8}}{2} \ &= 1\times10^{-8}\ \text{M} \ [\ce{SO4^{2-}}] &= \frac{2\times10^{-3}}{2} \ &= 1\times10^{-3}\ \text{M} \end{aligned} $$
The ionic product for the sparingly soluble salt is then
$$ \begin{aligned} [\ce{Ba^{2+}}][\ce{SO4^{2-}}] &= (1\times10^{-8})(1\times10^{-3}) \ &= 1\times10^{-11} \end{aligned} $$
Comparing this with the solubility product,
$$1\times10^{-11}<2\times10^{-10}$$
The ionic product is less than the solubility product, so the solution is unsaturated with respect to barium sulphate and no precipitate will occur. A precipitate appears only when the ionic product exceeds $K_{sp}$.
It is better to express concentration in molality rather than molarity, why? X gm of a metal (equivalent weight = 12) was completely dissolved in 100 cc of N/2 HCl. The volume was made up to 500 cc. It is found that 25 cc of the diluted acid solution required 17.5 cc of N/10 NaOH for complete neutralization. Find the value of X.
Why molality: molality is based on the mass of solvent, which does not change with temperature, whereas molarity uses volume, which changes as temperature (and hence density) changes. So molality gives a temperature-independent concentra...
State enthalpy of combustion. If the heats of formation of CO2, H2O and C6H12O6 are -395, -269.4 and -1169 kJ/mol respectively, calculate the heat of combustion of glucose.
Enthalpy of combustion: the enthalpy change when one mole of a substance is completely burnt in excess oxygen. For glucose: $$ \ce{C6H12O6 + 6O2 - 6CO2 + 6H2O} $$ . $$ \begin{aligned} \Delta Hc &= [6,\Delta Hf(\ce{CO2}) + 6,\Delta Hf(...
Define redox titration. 10 gm of NaOH was added to 200 cc of N/2 (f = 1.5) H2SO4. The volume was diluted to two litres. Predict whether the dilute solution is acidic, basic or neutral and also calculate the resulting molarity of the dilute solution.
Redox titration: a titration based on an oxidation-reduction reaction between the titrant and the analyte (e.g. $\ce{KMnO4}$ vs $\ce{FeSO4}$), the end point being detected by a change in oxidation state (often self-indicating).
Calculation:
$$ \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} $$
Give chemical reactions for the preparation of ethanal from i) 1,1-dibromoethane ii) ethyne iii) ethanoyl chloride. How is ethanal converted into propan-2-ol?
Ethanal from 1,1-dibromoethane by hydrolysis of the gem-dihalide with aqueous alkali: $$\ce{CH3CHBr2 + 2NaOH(aq) - CH3CHO + 2NaBr + H2O}$$ Ethanal from ethyne by hydration (Kucherov reaction): $$\ce{C2H2 + H2O -[HgSO4/H2SO4] CH3CHO}$$ Et...
Define heat of formation. The heats of combustion of methane, carbon and hydrogen are -210, -94 and -68 kcal respectively. Calculate the heat of formation of methane.
Heat of formation: the enthalpy change when one mole of a compound is formed from its elements in their standard states.
For
$$ \ce{C(s) + 2H2(g) -> CH4(g)} $$
, by Hess's law:
$$ \begin{aligned} \Delta H_f(\ce{CH4}) &= \Delta H_c(\ce{C}) + 2,\Delta H_c(\ce{H2}) - \Delta H_c(\ce{CH4}) \ &= (-94) + 2(-68) - (-210) \ &= -94 - 136 + 210 \ &= -20\ \text{kcal/mol} \end{aligned} $$
The heat of formation of methane is $-20$ kcal/mol (exothermic).
What is meant by normality factor? How many ml of conc. HNO3 of specific gravity 1.41 containing 69% by mass are required to prepare 500 ml of 0.5N HNO3?
Normality factor (f): the ratio of the actual normality of a solution to its intended (labelled) normality; it corrects for the difference between the calculated and true strength.
Calculation: equivalents of $\ce{HNO3}$ needed
$$ \begin{aligned} &= 0.5\times0.500 \ &= 0.25 \end{aligned} $$
eq. Since $\ce{HNO3}$ is monobasic, eq. wt. $=63$:
$$ \begin{aligned} \text{mass of HNO3 needed} &= 0.25\times63 \ &= 15.75\ \text{g} \end{aligned} $$
Mass of pure $\ce{HNO3}$ per mL of conc. acid
$$ \begin{aligned} &= 1.41\times0.69 \ &= 0.973\ \text{g/mL} \ V &= \frac{15.75}{0.973} \ &= 16.2\ \text{mL} \end{aligned} $$
About 16.2 mL of the concentrated acid is required (diluted to 500 mL).
Define normality. 0.8 gm of a divalent metal was dissolved in 100 ml of 1.28N HCl and the solution was diluted to 200 ml. Then 50 ml of the solution required 54.6 ml of 0.22N NaOH for neutralization. Find the atomic weight of the metal.
Normality: the number of gram-equivalents of solute dissolved per litre of solution.
Calculation:
Total $\ce{HCl}$ taken
$$ \begin{aligned} &= 1.28 \times 0.100 \ &= 0.128 \end{aligned} $$
eq (in 200 mL after dilution).
The 50 mL portion is one quarter of 200 mL, and its unreacted $\ce{HCl}$ is neutralized by NaOH:
$$ \begin{aligned} \text{NaOH} &= 0.22 \times \frac{54.6}{1000} \ &= 0.012012\ \text{eq} \ &= \text{excess HCl in 50 mL} \end{aligned} $$
Excess $\ce{HCl}$ in whole 200 mL
$$ \begin{aligned} &= 0.012012 \times 4 \ &= 0.048048\ \text{eq} \end{aligned} $$
$\ce{HCl}$ that reacted with the metal
$$ \begin{aligned} &= 0.128 - 0.048 \ &= 0.0800 \end{aligned} $$
eq $=$ equivalents of metal.
$$ \begin{aligned} \text{Equivalent weight} &= \frac{0.8}{0.0800} \ &= 10 \ &\Rightarrow \text{Atomic weight} = 10 \times 2 \ &= \textbf{20} \end{aligned} $$
What is meant by degree of ionization? 0.41 g of NaOH is placed in 100 ml of 0.1N H2SO4. Find the pH of the resulting solution.
Degree of ionization: the fraction of the total number of molecules of an electrolyte that actually dissociate into ions in solution ($\alpha = $ ionized fraction). Calculation: Moles of $$ \begin{aligned} \ce{NaOH} &= \dfrac{0.41}{40} ...
What is titration error? How is it minimized?
Titration error: the small difference between the observed end point (colour change of the indicator) and the true equivalence point of the reaction. It makes the measured volume slightly more or less than the exact volume needed. Minimi...
Calculate the pH of 1x10^-8 M HCl.
At this very low concentration the $\ce{H+}$ from water cannot be ignored. Solution: let total $[\ce{H+}] = x$. Charge/mass balance with $Kw=10^{-14}$ gives: $$ \begin{aligned} x &= [\ce{H+}]{acid} + [\ce{OH-}] \ &= 10^{-8} + \frac{10^{...
Define spontaneous process and give one example of it.
Spontaneous process: a physical or chemical change that takes place on its own under a given set of conditions without any continuous outside help. It proceeds in a definite direction accompanied by a decrease in Gibbs free energy ($\Delta G < 0$).
Example: flow of heat from a hot body to a cold body, or the rusting of iron: $$\ce{4Fe(s) + 3O2(g) -> 2Fe2O3(s)}$$
Calculate the enthalpy of formation in the following reactions: i) 2H2(g) + O2(g) -> 2H2O(l), H = -136 kcal ii) H2(g) + I2(g) -> 2HI(g), H = 24.8 kcal.
Enthalpy of formation is per one mole of the compound formed. (i) $$ \ce{2H2(g) + O2(g) - 2H2O(l)} $$ , $\Delta H = -136$ kcal for 2 mol water. $$ \begin{aligned} \Delta Hf(\ce{H2O}) &= \frac{-136}{2} \ &= -68\ \text{kcal/mol} \end{alig...
How would you convert sodium benzoate into acetophenone?
Heat sodium benzoate with soda-lime to get benzene, then carry out Friedel-Crafts acylation. $$ \begin{aligned} \ce{C6H5COONa + NaOH -[CaO][\Delta] C6H6 + Na2CO3} \ \ce{C6H6 + CH3COCl -[anhyd. AlCl3] C6H5COCH3 + HCl} \end{aligned} $$ Th...
Identify the major product X and mention its one important use (Methanal treated with NH3 gives X).
Methanal (formaldehyde) reacts with ammonia to give hexamethylenetetramine (urotropine), X = $\ce{(CH2)6N4}$: $$\ce{6HCHO + 4NH3 - (CH2)6N4 + 6H2O}$$ Use: it is used as a urinary antiseptic (urotropine) and as a hardener (with the explos...
Study every chapter with notes and solved questions
Open Chemistry notes and questionsLaboratory preparation of pure and dry nitrobenzene.
Nitrobenzene is prepared by nitrating benzene with a mixture of concentrated nitric acid and concentrated sulphuric acid, called the nitrating mixture. The sulphuric acid is there to generate the attacking electrophile, the nitronium ion:
Procedure. Concentrated sulphuric acid is added slowly to concentrated nitric acid in a round bottomed flask, with cooling, and benzene is then added in small portions from a dropping funnel, the flask being shaken constantly. The temperature is held between and ( to ) using a water bath, and the flask is fitted with a reflux condenser and a thermometer. The temperature must not be allowed to rise above about , or a second nitro group enters and m-dinitrobenzene is formed. The mixture is refluxed for about half an hour and then poured into a large volume of cold water, when nitrobenzene separates as a heavy pale yellow oil.
Purification. The crude oil is washed in a separating funnel, first with water, then with dilute sodium carbonate solution to remove the acids adhering to it, and again with water. It is then dried over anhydrous calcium chloride, and finally distilled, the fraction boiling at () being collected as pure nitrobenzene. It is a pale yellow oily liquid with the smell of bitter almonds, and it is poisonous.
The reaction sequence.
The clue settles the middle of the chain: heating phenol with zinc dust reduces it, removing the oxygen, so
and therefore is benzene, .
is obtained from by heating with soda lime, which is the decarboxylation of the salt of an aromatic acid, so is sodium benzoate, :
Benzene with methyl chloride on warming undergoes Friedel Crafts alkylation (anhydrous aluminium chloride being the catalyst), so is toluene, :
Toluene treated with the cerium(IV) oxidant in acid undergoes oxidation of the side chain only, which stops at the aldehyde, so is benzaldehyde, :
(This is the same conversion that the Etard reaction brings about with chromyl chloride, , followed by hydrolysis. A stronger oxidising agent such as acidified would have carried the side chain on to benzoic acid.)
In summary: = sodium benzoate, = benzene, = toluene, = benzaldehyde.
a) Propanone from: i) 2,2-dichloropropane (hydrolysis of gem-dihalide):
ii) isopropyl alcohol (oxidation):
iii) propyne (Markovnikov hydration):
Ethanoic acid conversions: i) to methane (decarboxylation with soda-lime):
ii) to methyl ethanoate (esterification):
b) Amines of :
Hoffmann's separation: treat the mixture with diethyl oxalate; the primary amine gives a solid oxamide, the secondary gives a liquid oxamic ester, the tertiary does not react. Distil off the tertiary amine first, then hydrolyse the residue with to regenerate the primary amine (from the solid) and the secondary amine (from the liquid), and separate them by distillation.
Ostwald's dilution law: for a weak electrolyte, (for very weak electrolytes ), so the degree of ionisation increases on dilution. Limitation: it holds only for weak electrolytes; it f...
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:
i) Common ion effect: the suppression of the ionization of a weak electrolyte when a strong electrolyte having a common ion is added (e.g. adding to lowers ).
ii) Solubility product (): for a sparingly soluble salt, the product of the molar concentrations of its ions in a saturated solution, each raised to its stoichiometric power.
In qualitative analysis: the common ion effect and control group precipitation. For group II, is passed in the presence of dilute ; the common suppresses ionization of , keeping low so that only the very insoluble sulphides (small ) precipitate. For group III, (common ) lowers from so only hydroxides with the smallest (Fe, Al, Cr) precipitate.
pH calculation:
from HCl: M, moles
from NaOH: M (pH 12 means pOH 2), moles
Excess
mol in total 500 mL:
a) Cannizzaro reaction: an aldehyde with no alpha-hydrogen, with conc. alkali, undergoes self oxidation-reduction: Perkin condensation: an aromatic aldehyde condenses with an acid anhydride in the p...
Molality (m): the number of moles of solute dissolved per kilogram of solvent. Calculation (basis: 1 litre = 1000 mL of solution): - 93% w/v means 930 g in 1...
(i) Ionic product of water (): the product of the molar concentrations of and ions in water, at 25 degrees. (ii) Solubility product (): for a sparingly soluble salt, the...
Standard solution: a solution whose concentration (normality/molarity) is exactly and accurately known.
Calculation: normality of the solution (equivalent weight of ):
Deci-normal is N. Using for neutralization:
End point and equivalence point.
The equivalence point is the stage in a titration at which the two reactants have been mixed in exactly equivalent amounts, that is when the number of gram equivalents of the acid added equals the number of gram equivalents of the base present. It is a theoretical, exact point fixed by the stoichiometry of the reaction, and it does not depend on the indicator.
The end point is the stage at which the indicator changes colour and the titration is stopped. It is what the experimenter actually observes.
The two are not identical: the end point is the experimental estimate of the equivalence point, and it coincides with it only if the indicator has been properly chosen, so that its range of colour change falls within the sharp change of pH at the equivalence point. A badly chosen indicator gives an end point measurably before or after the equivalence point, and that difference is the indicator error.
Numerical. A semi-normal solution is one of normality , so
Evaporating water removes solvent only, so the amount of solute is unchanged and the number of milliequivalents before and after is the same:
The solution must therefore be concentrated from down to , so the volume of water to be evaporated is
(i) Common ion effect. The common ion effect is the suppression of the ionisation of a weak electrolyte by adding to its solution a strong electrolyte that has an ion in common with it. By Le Chatelier's principle the added ion shifts the ionisation equilibrium backwards, so the degree of ionisation of the weak electrolyte falls. Adding ammonium chloride to ammonium hydroxide, for instance, supplies and pushes the equilibrium
to the left, lowering . The effect is used in qualitative analysis to control the concentration of the precipitating ion.
(ii) Ionic product of water. Water is a very weak electrolyte and ionises slightly as
. The product of the molar concentrations of the hydrogen and hydroxyl ions in water, or in any aqueous solution, at a given temperature is a constant called the ionic product of water:
It increases with temperature, and in pure water , giving pH .
Numerical. Equal volumes are mixed, so the total volume is doubled and each concentration is halved:
The ionic product for the sparingly soluble salt is then
Comparing this with the solubility product,
The ionic product is less than the solubility product, so the solution is unsaturated with respect to barium sulphate and no precipitate will occur. A precipitate appears only when the ionic product exceeds .
Enthalpy of combustion: the enthalpy change when one mole of a substance is completely burnt in excess oxygen. For glucose: . $$ \begin{aligned} \Delta Hc &= [6,\Delta Hf(\ce{CO2}) + 6,\Delta Hf(...
Redox titration: a titration based on an oxidation-reduction reaction between the titrant and the analyte (e.g. vs ), the end point being detected by a change in oxidation state (often self-indicating).
Calculation:
N; in 200 cc, eq
eq of acid.
Base (0.25 eq) exceeds acid (0.15 eq), so the solution is basic.
Excess
eq (mol) left after neutralization.
Diluted to 2 L, the resulting concentration of excess alkali:
Ethanal from 1,1-dibromoethane by hydrolysis of the gem-dihalide with aqueous alkali: Ethanal from ethyne by hydration (Kucherov reaction): Et...
Heat of formation: the enthalpy change when one mole of a compound is formed from its elements in their standard states.
For
, by Hess's law:
The heat of formation of methane is kcal/mol (exothermic).
Normality factor (f): the ratio of the actual normality of a solution to its intended (labelled) normality; it corrects for the difference between the calculated and true strength.
Calculation: equivalents of needed
eq. Since is monobasic, eq. wt. :
Mass of pure per mL of conc. acid
About 16.2 mL of the concentrated acid is required (diluted to 500 mL).
Normality: the number of gram-equivalents of solute dissolved per litre of solution.
Calculation:
Total taken
eq (in 200 mL after dilution).
The 50 mL portion is one quarter of 200 mL, and its unreacted is neutralized by NaOH:
Excess in whole 200 mL
that reacted with the metal
eq equivalents of metal.
Degree of ionization: the fraction of the total number of molecules of an electrolyte that actually dissociate into ions in solution ( ionized fraction). Calculation: Moles of $$ \begin{aligned} \ce{NaOH} &= \dfrac{0.41}{40} ...
At this very low concentration the from water cannot be ignored. Solution: let total . Charge/mass balance with gives: $$ \begin{aligned} x &= [\ce{H+}]{acid} + [\ce{OH-}] \ &= 10^{-8} + \frac{10^{...
Spontaneous process: a physical or chemical change that takes place on its own under a given set of conditions without any continuous outside help. It proceeds in a definite direction accompanied by a decrease in Gibbs free energy ().
Example: flow of heat from a hot body to a cold body, or the rusting of iron:
Enthalpy of formation is per one mole of the compound formed. (i) , kcal for 2 mol water. $$ \begin{aligned} \Delta Hf(\ce{H2O}) &= \frac{-136}{2} \ &= -68\ \text{kcal/mol} \end{alig...
Heat sodium benzoate with soda-lime to get benzene, then carry out Friedel-Crafts acylation. Th...
Methanal (formaldehyde) reacts with ammonia to give hexamethylenetetramine (urotropine), X = : Use: it is used as a urinary antiseptic (urotropine) and as a hardener (with the explos...