NEB Class 12 · Past paper
The complete NEB Class 12 2078 exam paper for Chemistry, all 33 questions with solved model answers.
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Give any two features of tetrahedral hybridization.
In tetrahedral ($sp^3$) hybridization one $s$ and three $p$ orbitals of the same shell mix to form four equivalent hybrid orbitals. Two of its features are:
Methane, $\ce{CH4}$, is the standard example, and $\ce{CCl4}$ and the carbon of ethane behave the same way.
Define the terms i) alkalimetry ii) standard solution
(i) Alkalimetry is the branch of volumetric analysis in which the strength (concentration) of an alkali is determined by titrating it against a standard solution of an acid, using a suitable indicator to find the end point. (The converse process, finding the strength of an acid with a standard alkali, is acidimetry.)
(ii) A standard solution is a solution whose concentration is exactly known, expressed in normality, molarity or in grams per litre. It is prepared by dissolving an accurately weighed quantity of a primary standard substance, such as oxalic acid or sodium carbonate, in water and making the volume up to the mark in a volumetric flask. It is the solution taken in the burette or the pipette against which the unknown is titrated.
Calculate pH of 1x10^-8 M HCl solution.
Hydrochloric acid is a strong acid and is completely ionised, so it supplies $1\times10^{-8}\ \text{M}$ of $\ce{H+}$. This concentration is smaller than the $10^{-7}\ \text{M}$ that water itself supplies, so the ionisation of water cannot be neglected here, and both sources must be added.
Let the total $[\ce{H+}]=x$. The $\ce{OH-}$ comes only from water, and $[\ce{OH-}]=K_w/x$. Since the solution is electrically neutral,
$$ \begin{aligned} [\ce{H+}] &= [\ce{Cl-}]+[\ce{OH-}] \ x &= 1\times10^{-8}+\frac{1\times10^{-14}}{x} \end{aligned} $$
Multiplying through by $x$ gives the quadratic
$$ \begin{aligned} x^{2}-1\times10^{-8}x-1\times10^{-14} &= 0 \ x &= \frac{1\times10^{-8}+\sqrt{(10^{-8})^{2}+4\times10^{-14}}}{2} \ &= \frac{1\times10^{-8}+2.0025\times10^{-7}}{2} \ [\ce{H+}] &= 1.051\times10^{-7}\ \text{M} \end{aligned} $$
Therefore
$$ \begin{aligned} \text{pH} &= -\log(1.051\times10^{-7}) \ &= 7-\log 1.051 \ \mathbf{pH=6.98} \end{aligned} $$
The answer is worth noticing: applying $\text{pH}=-\log(10^{-8})$ blindly would give $8$, which would make a solution of a strong acid alkaline, an impossible result. The correct value is just below $7$, as it must be for any acid, however dilute.
What products would you obtain at cathode and at anode when aq. NaCl is electrolysed using pt-electrode ?
In aqueous sodium chloride four ions are present, $\ce{Na+}$ and $\ce{H+}$ moving to the cathode and $\ce{Cl-}$ and $\ce{OH-}$ moving to the anode. At each electrode the ion that is more easily discharged is the one that reacts.
At the cathode, $\ce{H+}$ is discharged in preference to $\ce{Na+}$, because sodium stands much higher in the electrochemical series and its discharge potential is far greater, so hydrogen gas is liberated:
$$\ce{2H+ + 2e- -> H2 (g)}$$
At the anode, $\ce{Cl-}$ is discharged in preference to $\ce{OH-}$ when the solution is reasonably concentrated, so chlorine gas is liberated:
$$\ce{2Cl- -> Cl2 (g) + 2e-}$$
The $\ce{Na+}$ and $\ce{OH-}$ left in the solution accumulate as sodium hydroxide, which is why this electrolysis is the industrial source of caustic soda, chlorine and hydrogen. Platinum is used because it is inert and is not attacked at the anode.
What is the criteria of spontaneous process in light of i) Entropy change ii) Free energy change
(i) In terms of entropy change. A process is spontaneous if it increases the total entropy of the system and its surroundings taken together, that is
$$\Delta S_{total}=\Delta S_{system}+\Delta S_{surroundings}>0$$
The process is at equilibrium when $\Delta S_{total}=0$ and non spontaneous when $\Delta S_{total}<0$. The entropy of the system alone is not the criterion, since a system's entropy may fall in a spontaneous change provided the surroundings gain more.
(ii) In terms of free energy change. For a process at constant temperature and pressure, the criterion is stated for the system alone by the Gibbs free energy:
$$\Delta G=\Delta H-T\Delta S$$
The process is spontaneous when $\Delta G<0$ (negative, exergonic), at equilibrium when $\Delta G=0$, and non spontaneous when $\Delta G>0$. This is the more useful of the two criteria in practice, because it needs measurements on the system only.
M+N -> P is a hypothetical first order reaction. Write its possible rate law expression.
The reaction is stated to be of the first order overall, so the sum of the powers of the concentration terms in its rate law must be one. Since the order is found by experiment and not from the balanced equation, two rate laws are possible.
If the rate depends on $\ce{M}$ alone,
$$ \begin{aligned} \text{rate} &= -\frac{d[\ce{M}]}{dt} \ &= k[\ce{M}] \end{aligned} $$
and if it depends on $\ce{N}$ alone,
$$\text{rate}=k[\ce{N}]$$
In either case the reaction is first order with respect to one reactant and zero order with respect to the other, so that the overall order is $1+0=1$. The unit of $k$ is $\text{s}^{-1}$. A rate law such as $k[\ce{M}][\ce{N}]$ would make the reaction second order and is therefore ruled out.
State the first law of thermodynamics.
First law of thermodynamics: energy can neither be created nor destroyed, although it may be converted from one form into another. Equivalently, the total energy of the universe (a system together with its surroundings) remains constant, which is why it is also called the law of conservation of energy.
For a system that absorbs a quantity of heat $q$ and has work $w$ done on it, the law is written
$$\Delta U=q+w$$
where $\Delta U$ is the increase in the internal energy of the system. For work of expansion against a constant external pressure, $w=-P\Delta V$, so
$$\Delta U=q-P\Delta V$$
A useful consequence is that the internal energy of an isolated system, which exchanges neither heat nor work, is constant, and that a machine giving out more energy than it takes in, a perpetual motion machine of the first kind, is impossible.
Write reaction for the conversion of sodium benzoate into cyclohexane.
Sodium benzoate is first decarboxylated by heating with soda lime, a mixture of sodium hydroxide and quicklime, which removes the carboxyl group as sodium carbonate and gives benzene:
$$\ce{C6H5COONa + NaOH ->[CaO][\Delta] C6H6 + Na2CO3}$$
The benzene ring is then completely hydrogenated. Three molecules of hydrogen are added in the presence of finely divided nickel at about $473\ \text{K}$ under pressure, saturating the ring to give cyclohexane:
$$\ce{C6H6 + 3H2 ->[Ni][473 K] C6H12}$$
The two reactions together take sodium benzoate to cyclohexane.
Identify A and B in the following reaction sequence: CH3-Br --(Alc. AgCN, heat)--> A --(Ni/H2)--> B
$\ce{A}$ is methyl isocyanide, $\ce{CH3-NC}$. Silver cyanide is largely covalent, so the nitrogen of the cyanide group is the atom that attacks the carbon, and an alcoholic solution of $\ce{AgCN}$ therefore gives the isocyanide rather than the cyanide (potassium cyanide, being ionic, would have given methyl cyanide $\ce{CH3-CN}$):
$$\ce{CH3Br + AgCN ->[alc.][\Delta] CH3-NC + AgBr}$$
$\ce{B}$ is N-methylmethanamine, $\ce{CH3-NH-CH3}$, a secondary amine. On catalytic reduction the triple bond of the isocyanide takes up four hydrogen atoms, and because the alkyl group is already attached to the nitrogen, the product is a secondary amine:
$$\ce{CH3-NC + 4[H] ->[Ni/H2] CH3-NH-CH3}$$
This difference is the standard way of remembering the pair: a cyanide reduces to a primary amine ($\ce{CH3CH2NH2}$), while an isocyanide reduces to a secondary amine.
Which one of the following has higher boiling point and why ? propan-1-ol or propan-2-ol
Propan-1-ol has the higher boiling point ($\ce{CH3CH2CH2OH}$, about $97^\circ\text{C}$) compared with propan-2-ol ($\ce{CH3CH(OH)CH3}$, about $82^\circ\text{C}$).
Both are isomers of the same molecular formula $\ce{C3H8O}$ and both form intermolecular hydrogen bonds through the $\ce{-OH}$ group, so the difference lies in their shape. Propan-1-ol has a straight, unbranched chain with the hydroxyl group at the end, so its molecules pack closely, present a larger surface area of contact and have a freely exposed $\ce{-OH}$ group, giving strong van der Waals attraction and effective hydrogen bonding.
In propan-2-ol the chain is branched and the hydroxyl group sits on the middle carbon, where the two methyl groups crowd it. The molecule is more compact and nearly spherical, so the surface area of contact is smaller and the $\ce{-OH}$ group is partly shielded from its neighbours. The intermolecular forces are therefore weaker, less energy is needed to separate the molecules, and the boiling point is lower.
Starting from ethanol, how would you obtain methoxymethane ?
Methoxymethane, $\ce{CH3-O-CH3}$, has two carbon atoms but they are in separate methyl groups, so the two carbon chain of ethanol must first be cut down to one carbon. Ethanol is oxidised to ethanoic acid, the acid is decarboxylated to methane, and methane is carried through to methanol, two molecules of which are then joined by Williamson's synthesis.
$$ \begin{aligned} \ce{CH3CH2OH + 2[O] ->[K2Cr2O7/H2SO4] CH3COOH + H2O} \ \ce{CH3COOH + NaOH -> CH3COONa + H2O} \ \ce{CH3COONa + NaOH ->[CaO][\Delta] CH4 + Na2CO3} \ \ce{CH4 + Cl2 ->[h\nu] CH3Cl + HCl} \ \ce{CH3Cl + KOH(aq) ->[\Delta] CH3OH + KCl} \end{aligned} $$
Part of the methanol is converted into sodium methoxide, which then displaces the halide from a second molecule of methyl chloride:
$$ \begin{aligned} \ce{2CH3OH + 2Na -> 2CH3ONa + H2} \ \ce{CH3ONa + CH3Cl -> CH3-O-CH3 + NaCl} \end{aligned} $$
The last reaction is Williamson's synthesis, and since the halide here is primary the substitution is clean. (Simply dehydrating ethanol with concentrated sulphuric acid at $413\ \text{K}$ would give ethoxyethane, $\ce{C2H5-O-C2H5}$, not methoxymethane, which is why the chain has to be shortened first.)
Write an example of each of the following reaction: i) Rosenmmund's reduction ii) Cannizzaro's reaction
(i) Rosenmund's reduction. An acid chloride is reduced to an aldehyde by hydrogen over palladium supported on barium sulphate, the catalyst being deliberately poisoned with sulphur or quinoline so that the reduction stops at the aldehyde and does not go on to the alcohol:
$$\ce{CH3COCl + H2 ->[Pd/BaSO4] CH3CHO + HCl}$$
(ii) Cannizzaro's reaction. An aldehyde having no alpha hydrogen undergoes self oxidation and reduction (disproportionation) when warmed with concentrated alkali, one molecule being reduced to an alcohol and the other oxidised to the salt of an acid:
$$\ce{2HCHO + NaOH -> CH3OH + HCOONa}$$
Benzaldehyde behaves in the same way, giving benzyl alcohol and sodium benzoate.
Give any two reactions for the preparation of ethanoic acid.
(i) By the oxidation of ethanol. Ethanol is oxidised by acidified potassium dichromate, passing through ethanal to the acid:
$$\ce{CH3CH2OH + 2[O] ->[K2Cr2O7/H2SO4] CH3COOH + H2O}$$
(ii) By the hydrolysis of ethanenitrile. Methyl cyanide is boiled with dilute acid, when the cyano group is hydrolysed to a carboxyl group:
$$\ce{CH3CN + 2H2O ->[H+][\Delta] CH3COOH + NH3}$$
A third route, from a Grignard reagent, is equally acceptable: dry carbon dioxide is passed into methylmagnesium bromide and the adduct is hydrolysed,
$$\ce{CH3MgBr + CO2 -> CH3COOMgBr ->[H3O+] CH3COOH}$$
Give a chemical test reaction to distinguish ethanamine from N-methylmethanamine.
The two are distinguished by the carbylamine test (Hoffmann's isocyanide test), which is given by primary amines only.
Ethanamine, $\ce{CH3CH2NH2}$, is a primary amine. Warmed with chloroform and alcoholic potassium hydroxide it gives ethyl isocyanide, recognised at once by its extremely unpleasant smell:
$$\ce{CH3CH2NH2 + CHCl3 + 3KOH ->[\Delta] CH3CH2-NC + 3KCl + 3H2O}$$
N-methylmethanamine, $\ce{CH3-NH-CH3}$, is a secondary amine and has only one hydrogen on the nitrogen, so it cannot form the isocyanide. It gives no reaction and no foul smell under the same conditions.
The appearance of the offensive odour therefore identifies ethanamine, and its absence identifies N-methylmethanamine.
Why is amino group protected before nitration in aniline ?
Two things go wrong if aniline is nitrated directly.
First, the nitrating mixture is strongly acidic, and aniline, being basic, is protonated in it to the anilinium ion $\ce{C6H5NH3+}$. In this ion the nitrogen carries a positive charge and can no longer donate its lone pair to the ring, so the group changes character completely: instead of being activating and ortho para directing it becomes deactivating and meta directing. Direct nitration therefore yields a large proportion of m-nitroaniline along with the para product, instead of the p-nitroaniline wanted.
Second, aniline is very easily oxidised, and nitric acid is a powerful oxidising agent, so a good deal of the material is destroyed as dark tarry oxidation products.
The amino group is therefore protected by acetylation with acetic anhydride to give acetanilide. The lone pair of the nitrogen is now partly drawn towards the carbonyl group, which tempers the activation and prevents both the protonation and the oxidation, and the bulky acetamido group directs the incoming nitro group almost entirely to the para position. Hydrolysing the product with dilute acid or alkali removes the acetyl group and returns the free amine:
$$\ce{C6H5NH2 ->[(CH3CO)2O] C6H5NHCOCH3 ->[HNO3/H2SO4] p-O2N-C6H4-NHCOCH3 ->[H3O+] p-O2N-C6H4-NH2}$$
What is meant by: i) Zwitter ion ii) iso-electric point.
(i) Zwitter ion. An amino acid carries an acidic carboxyl group and a basic amino group in the same molecule, and in the solid state and in neutral solution the proton passes from the one to the other. The result is a dipolar ion that ca...
What are nitrogen bases ? Name the nitrogen bases present in RNA.
Nitrogen bases are the heterocyclic, nitrogen containing basic compounds that form one of the three units of a nucleotide, the others being a pentose sugar and a phosphate group. They are of two kinds, the purines, with two fused rings, ...
Define co-polymerization giving an example.
Co-polymerization is the polymerization in which two or more different monomers combine together to build a single polymer chain, so that the repeating unit contains residues of each of them. The product is a co-polymer, and its properties can be adjusted by changing the proportion of the monomers, which is the object of the process.
Example: Buna-S (styrene butadiene rubber, SBR). Buta-1,3-diene and styrene are polymerized together in the ratio of about three to one:
$$\ce{n CH2=CH-CH=CH2 + n C6H5-CH=CH2 ->} \text{[-CH2-CH=CH-CH2-CH(C6H5)-CH2-]}_n$$
Buna-S is tougher and more resistant to abrasion than natural rubber and is used for motor tyres. Nylon-6,6, made from hexamethylenediamine and adipic acid, and Terylene, from ethylene glycol and terephthalic acid, are equally good examples.
Write down the structure formula and a major use of insecticide.
A well known insecticide is DDT, p,p'-dichlorodiphenyltrichloroethane. Structural formula: $$\ce{Cl-C6H4-CH(-C6H4-Cl)-CCl3}$$ that is, a central $\ce{CH}$ carrying a $\ce{CCl3}$ group and two p-chlorophenyl rings: Major use: it is used t...
Why does silver nitrate produce a permanent black stain on the skin ?
Silver nitrate is a soluble salt of silver and the skin is made largely of protein. On contact the silver ion combines with the protein of the skin to give silver albuminate, and the organic matter of the skin then acts as a mild reducing agent. In the presence of light the bound silver ion is reduced to finely divided metallic silver, which is black:
$$\ce{Ag+ + e- -> Ag (black)}$$
The stain is permanent for two reasons. The metallic silver so formed is insoluble in water, in soap and in dilute acids, so it cannot be washed away; and it is deposited within the layers of the skin, held to the protein, not merely lying on the surface. It therefore disappears only as the stained cells are gradually shed in the natural renewal of the skin. This reduction of silver salts by light is the same reaction on which photography depends, and it is why silver nitrate is kept in dark coloured bottles.
What happens when ZnO is i) dissolved with caustic soda ii) heated with cobalt nitrate ?
(i) Dissolved with caustic soda. Zinc oxide is amphoteric, so besides dissolving in acids it also dissolves in a strong alkali, giving a soluble zincate: $$\ce{ZnO + 2NaOH - Na2ZnO2 + H2O}$$ The product is sodium zincate. (In solution it...
Name a major ore of each of iron and copper giving their formula.
Iron: the major ore is haematite, $\ce{Fe2O3}$ (red oxide of iron), which is the ore actually smelted in the blast furnace. (Magnetite, $\ce{Fe3O4}$, is the other important oxide ore.)
Copper: the major ore is copper pyrite, also called chalcopyrite, $\ce{CuFeS2}$, from which the greater part of the world's copper is extracted. (Cuprite, $\ce{Cu2O}$, and copper glance, $\ce{Cu2S}$, are other ores.)
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 $K_{sp}$.
Define standard hydrogen electrode and give its major use. The cost of electricity to deposit 10gm of Mg is Rs. 60. How much would it cost to deposit 100gm of copper from CuSO4 ? ( At. wt. of Cu = 63.5)
Standard hydrogen electrode. The standard hydrogen electrode (SHE) consists of a platinum foil coated with platinum black, dipped in a solution of hydrogen ions of unit activity (1 M $\ce{H+}$), around which pure hydrogen gas at one atmosphere pressure is bubbled at $298\ \text{K}$. It is represented as
$$\ce{Pt, H2 (1 atm) | H+ (1 M)}$$
and the half reaction is
$$ \ce{2H+ + 2e- <=> H2} $$
.
Major use. Its electrode potential is arbitrarily taken as exactly zero, so it serves as the primary reference electrode against which the standard electrode potential of every other electrode is measured. The unknown electrode is coupled with the SHE to form a cell, and the measured emf of that cell, with its sign, is by definition the standard electrode potential of the unknown electrode. The whole electrochemical series is built up in this way.
Numerical. The cost is proportional to the quantity of electricity, and by Faraday's first law the quantity of electricity is proportional to the number of gram equivalents deposited. So the two metals must be compared on an equivalent basis, not a mass basis.
The equivalent weight of a metal is its atomic weight divided by its valency, and both magnesium and copper (from $\ce{CuSO4}$) are divalent:
$$ \begin{aligned} E_{\ce{Mg}} &= \frac{24}{2} \ &= 12 \ \qquad E_{\ce{Cu}} &= \frac{63.5}{2} \ &= 31.75 \end{aligned} $$
Equivalents of magnesium actually deposited:
$$ \begin{aligned} n_{\ce{Mg}} &= \frac{10}{12} \ &= 0.8333\ \text{equivalent} \end{aligned} $$
This costs Rs. 60, so the cost of one equivalent is
$$\frac{60}{0.8333}=\text{Rs. }72\ \text{per equivalent}$$
Equivalents of copper to be deposited:
$$ \begin{aligned} n_{\ce{Cu}} &= \frac{100}{31.75} \ &= 3.1496\ \text{equivalent} \end{aligned} $$
Hence the cost is
$$3.1496\times72=\mathbf{Rs.\ 226.77}$$
that is about Rs. 227.
Write down the preparation, properties and uses of calomel.
Calomel is mercurous chloride, $\ce{Hg2Cl2}$, a white insoluble solid. Preparation. In the laboratory it is precipitated by adding a solution of a soluble chloride to a solution of a mercurous salt: $$\ce{Hg2(NO3)2 + 2NaCl - Hg2Cl2 (whit...
Mention an example of each of the following reactions: i) DNp test ii) Aldol condensation iii) Hoffmann's hypobromite reaction iv) Carbonylation reaction v) Coupling reaction.
(i) DNP test. 2,4-dinitrophenylhydrazine condenses with the carbonyl group of an aldehyde or a ketone to give an orange or yellow crystalline 2,4-dinitrophenylhydrazone, which is the test for a carbonyl compound:
$$\ce{CH3CHO + H2N-NH-C6H3(NO2)2 -> CH3CH=N-NH-C6H3(NO2)2 + H2O}$$
(ii) Aldol condensation. Two molecules of an aldehyde having an alpha hydrogen combine in dilute alkali, the alpha carbon of one adding to the carbonyl carbon of the other, to give a beta hydroxy aldehyde (an aldol):
$$\ce{2CH3CHO ->[dil. NaOH] CH3-CH(OH)-CH2-CHO}$$
(iii) Hoffmann's hypobromite reaction. An amide is degraded by bromine and alkali (that is by sodium hypobromite) to a primary amine having one carbon atom less, the carbonyl carbon leaving as carbonate:
$$\ce{CH3CONH2 + Br2 + 4NaOH -> CH3NH2 + 2NaBr + Na2CO3 + 2H2O}$$
(iv) Carbonylation reaction. Carbon monoxide is inserted into a molecule. Methanol is carbonylated to ethanoic acid over a rhodium catalyst, which is the industrial Monsanto process:
$$\ce{CH3OH + CO ->[Rh catalyst] CH3COOH}$$
(v) Coupling reaction. A diazonium salt joins to a phenol or an aromatic amine at the para position to give a brightly coloured azo compound, the basis of the azo dyes:
$$\ce{C6H5N2+Cl- + C6H5OH ->[NaOH][273-278 K] p-C6H5-N=N-C6H4-OH + HCl}$$
The product here is p-hydroxyazobenzene, an orange dye.
How is chloroform prepared in laboratory ? What happens when chloroform is heated with silver powder ?
Laboratory preparation of chloroform. Chloroform, $\ce{CHCl3}$, is prepared by the action of bleaching powder on ethanol or on acetone. Bleaching powder, $\ce{CaOCl2}$, plays three parts at once: with water it supplies chlorine, which ox...
Write down a structural formula with IUPAC name of each of primary and secondary alcohol having molecular formula C3H8O. How would you prepare these alcohols from CH3MgBr ?
The two alcohols of formula $\ce{C3H8O}$.
The primary alcohol is propan-1-ol, in which the hydroxyl group is on a carbon carrying two hydrogen atoms:
$$\ce{CH3-CH2-CH2-OH}$$
The secondary alcohol is propan-2-ol, in which the hydroxyl group is on the middle carbon, attached to two alkyl groups:
$$\ce{CH3-CH(OH)-CH3}$$
Preparation from methylmagnesium bromide. A Grignard reagent adds across the carbonyl group, its alkyl group going to the carbon and the $\ce{-MgBr}$ to the oxygen, and hydrolysis of the adduct then gives the alcohol. The class of alcohol obtained is decided by the carbonyl compound chosen, and since the methyl group contributes one carbon, the other partner must supply two.
For the primary alcohol, propan-1-ol, the reagent is treated with an epoxide (ethylene oxide), which opens to lengthen the chain by two carbons:
$$\ce{CH3MgBr + CH2(-O-)CH2 -> CH3CH2CH2OMgBr ->[H3O+] CH3CH2CH2OH}$$
For the secondary alcohol, propan-2-ol, the reagent is added to an aldehyde (ethanal), since an aldehyde always gives a secondary alcohol:
$$\ce{CH3MgBr + CH3CHO -> CH3-CH(CH3)-OMgBr ->[H3O+] CH3-CH(OH)-CH3}$$
The reactions are carried out in dry ether, and the water for the hydrolysis is added only at the end, since a Grignard reagent is destroyed by moisture.
Describe the laboratory method of preparation of pure and dry nitrobenzene. Identify A, B, C and D in the following reaction sequence: A --(NaOH+CaO)--> B --(CH3Cl/heat)--> C --(CeO2/H+)--> D. The compound B can be prepared by heating phenol with zinc-dust.
Laboratory preparation of pure and dry nitrobenzene.
Nitrobenzene is prepared by nitrating benzene with a mixture of concentrated nitric acid and concentrated sulphuric acid, called the nitrating mixture. The sulphuric acid is there to generate the attacking electrophile, the nitronium ion:
$$ \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}$$
(This is the same conversion that the Etard reaction brings about with chromyl chloride, $\ce{CrO2Cl2}$, followed by hydrolysis. A stronger oxidising agent such as acidified $\ce{KMnO4}$ would have carried the side chain on to benzoic acid.)
In summary: $\ce{A}$ = sodium benzoate, $\ce{B}$ = benzene, $\ce{C}$ = toluene, $\ce{D}$ = benzaldehyde.
a) An aliphatic haloalkane A gives compound B when heated with alc.KOH. Compound B reacts with HBr to give major product C. The compound C undergoes Wurtz reaction to produce 2,3-dimethyl butane. Identify A, B and C with their IUPAC names. Write the reactions involved. b) Apply Hoffmann's method to separate primary secondary and tertiary amines from their mixture.
(a) The chain is best read backwards from the known product.
Wurtz's reaction joins two molecules of an alkyl halide through sodium in dry ether,
$$ \ce{2R-X + 2Na -> R-R + 2NaX} $$
, so the product is a symmetrical alkane made of two identical halves. Splitting 2,3-dimethylbutane, $\ce{(CH3)2CH-CH(CH3)2}$, down the middle gives two isopropyl groups. Hence $\ce{C}$ must be the isopropyl halide, and since the reagent used to make it was $\ce{HBr}$, $\ce{C}$ is 2-bromopropane, $\ce{CH3-CHBr-CH3}$.
$\ce{C}$ is the major product of adding $\ce{HBr}$ to $\ce{B}$, and by Markovnikov's rule the bromine goes to the carbon carrying fewer hydrogen atoms, which is the middle carbon of propene. Hence $\ce{B}$ is propene, $\ce{CH3-CH=CH2}$.
$\ce{B}$ comes from $\ce{A}$ on heating with alcoholic potassium hydroxide, which is dehydrohalogenation. A haloalkane that loses $\ce{HBr}$ to give propene must be a bromopropane, and it cannot be 2-bromopropane, since that is $\ce{C}$ itself and the question makes $\ce{A}$ and $\ce{C}$ different compounds. Hence $\ce{A}$ is 1-bromopropane, $\ce{CH3CH2CH2Br}$.
The reactions are
$$ \begin{aligned} \ce{CH3CH2CH2Br + KOH ->[alcoholic][\Delta] CH3-CH=CH2 + KBr + H2O} \ \ce{CH3-CH=CH2 + HBr -> CH3-CHBr-CH3}\quad\text{(Markovnikov addition)} \ \ce{2CH3-CHBr-CH3 + 2Na ->[dry ether] (CH3)2CH-CH(CH3)2 + 2NaBr} \end{aligned} $$
The whole sequence is the standard way of converting a primary halide into the corresponding secondary halide, and it is worth noting that this change of position of the halogen is exactly what makes the final Wurtz product branched.
In summary: $\ce{A}$ = 1-bromopropane, $\ce{B}$ = propene, $\ce{C}$ = 2-bromopropane.
(b) Hoffmann's method for separating a mixture of primary, secondary and tertiary amines.
The mixture is treated with diethyl oxalate, $\ce{(COOC2H5)2}$, and the three classes of amine behave quite differently towards it, which is what makes the separation possible.
A primary amine reacts with both ester groups and gives a solid dialkyl oxamide:
$$\ce{2RNH2 + (COOC2H5)2 -> (CONHR)2 + 2C2H5OH}$$
A secondary amine has only one hydrogen on the nitrogen, so it can react at one ester group only and gives a liquid dialkyl oxamic ester:
$$\ce{R2NH + (COOC2H5)2 -> R2N-CO-COOC2H5 + C2H5OH}$$
A tertiary amine has no hydrogen on the nitrogen at all and therefore does not react; it remains as the free amine in the mixture.
The products are now separated in two stages. The reaction mixture is first filtered, which removes the crystalline dialkyl oxamide from the primary amine. What is left, the oxamic ester, the unreacted tertiary amine and the ethanol set free in the reaction, is then separated by fractional distillation, in which the tertiary amine, being volatile and unchanged, distils over first and is collected, leaving the oily oxamic ester behind.
Each derivative is then hydrolysed with alkali to recover its amine:
$$ \begin{aligned} \ce{(CONHR)2 + 2KOH -> 2RNH2 + (COOK)2} \ \ce{R2N-CO-COOC2H5 + 2KOH -> R2NH + (COOK)2 + C2H5OH} \end{aligned} $$
The three amines are thus obtained separately. (Hinsberg's method, using benzenesulphonyl chloride, achieves the same separation on a different principle.)
Define the terms: i) activation energy ii) order of a reaction iii) molecularity of a reaction iv) rate law. How does temperature affect the rate of chemical reaction ? The experimental data for the reaction P+Q -> Z are as below. Expt 1: P=0.50, Q=0.50, rate of formation of Z=1.6x10^-4; Expt 2: P=0.50, Q=1.00, rate=3.2x10^-4; Expt 3: P=1.00, Q=1.00, rate=3.2x10^-4 (all concentrations in molL^-1, rates in molL^-1 sec^-1). Calculate the rate of formation of Z when the initial concentration of P and Q are 2.00 molL^-1 and 4.00 molL^-1 respectively.
Definitions.
(i) Activation energy. The activation energy $E_a$ is the minimum extra energy, over and above their average energy, that the reactant molecules must possess before they can cross the energy barrier and be converted into products. Only those collisions in which the colliding molecules together carry at least this energy are effective. It is the difference between the energy of the transition state and that of the reactants, and the greater it is, the slower the reaction.
(ii) Order of a reaction. The order of a reaction is the sum of the powers to which the concentration terms are raised in the experimentally determined rate law. It is found only by experiment, it may be zero, fractional or even negative, and it need have no connection with the coefficients of the balanced equation.
(iii) Molecularity of a reaction. The molecularity is the number of molecules, atoms or ions that take part in the single elementary step by which the reaction occurs, as written in its balanced equation. It is a theoretical quantity, always a small whole number, and it is never zero or fractional. For a reaction that proceeds in several steps, molecularity has meaning only for each individual step, while the observed order belongs to the slowest of them.
(iv) Rate law. The rate law (rate equation) is the experimentally established expression relating the rate of a reaction to the concentrations of the reactants, each raised to the appropriate power, as in $\text{rate}=k[\ce{A}]^m[\ce{B}]^n$. The constant $k$ is the rate constant, or specific reaction rate, being the rate when each concentration is unity.
Effect of temperature on the rate. The rate of nearly every reaction rises sharply with temperature, and as a rough rule the rate is doubled or trebled for every $10^\circ$ rise. The ratio of the rate constants at two temperatures differing by ten degrees is called the temperature coefficient,
$$\frac{k_{t+10}}{k_t}\approx 2\text{ to }3$$
The reason is not simply that the molecules move faster and collide more often, which would account for only a very small part of the increase. It is that the fraction of molecules possessing energy equal to or greater than the activation energy grows very rapidly with temperature, as the Maxwell distribution of molecular energies broadens and its maximum shifts to the right. This is expressed by the Arrhenius equation
$$k=Ae^{-E_a/RT}$$
or in the form used for calculation,
$$\log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\left[\frac{T_2-T_1}{T_1T_2}\right]$$
which shows that $k$ increases exponentially as $T$ rises, and that a reaction with a large $E_a$ is the more sensitive to a change of temperature.
Numerical. Let the rate law be $\text{rate}=k[\ce{P}]^m[\ce{Q}]^n$. The orders are found by changing one concentration at a time.
Order with respect to $\ce{Q}$. Compare experiments 1 and 2, in which $[\ce{P}]$ is held at $0.50$ while $[\ce{Q}]$ is doubled:
$$ \begin{aligned} \frac{\text{rate}_2}{\text{rate}_1} &= \frac{3.2\times10^{-4}}{1.6\times10^{-4}} \ &= 2 \ &= \left(\frac{1.00}{0.50}\right)^{n} \ &= 2^{n} \ n &= 1 \end{aligned} $$
Order with respect to $\ce{P}$. Compare experiments 2 and 3, in which $[\ce{Q}]$ is held at $1.00$ while $[\ce{P}]$ is doubled:
$$ \begin{aligned} \frac{\text{rate}_3}{\text{rate}_2} &= \frac{3.2\times10^{-4}}{3.2\times10^{-4}} \ &= 1 \ &= \left(\frac{1.00}{0.50}\right)^{m} \ &= 2^{m} \ m &= 0 \end{aligned} $$
The rate is therefore independent of $[\ce{P}]$, and the rate law is
$$\text{rate}=k[\ce{Q}]$$
so the reaction is first order overall.
The rate constant, from experiment 1,
$$ \begin{aligned} k &= \frac{\text{rate}}{[\ce{Q}]} \ &= \frac{1.6\times10^{-4}}{0.50} \ &= 3.2\times10^{-4}\ \text{s}^{-1} \end{aligned} $$
The required rate, when $[\ce{P}]=2.00$ and $[\ce{Q}]=4.00\ \text{mol L}^{-1}$. Since $\ce{P}$ does not appear in the rate law its value is irrelevant:
$$ \begin{aligned} \text{rate} &= 3.2\times10^{-4}\times4.00 \ \text{rate} &= \mathbf{1.28\times10^{-3}\ mol\ L^{-1}s^{-1}} \end{aligned} $$
Write short notes on any two. i) Chemistry of Blue vitrol ii) Manufacture of steel by Open Hearth process iii) Hess's law of constant heat summation and its application iv) Laboratory preparation of anhydrous formic acid.
(i) Chemistry of blue vitriol.
Blue vitriol is copper sulphate pentahydrate, $\ce{CuSO4.5H2O}$, a blue triclinic crystalline solid. It is prepared by dissolving copper oxide, hydroxide or carbonate in dilute sulphuric acid and crystallising the solution, and on the large scale by the slow action of dilute sulphuric acid on scrap copper in the presence of air:
$$ \begin{aligned} \ce{CuO + H2SO4 -> CuSO4 + H2O} \ \ce{2Cu + 2H2SO4 + O2 -> 2CuSO4 + 2H2O} \end{aligned} $$
Of its five molecules of water, four are co-ordinated to the copper and the fifth is held to the sulphate by hydrogen bonding, which is why they are lost in stages on heating:
$$\ce{CuSO4.5H2O ->[373 K] CuSO4.H2O ->[503 K] CuSO4 (white) ->[1000 K] CuO + SO3}$$
The anhydrous salt is white, and because it turns blue again on taking up water it is used to test for the presence of water in a liquid. The crystals lose water slowly in dry air, that is they are efflorescent.
In solution it is acidic to litmus through hydrolysis. Adding ammonium hydroxide first precipitates a pale blue basic salt which then dissolves in excess to the deep blue cuprammonium ion, $\ce{[Cu(NH3)4]^{2+}}$. Potassium iodide gives white cuprous iodide with liberation of iodine, and hydrogen sulphide gives black copper sulphide. With potassium cyanide, and with excess iodide, the copper is reduced from the $+2$ to the $+1$ state.
It is used as a fungicide (with lime as Bordeaux mixture), in electroplating and electrorefining of copper, in Fehling's solution, in dyeing as a mordant, and in the purification of water.
(iii) Hess's law of constant heat summation.
Statement: the total heat change accompanying a chemical reaction is the same whether the reaction takes place in one step or in several steps, provided the initial and the final states are the same.
It follows directly from the first law of thermodynamics, since enthalpy is a state function and $\Delta H$ depends only on the initial and final states, not on the path. If it were untrue, energy could be created by going one way round a cycle and back the other.
If a reaction takes $\ce{A}$ to $\ce{D}$ directly with a heat change $\Delta H$, and also through the steps
$$ \ce{A -> B -> C -> D} $$
with heat changes $\Delta H_1$, $\Delta H_2$ and $\Delta H_3$, then
$$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$$
Applications. The great value of the law is that it lets thermochemical equations be added, subtracted and multiplied like algebraic equations, so that heats of reaction which cannot be measured directly may be calculated from those which can.
It is used to find the heat of formation of a compound that cannot be made directly from its elements. Carbon monoxide is the standard case: burning carbon in a limited supply of oxygen always gives some carbon dioxide as well, so $\Delta H_f$ of $\ce{CO}$ cannot be measured, but it follows at once from the two combustions that can be:
$$ \begin{aligned} \ce{C + O2 -> CO2},\qquad \Delta H &= -393.5\ \text{kJ} \ \ce{CO + 1/2O2 -> CO2},\qquad \Delta H &= -283.0\ \text{kJ} \end{aligned} $$
Subtracting the second from the first,
$$\ce{C + 1/2O2 -> CO},\qquad \Delta H=-393.5-(-283.0)=-110.5\ \text{kJ}$$
It is used besides to calculate heats of reaction from heats of formation, since $\Delta H_{reaction}=\sum\Delta H_f(\text{products})-\sum\Delta H_f(\text{reactants})$; to determine lattice energies through the Born Haber cycle, and bond energies; to find the heat of transition between two allotropic forms; and to obtain the heat of hydration of a salt.
In tetrahedral () hybridization one and three orbitals of the same shell mix to form four equivalent hybrid orbitals. Two of its features are:
Methane, , is the standard example, and and the carbon of ethane behave the same way.
Hydrochloric acid is a strong acid and is completely ionised, so it supplies of . This concentration is smaller than the that water itself supplies, so the ionisation of water cannot be neglected here, and both sources must be added.
Let the total . The comes only from water, and . Since the solution is electrically neutral,
Multiplying through by gives the quadratic
Therefore
The answer is worth noticing: applying blindly would give , which would make a solution of a strong acid alkaline, an impossible result. The correct value is just below , as it must be for any acid, however dilute.
In aqueous sodium chloride four ions are present, and moving to the cathode and and moving to the anode. At each electrode the ion that is more easily discharged is the one that reacts.
At the cathode, is discharged in preference to , because sodium stands much higher in the electrochemical series and its discharge potential is far greater, so hydrogen gas is liberated:
At the anode, is discharged in preference to when the solution is reasonably concentrated, so chlorine gas is liberated:
The and left in the solution accumulate as sodium hydroxide, which is why this electrolysis is the industrial source of caustic soda, chlorine and hydrogen. Platinum is used because it is inert and is not attacked at the anode.
(i) In terms of entropy change. A process is spontaneous if it increases the total entropy of the system and its surroundings taken together, that is
The process is at equilibrium when and non spontaneous when . The entropy of the system alone is not the criterion, since a system's entropy may fall in a spontaneous change provided the surroundings gain more.
(ii) In terms of free energy change. For a process at constant temperature and pressure, the criterion is stated for the system alone by the Gibbs free energy:
The process is spontaneous when (negative, exergonic), at equilibrium when , and non spontaneous when . This is the more useful of the two criteria in practice, because it needs measurements on the system only.
The reaction is stated to be of the first order overall, so the sum of the powers of the concentration terms in its rate law must be one. Since the order is found by experiment and not from the balanced equation, two rate laws are possible.
If the rate depends on alone,
and if it depends on alone,
In either case the reaction is first order with respect to one reactant and zero order with respect to the other, so that the overall order is . The unit of is . A rate law such as would make the reaction second order and is therefore ruled out.
First law of thermodynamics: energy can neither be created nor destroyed, although it may be converted from one form into another. Equivalently, the total energy of the universe (a system together with its surroundings) remains constant, which is why it is also called the law of conservation of energy.
For a system that absorbs a quantity of heat and has work done on it, the law is written
where is the increase in the internal energy of the system. For work of expansion against a constant external pressure, , so
A useful consequence is that the internal energy of an isolated system, which exchanges neither heat nor work, is constant, and that a machine giving out more energy than it takes in, a perpetual motion machine of the first kind, is impossible.
Sodium benzoate is first decarboxylated by heating with soda lime, a mixture of sodium hydroxide and quicklime, which removes the carboxyl group as sodium carbonate and gives benzene:
The benzene ring is then completely hydrogenated. Three molecules of hydrogen are added in the presence of finely divided nickel at about under pressure, saturating the ring to give cyclohexane:
The two reactions together take sodium benzoate to cyclohexane.
is methyl isocyanide, . Silver cyanide is largely covalent, so the nitrogen of the cyanide group is the atom that attacks the carbon, and an alcoholic solution of therefore gives the isocyanide rather than the cyanide (potassium cyanide, being ionic, would have given methyl cyanide ):
is N-methylmethanamine, , a secondary amine. On catalytic reduction the triple bond of the isocyanide takes up four hydrogen atoms, and because the alkyl group is already attached to the nitrogen, the product is a secondary amine:
This difference is the standard way of remembering the pair: a cyanide reduces to a primary amine (), while an isocyanide reduces to a secondary amine.
Propan-1-ol has the higher boiling point (, about ) compared with propan-2-ol (, about ).
Both are isomers of the same molecular formula and both form intermolecular hydrogen bonds through the group, so the difference lies in their shape. Propan-1-ol has a straight, unbranched chain with the hydroxyl group at the end, so its molecules pack closely, present a larger surface area of contact and have a freely exposed group, giving strong van der Waals attraction and effective hydrogen bonding.
In propan-2-ol the chain is branched and the hydroxyl group sits on the middle carbon, where the two methyl groups crowd it. The molecule is more compact and nearly spherical, so the surface area of contact is smaller and the group is partly shielded from its neighbours. The intermolecular forces are therefore weaker, less energy is needed to separate the molecules, and the boiling point is lower.
Methoxymethane, , has two carbon atoms but they are in separate methyl groups, so the two carbon chain of ethanol must first be cut down to one carbon. Ethanol is oxidised to ethanoic acid, the acid is decarboxylated to methane, and methane is carried through to methanol, two molecules of which are then joined by Williamson's synthesis.
Part of the methanol is converted into sodium methoxide, which then displaces the halide from a second molecule of methyl chloride:
The last reaction is Williamson's synthesis, and since the halide here is primary the substitution is clean. (Simply dehydrating ethanol with concentrated sulphuric acid at would give ethoxyethane, , not methoxymethane, which is why the chain has to be shortened first.)
(i) Rosenmund's reduction. An acid chloride is reduced to an aldehyde by hydrogen over palladium supported on barium sulphate, the catalyst being deliberately poisoned with sulphur or quinoline so that the reduction stops at the aldehyde and does not go on to the alcohol:
(ii) Cannizzaro's reaction. An aldehyde having no alpha hydrogen undergoes self oxidation and reduction (disproportionation) when warmed with concentrated alkali, one molecule being reduced to an alcohol and the other oxidised to the salt of an acid:
Benzaldehyde behaves in the same way, giving benzyl alcohol and sodium benzoate.
(i) By the oxidation of ethanol. Ethanol is oxidised by acidified potassium dichromate, passing through ethanal to the acid:
(ii) By the hydrolysis of ethanenitrile. Methyl cyanide is boiled with dilute acid, when the cyano group is hydrolysed to a carboxyl group:
A third route, from a Grignard reagent, is equally acceptable: dry carbon dioxide is passed into methylmagnesium bromide and the adduct is hydrolysed,
The two are distinguished by the carbylamine test (Hoffmann's isocyanide test), which is given by primary amines only.
Ethanamine, , is a primary amine. Warmed with chloroform and alcoholic potassium hydroxide it gives ethyl isocyanide, recognised at once by its extremely unpleasant smell:
N-methylmethanamine, , is a secondary amine and has only one hydrogen on the nitrogen, so it cannot form the isocyanide. It gives no reaction and no foul smell under the same conditions.
The appearance of the offensive odour therefore identifies ethanamine, and its absence identifies N-methylmethanamine.
Two things go wrong if aniline is nitrated directly.
First, the nitrating mixture is strongly acidic, and aniline, being basic, is protonated in it to the anilinium ion . In this ion the nitrogen carries a positive charge and can no longer donate its lone pair to the ring, so the group changes character completely: instead of being activating and ortho para directing it becomes deactivating and meta directing. Direct nitration therefore yields a large proportion of m-nitroaniline along with the para product, instead of the p-nitroaniline wanted.
Second, aniline is very easily oxidised, and nitric acid is a powerful oxidising agent, so a good deal of the material is destroyed as dark tarry oxidation products.
The amino group is therefore protected by acetylation with acetic anhydride to give acetanilide. The lone pair of the nitrogen is now partly drawn towards the carbonyl group, which tempers the activation and prevents both the protonation and the oxidation, and the bulky acetamido group directs the incoming nitro group almost entirely to the para position. Hydrolysing the product with dilute acid or alkali removes the acetyl group and returns the free amine:
Co-polymerization is the polymerization in which two or more different monomers combine together to build a single polymer chain, so that the repeating unit contains residues of each of them. The product is a co-polymer, and its properties can be adjusted by changing the proportion of the monomers, which is the object of the process.
Example: Buna-S (styrene butadiene rubber, SBR). Buta-1,3-diene and styrene are polymerized together in the ratio of about three to one:
Buna-S is tougher and more resistant to abrasion than natural rubber and is used for motor tyres. Nylon-6,6, made from hexamethylenediamine and adipic acid, and Terylene, from ethylene glycol and terephthalic acid, are equally good examples.
A well known insecticide is DDT, p,p'-dichlorodiphenyltrichloroethane. Structural formula: that is, a central carrying a group and two p-chlorophenyl rings: Major use: it is used t...
Silver nitrate is a soluble salt of silver and the skin is made largely of protein. On contact the silver ion combines with the protein of the skin to give silver albuminate, and the organic matter of the skin then acts as a mild reducing agent. In the presence of light the bound silver ion is reduced to finely divided metallic silver, which is black:
The stain is permanent for two reasons. The metallic silver so formed is insoluble in water, in soap and in dilute acids, so it cannot be washed away; and it is deposited within the layers of the skin, held to the protein, not merely lying on the surface. It therefore disappears only as the stained cells are gradually shed in the natural renewal of the skin. This reduction of silver salts by light is the same reaction on which photography depends, and it is why silver nitrate is kept in dark coloured bottles.
(i) Dissolved with caustic soda. Zinc oxide is amphoteric, so besides dissolving in acids it also dissolves in a strong alkali, giving a soluble zincate: The product is sodium zincate. (In solution it...
Iron: the major ore is haematite, (red oxide of iron), which is the ore actually smelted in the blast furnace. (Magnetite, , is the other important oxide ore.)
Copper: the major ore is copper pyrite, also called chalcopyrite, , from which the greater part of the world's copper is extracted. (Cuprite, , and copper glance, , are other ores.)
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 .
Standard hydrogen electrode. The standard hydrogen electrode (SHE) consists of a platinum foil coated with platinum black, dipped in a solution of hydrogen ions of unit activity (1 M ), around which pure hydrogen gas at one atmosphere pressure is bubbled at . It is represented as
and the half reaction is
.
Major use. Its electrode potential is arbitrarily taken as exactly zero, so it serves as the primary reference electrode against which the standard electrode potential of every other electrode is measured. The unknown electrode is coupled with the SHE to form a cell, and the measured emf of that cell, with its sign, is by definition the standard electrode potential of the unknown electrode. The whole electrochemical series is built up in this way.
Numerical. The cost is proportional to the quantity of electricity, and by Faraday's first law the quantity of electricity is proportional to the number of gram equivalents deposited. So the two metals must be compared on an equivalent basis, not a mass basis.
The equivalent weight of a metal is its atomic weight divided by its valency, and both magnesium and copper (from ) are divalent:
Equivalents of magnesium actually deposited:
This costs Rs. 60, so the cost of one equivalent is
Equivalents of copper to be deposited:
Hence the cost is
that is about Rs. 227.
Calomel is mercurous chloride, , a white insoluble solid. Preparation. In the laboratory it is precipitated by adding a solution of a soluble chloride to a solution of a mercurous salt: $$\ce{Hg2(NO3)2 + 2NaCl - Hg2Cl2 (whit...
(i) DNP test. 2,4-dinitrophenylhydrazine condenses with the carbonyl group of an aldehyde or a ketone to give an orange or yellow crystalline 2,4-dinitrophenylhydrazone, which is the test for a carbonyl compound:
(ii) Aldol condensation. Two molecules of an aldehyde having an alpha hydrogen combine in dilute alkali, the alpha carbon of one adding to the carbonyl carbon of the other, to give a beta hydroxy aldehyde (an aldol):
(iii) Hoffmann's hypobromite reaction. An amide is degraded by bromine and alkali (that is by sodium hypobromite) to a primary amine having one carbon atom less, the carbonyl carbon leaving as carbonate:
(iv) Carbonylation reaction. Carbon monoxide is inserted into a molecule. Methanol is carbonylated to ethanoic acid over a rhodium catalyst, which is the industrial Monsanto process:
(v) Coupling reaction. A diazonium salt joins to a phenol or an aromatic amine at the para position to give a brightly coloured azo compound, the basis of the azo dyes:
The product here is p-hydroxyazobenzene, an orange dye.
Laboratory preparation of chloroform. Chloroform, , is prepared by the action of bleaching powder on ethanol or on acetone. Bleaching powder, , plays three parts at once: with water it supplies chlorine, which ox...
The two alcohols of formula .
The primary alcohol is propan-1-ol, in which the hydroxyl group is on a carbon carrying two hydrogen atoms:
The secondary alcohol is propan-2-ol, in which the hydroxyl group is on the middle carbon, attached to two alkyl groups:
Preparation from methylmagnesium bromide. A Grignard reagent adds across the carbonyl group, its alkyl group going to the carbon and the to the oxygen, and hydrolysis of the adduct then gives the alcohol. The class of alcohol obtained is decided by the carbonyl compound chosen, and since the methyl group contributes one carbon, the other partner must supply two.
For the primary alcohol, propan-1-ol, the reagent is treated with an epoxide (ethylene oxide), which opens to lengthen the chain by two carbons:
For the secondary alcohol, propan-2-ol, the reagent is added to an aldehyde (ethanal), since an aldehyde always gives a secondary alcohol:
The reactions are carried out in dry ether, and the water for the hydrolysis is added only at the end, since a Grignard reagent is destroyed by moisture.
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:
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) The chain is best read backwards from the known product.
Wurtz's reaction joins two molecules of an alkyl halide through sodium in dry ether,
, so the product is a symmetrical alkane made of two identical halves. Splitting 2,3-dimethylbutane, , down the middle gives two isopropyl groups. Hence must be the isopropyl halide, and since the reagent used to make it was , is 2-bromopropane, .
is the major product of adding to , and by Markovnikov's rule the bromine goes to the carbon carrying fewer hydrogen atoms, which is the middle carbon of propene. Hence is propene, .
comes from on heating with alcoholic potassium hydroxide, which is dehydrohalogenation. A haloalkane that loses to give propene must be a bromopropane, and it cannot be 2-bromopropane, since that is itself and the question makes and different compounds. Hence is 1-bromopropane, .
The reactions are
The whole sequence is the standard way of converting a primary halide into the corresponding secondary halide, and it is worth noting that this change of position of the halogen is exactly what makes the final Wurtz product branched.
In summary: = 1-bromopropane, = propene, = 2-bromopropane.
(b) Hoffmann's method for separating a mixture of primary, secondary and tertiary amines.
The mixture is treated with diethyl oxalate, , and the three classes of amine behave quite differently towards it, which is what makes the separation possible.
A primary amine reacts with both ester groups and gives a solid dialkyl oxamide:
A secondary amine has only one hydrogen on the nitrogen, so it can react at one ester group only and gives a liquid dialkyl oxamic ester:
A tertiary amine has no hydrogen on the nitrogen at all and therefore does not react; it remains as the free amine in the mixture.
The products are now separated in two stages. The reaction mixture is first filtered, which removes the crystalline dialkyl oxamide from the primary amine. What is left, the oxamic ester, the unreacted tertiary amine and the ethanol set free in the reaction, is then separated by fractional distillation, in which the tertiary amine, being volatile and unchanged, distils over first and is collected, leaving the oily oxamic ester behind.
Each derivative is then hydrolysed with alkali to recover its amine:
The three amines are thus obtained separately. (Hinsberg's method, using benzenesulphonyl chloride, achieves the same separation on a different principle.)
Definitions.
(i) Activation energy. The activation energy is the minimum extra energy, over and above their average energy, that the reactant molecules must possess before they can cross the energy barrier and be converted into products. Only those collisions in which the colliding molecules together carry at least this energy are effective. It is the difference between the energy of the transition state and that of the reactants, and the greater it is, the slower the reaction.
(ii) Order of a reaction. The order of a reaction is the sum of the powers to which the concentration terms are raised in the experimentally determined rate law. It is found only by experiment, it may be zero, fractional or even negative, and it need have no connection with the coefficients of the balanced equation.
(iii) Molecularity of a reaction. The molecularity is the number of molecules, atoms or ions that take part in the single elementary step by which the reaction occurs, as written in its balanced equation. It is a theoretical quantity, always a small whole number, and it is never zero or fractional. For a reaction that proceeds in several steps, molecularity has meaning only for each individual step, while the observed order belongs to the slowest of them.
(iv) Rate law. The rate law (rate equation) is the experimentally established expression relating the rate of a reaction to the concentrations of the reactants, each raised to the appropriate power, as in . The constant is the rate constant, or specific reaction rate, being the rate when each concentration is unity.
Effect of temperature on the rate. The rate of nearly every reaction rises sharply with temperature, and as a rough rule the rate is doubled or trebled for every rise. The ratio of the rate constants at two temperatures differing by ten degrees is called the temperature coefficient,
The reason is not simply that the molecules move faster and collide more often, which would account for only a very small part of the increase. It is that the fraction of molecules possessing energy equal to or greater than the activation energy grows very rapidly with temperature, as the Maxwell distribution of molecular energies broadens and its maximum shifts to the right. This is expressed by the Arrhenius equation
or in the form used for calculation,
which shows that increases exponentially as rises, and that a reaction with a large is the more sensitive to a change of temperature.
Numerical. Let the rate law be . The orders are found by changing one concentration at a time.
Order with respect to . Compare experiments 1 and 2, in which is held at while is doubled:
Order with respect to . Compare experiments 2 and 3, in which is held at while is doubled:
The rate is therefore independent of , and the rate law is
so the reaction is first order overall.
The rate constant, from experiment 1,
The required rate, when and . Since does not appear in the rate law its value is irrelevant:
(i) Chemistry of blue vitriol.
Blue vitriol is copper sulphate pentahydrate, , a blue triclinic crystalline solid. It is prepared by dissolving copper oxide, hydroxide or carbonate in dilute sulphuric acid and crystallising the solution, and on the large scale by the slow action of dilute sulphuric acid on scrap copper in the presence of air:
Of its five molecules of water, four are co-ordinated to the copper and the fifth is held to the sulphate by hydrogen bonding, which is why they are lost in stages on heating:
The anhydrous salt is white, and because it turns blue again on taking up water it is used to test for the presence of water in a liquid. The crystals lose water slowly in dry air, that is they are efflorescent.
In solution it is acidic to litmus through hydrolysis. Adding ammonium hydroxide first precipitates a pale blue basic salt which then dissolves in excess to the deep blue cuprammonium ion, . Potassium iodide gives white cuprous iodide with liberation of iodine, and hydrogen sulphide gives black copper sulphide. With potassium cyanide, and with excess iodide, the copper is reduced from the to the state.
It is used as a fungicide (with lime as Bordeaux mixture), in electroplating and electrorefining of copper, in Fehling's solution, in dyeing as a mordant, and in the purification of water.
(iii) Hess's law of constant heat summation.
Statement: the total heat change accompanying a chemical reaction is the same whether the reaction takes place in one step or in several steps, provided the initial and the final states are the same.
It follows directly from the first law of thermodynamics, since enthalpy is a state function and depends only on the initial and final states, not on the path. If it were untrue, energy could be created by going one way round a cycle and back the other.
If a reaction takes to directly with a heat change , and also through the steps
with heat changes , and , then
Applications. The great value of the law is that it lets thermochemical equations be added, subtracted and multiplied like algebraic equations, so that heats of reaction which cannot be measured directly may be calculated from those which can.
It is used to find the heat of formation of a compound that cannot be made directly from its elements. Carbon monoxide is the standard case: burning carbon in a limited supply of oxygen always gives some carbon dioxide as well, so of cannot be measured, but it follows at once from the two combustions that can be:
Subtracting the second from the first,
It is used besides to calculate heats of reaction from heats of formation, since ; to determine lattice energies through the Born Haber cycle, and bond energies; to find the heat of transition between two allotropic forms; and to obtain the heat of hydration of a salt.