NEB Class 12 Science Chemistry Question Paper 2080 (Set A) Nepal

The reaction of phenol with chloroform ($CHCl_3$) and aqueous potassium hydroxide ($KOH$) to introduce a formyl group ortho to the hydroxyl group, yielding salicylaldehyde, is known as the Reimer-Tiemann reaction.

Carboxylic acids react with sodium bicarbonate ($NaHCO_3$) to evolve carbon dioxide gas with brisk effervescence, a reaction not typically shown by simpler alcohols or phenols (except highly nitrated ones), making it a standard diagnostic test.

Silver reacts with trace amounts of hydrogen sulfide ($H_2S$) present in the air to form a black layer of silver sulfide ($Ag_2S$), which causes tarnishing.

Lithopone is a white pigment consisting of a mixture of zinc sulfide ($ZnS$) and barium sulfate ($BaSO_4$), typically prepared by coprecipitation.

The electron-withdrawing inductive effect ($-I$ effect) increases the acidity of carboxylic acids. Fluorine is more electronegative than chlorine, and having two fluorine atoms exerts a stronger withdrawal effect than one. Thus, the acidity follows the order: $F_2CHCOOH > FCH_2COOH > ClCH_2COOH > CH_3COOH$.

Ethanol (ethyl alcohol) is commonly referred to as grain alcohol because it is frequently produced via the fermentation of grains like corn, wheat, or barley.

Nitrous acid ($HNO_2$) reacts with primary aliphatic amines to produce alcohol accompanied by the evolution of nitrogen gas bubbles. With secondary amines, it forms a yellow, oily nitrosamine compound, providing an easy distinction.

The specific rate constant ($k$) for a reaction is independent of the concentration of reactants or products and independent of elapsed time, but it changes significantly with a change in temperature according to the Arrhenius equation.

Given $pH = 13$, the concentration of $H^+$ ions is $[H^+] = 10^{-pH} = 10^{-13}\,\text{mol/L}$. In $1\,\text{mL}$ ($10^{-3}\,\text{L}$), the number of moles of $H^+$ ions is $10^{-13}\,\text{mol/L} \times 10^{-3}\,\text{L} = 10^{-16}\,\text{mol}$. The total number of $H^+$ ions is $10^{-16}\,\text{mol} \times 6.022 \times 10^{23}\,\text{ions/mol} = 6.022 \times 10^{7}\,\text{ions}$.

At normal equilibrium melting point, ice and water coexist at $T = 0^\circ\text{C} = 273.15\,\text{K}$. $\Delta S = \dfrac{\Delta H}{T} = \dfrac{6\,\text{KJ/mol}}{273.15\,\text{K}} \approx 0.02196\,\text{KJ K}^{-1}\text{mol}^{-1} \approx 0.022\,\text{KJ K}^{-1}\text{mol}^{-1}$.

Using the dilution equation $N_1V_1 = N_2V_2$: Here $N_1 = 2\,\text{N}$, $V_1 = 20\,\text{mL}$, and decinormal means $N_2 = 0.1\,\text{N}$. $2 \times 20 = 0.1 \times V_2 \implies V_2 = \dfrac{40}{0.1} = 400\,\text{mL}$. Volume of water to be added = Final volume ($V_2$) - Initial volume ($V_1$) = $400 - 20 = 380\,\text{mL}$.

i) Coupling reaction: Benzene diazonium chloride reacts with phenol in a mildly alkaline medium to yield p-hydroxyazobenzene (an orange dye): $C_6H_5N_2^+Cl^- + C_6H_5OH \rightarrow C_6H_5-N=N-C_6H_4-OH + HCl$

ii) Hydroboration oxidation: Propene reacts with diborane ($BH_3/THF$) followed by alkaline hydrogen peroxide ($H_2O_2/OH^-$) to yield propan-1-ol: $CH_3-CH=CH_2 \xrightarrow[2. H_2O_2, OH^-]{1. BH_3} CH_3-CH_2-CH_2-OH$

iii) Clemmensen's reduction: Ethanal is reduced with zinc amalgam ($Zn/Hg$) and concentrated $HCl$ to form ethane: $CH_3CHO + 4[H] \xrightarrow{Zn-Hg / conc. HCl} CH_3-CH_3 + H_2O$

iv) Fehling test: Aliphatic aldehydes like ethanal reduce Fehling's solution to form a red precipitate of cuprous oxide ($Cu_2O$): $CH_3CHO + 2Cu^{2+} + 5OH^- \rightarrow CH_3COO^- + Cu_2O\downarrow + 3H_2O$

v) Carbonylation reaction: Methanol reacts with carbon monoxide in the presence of a rhodium catalyst to produce acetic acid: $CH_3OH + CO \xrightarrow{\text{Rh catalyst}} CH_3COOH$

i) Since compound (A) with formula $C_4H_8O_2$ yields an acid (B) and an alcohol (C) upon hydrolysis, it must be an ester. Reduction of acid (B) with $LiAlH_4$ produces the exact same alcohol (C), meaning both the carboxylic acid segment and the alcohol portion contain equal number of carbon atoms (2 carbons each). Thus:

• Compound (A) is Ethyl ethanoate: $CH_3COOCH_2CH_3$
• Compound (B) is Ethanoic acid: $CH_3COOH$
• Compound (C) is Ethanol: $CH_3CH_2OH$

Reaction of Hydrolysis: $CH_3COOCH_2CH_3 + H_2O \xrightarrow{H^+} CH_3COOH + CH_3CH_2OH$

Reaction of Reduction of B: $CH_3COOH \xrightarrow[2. H_2O]{1. LiAlH_4} CH_3CH_2OH$

ii) When ester (A) is reduced with $LiAlH_4$, it undergoes reductive cleavage to produce two molecules of ethanol: $CH_3COOCH_2CH_3 \xrightarrow[2. H_2O]{1. LiAlH_4} 2 CH_3CH_2OH$

In Hoffmann's method, diethyl oxalate is added to the mixture of primary ($R-NH_2$), secondary ($R_2NH$), and tertiary ($R_3N$) amines:

1. Primary amine reacts to form a solid dialkyl oxamide: $(COOC_2H_5)_2 + 2R-NH_2 \rightarrow (CONHR)_2\downarrow + 2C_2H_5OH$
2. Secondary amine reacts to form a liquid oxamic ester: $(COOC_2H_5)_2 + R_2NH \rightarrow COONR_2-COOC_2H_5 + C_2H_5OH$
3. Tertiary amine does not react due to the absence of a replaceable hydrogen atom on nitrogen.

Separation Process:

• The unreacted tertiary amine ($R_3N$) is separated out by simple distillation because it has a lower boiling point.
• The residue containing the solid dialkyl oxamide and liquid oxamic ester is filtered. The solid oxamide is separated from the liquid ester.
• The separated dialkyl oxamide is heated with a strong alkali (like aqueous $KOH$) to recover the pure primary amine via distillation: $(CONHR)_2 + 2KOH \rightarrow (COOK)_2 + 2R-NH_2\uparrow$
• The liquid oxamic ester is similarly distilled with a strong alkali to recover the pure secondary amine: $COONR_2-COOC_2H_5 + 2KOH \rightarrow (COOK)_2 + R_2NH\uparrow + C_2H_5OH$

According to the electrochemical theory of rusting, a drop of water containing dissolved $O_2$ and $CO_2$ acts as a miniature electrochemical cell on the iron surface.

• At Anode: Iron undergoes oxidation: $Fe \rightarrow Fe^{2+} + 2e^-$
• At Cathode: Electrons migrate through the metal to another point where oxygen is reduced in the presence of $H^+$ ions (from $H_2CO_3$): $O_2 + 4H^+ + 4e^- \rightarrow 2H_2O$
• Overall reaction: $2Fe + O_2 + 4H^+ \rightarrow 2Fe^{2+} + 2H_2O$

The $Fe^{2+}$ ions are further oxidized by atmospheric oxygen to form hydrated ferric oxide, which is rust: $4Fe^{2+} + O_2 + 4H_2O + 2xH_2O \rightarrow 2Fe_2O_3\cdot xH_2O\downarrow + 8H^+$

Prevention Processes: Rusting can be prevented by methods like Galvanization (coating iron with a protective layer of zinc), sacrificial protection, cathodic protection, painting, or electroplating.

i) Transition metal: A transition metal is defined as an element that forms at least one stable ion containing a partially filled or incomplete d-subshell.

ii) a) Coloured compounds: Transition metals possess partially filled d-orbitals. When ligands approach, the degenerate d-orbitals split into groups of different energy. Electrons can absorb visible light to jump from a lower energy d-orbital to a higher one ($d-d$ transition), emitting the complementary colour. b) Zinc as non-typical: Zinc ($Zn$) has the electronic configuration $[Ar]3d^{10}4s^2$. Neither in its elemental ground state nor in its common stable oxidation state ($Zn^{2+}: [Ar]3d^{10}$) does it have a partially filled d-subshell. Therefore, it does not display typical characteristics like colored complexes or variable oxidation states.

iii) d-orbital splitting in Octahedral Field: In an octahedral field, the five degenerate d-orbitals split into two sets due to ligand repulsion along the axes:

• Lower energy triply degenerate set: $t_{2g}$ ($d_{xy}, d_{yz}, d_{zx}$)
• Higher energy doubly degenerate set: $e_g$ ($d_{x^2-y^2}, d_{z^2}$)

i) Aluminum has a lower standard reduction potential ($-1.67\,\text{V}$), so it undergoes oxidation at the anode. Silver has a higher potential ($+0.80\,\text{V}$), acting as the cathode where reduction happens. Cell notation:

Anode is $Al$ electrode, Cathode is $Ag$ electrode.

ii) Half-reactions:

• Oxidation Half Reaction (at Anode): $Al(s) \rightarrow Al^{3+}(aq) + 3e^-$
• Reduction Half Reaction (at Cathode): $3Ag^+(aq) + 3e^- \rightarrow 3Ag(s)$

iii) E.M.F. Calculation:

i) Enthalpy of formation: It is the change in enthalpy when one mole of a substance is formed directly from its constituent elements in their standard reference states under standard conditions.

ii) Balanced thermochemical equations:

1. $H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta H_f^circ = -286\,\text{KJ/mol}$
2. $C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta H_f^circ = -394\,\text{KJ/mol}$
3. $3C(s) + 3H_2(g) + \frac{1}{2}O_2(g) \rightarrow C_3H_5OH(l) \quad \Delta H_f^circ = -2010\,\text{KJ/mol}$ (Note: Formula in scanning text $C_3H_5OH$ is evaluated based on normal balanced structure layout or treated as is)

iii) Enthalpy of combustion calculation: The chemical equation for complete combustion of $C_3H_5OH(l)$ is:

i) Relation between pH and pOH: We know that the ionic product of water ($K_w$) at $25^circ\text{C}$ is:

Taking negative logarithm ($-\log$) on both sides:

Since $-\log[H^+] = pH$ and $-\log[OH^-] = pOH$, we get:

ii) Solubility Product Calculation: Let the volume of the solution be $1\,\text{L}$. When $0.2\,\text{mol}$ of $HCl$ is added, it dissociates completely providing $[Cl^-] = 0.2\,\text{M}$ (neglecting the minute contribution from $AgCl$ due to the common ion effect).

This represents the remaining concentration of $Ag^+$ in solution. In a pure saturated solution before adding $HCl$, the solubility was $S_0 = \sqrt{K_{sp}} = \sqrt{1.8 \times 10^{-10}} \approx 1.34 \times 10^{-5}\,\text{mol/L}$.

The amount of $AgCl$ precipitated equals the loss in dissolved ions:

A sequential conversion path linking these compounds is as follows:

1. Conversion of 1-Chloropropane to Propan-1-ol: $CH_3CH_2CH_2Cl + KOH (aq) \xrightarrow{\Delta} CH_3CH_2CH_2OH + KCl$
2. Conversion of Propan-1-ol to Propanoic acid: $CH_3CH_2CH_2OH \xrightarrow{K_2Cr_2O_7 / dil.H_2SO_4} CH_3CH_2COOH$
3. Conversion of Propanoic acid to Propanamide: $CH_3CH_2COOH + NH_3 \xrightarrow{\Delta} CH_3CH_2CONH_{2} + H_2O$
4. Conversion of Propanamide to Ethylamine (Hoffmann Bromamide Degradation): $CH_3CH_2CONH_2 + Br_2 + 4KOH \rightarrow CH_3CH_2NH_2 + K_2CO_3 + 2KBr + 2H_2O$

i) Molecular formula: $HCOOH$ IUPAC Name: Methanoic acid

ii) a) From Chloroform: Chloroform is heated with aqueous $KOH$ solution to give potassium formate, which on acidification gives formic acid: $CHCl_3 + 4KOH (aq) \rightarrow HCOOK + 3KCl + 2H_2O \xrightarrow{HCl} HCOOH$ b) From Ethane-1,2-dioic acid (Oxalic acid): Oxalic acid is heated with glycerol at around $110^\circ\text{C}$ to give formic acid via decarboxylation: $(COOH)_2 \xrightarrow{\text{Glycerol, } 110^\circ\text{C}} HCOOH + CO_2\uparrow$

iii) Distinguishing Test: Formic acid contains a formyl hydrogen atom, allowing it to reduce Tollens' reagent to a shiny silver mirror, whereas acetic acid does not react: $HCOOH + 2[Ag(NH_3)_2]^+ + 2OH^- \rightarrow 2Ag\downarrow (silver\;mirror) + CO_2 + 4NH_3 + 2H_2O$

iv) Acidity explanation: In acetic acid ($CH_3COOH$), the methyl group ($CH_3-$) is an electron-donating group due to its $+I$ effect, which increases electron density on the carboxyl group and destabilizes the acetate ion. In formic acid ($HCOOH$), there is only a hydrogen atom which lacks any significant $+I$ effect, leaving the formate anion more stable and making formic acid more acidic.

i) Distinction:

• Normality Factor ($f$): It is a correction factor defined as the ratio of the actual weight of the solute taken to the theoretical weight required to prepare a solution of exact desired normality ($f = \frac{\text{Actual Normality}}{\text{Desired Normality}}$).
• Titration Error: It is the difference between the experimental end-point (where indicator changes color) and the theoretical equivalence point (where stoichiometric reaction finishes) in a volumetric titration.

ii) a) Actual Strength & Normality Factor: Equivalent mass of $Na_2CO_3 = \frac{106}{2} = 53\,\text{g/eq}$. Theoretical mass required for $250\,\text{mL}$ of decinormal ($0.1\,\text{N}$) solution:

Normality factor ($f$) = $\frac{\text{Actual Mass}}{\text{Theoretical Mass}} = \frac{1.31}{1.325} \approx 0.9887$ Actual Normality = $0.1 \times 0.9887 = 0.09887\,\text{N}$.

b) Volume to be evaporated: To convert $250\,\text{mL}$ of $0.09887\,\text{N}$ solution to exactly $1\,\text{N}$ solution:

Volume of water to be evaporated = $250 - 24.72 = 225.28\,\text{mL}$.

c) Mass of Oxalic acid crystal ($H_2C_2O_4 \cdot 2H_2O$): Equivalent mass of oxalic acid crystal = $\frac{126}{2} = 63\,\text{g/eq}$. Number of gram equivalents of $Na_2CO_3$ present = $\frac{1.31}{53} \approx 0.02472\,\text{eq}$. For complete neutralization, equivalents of Oxalic Acid = Equivalents of $Na_2CO_3 = 0.02472\,\text{eq}$.

iv) Application in Pharmaceutical Industry: Titration is widely applied to determine the purity and exact concentration of active pharmaceutical ingredients (APIs) in raw materials and finished tablets/formulations, ensuring regulatory quality standards.

i) Half-life period: It is the time required for the reactant concentration to decrease to exactly half of its initial value.

ii) Independence for 1st order: The integrated rate equation for a first-order reaction is $k = \frac{2.303}{t}\log\frac{[A]_0}{[A]}$. At $t = t_{1/2}$, $[A] = \frac{[A]_0}{2}$. Substituting this gives $k = \frac{2.303}{t_{1/2}}\log(2) \implies t_{1/2} = \frac{0.693}{k}$. Since initial concentration $[A]_0$ cancels out completely, $t_{1/2}$ is independent of it.

iii) Biological Application: It is used extensively in radiocarbon dating to determine the age of organic biological fossils, and in pharmacokinetics to estimate how long a drug remains active inside an organism.

iv) a) Rate Constant & Half-life: $a = 100$, $x = 60$, remaining $(a-x) = 40$ at $t = 30\,\text{min}$.

b) Time for 99% completion: $x = 99$, remaining $(a-x) = 1$.

v) Surface Area Effect Example: Finely powdered calcium carbonate ($CaCO_3$) reacts much faster with dilute hydrochloric acid ($HCl$) to release $CO_2$ gas than a single solid large marble chip of the same weight, due to the increased surface area available for collision.

i) An alcohol with molecular formula $C_3H_8O$ that gives a blue color in the Victor Meyer's test must be a secondary alcohol. Therefore:

• Structural Formula: $CH_3-CH(OH)-CH_3$
• IUPAC Name: Propan-2-ol

ii) Victor-Meyer's Test Reactions:

1. $CH_3CH(OH)CH_3 + PI_3 \rightarrow CH_3CH(I)CH_3$
2. $CH_3CH(I)CH_3 + AgNO_2 \rightarrow CH_3CH(NO_2)CH_3$
3. $CH_3CH(NO_2)CH_3 + HNO_2 \rightarrow (CH_3)_2C(NO_2)NO$ (Pseudonitrole)
4. $(CH_3)_2C(NO_2)NO + KOH \rightarrow \text{Blue colored solution}$

iii) Preparation from $CH_3MgBr$ (Grignard reagent): Reacting methyl magnesium bromide with ethanal ($CH_3CHO$) followed by acid hydrolysis yields propan-2-ol: $CH_3CHO + CH_3MgBr \rightarrow CH_3CH(OMgBr)CH_3 \xrightarrow{H^+/H_2O} CH_3CH(OH)CH_3 + Mg(OH)Br$

iv) Oxidation: When propan-2-ol is oxidized with an oxidizing agent like acidified $K_2Cr_2O_7$, it forms propanone (acetone): $CH_3CH(OH)CH_3 \xrightarrow{[O]} CH_3COCH_3 + H_2O$

v) Conversion into Propene: Heating propan-2-ol with concentrated sulfuric acid ($H_2SO_4$) at $170^\circ\text{C}$ causes dehydration to form propene: $CH_3CH(OH)CH_3 \xrightarrow{\text{conc. } H_2SO_4, 170^\circ\text{C}} CH_3-CH=CH_2 + H_2O$

i) Preparation from Toluene (Etard's Reaction): Toluene is oxidized using chromyl chloride ($CrO_2Cl_2$) in $CS_2$ followed by hydrolysis to form benzaldehyde: $C_6H_5CH_3 + 2CrO_2Cl_2 \rightarrow C_6H_5CH(OCrOHCl_2)_2 \xrightarrow{H_3O^+} C_6H_5CHO$

ii) a) Conversion to Benzyl chloride: First, reduce benzaldehyde to benzyl alcohol using $LiAlH_4$, then treat it with $PCl_5$: $C_6H_5CHO \xrightarrow{LiAlH_4} C_6H_5CH_2OH \xrightarrow{PCl_5} C_6H_5CH_2Cl$

b) Conversion to Toluene (Clemmensen Reduction): Benzaldehyde can be reduced directly to toluene using zinc amalgam and concentrated hydrochloric acid: $C_6H_5CHO + 4[H] \xrightarrow{Zn-Hg / conc. HCl} C_6H_5CH_3 + H_2O$

c) Conversion to Cinnamic acid (Perkin's Reaction): Heating benzaldehyde with acetic anhydride in the presence of sodium acetate yields cinnamic acid: $C_6H_5CHO + (CH_3CO)_2O \xrightarrow{CH_3COONa, \Delta} C_6H_5CH=CHCOOH + CH_3COOH$