Maharashtra Board HSC Chemistry
July 2022 Question Paper - Full Detailed Solutions
SECTION − A
Q.1Select and write the correct answer for the following multiple choice type of questions:
(i) Cyclohexyl chloride + Mg \(\xrightarrow{\text{Dry ether}}\) A \(\xrightarrow{\text{H}_2\text{O}}\) B. The product 'B' is _______.
Answer: (a)
Explanation:
Step 1: Reaction of Cyclohexyl chloride with Mg in dry ether forms Cyclohexyl magnesium chloride (Grignard reagent 'A').
Step 2: Hydrolysis of Grignard reagent (A) yields the corresponding alkane.
Reaction: \(\text{C}_6\text{H}_{11}\text{Cl} + \text{Mg} \xrightarrow{\text{ether}} \text{C}_6\text{H}_{11}\text{MgCl} \xrightarrow{\text{H}_2\text{O}} \text{C}_6\text{H}_{12} (\text{Cyclohexane}) + \text{Mg(OH)Cl}\).
Step 1: Reaction of Cyclohexyl chloride with Mg in dry ether forms Cyclohexyl magnesium chloride (Grignard reagent 'A').
Step 2: Hydrolysis of Grignard reagent (A) yields the corresponding alkane.
Reaction: \(\text{C}_6\text{H}_{11}\text{Cl} + \text{Mg} \xrightarrow{\text{ether}} \text{C}_6\text{H}_{11}\text{MgCl} \xrightarrow{\text{H}_2\text{O}} \text{C}_6\text{H}_{12} (\text{Cyclohexane}) + \text{Mg(OH)Cl}\).
(ii) General electronic configuration of 3d series of ‘d’ block elements is _______.
Answer: (a)
\([\text{Ar}] 3d^{1-10} 4s^{2}\) (Note: While some elements like Cr and Cu have \(4s^1\), the most standard generalized option provided is (a) utilizing Argon [Ar] core).
(iii) Correct IUPAC name of tert-butyl alcohol is _______.
Answer: (c)
2-Methyl propan-2-ol
Structure: \(\text{(CH}_3)_3\text{C-OH}\). The longest chain is propane (3 carbons), with a methyl group and OH group at carbon 2.
Structure: \(\text{(CH}_3)_3\text{C-OH}\). The longest chain is propane (3 carbons), with a methyl group and OH group at carbon 2.
(iv) The standard emf of the following cell at 298K is _______.
\(\text{Zn(s)}|\text{Zn}^{+2}(1\text{M})||\text{Cr}^{+3}(0.1\text{M})|\text{Cr(s)}\)
\(\text{E}^\circ_{\text{Zn}} = -0.76\text{V}, \text{E}^\circ_{\text{Cr}} = -0.74\text{V}\)
Answer: (b)
+0.02V
Calculation: \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\)
\(E^\circ_{\text{cell}} = E^\circ_{\text{Cr}} - E^\circ_{\text{Zn}} = (-0.74\text{V}) - (-0.76\text{V}) = +0.02\text{V}\).
Calculation: \(E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}}\)
\(E^\circ_{\text{cell}} = E^\circ_{\text{Cr}} - E^\circ_{\text{Zn}} = (-0.74\text{V}) - (-0.76\text{V}) = +0.02\text{V}\).
(v) In the following oxyacid, chlorine has +7 oxidation state:
Answer: (d)
\(\text{HClO}_4\) (Perchloric acid)
Calculation: \(1(+1) + x + 4(-2) = 0 \Rightarrow 1 + x - 8 = 0 \Rightarrow x = +7\).
Calculation: \(1(+1) + x + 4(-2) = 0 \Rightarrow 1 + x - 8 = 0 \Rightarrow x = +7\).
(vi) The work done during isothermal irreversible expansion of 2 moles of helium from 2dm³ to 4 dm³ at 1 bar pressure and at 298K is _______.
Answer: (d)
–0.2 kJ
Calculation: Work \(W = -P_{\text{ext}} \Delta V\)
\(W = -1 \text{ bar} \times (4 - 2) \text{ dm}^3 = -2 \text{ bar dm}^3\)
\(1 \text{ bar dm}^3 = 100 \text{ J}\)
\(W = -200 \text{ J} = -0.2 \text{ kJ}\).
Calculation: Work \(W = -P_{\text{ext}} \Delta V\)
\(W = -1 \text{ bar} \times (4 - 2) \text{ dm}^3 = -2 \text{ bar dm}^3\)
\(1 \text{ bar dm}^3 = 100 \text{ J}\)
\(W = -200 \text{ J} = -0.2 \text{ kJ}\).
(vii) The correct relation between edge length and radius of an atom in simple cubic lattice is _______.
Answer: (c)
\(a = 2r\)
(viii) Lactose on hydrolysis gives _______.
Answer: (a)
galactose + glucose
(ix) ZWT in green chemistry stands for:
Answer: (b)
zero waste technology
(x) The most basic amine amongst the following is _______.
Answer: (b)
\(\text{(CH}_3)_2\text{NH}\)
Reason: In aqueous solution, secondary amines are more basic than tertiary and primary due to a combination of inductive effect, solvation effect, and steric hindrance. For methyl substituted amines: \(2^\circ > 1^\circ > 3^\circ > \text{NH}_3\).
Reason: In aqueous solution, secondary amines are more basic than tertiary and primary due to a combination of inductive effect, solvation effect, and steric hindrance. For methyl substituted amines: \(2^\circ > 1^\circ > 3^\circ > \text{NH}_3\).
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Q.2Answer the following questions:
(i) Write relation between molar conductivity and conductivity of solution.
\(\Lambda_m = \frac{1000k}{C}\)
Where:
\(\Lambda_m\) is molar conductivity (S cm² mol⁻¹)
\(k\) is conductivity (S cm⁻¹)
\(C\) is concentration in mol L⁻¹ (Molarity)
Where:
\(\Lambda_m\) is molar conductivity (S cm² mol⁻¹)
\(k\) is conductivity (S cm⁻¹)
\(C\) is concentration in mol L⁻¹ (Molarity)
(ii) Calculate effective atomic number of Co⁺³ in [Co(NH₃)₆]³⁺ complex.
Formula: \(\text{EAN} = Z - X + Y\)
\(Z\) (Atomic no. of Co) = 27
\(X\) (Oxidation state) = +3 (So, electrons lost = 3)
\(Y\) (Electrons donated by ligands) = \(6 \times 2 = 12\)
\(\text{EAN} = 27 - 3 + 12 = 36\).
\(Z\) (Atomic no. of Co) = 27
\(X\) (Oxidation state) = +3 (So, electrons lost = 3)
\(Y\) (Electrons donated by ligands) = \(6 \times 2 = 12\)
\(\text{EAN} = 27 - 3 + 12 = 36\).
(iii) Write the name of reaction during conversion of phenol to salicylic acid.
Kolbe’s reaction (or Kolbe-Schmitt reaction).
(iv) Write the IUPAC name of α-methylpropionic acid.
Structure: \(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{COOH}\)
IUPAC Name: 2-Methylpropanoic acid.
IUPAC Name: 2-Methylpropanoic acid.
(v) Write the formula of Hinsberg’s reagent.
\(\text{C}_6\text{H}_5\text{SO}_2\text{Cl}\) (Benzene sulfonyl chloride).
(vi) Write the name of monomer used for preparation of Nylon 6.
\(\varepsilon\)-Caprolactam.
(vii) Write cell representation of standard hydrogen electrode.
\(\text{Pt(s)} \mid \text{H}_2(\text{g, 1 atm}) \mid \text{H}^+(\text{aq, 1 M})\).
(viii) Write chemical composition of Zieglar-Natta catalyst.
It is a mixture of Titanium tetrachloride (\(\text{TiCl}_4\)) and Triethylaluminium ([\(\text{Al(C}_2\text{H}_5)_3\)]).
SECTION − B
Q.3Define: (i) Osmotic pressure (ii) Ebullioscopic constant
(i) Osmotic pressure: It is defined as the excess hydrostatic pressure that must be applied to the solution side to just stop the flow of solvent molecules into the solution through a semipermeable membrane.
(ii) Ebullioscopic constant (\(K_b\)): It is defined as the elevation in boiling point produced when 1 mole of non-volatile solute is dissolved in 1 kg (1000 g) of the solvent.
(ii) Ebullioscopic constant (\(K_b\)): It is defined as the elevation in boiling point produced when 1 mole of non-volatile solute is dissolved in 1 kg (1000 g) of the solvent.
Q.4The pH of solution is 3.12. Calculate the concentration of H₃O⁺ ion.
Given: \(\text{pH} = 3.12\)
Formula: \(\text{pH} = -\log_{10}[\text{H}_3\text{O}^+]\)
\(\therefore [\text{H}_3\text{O}^+] = \text{antilog}(-\text{pH})\)
\([\text{H}_3\text{O}^+] = \text{antilog}(-3.12)\)
\([\text{H}_3\text{O}^+] = \text{antilog}(\bar{4}.88)\)
From antilog table for 0.88: Value is 7586.
Answer: \([\text{H}_3\text{O}^+] = 7.586 \times 10^{-4} \text{ M}\).
Formula: \(\text{pH} = -\log_{10}[\text{H}_3\text{O}^+]\)
\(\therefore [\text{H}_3\text{O}^+] = \text{antilog}(-\text{pH})\)
\([\text{H}_3\text{O}^+] = \text{antilog}(-3.12)\)
\([\text{H}_3\text{O}^+] = \text{antilog}(\bar{4}.88)\)
From antilog table for 0.88: Value is 7586.
Answer: \([\text{H}_3\text{O}^+] = 7.586 \times 10^{-4} \text{ M}\).
Q.5State Kohlrausch Law of independent migration of ions. Write one application.
Statement: Kohlrausch law states that at infinite dilution, each ion migrates independently of its co-ion and makes its own contribution to the total molar conductivity of an electrolyte irrespective of the nature of the other ion with which it is associated.
\(\Lambda_0 = \lambda_+^0 + \lambda_-^0\)
Application: It is used to calculate the molar conductivity of weak electrolytes at infinite dilution, which cannot be determined experimentally by extrapolation.
\(\Lambda_0 = \lambda_+^0 + \lambda_-^0\)
Application: It is used to calculate the molar conductivity of weak electrolytes at infinite dilution, which cannot be determined experimentally by extrapolation.
Q.6Distinguish between Schottky and Frenkel defect.
| Schottky Defect | Frenkel Defect |
|---|---|
| It is produced due to the absence of an equal number of cations and anions from lattice sites. | It is produced when an ion (usually cation) leaves its lattice site and occupies an interstitial site. |
| Density of the crystal decreases. | Density of the crystal remains constant. |
| Usually shown by ionic compounds with high coordination numbers (e.g., NaCl, KCl). | Usually shown by ionic compounds with low coordination numbers (e.g., AgCl, ZnS). |
Q.7Derive the relationship between ∆H and ∆U for gas phase reactions.
Derivation:
1. Enthalpy is defined as \(H = U + PV\).
2. For a change in state at constant pressure:
\(\Delta H = \Delta U + P\Delta V\) ...(1)
3. For an ideal gas reaction, \(PV = nRT\).
Let \(n_1\) be moles of gaseous reactants and \(n_2\) be moles of gaseous products.
\(PV_1 = n_1RT\) and \(PV_2 = n_2RT\).
\(P\Delta V = P(V_2 - V_1) = PV_2 - PV_1\)
\(P\Delta V = n_2RT - n_1RT = (n_2 - n_1)RT = \Delta n_g RT\).
4. Substituting this value in equation (1):
\(\Delta H = \Delta U + \Delta n_g RT\)
1. Enthalpy is defined as \(H = U + PV\).
2. For a change in state at constant pressure:
\(\Delta H = \Delta U + P\Delta V\) ...(1)
3. For an ideal gas reaction, \(PV = nRT\).
Let \(n_1\) be moles of gaseous reactants and \(n_2\) be moles of gaseous products.
\(PV_1 = n_1RT\) and \(PV_2 = n_2RT\).
\(P\Delta V = P(V_2 - V_1) = PV_2 - PV_1\)
\(P\Delta V = n_2RT - n_1RT = (n_2 - n_1)RT = \Delta n_g RT\).
4. Substituting this value in equation (1):
\(\Delta H = \Delta U + \Delta n_g RT\)
Q.8What is the action of chlorine on the following: (i) NH₃ (excess) (ii) phosphorous?
(i) Action on Excess Ammonia:
When chlorine reacts with excess ammonia, ammonium chloride and nitrogen gas are formed.
\(8\text{NH}_3 \text{ (excess)} + 3\text{Cl}_2 \rightarrow 6\text{NH}_4\text{Cl} + \text{N}_2\)
(ii) Action on Phosphorous:
Chlorine reacts with phosphorus to form phosphorus trichloride (limited Cl₂) or phosphorus pentachloride (excess Cl₂).
\(\text{P}_4 + 6\text{Cl}_2 \rightarrow 4\text{PCl}_3\)
(or \(\text{P}_4 + 10\text{Cl}_2 \rightarrow 4\text{PCl}_5\)).
When chlorine reacts with excess ammonia, ammonium chloride and nitrogen gas are formed.
\(8\text{NH}_3 \text{ (excess)} + 3\text{Cl}_2 \rightarrow 6\text{NH}_4\text{Cl} + \text{N}_2\)
(ii) Action on Phosphorous:
Chlorine reacts with phosphorus to form phosphorus trichloride (limited Cl₂) or phosphorus pentachloride (excess Cl₂).
\(\text{P}_4 + 6\text{Cl}_2 \rightarrow 4\text{PCl}_3\)
(or \(\text{P}_4 + 10\text{Cl}_2 \rightarrow 4\text{PCl}_5\)).
Q.9Write the molecular formula of the following minerals: (i) chalcopyrite (ii) calamine
(i) Chalcopyrite: \(\text{CuFeS}_2\)
(ii) Calamine: \(\text{ZnCO}_3\)
(ii) Calamine: \(\text{ZnCO}_3\)
Q.10Show that time required for 99.9% completion of a first order reaction is three times the time required for 90% completion.
For first order reaction: \(t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}\)
Case 1: 99.9% completion
\([A]_0 = 100\), \([A]_t = 100 - 99.9 = 0.1\)
\(t_{99.9\%} = \frac{2.303}{k} \log \frac{100}{0.1} = \frac{2.303}{k} \log(1000)\)
\(t_{99.9\%} = \frac{2.303}{k} \times 3\) ...(1)
Case 2: 90% completion
\([A]_0 = 100\), \([A]_t = 100 - 90 = 10\)
\(t_{90\%} = \frac{2.303}{k} \log \frac{100}{10} = \frac{2.303}{k} \log(10)\)
\(t_{90\%} = \frac{2.303}{k} \times 1\) ...(2)
Taking ratio of (1) and (2):
\(\frac{t_{99.9\%}}{t_{90\%}} = \frac{3 \times (2.303/k)}{1 \times (2.303/k)} = 3\)
\(\therefore t_{99.9\%} = 3 \times t_{90\%}\). (Hence Proved)
Case 1: 99.9% completion
\([A]_0 = 100\), \([A]_t = 100 - 99.9 = 0.1\)
\(t_{99.9\%} = \frac{2.303}{k} \log \frac{100}{0.1} = \frac{2.303}{k} \log(1000)\)
\(t_{99.9\%} = \frac{2.303}{k} \times 3\) ...(1)
Case 2: 90% completion
\([A]_0 = 100\), \([A]_t = 100 - 90 = 10\)
\(t_{90\%} = \frac{2.303}{k} \log \frac{100}{10} = \frac{2.303}{k} \log(10)\)
\(t_{90\%} = \frac{2.303}{k} \times 1\) ...(2)
Taking ratio of (1) and (2):
\(\frac{t_{99.9\%}}{t_{90\%}} = \frac{3 \times (2.303/k)}{1 \times (2.303/k)} = 3\)
\(\therefore t_{99.9\%} = 3 \times t_{90\%}\). (Hence Proved)
Q.11Convert ethyl bromide to: (i) ethyl iodide (ii) ethyl fluoride
(i) Ethyl iodide (Finkelstein Reaction):
\(\text{C}_2\text{H}_5\text{Br} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{C}_2\text{H}_5\text{I} + \text{NaBr}\)
(ii) Ethyl fluoride (Swarts Reaction):
\(\text{C}_2\text{H}_5\text{Br} + \text{AgF} \rightarrow \text{C}_2\text{H}_5\text{F} + \text{AgBr}\)
(Note: Other metal fluorides like \(\text{Hg}_2\text{F}_2, \text{CoF}_2\) can also be used).
\(\text{C}_2\text{H}_5\text{Br} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{C}_2\text{H}_5\text{I} + \text{NaBr}\)
(ii) Ethyl fluoride (Swarts Reaction):
\(\text{C}_2\text{H}_5\text{Br} + \text{AgF} \rightarrow \text{C}_2\text{H}_5\text{F} + \text{AgBr}\)
(Note: Other metal fluorides like \(\text{Hg}_2\text{F}_2, \text{CoF}_2\) can also be used).
Q.12Explain linkage isomerism in complexes with one example.
Definition: Linkage isomerism arises in coordination compounds containing ambidentate ligands (ligands that can coordinate through two different atoms).
Example:
The nitrite ion (\(\text{NO}_2^-\)) can bind to the central metal via nitrogen or oxygen.
1. \([Co(NH_3)_5(NO_2)]^{2+}\) - Nitro complex (yellow), linked through N.
2. \([Co(NH_3)_5(ONO)]^{2+}\) - Nitrito complex (red), linked through O.
Example:
The nitrite ion (\(\text{NO}_2^-\)) can bind to the central metal via nitrogen or oxygen.
1. \([Co(NH_3)_5(NO_2)]^{2+}\) - Nitro complex (yellow), linked through N.
2. \([Co(NH_3)_5(ONO)]^{2+}\) - Nitrito complex (red), linked through O.
Q.13What is the action of the following on carboxylic acid: (i) SOCl₂ (ii) P₂O₅?
(i) Action of Thionyl Chloride (SOCl₂):
It forms acyl chloride (acid chloride).
\(\text{R-COOH} + \text{SOCl}_2 \rightarrow \text{R-COCl} + \text{SO}_2 \uparrow + \text{HCl} \uparrow\)
(ii) Action of Phosphorus Pentoxide (P₂O₅):
It acts as a dehydrating agent to form acid anhydride.
\(2\text{R-COOH} \xrightarrow{\text{P}_2\text{O}_5, \Delta} (\text{RCO})_2\text{O} + \text{H}_2\text{O}\)
It forms acyl chloride (acid chloride).
\(\text{R-COOH} + \text{SOCl}_2 \rightarrow \text{R-COCl} + \text{SO}_2 \uparrow + \text{HCl} \uparrow\)
(ii) Action of Phosphorus Pentoxide (P₂O₅):
It acts as a dehydrating agent to form acid anhydride.
\(2\text{R-COOH} \xrightarrow{\text{P}_2\text{O}_5, \Delta} (\text{RCO})_2\text{O} + \text{H}_2\text{O}\)
Q.14Write balanced chemical reactions of the following reagents on carbolic acid: (i) Br₂ water (ii) Concentrated HNO₃
Carbolic acid is Phenol.
(i) Action of Bromine water:
Phenol reacts with bromine water to give a white precipitate of 2,4,6-Tribromophenol.
\(\text{Phenol} + 3\text{Br}_2(\text{aq}) \rightarrow 2,4,6\text{-Tribromophenol} + 3\text{HBr}\)
(ii) Action of Conc. Nitric Acid:
Phenol reacts with concentrated nitric acid in the presence of conc. sulfuric acid to form 2,4,6-Trinitrophenol (Picric acid).
\(\text{Phenol} + 3\text{HNO}_3 (\text{conc}) \xrightarrow{\text{conc. H}_2\text{SO}_4} 2,4,6\text{-Trinitrophenol} + 3\text{H}_2\text{O}\)
(i) Action of Bromine water:
Phenol reacts with bromine water to give a white precipitate of 2,4,6-Tribromophenol.
\(\text{Phenol} + 3\text{Br}_2(\text{aq}) \rightarrow 2,4,6\text{-Tribromophenol} + 3\text{HBr}\)
(ii) Action of Conc. Nitric Acid:
Phenol reacts with concentrated nitric acid in the presence of conc. sulfuric acid to form 2,4,6-Trinitrophenol (Picric acid).
\(\text{Phenol} + 3\text{HNO}_3 (\text{conc}) \xrightarrow{\text{conc. H}_2\text{SO}_4} 2,4,6\text{-Trinitrophenol} + 3\text{H}_2\text{O}\)
SECTION − C
Q.15Write a note on ‘aldol’ condensation.
Aldol Condensation:
Aldehydes or ketones containing at least one \(\alpha\)-hydrogen atom undergo a reaction in the presence of a dilute alkali (like NaOH, KOH, or \(\text{Na}_2\text{CO}_3\)) as catalyst to form \(\beta\)-hydroxy aldehydes (aldol) or \(\beta\)-hydroxy ketones (ketol). This reaction is called Aldol condensation.
General Reaction (for Acetaldehyde):
\(2\text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{-CH(OH)-CH}_2\text{-CHO}\) (3-Hydroxybutanal / Aldol)
Upon heating, the aldol loses water to form an \(\alpha,\beta\)-unsaturated aldehyde:
\(\text{CH}_3\text{-CH(OH)-CH}_2\text{-CHO} \xrightarrow{\Delta} \text{CH}_3\text{-CH=CH-CHO} + \text{H}_2\text{O}\) (But-2-enal)
Aldehydes or ketones containing at least one \(\alpha\)-hydrogen atom undergo a reaction in the presence of a dilute alkali (like NaOH, KOH, or \(\text{Na}_2\text{CO}_3\)) as catalyst to form \(\beta\)-hydroxy aldehydes (aldol) or \(\beta\)-hydroxy ketones (ketol). This reaction is called Aldol condensation.
General Reaction (for Acetaldehyde):
\(2\text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{-CH(OH)-CH}_2\text{-CHO}\) (3-Hydroxybutanal / Aldol)
Upon heating, the aldol loses water to form an \(\alpha,\beta\)-unsaturated aldehyde:
\(\text{CH}_3\text{-CH(OH)-CH}_2\text{-CHO} \xrightarrow{\Delta} \text{CH}_3\text{-CH=CH-CHO} + \text{H}_2\text{O}\) (But-2-enal)
Q.16What is a Lanthanoid contraction? Write similarities between lanthanoids and actinoids.
Lanthanoid Contraction:
The steady decrease in the atomic and ionic radii of lanthanoid elements with increasing atomic number is called lanthanoid contraction. This is due to the poor shielding effect of 4f electrons.
Similarities between Lanthanoids and Actinoids:
1. Both series involve the filling of f-orbitals (4f for lanthanoids, 5f for actinoids).
2. Both show a common oxidation state of +3.
3. Both are electropositive metals and act as strong reducing agents.
4. Both show contraction in atomic and ionic sizes (Lanthanoid contraction and Actinoid contraction).
The steady decrease in the atomic and ionic radii of lanthanoid elements with increasing atomic number is called lanthanoid contraction. This is due to the poor shielding effect of 4f electrons.
Similarities between Lanthanoids and Actinoids:
1. Both series involve the filling of f-orbitals (4f for lanthanoids, 5f for actinoids).
2. Both show a common oxidation state of +3.
3. Both are electropositive metals and act as strong reducing agents.
4. Both show contraction in atomic and ionic sizes (Lanthanoid contraction and Actinoid contraction).
Q.17Calculate the standard enthalpy of formation of CH₃–OH, if standard heat of combustion of methyl alcohol is –726 kJ mol⁻¹.
Goal: Find \(\Delta_f H^\circ (\text{CH}_3\text{OH})\).
Reaction: \(\text{C(s)} + 2\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{OH(l)}\)
Given Data:
1. \(\text{CH}_3\text{OH} + \frac{3}{2}\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \quad \Delta H_1 = -726 \text{ kJ}\)
2. \(\text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad \Delta H_2 = -393 \text{ kJ}\)
3. \(\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O} \quad \Delta H_3 = -286 \text{ kJ}\)
Calculation:
To get the target equation:
Keep eq(2) as is: \(\text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad (-393)\)
Multiply eq(3) by 2: \(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \quad (2 \times -286 = -572)\)
Reverse eq(1): \(\text{CO}_2 + 2\text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \frac{3}{2}\text{O}_2 \quad (+726)\)
Summing these:
\(\Delta H = -393 - 572 + 726\)
\(\Delta H = -965 + 726 = -239 \text{ kJ mol}^{-1}\).
Reaction: \(\text{C(s)} + 2\text{H}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g}) \rightarrow \text{CH}_3\text{OH(l)}\)
Given Data:
1. \(\text{CH}_3\text{OH} + \frac{3}{2}\text{O}_2 \rightarrow \text{CO}_2 + 2\text{H}_2\text{O} \quad \Delta H_1 = -726 \text{ kJ}\)
2. \(\text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad \Delta H_2 = -393 \text{ kJ}\)
3. \(\text{H}_2 + \frac{1}{2}\text{O}_2 \rightarrow \text{H}_2\text{O} \quad \Delta H_3 = -286 \text{ kJ}\)
Calculation:
To get the target equation:
Keep eq(2) as is: \(\text{C} + \text{O}_2 \rightarrow \text{CO}_2 \quad (-393)\)
Multiply eq(3) by 2: \(2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O} \quad (2 \times -286 = -572)\)
Reverse eq(1): \(\text{CO}_2 + 2\text{H}_2\text{O} \rightarrow \text{CH}_3\text{OH} + \frac{3}{2}\text{O}_2 \quad (+726)\)
Summing these:
\(\Delta H = -393 - 572 + 726\)
\(\Delta H = -965 + 726 = -239 \text{ kJ mol}^{-1}\).
Q.18What happens when: (i) Ethene reacts with iodine monochloride (ii) Sulphur dioxide is oxidised in presence of V₂O₅ (iii) Cu heated with concentrated H₂SO₄
(i) Ethene + Iodine monochloride (ICl):
\(\text{CH}_2=\text{CH}_2 + \text{I-Cl} \rightarrow \text{CH}_2(\text{I})-\text{CH}_2(\text{Cl})\)
Product: 1-Chloro-2-iodoethane.
(ii) SO₂ oxidised with V₂O₅:
\(2\text{SO}_2 + \text{O}_2 \xrightarrow{\text{V}_2\text{O}_5} 2\text{SO}_3\)
Product: Sulphur trioxide. (This is a key step in Contact Process).
(iii) Cu + Conc. H₂SO₄:
\(\text{Cu} + 2\text{H}_2\text{SO}_4 (\text{conc}) \rightarrow \text{CuSO}_4 + \text{SO}_2 \uparrow + 2\text{H}_2\text{O}\)
Product: Copper sulfate, sulfur dioxide, and water.
\(\text{CH}_2=\text{CH}_2 + \text{I-Cl} \rightarrow \text{CH}_2(\text{I})-\text{CH}_2(\text{Cl})\)
Product: 1-Chloro-2-iodoethane.
(ii) SO₂ oxidised with V₂O₅:
\(2\text{SO}_2 + \text{O}_2 \xrightarrow{\text{V}_2\text{O}_5} 2\text{SO}_3\)
Product: Sulphur trioxide. (This is a key step in Contact Process).
(iii) Cu + Conc. H₂SO₄:
\(\text{Cu} + 2\text{H}_2\text{SO}_4 (\text{conc}) \rightarrow \text{CuSO}_4 + \text{SO}_2 \uparrow + 2\text{H}_2\text{O}\)
Product: Copper sulfate, sulfur dioxide, and water.
Q.19Calculate the number of atoms and unit cell present in 0.5 g of Niobium if it forms body centred cubic structure. The density of Niobium is 8.55 g cm⁻³ and edge length of unit cell is 330.6 pm. Write preparation of glucose from sucrose.
Part A: Numerical
Given: Mass \(x = 0.5 \text{ g}\), \(\rho = 8.55 \text{ g cm}^{-3}\), \(a = 330.6 \text{ pm} = 3.306 \times 10^{-8} \text{ cm}\).
1. Calculate Volume of sample:
\(V_{\text{total}} = \text{Mass} / \text{Density} = 0.5 / 8.55 = 0.05848 \text{ cm}^3\).
2. Calculate Volume of one unit cell:
\(V_{\text{cell}} = a^3 = (3.306 \times 10^{-8})^3 = 36.13 \times 10^{-24} \text{ cm}^3\).
3. Number of unit cells:
\(N_{\text{cells}} = V_{\text{total}} / V_{\text{cell}} = \frac{0.05848}{36.13 \times 10^{-24}} = 1.62 \times 10^{21}\) unit cells.
4. Number of atoms:
For BCC, \(Z = 2\) atoms/cell.
\(\text{Total atoms} = 2 \times 1.62 \times 10^{21} = 3.24 \times 10^{21}\) atoms.
Part B: Preparation of glucose from sucrose
Sucrose is hydrolyzed by boiling with dilute HCl or H₂SO₄ in alcoholic solution to give glucose and fructose.
\(\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_{12}\text{O}_6 (\text{Glucose}) + \text{C}_6\text{H}_{12}\text{O}_6 (\text{Fructose})\).
Given: Mass \(x = 0.5 \text{ g}\), \(\rho = 8.55 \text{ g cm}^{-3}\), \(a = 330.6 \text{ pm} = 3.306 \times 10^{-8} \text{ cm}\).
1. Calculate Volume of sample:
\(V_{\text{total}} = \text{Mass} / \text{Density} = 0.5 / 8.55 = 0.05848 \text{ cm}^3\).
2. Calculate Volume of one unit cell:
\(V_{\text{cell}} = a^3 = (3.306 \times 10^{-8})^3 = 36.13 \times 10^{-24} \text{ cm}^3\).
3. Number of unit cells:
\(N_{\text{cells}} = V_{\text{total}} / V_{\text{cell}} = \frac{0.05848}{36.13 \times 10^{-24}} = 1.62 \times 10^{21}\) unit cells.
4. Number of atoms:
For BCC, \(Z = 2\) atoms/cell.
\(\text{Total atoms} = 2 \times 1.62 \times 10^{21} = 3.24 \times 10^{21}\) atoms.
Part B: Preparation of glucose from sucrose
Sucrose is hydrolyzed by boiling with dilute HCl or H₂SO₄ in alcoholic solution to give glucose and fructose.
\(\text{C}_{12}\text{H}_{22}\text{O}_{11} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_{12}\text{O}_6 (\text{Glucose}) + \text{C}_6\text{H}_{12}\text{O}_6 (\text{Fructose})\).
Q.20Define: Nanochemistry. What happens when vapours of 1° and 2° alcohols are passed over hot Cu metal?
Nanochemistry: It is the branch of chemistry that deals with the study of synthesis, characterization, properties, and applications of materials at the nanoscale (1-100 nm).
Action of Hot Cu (573 K):
1. 1° Alcohol: Undergoes dehydrogenation to form Aldehydes.
\(\text{R-CH}_2\text{OH} \xrightarrow{\text{Cu, 573K}} \text{R-CHO} + \text{H}_2\)
2. 2° Alcohol: Undergoes dehydrogenation to form Ketones.
\(\text{R}_2\text{CH-OH} \xrightarrow{\text{Cu, 573K}} \text{R}_2\text{C=O} + \text{H}_2\)
Action of Hot Cu (573 K):
1. 1° Alcohol: Undergoes dehydrogenation to form Aldehydes.
\(\text{R-CH}_2\text{OH} \xrightarrow{\text{Cu, 573K}} \text{R-CHO} + \text{H}_2\)
2. 2° Alcohol: Undergoes dehydrogenation to form Ketones.
\(\text{R}_2\text{CH-OH} \xrightarrow{\text{Cu, 573K}} \text{R}_2\text{C=O} + \text{H}_2\)
Q.215% aqueous solution of cane sugar has freezing point of 271 K. Calculate freezing point of 5% glucose solution. (Molar mass sugar=342). Complete reaction: R'–NO₂ + Sn/HCl...
Part A: Freezing Point Calculation
Assumed Freezing point of pure water (\(T_f^0\)) = 273.15 K (or approx 273 K).
Given: \(T_f\) (sugar) = 271 K.
\(\Delta T_f\) (sugar) = \(273.15 - 271 = 2.15 \text{ K}\) (Using 273.15 is standard).
Formula: \(\Delta T_f = K_f \times \frac{W_2 \times 1000}{M_2 \times W_1}\)
For both solutions, \(W_2\) (5g), \(W_1\) (95g), and \(K_f\) are same.
\(\Delta T_f \propto \frac{1}{M_2}\)
\(\frac{\Delta T_f(\text{glucose})}{\Delta T_f(\text{sugar})} = \frac{M(\text{sugar})}{M(\text{glucose})}\)
\(\frac{\Delta T_f(\text{glucose})}{2.15} = \frac{342}{180}\)
\(\Delta T_f(\text{glucose}) = 2.15 \times 1.9 = 4.085 \text{ K}\)
Freezing Point of Glucose = \(273.15 - 4.085 = 269.065 \text{ K}\).
Part B: Complete Reaction
\(\text{R}'\text{-NO}_2 + 6[\text{H}] \xrightarrow{\text{Sn/Conc. HCl}} \text{R}'\text{-NH}_2 + 2\text{H}_2\text{O}\)
Product is Primary Amine (\(\text{R}'\text{-NH}_2\)).
Assumed Freezing point of pure water (\(T_f^0\)) = 273.15 K (or approx 273 K).
Given: \(T_f\) (sugar) = 271 K.
\(\Delta T_f\) (sugar) = \(273.15 - 271 = 2.15 \text{ K}\) (Using 273.15 is standard).
Formula: \(\Delta T_f = K_f \times \frac{W_2 \times 1000}{M_2 \times W_1}\)
For both solutions, \(W_2\) (5g), \(W_1\) (95g), and \(K_f\) are same.
\(\Delta T_f \propto \frac{1}{M_2}\)
\(\frac{\Delta T_f(\text{glucose})}{\Delta T_f(\text{sugar})} = \frac{M(\text{sugar})}{M(\text{glucose})}\)
\(\frac{\Delta T_f(\text{glucose})}{2.15} = \frac{342}{180}\)
\(\Delta T_f(\text{glucose}) = 2.15 \times 1.9 = 4.085 \text{ K}\)
Freezing Point of Glucose = \(273.15 - 4.085 = 269.065 \text{ K}\).
Part B: Complete Reaction
\(\text{R}'\text{-NO}_2 + 6[\text{H}] \xrightarrow{\text{Sn/Conc. HCl}} \text{R}'\text{-NH}_2 + 2\text{H}_2\text{O}\)
Product is Primary Amine (\(\text{R}'\text{-NH}_2\)).
Q.22What is denaturation of protein? Derive an expression of Ostwald’s dilution law for weak acid.
Denaturation of protein: It is a process in which the quaternary, tertiary, and secondary structures of proteins are destroyed by physical changes (like heat, pH change) while the primary structure (sequence of amino acids) remains intact. It results in the loss of biological activity.
Derivation for Weak Acid (HA):
Consider a weak acid HA dissociated in water:
\(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\)
Let \(C\) be the initial concentration and \(\alpha\) be degree of dissociation.
At eq: \([\text{HA}] = C(1-\alpha)\), \([\text{H}^+] = C\alpha\), \([\text{A}^-] = C\alpha\).
Dissociation constant \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\)
\(K_a = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}\)
For weak acids, \(\alpha \ll 1\), so \(1-\alpha \approx 1\).
\(K_a = C\alpha^2\) or \(\alpha = \sqrt{K_a/C}\).
Derivation for Weak Acid (HA):
Consider a weak acid HA dissociated in water:
\(\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-\)
Let \(C\) be the initial concentration and \(\alpha\) be degree of dissociation.
At eq: \([\text{HA}] = C(1-\alpha)\), \([\text{H}^+] = C\alpha\), \([\text{A}^-] = C\alpha\).
Dissociation constant \(K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}\)
\(K_a = \frac{(C\alpha)(C\alpha)}{C(1-\alpha)} = \frac{C\alpha^2}{1-\alpha}\)
For weak acids, \(\alpha \ll 1\), so \(1-\alpha \approx 1\).
\(K_a = C\alpha^2\) or \(\alpha = \sqrt{K_a/C}\).
Q.23Define: Nanotechnology. Write any two applications of electrochemical series.
Nanotechnology: It is the design, characterization, production, and application of structures, devices, and systems by controlling shape and size at the nanometer scale.
Applications of Electrochemical Series:
1. To compare the relative strength of oxidizing and reducing agents.
2. To predict the spontaneity of a redox reaction (feasible if \(E^\circ_{\text{cell}}\) is positive).
Applications of Electrochemical Series:
1. To compare the relative strength of oxidizing and reducing agents.
2. To predict the spontaneity of a redox reaction (feasible if \(E^\circ_{\text{cell}}\) is positive).
Q.24Reaction steps: (i) NO₂ + F₂ → NO₂F + F (slow), (ii) F + NO₂ → NO₂F (fast). Answer subquestions. Write reaction for Teflon.
(a) Overall reaction: Add steps (i) and (ii).
\(2\text{NO}_2(\text{g}) + \text{F}_2(\text{g}) \rightarrow 2\text{NO}_2\text{F}(\text{g})\)
(b) Rate Law: Rate depends on the slow step.
\(\text{Rate} = k[\text{NO}_2][\text{F}_2]\)
(c) Reaction Intermediate: \(\text{F}(\text{g})\) (produced in step 1, consumed in step 2).
Preparation of Teflon:
Polymerization of tetrafluoroethylene under high pressure with peroxide catalyst.
\(n \text{CF}_2=\text{CF}_2 \xrightarrow{\text{Polymerisation}} -(\text{CF}_2-\text{CF}_2)_n- \text{ (Teflon/PTFE)}\)
\(2\text{NO}_2(\text{g}) + \text{F}_2(\text{g}) \rightarrow 2\text{NO}_2\text{F}(\text{g})\)
(b) Rate Law: Rate depends on the slow step.
\(\text{Rate} = k[\text{NO}_2][\text{F}_2]\)
(c) Reaction Intermediate: \(\text{F}(\text{g})\) (produced in step 1, consumed in step 2).
Preparation of Teflon:
Polymerization of tetrafluoroethylene under high pressure with peroxide catalyst.
\(n \text{CF}_2=\text{CF}_2 \xrightarrow{\text{Polymerisation}} -(\text{CF}_2-\text{CF}_2)_n- \text{ (Teflon/PTFE)}\)
Q.25Define: Elastomer. Write two postulates of Werner theory of coordinate complexes.
Elastomer: Polymers that possess elastic character (can be stretched and return to original shape) due to weak intermolecular forces (van der Waals) and few crosslinks (e.g., Rubber).
Werner Theory Postulates:
1. Metals possess two types of valencies: Primary valency (ionizable, corresponds to oxidation state) and Secondary valency (non-ionizable, corresponds to coordination number).
2. Secondary valencies are directional in space (giving geometry), whereas primary valencies are non-directional.
Werner Theory Postulates:
1. Metals possess two types of valencies: Primary valency (ionizable, corresponds to oxidation state) and Secondary valency (non-ionizable, corresponds to coordination number).
2. Secondary valencies are directional in space (giving geometry), whereas primary valencies are non-directional.
Q.26Write four salient features of SN1 mechanism. Write chemical reaction for carbylamine test.
Features of SN1 Mechanism:
1. It is a unimolecular reaction (First order kinetics), Rate \(\propto [\text{Substrate}]\).
2. It takes place in two steps. Step 1 is formation of carbocation (slow/RDS).
3. It proceeds via a planar carbocation intermediate.
4. It results in racemization (mixture of retention and inversion products).
Carbylamine Test Reaction:
Aliphatic or aromatic primary amines heat with chloroform and alcoholic KOH to form foul-smelling isocyanides.
\(\text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH(alc)} \xrightarrow{\Delta} \text{R-NC} \text{ (Isocyanide)} + 3\text{KCl} + 3\text{H}_2\text{O}\).
1. It is a unimolecular reaction (First order kinetics), Rate \(\propto [\text{Substrate}]\).
2. It takes place in two steps. Step 1 is formation of carbocation (slow/RDS).
3. It proceeds via a planar carbocation intermediate.
4. It results in racemization (mixture of retention and inversion products).
Carbylamine Test Reaction:
Aliphatic or aromatic primary amines heat with chloroform and alcoholic KOH to form foul-smelling isocyanides.
\(\text{R-NH}_2 + \text{CHCl}_3 + 3\text{KOH(alc)} \xrightarrow{\Delta} \text{R-NC} \text{ (Isocyanide)} + 3\text{KCl} + 3\text{H}_2\text{O}\).
SECTION − D
Q.27Boiling point problem (Ethyl acetate). Explain pseudo first order reaction.
Numerical:
Given:
\(T_b^\circ = 77.06^\circ \text{C}\)
\(T_b = 84.27^\circ \text{C}\)
\(\Delta T_b = 84.27 - 77.06 = 7.21 \text{ K}\) (or \(^\circ \text{C}\), difference is same)
\(W_2 = 50 \text{ g}, W_1 = 150 \text{ g}, K_b = 2.77 \text{ K kg mol}^{-1}\)
Formula: \(M_2 = \frac{1000 \times K_b \times W_2}{\Delta T_b \times W_1}\)
\(M_2 = \frac{1000 \times 2.77 \times 50}{7.21 \times 150}\)
\(M_2 = \frac{138500}{1081.5} \approx 128.06 \text{ g mol}^{-1}\).
Pseudo First Order Reaction:
A chemical reaction which is not first order (usually higher order, like bimolecular) but behaves as a first order reaction because one of the reactants is present in large excess.
Example: Acid hydrolysis of ethyl acetate.
\(\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O (excess)} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}\).
Rate = \(k'[\text{Ester}][\text{H}_2\text{O}]\), since \([\text{H}_2\text{O}]\) is constant, Rate = \(k[\text{Ester}]\).
Given:
\(T_b^\circ = 77.06^\circ \text{C}\)
\(T_b = 84.27^\circ \text{C}\)
\(\Delta T_b = 84.27 - 77.06 = 7.21 \text{ K}\) (or \(^\circ \text{C}\), difference is same)
\(W_2 = 50 \text{ g}, W_1 = 150 \text{ g}, K_b = 2.77 \text{ K kg mol}^{-1}\)
Formula: \(M_2 = \frac{1000 \times K_b \times W_2}{\Delta T_b \times W_1}\)
\(M_2 = \frac{1000 \times 2.77 \times 50}{7.21 \times 150}\)
\(M_2 = \frac{138500}{1081.5} \approx 128.06 \text{ g mol}^{-1}\).
Pseudo First Order Reaction:
A chemical reaction which is not first order (usually higher order, like bimolecular) but behaves as a first order reaction because one of the reactants is present in large excess.
Example: Acid hydrolysis of ethyl acetate.
\(\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O (excess)} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}\).
Rate = \(k'[\text{Ester}][\text{H}_2\text{O}]\), since \([\text{H}_2\text{O}]\) is constant, Rate = \(k[\text{Ester}]\).
Q.28Why does aq. CuSO₄ solution turn blue litmus red? Why compounds of transition metal ions are coloured?
Aq. CuSO₄ turns litmus red:
Copper sulfate is a salt of a strong acid (H₂SO₄) and a weak base (Cu(OH)₂). In aqueous solution, it undergoes hydrolysis to form an acidic solution due to the presence of free \(\text{H}^+\) ions.
\(\text{Cu}^{2+} + 2\text{H}_2\text{O} \rightleftharpoons \text{Cu(OH)}_2 + 2\text{H}^+\)
Colour of Transition Metal Ions:
1. Transition metal ions often have incomplete d-orbitals (\(d^{1-9}\)).
2. When ligands approach, the d-orbitals split into two energy levels (crystal field splitting).
3. Electrons absorb energy from the visible region of light to jump from lower d-orbitals to higher d-orbitals (d-d transition).
4. The transmitted light (complementary color) is observed as the color of the compound.
Copper sulfate is a salt of a strong acid (H₂SO₄) and a weak base (Cu(OH)₂). In aqueous solution, it undergoes hydrolysis to form an acidic solution due to the presence of free \(\text{H}^+\) ions.
\(\text{Cu}^{2+} + 2\text{H}_2\text{O} \rightleftharpoons \text{Cu(OH)}_2 + 2\text{H}^+\)
Colour of Transition Metal Ions:
1. Transition metal ions often have incomplete d-orbitals (\(d^{1-9}\)).
2. When ligands approach, the d-orbitals split into two energy levels (crystal field splitting).
3. Electrons absorb energy from the visible region of light to jump from lower d-orbitals to higher d-orbitals (d-d transition).
4. The transmitted light (complementary color) is observed as the color of the compound.
Q.29State and explain Hess’s law. Interhalogen compounds. Uses of Neon.
Hess's Law of Constant Heat Summation:
The enthalpy change for a chemical reaction is the same regardless of the path by which the reaction occurs (i.e., whether it takes place in one step or multiple steps). \(\Delta H = \Delta H_1 + \Delta H_2 + ...\)
Interhalogen Compounds:
Compounds formed by the reaction of two or more different halogens are called interhalogen compounds. General formula \(XY_n\) where X is a larger halogen and Y is a smaller halogen. (e.g., \(\text{ICl}, \text{BrF}_3\)).
Uses of Neon:
1. Used in neon discharge lamps and advertising signs (glows orange-red).
2. Used in beacon lights for safety of air navigation as the light penetrates fog.
The enthalpy change for a chemical reaction is the same regardless of the path by which the reaction occurs (i.e., whether it takes place in one step or multiple steps). \(\Delta H = \Delta H_1 + \Delta H_2 + ...\)
Interhalogen Compounds:
Compounds formed by the reaction of two or more different halogens are called interhalogen compounds. General formula \(XY_n\) where X is a larger halogen and Y is a smaller halogen. (e.g., \(\text{ICl}, \text{BrF}_3\)).
Uses of Neon:
1. Used in neon discharge lamps and advertising signs (glows orange-red).
2. Used in beacon lights for safety of air navigation as the light penetrates fog.
Q.30Explain homoleptic and heteroleptic complexes. Convert carboxylic acids to (i) ester (ii) acid amide.
Homoleptic Complexes: Complexes in which the metal is bound to only one kind of donor group (ligand). Ex: \([\text{Co(NH}_3)_6]^{3+}\).
Heteroleptic Complexes: Complexes in which the metal is bound to more than one kind of donor group. Ex: \([\text{Co(NH}_3)_4\text{Cl}_2]^{+}\).
Conversions:
(i) Carboxylic acid to Ester (Esterification):
\(\text{R-COOH} + \text{R'-OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{R-COOR'} + \text{H}_2\text{O}\)
(ii) Carboxylic acid to Acid Amide:
\(\text{R-COOH} + \text{NH}_3 \rightarrow \text{R-COONH}_4 \xrightarrow{\Delta} \text{R-CONH}_2 + \text{H}_2\text{O}\)
Heteroleptic Complexes: Complexes in which the metal is bound to more than one kind of donor group. Ex: \([\text{Co(NH}_3)_4\text{Cl}_2]^{+}\).
Conversions:
(i) Carboxylic acid to Ester (Esterification):
\(\text{R-COOH} + \text{R'-OH} \xrightarrow{\text{conc. H}_2\text{SO}_4} \text{R-COOR'} + \text{H}_2\text{O}\)
(ii) Carboxylic acid to Acid Amide:
\(\text{R-COOH} + \text{NH}_3 \rightarrow \text{R-COONH}_4 \xrightarrow{\Delta} \text{R-CONH}_2 + \text{H}_2\text{O}\)
Q.31Define: Green chemistry. Identify A and B in reaction. Action of hot HI on glucose.
Green Chemistry: It is the design of chemical products and processes that reduce or eliminate the use or generation of hazardous substances.
Identify A and B:
Reaction: Isopropyl alcohol \(\xrightarrow{\text{PBr}_3}\) A \(\xrightarrow{\text{NH}_3 \text{ excess}}\) B
1. \(\text{CH}_3\text{-CH(OH)-CH}_3 + \text{PBr}_3 \rightarrow \text{CH}_3\text{-CH(Br)-CH}_3\) (A = Isopropyl bromide / 2-Bromopropane).
2. \(\text{CH}_3\text{-CH(Br)-CH}_3 + \text{NH}_3 (\text{excess}) \rightarrow \text{CH}_3\text{-CH(NH}_2)\text{-CH}_3\) (B = Isopropyl amine / Propan-2-amine).
Action of hot HI on Glucose:
When glucose is heated with HI for a long time, it undergoes reduction to form n-hexane. This indicates all 6 carbons are in a straight chain.
\(\text{Glucose} \xrightarrow{\text{HI, }\Delta} \text{n-Hexane}\).
Identify A and B:
Reaction: Isopropyl alcohol \(\xrightarrow{\text{PBr}_3}\) A \(\xrightarrow{\text{NH}_3 \text{ excess}}\) B
1. \(\text{CH}_3\text{-CH(OH)-CH}_3 + \text{PBr}_3 \rightarrow \text{CH}_3\text{-CH(Br)-CH}_3\) (A = Isopropyl bromide / 2-Bromopropane).
2. \(\text{CH}_3\text{-CH(Br)-CH}_3 + \text{NH}_3 (\text{excess}) \rightarrow \text{CH}_3\text{-CH(NH}_2)\text{-CH}_3\) (B = Isopropyl amine / Propan-2-amine).
Action of hot HI on Glucose:
When glucose is heated with HI for a long time, it undergoes reduction to form n-hexane. This indicates all 6 carbons are in a straight chain.
\(\text{Glucose} \xrightarrow{\text{HI, }\Delta} \text{n-Hexane}\).
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