12th CHEMISTRY - PUBLIC COMPULSORY QUESTIONS AND PROBLEMS
1. A hydride of 2nd period alkali metal (A) on reaction with compound of Boron (B) to give a reducing agent (C). Identify (A), (B) and (C). (Jun-20)
Answer:
$2LiH + B_2H_6 \xrightarrow{\text{Ether}} 2LiBH_4$
(A) + (B) $\rightarrow$ (C)
Result:
(A) $LiH$ - Lithium Hydride
(B) $B_2H_6$ - Diborane
(C) $LiBH_4$ - Lithium Borohydride
2. Aluminium to thallium ionisation enthalpy decreases is only a marginal difference. Why and explain. (Mar-20)
Answer:
Due to the presence of inner d and f-electrons which has poor shielding effect compared to s and p-electrons.
5. Coordination Chemistry
1. Write the ligand, central metal ion and IUPAC name for the $[Ag(NH_3)_2]^+$ complex. (May 22)
Answer:
Ligand: $NH_3$ (ammine)
Central metal ion: $Ag^+$ (silver)
IUPAC name: Diamminesilver (I) ion
2. Write the ligand, central metal ion and IUPAC name for the $[Co(CN)_2Cl_2]Cl$ complex. (May 22)
Answer:
Ligand: $CN^-$ (cyanido), $Cl^-$ (chlorido)
Central metal ion: $Co^{3+}$ (Cobalt)
IUPAC name: Dichloridodicyanidocobalt(III) chloride
3. Write the IUPAC ligand name for the following (Aug 22)
1) $C_2O_4^{2-}$ 2) $H_2O$ 3) $Cl^-$ 4) $NH_3$
Answer:
1) $C_2O_4^{2-}$ - Oxalato
2) $H_2O$ - Aqua
3) $Cl^-$ - Chlorido
4) $NH_3$ - Ammine
4. For the complex, $[Pt(NO_2)(H_2O)(NH_3)_2]Br$, identify the following.
(a) Central metal atom / ion
(b) Coordination number
(c) Ligand
(d) Oxidation number of central metal ion
Answer:
(a) Central metal atom/ion: $Pt^{2+}$
(b) Coordination number: 4
(c) Ligand: $NO_2^-$ (Nitrito-N), $H_2O$ (Aqua), $NH_3$ (Ammine)
(d) Oxidation number of central metal ion: +2
5. Write the IUPAC name for the following compounds
A) $[Ag(NH_3)_2]^+$
B) $[Co(NH_3)_5Cl]^{2+}$
Answer:
A) $[Ag(NH_3)_2]^+$ - Diamminesilver(I) ion
B) $[Co(NH_3)_5Cl]^{2+}$ - Pentaamminechloridocobalt(III) ion
1. Classify the following solids (Aug 21)
a) Naphthalene b) Brass c) Diamond d) NaCl e) Glucose g) $SiO_2$
Answer:
Covalent solids: Diamond, $SiO_2$
Molecular solids: Naphthalene, Glucose
Ionic solids: NaCl
Metallic solids: Brass
2. Calculate the number of atoms in a FCC unit cell. (Mar 23)
Answer:
The number of atoms in a FCC unit cell is calculated as:
$FCC = \frac{N_c}{8} + \frac{N_f}{2} = \frac{8}{8} + \frac{6}{2} = 1 + 3 = 4$
3. Aluminium crystallizes in a cubic close packed structure. Its metallic radius is 125pm. calculate the edge length of unit cell. (Mar 24)
Answer:
Given: $CCP = FCC$, $r = 125\text{ pm}$
Formula for FCC: $4r = \sqrt{2}a \implies a = \frac{4r}{\sqrt{2}}$
$a = \frac{4 \times 125}{1.414} = 353.5 \text{ pm}$
Result: Edge length of unit cell (a) = 353.5 pm.
1. What is an order of a reaction? (Mar 24)
Answer:
It is the sum of the powers of concentration terms involved in the experimentally determined rate law.
2. Powdered $CaCO_3$ reacts much faster with dilute HCl than with the same mass of $CaCO_3$ as marble. Give reason (Jun 20)
Answer:
When the $CaCO_3$ particles are smaller, the surface area exposed to the acid is higher.
$\rightarrow$ (Rate of a reaction increases when the surface area of a solid reactant is increased).
3. The rate constant for a first order reaction is $1.54 \times 10^{-3} s^{-1}$. Calculate its half life time. (Jun 20, 24 Mar 25)
Solution:
Given: $k = 1.54 \times 10^{-3} s^{-1}$, $t_{1/2} = ?$
$t_{1/2} = \frac{0.693}{k}$
$t_{1/2} = \frac{0.693}{1.54 \times 10^{-3}} = 450 \text{ s}$
4. Show that in case of first order reaction, the time required for 99% completion is twice times the time required for the completion of 90% of the reaction. (Mar 23)
Solution:
$t = \frac{2.303}{k} \log \frac{[A_o]}{[A]}$
For 90% completion: $A_o = 100$, $A = 100 - 90 = 10$
$t_{90\%} = \frac{2.303}{k} \log \frac{100}{10}$ (or) $t_{90\%} = \frac{2.303}{k} \log 10$
For 99% completion: $A_o = 100$, $A = 100 - 99 = 1$
$t_{99\%} = \frac{2.303}{k} \log \frac{100}{1}$ (or) $t_{99\%} = \frac{2.303}{k} \log 100$
Ratio: $\frac{t_{99\%}}{t_{90\%}} = \frac{\log 100}{\log 10} = \frac{2}{1} = 2$
Therefore, $t_{99\%} = 2 \times t_{90\%}$
5. The rate of the reaction $x + 2y \rightarrow$ product is $4 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1}$ if $[x]=[y]=0.2\text{M}$ and rate constant at 400K is $2 \times 10^{-2}\text{s}^{-1}$. What is the overall order of the reaction. (Sep 20)
Solution:
$\text{Rate} = k[x]^m[y]^n$
$4 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1} = 2 \times 10^{-2} \text{s}^{-1} (0.2 \text{ mol L}^{-1})^m (0.2 \text{ mol L}^{-1})^n$
$\frac{4 \times 10^{-3}}{2 \times 10^{-2}} = (0.2)^{m+n}$
$0.2 = (0.2)^{m+n}$
Comparing the powers on both sides: $m+n = 1$
$\rightarrow$ Overall order of the reaction = 1
6. In a first order reaction $x \rightarrow y$ if k is the rate constant and the initial concentration of the reactant x is 0.1 M. What is the value of $t_{1/2}$ in the reaction? (Jun 25)
Solution:
$\rightarrow$ The half-life of a first order reaction is independent of the initial concentration.
$t_{1/2} = \frac{0.693}{k}$
1. Calculate the concentration of $OH^-$ in a fruit juice which contains $2 \times 10^{-3}\text{M}$, $H_3O^+$ ion. Identify the nature of the solution. (Jun 23)
Solution:
$K_w = [H_3O^+][OH^-] = 1 \times 10^{-14}$
$[OH^-] = \frac{K_w}{[H_3O^+]} = \frac{1 \times 10^{-14}}{2 \times 10^{-3}} = 5 \times 10^{-12} \text{ M}$
Since $[H_3O^+] (2 \times 10^{-3}\text{M}) \gg [OH^-] (5 \times 10^{-12}\text{M})$, the solution is acidic in nature.
2. Calculate the pH of 0.1M $CH_3COONa$ solution. (pKa for $CH_3COOH$ is 4.74). (Sep 20)
Solution:
$pH = 7 + \frac{pK_a}{2} + \frac{\log c}{2}$
$pH = 7 + \frac{4.74}{2} + \frac{\log 0.1}{2}$
$pH = 7 + 2.37 - 0.5 = 8.87$
3. Find the pH of a buffer solution containing 0.20 mole per litre sodium acetate and 0.18 mole per litre acetic acid. $K_a$ is $1.8 \times 10^{-5}$. (Aug 21)
Solution:
$pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 5 - \log 1.8 = 5 - 0.26 = 4.74$
Using Henderson-Hasselbalch equation:
$pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$
$pH = 4.74 + \log\frac{0.20}{0.18} = 4.74 + \log\frac{10}{9}$
$pH = 4.74 + \log 10 - \log 9 = 4.74 + 1 - 0.95 = 4.79$
4. Calculate the pH of 0.1M $CH_3COOH$ solution. Dissociation constant of acetic acid is $1.8 \times 10^{-5}$. (Mar 25)
Solution:
$[H^+] = \sqrt{K_a \times C} = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \text{ M}$
$pH = -\log[H^+] = -\log(1.34 \times 10^{-3}) = 3 - \log 1.34 = 3 - 0.1271 = 2.87$
1. A solution of silver nitrate is electrolysed for 30 minutes with a current of 2 amperes. Calculate the mass of silver deposited at the cathode. (Sep 20)
Solution:
Given: $I = 2\text{A}$, $t = 30 \times 60 \text{ s}$, $Z = \frac{108 \text{ g/mol}}{96500 \text{ C/mol}}$
Using Faraday's First Law, $m = Z I t$
$m = \frac{108}{96500} \times 2 \times 30 \times 60 = \frac{108 \times 36}{965} = 4.03 \text{ g}$
2. A solution of silver nitrate is electrolysed for 20 minutes with a current of 2 amperes. Calculate the mass of silver deposited at the cathode. (July 22)
Solution:
Given: $I = 2\text{A}$, $t = 20 \times 60 \text{ s}$
$m = Z I t = \frac{108}{96500} \times 2 \times 20 \times 60 = \frac{108 \times 24}{965} = 2.68 \text{ g}$
3. A conductivity cell has two platinum electrodes separated by a distance 1.5 cm and the cross sectional area is 4.5 sq cm. Using this cell, the resistance of 0.5 N electrolytic solution was measured as $15\ \Omega$. Find the specific conductance of the solution. (Mar 20)
Solution:
Given: $l = 1.5 \text{ cm} = 1.5 \times 10^{-2} \text{ m}$, $A = 4.5 \text{ cm}^2 = 4.5 \times 10^{-4} \text{ m}^2$, $R = 15\ \Omega$
Specific conductance $\kappa = \frac{1}{R} \times \frac{l}{A}$
$\kappa = \frac{1}{15} \times \frac{1.5 \times 10^{-2}}{4.5 \times 10^{-4}} = 22.2 \text{ S m}^{-1}$
4. Can $Fe^{3+}$ oxidises bromide to bromine under standard conditions? Given: $E^\circ_{Fe^{3+}/Fe^{2+}} = 0.771 \text{ V}$; $E^\circ_{Br_2/Br^-} = 1.09 \text{ V}$. (Mar 24)
Solution:
Anode (Oxidation): $2Br^- \rightarrow Br_2 + 2e^-$ ($E^\circ_{ox} = -1.09 \text{ V}$)
Cathode (Reduction): $2Fe^{3+} + 2e^- \rightarrow 2Fe^{2+}$ ($E^\circ_{red} = +0.771 \text{ V}$)
Overall: $2Fe^{3+} + 2Br^- \rightarrow 2Fe^{2+} + Br_2$
$E^\circ_{cell} = E^\circ_{ox} + E^\circ_{red} = -1.09 + 0.771 = -0.319 \text{ V}$
Since $E^\circ_{cell}$ is negative, $\Delta G^\circ$ is positive and the cell reaction is non-spontaneous. Hence, $Fe^{3+}$ cannot oxidise $Br^-$ to $Br_2$.
5. Reduction potential of two metals $M_1$ and $M_2$ are $E^\circ(M_1^{2+}/M_1) = -2.35 \text{ V}$ and $E^\circ(M_2^{2+}/M_2) = 0.2 \text{ V}$. Predict which one is better for coating the surface of iron. Given: $E^\circ(Fe^{2+}/Fe) = -0.44 \text{ V}$. (Jun 23)
Solution:
Oxidation potential of $M_1$ ($+2.35 \text{ V}$) is more positive than the oxidation potential of Fe ($+0.44 \text{ V}$). This indicates $M_1$ acts as a sacrificial anode and will prevent iron from rusting. Thus, $M_1$ is better for coating.
11. Hydroxy compounds and Ethers
1. Why the C-O-C bond angle is slightly greater than the tetrahedral bond angle.
Answer:
➤ Due to the repulsive interaction between the two bulkier alkyl groups.
2. Compound (A) of molecular formula $C_6H_6O$ gives purple colouration with neutral $FeCl_3$. Compound (A) reacts with ammonia to give Compound (B) and it also reacts with Zn dust to give Compound (C). Identify the Compounds A, B, C and write down the equations.
Answer:
$C_6H_5OH + NH_3 \xrightarrow{\text{Anhydrous } ZnCl_2} C_6H_5NH_2 + H_2O$
(A) $\rightarrow$ (B)
$C_6H_5OH + Zn \xrightarrow{\Delta} C_6H_6 + ZnO$
(A) $\rightarrow$ (C)
Result:
(A) $C_6H_5OH$ - Phenol
(B) $C_6H_5NH_2$ - Aniline
(C) $C_6H_6$ - Benzene
3. Find the products X and Y in the following reactions.
Phenol $\xrightarrow{Zn, \Delta}$ (X)
Phenol $\xrightarrow{NH_3, \Delta, \text{Anhydrous } ZnCl_2}$ (Y)
Answer:
$C_6H_5OH \xrightarrow{Zn, \Delta} C_6H_6 + ZnO$
(X) = $C_6H_6$ Benzene
$C_6H_5OH \xrightarrow{NH_3, \Delta, \text{Anhydrous } ZnCl_2} C_6H_5NH_2 + H_2O$
(Y) = $C_6H_5NH_2$ Aniline
4. How is phenol prepared from chlorobenzene?
Answer:
$C_6H_5-Cl \text{ (chlorobenzene)} + NaOH \xrightarrow{623K, 300\text{ bar}} C_6H_5ONa \xrightarrow{HCl} C_6H_5OH \text{ (phenol)} + NaCl$
5. $C_2H_6O_2$ (A) $\xrightarrow{\text{(i) Anhydrous } ZnCl_2 \text{ (ii) Tautomerisation}}$ (B) $\xrightarrow{\text{Zn-Hg/Conc HCl}}$ (C). Identify A, B, and C in reactions.
Answer:
$HO-CH_2-CH_2-OH \text{ (A)} \xrightarrow{\text{(i) Anhydrous } ZnCl_2 \text{ (ii) Tautomerisation}} CH_3-CHO \text{ (B)}$
$CH_3-CHO \text{ (B)} \xrightarrow{\text{Zn-Hg/Conc HCl (Clemmensen Reduction)}} CH_3-CH_3 \text{ (C)}$
Result:
(A) $HO-CH_2-CH_2-OH$ - Ethylene glycol
(B) $CH_3-CHO$ - Acetaldehyde (Ethanal)
(C) $CH_3-CH_3$ - Ethane
11. Hydroxy compounds and Ethers (Part 2)
1. Identify the compounds A, B and C in the following sequence of reactions.
$C_6H_5-OH \xrightarrow{\text{Zn dust}} A \xrightarrow[\text{anhydrous } AlCl_3]{CH_3Cl} B \xrightarrow{H^+/KMnO_4} C$
Answer:
$C_6H_5-OH \xrightarrow{\text{Zn dust}} C_6H_6 \text{ (A)} \xrightarrow[\text{anhydrous } AlCl_3]{CH_3Cl} C_6H_5-CH_3 \text{ (B)} \xrightarrow{H^+/KMnO_4} C_6H_5-COOH \text{ (C)}$
Result:
(A) $C_6H_6$ - Benzene
(B) $C_6H_5-CH_3$ - Toluene
(C) $C_6H_5-COOH$ - Benzoic acid
2. Identify the compounds A, B and C in the following sequence of reactions.
$C_6H_5-OH \xrightarrow{PBr_3} A \xrightarrow{\text{Aq.NaOH}} B \rightarrow C$
Answer:
$CH_3CH_2-OH \xrightarrow{PBr_3} CH_3CH_2Br \text{ (A)} \xrightarrow{\text{Aq.NaOH}} CH_3CH_2-OH \text{ (B)} \rightarrow CH_3CH_2-ONa \text{ (C)}$
Result:
(A) $CH_3CH_2-Br$ - Ethyl bromide (Bromo ethane)
(B) $CH_3CH_2-OH$ - Ethyl alcohol (Ethanol)
(C) $CH_3CH_2-ONa$ - Sodium ethoxide
3. Phenol is distilled with Zn dust followed by Friedel - Crafts alkylation with propyl chloride to give a compound A, A on oxidation gives (B) Identify A and B.
Answer:
$C_6H_5-OH \text{ (phenol)} \xrightarrow{\text{Zn dust}} C_6H_6 \text{ (A)} \xrightarrow[\text{anhydrous } AlCl_3]{CH_3CH_2CH_2Cl} C_6H_5CH_2CH_2CH_3 \text{ (B)} \xrightarrow{H^+/KMnO_4} C_6H_5-COOH \text{ (C)}$
Result:
(A) $C_6H_6$ - Benzene
(B) $C_6H_5CH_2CH_2CH_3$ - n-propyl benzene
(C) $C_6H_5-COOH$ - Benzoic acid
4. Identify the compounds A, B and C in the following sequence of reactions.
$C_6H_5-OH \xrightarrow{\text{NaOH}} A \xrightarrow[\text{ii) 4-7bar}]{\text{i) } CO_2, 400K} B \xrightarrow{H^+/H_2O} C$
Answer:
Phenol + $NaOH \rightarrow$ Sodium phenoxide (A)
Sodium phenoxide (A) + $CO_2 \xrightarrow{400K, 4-7\text{bar}}$ Sodium salicylate (B)
Sodium salicylate (B) $\xrightarrow{H^+/H_2O}$ Salicylic acid (C)
Result:
(A) Sodium phenoxide
(B) Sodium salicylate
(C) Salicylic acid
13. Organic Nitrogen Compounds
1. There are two isomers with the formula $CH_3NO_2$. How will you distinguish between them? (Jun 24, Mar 25)
Answer:
| Nitro form |
Aci form |
| 1. Less acidic |
1. More acidic |
| 2. Dissolve in NaOH slowly |
2. Dissolve in NaOH instantly |
| 3. Decolourises $FeCl_3$ solution |
3. With $FeCl_3$ gives reddish brown colour |
| 4. Electrical conductivity is low |
4. Electrical conductivity is high |
2. Identify the compounds A, B, and C in the following sequence of reaction. (Jun 23)
$C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{HNO_2/273K} B \xrightarrow{C_6H_5OH} C$
Answer:
$C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2 \text{ (A)} + 2H_2O$
$C_6H_5NH_2 \xrightarrow{HNO_2/273K} C_6H_5N_2Cl \text{ (B)}$
$C_6H_5N_2Cl + C_6H_5OH \rightarrow p\text{-Hydroxy azobenzene (C)}$
Result:
(A) $C_6H_5NH_2$ - Aniline
(B) $C_6H_5N_2Cl$ - Benzene diazonium chloride
(C) $C_6H_5-N=N-C_6H_4OH$ - p-Hydroxy azobenzene
3. How is aryl halide prepared by using $Cu_2Cl_2/HCl$ (or) $Cu_2Br_2/HBr$? (or) (Write a note on Sandmeyer reaction.) (Sep 20, Mar 20)
Answer:
$C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2$
4. Identify the compounds A, and B in the following sequence of reactions. (May 22)
$CH_3-Br \xrightarrow{NaN_3} A \xrightarrow{LiAlH_4} B$
Answer:
$CH_3-Br \xrightarrow{NaN_3} CH_3N_3 \text{ (A)} \xrightarrow{LiAlH_4} CH_3-NH_2 \text{ (B)} + N_2$
Result:
(A) $CH_3N_3$ - Methyl azide
(B) $CH_3-NH_2$ - Methyl amine
5. Identify the compounds A, and B in the following sequence of reactions. (May 22)
$CH_3CH_2-NO_2 \xrightarrow{Sn/HCl} A \xrightarrow{CH_3COCl} B$
Answer:
$CH_3CH_2-NO_2 \xrightarrow{Sn/HCl} CH_3CH_2NH_2 \text{ (A)}$
$CH_3CH_2NH_2 \xrightarrow{CH_3COCl} CH_3CH_2-NH-COCH_3 \text{ (B)}$
Result:
(A) $CH_3CH_2NH_2$ - Ethyl amine
(B) $CH_3CH_2-NH-COCH_3$ - N-Ethyl acetamide
6. From the following reaction, Identify A and B. (Sep 21)
$CH_3-NO_2 \xrightarrow{Sn/HCl, 6(H)} A$
$CH_3-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} B$
Answer:
$CH_3-NO_2 \xrightarrow{Sn/HCl, 6(H)} CH_3-NH_2 \text{ (A)}$
$CH_3-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} CH_3-NHOH \text{ (B)}$
Result:
(A) $CH_3-NH_2$ - Methyl amine
(B) $CH_3-NHOH$ - Methyl hydroxylamine
7. Identify the compounds A, B, and C in the following reaction. (Jun 22)
$C_6H_5-NO_2 \xrightarrow{Sn/HCl} A$
$C_6H_5-NO_2 \xrightarrow{Zn/NH_4Cl} B$
$C_6H_5-NO_2 \xrightarrow{Zn/NaOH} C$
Answer:
$C_6H_5-NO_2 \xrightarrow{Sn/HCl, 6(H)} C_6H_5-NH_2 \text{ (A)}$
$C_6H_5-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} C_6H_5-NHOH \text{ (B)}$
$C_6H_5-NO_2 \xrightarrow{Zn/NaOH} C_6H_5-NH-NH-C_6H_5 \text{ (C)}$
Result:
(A) $C_6H_5-NH_2$ - Aniline
(B) $C_6H_5-NHOH$ - Phenyl hydroxylamine
(C) $C_6H_5-NH-NH-C_6H_5$ - Hydrazobenzene
8. Identify A and B. (Mar 20)
A $\xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-CH_2-NH_2$
B $\xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-NH-CH_3$
Answer:
$CH_3CN \text{ (A)} \xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-CH_2-NH_2$
$CH_3NC \text{ (B)} \xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-NH-CH_3$
Result:
(A) $CH_3CN$ - Methyl cyanide
(B) $CH_3NC$ - Methyl isocyanide
9. Identify A and B. (Sep 20)
Dicarboxylic acid $\xrightarrow{SOCl_2}$ A $\xrightarrow{+2NH_3}$ B
Answer:
$HOOC-(CH_2)_n-COOH \xrightarrow{SOCl_2} ClOC-(CH_2)_n-COCl \text{ (A)} + 2NH_3 \rightarrow H_2NOC-(CH_2)_n-CONH_2 \text{ (B)}$
Result:
(A) Acid Chloride derivative
(B) Amide derivative
10. Compound 'A' of molecular $C_2H_4O$ reacts with $Na(Hg)/C_2H_5OH$ gives compound 'B' of molecular formula $C_2H_7N$ which undergo carbylamines test. Compound 'B' on reduction with nitrous acid gives compound 'C' of molecular formula $C_2H_6O$ by liberating nitrogen. Identify A, B and C and write the reactions involved. (Aug 21)
Answer:
$CH_3CN \text{ (A)} \xrightarrow{Na/Hg, C_2H_5OH} CH_3-CH_2-NH_2 \text{ (B)}$
$CH_3-CH_2-NH_2 \text{ (B)} + HNO_2 \rightarrow CH_3-CH_2-OH \text{ (C)}$
Result:
(A) $CH_3-CN$ - Methyl cyanide (Note: Formula in question is a typo for $C_2H_3N$)
(B) $CH_3-CH_2-NH_2$ - Ethyl amine
(C) $CH_3-CH_2-OH$ - Ethyl alcohol (Ethanol)
11. An organic compound (A) of molecular $C_2H_3N$ on reacts with Zn - (Hg)/Conc.HCl to give compound (B) which reacts with $HNO_3$ forming compound C (as major product) and compound (D). Compound C reacts with conc.HCl to give compound (E) (Table vinegar) and hydroxylamine. Identify A, B, C, D and E with suitable reactions. (Mar 24)
Answer:
$CH_3-CHO \text{ (A)} + 4(H) \xrightarrow{Zn-Hg/\text{conc.HCl}} CH_3-CH_3 \text{ (B)}$
$CH_3-CH_3 \text{ (B)} + HNO_3 \xrightarrow{675K} CH_3-CH_2-NO_2 \text{ (C)} + CH_3-NO_2 \text{ (D)}$
$CH_3-CH_2-NO_2 \text{ (C)} \xrightarrow{HCl/H_2O, \text{Boil}} CH_3-COOH \text{ (E)} + NH_2-OH$
Result:
(A) $CH_3-CHO$ - Acetaldehyde (Note: Formula in question is a typo for $C_2H_4O$)
(B) $CH_3-CH_3$ - Ethane
(C) $CH_3-CH_2-NO_2$ - Nitroethane
(D) $CH_3-NO_2$ - Nitromethane
(E) $CH_3-COOH$ - Acetic acid
12. Identify the compounds A, B, and C in the following reaction. (Mar 25)
$C_6H_5NO_2 \xrightarrow{Sn/HCl} A \xrightarrow{NaNO_2/HCl, 0^\circ C} B \xrightarrow{CuCN} C$
Answer:
$C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2 \text{ (A)} \xrightarrow{NaNO_2/HCl, 0^\circ C} C_6H_5N_2Cl \text{ (B)} \xrightarrow{CuCN} C_6H_5CN \text{ (C)}$
Result:
(A) $C_6H_5NH_2$ - Aniline
(B) $C_6H_5N_2Cl$ - Benzene diazonium chloride
(C) $C_6H_5CN$ - Phenyl cyanide (cyanobenzene)
13. Organic Nitrogen Compounds (Part 2)
1. Identify the compounds A, B and C in the following sequence of reactions.
$C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{HNO_2/273K} B \xrightarrow{H_2O/\Delta} C$
Answer:
$C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2 \text{ (A)} \xrightarrow{HNO_2/273K} C_6H_5N_2Cl \text{ (B)} \xrightarrow{H_2O/\Delta} C_6H_5OH \text{ (C)}$
Result:
(A) $C_6H_5NH_2$ - Aniline
(B) $C_6H_5N_2Cl$ - Benzenediazoniumchloride
(C) $C_6H_5OH$ - Phenol
2. Identify the compounds A and B in the following sequence of reactions.
$C_6H_5N_2Cl \xrightarrow{Cu/HCl} A \xrightarrow{NH_3, Cu_2O/200^\circ C} B$
Answer:
$C_6H_5N_2Cl \xrightarrow{Cu/HCl} C_6H_5Cl \text{ (A)} \xrightarrow{NH_3, Cu_2O/200^\circ C} C_6H_5NH_2 \text{ (B)}$
Result:
(A) $C_6H_5Cl$ - Chlorobenzene
(B) $C_6H_5NH_2$ - Aniline
3. Identify the compounds A, B and C in the following sequence of reactions.
$C_6H_5N_2Cl \xrightarrow{CuCN} A \xrightarrow{H_2O/H^+} B \xrightarrow{NH_3} C$
Answer:
$C_6H_5N_2Cl \xrightarrow{CuCN} C_6H_5CN \text{ (A)} \xrightarrow{H_2O/H^+} C_6H_5COOH \text{ (B)} \xrightarrow{NH_3} C_6H_5CONH_2 \text{ (C)}$
Result:
(A) $C_6H_5CN$ - Phenylcyanide (Cyanobenzene)
(B) $C_6H_5COOH$ - Benzoic acid
(C) $C_6H_5CONH_2$ - Benzamide
4. Identify the compounds A and B in the following sequence of reactions.
$C_6H_5N_2Cl \xrightarrow{H_2O/\Delta} A \xrightarrow{NH_3, \text{anhydrous } ZnCl_2/300^\circ C} B$
Answer:
$C_6H_5N_2Cl \xrightarrow{H_2O/\Delta} C_6H_5OH \text{ (A)} \xrightarrow{NH_3, \text{anhydrous } ZnCl_2/300^\circ C} C_6H_5NH_2 \text{ (B)}$
Result:
(A) $C_6H_5OH$ - Phenol
(B) $C_6H_5NH_2$ - Aniline
5. Identify the compounds A, B and C in the following sequence of reactions.
$CH_3-NO_2 \xrightarrow{LiAlH_4} A \xrightarrow{2CH_3CH_2Br} B \xrightarrow{H_2SO_4} C$
Answer:
$CH_3-NO_2 \xrightarrow{LiAlH_4} CH_3NH_2 \text{ (A)} \xrightarrow{2CH_3CH_2Br} CH_3-N(CH_2CH_3)_2 \text{ (B)} \xrightarrow{H_2SO_4} [CH_3-NH(CH_2CH_3)_2]^+HSO_4^- \text{ (C)}$
Result:
(A) Methylamine (Methanamine)
(B) N-Ethyl-N-methylethane-1-amine
(C) Ethyl methyl ammonium hydrogen sulphate
6. Identify the compounds A, B and C in the following sequence of reactions.
Butanoic acid $\xrightarrow{liq.NH_3} \dots \xrightarrow{Br_2/KOH} A \xrightarrow{NaNO_2/HCl, 0^\circ C} B \xrightarrow{H^+/KMnO_4} C$
Answer:
$CH_3CH_2CH_2COOH \xrightarrow{liq.NH_3} CH_3CH_2CH_2CONH_2 \xrightarrow{Br_2/KOH} CH_3CH_2CH_2NH_2 \text{ (A)}$
$CH_3CH_2CH_2NH_2 \text{ (A)} \xrightarrow{NaNO_2/HCl, 0^\circ C} CH_3CH_2CH_2OH \text{ (B)}$
$CH_3CH_2CH_2OH \text{ (B)} \xrightarrow{H^+/KMnO_4} CH_3CH_2COOH \text{ (C)}$
Result:
(A) $CH_3CH_2CH_2NH_2$ - Propanamine
(B) $CH_3CH_2CH_2OH$ - Propanol
(C) $CH_3CH_2COOH$ - Propanoic acid
13. Organic Nitrogen Compounds
1. There are two isomers with the formula $CH_3NO_2$. How will you distinguish between them? (Jun 24, Mar 25)
Answer:
| Nitro form ($CH_3NO_2$) |
Aci form ($CH_2=N^+(O^-)OH$) |
| 1. Less acidic |
1. More acidic |
| 2. Dissolve in NaOH slowly |
2. Dissolve in NaOH instantly |
| 3. Decolourises $FeCl_3$ solution |
3. With $FeCl_3$ gives reddish brown colour |
| 4. Electrical conductivity is low |
4. Electrical conductivity is high |
3. How is aryl halide prepared by using $Cu_2Cl_2/HCl$ (or) $Cu_2Br_2/HBr$? (or) (Write a note on Sandmeyer reaction.) (Sep 20, Mar 20)
Answer:
Benzene diazonium chloride reacts with cuprous chloride in HCl to form chlorobenzene.
$C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2 \uparrow$
Additional Important Problems (Mixed)
1. A double salt which contains fourth period alkali metal (A) on heating at 500K gives (B). Aqueous solution of (B) gives white precipitate with $BaCl_2$ and gives a red colour compound with alizarin. Identify A and B.
Answer:
$K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O \xrightarrow{500 \text{ K}} K_2SO_4 \cdot Al_2(SO_4)_3 + 24H_2O$
(A) $\rightarrow$ (B)
Result:
(A) Potash alum
(B) Burnt alum
2. Barium has a body centered cubic unit cell with a length of 508pm along an edge. What is the density of barium in $\text{g cm}^{-3}$?
Solution:
Given: For BCC, $n = 2$. $M = 137.3 \text{ g mol}^{-1}$. $a = 508 \text{ pm} = 5.08 \times 10^{-8} \text{ cm}$.
$\rho = \frac{n \times M}{a^3 \times N_A}$
$\rho = \frac{2 \times 137.3}{(5.08 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$
$\rho \approx 3.5 \text{ g cm}^{-3}$
