12th CHEMISTRY - PUBLIC COMPULSORY QUESTIONS AND PROBLEMS
2. P- Block Elements I
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)
$2LiH + B_2H_6 \xrightarrow{\text{Ether}} 2LiBH_4$
(A) + (B) $\rightarrow$ (C)
Result:
(A) $LiH$ - Lithium Hydride
(B) $B_2H_6$ - Diborane
(C) $LiBH_4$ - Lithium Borohydride
(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.
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
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
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) $NH_3$
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
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
(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
(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) $[Co(NH_3)_5Cl]^{2+}$
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
A) $[Ag(NH_3)_2]^+$ - Diamminesilver(I) ion
B) $[Co(NH_3)_5Cl]^{2+}$ - Pentaamminechloridocobalt(III) ion
6. Solid State
1. Classify the following solids (Aug 21)
a) Naphthalene b) Brass c) Diamond d) NaCl e) Glucose g) $SiO_2$
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
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:
$FCC = \frac{N_c}{8} + \frac{N_f}{2} = \frac{8}{8} + \frac{6}{2} = 1 + 3 = 4$
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.
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.
7. Chemical Kinetics
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.
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).
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: $k = 1.54 \times 10^{-3} s^{-1}$, $t_{1/2} = ?$
$t_{1/2} = \frac{0.693}{k}$
$t_{1/2} = \frac{0.693}{1.54 \times 10^{-3}} = 450 \text{ s}$
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, $t_{99\%} = 2 \times t_{90\%}$
$t = \frac{2.303}{k} \log \frac{[A_o]}{[A]}$
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
$\text{Rate} = k[x]^m[y]^n$
$4 \times 10^{-3} \text{ mol L}^{-1}\text{s}^{-1} = 2 \times 10^{-2} \text{s}^{-1} (0.2 \text{ mol L}^{-1})^m (0.2 \text{ mol L}^{-1})^n$
$\frac{4 \times 10^{-3}}{2 \times 10^{-2}} = (0.2)^{m+n}$
$0.2 = (0.2)^{m+n}$
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.
$t_{1/2} = \frac{0.693}{k}$
$\rightarrow$ The half-life of a first order reaction is independent of the initial concentration.
$t_{1/2} = \frac{0.693}{k}$
8. Ionic Equilibrium
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.
$K_w = [H_3O^+][OH^-] = 1 \times 10^{-14}$
$[OH^-] = \frac{K_w}{[H_3O^+]} = \frac{1 \times 10^{-14}}{2 \times 10^{-3}} = 5 \times 10^{-12} \text{ M}$
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:
$pH = 7 + \frac{pK_a}{2} + \frac{\log c}{2}$
$pH = 7 + \frac{4.74}{2} + \frac{\log 0.1}{2}$
$pH = 7 + 2.37 - 0.5 = 8.87$
$pH = 7 + \frac{pK_a}{2} + \frac{\log c}{2}$
$pH = 7 + \frac{4.74}{2} + \frac{\log 0.1}{2}$
$pH = 7 + 2.37 - 0.5 = 8.87$
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:
$pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}$
$pH = 4.74 + \log\frac{0.20}{0.18} = 4.74 + \log\frac{10}{9}$
$pH = 4.74 + \log 10 - \log 9 = 4.74 + 1 - 0.95 = 4.79$
$pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 5 - \log 1.8 = 5 - 0.26 = 4.74$
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:
$[H^+] = \sqrt{K_a \times C} = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \text{ M}$
$pH = -\log[H^+] = -\log(1.34 \times 10^{-3}) = 3 - \log 1.34 = 3 - 0.1271 = 2.87$
$[H^+] = \sqrt{K_a \times C} = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \text{ M}$
$pH = -\log[H^+] = -\log(1.34 \times 10^{-3}) = 3 - \log 1.34 = 3 - 0.1271 = 2.87$
9. Electrochemistry
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, $m = Z I t$
$m = \frac{108}{96500} \times 2 \times 30 \times 60 = \frac{108 \times 36}{965} = 4.03 \text{ g}$
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: $I = 2\text{A}$, $t = 20 \times 60 \text{ s}$
$m = Z I t = \frac{108}{96500} \times 2 \times 20 \times 60 = \frac{108 \times 24}{965} = 2.68 \text{ g}$
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 $\kappa = \frac{1}{R} \times \frac{l}{A}$
$\kappa = \frac{1}{15} \times \frac{1.5 \times 10^{-2}}{4.5 \times 10^{-4}} = 22.2 \text{ S m}^{-1}$
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 $Br_2$.
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.
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.
➤ 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)
$C_6H_5OH + NH_3 \xrightarrow{\text{Anhydrous } ZnCl_2} C_6H_5NH_2 + H_2O$
(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
(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)
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
$C_6H_5OH \xrightarrow{Zn, \Delta} C_6H_6 + ZnO$
(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:
$C_6H_5-Cl \text{ (chlorobenzene)} + NaOH \xrightarrow{623K, 300\text{ bar}} C_6H_5ONa \xrightarrow{HCl} C_6H_5OH \text{ (phenol)} + NaCl$
$C_6H_5-Cl \text{ (chlorobenzene)} + NaOH \xrightarrow{623K, 300\text{ bar}} C_6H_5ONa \xrightarrow{HCl} C_6H_5OH \text{ (phenol)} + NaCl$
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:
$HO-CH_2-CH_2-OH \text{ (A)} \xrightarrow{\text{(i) Anhydrous } ZnCl_2 \text{ (ii) Tautomerisation}} CH_3-CHO \text{ (B)}$
$CH_3-CHO \text{ (B)} \xrightarrow{\text{Zn-Hg/Conc HCl (Clemmensen Reduction)}} CH_3-CH_3 \text{ (C)}$
$HO-CH_2-CH_2-OH \text{ (A)} \xrightarrow{\text{(i) Anhydrous } ZnCl_2 \text{ (ii) Tautomerisation}} CH_3-CHO \text{ (B)}$
$CH_3-CHO \text{ (B)} \xrightarrow{\text{Zn-Hg/Conc HCl (Clemmensen Reduction)}} CH_3-CH_3 \text{ (C)}$
Result:
(A) $HO-CH_2-CH_2-OH$ - Ethylene glycol
(B) $CH_3-CHO$ - Acetaldehyde (Ethanal)
(C) $CH_3-CH_3$ - Ethane
(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.
$C_6H_5-OH \xrightarrow{\text{Zn dust}} A \xrightarrow[\text{anhydrous } AlCl_3]{CH_3Cl} B \xrightarrow{H^+/KMnO_4} C$
$C_6H_5-OH \xrightarrow{\text{Zn dust}} A \xrightarrow[\text{anhydrous } AlCl_3]{CH_3Cl} B \xrightarrow{H^+/KMnO_4} C$
Answer:
$C_6H_5-OH \xrightarrow{\text{Zn dust}} C_6H_6 \text{ (A)} \xrightarrow[\text{anhydrous } AlCl_3]{CH_3Cl} C_6H_5-CH_3 \text{ (B)} \xrightarrow{H^+/KMnO_4} C_6H_5-COOH \text{ (C)}$
$C_6H_5-OH \xrightarrow{\text{Zn dust}} C_6H_6 \text{ (A)} \xrightarrow[\text{anhydrous } AlCl_3]{CH_3Cl} C_6H_5-CH_3 \text{ (B)} \xrightarrow{H^+/KMnO_4} C_6H_5-COOH \text{ (C)}$
Result:
(A) $C_6H_6$ - Benzene
(B) $C_6H_5-CH_3$ - Toluene
(C) $C_6H_5-COOH$ - Benzoic acid
(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.
$C_6H_5-OH \xrightarrow{PBr_3} A \xrightarrow{\text{Aq.NaOH}} B \rightarrow C$
$C_6H_5-OH \xrightarrow{PBr_3} A \xrightarrow{\text{Aq.NaOH}} B \rightarrow C$
Answer:
$CH_3CH_2-OH \xrightarrow{PBr_3} CH_3CH_2Br \text{ (A)} \xrightarrow{\text{Aq.NaOH}} CH_3CH_2-OH \text{ (B)} \rightarrow CH_3CH_2-ONa \text{ (C)}$
$CH_3CH_2-OH \xrightarrow{PBr_3} CH_3CH_2Br \text{ (A)} \xrightarrow{\text{Aq.NaOH}} CH_3CH_2-OH \text{ (B)} \rightarrow CH_3CH_2-ONa \text{ (C)}$
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
(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:
$C_6H_5-OH \text{ (phenol)} \xrightarrow{\text{Zn dust}} C_6H_6 \text{ (A)} \xrightarrow[\text{anhydrous } AlCl_3]{CH_3CH_2CH_2Cl} C_6H_5CH_2CH_2CH_3 \text{ (B)} \xrightarrow{H^+/KMnO_4} C_6H_5-COOH \text{ (C)}$
$C_6H_5-OH \text{ (phenol)} \xrightarrow{\text{Zn dust}} C_6H_6 \text{ (A)} \xrightarrow[\text{anhydrous } AlCl_3]{CH_3CH_2CH_2Cl} C_6H_5CH_2CH_2CH_3 \text{ (B)} \xrightarrow{H^+/KMnO_4} C_6H_5-COOH \text{ (C)}$
Result:
(A) $C_6H_6$ - Benzene
(B) $C_6H_5CH_2CH_2CH_3$ - n-propyl benzene
(C) $C_6H_5-COOH$ - Benzoic acid
(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.
$C_6H_5-OH \xrightarrow{\text{NaOH}} A \xrightarrow[\text{ii) 4-7bar}]{\text{i) } CO_2, 400K} B \xrightarrow{H^+/H_2O} C$
$C_6H_5-OH \xrightarrow{\text{NaOH}} A \xrightarrow[\text{ii) 4-7bar}]{\text{i) } CO_2, 400K} B \xrightarrow{H^+/H_2O} C$
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)
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
(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)
$C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{HNO_2/273K} B \xrightarrow{C_6H_5OH} C$
$C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{HNO_2/273K} B \xrightarrow{C_6H_5OH} C$
Answer:
$C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2 \text{ (A)} + 2H_2O$
$C_6H_5NH_2 \xrightarrow{HNO_2/273K} C_6H_5N_2Cl \text{ (B)}$
$C_6H_5N_2Cl + C_6H_5OH \rightarrow p\text{-Hydroxy azobenzene (C)}$
$C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2 \text{ (A)} + 2H_2O$
$C_6H_5NH_2 \xrightarrow{HNO_2/273K} C_6H_5N_2Cl \text{ (B)}$
$C_6H_5N_2Cl + C_6H_5OH \rightarrow p\text{-Hydroxy azobenzene (C)}$
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
(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:
$C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2$
$C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2$
4. Identify the compounds A, and B in the following sequence of reactions. (May 22)
$CH_3-Br \xrightarrow{NaN_3} A \xrightarrow{LiAlH_4} B$
$CH_3-Br \xrightarrow{NaN_3} A \xrightarrow{LiAlH_4} B$
Answer:
$CH_3-Br \xrightarrow{NaN_3} CH_3N_3 \text{ (A)} \xrightarrow{LiAlH_4} CH_3-NH_2 \text{ (B)} + N_2$
$CH_3-Br \xrightarrow{NaN_3} CH_3N_3 \text{ (A)} \xrightarrow{LiAlH_4} CH_3-NH_2 \text{ (B)} + N_2$
Result:
(A) $CH_3N_3$ - Methyl azide
(B) $CH_3-NH_2$ - Methyl amine
(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)
$CH_3CH_2-NO_2 \xrightarrow{Sn/HCl} A \xrightarrow{CH_3COCl} B$
$CH_3CH_2-NO_2 \xrightarrow{Sn/HCl} A \xrightarrow{CH_3COCl} B$
Answer:
$CH_3CH_2-NO_2 \xrightarrow{Sn/HCl} CH_3CH_2NH_2 \text{ (A)}$
$CH_3CH_2NH_2 \xrightarrow{CH_3COCl} CH_3CH_2-NH-COCH_3 \text{ (B)}$
$CH_3CH_2-NO_2 \xrightarrow{Sn/HCl} CH_3CH_2NH_2 \text{ (A)}$
$CH_3CH_2NH_2 \xrightarrow{CH_3COCl} CH_3CH_2-NH-COCH_3 \text{ (B)}$
Result:
(A) $CH_3CH_2NH_2$ - Ethyl amine
(B) $CH_3CH_2-NH-COCH_3$ - N-Ethyl acetamide
(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)
$CH_3-NO_2 \xrightarrow{Sn/HCl, 6(H)} A$
$CH_3-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} B$
$CH_3-NO_2 \xrightarrow{Sn/HCl, 6(H)} A$
$CH_3-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} B$
Answer:
$CH_3-NO_2 \xrightarrow{Sn/HCl, 6(H)} CH_3-NH_2 \text{ (A)}$
$CH_3-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} CH_3-NHOH \text{ (B)}$
$CH_3-NO_2 \xrightarrow{Sn/HCl, 6(H)} CH_3-NH_2 \text{ (A)}$
$CH_3-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} CH_3-NHOH \text{ (B)}$
Result:
(A) $CH_3-NH_2$ - Methyl amine
(B) $CH_3-NHOH$ - Methyl hydroxylamine
(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)
$C_6H_5-NO_2 \xrightarrow{Sn/HCl} A$
$C_6H_5-NO_2 \xrightarrow{Zn/NH_4Cl} B$
$C_6H_5-NO_2 \xrightarrow{Zn/NaOH} C$
$C_6H_5-NO_2 \xrightarrow{Sn/HCl} A$
$C_6H_5-NO_2 \xrightarrow{Zn/NH_4Cl} B$
$C_6H_5-NO_2 \xrightarrow{Zn/NaOH} C$
Answer:
$C_6H_5-NO_2 \xrightarrow{Sn/HCl, 6(H)} C_6H_5-NH_2 \text{ (A)}$
$C_6H_5-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} C_6H_5-NHOH \text{ (B)}$
$C_6H_5-NO_2 \xrightarrow{Zn/NaOH} C_6H_5-NH-NH-C_6H_5 \text{ (C)}$
$C_6H_5-NO_2 \xrightarrow{Sn/HCl, 6(H)} C_6H_5-NH_2 \text{ (A)}$
$C_6H_5-NO_2 \xrightarrow{Zn/NH_4Cl, 4(H)} C_6H_5-NHOH \text{ (B)}$
$C_6H_5-NO_2 \xrightarrow{Zn/NaOH} C_6H_5-NH-NH-C_6H_5 \text{ (C)}$
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
(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 $\xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-NH-CH_3$
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:
$CH_3CN \text{ (A)} \xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-CH_2-NH_2$
$CH_3NC \text{ (B)} \xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-NH-CH_3$
$CH_3CN \text{ (A)} \xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-CH_2-NH_2$
$CH_3NC \text{ (B)} \xrightarrow{Na-Hg/C_2H_5OH, 4H} CH_3-NH-CH_3$
Result:
(A) $CH_3CN$ - Methyl cyanide
(B) $CH_3NC$ - Methyl isocyanide
(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
Dicarboxylic acid $\xrightarrow{SOCl_2}$ A $\xrightarrow{+2NH_3}$ B
Answer:
$HOOC-(CH_2)_n-COOH \xrightarrow{SOCl_2} ClOC-(CH_2)_n-COCl \text{ (A)} + 2NH_3 \rightarrow H_2NOC-(CH_2)_n-CONH_2 \text{ (B)}$
$HOOC-(CH_2)_n-COOH \xrightarrow{SOCl_2} ClOC-(CH_2)_n-COCl \text{ (A)} + 2NH_3 \rightarrow H_2NOC-(CH_2)_n-CONH_2 \text{ (B)}$
Result:
(A) Acid Chloride derivative
(B) Amide derivative
(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:
$CH_3CN \text{ (A)} \xrightarrow{Na/Hg, C_2H_5OH} CH_3-CH_2-NH_2 \text{ (B)}$
$CH_3-CH_2-NH_2 \text{ (B)} + HNO_2 \rightarrow CH_3-CH_2-OH \text{ (C)}$
$CH_3CN \text{ (A)} \xrightarrow{Na/Hg, C_2H_5OH} CH_3-CH_2-NH_2 \text{ (B)}$
$CH_3-CH_2-NH_2 \text{ (B)} + HNO_2 \rightarrow CH_3-CH_2-OH \text{ (C)}$
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)
(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:
$CH_3-CHO \text{ (A)} + 4(H) \xrightarrow{Zn-Hg/\text{conc.HCl}} CH_3-CH_3 \text{ (B)}$
$CH_3-CH_3 \text{ (B)} + HNO_3 \xrightarrow{675K} CH_3-CH_2-NO_2 \text{ (C)} + CH_3-NO_2 \text{ (D)}$
$CH_3-CH_2-NO_2 \text{ (C)} \xrightarrow{HCl/H_2O, \text{Boil}} CH_3-COOH \text{ (E)} + NH_2-OH$
$CH_3-CHO \text{ (A)} + 4(H) \xrightarrow{Zn-Hg/\text{conc.HCl}} CH_3-CH_3 \text{ (B)}$
$CH_3-CH_3 \text{ (B)} + HNO_3 \xrightarrow{675K} CH_3-CH_2-NO_2 \text{ (C)} + CH_3-NO_2 \text{ (D)}$
$CH_3-CH_2-NO_2 \text{ (C)} \xrightarrow{HCl/H_2O, \text{Boil}} CH_3-COOH \text{ (E)} + NH_2-OH$
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
(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)
$C_6H_5NO_2 \xrightarrow{Sn/HCl} A \xrightarrow{NaNO_2/HCl, 0^\circ C} B \xrightarrow{CuCN} C$
$C_6H_5NO_2 \xrightarrow{Sn/HCl} A \xrightarrow{NaNO_2/HCl, 0^\circ C} B \xrightarrow{CuCN} C$
Answer:
$C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2 \text{ (A)} \xrightarrow{NaNO_2/HCl, 0^\circ C} C_6H_5N_2Cl \text{ (B)} \xrightarrow{CuCN} C_6H_5CN \text{ (C)}$
$C_6H_5NO_2 \xrightarrow{Sn/HCl} C_6H_5NH_2 \text{ (A)} \xrightarrow{NaNO_2/HCl, 0^\circ C} C_6H_5N_2Cl \text{ (B)} \xrightarrow{CuCN} C_6H_5CN \text{ (C)}$
Result:
(A) $C_6H_5NH_2$ - Aniline
(B) $C_6H_5N_2Cl$ - Benzene diazonium chloride
(C) $C_6H_5CN$ - Phenyl cyanide (cyanobenzene)
(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.
$C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{HNO_2/273K} B \xrightarrow{H_2O/\Delta} C$
$C_6H_5NO_2 \xrightarrow{Fe/HCl} A \xrightarrow{HNO_2/273K} B \xrightarrow{H_2O/\Delta} C$
Answer:
$C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2 \text{ (A)} \xrightarrow{HNO_2/273K} C_6H_5N_2Cl \text{ (B)} \xrightarrow{H_2O/\Delta} C_6H_5OH \text{ (C)}$
$C_6H_5NO_2 \xrightarrow{Fe/HCl} C_6H_5NH_2 \text{ (A)} \xrightarrow{HNO_2/273K} C_6H_5N_2Cl \text{ (B)} \xrightarrow{H_2O/\Delta} C_6H_5OH \text{ (C)}$
Result:
(A) $C_6H_5NH_2$ - Aniline
(B) $C_6H_5N_2Cl$ - Benzenediazoniumchloride
(C) $C_6H_5OH$ - Phenol
(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.
$C_6H_5N_2Cl \xrightarrow{Cu/HCl} A \xrightarrow{NH_3, Cu_2O/200^\circ C} B$
$C_6H_5N_2Cl \xrightarrow{Cu/HCl} A \xrightarrow{NH_3, Cu_2O/200^\circ C} B$
Answer:
$C_6H_5N_2Cl \xrightarrow{Cu/HCl} C_6H_5Cl \text{ (A)} \xrightarrow{NH_3, Cu_2O/200^\circ C} C_6H_5NH_2 \text{ (B)}$
$C_6H_5N_2Cl \xrightarrow{Cu/HCl} C_6H_5Cl \text{ (A)} \xrightarrow{NH_3, Cu_2O/200^\circ C} C_6H_5NH_2 \text{ (B)}$
Result:
(A) $C_6H_5Cl$ - Chlorobenzene
(B) $C_6H_5NH_2$ - Aniline
(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.
$C_6H_5N_2Cl \xrightarrow{CuCN} A \xrightarrow{H_2O/H^+} B \xrightarrow{NH_3} C$
$C_6H_5N_2Cl \xrightarrow{CuCN} A \xrightarrow{H_2O/H^+} B \xrightarrow{NH_3} C$
Answer:
$C_6H_5N_2Cl \xrightarrow{CuCN} C_6H_5CN \text{ (A)} \xrightarrow{H_2O/H^+} C_6H_5COOH \text{ (B)} \xrightarrow{NH_3} C_6H_5CONH_2 \text{ (C)}$
$C_6H_5N_2Cl \xrightarrow{CuCN} C_6H_5CN \text{ (A)} \xrightarrow{H_2O/H^+} C_6H_5COOH \text{ (B)} \xrightarrow{NH_3} C_6H_5CONH_2 \text{ (C)}$
Result:
(A) $C_6H_5CN$ - Phenylcyanide (Cyanobenzene)
(B) $C_6H_5COOH$ - Benzoic acid
(C) $C_6H_5CONH_2$ - Benzamide
(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.
$C_6H_5N_2Cl \xrightarrow{H_2O/\Delta} A \xrightarrow{NH_3, \text{anhydrous } ZnCl_2/300^\circ C} B$
$C_6H_5N_2Cl \xrightarrow{H_2O/\Delta} A \xrightarrow{NH_3, \text{anhydrous } ZnCl_2/300^\circ C} B$
Answer:
$C_6H_5N_2Cl \xrightarrow{H_2O/\Delta} C_6H_5OH \text{ (A)} \xrightarrow{NH_3, \text{anhydrous } ZnCl_2/300^\circ C} C_6H_5NH_2 \text{ (B)}$
$C_6H_5N_2Cl \xrightarrow{H_2O/\Delta} C_6H_5OH \text{ (A)} \xrightarrow{NH_3, \text{anhydrous } ZnCl_2/300^\circ C} C_6H_5NH_2 \text{ (B)}$
Result:
(A) $C_6H_5OH$ - Phenol
(B) $C_6H_5NH_2$ - Aniline
(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.
$CH_3-NO_2 \xrightarrow{LiAlH_4} A \xrightarrow{2CH_3CH_2Br} B \xrightarrow{H_2SO_4} C$
$CH_3-NO_2 \xrightarrow{LiAlH_4} A \xrightarrow{2CH_3CH_2Br} B \xrightarrow{H_2SO_4} C$
Answer:
$CH_3-NO_2 \xrightarrow{LiAlH_4} CH_3NH_2 \text{ (A)} \xrightarrow{2CH_3CH_2Br} CH_3-N(CH_2CH_3)_2 \text{ (B)} \xrightarrow{H_2SO_4} [CH_3-NH(CH_2CH_3)_2]^+HSO_4^- \text{ (C)}$
$CH_3-NO_2 \xrightarrow{LiAlH_4} CH_3NH_2 \text{ (A)} \xrightarrow{2CH_3CH_2Br} CH_3-N(CH_2CH_3)_2 \text{ (B)} \xrightarrow{H_2SO_4} [CH_3-NH(CH_2CH_3)_2]^+HSO_4^- \text{ (C)}$
Result:
(A) Methylamine (Methanamine)
(B) N-Ethyl-N-methylethane-1-amine
(C) Ethyl methyl ammonium hydrogen sulphate
(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 $\xrightarrow{liq.NH_3} \dots \xrightarrow{Br_2/KOH} A \xrightarrow{NaNO_2/HCl, 0^\circ C} B \xrightarrow{H^+/KMnO_4} C$
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:
$CH_3CH_2CH_2COOH \xrightarrow{liq.NH_3} CH_3CH_2CH_2CONH_2 \xrightarrow{Br_2/KOH} CH_3CH_2CH_2NH_2 \text{ (A)}$
$CH_3CH_2CH_2NH_2 \text{ (A)} \xrightarrow{NaNO_2/HCl, 0^\circ C} CH_3CH_2CH_2OH \text{ (B)}$
$CH_3CH_2CH_2OH \text{ (B)} \xrightarrow{H^+/KMnO_4} CH_3CH_2COOH \text{ (C)}$
$CH_3CH_2CH_2COOH \xrightarrow{liq.NH_3} CH_3CH_2CH_2CONH_2 \xrightarrow{Br_2/KOH} CH_3CH_2CH_2NH_2 \text{ (A)}$
$CH_3CH_2CH_2NH_2 \text{ (A)} \xrightarrow{NaNO_2/HCl, 0^\circ C} CH_3CH_2CH_2OH \text{ (B)}$
$CH_3CH_2CH_2OH \text{ (B)} \xrightarrow{H^+/KMnO_4} CH_3CH_2COOH \text{ (C)}$
Result:
(A) $CH_3CH_2CH_2NH_2$ - Propanamine
(B) $CH_3CH_2CH_2OH$ - Propanol
(C) $CH_3CH_2COOH$ - Propanoic acid
(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.
$C_6H_5N_2Cl \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2 \uparrow$
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)
$K_2SO_4 \cdot Al_2(SO_4)_3 \cdot 24H_2O \xrightarrow{500 \text{ K}} K_2SO_4 \cdot Al_2(SO_4)_3 + 24H_2O$
(A) $\rightarrow$ (B)
Result:
(A) Potash alum
(B) Burnt alum
(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, $n = 2$. $M = 137.3 \text{ g mol}^{-1}$. $a = 508 \text{ pm} = 5.08 \times 10^{-8} \text{ cm}$.
$\rho = \frac{n \times M}{a^3 \times N_A}$
$\rho = \frac{2 \times 137.3}{(5.08 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$
$\rho \approx 3.5 \text{ g cm}^{-3}$
Given: For BCC, $n = 2$. $M = 137.3 \text{ g mol}^{-1}$. $a = 508 \text{ pm} = 5.08 \times 10^{-8} \text{ cm}$.
$\rho = \frac{n \times M}{a^3 \times N_A}$
$\rho = \frac{2 \times 137.3}{(5.08 \times 10^{-8})^3 \times 6.022 \times 10^{23}}$
$\rho \approx 3.5 \text{ g cm}^{-3}$
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