Accomplish the following conversions:
Question: Accomplish the following conversions: (i)Nitrobenzene to benzoic acid (ii)Benzene tom-bromophenol (iii)Benzoic acid to aniline (iv)Aniline to 2,4,6-tribromofluorobenzene (v)Benzyl chloride to 2-phenylethanamine (vi)Chlorobenzene top-chloroaniline (vii)Aniline top-bromoaniline (viii)Benzamide to toluene (ix)Aniline to benzyl alcohol. Solution: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii) (ix)...
Read More →Write short notes on the following:
Question: Write short notes on the following: (i)Carbylamine reaction(ii)Diazotisation (iii)Hofmanns bromamide reaction(iv)Coupling reaction (v)Ammonolysis(vi)Acetylation (vii)Gabriel phthalimide synthesis. Solution: (i)Carbylamine reaction Carbylamine reaction is used as a test for the identification of primary amines. When aliphatic and aromatic primary amines are heated with chloroform and ethanolic potassium hydroxide, carbylamines (or isocyanides) are formed. These carbylamines have very un...
Read More →A calorimter of water equivalent 20 g contains 180 g
Question: A calorimter of water equivalent $20 \mathrm{~g}$ contains $180 \mathrm{~g}$ of water at $25^{\circ} \mathrm{C} . ~ ' m$ ' grams of steam at $100^{\circ} \mathrm{C}$ is mixed in it till the temperature of the mixture is $31^{\circ} \mathrm{C}$. The value of $' m^{\prime}$ is close to (Latent heat of water $=540 \mathrm{cal} \mathrm{g}^{-1}$, specific heat of water $=1 \mathrm{cal} \mathrm{g}^{-1^{\circ}}{ }^{\circ}{ }^{-1}$ )24$3.2$$2.6$Correct Option: 1 Solution: (1) Heat given by wat...
Read More →Describe a method for the identification of primary,
Question: Describe a method for the identification of primary, secondary and tertiary amines. Also write chemical equations of the reactions involved. Solution: Primary, secondary and tertiary amines can be identified and distinguished by Hinsbergs test. In this test, the amines are allowed to react with Hinsbergs reagent, benzenesulphonyl chloride (C6H5SO2Cl). The three types of amines react differently with Hinsbergs reagent. Therefore, they can be easily identified using Hinsbergs reagent. Pr...
Read More →How will you convert:
Question: How will you convert: (i)Ethanoic acid into methanamine (ii)Hexanenitrile into 1-aminopentane (iii)Methanol to ethanoic acid (iv)Ethanamine into methanamine (v)Ethanoic acid into propanoic acid (vi)Methanamine into ethanamine (vii)Nitromethane into dimethylamine (viii)Propanoic acid into ethanoic acid Solution: (i) (ii) (iii) (iv) (v) (vi) (vii) (viii)...
Read More →Find the number of coins,
Question: Find the number of coins, 1.5 cm is diameter and 0.2 cm thick, to be melted to form a right circular cylinder of height 10 cm and diameter 4.5 cm. Solution: Volume of one coin $=\pi \times(1.5)^{2 / 4} \times 0.2$ Volume of cylinder $=\pi \times(4.5)^{2} \times \frac{10}{4}$ So number of coins to be melted $=\frac{\text { volume of cylinder }}{\text { volume of each coin }}$ $=\left(\frac{4.5}{1.5}\right)^{2} \times \frac{10}{0.2}$ $=9 \times 50$ $=450$...
Read More →Arrange the following:
Question: Arrange the following: (i)In decreasing order of the pKbvalues: C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2 (ii)In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2 (iii)In increasing order of basic strength: (a)Aniline,p-nitroaniline andp-toluidine (b)C6H5NH2, C6H5NHCH3, C6H5CH2NH2. (iv)In decreasing order of basic strength in gas phase: C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3 (v)In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2 (vi)In increasing ...
Read More →Arrange the following:
Question: Arrange the following: (i)In decreasing order of the pKbvalues: C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2 (ii)In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2 (iii)In increasing order of basic strength: (a)Aniline,p-nitroaniline andp-toluidine (b)C6H5NH2, C6H5NHCH3, C6H5CH2NH2. (iv)In decreasing order of basic strength in gas phase: C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3 (v)In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2 (vi)In increasing ...
Read More →In the given figure, PA, QB and RC are perpendicular to AC. If AP = x, QB = z, RC = y,
Question: In the given figure, $P A, Q B$ and $R C$ are perpendicular to $A C$. If $A P=x, Q B=z, R C=y, A B=a$ and $B C=b$, show that $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}$. Solution: In $\triangle P A C$ and $\triangle Q B C$, we have : $\angle A=\angle B \quad\left(\right.$ Both angles are $\left.90^{\circ}\right)$ $\angle P=\angle Q \quad$ (Correspond ing angles) and $\angle C=\angle C \quad$ (Common angles) Therefore, $\triangle P A C \sim \triangle Q B C$ $\frac{A P}{B Q}=\frac{A C}{B C}$ $...
Read More →A hemispherical tank full of water is emptied by a pipe
Question: A hemispherical tank full of water is emptied by a pipe at the rate of $\frac{25}{7}$ litres per second. How much time will it take to half-empty the tank, If the tank is 3 metres in diameter? Solution: Volume of half of hemispherical tank $=\frac{2}{3} \pi r^{3}$ $=\frac{2}{3} \times \frac{22}{7} \times\left(\frac{3}{2}\right)^{3}$ $=\frac{49500}{7} \mathrm{ltr} .$ Amount of water emptied by pipe in $1 \mathrm{sec} .=\frac{25}{7} \mathrm{ltr}$. So, time taken $=\frac{7}{25} \times \fr...
Read More →Account for the following:
Question: Account for the following: (i)pKbof aniline is more than that of methylamine. (ii)Ethylamine is soluble in water whereas aniline is not. (iii)Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (iv)Although amino group iso,pdirecting in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount ofm-nitroaniline. (v)Aniline does not undergo Friedel-Crafts reaction. (vi)Diazonium salts of aromatic amines are more stab...
Read More →A metallic sphere cools from
Question: A metallic sphere cools from $50^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ in $300 \mathrm{~s}$. If atmospheric temperature around is $20^{\circ} \mathrm{C}$, then the sphere's temperature after the next 5 minutes will be close to :$31^{\circ} \mathrm{C}$$33^{\circ} \mathrm{C}$$28^{\circ} \mathrm{C}$$35^{\circ} \mathrm{C}$Correct Option: , 2 Solution: (2) From Newton's Law of cooling, $\frac{T_{1}-T_{2}}{t}=K\left[\frac{T_{1}+T_{2}}{2}-T_{0}\right]$ Here, $\mathrm{T}_{1}=50^{\circ...
Read More →Give one chemical test to distinguish between the following pairs of compounds.
Question: Give one chemical test to distinguish between the following pairs of compounds. (i)Methylamine and dimethylamine (ii)Secondary and tertiary amines (iii)Ethylamine and aniline (iv)Aniline and benzylamine (v)Aniline and N-methylaniline. Solution: (i)Methylamine and dimethylamine can be distinguished by the carbylamine test. Carbylamine test: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form foul-smelling isocyanides or carbylamines. M...
Read More →A solid metal sphere of 6 cm diameter
Question: A solid metal sphere of 6 cm diameter is melted and a circular sheet of thickness 1 cm is prepared. Determine the diameter of the sheet. Solution: Diameter of sphere = 6 cm Therefore, Radius = 3 cm. Therefore, Surface area of sphere $=4 \pi r^{2}$ $=4 \times \pi \times(3)^{2}$ $=36 \pi \mathrm{cm}^{2}$ Area of the circular sheet $=\pi r^{2}$ Therefore, Surface area of sphere = area of the circular sheet $r=6 \mathrm{~cm}$ Therefore, Diameter of the sheet = 2 6 = 12 cm...
Read More →Write IUPAC names of the following compounds and classify them into primary,
Question: Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i)(CH3)2CHNH2(ii)CH3(CH2)2NH2 (iii)CH3NHCH(CH3)2(iv)(CH3)3CNH2 (v)C6H5NHCH3(vi)(CH3CH2)2NCH3 (vii)mBrC6H4NH2 Solution: (i)1-Methylethanamine (10amine) (ii)Propan-1-amine (10amine) (iii)NMethyl-2-methylethanamine (20amine) (iv)2-Methylpropan-2-amine (10amine) (v)NMethylbenzamine or N-methylaniline (20amine) (vi)N-Ethyl-N-methylethanamine (30amine) (vii)3-Bromobenzenamine or 3-bro...
Read More →If r1 and r2 be the radii of two solid metallic
Question: If $r_{1}$ and $r_{2}$ be the radii of two solid metallic spheres and if they are melted into one solid sphere, prove that the radius of the new sphere is $\left(r_{1}^{3}+r_{2}^{3}\right) .^{1 / 3}$ Solution: Volume of first sphere $=\frac{4}{3} \pi r_{1}^{3}$ Volume of second sphere $=\frac{4}{3} \pi r_{2}^{3}$ Total volume of new sphere $=\left(\frac{4}{3} \pi r_{1}^{3}+\frac{4}{3} \pi r_{2}^{3}\right)$ Say of radius of new sphere =r3 Volume of new sphere $=\frac{4}{3} \pi r_{3}^{3}...
Read More →Convert
Question: Convert (i)3-Methylaniline into 3-nitrotoluene. (ii)Aniline into 1,3,5-tribromobenzene. Solution: (i) (ii)...
Read More →Write structures of different isomers corresponding to the molecular formula,
Question: Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid. Solution: The structures of different isomers corresponding to the molecular formula, C3H9N are given below: (a) $\mathrm{CH}_{3}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$ Propan-1-amine (10) (b) Propan-2-amine (10) (c) (d) N,N-Dimethylmethanamine (30) 10amines, (a) propan-1-amine, and (b) Propan-2-...
Read More →A bakelite beaker has volume capacity of 500cc
Question: A bakelite beaker has volume capacity of $500 \mathrm{cc}$ at $30^{\circ} \mathrm{C}$. When it is partially filled with $V_{m}$ volume (at $30^{\circ} \mathrm{C}$ ) of mercury, it is found that the unfilled volume of the beaker remains constant as temperature is varied. If $\gamma_{\text {(beaker) }}=6 \times$ $10^{-6}{ }^{\circ} \mathrm{C}^{-1}$ and $\gamma_{\text {(mercury) }}=1.5 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$, where $\gamma$ is the coefficient of volume expansion, then ...
Read More →In the given figure, ∆ABC is an obtuse triangle, obtuse-angles at B
Question: In the given figure, ∆ABCis an obtuse triangle, obtuse-angles atB. IfADCB,Prove thatAC2=AB2+BC2+ 2BCBD. Solution: Applying Pythagoras theorem in right-angled triangle ADC, we get: $A C^{2}=A D^{2}+D C^{2}$ $\Rightarrow A C^{2}-D C^{2}=A D^{2}$ $\Rightarrow A D^{2}=A C^{2}-D C^{2}$ ..........(1) Applying Pythagoras theorem in right-angled triangle ADB, we get: $A B^{2}=A D^{2}+D B^{2}$ $\Rightarrow A B^{2}-D B^{2}=A D^{2}$ $\Rightarrow A D^{2}=A B^{2}-D B^{2} \quad \ldots(2)$ From equat...
Read More →Write chemical reaction of aniline with benzoyl chloride and write the name of
Question: Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained. Solution:...
Read More →A rectangular vessel of dimensions 20 cm × 16 cm × 11 cm
Question: A rectangular vessel of dimensions 20 cm 16 cm 11 cm is full of water. This water is poured into a conical vessel. The top of the conical vessel has its radius 10 cm. If the conical vessel is filled completely, determine its height. Solution: Volume of cone $=\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}$ $\frac{1}{3} \pi \mathrm{r}^{2} \mathrm{~h}=20 \times 16 \times 11$ $=33.6 \mathrm{~cm}$...
Read More →Write reactions of the final alkylation product of aniline with excess of
Question: Write reactions of the final alkylation product of aniline with excess of methyliodide in the presence of sodium carbonate solution. Solution: Aniline reacts with methyl iodide to produce N, N-dimethylaniline. With excess methyl iodide, in the presence of Na2CO3solution, N, N-dimethylaniline produces N, N, Ntrimethylanilinium carbonate....
Read More →A balloon filled with helium
Question: A balloon filled with helium $\left(32^{\circ} \mathrm{C}\right.$ and $\left.1.7 \mathrm{~atm} .\right)$ bursts. Immediately afterwards the expansion of helium can be considered as :irreversible isothermalirreversible adiabaticreversible adiabaticreversible isothermalCorrect Option: , 2 Solution: (2) Bursting of helium balloon is irreversible and in this\ process $\Delta Q=0$, so adiabatic....
Read More →Complete the following acid-base reactions and name the products:
Question: Complete the following acid-base reactions and name the products: (i)CH3CH2CH2NH2+ HCl (ii)(C2H5)3N + HCl Solution:...
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