Arrange the following compounds in increasing order of their boiling points.
Question: Arrange the following compounds in increasing order of their boiling points. CH3CHO, CH3CH2OH, CH3OCH3, CH3CH2CH3 Solution: The molecular masses of the given compounds are in the range 44 to 46.CH3CH2OH undergoes extensive intermolecular H-bonding, resulting in the association of molecules. Therefore, it has the highest boiling point. CH3CHO is more polar than CH3OCH3and so CH3CHO has stronger intermolecular dipole dipole attraction than CH3OCH3CH3CH2CH3has only weak van der Waals forc...
Read More →Write the structures of products of the following reactions;
Question: Write the structures of products of the following reactions; (i) (ii) (iii) (iv) Solution: (i) (ii) (iii) (iv)...
Read More →A sphere of diameter 5 cm is dropped into a cylindrical
Question: A sphere of diameter 5 cm is dropped into a cylindrical vessel partly filled with water. The diameter of the base of the vessel is 10 cm. If the sphere is completely submerged, by how much will the level of water rise? Solution: Radius of sphere $r=\frac{5}{2} \mathrm{~cm}$, radius of cylindrical vessel $r_{1}^{2}=\frac{10}{2} \mathrm{~cm}$. When the sphere is completely submerged into the vessel. The level of water will be raised letxbe height of level of raised water. Therefore, The ...
Read More →An ideal gas in a cylinder is separated by a piston
Question: An ideal gas in a cylinder is separated by a piston in such a way that the entropy of one part is $S_{1}$ and that of the other part is $S_{2}$. Given that $\mathrm{S}_{1}\mathrm{S}_{2}$. If the piston is removed then the total entropy of the system will be:$\mathrm{S}_{1} \times \mathrm{S}_{2}$$\mathrm{S}_{1}-\mathrm{S}_{2}$$\frac{S_{1}}{S_{2}}$$\mathrm{S}_{1}+\mathrm{S}_{2}$Correct Option: 4, Solution:...
Read More →Write the structures of the following compounds.
Question: Write the structures of the following compounds. (i)-Methoxypropionaldehyde (ii)3-Hydroxybutanal (iii)2-Hydroxycyclopentane carbaldehyde (iv)4-Oxopentanal (v)Di-sec-butyl ketone (vi)4-Fluoroacetophenone Solution: (i) (ii) (iii) (iv) (v) (vi)...
Read More →A right circular cylinder and a right circular cone have equal
Question: A right circular cylinder and a right circular cone have equal bases and equal heights. If their curved surfaces are in the ratio 8 : 5, determine the ratio of the radius of the base to the height of either of them. Solution: For right circular cylinder, letr1= r,h1= h. Then, curved surface area, $s_{1}$ of cylinder $=2 \pi r_{1} h_{1}=2 \pi r h$ (i) For right circular cone, let $r_{2}=r, h_{2}=h$ Then, curved surface area, $s_{2}$ of cone $=\pi r_{2} l$ where $l=\sqrt{r_{2}{ }^{2}+h^{...
Read More →Write the structures of the following compounds.
Question: Write the structures of the following compounds. (i)-Methoxypropionaldehyde (ii)3-Hydroxybutanal (iii)2-Hydroxycyclopentane carbaldehyde (iv)4-Oxopentanal (v)Di-sec-butyl ketone (vi)4-Fluoroacetophenone Solution: (i) (ii) (iii) (iv) (v) (vi)...
Read More →For an adiabatic expansion of an ideal gas,
Question: For an adiabatic expansion of an ideal gas, the fractional change in its pressure is equal to (where $\gamma$ is the ratio of specific heats):$-\gamma \frac{\mathrm{dV}}{\mathrm{V}}$$-\gamma \frac{\mathrm{V}}{\mathrm{dV}}$$-\frac{1}{\gamma} \frac{\mathrm{dV}}{\mathrm{V}}$$\frac{\mathrm{dV}}{\mathrm{V}}$Correct Option: 1 Solution: (1) $\mathrm{PV} \gamma=$ constant Differentiating $\frac{\mathrm{dP}}{\mathrm{dV}}=-\frac{\gamma \mathrm{P}}{\mathrm{V}}$ $\frac{\mathrm{dP}}{\mathrm{P}}=-\f...
Read More →The vertical height of a conical tent is 42 dm and the diameter of its base is 5.4 m.
Question: The vertical height of a conical tent is 42 dm and the diameter of its base is 5.4 m. Find the number of persons it can accommodate if each person is to be allowed 29.16 cubic dm. Solution: Radius of conicaltent, $r=\frac{5.4}{2}$ $=2.7 \mathrm{~m}$ $=27 \mathrm{dm}$ Height of conical tenth= 42 dm The volume of conical tent $=\frac{1}{3} \pi r^{2} h$ $=22 \times 27 \times 27 \times 2$ $=32076 \mathrm{dm}^{3}$ Since, each person is to be allowed $29.16 \mathrm{dm}^{3}$, Therefore, $=\fr...
Read More →When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:
Question: When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: Give a mechanism for this reaction. (Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom. Solution: The mechanism of the given reaction involves the following steps: Step 1:Protonation Step 2:Formation of 2 carbocation by the elimination of a water molecule Step 3:Re-arrangement by the hydride-ion shift Step 4:Nucle...
Read More →When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place:
Question: When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place: Give a mechanism for this reaction. (Hint : The secondary carbocation formed in step II rearranges to a more stable tertiary carbocation by a hydride ion shift from 3rd carbon atom. Solution: The mechanism of the given reaction involves the following steps: Step 1:Protonation Step 2:Formation of 2 carbocation by the elimination of a water molecule Step 3:Re-arrangement by the hydride-ion shift Step 4:Nucle...
Read More →Match the following columns:
Question: Match the following columns: Solution: (a) - (r)Let the man starts from A and goes 10 m due east at B and then 20 m due north at C.Then, in right-angled triangle ABC, we have: $A B^{2}+B C^{2}=A C^{2}$ $\Rightarrow A C=\sqrt{10^{2}+20^{2}}=\sqrt{100+200}=10 \sqrt{3}$ Hence, the man is $10 \sqrt{3} \mathrm{~m}$ away from the starting point. (b) - (q)Let the triangle be ABC with altitude AD.In right-angled triangle ABD, we have: $A B^{2}=A D^{2}+B D^{2}$ $\Rightarrow A D^{2}=10^{2}-5^{2}...
Read More →A conical vessel whose internal radius is 10 cm and height 48 cm
Question: A conical vessel whose internal radius is 10 cm and height 48 cm is full of water. Find the volume of water. If this water is poured into a cylindrical vessel with internal radius 20 cm, find the height to which the water level rises in it. Solution: Radius of conical vesselr= 10 cm Height of conical vesselh= 48 cm The volume of water = volume of conical vessel. $=\frac{1}{3} \pi r^{2} h$ $=1600 \pi \mathrm{cm}^{3}$ $=1600 \times 3.14$ $=5024 \mathrm{~cm}^{3}$ Leth' be the height of cy...
Read More →Show how would you synthesise the following alcohols from appropriate alkenes?
Question: Show how would you synthesise the following alcohols from appropriate alkenes? (i) (ii) (iii) (iv) Solution: The given alcohols can be synthesized by applying Markovnikovs rule of acid-catalyzed hydration of appropriate alkenes. (i) (ii) (iii) Acid-catalyzed hydration of pent-2-ene also produces pentan-2-ol but along with pentan-3-ol. Thus, the first reaction is preferred over the second one to get pentan-2-ol. (iv)...
Read More →The P-V diagram of a diatomic ideal gas system going under cyclic process as shown in figure.
Question: The $\mathrm{P}-\mathrm{V}$ diagram of a diatomic ideal gas system going under cyclic process as shown in figure. The work done during an adiabatic process $\mathrm{CD}$ is (use $\gamma=1.4)$ $-500 \mathrm{~J}$$-400 \mathrm{~J}$$400 \mathrm{~J}$$200 \mathrm{~J}$Correct Option: 1 Solution: (1) Adiabatic process is from $\mathrm{C}$ to $\mathrm{D}$ $\mathrm{WD}=\frac{\mathrm{P}_{2} \mathrm{~V}_{2}-\mathrm{P}_{1} \mathrm{~V}_{1}}{1-\gamma}$ $=\frac{P_{\mathrm{D}} V_{\mathrm{D}}-\mathrm{P}...
Read More →Write equations of the following reactions:
Question: Write equations of the following reactions: (i)Friedel-Crafts reactionalkylation of anisole. (ii)Nitration of anisole. (iii)Bromination of anisole in ethanoic acid medium. (iv)Friedel-Crafts acetylation of anisole. Solution: (i) (ii) (iii) (iv)...
Read More →A well of diameter 3 m is dug 14 m deep.
Question: A well of diameter 3 m is dug 14 m deep. The earth taken out of it is spread evenly all around it to a width of 4 m to form an embankment. Find the height of the embankment. Solution: We have, Radius of well $r=\frac{3}{2} \mathrm{~m}$ Depth of well = 14 m The volume of well $=\pi \times\left(\frac{3}{2}\right)^{2} \times 14$ $=\pi \times \frac{9}{4} \times 14$ $=\frac{63}{2} \pi \mathrm{cm}^{3}$ Therefore, Volume of earth dugout $=$ volume of well $=\frac{63}{2} \pi \mathrm{m}^{3}$ Le...
Read More →Write the mechanism of the reaction of HI with methoxymethane.
Question: Write the mechanism of the reaction of HI with methoxymethane. Solution: he mechanism of the reaction of HI with methoxymethane involves the following steps: Step1:Protonation of methoxymethane: Step2:Nucleophilic attack of I: Step3: When HI is in excess and the reaction is carried out at a high temperature, the methanol formed in the second step reacts with another HI molecule and gets converted to methyl iodide...
Read More →Explain the fact that in aryl alkyl ethers
Question: Explain the fact that in aryl alkyl ethers (i)The alkoxy group activates the benzene ring towards electrophilic substitution and (ii)It directs the incoming substituents to ortho and para positions in benzene ring. Solution: (i) In aryl alkyl ethers, due to the +R effect of the alkoxy group, the electron density in the benzene ring increases as shown in the following resonance structure. Thus, benzene is activated towards electrophilic substitution by the alkoxy group. (ii)It can also ...
Read More →Write the equation of the reaction of hydrogen iodide with:
Question: Write the equation of the reaction of hydrogen iodide with: (i)1-propoxypropane (ii)Methoxybenzene and (iii)Benzyl ethyl ether Solution: (i) (ii) (iii)...
Read More →Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method.
Question: Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason. Solution: The formation of ethers by dehydration of alcohol is a bimolecular reaction (SN2) involving the attack of an alcohol molecule on a protonated alcohol molecule. In the method, the alkyl group should be unhindered. In case of secondary or tertiary alcohols, the alkyl group is hindered. As a result, elimination dominates substitution. Hence, in place of ethers, alke...
Read More →A cubic cm of gold is drawn into a wire 0.1 mm in diameter,
Question: A cubic cm of gold is drawn into a wire 0.1 mm in diameter, find the length of the wire. Solution: Letxbe the length of wire. Clearly, the volume of gold = volume of wire. {Volume of gold = 1 cm3= 1000 mm} $1000=\pi\left(\frac{0.1}{2}\right)^{2} \times x$ $\frac{22}{7} \times \frac{1}{400} \times x=1000$ $=\frac{1400 \times 1000}{11}$ $=127300 \mathrm{~mm}$ $x=127.3 \mathrm{~m}$ Hence, the length of wire is 127.3 m....
Read More →Which one is the correct option for the two different thermodynamic processes?
Question: Which one is the correct option for the two different thermodynamic processes? (C) and (A)(C) and (D)(A) only$(B)$ and $(C)$Correct Option: , 2 Solution: (2) Option (a) is wrong; since in adiabatic process $V \neq$ constant. Option (b) is wrong, since in isothermal process $T=$ constant Option (c) $\backslash \$ (d) matches isothermes $\backslash \$ adiabatic formula: constant $\backslash \ \frac{\mathrm{T}^{\gamma}}{\mathrm{p}^{\gamma-1}}=$ constant...
Read More →How is 1-propoxypropane synthesised from propan-1-ol?
Question: How is 1-propoxypropane synthesised from propan-1-ol? Write mechanism of this reaction. Solution: 1-propoxypropane can be synthesized from propan-1-ol by dehydration. Propan-1-ol undergoes dehydration in the presence of protic acids (such as H2SO4, H3PO4) to give 1-propoxypropane. The mechanism of this reaction involves the following three steps: Step 1:Protonation Step 2:Nucleophilic attack Step 3:Deprotonation...
Read More →Illustrate with examples the limitations of
Question: Illustrate with examples the limitations of Williamson synthesis for the preparation of certain types of ethers. Solution: The reaction of Williamson synthesis involves SN2 attack of an alkoxide ion on a primary alkyl halide. But if secondary or tertiary alkyl halides are taken in place of primary alkyl halides, then elimination would compete over substitution. As a result, alkenes would be produced. This is because alkoxides are nucleophiles as well as strong bases. Hence, they react ...
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