Out of benzene, m–dinitrobenzene and toluene
Question: Out of benzene,mdinitrobenzene and toluene which will undergo nitration most easily and why? Solution: The ease of nitration depends on the presence of electron density on the compound to form nitrates. Nitration reactions are examples of electrophilic substitution reactions where an electron-ich species is attacked by a nitronium ion $\left(\mathrm{NO}_{2}-\right)$. Now, $\mathrm{CH}_{3}-$ group is electron donating and $\mathrm{NO}_{2}-$ is electron withdrawing. Therefore, toluene wi...
Read More →Solve the system of the following equations
Question: Solve the system of the following equations $\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4$ $\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1$ $\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2$ Solution: Let $\frac{1}{x}=p, \frac{1}{y}=q, \frac{1}{z}=r$. Then the given system of equations is as follows: $2 p+3 q+10 r=4$ $4 p-6 q+5 r=1$ $6 p+9 q-20 r=2$ This system can be written in the form ofAX=B, where $A=\left[\begin{array}{ccc}2 3 10 \\ 4 -6 5 \\ 6 9 -20\end{array}\right], X=\left[\begin{array}{l}p \\ q \\ r\...
Read More →Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+
Question: Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+ (a) Chlorobenzene, 2,4-dinitrochlorobenzene,p-nitrochlorobenzene (b) Toluene, $p-\mathrm{H}_{3} \mathrm{C}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NO}_{2}, p-\mathrm{O}_{2} \mathrm{~N}-\mathrm{C}_{6} \mathrm{H}_{4}-\mathrm{NO}_{2}$. Solution: Electrophiles are reagents that participate in a reaction by accepting an electron pair in order to bond to nucleophiles. The higher the...
Read More →Using properties of determinants, prove that:
Question: Using properties of determinants, prove that: $\left|\begin{array}{ccc}\sin \alpha \cos \alpha \cos (\alpha+\delta) \\ \sin \beta \cos \beta \cos (\beta+\delta) \\ \sin \gamma \cos \gamma \cos (\gamma+\delta)\end{array}\right|=0$ Solution: $\Delta=\left|\begin{array}{lll}\sin \alpha \cos \alpha \cos (\alpha+\delta) \\ \sin \beta \cos \beta \cos (\beta+\delta) \\ \sin \gamma \cos \gamma \cos (\gamma+\delta)\end{array}\right|$ $=\frac{1}{\sin \delta \cos \delta}$$\left|\begin{array}{lll}...
Read More →Write structures of all the alkenes which on hydrogenation give 2-methylbutane.
Question: Write structures of all the alkenes which on hydrogenation give 2-methylbutane. Solution: The basic skeleton of 2-methylbutane is shown below: On the basis of this structure, various alkenes that will give 2-methylbutane on hydrogenation are: (a) (b) (c)...
Read More →Using properties of determinants, prove that:
Question: Using properties of determinants, prove that: $\left|\begin{array}{lll}1 1+p 1+p+q \\ 2 3+2 p 4+3 p+2 q \\ 3 6+3 p 10+6 p+3 q\end{array}\right|=1$ Solution: $\Delta=\left|\begin{array}{lll}1 1+p 1+p+q \\ 2 3+2 p 4+3 p+2 q \\ 3 6+3 p 10+6 p+3 q\end{array}\right|$ Applying $R_{2} \rightarrow R_{2}-2 R_{1}$ and $R_{3} \rightarrow R_{3}-3 R_{1}$, we have: $\Delta=\left|\begin{array}{ccc}1 1+p 1+p+q \\ 0 1 2+p \\ 0 3 7+3 p\end{array}\right|$ Applying $R_{3} \rightarrow R_{3}-3 R_{2}$, we ha...
Read More →How would you convert the following compounds into benzene?
Question: How would you convert the following compounds into benzene? (i) Ethyne (ii) Ethene (iii) Hexane Solution: (i)Benzene from Ethyne: .(ii) Benzene from Ethene: (iii)Hexane to Benzene...
Read More →If aare in A.P., prove that a, b, c are in A.P.
Question: If $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P., prove that $a, b, c$ are in A.P. Solution: It is given that $a\left(\frac{1}{b}+\frac{1}{c}\right), b\left(\frac{1}{c}+\frac{1}{a}\right), c\left(\frac{1}{a}+\frac{1}{b}\right)$ are in A.P. $\therefore b\left(\frac{1}{c}+\frac{1}{a}\right)-a\left(\frac{1}{b}+\frac{1}{c}\right)=c\left(\frac{1}{a}+\frac{1}{b}\right)-b\left(\frac{1}{c}+\frac{1}{a}\right)$ $...
Read More →Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Question: Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty? Solution: Benzene is a planar molecule having delocalized electrons above and below the plane of ring. Hence, it is electron-rich. As a result, it is highly attractive to electron deficient species i.e., electrophiles. Therefore, it undergoes electrophilic substitution reactions very easily. Nucleophiles are electron-rich. Hence, they are repelled by benzene. Hence, benz...
Read More →Using properties of determinants, prove that:
Question: Using properties of determinants, prove that: $\left|\begin{array}{ccc}3 a -a+b -a+c \\ -b+a 3 b -b+c \\ -c+a -c+b 3 c\end{array}\right|=3(a+b+c)(a b+b c+c a)$ Solution: $\Delta=\left|\begin{array}{ccc}3 a -a+b -a+c \\ -b+a 3 b -b+c \\ -c+a -c+b 3 c\end{array}\right|$ Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}$, we have; $\Delta=\left|\begin{array}{ccc}a+b+c -a+b -a+c \\ a+b+c 3 b -b+c \\ a+b+c -c+b 3 c\end{array}\right|$ $=(a+b+c)\left|\begin{arr...
Read More →Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom
Question: Obtain the first Bohrs radius and the ground state energy of amuonichydrogen atom[i.e., an atom in which a negatively charged muon () of mass about 207meorbits around a proton]. Solution: Mass of a negatively charged muon, $m_{\mu}=207 m_{e}$ According to Bohrs model, Bohr radius, $r_{e} \propto\left(\frac{1}{m_{e}}\right)$ And, energy of a ground state electronic hydrogen atom, $E_{e} \propto m_{e}$. Also, energy of a ground state muonic hydrogen atom, $E_{\mu} \propto m_{\mu}$. We ha...
Read More →Arrange benzene, n-hexane and ethyne in decreasing order of acidic behaviour.
Question: Arrange benzene,n-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour. Solution: Acidic character of a species is defined on the basis of ease with which it can lose its Hatoms. The hybridization state of carbon in the given compound is: As thescharacter increases, the electronegativity of carbon increases and the electrons of CH bond pair lie closer to the carbon atom. As a result, partial positive charge of Hatom increases and H+ions are set...
Read More →The pth, qth and rth terms of an A.P. are a, b, c respectively.
Question: The $p^{\text {th }}, q^{\text {th }}$ and $t^{\text {th }}$ terms of an A.P. are $a, b, c$ respectively. Show that $(q-r) a+(r-p) b+(p-q) c=0$ Solution: Lettanddbe the first term and the common difference of the A.P. respectively. The $n^{\text {th }}$ term of an A.P. is given by, $a_{n}=t+(n-1) d$ Therefore, $a_{p}=t+(p-1) d=a \ldots$ $a_{q}=t+(q-1) d=b \ldots(2)$ $a_{r}=t+(r-1) d=c \ldots(3)$ Subtracting equation (2) from (1), we obtain $(p-1-q+1) d=a-b$ $\Rightarrow(p-q) d=a-b$ $\t...
Read More →Write down the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene).
Question: Write down the products of ozonolysis of 1, 2-dimethylbenzene (o-xylene). How does the result support Kekul structure for benzene? Solution: o-xylene has two resonance structures: All three products, i.e., methyl glyoxal, 1, 2-dimethylglyoxal, and glyoxal are obtained from two Kekule structures. Since all three products cannot be obtained from any one of the two structures, this proves thato-xylene is a resonance hybrid of two Kekule structures (I and II)....
Read More →Using properties of determinants, prove that:
Question: Using properties of determinants, prove that: $\left|\begin{array}{lll}x x^{2} 1+p x^{3} \\ y y^{2} 1+p y^{3} \\ z z^{2} 1+p z^{3}\end{array}\right|=(1+p x y z)(x-y)(y-z)(z-x)$ Solution: $\Delta=\left|\begin{array}{lll}x x^{2} 1+p x^{3} \\ y y^{2} 1+p y^{3} \\ z z^{2} 1+p z^{3}\end{array}\right|$ Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, we have: $\Delta=\left|\begin{array}{lll}x x^{2} 1+p x^{3} \\ y-x y^{2}-x^{2} p\left(y^{3}-x^{3}\right) \\ z-x z^{...
Read More →If Bohr’s quantisation postulate
Question: If Bohrs quantisation postulate (angular momentum =nh/2) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantisation of orbits of planets around the sun? Solution: We never speak of quantization of orbits of planets around the Sun because the angular momentum associated with planetary motion is largely relative to the value of Plancks constant(h). The angular momentum of the Earth in its orbit is of the order of...
Read More →Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide,
Question: Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism. Solution: Addition of HBr to propene is an example of an electrophilic substitution reaction. Hydrogen bromide provides an electrophile, H+. This electrophile attacks the double bond to form 1 and 2 carbocations as shown: Secondary carbocations are more stable than primary carbocations. Hence, the former predominates since it ...
Read More →The total energy of an electron in the first excited state of the hydrogen atom is about
Question: The total energy of an electron in the first excited state of the hydrogen atom is about 3.4 eV. (a)What is the kinetic energy of the electron in this state? (b)What is the potential energy of the electron in this state? (c)Which of the answers above would change if the choice of the zero of potential energy is changed? Solution: (a)Total energy of the electron,E= 3.4 eV Kinetic energy of the electron is equal to the negative of the total energy. K= E = ( 3.4) = +3.4 eV Hence, the kine...
Read More →Let S be the sum,
Question: Let S be the sum, P the product and R the sum of reciprocals ofnterms in a G.P. Prove that$P^{2} R^{n}=S^{n}$ Solution: Let the G.P. be $a, a r, a r^{2}, a r^{3}, \ldots a r^{n-1} \ldots$ According to the given information, $\mathrm{S}=\frac{a\left(r^{n}-1\right)}{r-1}$ $\mathrm{P}=a^{n} \times v^{1+2+\ldots+n-1}$ $=a^{n} r^{\frac{a(n-1)}{2}}$ $\left[\because\right.$ Sum of first $n$ natural numbers is $\left.n \frac{(n+1)}{2}\right]$ $\mathrm{R}=\frac{1}{a}+\frac{1}{a r}+\ldots+\frac{...
Read More →Classically, an electron can be in any orbit around the nucleus of an atom.
Question: Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of...
Read More →Using properties of determinants, prove that:
Question: Using properties of determinants, prove that: $\left|\begin{array}{lll}\alpha \alpha^{2} \beta+\gamma \\ \beta \beta^{2} \gamma+\alpha \\ \gamma \gamma^{2} \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$ Solution: $\Delta=\left|\begin{array}{lll}\alpha \alpha^{2} \beta+\gamma \\ \beta \beta^{2} \gamma+\alpha \\ \gamma \gamma^{2} \alpha+\beta\end{array}\right|$ Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, ...
Read More →then show that a, b, c and d are in G.P.
Question: If $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}(x \neq 0)$, then show that $a, b, c$ and $d$ are in G.P. Solution: It is given that, $\frac{a+b x}{a-b x}=\frac{b+c x}{b-c x}$ $\Rightarrow(a+b x)(b-c x)=(b+c x)(a-b x)$ $\Rightarrow a b-a c x+b^{2} x-b c x^{2}=a b-b^{2} x+a c x-b c x^{2}$ $\Rightarrow 2 b^{2} x=2 a c x$ $\Rightarrow b^{2}=a c$ $\Rightarrow \frac{b}{a}=\frac{c}{b}$ $\ldots(1)$ Also, $\frac{b+c x}{b-c x}=\frac{c+d x}{c-d x}$ $\Rightarrow(b+c x)(c-d x)=(b-c ...
Read More →What effect does branching of an alkane chain has on its boiling point?
Question: What effect does branching of an alkane chain has on its boiling point? Solution: Alkanes experience inter-molecular Van der Waals forces. The stronger the force, the greater will be the boiling point of the alkane. As branching increases, the surface area of the molecule decreases which results in a small area of contact. As a result, the Van der Waals force also decreases which can be overcome at a relatively lower temperature. Hence, the boiling point of an alkane chain decreases wi...
Read More →Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n−1). F
Question: Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from levelnto level (n1). For largen, show that this frequency equals the classical frequency of revolution of the electron in the orbit. Solution: It is given that a hydrogen atom de-excites from an upper level (n)to a lower level (n1). We have the relation for energy (E1) of radiation at levelnas: $E_{1}=h v_{1}=\frac{h m e^{4}}{(4 \pi)^{3} \in_{0}^{2}\left(\frac{h}{2 \pi}\right)^{3}} \times\l...
Read More →Using properties of determinants, prove that:
Question: Using properties of determinants, prove that: $\left|\begin{array}{lll}\alpha \alpha^{2} \beta+\gamma \\ \beta \beta^{2} \gamma+\alpha \\ \gamma \gamma^{2} \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$ Solution: $\Delta=\left|\begin{array}{lll}\alpha \alpha^{2} \beta+\gamma \\ \beta \beta^{2} \gamma+\alpha \\ \gamma \gamma^{2} \alpha+\beta\end{array}\right|$ Applying $R_{2} \rightarrow R_{2}-R_{1}$ and $R_{3} \rightarrow R_{3}-R_{1}$, ...
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