Chemistry NCERT Exemplar Solutions Class 11th Chapter Thirteen: Overview, Questions, Preparation

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The structure of 2-methylbutane is 

2-methylbutane

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It is a Wurtz reaction where alkane is formed.

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(iv) The two substituents attached to the same C atom of the C=C bond should be different to exhibit geometrical isomerisms.

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This is a short answer type question as classified in NCERT Exemplar

The 2-methylpropane would lead to two types of radicals.

Among the two 3^o radical is most stable as it contains 9 ∝ -hydrogen whereas 1^o radical contains only 1 ∝ -hydrogen.

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(iv) C > B > D > A

The boiling point increases with the increase in the length of the side chain. However, with the branching, the boiling point decreases due to the decrease in the surface area.

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(iii) The addition of alkyl halides to propene depends upon the H-X bond strength i.e. lesser the bond strength higher will be the reactivity of the alkyl halide. As down the group size of the halogens increases thus the bond strength decreases.

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This is a long answer type question as classified in NCERT Exemplar

The molar mass of the compound = 3.38 x 22400/896 =831gmol^-1

Thus the empirical formula of the compound is C₃H₅

Molecular formula =n x empirical formula = 831/41 x C₃H₅ = C₆H₁₀

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The structure of A is

The reaction involved are as follows:

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The bond energy of HCl is higher than that of HBr thus it is not cleaved by free radical mechanism to exhibit peroxide effect. However in case of HI the bond energy is so low that the iodine radical forms readily and after formation it combines to form an iodine molecule.

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If arenes would undergo electrophilic addition reaction then they would lose their aromaticity which would lead to comparatively less stability. However, alkenes undergo electrophilic addition reaction via carbocation intermediate.

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Yes, it will execute geometrical isomerism as the product formed can be either cis-2-butene or trans-2-butene.

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The rotation around C-C single bond is possible but it is not completely free due to the torsional strain which is about 1^-20 KJ mol^-1.

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The rate-determining step involved in the reaction is 

CH₃-CH=CH₂ + HX → CH₃-CH⁺-CH₃ + X^-

So the rate-determining step depends on the bond energy of HX i.e. higher the bond energy lesser would be reactivity.

Thus the order of reactivity of these halogen acids is

HI>HBr>HCl

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The benzene here would undergo Friedel-Crafts alkylation reaction where carbocation is formed as an intermediate, thus secondary carbocation is formed as an intermediate due to its greater stability than that of the primary carbocation.

The final product obtained is 

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This is a short answer type question as classified in NCERT Exemplar

The electron donating group increases the reactivity in electrophilic substitution reaction whereas the electron withdrawing group decreases the reactivity in electrophilic substitution.

Thus, the order of reactivity

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Due to the +R effect halogens are ortho and para directing groups.

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The nitro group strongly deactivate the benzene towards the electrophilic substitution reaction due to the -R and -I effect

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This is a short answer type question as classified in NCERT Exemplar

In presence of (Ph-CO-O)₂it lead to CH₃-CH₂-CH₂Br as the reaction undergo by free radical mechanism

However in absence of (Ph-CO-O)₂ the reaction undergo by carbocation intermediate thus lead to give CH₃-CHBr-CH₃

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The reaction is

The alkane is

The tertiary bromide is

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(i) I₂< Br₂< Cl₂< F₂

The reactivity of halogens with alkanes depends on the electronegativity i.e. higher the electronegativity higher would be the reaction. As down the group reactivity decrease so thus its reactivity.

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(ii) R–Cl< R–Br < R–I 

The reduction of alkyl halides depends upon the C-X bond strength i.e. lesser the bond strength higher will be the reactivity of alkyl halide. As down the group size of the halogens increases thus the bond strength decreases.

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(i) 3, 6 – Diethyl – 2 – methyl octane

Octane is the longest chain here and the side chain follows the lowest sum rule.

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(i) The reaction follows Markovnikov's rule leading to giving 2^o carbocation as an intermediate.

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(ii) The electronegativity of sp hybridization is higher than that of sp³ hybridization.

The +I effect destabilizes the carbanion.

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(iv) The order of reactivity of β– elimination reaction is 3^o>2^o>1^o

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(iii) As the unburn carbon left out in this reaction thus it is an incomplete reaction

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(iii) & (iv) Methane is a controlled oxidation reaction that leads to giving formaldehyde and methanol.

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(iii) & (iv) When only the alkyl group is attached to the C atom of the C=C bond then only it would lead to giving only ketone on ozonolysis.

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(iii) & (iv) The longest carbon chain is of 10 carbon atom

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(i) & (iv) The longest carbon chain is of 10 carbon atom

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(i) & (iii)

The halogens atom weakly deactivate the ring due to the -I effect and directs the incoming group to the ortho and para position due to the +R effect.

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(i) & (iii) The nitro group strongly deactivate the ring due to the -R effect and -I effect

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(i) & (iii)

The +I effect of the -OCH₃ group stabilizes the carbocation.

The two resonating structure of CH₂=CH-CH₂^⊕_↔^⊕CH₂-CH=CH₂ makes it more stable

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(i) & (iii)

The structures that follow the (4n+2) Huckle rule of aromaticity is considered to be aromatic

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(ii) & (iii)   

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(i) → (d); (ii) → (a); (iii) → (e) : (iv) → (c) : (v) → (b)

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(i) → (b); (ii) → (c); (iii) → (a)

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(i) → (d); (ii) → (c); (iii) → (b) : (iv) → (a)

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(i) → (d); (ii) → (a); (iii) → (b) : (iv) → (c)

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(i) Because cyclooctatetraene is tubbed shaped.

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(i) The -CH₃ group is an activating group and it directs the upcoming substituent to the ortho and para position.

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(i) The sulphuric acid helps in furnishing NO₂⁺ (electrophile).

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(iii) With the branching the boiling decreases due to the decrease in surface area which leads to decrease its van there Waals force.

$\to$ ${\pi }_{{\mathrm{K}\mathrm{N}\mathrm{O}}_3}={\pi }_{\text{glucose.}}$

$(\mathrm{i}\mathrm{C}\mathrm{R}\mathrm{T}{)}_{{\mathrm{K}\mathrm{N}\mathrm{O}}_3}=(\mathrm{C}\mathrm{R}\mathrm{T}{)}_{\text{glucose}}$

 $\frac{}{}\frac{}{}$ $\frac{\mathrm{i}\times 1.34}{101.1}\times \mathrm{R}\mathrm{T}=\frac{4.77}{180}\times \mathrm{R}\times \mathrm{T}$

$\mathrm{i}=\frac{4.77}{180}\times \frac{101.1}{1.34}=2$

Hence, extent of dissociation is $100\mathrm{\%}$

Charge (q)   $=it=2\times 50\times 60=6000\mathrm{C}$

$\Rightarrow \mathrm{}6000C=\left(\frac{6000}{96500}F\right)$

${\mathrm{M}\mathrm{n}\mathrm{O}}_4^{-}+3{\mathrm{e}}^{(-)}+2{\mathrm{H}}_2\mathrm{O}\stackrel{\text{weakly alkaline medium}}{?}{\mathrm{M}\mathrm{n}\mathrm{O}}_2+4{\mathrm{O}\mathrm{H}}^{-}$

Mass of ${\mathrm{M}\mathrm{n}\mathrm{O}}_2$    deposited $=\frac{1}{3}\times \frac{6000}{96500}\times 87=1.8$

The number of >C = 0 groups in tripeptide is 0 .

Molarity $=\frac{\text{moles of}\text{solute}}{\text{volume of solution in}L}=\frac{1*1.15}{1342}*1000=0.86\mathrm{M}$

$N_1{V}_1=N_2{V}_2$

$0.01*V_1=0.01*20*3$

$V_1=60\mathrm{m}\mathrm{L}$

${\mathrm{C}\mathrm{r}\mathrm{O}}_4^{2-}+2{\mathrm{H}}^{+}+2{\mathrm{H}}_2{\mathrm{O}}_2?\underset{\text{Chromium Pentoxide}}{{\mathrm{C}\mathrm{r}\mathrm{O}}_5+3{\mathrm{H}}_2\mathrm{O}}$

$\stackrel{\stackrel{+4}{{\mathrm{N}\mathrm{O}}_2}}{2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+3}{{\mathrm{H}\mathrm{N}\mathrm{O}}_2}+\stackrel{+5}{\mathrm{H}\mathrm{N}\mathrm{O}}{}_3$

$\stackrel{0}{{\mathrm{C}\mathrm{l}}_2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{-1}{\mathrm{H}\mathrm{C}\mathrm{l}}+\mathrm{H}\mathrm{O}\mathrm{C}\mathrm{l}$

$2\stackrel{+4}{{\mathrm{C}\mathrm{l}\mathrm{O}}_2}+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+3}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_2}+\stackrel{+5}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_3}$

$\stackrel{+6}{{\mathrm{C}\mathrm{l}}_2}{\mathrm{O}}_6+{\mathrm{H}}_2\mathrm{O}\to \stackrel{+5}{{\mathrm{H}\mathrm{C}\mathrm{l}\mathrm{O}}_3}+\stackrel{+{+}^{+}}{{\mathrm{C}\mathrm{l}\mathrm{O}}_4}$

Kindly consider the following Image 

Amphiprotic species are those which behave like acid and base both. They can donate and accept a proton.

${\mathrm{M}\mathrm{a}}_2{\mathrm{b}}_2{\mathrm{c}}_2=5$ Geometrical Isomers

M (AA)b2c2:3  $\mathrm{M}\left(\mathrm{A}\mathrm{A}\right){\mathrm{b}}_2{\mathrm{c}}_2:3$  Geometrical Isomers

$logK=logA-\frac{Ea}{2.303RT}$

$Slope=\frac{Ea}{2.303R}=-10,000K$  

$\frac{Ea}{2.303R}=10^4$            

$log\left(10^{-5}\right)=logA-10^4\times \frac{1}{500}$   (at 500 K temperature)

$\therefore T=\frac{10^4}{19}K=\frac{100}{19}\times 10^{2}K=5.2631\times 10^{2}K=526.31K\approx 526K$

$\begin{array}{l}\left(+6\right)\left(+2\right)\left(+3\right)\left(+6\right)\\ Cr{O_4}^{-2}+\underset{}{\underset{?}{S_2{O_3}^{-2}\to Cr{{\left(OH\right)}_4}^{-}+S{O_4}^{-2}}}\end{array}$  

O.N.C = 4

n factor of S₂O₃^-- = 8

n factor of CrO₄^-- = 3

$m.eCr{O}_4^{-2}=m.e.S_2{O}_3^{-2}$        

$V\times \left(0.154\times 3\right)=40\times \left(0.25\times 8\right)\therefore v=\frac{80}{0.462}$       

= 173.16 ml $\approx$ 173 ml

NaOH + Na₂CO₃

(1) When Hph is added

m.e NaOH + $\frac{1}{2}m.eN{a}_{2}C{O}_3=m.e.HCl=17.5\times \frac{1}{10}=1.75$  (m.e = milli equivalents)

(2) When MeOH is added after Hph

$\frac{1}{2}meN{a}_{2}C{O}_3=m.eHCl=1.5\times \frac{1}{10}=0.15$

$\begin{array}{l}\therefore m.eNaOH=1.75-0.15=1.6\\ m.eN{a}_{2}C{O}_3=0.15\times 2=0.3\end{array}$  

$\frac{W_{N{a}_{2}C{O}_3}}{E_{N{a}_{2}C{O}_3}}\times 1000=0.3$   

$Weight\mathrm{\%}ofN{a}_{2}C{O}_3=\left(\frac{\frac{0.3\times 53}{1000}}{0.4}\right)\times 100=\frac{0.3\times 53}{10\times 0.4}=\frac{15.9}{4}=3.975\mathrm{\%}\approx 4\mathrm{\%}.$               

  $?U=?742.24\text{?}kJ/mole?H_{298}=?$

NH₂CN (s) + $\frac{3}{2}$ O₂ (g) ® N₂ (g) + O₂ (g) + H₂O (l)

 = $?742.24+\frac{1}{2}*\frac{8.314}{1000}*298$  

$?742.24+1.238=?741.002??741kJ/mol$                  

So, magnitude of  $?H$ = 741 kJ/mol.

$A_2{B}_3?2A^{+3}+3B^{-2}$

1-α        2α        3α

$\therefore i=1+4\alpha =1+4\times 0.6=1+2.4=3.4$

$\Delta T_b=i{k}_{b}m=3.4\times 0.52\times 1=1.768\approx 1.77K$

$T_b=374.77K\approx 375K.$         

$\frac{P_1}{T_1}=\frac{P_2}{T_2}\Rightarrow \frac{35}{300}=\frac{40}{T_2}$

$\therefore T_2=\frac{40\times 300}{35}=342.85K$

$T_2\left(°C\right)=69.707\approx 70°C$

$\Delta H_{sub}ofNametal=496kJ/mole$

Bond dissociation energy of Br₂ = 228 kJ/mole

$\Delta H_f=495.8+-325-728.4=-557.6kJ/mole=-5576\times 10^{-1}kJ/mole$       

Note : In question is neglect ${\Delta }_a{H}^{0}ofNa,{\Delta }_{vap}{H}^{O}ofB{r}_2\left(l\right)and{\Delta }_{bond}{H}^{0}ofB{r}_2\left(g\right)$

$R_{f}value=\frac{Distancemovedbysubstancefrombaseline}{Distancemovedbythesolventfrombaseline}=\frac{2}{5}=0.4=4.0\times 10^{-1}$