Chemistry Ncert Exemplar 2019 Solutions for Class 11 Science Chemistry Chapter 13 - Hydrocarbons

Answer:

Boiling point increases with increase in chain length and decreases with decrease in surface area and branching.
Hence, the correct answer is option (iv).

Answer:

The reactivity of halogens decreases down the group as Fluorine is the most electronegative element and electronegativity decreases down the group.
Hence, the correct answer is option (i).

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The reduction of alkyl halides increases with decreasing bond strength. As the size of halogen increases, bond strength decreases.
Hence, the correct answer is option (ii).

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While naming a branched alkane, parent chain should be selected in such a way that it has maximum chain length and maximum number of substituents.
Hence, the correct answer is option (i).

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According to Markovnikov rule for addition to unsymmetrical alkenes (1-butene), negative part (Br⁻) of attacking reagent (HBr) will attach to carbon atom with lesser number of hydrogen atoms and positive part (H⁺) to carbon atom with more number of hydrogen atoms. Thus, A and B are obtained as major products because they are enantiomers while C is obtained as minor product.
Hence, the correct answer is option (i).

Answer:

A molecule shows geometrical isomersim when both the groups on double bonded carbon atoms are different. But in option (iv), one of the double bonded carbon atoms has two methyl groups attached to it. So, it cannot show geometrical isomerism.
Hence, the correct answer is option (iv).

Answer:

The bond dissociation energy of HI is very low due to large size of I. Hence, it has highest reactivity with alkenes. As the size of halogens decreases, bond dissociation energy of hydrogen halides increases and reactivity with alkenes decreases.
Hence, the correct answer is option (iii).

Answer:

The stability of carbanion increases with increase in s-character and +I (Inductive) effect decreases the stabilty.
Hence, the correct answer is option (ii).

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The stability of alkene formed after β–elimination reaction of alkyl halides with alcoholic KOH dtermines the rate of reaction. More the +I (Inductive) efffect, more stable will be the alkene and hence, more will be the reactivity of alkyl halide.
Hence, the correct answer is option (iv).

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Incomplete combustion of methane leads to formation of carbon black.
Hence, the correct answer is option (iii).

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Controlled oxidation of alkanes is heating alkanes in regulated supply of oxygen at high pressure and in presence of catalysts.
Hence, the correct answers are options (iii) and (iv).

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Disubstituted alkenes on ozonolysis yield ketones only.

Hence, the correct answers are options (iii) and (iv).

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The common name of 1-methylethyl is iso-propyl and that of 1-methylpropyl is sec-Butyl. IUPAC naming is done in alphabetical order of substituents.
Hence, the correct answers are options (iii) and (iv).

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The chain with maximum number of carbon atoms (ten here) is selected as parent chain. The neo-Pentyl group is 2′, 2′–Dimethylpropyl.
Hence, the correct answers are options (i) and (iv).

Answer:

As halogens are electronegative, so, the ring gets somewhat deactivated by −I (Inductive) effect of halogens. But due to +R (Resonance) effect, halogens make their lone pairs of electrons participate in resonance and increase electron charge density at ortho and para positions of ring. Thus, an electrophile can easily attack at these positions.
Hence, the correct answers are options (i) and (iii).
 

Answer:

As nitro group is electron withdrawing group, so, the ring gets deactivated by −I (Inductive) effect. Due to −R (Resonance) effect, nitro group withdraws electron density away from ring and electron charge density at ortho and para positions of ring is decreased. Thus, an electrophile attacks at meta position.
Hence, the correct answers are options (i) and (iii).

Answer:

The stability of carbocation increases in the order: Vinyl < 1⁰ < 2⁰ < 3⁰ < Allyl < Benzyl < Resonance.
Hence, the correct answers are options (i) and (iii).

Answer:

According to Huckel rule, a molecule is aromatic if it is planar, complete delocalisation of π electrons takes place in ring and it has 4n+2 π electrons in ring, where n is an integer. All these conditions are satisfied by molecules in options (i) and (iii). The value of n is 0 in option (i) and 1 for each ring in option (iii).
Hence, the correct answers are options (i) and (iii).

Answer:

The dipole moment of trans-Pent-2-ene is small but not zero because of the unequal +I (Inductive) effect of two substituents CH₃ and Câ‚‚Hâ‚… on double bonded carbon. In cis-Hex-3-ene, +I (Inductive) effect of two Câ‚‚Hâ‚… groups is equal but the dipole moment is not zero as they are inclined at an angle of 60⁰ to each other. In other two options, dipole moment is zero.
Hence, the correct answers are options (ii) and (iii).

Answer:

Both alkenes and arenes are electron rich and hence, electrophile easily attacks these sites. Alkenes undergo electrophilic addition reaction because energy released due to formation of two new σ bonds is more than energy required to break one π bond. But in the case of arenes,  addition reaction would result in loss of aromaticity and resonance energy. Therefore, arenes prefer electrophilic substitution reactions in which aromaticity is retained.

Answer:

The reduction of 2-butyne will yield 2-butene and it will show geometrical isomerism. This is because each of the two doubly bonded carbons have two different substituents−H and CH₃. This is the requirement of geometrical isomerism.

Answer:

Rotation around carbon-carbon single bond of ethane is restricted due to an energy barrier of 1−20 kJ mol⁻¹. This is because of the weak repulsive interaction known as torsional strain between adjacent bonds.

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Staggered conformation of ethane is more stable than eclipsed conformation because staggered conformation has carbon-hydrogen bonds as far apart as possible. Thus, minimum repulsive forces, minimum torsional strain, minimum energy and maximum stability.

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The reactivity of halogen acids is inversely proportional to their bond energies. The bond energy of HI is least due to large size of I. Hence, it has highest reactivity with propene. As the size of halogens decreases, bond energy of halogen acids increases and reactivity with propene decreases. So, the order of reactivity of these halogen acids is HI > HBr > HCl.

Answer:

The primary carbocation formed is less stable and rearranges to form more stable secondary carbocation. This attacks as an electrophile on benzene ring to form isopropyl benzene as major product.

 

Answer:

(i)
 
(ii)

Answer:

Anisole has electron donating OCH₃ group which shows +I (Inductive) effect and +R (Resonance) effect leading to highest reactivity with an electrophile. Chlorobenzene has Cl group which is electronegative and shows −I (Inductive) effect and +R (Resonance) effect leading to decreased reactivity. But nitrobenzene has electron withdrawing NOâ‚‚ group which −I â€‹(Inductive) effect and −R (Resonance) effect leading to least reactivity. Thus, the reactivity order is Anisole > Chlorobenzene > Nitrobenzene.

Answer:

Halogens are electronegative, so, they deactivate the ring by their −I (Inductive) effect. But they increase the electron density at o- and p- positions due to delocalisation of their lone pairs of electrons into the ring by +R (Resonance) effect. Hence, they are o- and p-directing in haloarenes.

Answer:

Nitro group is an electron withdrawing group which shows −I (Inductive) effect and −R (Resonance) effect. Due to these two effects, benzene ring is deactivated and electron density is decreased at o- and p- positions. This makes the benzene ring less reactive in comparison to the unsubstituted benzene ring. 

Answer:

The conversion of acetylene to nitrobenzene is as follows: 

Answer:

${\mathrm{H}}_3\mathrm{C}-\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2\underset{\mathrm{HBr}}{\overset{{(\mathrm{Ph}-\mathrm{CO}-\mathrm{O})}_2}{\to }}{\mathrm{H}}_3\mathrm{C}-{\mathrm{CH}}_2-{\mathrm{CH}}_2\mathrm{Br}$
This reaction follows the anti-Markovnikov rule which states that in presence of peroxide, the negative part of HBr adds to the carbon of unsymmetrical alkene which has more number of hydrogen atoms. It takes place by a free radical mechanism.

$$\begin{gathered} {\mathrm{H}}_3\mathrm{C}-\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2\stackrel{\mathrm{HBr}}{\to }{\mathrm{H}}_3\mathrm{C}-\underset{|}{\mathrm{C}}\mathrm{H}-{\mathrm{CH}}_3 \\ \mathrm{Br} \end{gathered}$$
This reaction follows the Markovnikov rule which states that the negative part of HBr adds to that carbon of unsymmetrical alkene which has less number of hydrogen atoms.

Answer:

Nucleophiles	Electrophiles
(i) H₃CO⁻	(iii)
(ii)	(iv)
(vi) Br⁻	(v) (H₃C)₃C+
(vii) H₃COH

Answer:

Three monochlorinated products can be obtained from 2-methylbutane.
$$\begin{gathered} \left(A\right) Cl-C{H}_2-\underset{|}{C}H-C{H}_2-C{H}_3 \\ C{H}_3 \\ \left(B\right) C{H}_3-\underset{|}{C}H-\underset{|}{C}H-C{H}_3 \\ C{H}_3 Cl \\ \left(C\right) Cl \\ H_{3}C-\underset{\left|}{\overset{\right|}{C}}-C{H}_2-C{H}_3 \\ C{H}_3 \end{gathered}$$

Answer:

Alkyl halides on reaction with sodium in presence of dry ether yield higher alkanes. This reaction is called Wurtz reaction. The following products are formed.
$$\begin{gathered} {\mathrm{H}}_3\mathrm{C}-\underset{|}{\mathrm{CH}}-{\mathrm{CH}}_2\mathrm{I}+2\mathrm{Na}+{\mathrm{ICH}}_2-\underset{|}{\mathrm{C}}\mathrm{H}-\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{Dry} \mathrm{Ether}}{\to }{\mathrm{H}}_3\mathrm{C}-\underset{|}{\mathrm{CH}}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\underset{|}{\mathrm{C}}\mathrm{H}-{\mathrm{CH}}_3 \\ {\mathrm{CH}}_3 {\mathrm{CH}}_3 {\mathrm{CH}}_3 {\mathrm{CH}}_3 \end{gathered}$$
1-Iodo-2-methylpropane                                                    2,5-Dimethylhexane

$$\begin{gathered} {\mathrm{H}}_3\mathrm{C}-\underset{|}{\mathrm{CH}}-\mathrm{I}+2\mathrm{Na}+\mathrm{I}-\underset{|}{\mathrm{C}}\mathrm{H}-\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{Dry} \mathrm{Ether}}{\to }{\mathrm{H}}_3\mathrm{C}-\underset{|}{\mathrm{CH}}-\underset{|}{\mathrm{C}}\mathrm{H}-{\mathrm{CH}}_3 \\ {\mathrm{CH}}_3 {\mathrm{CH}}_3 {\mathrm{CH}}_3 {\mathrm{CH}}_3 \end{gathered}$$
2-Iodopropane                                              2,3-Dimethylbutane

$$\begin{gathered} {\mathrm{H}}_3\mathrm{C}-\underset{|}{\mathrm{CH}}-{\mathrm{CH}}_2\mathrm{I}+2\mathrm{Na}+\mathrm{I}-\underset{|}{\mathrm{C}}\mathrm{H}-\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{Dry} \mathrm{Ether}}{\to }{\mathrm{H}}_3\mathrm{C}-\underset{|}{\mathrm{CH}}-{\mathrm{CH}}_2-\underset{|}{\mathrm{C}}\mathrm{H}-{\mathrm{CH}}_3 \\ {\mathrm{CH}}_3 {\mathrm{CH}}_3 {\mathrm{CH}}_3 {\mathrm{CH}}_3 \end{gathered}$$
1-Iodo-2-methylpropane                                     2,4-Dimethylpentane

 

Answer:

Two hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane are as follows:

Radical (A) is more stable because it is 3⁰ radical and has 9αH. Thus, it is more stabilized by hyperconjugation than 1⁰ radical which has only 1αH.

Answer:

The structure of alkane is $$\begin{gathered} H_{3}C-\underset{|}{C}H-C{H}_2-\underset{ }{C{H}_2-\underset{|}{C}}H-C{H}_3 \\ C{H}_3 C{H}_3 \end{gathered}$$  and the structure of tertiary bromide is $$\begin{gathered} Br \\ {H}_{3}C-\stackrel{|}{\underset{|}{C}}-C{H}_2-\underset{ }{C{H}_2-\underset{|}{C}}H-C{H}_3 \\ C{H}_3 C{H}_3 \end{gathered}$$.

Answer:

Answer:

A compound is aromatic according to Huckel rule if it has (4n + 2) π−electrons in the ring where n is an integer (n = 0, 1, 2,...........). Also, another condition for aromaticity is that the ring should be planar. Hence, compounds (B), (D) and (F) are aromatic because of the presence of 6π, 10π and 14π−electrons delocalized in the ring respectively.

Answer:

The following steps will convert ethanol to ethyl hydrogensulphate.

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Answer:

The boiling point of alkanes decreases on branching. Hence, the correct matching of hydrocarbons with their boiling points is as follows.
 

Answer:

S.No.	Column I	Column II
(i)		(d) Chlorobenzene
(ii)		(c) Toulene
(iii)		(b) Methyl phenyl ketone
(iv)		(a) Benzoic acid

Answer:

Answer:

A compound is said to be aromatic if it has the following characteristics:
(i) Planar ring containing conjugated π bonds.
(ii) Complete delocalisation of the π−electrons in ring system i.e. each atom in the ring has unhybridised p-orbital, and
(iii) Presence of (4n + 2) π−electrons in the ring where n is an integer (n = 0, 1, 2,...........) [Huckel rule].
Hence, the correct answer is option (i).

Answer:

Toulene has methyl group attached to benzene ring and methyl group shows +R (Resonance) effect. Thus, it increases the electron density at o– and p– positions and gives o– and p–xylene on Friedal Crafts methylation.
Hence, the correct answer is option (i).

Answer:

The process of generation of electrophile, NO₂⁺ for nitration of benzene is a simple acid-base equilibrium. Here, concentrated sulphuric acid acts as an acid by transfer of proton to concentrated nitric acid which acts as a base.
Hence, the correct answer is option (i).

Answer:

The boiling point of alkanes decreases on branching and among isomeric pentanes, n-pentane has the highest boiling point because of straight chain while 2, 2-dimethylpentane has the lowest boiling point due to maximum branching.
Hence, the correct answer is option (iii).

Answer:

${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{Br}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\stackrel{\mathrm{alc}.\mathrm{KOH}}{\to }{\mathrm{C}}_5{\mathrm{H}}_{10}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\stackrel{{\mathrm{Br}}_2 \mathrm{in} {\mathrm{CS}}_2}{\to }{\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{Br}}_2\mathbf{\left(}\mathbf{C}\mathbf{\right)}\underset{-2\mathrm{HBr}}{\overset{\mathrm{alc}.\mathrm{KOH}}{\to }}{\mathrm{C}}_5{\mathrm{H}}_8\mathbf{\left(}\mathbf{D}\mathbf{\right)}\stackrel{\mathrm{Na}-\mathrm{liq}.{\mathrm{NH}}_3}{\to }{\mathrm{C}}_5{\mathrm{H}}_7-\mathrm{Na}+\frac{1}{2}{\mathrm{H}}_2$

Since hydrogenation of (D) yields a straight-chain alkane, the structures of (A), (B), (C) and (D) are as follows.

$$\begin{gathered} \mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{ }{\mathrm{H}}_3\mathrm{C}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_2\mathrm{Br} \\ \mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{ }{\mathrm{H}}_3\mathrm{C}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{CH}={\mathrm{CH}}_2 \\ \mathbf{\left(}\mathbf{C}\mathbf{\right)}\mathbf{ }{\mathrm{H}}_3\mathrm{C}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{CH}\left(\mathrm{Br}\right)-{\mathrm{CH}}_2\mathrm{Br} \\ \mathbf{\left(}\mathbf{D}\mathbf{\right)}\mathbf{ }{\mathrm{H}}_3\mathrm{C}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{C}\equiv \mathrm{CH} \end{gathered}$$

Answer:

Let us first find out the molecular mass of hydrocarbon ‘A’.

 

Now, let us find out the empirical formula of hydrocarbon ‘A’.
 

Answer:

The production of a dialdehyde, a ketone and an aldehyde suggests that the hydrocarbon 'A' is dialkene. Hence, the structure of 'A' is as follows:


The IUPAC name of 'A' is 2-methylocta-2,6-diene.

Answer:

The peroxide effect in case of HBr proceeds via free radical chain mechanism.

This effect is not known in case of HCl because of the higher bond dissociation enthalpy (430.5 kJ mol⁻1) of H-Cl bond as compared to bond dissociation enthalpy (363.7 kJ mol⁻1) of H-Br bond. Therefore, H-Cl bond is not cleaved by free radical.
Peroxide effect is not seen in case of HI because of quite low bond dissociation enthalpy (296.8 kJ mol⁻1) of H-I bond. So, iodine free radicals combine to form iodine molecules instead of adding to propene.