NCERT Exemplar

Hint

(d) As the number of carbon atom increases, boiling point increases. Boiling point decreases with branching.

Hint: Boiling point inversely proportional to branching
Boiling point of an alkane depends on number of carbon atoms, as the number of carbon atom increases boiling point also increases. It also depends on branching in the alkane, as branching increases boiling point of alkane decreases.

Alkanes with even number of carbon atoms have greater melting points than those with odd number of carbon atoms As per both factor the order of boiling point is as follows:
n-butane \(<2,2\)-dimethylpropane \(<2\)-methylbutane \(<\) n-pentane

Hint

(a) The reactivity order of halogens with alkanes is \(\mathrm{I}_2<\mathrm{Br}_2<\mathrm{Cl}_2<\mathrm{F}_2\)

HINT- High electronegativity means high reactivity with alkanes
EXPLAINATION:
Alkane react with \(\mathrm{F}_2\) is vigorously and with \(\mathrm{I}_2\) the reaction is too slow that it requires a catalyst. It is because of high electronegativity of fluorine. Reactivity decreases with decrease in electronegativity and electronegativity decreases down the group.

Hint

(b) The reactivity of reduction of alkyl halides with \(\mathrm{Zn} / \mathrm{HCl}\) increases as the strength of the \(\mathrm{C}-\mathrm{X}\) bond decreases, i.e., \(\mathrm{R}-\mathrm{Cl}<\mathrm{R}-\mathrm{Br}<\mathrm{R}-\mathrm{I}\)

HINT: Weaker bond of halogen with carbon gets easily reduced.

Hint

(a)

Hint

Hint: Markownikoff’s rule

(a) The alkene is unsymmetrical, hence will follow Markovnikov rule to give major product.

The alkene is unsymmetrical, hence will follow Markownikoff’s rule to give major product. The major product is 2- bromobutane and minor product is 1bromobutane. The 2 -bromobutane contains a chiral center hence racemic mixture is obtained. So, products A and B are major and product \(\mathrm{C}\) is minor. The reaction is as follows:

Hint

HINT: If same groups are attached with any double bonded carbon atom then compound does not show geometrical isomerism.

In option (d), a carbon with double bond has two same functional groups \(\left(\mathrm{CH}_3\right)\) attached. The rotation around carbon will not produce a new compound. Hence, geometrical isomerism is not possible.

Hint

Hint: Reactivity of an acid inversely proportional to the bond energy.

(c) The decreasing order of reactivity of hydrogen halides with propene is \(\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}\) . As the size of halogen increases, the strength of \(\mathrm{H}-\mathrm{X}\) bond decreases and hence, reactivity increases.

Explanation:

The order of reactivity parallels the order of acidity. The stronger the acid, the more readily it provides \(\mathrm{H}^{+}\)ions for addition to the double bond. The stability of hydrogen halides decreases down the group due to decrease in bond \((\mathrm{H}-\mathrm{X})\) dissociation energy in the order: \(\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}\)

Bond energy of \(\mathrm{HI}\) is \(296.8 \mathrm{~kJ} / \mathrm{mol}, \mathrm{HBr}\) is \(362.7 \mathrm{~kJ} / \mathrm{mol}\) and \(\mathrm{HCl}\) is \(430.5 \mathrm{~kJ} / \mathrm{mol}\). As bond energy decreases the reactivity of hydrogen halide compound increases. In the case of hydrogen halide, down the group reactivity increases with propene.

Hence, \(\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}\) is the order of reactivity with propene.

Hint

(b) Hint: As the percentage of s character increases electronegativity of carbon increase
EXPLANATION
+ I effect decreases the stability of carbon anion. Since \(\mathrm{CH}_3\) – group has + I-effect, therefore, it intensifies the negative charge and hence destabilises (A) relative to (B). \(\mathrm{sp}\) hybridised carbanion is more stabilised than \(\mathrm{sp}^3\) hybridized carbanion.

Thus, the stability order is as follows:

Hint

(d) Hint: Stability of alkene decide the rate of reaction
Alkyl halides on heating with alcoholic potash elminates one molecule of \(\mathrm{HX}\) to form an alkene. Hydrogen is eliminated from \(\beta\)-carbon atom, thus this reaction is know as \(\beta\)-elimination reaction. Stability of product that is alkene determines rate of reaction.

More the number of \(\beta\)-substituents (alkyl groups), more stable alkene will be formed on \(\beta\)-elimination and more will be the reactivity. Thus, the decreasing order of the rate of \(\beta\)-elimination reaction with alcoholic \(K O H\) is: \(A>C>B\).

Hint

(c) HINT: Incomplete combustion forms carbon black.

Explanation:

During incomplete combustion of alkanes with insufficient amount of air or dioxygen carbon black is formed which is used in the manufacturing of ink, printer ink, black pigments and as filters. Thus, The incomplete combustion of methane can result in a slew of partially oxidized products, including COCO, but also methanol, formic acid, formaldehyde, and higher hydrocarbons also.
(c) Dining incomplete combustion of alkanes with insufficient amount of air or dioxygen, carbon black is formed which is used in the manufacture of ink, printer ink, black pigments and as filters. Thus,
\(
\mathrm{CH}_4(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \xrightarrow[\text { combustion }]{\text { Incomplete }} \mathrm{C}(\mathrm{s})+2 \mathrm{H}_2 \mathrm{O}(l)
\)

Hint

(c, d)

\(
\mathrm{CH}_4+\mathrm{O}_2 \xrightarrow{\mathrm{Mo}_2 \mathrm{O}_3} \mathrm{HCHO}+\mathrm{H}_2 \mathrm{O}
\)

\(
2 \mathrm{CH}_4+\mathrm{O}_2 \xrightarrow[100 \mathrm{~atm}]{\mathrm{Cu} / 523 \mathrm{~K}} 2 \mathrm{CH}_3 \mathrm{OH}
\)

Reactions in which methane does not undergo complete combustion to give carbon dioxide and water or incomplete combustion to give carbon and water are controlled oxidation reactions.

Hint

(c, d) Alkenes which have two substituents on each carbon atom of the double bond give a mixture of ketones on ozonolysis. Thus, options (c) and (d) give mixture of ketones.

Hint

(c,d)

Hint

(a,d) The answer is the option (i) 5-(\(2^{\prime}\), \(2^{\prime}\)-Dimethylpropyl)-decane & (iv) 5-neo-Pentyldecane
Explanation: The longest carbon chain has 10 carbon atoms, hence ‘decane’. Sidechain is present on
5th carbon atom, viz., neo-Pentyl or 2′,2′-Dimethylpropyl according to IUPAC. Hence option (i) & (iv).

Hint

(a, c) For an electrophilic substitution reaction, the presence of halogen atom in the benzene ring deactivates the ring by inductive effect and increases the charge density at ortho- and para-position relative to meta-position by resonance. –
When chlorine is attached to benzene ring, chlorine being more electronegative pulls the electrons because of its -I-effect. The electron cloud of benzene becomes less dense. Thus, chlorine makes the benzene ring in aryl halide somewhat deactivated. But due to resonance, the electron density on ortho- and parapositions is greater than on meta-position.

The last structure contributes more to the orientation and hence halogens are \(o\) – and \(p\)-directors.

Hint

(a, c) Nitro group by virtue of-I-effect withdraws electrons from the ring and increases the charge and destabilizes carbocation.

In ortho, para-attack of electrophile on nitrobenzene, we are getting two structures (A) and (B) in which positive charge is appearing on the carbon atom directly attached to the nitro group.
As nitro group is electron withdrawing by nature, it decreases the stability of such product and hence meta attack is more feasible when electron withdrawing substituents are attached.

Hint

(a, c)

\(
\mathrm{CH}_3-\mathrm{O}-\mathrm{CH}_2^{\oplus} \text { is more stable than } \mathrm{CH}_3-\mathrm{CH}_2^{\oplus}
\)
\(
\text { because }+\mathrm{R} \text { effect of } \mathrm{CH}_3-\mathrm{O} \text { is } \text { greater than }-\mathrm{CH}_3 \text { (stability of carbocation increases due to the }+\mathrm{R} \text { effect). }
\)
(iii) We know that the resonance effect is stronger than +I effect. Here,
\(\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2^{\oplus}\)
is stabilised by resonance effect; whereas \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2^{\oplus}\)
\(\text { is stabilised by }+I \text { effect only . Hence, }\)
\(
\mathrm{CH}_2=\mathrm{CH}-\mathrm{CH}_2^{\oplus} \text { is more stable than } \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2^{\oplus}
\)

Hint

(a, c) Aromaticity requres following condition
(i) Planarity
(ii) Complete delocalisation of \(\pi\) electrons in the ring.
(iii) Presence of \((4 n+2) \pi\) electrons in the ring.

Cyclooctatetraene is non-planar and has \(8 \pi\)-electrons. It is not aromatic. Cyclopropenyl anion is planar but has \(4 \pi\)-electrons. It is not aromatic.

Hint

(b,c)

Since, the +I effect of \(\mathrm{CH}_2 \mathrm{CH}_3\) group is higher than that of \(\mathrm{CH}_3\) group, therefore, the dipole moments of \(\mathrm{C-}\) \(\mathrm{CH}_3\) and \(\mathrm{C}-\mathrm{CH}_2 \mathrm{CH}_3\) bonds are unequal. Although these two dipoles oppose each other, yet they do not exactly cancel out each other and hence trans-z-pentene has small but finite dipole moment. In cis-hex-3-ene, although the dipole moments of the two \(\mathrm{C}-\mathrm{CH}_2 \mathrm{CH}_3\) bond are equal, but they are inclined to each other at an angle of \(60^{\circ}\) and hence have a finite dipole moment.

Hint

(a)

Hint

(b) Hint: Boiling point inversely proportional to the branching in the molecule
In hydrocarbon as branching increases in the molecule, the boiling point of hydrocarbon decreases in comparable molecular masses.
Explanation:
(i) There are more Vander Waal’s forces in n-pentane, and its boiling point is also high since there is no branching and surface area.
(ii) The boiling point of iso-pentane is less because the molar mass is the same except there’s one brach which reduces the surface area.
(iii) The boiling point of neo-pentane is the lowest because it has two side chains which have the same molar mass.

Hint

(c) The first reaction shows benzene undergoes chlorination.
The second reaction is Friedel-Craft alkylation.
The third reaction exhibits friedel-Craft acylation of benzene.
In the fourth reaction shown below, toluene undergoes oxidation with \(\mathrm{KMnO}_4\) to form benzoic acid.

Hint

(d) Hint: Chemical properties of alkenes.

(i) \(\rightarrow\) (d) (ii) \(\rightarrow\) (a) (iii) \(\rightarrow\) (b) (iv) \(\rightarrow\) (c)

Hint

(a) Hint: cyclooctatetraene is not a planar compound

According to Huckel rule Aromaticity is shown by compounds possessing the following characteristics
(i) Compound must be planar and cyclic
(ii) Complete delocalization of \(\pi\) electrons in the ring
(iii) Presence of conjugated \((4 n+2) \pi\) electrons in the ring where \(n\) is an integer \((n=0,1,2, \ldots\).
cyclooctatetraene (given) has a tub-like structure. It loses planarity. The number of \(\pi\)-electron delocalized is 8 and \(\mathrm{n}\) is not an integer. Hence, cycloctatetraene is a non-aromatic compound.
Thus, both assertion and reason are true and the reason is the correct explanation of assertion.

Hint

(a) Hint: \(\mathrm{CH}_3\) group is an electron-donating group
Toluene has \(-\mathrm{CH}_3\) group attached to benzene. \(-\mathrm{CH}_3\) group activates the benzene for the attack of an electrophile.
In the resonating structure of toluene, electronic density is more on ortho and para position.
Hence, substitution takes place mainly at these positions. Thus, both assertion and reason are true and the reason is the correct explanation of assertion.

Hint

(a) Hint: Nitration mixture contains conc. \(\mathrm{H}_2 \mathrm{SO}_4\) and conc. \(\mathrm{HNO}_3\)
In the nitration of benzene with nitric acid, sulphuric acid acts as a catalyst. It helps in the formation of electrophile i.e., nitronium ion \(\mathrm{NO}_2^{+}\).
The reaction is as follows:
\(
\mathrm{HNO}_3+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{NO}_2^{+}+2 \mathrm{HSO}_4^{-}+\mathrm{H}_3 \mathrm{O}^{+}
\)

Hint

(c) Hint: Boiling point is inversely proportional to branching
As branching increases, the boiling point of alkane decreases in comparable molecular masses. The structures of pentane and 2,2-dimethylpropane are as follows:

As branching increases the boiling point of the compound decreases but in compounds with comparable molar masses. Hence, 2,2-dimethylpropane boiling point is lower than pentane because it has a higher branching than pentane.
Thus, both assertion and reason are not correct.