9.8 Exercise Problems and Solutions

Q1. How do you account for the formation of ethane during chlorination of methane?

Answer: Initiation : The chlorination of methane follows a free radical mechanism. Initially, by the action of heat or sunlight on \(\mathrm{Cl}_2\), the \(\mathrm{Cl}-\mathrm{Cl}\) bond breaks

Propagation : This produces two \(\dot{\mathrm{Cl}}\) free radicals which abstract a hydrogen free radical from the methane molecule thereby generating a methyl free radical.

Termination : Now this methyl free radical may either combine with a chlorine free radical to form the desired product – chloromethane or may combine with another methyl radical to form an ethane molecule.

Q2. Write IUPAC names of the following compounds :

Answer:

Q3. For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated :
(a) \(\mathrm{C}_4 \mathrm{H}_8\) (one double bond)
(b) \(\mathrm{C}_5 \mathrm{H}_8\) (one triple bond)

Answer: (a) \(\mathrm{C}_4 \mathrm{H}_8\) can have the following structures with one single double bond :

(b) \(\mathrm{C}_5 \mathrm{H}_8\) can have the following structures with one triple bond :

Q4. Write IUPAC names of the products obtained by the ozonolysis of the following compounds :
(i) Pent-2-ene
(ii) 3,4-Dimethylhept-3-ene
(iii) 2-Ethylbut-1-ene
(iv) 1-Phenylbut-1-ene

Answer:

Q5. An alkene ‘ A ‘ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’.

Answer: Step 1: Write the structure of the products side by side with their oxygen atoms pointing towards each other.

Step 2: Remove the oxygen atoms and join the two ends by a double bond, the structure of the alkene ‘A’ is

Q6. An alkene ‘A’ contains three \(\mathrm{C}-\mathrm{C}\), eight \(\mathrm{C}-\mathrm{H~} \sigma\) bonds and one \(\mathrm{C}-\mathrm{C~} \pi\) bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.

Answer:\(A\) is an alkene. \(A\) contains
(i) \(3 C-C~ \sigma\) bonds,
(ii) \(8 \mathrm{C}-\mathrm{H~} \sigma\) bonds and
(iii) \(1 \mathrm{C}-\mathrm{C~} \pi\) bond.

\(
(A) \xrightarrow{\mathrm{O}_3} 2 \mathrm{RCHO}
\)

Molar mass of \(\mathrm{RCHO}=44 \mathrm{u}\)

(i) The clue to finding the structure and therefore the IUPAC name of \(A\) lies in the aldehyde \(\mathrm{RCHO}\).
The mass of – \(\mathrm{CHO}\) group is \(29 \mathrm{~u}\). This implies that the alkyl group attached has a mass of \(44-29=15\). This can be the mass of \(-\mathrm{CH}_3\) group only.
\(\therefore \quad \mathrm{RCHO}\) is \(\mathrm{CH}_3 \mathrm{CHO}\).
(ii) Since 2 moles of only \(\mathrm{CH}_3 \mathrm{CHO}\) is obtained. Therefore, the alkene is

Thus, \((A)\) is \(\mathrm{CH}_3-\mathrm{HC}=\mathrm{CH}-\mathrm{CH}_3\) and its IUPAC name is but-2-ene.
This structure also confirms the given number of \(\sigma\) and \(\pi\) bonds.

Q7. Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?

Answer:

Q8. Write chemical equations for combustion reaction of the following hydrocarbons:
(i) Butane
(ii) Pentene
(iii) Hexyne
(iv) Toluene

Answer: Upon combustion, any hydrocarbon produces \(\mathrm{CO}_2\) and \(\mathrm{H}_2 \mathrm{O}\). For the given hydrocarbons the reactions may be represented as :

(i) Butane: \(\mathrm{C}_4 \mathrm{H}_{10}+\frac{13}{2} \mathrm{O}_2 \xrightarrow{\Delta} 4 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O}\)
(ii) Pentene: \(\mathrm{C}_5 \mathrm{H}_{10}+\frac{15}{2} \mathrm{O}_2 \xrightarrow{\Delta} 5 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O}\)
(iii) Hexyne: \(\mathrm{C}_6 \mathrm{H}_{10}+\frac{17}{2} \mathrm{O}_2 \stackrel{\Delta}{\longrightarrow} 6 \mathrm{CO}_2+5 \mathrm{H}_2 \mathrm{O}\)
(iv) Toluene: \(\mathrm{C}_7 \mathrm{H}_8+9 \mathrm{O}_2 \stackrel{\Delta}{\longrightarrow} 7 \mathrm{CO}_2+4 \mathrm{H}_2 \mathrm{O}\)

Q9. Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?

Answer: The structures of cis- and trans-isomer of hex-2-ene are:

Of the given isomers, the cis isomer has a higher boiling point. This difference arises due to higher dipole moment of the cis isomer which introduces a somewhat ionic character in the compound. In the trans isomer, the dipoles cancel each other resulting in a small dipole moment as the case may be.

Q10. Why is benzene extra ordinarily stable though it contains three double bonds?

Answer: 1. The extraordinary stability of benzene molecule may be attributed to resonance in the molecule. In benzene, each of the \(6 \mathrm{C}\) atoms is \(\mathrm{sp}^2\) hybridised with one p-orbital on each carbon atom left unhybridised. While 2 of the \(\mathrm{sp}^2\) orbitals form bonds with \(2 \mathrm{C}\)-atoms, the third one is involved in bonding with hvdrogen atom. Thus, 3 of the valencies of \(\mathrm{C}\) are satisfied.
2. This leaves the unhybridised p-orbital containing 1 electron each for bonding. Each of these p-orbitals can overlap with the adjacent \(\mathrm{C}\) atom and thus, results in bonding.
3. Since the probability for each p-orbital to overlap with either of the two immediate neighbours is equal. Therefore, it alternately does so.
4. This increased attraction is the reason for the ‘extraordinary’ stability of the benzene molecule with 3 double bonds.
5. These \(3 \pi\)-bonds are not localised but are spread over the entire molecule.

The six electrons of the \(p\)-orbitals cover all the six carbon atoms, and are said to be delocalized. As a result of delocalization there formed a stronger \(n\)-bond and a more stable molecule.

Q11. What are the necessary conditions for any system to be aromatic?

Answer: The necessary and sufficient condition for any system to be aromatic is given by Huckel’s rule. A compound is said to be aromatic if it satisfies the following three conditions:
(i) It should have a planar structure.
(ii) The \(\pi\) – electrons of the compound are completely delocalized in the ring.
(iii) The total number of \(\pi\) – electrons present in the ring should be equal to ( \(4 n\) +2 ),
where \(n=0,1,2 \ldots\) etc. This is known as Huckel’s rule.

Q12. Explain why the following systems are not aromatic?

Answer: One of the conditions stated by the Huckel’s rule for any system to be aromatic is that of planarity i.e., all atoms of the molecule must be present on the same plane. This rule is violated in structure (i) and (ii). The carbon atom indicated below are \(\mathrm{sp}^3\) hybridised which disallow planarity \(\left(\mathrm{sp}^3\right.\) hybridised carbon is tetrahedral in geometry).

(i) Due to the presence of a sp3-hybridized carbon, the system is not planar. It does contain six \(\pi\)-electrons but the system is not fully conjugated since all the six \(\pi\)-electrons do not form a single cyclic electron cloud which surrounds all the atoms of the ring. Therefore, it is not an aromatic compound.

(ii) Due to the presence of a sp3 -carbon, the system is not planar. Further, it contains only four n-electrons, therefore, the system is not aromatic because it does not contain planar cyclic cloud having \((4 n+2) \pi\) electrons.

(iii) Cyclo-octatetraene is not planar but is tub shaped. It is, therefore, a non-planar system having \(8 \mathrm{~\pi}\)-electrons. Therefore, the molecule is not aromatic since it does not contain a planar cyclic cloud having \((4 n+2) \pi\) electrons.

Note: In (iii) the number of \(\pi\) electrons is 8 ( 2 per double bond). The Huckel’s rule allows \(2,6,10,14, \ldots\) etc. \(n\) electrons for any aromatic system. Since (iii) does not have \((4 n+2) \pi\) electrons, therefore, it is not aromatic.

Q13. How will you convert benzene into
(i) \(p\)-nitrobromobenzene
(ii) \(m\)-nitrochlorobenzene
(iii) \(p\) – nitrotoluene
(iv) acetophenone?

Answer: (i) The two substituents in the benzene ring are present at p-positions. Therefore, the sequence of reactions should be such that first an o, p-directing group, i.e., Br atom should be introduced in the benzene ring and this should be followed by nitration. Thus,

(ii) Here since the two substituents are at p-position w.r.t. each other, therefore, the first substituent in the benzene ring should be a o, p-directing group (i.e., \(\mathrm{CH}_3\) ) and then the other group (i.e., \(\mathrm{NO}_2\) ) should be introduced. Therefore, the sequence of reactions is:

(iii) Here since the two substituents are at m-position w.r.t. each other, therefore, the first substituent in the benzene ring should be a m-directing group (i.e., \(\mathrm{NO}_2\) ) and then other group (i.e., \(\mathrm{Cl}\) ) should be introduced.

(iv)Acetophenone can be prepared by F.C. acylation using either acetyl chloride or acetic anhydride.

Q14. In the alkane \(\mathrm{H}_3 \mathrm{C}-\mathrm{CH}_2-\mathrm{C}\left(\mathrm{CH}_3\right)_2-\mathrm{CH}_2-\mathrm{CH}\left(\mathrm{CH}_3\right)_2\), identify \(1^{\circ}, 2^{\circ}, 3^{\circ}\) carbon atoms and give the number of \(\mathrm{H}\) atoms bonded to each one of these.

Answer: The expanded formula of the given compound is

\(15 \mathrm{H}\) attached to five \(1^{\circ}\) carbon. \(4 \mathrm{H}\) attached to two \(2^{\circ}\) carbon. \(1 \mathrm{H}\) attached to one \(3^{\circ}\) carbon.

Q15. What effect does branching of an alkane chain has on its boiling point?

Answer: Branching of carbon atom chain decreases the boiling point of alkane. As the branching in an alkane increases, the shape of the molecule approaches a sphere and size of the branched chain alkane becomes less than that of its straight chain counterpart. The reduced surface area results in decreased van der Waals’ interaction and finally leads to lower boiling point as compared to straight chain alkanes of comparable molar mass.

Q16. Addition of \(\mathrm{HBr}\) to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

Answer: Addition of \(\mathrm{HBr}\) to propene takes place as per the Markownikoff’s rule via the formation of most stable carbocation. Addition of \(\mathrm{HBr}\) to propene is an ionic electrophilic addition reaction in which the electrophile, i.e., \(\mathrm{H}^{+}\)first adds to give a more stable \(2^{\circ}\) carbocation. In the 2nd step, the carbocation is rapidly attacked by the nucleophile \(\mathrm{Br}\) ion to give 2-bromopropane.

Thus, it is an example of electrophilic addition reaction.

But, when the same reaction is carried out in the presence of peroxide, which is a free radical generator, the mechanism follows free radical mechanism where the intermediate is a carbon free radical. In presence of benzoyl peroxide, the reaction is still electrophilic but the electrophile here is a \(\mathrm{Br}\) free radical which is obtained by the action of benzoyl peroxide on \(\mathrm{HBr}\)

Here, the free radical is the \(2^{\circ}\) free radical and therefore it is the one which finally forms 1-bromopropane.

This explains why different products are formed under different reaction conditions. From the above discussion, it is evident that although both reactions are electrophilic addition reactions but it is due to different order of addition of \(\mathrm{H}\) and \(\mathrm{Br}\) atoms which gives different products.

Q17. Write down the products of ozonolysis of 1,2-dimethylbenzene (o-xylene). How does the result support Kekule structure for benzene?

Answer: o-Xylene may be regarded as a resonance hybrid of the following two Kekule structures. Ozonolysis of each one of these gives two products. The products of ozonolysis of \(o\)-xylene are:

These three products cannot be obtained if we assume a triene structure of benzene. They can be obtained only if it is assumed that the double bonds in benzene are delocalised. If the double bonds are assumed to be between \(\mathrm{C_1-C_2, C_3-C_4}\) and \(\mathrm{C_5-C_6}\) then

If the double bonds are assumed to be localized between \(\mathrm{C_2-C_3, C_4-C_3, C_6-C_1}\), then

But since all the above products are obtained, it indicates that only one structure is not possible and the resonance structures proposed by Kekule are the only possibilities.

Thus, in all, three products are formed. Since all the three products cannot be obtained from any one of the two Kekule structures, this’ shows that o-xylene is a resonance hybrid of the two Kekule structures (I and II).

Q18. Arrange benzene, \(n\)-hexane and ethyne in decreasing order of acidic behaviour. Also give reason for this behaviour.

Answer: The hybridization state of carbon in these three compounds is:

Since \(s\)-electrons are closer to the nucleus, therefore, as the \(s\)-character of the orbital making the \(\mathrm{C}-\mathrm{H}\) bond increases, the electrons of \(\mathrm{C}-\mathrm{H}\) bond lie closer and closer to the carbon atom. In other words, the partial +ve charge on the \(\mathrm{H}\)-atom and hence the acidic character increases as the \(s\)-character of the orbital increases. Thus, the acidic character decreases in the order: Ethyne \(>\) Benzene \(>\) Hexane.

Note: Acidic strength of any compound is defined as its ability to release \(\mathrm{H}\) atom as \(\stackrel{+}{\mathrm{H}}\). This atom to which \(\mathrm{H}\) is attached is highly electronegative. Such atoms tend to pull the bond pair of electrons towards themselves thereby polarising the bond and creating \(\delta^{+}\) charge on \(\mathrm{H}\).

In the compounds in question, the \(\mathrm{H}\) is attached to carbon atom which does not have a very high electronegativity value. Moreover, the electronegativity of carbon is somewhat increased if the \(s\)-character of its hybridised orbitals is high. Greater is the \(s\)-character of carbon, greater is its electronegativity as a result of which, the bond pair of electrons of the \(\mathrm{C}-\mathrm{H}\) bond moves towards carbon atom, creating a partial positive charge on \(\mathrm{H}\) and facilitating its release as \(\mathrm{H}^{+}\).

Q19. Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?

Answer: Electrophiles are species that are electron deficient and hence, seek electron rich molecules. Benzene is one such molecule which is rich in electrons. It is so because benzene has \(6 {~n}\) electrons delocalised over the entire molecule which acts as a good host for electrophiles. Another point working in favour of electrophilic substitution reactions is the retention of aromaticity. A benzene molecule is highly stable owing to its aromatic character. Therefore, it would not want to lose its aromaticity. Upon undergoing electrophilic substitution reaction, this aromaticity is not lost, it is retained and hence, benzene undergoes electrophilic substitution.

Contrast this with a nucleophilic substitution reaction where the nucleophile attacks. A nucleophile \(\left(\mathrm{Nu}^{-}\right)\)is a species that seeks a positive centre or an electron deficient species. Obviously, benzene is not electron deficient and therefore, will not be a welcome site for a \(\mathrm{Nu}^{-}\). This is the major reason why benzene does not undergo a nucleophilic substitution reaction. Another reason working against these reactions is the difficulty with which the transition state is formed. The transition state benzyne involved here is formed with great difficulty and hence, these reactions are difficult to bring about.

Note: Due to the presence of an electron cloud containing \(6 {~n}\)-electrons above and below the plane of the ring, benzene is a rich source of electrons. Consequently, it attracts the electrophiles (electron-deficient) reagents towards it and repels nucleophiles (electron- rich) reagents. As a result, benzene undergoes electrophilic substitution reactions easily and nucleophilic substitutions with difficulty.

Q20. How would you convert the following compounds into benzene?
(i) Ethyne
(ii) Ethene
(iii) Hexane

Answer: (i) When acetylene is heated in red hot iron tube at \(873 \mathrm{~K}\), benzene is obtained.

(ii) Ethene is first converted into ethyne and then to benzene as shown above.

(iii) When vapours of hexane are passed over heated catalyst consisting of \(\mathrm{Cr}_2 \mathrm{O}_3\), \(\mathrm{MO}_2 \mathrm{O}_3\) and \(\mathrm{V}_2 \mathrm{O}_5\) at \(773 \mathrm{~K}\) under \(10-20\) atm pressure, cyclization and aromatization occurs simultaneously to afford benzene

Q21. Write structures of all the alkenes which on hydrogenation give 2-methylbutane.

Answer: The basic skeleton 2-methylbutane is

Putting double bonds at various different positions and satisfying the tetracovalency of each carbon, the structures of various alkenes which give 2-methylbutane on hydrogenation are:

Q22. Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, \(\mathrm{E}^{+}\)
(a) Chlorobenzene, 2,4-dinitrochlorobenzene, p-nitrochlorobenzene
(b) Toluene, \(p-\mathrm{H}_3 \mathrm{C}-\mathrm{C}_6 \mathrm{H}_4-\mathrm{NO}_2, p-\mathrm{O}_2 \mathrm{~N}-\mathrm{C}_6 \mathrm{H}_4-\mathrm{NO}_2\).

Answer: Any substituted benzene compound is said to be more reactive towards an electrophile if the substituent on benzene increases electron density on it by electron donation. Based on this, the given compounds may be arranged as :

(a) The typical reactions of benzene are electrophilic substitution reactions. Higher the electrondensity in the benzene ring, more reactive is the compound towards these reactions. Since \(\mathrm{NO_2}\) is a more powerful electron-withdrawing group than \(\mathrm{Cl}\), therefore, more the number of nitro groups, less reactive is the compound. Thus, the overall reactivity decreases in the order:
Chlorobenzene \(>\mathrm{p}\)-nitrochlorobenzene \(>2\), 4-dinitrochlorobenzene

(b) Here, \(\mathrm{CH} 3\) group is electron donating but \(\mathrm{NO_2}\) group is electron-withdrawing. Therefore, the maximum electron-density will be in toluene, followed by p-nitrotoluene followed by p-dinitrobenzene. Thus, the overall reactivity decreases in the order:

Q23. Out of benzene, \(m\)-dinitrobenzene and toluene which will undergo nitration most easily and why?

Answer: During nitration, the electrophile \(\mathrm{NO}_2^{+}\) attacks the benzene ring. Nitration will be easier if the benzene ring shows increased electron density. This happens when electron releasing groups such as \(-\mathrm{R},-\mathrm{NH}_2\), \(-\mathrm{NHCOCH}_3,-\mathrm{OH},-\mathrm{OMe}\) etc. are attached to the ring whereas, the attachment of electron withdrawing groups such as \(-\mathrm{NO}_2\), \(-\mathrm{CHO},-\mathrm{COR},-\mathrm{COOH}\) reduces the electron availability for \(\mathrm{NO}_2^{+}\)and nitration becomes difficult.
Therefore, relative ease of nitration of given molecules may be arranged as:

Note: \(\mathrm{CH}_3\) group is electron-donating while \(-\mathrm{NO}_2\) group is electron-withdrawing. Therefore, maximum electron density will be in toluene, followed by benzene and least in \(\mathrm{m}\)-dinitrobenzene. Therefore, the ease of nitration decreases in the order: toluene > benzene > m-dinitrobenzene.

Q24. Suggest the name of a Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene.

Answer: Anhydrous Ferric Chloride \(\left(\mathrm{FeCl}_3\right)\) is another Lewis acid which can be used.

Q25. Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.

Answer: Wurtz reaction is used for preparation of alkanes. During this reaction, an alkyl halide with half the number of carbon atoms than the desired alkane is made to react with sodium metal in acetone. This leads to the formation of the desired alkane. e.g., If the desired alkane is ethane, methyl iodide is taken.

While this method is highly successful for producing alkanes with even number of carbon atoms but it gives a mixture of alkanes when odd numbered alkanes are to be formed. This happens because two different alkyl halides not only react with each other but also with themselves. This can be better illustrated with an example – propane is prepared by taking methyl iodide and ethyl iodide.

The side reactions that take place in the reaction mixture are:

EXEMPLAR

Q26. Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reactions? Explain.

Answer: Both alkenes and arenes are electron-rich. Therefore undergo electrophilic reactions. Olefins undergo addition reactions because addition of a reagent to an olefin gives a more stable product as \(s p^2\) hybridisation changes to \(s p^3\) hybridisation. Addition to the double bond of an arene would give a product with less or no resonance stability hence addition is difficult arenes. On the other hand in substitution reaction resonance stabilisation is retained therefore, arenes undergo substitution reaction. For example,

Arenes, on the other hand, cannot undergo electrophilic addition reactions. This is because benzene has a large resonance energy of \(150.4 \mathrm{~kJ} \mathrm{~mol}^{-1}\). During electrophilic addition reactions, two new \(\sigma\)-bonds are formed but the aromatic character of benzene gets destroyed and, therefore, resonance energy of benzene ring is lost. Hence, electrophilic addition reactions of arenes are not energetically favourable. Arenes, in contrast, undergo electrophilic substitution reactions in which \(\sigma \mathrm{~C}-\mathrm{H}\) bond is broken and new \(\sigma \mathrm{~C}-\mathrm{X}\) bond is formed: The aromatic character of benzene ring is not destroyed and benzene retains its resonance energy. Hence, arenes undergo electrophilic substitution reactions.

Note: Arenes are stabilized by resonance with delocalization of \(\pi\) electrons. On addition reaction to the double bond of arene, a product is obtained which is not resonance stabilized whereas on substitution the resonance stability of arene is maintained. Thus, arenes prefer to undergo substitution reaction while alkenes prefer to undergo addition reaction.

Q27. Alkynes on reduction with sodium in liquid ammonia form trans alkenes. Will the butene thus formed on reduction of 2-butyne show the geometrical isomerism?

Answer: Trans-2-butene formed by the reduction of 2-butyne is capable of showing geometrical isomerism.

Q28. Rotation around carbon-carbon single bond of ethane is not completely free. Justify the statement.

Answer:The rotation about \(\mathrm{C}-\mathrm{C}\) bond is restricted because of repulsion between electron cloud of \(\mathrm{C}-\mathrm{H}\) bonds on either carbon atoms.

Ethane contains carbon-carbon sigma \((\sigma)\) bond. Electron distribution of the sigma molecular orbital is symmetrical around the intemuclear axis of the \(\mathrm{C}-\mathrm{C}\) bond which is not disturbed due to rotation about its axis. This permits free rotation around a \(\mathrm{C}-\mathrm{C}\) single bond. However, rotation around a \(\mathrm{C}-\mathrm{C}\) single bond is not completely free. It is hindered by a small energy barrier due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive interaction is called torsional strain. Of all the conformations of ethane, the staggered form has the least torsional strain and the eclipsed form has the maximum torsional strain. The energy difference between the two extreme forms is of the order of \(12.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\), which is very small. It has not been possible to separate and isolate different conformational isomers of ethane.

Alkanes can have infinite number of conformations by rotation around \(\mathrm{C}-\mathrm{C}\) single bonds. This rotation around a \(\mathrm{C}-\mathrm{C}\) single bond is hindered by a small energy barrier of \(1-20 \mathrm{~kJ} \mathrm{~mol}^{-1}\)due to weak repulsive interaction between the adjacent bonds. Such a type of repulsive interaction is called torsional strain. In staggered form of ethane, the electron cloud of carbon hydrogen bonds are far apart.
Hence, minimum replusive force. In eclipsed electron cloud of carbon-hydrogen become close resulting in increase in electron could repulsion. This repulsion affects stability of a conformer.
In all the conformations of ethane the staggered form has least torsional strain and the eclipsed form has the maximum torsional strain. Hence, rotation around \(\mathrm{C}-\mathrm{C}\) bond in ethane is not completely free.

Q29. Draw Newman and Sawhorse projections for the eclipsed and staggered conformations of ethane. Which of these conformations is more stable and why?

Answer:

In staggered form of ethane, the electron clouds of carbon-hydrogen bonds are as far apart as possible. Thus, there are minimum repulsive forces, minimum energy and maximum stability of the molecule. On the other hand, when the staggered form changes into the eclipsed form, the electron clouds of the carbon-hydrogen bonds come closer to each other resulting in increase in electron cloud repulsions. To check the increased repulsive forces, molecule will have to possess more energy and thus has lesser stability.

Note: Staggered form of ethane is more stable than the eclipsed conformation, by about \(12.55 \mathrm{~kJ} / \mathrm{mol}\). This is because any two hydrogen atom on adjacent carbon atoms of staggered conformation are maximum apart while in eclipsed conformation, they cover or eclipse each other in space. Thus, in staggered form, there is minimum repulsive forces, minimum energy and maximum stability of the molecule.

Q30. The intermediate carbocation formed in the reactions of \(\mathrm{HI}, \mathrm{HBr}\) and \(\mathrm{HCl}\) with propene is the same and the bond energy of \(\mathrm{HCl}, \mathrm{HBr}\) and \(\mathrm{HI}\) is \(430.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\), \(363.7 \mathrm{~kJ} \mathrm{~mol}^{-1}\) and \(296.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\) respectively. What will be the order of reactivity of these halogen acids?

Answer: Bond dissociation energy is least for \(\mathrm{HI}\) and maximum for \(\mathrm{HCl}\) therefore, order of reactivity will be \(\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}\).

Q31. What will be the product obtained as a result of the following reaction and why?

Answer: Propyl chloride forms less stable \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2^{\oplus}\) carbocation with anhydrous \(\mathrm{AlCl}_3\) which rearranges to a more stable \(\mathrm{CH}_3-\stackrel{\oplus}{\mathrm{C}} \mathrm{H}-\mathrm{CH}_3\) carbocation and gives isopropylbenzene as the product of the reaction.

Propyl chloride forms \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH_2}^{+}\) with anhydrous \(\mathrm{Al} \mathrm{Cl}_3\) which is less stable. This rearranges to a more stable carbocation as:

Q32. How will you convert benzene into (i) \(p\) – nitrobromobenzene
(ii) \(m\)-nitrobromobenzene

Answer:

Q33. Arrange the following set of compounds in the order of their decreasing relative reactivity with an electrophile. Give reason.

Answer: The methoxy group \(\left(-\mathrm{OCH}_3\right)\) is electron releasing group. It increases the electron density in benzene nucleus due to resonance effect (+R-effect). Hence, it makes anisole more reactive than benzene towards the electrophile.

In case of alkyl halides, the electron density increases at ortho and para positions due to \(+R\) effect. However, the halogen atom also withdraws electrons from the ring because of its -I effect. Since the -I effect is stronger than the \(+R\) effect, the halogens are moderately deactivating. Thus, overall electron density on benzene ring decreases, which makes further substitution difficult. \(-\mathrm{NO}_2\) group is electron withdrawing group. It decreases the electron density in benzene nucleus due to its strong -R-effect and strong -I-effect. Hence, it makes nitrobenzene less reactive. Therefore, overall reactivity of these three compounds towards electrophiles decreases in the following order:

The \(+\mathrm{R}\) effect of \(-\mathrm{OCH}_3>-\mathrm{Cl}\) and \(-\mathrm{NO}_2\) has a \(-\mathrm{R}\) effect. Relative reactivity of the substituted benzene rings is as follows :
\(
\mathrm{C}_6 \mathrm{H}_5-\mathrm{OCH}_3>\mathrm{C}_6 \mathrm{H}_5-\mathrm{Cl}>\mathrm{C}_6 \mathrm{H}_5-\mathrm{NO}_2
\)

Q34. Despite their – I effect, halogens are \(o\) – and \(p\)-directing in haloarenes. Explain.

Answer: Halogens attached to benzene rings exert \(-I\) and \(+R\) effect. \(+R\) effect dominates \(-\mathrm{I}\) effect and increases the electron density at ortho and para positions of the benzene ring with respect to halogens.

In case of aryl halides, halogens are little deactivating because of their strong -I effect. Therefore, overall electron density on the benzene ring decreases. In other words, halogens are deactivating due to -I effect. However, because of the +R-effect, i.e., participation of lone pairs of electrons on the halogen atom with the \(\pi\)-electrons of the benzene ring, the electron density increases more at \(o\) – and \(p\)-positions than at \(m\)-positions.

As a result, halogens are \(o\)-, \(p\)-directing. The combined result of \(+R\)-effect and -I -effect of halogens is that halogens are deactivating but \({o}, {p}\)-directing.

Q35. Why does presence of a nitro group make the benzene ring less reactive in comparison to the unsubstituted benzene ring. Explain.

Answer: Nitro group is an electron withdrawing group ( \(-R\) and \(-I\) effects). It deactivates the ring by decreasing nucleophilicity for further substitution.

Q36. Suggest a route for the preparation of nitrobenzene starting from acetylene?

Answer: Acetylene when passed through red hot iron tube at \(500^{\circ} \mathrm{C}\) undergoes cyclic polymerisation to give benzene which upon nitration gives nitrobenzene.

Q37. Predict the major product (s) of the following reactions and explain their formation.

\(
\mathrm{H}_3 \mathrm{C}-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow[\mathrm{HBr}]{(\mathrm{Ph}-\mathrm{CO}-\mathrm{O})_2}
\)
\(
\mathrm{H}_3 \mathrm{C}-\mathrm{CH}=\mathrm{CH}_2 \xrightarrow{\mathrm{HBr}}
\)

Answer: Addition of \(\mathrm{HBr}\) to unsymmetrical alkenes follows Markonikov rule. It states that negative part of the addendum (adding molecule) gets attached to that carbon atom which possesses lesser number of hydrogen atoms. Mechanism: Hydrogen bromide provides an electrophile, \(\mathrm{H}^{+}\), which attacks the double bond to form carbocation as shown below:

Addition reaction of \(\mathrm{HBr}\) to unsymmetrical alkenes in the presence of peroxide follows anti-Markovnikov rule. Mechanism: Peroxide effect proceeds via free radical chain mechanism as given below:

Q38. Nucleophiles and electrophiles are reaction intermediates having electron rich and electron deficient centres respectively. Hence, they tend to attack electron deficient and electron rich centres respectively. Classify the following species as electrophiles and nucleophiles.

Answer: Electrophiles are electron deficient species. They may be natural or positively charged. Nucleophiles are electron rich species. They may be neutral or negatively charged.

Q39. The relative reactivity of \(1^{\circ}, 2^{\circ}, 3^{\circ}\) hydrogen’s towards chlorinaltion is \(1: 3.8: 5\). Calculate the percentages of all monochlorinated products obtained from 2-methylbutane.

Answer: 2 -Methyl butane is

Possible compounds are A, B and C given below :

Relative amounts of \(\mathrm{A}, \mathrm{B}=\) and \(\mathrm{C}\) compounds = number of hydrogen \(\times\) relative reactivity
\(
\begin{array}{llll}
& \mathbf{A}\left(\mathbf{1}^{\circ}\right) & \mathbf{B}\left(\mathbf{2}^{\circ}\right) & \mathbf{C}\left(\mathbf{3}^{\circ}\right) \\
\text { Relative amount } & 9 \times 1=9 & 2 \times 3.8=7.6 & 1 \times 5=5
\end{array}
\)
Total Amount of monohaloginated compounds \(=9+7.6+5=21.6\)
Percentage of A \(=\frac{9}{21.6} \times 100=41.7 \%\)
Percentage of B \(=\frac{7.6}{21.6} \times 100=35.2 \%\)
Percentage of C \(=\frac{5}{21.6} \times 100=23.1 \%\)

Q40. Write the structures and names of products obtained in the reactions of sodium with a mixture of 1-iodo-2-methylpropane and 2-iodopropane.

Answer:

Q41. Write hydrocarbon radicals that can be formed as intermediates during monochlorination of 2-methylpropane? Which of them is more stable? Give reasons.

Answer: 2-methylpropane gives two types of radicals.

Radical I is tertiary where as radical II is primary. Radical I is more stable due to hyperconjugation. Radical (I) is more stable because it is \(3^{\circ}\) is free radical and stabilised by nine hyperconjugative structures (as it has \(9 \alpha\)-hydrogens) Radical (II) is less stable because it is \(1^{\circ}\) free radical and stabilised by only one hyperconjugative structure (as it has only \(1 \alpha\)-hydrogen).

Q42. An alkane \(\mathrm{C}_8 \mathrm{H}_{18}\) is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide.

Answer: Since the alkanes upon monobromination forms a single isomeric tertiary alkyl bromide, the alkane \(\mathrm{C}_8 \mathrm{H}_{18}\) is a symmetrical secondary alkane of the corresponding primary alkyl halide which gives alkane by Wurtz reaction, is a branched chain. The reaction is as follows:

Q43. The ring systems having following characteristics are aromatic.
(i) Planar ring containing conjugated \(\pi\) bonds.
(ii) Complete delocalisation of the \(\pi\)-electrons in ring system i.e. each atom in the ring has unhybridised \(p\)-orbital, and
(iii) Presence of \((4 n+2) \pi\)-electrons in the ring where \(n\) is an integer \((\mathrm{n}=0,1,2, \ldots \ldots \ldots . ..\)) [Huckel rule]
Using this information classify the following compounds as aromatic/nonaromatic.

Answer: \(\mathrm{A}=\) Planar ring, all atoms of the ring \(s p^2\) hybridised, has six delocalised
\(\pi\) electrons, follows Huckel rule. It is aromatic.
\(\mathrm{B}=\) Has six \(\pi\) electrons, but the delocalisation stops at \(s p^3\) hybridised
\(\mathrm{CH}_2\) – carbon. Hence, not aromatic.
\(\mathrm{C}=\) Six delocalised \(\pi\)-electrons ( \(4 \pi\) electrons +2 unshared electrons on
negatively charged carbon) in a planar ring, follows Huckel’s rule.
It is aromatic.
\(\mathrm{D}=\) Has only four delocalised \(\pi\)-electrons. It is non aromatic.
\(\mathrm{E}=\) Six delocalised \(\pi\)-electrons follows Huckel’s rule. \(\pi\) electrons are in \(s p^2\) hybridised orbitals, conjugation all over the ring because of positively charged carbon. The ring is planar hence is aromatic.
\(\mathrm{F}=\) Follows Huckel’s rule, has \(2 \pi\) electrons i.e. \((4 n+2) \pi\)-electrons where \((n=0)\), delocalised \(\pi\)-electrons. It is aromatic.
G= for aromatic and non-aromatic compound first condition is all carbon should be \(sp^2\) in a ring, but there is one carbon which is \(sp^3\) hybridised, so it will be not aromatic.  \(8 \pi\) electrons, does not follow Huckel’s rule i.e., \((4 n+2) \pi\)-electrons rule.

Q44. Which of the following compounds are aromatic according to Huckel’s rule?

Answer: (B, D, E & F) are aromatic.

\(\mathrm{A}=\) Has \(8 \pi\) electrons, does not follow Huckel rule. Orbitals of one carbon atom are not in conjugation. It is not aromatic.
\(\mathrm{B}=\) Has \(6 \pi\) delocalised electrons. Hence, is aromatic.
\(\mathrm{C}=\) Has \(6 \pi\) electrons in conjugation but not in the ring. Non aromatic.
\(\mathrm{D}=10 \pi\) electrons in planar rings, aromatic.
\(\mathrm{E}=\) Out of \(8 \pi\) electrons it has delocalised \(6 \pi\) electrons in one six membered planar ring, which follows Huckel’s rule due to which it will be aromatic.
\(\mathrm{F}=14 \pi\) electrons are in conjugation and are present in a ring. Huckel’s rule is being followed. Compound will be aromatic if ring is planar. But the molecule also has to be able to assume a fully planar conformation (with all the pi orbitals parallel) to be fully aromatic. While we can draw a configuration where all the C-C bond angles are \(120^{\circ}\).

Q45. Suggest a route to prepare ethyl hydrogensulphate \(\left(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OSO}_2-\mathrm{OH}\right)\) starting from ethanol \(\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)\).

Answer:For preparation of ethyl hydrogensulphate \(\left(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{OSO}_2-\mathrm{OH}\right)\) starting from ethanol \(\left(\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}\right)\), it is the two steps mechanism. Step I Protonation of alcohol
\(
\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{H}^{+}+{ }_{-} \mathrm{OSO}_2 \mathrm{OH}
\)

Step II Attack of nucleophile

Temperature should not be allowed to rise above \(383 \mathrm{~K}\), otherwise diethyl ether will be produced at \(413 \mathrm{~K}\) or ethene at \(433 \mathrm{~K}\).

Q46. An alkyl halide \(\mathrm{C}_5 \mathrm{H}_{11} \mathrm{Br}(\mathrm{A})\) reacts with ethanolic \(\mathrm{KOH}\) to give an alkene ‘ \(\mathrm{B}\) ‘, which reacts with \(\mathrm{Br}_2\) to give a compound ‘ \(\mathrm{C}\) ‘, which on dehydrobromination gives an alkyne ‘ \(D\) ‘. On treatment with sodium metal in liquid ammonia one mole of ‘ \(\mathrm{D}\) ‘ gives one mole of the sodium salt of ‘ \(\mathrm{D}\) ‘ and half a mole of hydrogen gas. Complete hydrogenation of ‘ \(D\) ‘ yields a straight chain alkane. Identify \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and \(\mathrm{D}\). Give the reactions invovled.

Answer: The reaction scheme involved in the problem is

The reactions suggest that (D) is a terminal alkyne. This means triple bond is at the end of the chain. It could be either (I) or (II).

Since alkyne ‘D’ on hydrogenation yields straight chain alkane, therefore structure I is the structure of alkyne (D). Hence, the structures of A, B and C are as follows:

(A) \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}\)
(B) \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2\)
(C) \(\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CH}(\mathrm{Br})-\mathrm{CH}_2 \mathrm{Br}\)

Q47. \(896 \mathrm{~mL}\) vapour of a hydrocarbon ‘A’ having carbon \(87.80 \%\) and hydrogen \(12.19 \%\) weighs \(3.28 \mathrm{~g}\) at STP. Hydrogenation of ‘A’ gives 2-methylpentane. Also ‘A’ on hydration in the presence of \(\mathrm{H}_2 \mathrm{SO}_4\) and \(\mathrm{HgSO}_4\) gives a ketone ‘B’ having molecular formula \(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}\). The ketone ‘ \(\mathrm{B}\) ‘ gives a positive iodoform test. Find the structure of ‘A’ and give the reactions involved.

Answer:Step I:
\(896 \mathrm{~mL}\) vapour of \(\mathrm{C}_x \mathrm{H}_y\) (A) weighs \(3.28 \mathrm{~g}\)
\(22700 \mathrm{~mL}\) vapour of \(\mathrm{C}_x \mathrm{H}_y(\mathrm{~A})\) weighs \(\frac{3.28 \times 22700}{896} \mathrm{~g} \mathrm{~mol}^{-1}=83.1 \mathrm{~g} \mathrm{~mol}^{-1}\)
Step II:
\(
\begin{array}{llcccc}
\begin{array}{l}
\text { Element } \\
\end{array} & (\%) & \begin{array}{c}
\text { Atomic } \\
\text { mass }
\end{array} & \text { Relative ratio } & \begin{array}{c}
\text { Relative no. } \\
\text { of atoms }
\end{array} & \begin{array}{c}
\text { Simplest } \\
\text { ratio }
\end{array} \\
\mathrm{C} & 87.8 & 12 & 7.31 & 1 & 3 \\
\mathrm{H} & 12.19 & 1 & 12.19 & 1.66 & 4.98 \approx 5
\end{array}
\)
Empirical formula of ‘ \(\mathrm{A}\) ‘ \(\mathrm{C}_3 \mathrm{H}_5\)
Empirical formula mass \(=35+5=41 \mathrm{u}\)
\(
\mathrm{n}=\frac{\text { Molecular mass }}{\text { Empirical formula mass }}=\frac{83.1}{41}=2.02 \approx 2
\)
\(\Rightarrow\) Molecular mass is double of the empirical formula mass.
\(\therefore\) Molecular Formula is \(\mathrm{C}_6 \mathrm{H}_{10}\)
Step III:
\(
\mathrm{C}_6 \mathrm{H}_{10} \xrightarrow{2 \mathrm{H}_2} \text { 2-methylpentane }
\)
\(\text { (A) }\)
Structure of 2-methylpentane is

Hence, the molecule has a five carbon chain with a methyl group at the second carbon atom.

‘ \(A\) ‘ adds a molecule of \(\mathrm{H}_2 \mathrm{O}\) in the presence of \(\mathrm{Hg}^{2+}\) and \(\mathrm{H}^{+}\), it should be an alkyne. Two possible structures for ‘ \(\mathrm{A}\) ‘ are:

Since the ketone (B) gives a positive iodoform test, it should contain a \(-\mathrm{COCH}_3\) group. Hence the structure of ketone is as follows:

Therefore structure of alkyne is II.

Q48. An unsaturated hydrocarbon ‘A’ adds two molecules of \(\mathrm{H}_2\) and on reductive ozonolysis gives butane-1,4-dial, ethanal and propanone. Give the structure of ‘A’, write its IUPAC name and explain the reactions involved.

Answer: Two molecules of hydrogen add on ‘ \(A\) ‘ this shows that ‘ \(A\) ‘ is either an alkadiene or an alkyne.
On reductive ozonolysis ‘ \(A\) ‘ gives three fragments, one of which is dialdehyde. Hence, the molecule has broken down at two sites. Therefore, ‘A’ has two double bonds. It gives the following three fragments:
\(
\mathrm{OHC}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CHO}, \mathrm{CH}_3 \mathrm{CHO} \text { and } \mathrm{CH}_3-\mathrm{CO}-\mathrm{CH}_3
\)
Hence, its structure as deduced from the three fragments must be

Q49. In the presence of peroxide addition of \(\mathrm{HBr}\) to propene takes place according to anti Markovnikov’s rule but peroxide effect is not seen in the case of \(\mathrm{HCl}\) and HI. Explain.

Answer: \(
\underset{\text { propene }}{\mathrm{CH}_3-\mathrm{CH}}=\mathrm{CH}_2+\mathrm{HBr} \xrightarrow{\text { Peroxide }} \underset{\text { n-propyl bromide }}{\mathrm{CH}_3-\mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br}}
\)
The mechanism of the reaction is

Peroxide effect is effective only in the case of \(\mathrm{HBr}\) and not seen in the case of \(\mathrm{HCl}\) and \(\mathrm{HI}\). This is due to the following reasons
(i) \(\mathrm{H}-\mathrm{Cl}\) bond \((103 \mathrm{kcal} / \mathrm{mol})\) is stronger than \(\mathrm{H}-\mathrm{Br}\) bond \((87 \mathrm{kcal} / \mathrm{mol})\)
\(\mathrm{H}-\mathrm{Cl}\) bond is not decomposed by the peroxide free radical whereas the \(\mathrm{H}-\mathrm{I}\) bond is weaker \((71 \mathrm{kcal} / \mathrm{mol})\) form iodine free radicals.
(ii) lodine free radical \(\left(\mathrm{I}^{\circ}\right)\) formed as \(\mathrm{H}-\mathrm{I}\) bond is weaker but iodine free radicals readily combine with each other to form iodine molecules rather attacking the double bond.