Hydrocarbons Class 11 Chemistry NCERT Solutions

Step-by-step solution

1. 1Methane and chlorine react in sunlight or at a high temperature by a free-radical chain mechanism. Chlorine first dissociates into chlorine atoms.
2. 2
   $$\mathrm{C}{\mathrm{l}}_2\stackrel{\mathrm{h}\nu }{\to }2Cl\bullet$$
3. 3A chlorine atom abstracts a hydrogen atom from methane, producing hydrogen chloride and a methyl radical.
4. 4
   $$\mathrm{Cl}\bullet +\mathrm{C}{\mathrm{H}}_4\to \mathrm{HCl}+\mathrm{C}{\mathrm{H}}_3\bullet$$
5. 5The methyl radical reacts with another chlorine molecule, producing chloromethane and a new chlorine atom. The propagation steps repeat while methane and chlorine remain.
6. 6
   $$\mathrm{C}{\mathrm{H}}_3\bullet +\mathrm{C}{\mathrm{l}}_2\to \mathrm{C}{\mathrm{H}}_3\mathrm{Cl}+\mathrm{Cl}\bullet$$
7. 7Two methyl radicals can also combine in a termination step. Their combination forms a C–C σ bond and gives ethane.
8. 8
   $$\mathrm{C}{\mathrm{H}}_3\bullet +\mathrm{C}{\mathrm{H}}_3\bullet \to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_3$$

Step-by-step solution

1. 1In (a), the longest chain containing the double bond has four carbons; numbering from the end nearest the double bond places it at C-2 and the methyl group at C-2.
2. 2In (b), the five-carbon parent contains both multiple bonds. The numbering that gives the double bond the lower locant gives pent-1-en-3-yne, also written pent-1-ene-3-yne in some older answer formats.
3. 3In (c), the four-carbon chain has double bonds at C-1 and C-3, giving buta-1,3-diene.
4. 4In (d), benzene is treated as a phenyl substituent because the principal chain is the four-carbon chain containing the C=C bond; the double bond is at C-1 and phenyl is at C-4.
5. 5In (e), the hydroxyl group is the principal group, so benzene is the parent and the adjacent methyl substituent is at C-2: 2-methylphenol, or o-cresol.
6. 6In (f), the longest continuous parent chain has ten carbons. The attached branched group is a 2-methylpropyl group at C-5.
7. 7In (g), the longest chain containing the maximum number of double bonds has ten carbons. Numbering from the terminal alkene end gives double bonds at 1, 5 and 8 and an ethyl substituent at C-4.

Step-by-step solution

1. 1For C₄H₈ with one C=C bond, distribute the four carbon atoms between a terminal alkene, an internal alkene and a branched alkene.
2. 2The three structural formulas are CH₂=CH–CH₂–CH₃, CH₃–CH=CH–CH₃ and CH₂=C(CH₃)₂.
3. 3The first two are position isomers, while the third is a chain isomer of them. But-2-ene also has cis and trans geometrical forms because each double-bond carbon has two different groups.
4. 4For C₅H₈ with one C≡C bond, the possible placements of the triple bond give pent-1-yne and pent-2-yne.
5. 5The branched skeleton gives HC≡C–CH(CH₃)–CH₃, which is 3-methylbut-1-yne.

Step-by-step solution

1. 1Reductive ozonolysis cleaves the C=C bond and converts each alkene carbon into a carbonyl carbon. An alkene carbon carrying H gives an aldehyde; one carrying two carbon groups gives a ketone.
2. 2Pent-2-ene, CH₃–CH=CH–CH₂–CH₃, therefore gives ethanal and propanal.
3. 3In 3,4-dimethylhept-3-ene, the left double-bond carbon has an ethyl and a methyl group, while the right one has a methyl and a propyl group. The products are butan-2-one and pentan-2-one.
4. 42-Ethylbut-1-ene, CH₂=C(C₂H₅)–CH₂–CH₃, gives methanal from the terminal CH₂ carbon and pentan-3-one from the substituted carbon.
5. 51-Phenylbut-1-ene, C₆H₅–CH=CH–CH₂–CH₃, gives benzaldehyde and propanal.

Step-by-step solution

1. 1Ethanal, CH₃CHO, means that one carbon of the original double bond was attached to CH₃ and H.
2. 2Pentan-3-one, CH₃CH₂COCH₂CH₃, means that the other alkene carbon was attached to two ethyl groups.
3. 3Join the two carbonyl carbons after removing their oxygen atoms. The resulting structure is CH₃–CH=C(CH₂CH₃)₂.
4. 4The longest chain containing the double bond has five carbons. Numbering gives the double bond at C-2 and a methyl-derived ethyl substituent at C-3.

Step-by-step solution

1. 1The aldehyde with molar mass 44 u is ethanal: its formula is CH₃CHO and its molar mass is 2(12) + 4(1) + 16 = 44 u.
2. 2Two molecules of ethanal arise when the two carbons of the original C=C bond each carry CH₃ and H. Rejoining the carbonyl carbons gives CH₃–CH=CH–CH₃.
3. 3This structure has three C–C σ bonds, eight C–H σ bonds and one C–C π bond, matching the data.

Step-by-step solution

1. 1Propanal, CH₃CH₂CHO, identifies one alkene carbon as attached to an ethyl group and H.
2. 2Pentan-3-one, CH₃CH₂COCH₂CH₃, identifies the other alkene carbon as attached to two ethyl groups.
3. 3Join these two carbonyl carbons by a C=C bond and remove the two oxygen atoms: CH₃CH₂–CH=C(CH₂CH₃)₂.
4. 4The longest chain has six carbons, the double bond is at C-3 and the second ethyl group is a substituent at C-4.

Step-by-step solution

1. 1Complete combustion converts every carbon atom into CO₂ and every hydrogen atom into H₂O, while O₂ supplies oxygen and heat is released.
2. 2
   $$2{\mathrm{C}}_4{\mathrm{H}}_{10}+13{\mathrm{O}}_2\to 8C{\mathrm{O}}_2+10{\mathrm{H}}_2\mathrm{O}+\mathrm{heat}$$
3. 3
   $${\mathrm{C}}_5{\mathrm{H}}_{10}+\frac{15}{2}{\mathrm{O}}_2\to 5C{\mathrm{O}}_2+5{\mathrm{H}}_2\mathrm{O}+\mathrm{heat}$$
4. 4
   $$2{\mathrm{C}}_6{\mathrm{H}}_{10}+17{\mathrm{O}}_2\to 12C{\mathrm{O}}_2+10{\mathrm{H}}_2\mathrm{O}+\mathrm{heat}$$
5. 5
   $${\mathrm{C}}_7{\mathrm{H}}_8+9{\mathrm{O}}_2\to 7C{\mathrm{O}}_2+4{\mathrm{H}}_2\mathrm{O}+\mathrm{heat}$$
6. 6Pentene and hexyne are represented by their molecular formulas because the position of the multiple bond does not affect the complete-combustion atom balance.

Step-by-step solution

1. 1Number the parent chain CH₃–CH=CH–CH₂–CH₂–CH₃. Rotation about the C=C bond is restricted, so the CH₃ group on C-2 and the propyl group on C-3 can be on the same or opposite sides.
2. 2In cis-hex-2-ene, the CH₃ and propyl groups are on the same side of the double bond.
3. 3In trans-hex-2-ene, the CH₃ and propyl groups are on opposite sides of the double bond.
4. 4The cis form has a larger net dipole moment because the bond dipoles do not cancel as effectively. Stronger dipole–dipole attraction gives it the higher boiling point.

Step-by-step solution

1. 1The three double bonds drawn in one Kekulé structure are not localised in three fixed places. The six π electrons are delocalised over all six sp² carbon atoms.
2. 2The two Kekulé contributors and the other equivalent resonance contributors describe one resonance hybrid. The hybrid is lower in energy than any single contributor.
3. 3All six carbon–carbon bonds are consequently equivalent and have a length intermediate between a normal single and a normal double bond.
4. 4The delocalised six-π-electron system is aromatic and strongly stabilised, so benzene resists the addition reactions expected of an isolated triene.

Step-by-step solution

1. 1The system must be cyclic, so that a continuous ring of p orbitals can form.
2. 2The atoms of the ring must be planar, allowing parallel p orbitals to overlap around the entire cycle.
3. 3Every ring atom must have a p orbital in the continuously conjugated system; an sp³ atom interrupts the cyclic π cloud.
4. 4The delocalised cyclic system must contain 4n+2 π electrons, where n is a non-negative integer. This is Hückel's rule.

Step-by-step solution

1. 1For (i), the total of six π-electrons would satisfy the 4n+2 count, but the sp³ carbon is tetrahedral and has no p orbital. The ring is not fully conjugated and cannot maintain one continuous planar π cloud.
2. 2For (ii), the sp³ carbon breaks planarity and cyclic conjugation, and four π-electrons do not equal 4n+2 for an integer n.
3. 3For (iii), cyclooctatetraene has eight π-electrons. It avoids antiaromaticity by adopting a non-planar tub conformation, so the required planar continuous π system is absent; eight is also a 4n count, not a 4n+2 count.

Step-by-step solution

1. 1For (i), brominate benzene first. Br is an ortho/para director, so nitration of bromobenzene gives ortho and para products; isolate the para isomer.
2. 2
   $${\mathrm{C}}_6{\mathrm{H}}_6\stackrel{\mathrm{B}{\mathrm{r}}_2/FeB{\mathrm{r}}_3}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{Br}\stackrel{conc.\mathrm{HN}{\mathrm{O}}_3/conc.{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }\mathrm{p}-\mathrm{Br}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{N}{\mathrm{O}}_2$$
3. 3For (ii), nitrate benzene first. NO₂ is a meta director, so chlorination of nitrobenzene gives the meta product.
4. 4
   $${\mathrm{C}}_6{\mathrm{H}}_6\stackrel{\mathrm{HN}{\mathrm{O}}_3/{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{O}}_2\stackrel{\mathrm{C}{\mathrm{l}}_2/FeC{\mathrm{l}}_3}{\to }\mathrm{m}-\mathrm{Cl}{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{N}{\mathrm{O}}_2$$
5. 5For (iii), use Friedel–Crafts methylation to obtain toluene. CH₃ is ortho/para directing, so nitration gives the para isomer along with the ortho isomer, which can be separated.
6. 6
   $${\mathrm{C}}_6{\mathrm{H}}_6\stackrel{\mathrm{C}{\mathrm{H}}_{3}Cl/AlC{\mathrm{l}}_3}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{HN}{\mathrm{O}}_3/{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }\mathrm{p}-\mathrm{C}{\mathrm{H}}_3{\mathrm{C}}_6{\mathrm{H}}_4\mathrm{N}{\mathrm{O}}_2$$
7. 7For (iv), carry out Friedel–Crafts acylation with acetyl chloride and anhydrous aluminium chloride.
8. 8
   $${\mathrm{C}}_6{\mathrm{H}}_6+\mathrm{C}{\mathrm{H}}_3\mathrm{COCl}\stackrel{\mathrm{AlC}{\mathrm{l}}_3}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COC}{\mathrm{H}}_3+\mathrm{HCl}$$

Step-by-step solution

1. 1Classify a carbon by the number of other carbon atoms directly attached to it: one gives primary, two secondary, three tertiary and four quaternary.
2. 2The five terminal CH₃ groups are primary carbons. Each is bonded to three H atoms, so the primary set carries 15 H atoms in total.
3. 3The two CH₂ carbons in the main chain are secondary carbons. Each is bonded to two H atoms, giving 4 H in total.
4. 4The CH carbon at the right-hand branch point is tertiary and carries one H. The central C(CH₃)₂ carbon is quaternary and carries no H.

Step-by-step solution

1. 1Alkanes are held together mainly by London dispersion forces, whose strength increases with the area of contact between molecules.
2. 2Branching makes a molecule more compact and generally reduces its effective surface area, so isomers with greater branching have weaker intermolecular contact.
3. 3Less energy is therefore needed to separate branched molecules, and their boiling points are lower than those of the corresponding less-branched isomers.

Step-by-step solution

1. 1Without peroxide, HBr adds by an ionic electrophilic mechanism. Protonation at the terminal carbon produces the more stable secondary carbocation at C-2.
2. 2
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2+{\mathrm{H}}^{+}\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}^{+}\mathrm{C}{\mathrm{H}}_3\stackrel{\mathrm{B}{\mathrm{r}}^{-}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{CHBrC}{\mathrm{H}}_3$$
3. 3This is Markovnikov orientation: hydrogen adds to the carbon with more hydrogens and bromine to the more substituted carbon.
4. 4Benzoyl peroxide initiates a radical chain. It decomposes to benzoyloxy radicals, which ultimately generate bromine radicals.
5. 5
   $$({\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COO}{)}_2\stackrel{\mathrm{\Delta }}{\to }2{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COO}\bullet \to 2{\mathrm{C}}_6{\mathrm{H}}_5\bullet +2C{\mathrm{O}}_2$$
6. 6A bromine radical adds to the terminal carbon so that the radical left on the middle carbon is secondary and more stable.
7. 7
   $$\mathrm{Br}\bullet +\mathrm{C}{\mathrm{H}}_3\mathrm{CH}=\mathrm{C}{\mathrm{H}}_2\to \mathrm{C}{\mathrm{H}}_3\mathrm{CH}\bullet \mathrm{C}{\mathrm{H}}_2\mathrm{Br}$$
8. 8That radical abstracts hydrogen from HBr, forming 1-bromopropane and regenerating Br•. The chain continues.
9. 9
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{CH}\bullet \mathrm{C}{\mathrm{H}}_2\mathrm{Br}+\mathrm{HBr}\to \mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Br}+\mathrm{Br}\bullet$$

Step-by-step solution

1. 1Ozonolysis cleaves the ring double bonds and turns the ring carbons into carbonyl groups. The possible fragments are glyoxal, methylglyoxal and butane-2,3-dione (diacetyl).
2. 2Glyoxal is OHC–CHO; methylglyoxal is CH₃CO–CHO; and butane-2,3-dione is CH₃CO–COCH₃.
3. 3A single fixed Kekulé drawing would suggest a particular pair of cleavages. The collection of products is instead consistent with the equivalent double-bond arrangements of a resonance hybrid.
4. 4Thus ozonolysis supports delocalisation of the π electrons rather than one permanently fixed set of three isolated double bonds.

Step-by-step solution

1. 1The relevant carbon–hydrogen bonds are sp, sp² and sp³ in ethyne, benzene and n-hexane respectively.
2. 2Greater s-character pulls the bonding electron pair closer to carbon and makes the corresponding hydrogen easier to remove as H⁺.
3. 3Therefore the sp C–H bond of ethyne is the most acidic, the sp² C–H bond of benzene is intermediate, and the sp³ C–H bonds of n-hexane are least acidic.

Step-by-step solution

1. 1Benzene has a delocalised π-electron cloud above and below the ring. Its electron-rich π system attracts an electrophile and forms a resonance-stabilised σ-complex.
2. 2Loss of H⁺ from the σ-complex restores aromaticity, so substitution is strongly favoured over addition.
3. 3A nucleophile is electron-rich and is repelled by the electron-rich aromatic ring. Ordinary nucleophilic substitution would also require loss of aromaticity or a poor leaving group.
4. 4Special electron-withdrawing substituents and activation can make nucleophilic substitution possible, but benzene itself reacts much more readily with electrophiles.

Step-by-step solution

1. 1For (i), pass ethyne through a red-hot iron tube at about 873 K. Three ethyne molecules undergo cyclic polymerisation.
2. 2
   $$3HC\equiv \mathrm{CH}\underset{\mathrm{Fe}\mathrm{tube}}{\overset{873\mathrm{K}}{\to }}{\mathrm{C}}_6{\mathrm{H}}_6$$
3. 3For (ii), pass ethene through the same red-hot iron tube; three ethene molecules cyclise to benzene.
4. 4
   $$3C{\mathrm{H}}_2=\mathrm{C}{\mathrm{H}}_2\underset{\mathrm{Fe}\mathrm{tube}}{\overset{873\mathrm{K}}{\to }}{\mathrm{C}}_6{\mathrm{H}}_6$$
5. 5For (iii), aromatise hexane over heated chromium(oxide)/aluminium(oxide), commonly at 773 K and elevated pressure.
6. 6
   $${\mathrm{C}}_6{\mathrm{H}}_{14}\underset{773\mathrm{K},\mathrm{C}{\mathrm{r}}_2{\mathrm{O}}_3/A{\mathrm{l}}_2{\mathrm{O}}_3}{\overset{10\text{–}20\mathrm{atm}}{\to }}{\mathrm{C}}_6{\mathrm{H}}_6+4{\mathrm{H}}_2$$

Step-by-step solution

1. 1The carbon skeleton of 2-methylbutane is CH₃–CH(CH₃)–CH₂–CH₃. Place a C=C bond between any two adjacent carbons of this skeleton without changing the skeleton.
2. 2Putting the double bond between C-1 and C-2 gives CH₂=C(CH₃)–CH₂–CH₃, 2-methylbut-1-ene.
3. 3Putting it between C-2 and C-3 gives CH₃–C(CH₃)=CH–CH₃, 2-methylbut-2-ene.
4. 4Writing the same skeleton from the other end places the double bond between the terminal carbon and the adjacent branched carbon, giving CH₂=CH–CH(CH₃)–CH₃, 3-methylbut-1-ene.

Step-by-step solution

1. 1An electron-donating group increases electron density in the ring and activates it towards electrophilic attack. An electron-withdrawing nitro group decreases electron density and deactivates the ring.
2. 2In (a), chlorobenzene has no ring nitro group, p-nitrochlorobenzene has one deactivating NO₂ group, and 2,4-dinitrochlorobenzene has two strong deactivating groups.
3. 3In (b), CH₃ activates the ring, one NO₂ group deactivates it, and two NO₂ groups deactivate it most strongly.

Step-by-step solution

1. 1Nitration is an electrophilic aromatic substitution and is favoured by groups that increase electron density in the ring.
2. 2The methyl group of toluene donates electron density by hyperconjugation and the +I effect, so it activates the ring.
3. 3Each nitro group withdraws electron density strongly. The two NO₂ groups in m-dinitrobenzene make that ring the least reactive towards an electrophile.

Step-by-step solution

1. 1Friedel–Crafts alkylation needs a Lewis acid to generate the electrophilic alkylating species from an alkyl halide. Anhydrous FeCl₃ is one suitable alternative, and BF₃ is another.
2. 2With ethyl chloride, benzene undergoes Friedel–Crafts alkylation to ethylbenzene.
3. 3
   $${\mathrm{C}}_6{\mathrm{H}}_6+{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{Cl}\stackrel{\mathrm{FeC}{\mathrm{l}}_3}{\to }{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{C}}_2{\mathrm{H}}_5+\mathrm{HCl}$$

Step-by-step solution

1. 1Wurtz coupling joins two alkyl groups from two alkyl-halide molecules. If two different alkyl halides are used, each can also couple with a molecule of its own kind.
2. 2To make an odd-carbon alkane by this method, the two different alkyl groups must have different numbers of carbon atoms, so self-coupling products are unavoidable.
3. 3For example, ethyl bromide and propyl bromide can form the desired pentane, but they also form butane and hexane.
4. 4
   $$\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{Br}+\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{Br}\stackrel{2Na,\mathrm{ether}}{\to }\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_2\mathrm{C}{\mathrm{H}}_3$$
5. 5The desired pentane is therefore contaminated with butane and hexane, making the reaction unsuitable for a clean preparation.