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Important Questions for CBSE Class 11 Chemistry Chapter 9 - Hydrocarbons

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CBSE Class 11 Chemistry Chapter 9 Important Questions - Free PDF Download

Free PDF download of Important Questions for CBSE Class 11 Chemistry Chapter 9 - Hydrocarbons prepared by expert Chemistry teachers from latest edition of CBSE(NCERT) books. Register online for Chemistry tuition on Vedantu.com to score more marks in your examination. To learn the other chapters of this subject, refer to CBSE Class 11 Chemistry Important Questions.

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Study Important Questions for Class 11 Chemistry Chapter 9 – Hydrocarbons

1 Mark Questions

1. Classify the hydrocarbons according to the carbon – carbon bond

Ans: According to the carbon–carbon bond that occurs between them, hydrocarbons are divided into three categories: 

(a) saturated hydrocarbon 

(b) unsaturated hydrocarbon 

(c) aromatic hydrocarbon.


2. What are cycloalkanes?

Ans: Cycloalkanes are formed when carbon atoms form a closed chain or ring.


3. Why does carbon have a larger tendency of catenation than silicon although they have the same number of electrons?

Ans: It's because the C-C bond is smaller but stronger ($\text{335 KJ mo}{{\text{l}}^{\text{-1}}}$) than the Si bond ($\text{225}\text{.7 KJ mo}{{\text{l}}^{\text{-1}}}$).


4. Write IUPAC name of the following


5-(2-Methylpropyl)-decane


Ans:  The IUPAC name of the compound is 5-(2-Methylpropyl)-decane.


5. What is hydrogenation?

Ans: To produce alkanes, dihydrogen gas is added to alkenes and alkenes in the presence of finely split catalysts such as Pt, Pd, or Ni. This is referred to as hydrogenation.


6. How would you convert ethene to ethane molecules?

Ans:  The reaction is given below:

\[\text{C}{{\text{H}}_{\text{2}}}\text{=C}{{\text{H}}_{\text{2}}}\text{ + }{{\text{H}}_{\text{2}}}\xrightarrow{\text{Pd/Pt/Ni}}\text{ C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{3}}}\] 


7. Give the IUPAC name of the lowest molecular weight alkane that contains a quaternary carbon.

Ans: 2, 2-dimethylpropane.


8. Methane does not react with chlorine in the dark. Why?

Ans: Chlorination of methane is a substitution process involving free radicals. Chlorine cannot be transformed into free radicals in the dark, therefore the reaction does not take place.


9. Which conformation of ethane is more stable?

Ans: Staggered conformation.


10. State Le chatelier’s principle.

Ans: It asserts that a change in any of the elements that define a system's equilibrium conditions will cause the system to alter in such a way that the effect of the change is reduced or counteracted.


11. Can a catalyst change the position of equilibrium in a reaction?

Ans: No, a catalyst cannot alter the equilibrium position of a chemical process. A catalyst, on the other hand, has an effect on the reaction rate.


12. What is the effect of reducing the volume on the system described below?

\[\text{2C(s) + }{{\text{O}}_{\text{2}}}\text{(g) }\rightleftharpoons \text{2 CO(g)}\] 

Ans: A rise in volume occurs in tandem with the forward response. As a result, according to Chatelier's principle, lowering the volume will cause the equilibrium to shift forward.


13. What happens when temperature increases for a reaction?

Ans: As the temperature rises, the equilibrium constant for an exothermic process ($\text{ }\!\!\Delta\!\!\text{ H = -ve}$) drops.


14. Can a catalyst change the position of equilibrium in a reaction?

Ans: No, a catalyst cannot alter the equilibrium position of a chemical process. A catalyst, on the other hand, has an effect on the reaction rate.


15. If ${{\text{Q}}_{\text{C}}}$ <  ${{\text{K}}_{\text{C}}}$, when we continuously remove the product, what would be the direction of the reaction?

Ans: When a product is continuously removed, ${{\text{Q}}_{\text{C}}}$ remains smaller than ${{\text{K}}_{\text{C}}}$, and the reaction continues to go ahead.


16. What is a Lindlars’ catalyst?

Ans: The Lindlar catalyst is characterised as partly deactivated palletized charcoal.


17. How is alkene produced by vicinal dihalide?

Ans: Vicinal dihalide loses a $\text{Zn}{{\text{X}}_{\text{2}}}$ molecule from the alkene upon treatment with Zn metal. This is called a dehalogenation reaction. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{2}}}\text{Br-C}{{\text{H}}_{\text{2}}}\text{Br + Zn }\to \text{ C}{{\text{H}}_{\text{2}}}\text{=C}{{\text{H}}_{\text{2}}}\text{ + ZnB}{{\text{r}}_{\text{2}}}\] 


18. Arrange the following halogen atom to determine rate of the reaction. Iodine, Chlorine, Bromine.

Ans: The order will be Iodine > Bromine > Chlorine


19. What is $\text{ }\!\!\beta\!\!\text{ -elimination}$ reaction?

Ans: When the atom of hydrogen is removed from $\text{ }\!\!\beta\!\!\text{ -carbon}$ (carbon atom next to the carbon to which halogen is attached).


20. What is the number of $\text{ }\!\!\sigma\!\!\text{ }$ and $\text{ }\!\!\pi\!\!\text{ }$ bond in $\text{N}\equiv \text{C-CH-C}\equiv \text{N}$?

Ans:  The numbers of $\text{ }\!\!\sigma\!\!\text{ }$ bonds are 7 and the numbers of $\text{ }\!\!\pi\!\!\text{ }$ bonds are 5.


21. Name the type of hybridization in C (2) and C (3) in the following molecule:


Hybridization of C(2) is sp and the Hybridization of C(3) Molecule


Ans: The hybridization of C(2) is sp and the hybridization of C(3) is $\text{s}{{\text{p}}^{\text{2}}}$.


22. Why do alkynes not show geometrical isomerism?

Ans: Alkynes' structure is linear. They cannot therefore demonstrate geometric isomerism.


23. Write the general formula of alkynes.

Ans: The general formula of alkynes is ${{\text{C}}_{\text{n}}}{{\text{H}}_{\text{2n-2}}}$.


24. Name the simplest alkyne.

Ans: Ethyne (${{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}$)


25. Write combustion reaction for hexyne.

Ans: Combustion reaction for hexyne is given below:

\[\text{HC}\equiv \text{C-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{3}}}\text{ + }\frac{\text{17}}{\text{2}}{{\text{O}}_{\text{2}}}\text{ }\to \text{ 6C}{{\text{O}}_{\text{2}}}\text{ + 5}{{\text{H}}_{\text{2}}}\text{O}\] 

 

26. How will you convert ethyne to benzene?

Ans: The reaction is given below:


Ethyne to Benzene


27. What are benzenoids?

Ans: Benzene ring is recognised as an aromatic hydrocarbon compound.


28. Although benzene is highly unsaturated; it does not undergo addition reactions. Give a reason.

Ans: Contrary to olefins, benzene $\text{ }\!\!\pi\!\!\text{ -electrons}$are relocated (resonance), and hence they are not reactive in terms of further reactions.


29. How will you convert the following compounds into benzene?

(i) Ethene

Ans: The reaction is given below:


Ethene into Benzene


(ii) Hexane

Ans: The reaction is given below:


Hexane into Benzene


2 Marks Questions

30. The boiling point of hydrocarbons decreases with increase in branching. Give a reason.

Ans: The outcome is a more compact structure (almost spherical). This decreases the efficient surface area and therefore the strength of the forces of the Vander wall, which reduces the boiling point.


31. Unsaturated compounds undergo additional reactions. Why?

Ans: Unsaturated hydrocarbon molecules include two- or three-fold bonds of carbon. The $\text{ }\!\!\pi\!\!\text{ -bond}$is a multiple bond, which becomes unstable and hence adds across numerous bonds.


32. To which category of compounds does cyclohexane belong?

Ans: Saturated alicyclic hydrocarbons.


33. Draw the structure of the following compounds all showing C and H atoms.

(a) 2-methyl -3-isopropyl heptanes

Ans: The compound is drawn below:


2-methyl -3-isopropyl heptanes


(b) Dicyclopropyl methane

Ans: The compound is drawn below:


Dicyclopropyl methane


34. Draw all the possible structural isomers with the molecular formula ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{14}}}$, Name them.

Ans: These are given below:

n- Hexane: $\text{C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{3}}}$ 

2- Methyl pentane:


Structural Isomers


3-Methyl pentane:


3-Methyl pentane:


2, 2-Dimethylbutane:


2, 2-Dimethylbutane:


2, 3-Dimethyl butane:


2, 3-Dimethyl butane:


35. Sodium salt of which acid will be needed for the preparation of propane? Write a chemical equation for the reaction.

Ans: Butanoic acid. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{2}}}\text{CO}{{\text{O}}^{\text{-}}}\text{N}{{\text{a}}^{\text{+}}}\text{ + NaOH }\to \text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}{{\text{H}}_{\text{3}}}\text{ + N}{{\text{a}}_{\text{2}}}\text{C}{{\text{O}}_{\text{3}}}\]


36. Cyclobutane is less reactive than cyclopropane. Justify.

Ans: The C-C-C bond angle in the cyclobutane molecule is $\text{9}{{\text{0}}^{\text{o}}}$ and in cyclopropane $\text{6}{{\text{0}}^{\text{o}}}$. It demonstrates that in cyclobutane the difference between the bond angle of the tetrahedral ($\text{10}{{\text{9}}^{\text{o}}}\text{28 }\!\!'\!\!\text{ }$) and cyclopropane is smaller. In other words, compared with cyclobutane, cyclopropane is under severe strain and is hence more reactive.


37. How will you prepare isobutane?

Ans: Isobutane comes from 3-methyl butanoic acid decarboxylation with 630K of soda lime. The reaction is given below:


Isobutane


38. The boiling point of alkanes shows a steady increase with increase in molecular mass. Why?

Ans: This is because the intermolecular van der waals increase forces when the molecular size or the surface of the molecule increases.


39. Pentane has three isomers i.e; pentane, 2-methyl butane and 2,2-dimethylpropane . The b.p of pentane is 309.1K whereas 2,2-dimethylpropane shows a b.p of 282.5K. Why?

Ans: The molecule becomes spherical by increasing the number of branch chains. This leads to less contact and hence a weak molecular inter-forest, which is addressed by comparatively lower temperature between spherical molecules.


40.  Draw the New man’s projection formula of the staggered form of 1,2-dichloroethane.

Ans: The figure is given below:


New man’s projection formula


41.  All the four C-H bonds in methane are identical. Give reasons.

Ans: The four C-H methane bonds are similar as they originate by overlapping the same sort of orbital: carbon and hydrogen hybrid orbital.


42.  When alkanes are heated, the C-C bonds rather than the C-H bonds break. Give a reason.

Ans: The C-C connection instead of the C-H bond breaks when the alkanes are warmed up because the bond C-C has a lower energy bond ($\text{ }\!\!\Delta\!\!\text{ H = 83 Kcal/ mole}$) than the bond C-H ($\text{ }\!\!\Delta\!\!\text{ H = 99 Kcal/ mole}$).


43. How would you convert cyclohexane to benzene?

Ans: In a red hot tube, cyclohexane is oxidised in order to produce benzene when handled with iron or quartz. The reaction is given below:


Cyclohexane to benzene


44.  How is iso-butane prepared?

Ans: By decarboxylation of 3 – methyl butanoic acid with soda lime at 630 K. The reaction is given below:


Iso-butane prepared


45. Why does the addition of inert gas does does not change the equilibrium?

Ans: The inert gas does not affect the partial pressure or molar concentration of the material involved in the reaction by adding it in a continuous volume.


46.  The equilibrium constant of a reaction increases with rise in temperature. Is the reaction exo – or endothermic?

Ans: As the temperature rises, the equilibrium constant rises with it. As a result, the reaction is an endothermic one.


47.  Using Le – chatelier principle, predict the effect of

(a) decreasing the temperature

(b) increasing the temperature

in each of the following equilibrium systems:

i. ${{\text{N}}_{\text{2}}}\text{(g) + 3}{{\text{H}}_{\text{2}}}\text{(g) }\rightleftharpoons \text{2N}{{\text{H}}_{\text{3}}}\text{(g) +  }\!\!\Delta\!\!\text{ }$

Ans: For an exothermic reaction increase in temperature shifts the equilibrium to the left and decrease in temperature shifts it to the left. 

ii. ${{\text{N}}_{\text{2}}}\text{(g) + }{{\text{O}}_{\text{2}}}\text{(g) +  }\!\!\Delta\!\!\text{  }\rightleftharpoons \text{2NO(g)}$ 

Ans:  For an endothermic reaction increase in temperature shifts the equilibrium to the right and decrease in temperature shifts it to the right.


48. In the reaction equilibrium

\[\text{A+B}\rightleftharpoons \text{C+D}\]

What will happen to the concentrations of A, B and D if concentration of C is increased. 

Ans: For the given reaction: \[\text{A+B}\rightleftharpoons \text{C+D}\] 

\[{{\text{K}}_{\text{C}}}\text{=}\frac{\left[ \text{C} \right]\left[ \text{D} \right]}{\left[ \text{A} \right]\left[ \text{B} \right]}\]

When the concentration of a product is raised, the concentration of other components varies, resulting in a reduction in the concentration of C, and vice versa.

If the conc of C is increased, the conc of D will drop, but the concs of A and B will concurrently rise, resulting in the same numerical value of ${{\text{K}}_{\text{C}}}$, and vice versa. The balance of power shifts to the left.

(i) What will happen if concentration of A is increased?

Ans: When the conc. of A is increased, the conc of B decreases, but the conc. of C and D rise at the same time, resulting in the numerical value of Kc remaining the same, and vice versa. The balance shifts rightward. 


49. How is alkene produced by Kolbe’s electrolytic method?

Ans:  The reaction is given below:


Reaction of alkene produced by Kolbe’s electrolytic method


Reaction of alkene produced by Kolbe’s electrolytic method


50.  How is alkene prepared from alcohol by acidic dehydration?

Ans: When alcohols are heated with concentrated sulfuric acid, one water molecule is removed, resulting in alkenes. The reaction is given below:


Reaction of alkene preparing from alcohol by acidic dehydration


51.  How are trans alkenes formed by alkynes?

Ans: Trans alkenes are formed when alkynes are reduced with sodium in liquid ammonia. The reaction is given below:


Formation of Trans alkenes


52. How are cis – alkenes formed by alkynes?

Ans:  Cis-alkene is formed when alkenes are partially reduced with a specified amount of dihydrogen in the presence of palladised charcoal and then partially deactivated with poisons such as sulphur compounds or quinoline. The reaction is given below:


Formation of Cis – alkenes


53.  State Markownikov’s Rule.

Ans: It states that when a polar compound is added to an unsymmetrical alkenes, or alkynes positive part goes to the most substituted carbon atom and negative part goes to the least substituted carbon atom.


54.  Write the chemical equations of reactions involved in ozonolysis of alkenes.

Ans: Alkenes react with ozone to generate ozonide, which is then reduced in the presence of Zn to produce aldehyde and ketones. The reaction of an example is given below:


Chemical equations of reactions involved in ozonolysis of alkenes


55. How will you distinguish between butene – 1 and butene – 2?

Ans: Butenes 1 and 2 can be differentiated by ozonolysis or oxidation with an acidic KMnO4 solution, resulting in distinct carbonyl products. The reactions are given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-CH=C}{{\text{H}}_{\text{2}}}\text{ }\xrightarrow{{{\text{O}}_{\text{3}}}}\text{ C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{CHO + HCHO}\] 

\[\text{C}{{\text{H}}_{\text{3}}}\text{-CH=CH-C}{{\text{H}}_{\text{3}}}\text{ }\xrightarrow{{{\text{O}}_{\text{3}}}}\text{ C}{{\text{H}}_{\text{3}}}\text{CHO + C}{{\text{H}}_{\text{3}}}\text{CHO}\] 


56.  State kharasch effect.

Ans: It claims that when peroxides such as benzoyl peroxide are present, the addition of HBr (but not HCl or HI) to unsymmetrical alkenes violates Markontkov's rule. The reaction is given below:

\[\text{C}{{\text{H}}_{\text{3}}}\text{CH=C}{{\text{H}}_{\text{2}}}\text{ + HBr }\xrightarrow{\text{Peroxide}}\text{ C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{3}}}\] 


57.  How is alkyne prepared from calcium carbide?

Ans: Ethyne is produced by treating calcium carbide with water. The reaction is given below:

\[\text{Ca}{{\text{C}}_{\text{2}}}\text{ + 2}{{\text{H}}_{\text{2}}}\text{O }\to \text{ Ca(OH}{{\text{)}}_{\text{2}}}\text{ + }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\] 


58. How is alkyne prepared by Kolbe’s method?

Ans: The reaction is given below:


Reaction of Alkyne prepared by Kolbe’s method


Reaction of Alkyne prepared by Kolbe’s method


59. How is alkyne prepared from vicinal dihalides?

Ans: Vicinal dihalides are dehydrohalogenation when exposed to alcoholic potassium hydroxide. One hydrogen halide molecule is removed to generate alkenyl halide, which is then treated with sodium amide to produce alkynes. The reaction is given below:


Alkyne prepared from vicinal dihalides


60. How will you distinguish between ethylene and methane?

Ans: Methane does not produce bromine water colour or Baeyer's reagent colour, but ethylene does.


61. Although acetylene is acidic in nature, it does not react with NaOH or KOH. Give a reason?

Ans: Because acetylene is a weak acid ($\text{p}{{\text{K}}_{\text{a}}}\text{=25}$), only a very strong base, such as the amide ion ($\text{NH}_{\text{2}}^{\text{-}}$), can efficiently remove a proton.


62. Write the conversion of ethene to ethyne.

Ans: The reaction is given below:

\[\text{C}{{\text{H}}_{\text{2}}}\text{=C}{{\text{H}}_{\text{2}}}\xrightarrow{\text{B}{{\text{r}}_{\text{2}}}}\text{ Br-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-Br }\xrightarrow{\text{KOH(alc}\text{.)}}\text{CH}\equiv \text{CH}\] 


63.  How will you distinguish between butyne – 1 and butyne – 2?

Ans: With an acetylene hydrogen atom, butyne – 1 ($\text{C}{{\text{H}}_{\text{3}}}\text{C}{{\text{H}}_{\text{2}}}\text{C}\equiv \text{CH}$) produces white precipitate with ammoniacal silver nitrate and red precipitate with ammoniacal cuprous chloride. However, butyne – 2 ($\text{C}{{\text{H}}_{\text{3}}}\text{-C}\equiv \text{C-C}{{\text{H}}_{\text{3}}}$), which lacks an acetylene hydrogen atom, is unresponsive to either reagent.


64.  How would you carry out the following conversion propene to ethyne.

Ans: The reaction is given below:


Conversion propene to ethyne


65.  How will you convert propyne to propanone?

Ans: The reaction is given below:


Conversion of propyne to propanone


66.  How will you convert ethyne to ethane?

Ans: The reaction is given below:

\[\text{HC}\equiv \text{CH + }{{\text{H}}_{\text{2}}}\xrightarrow{\text{Pt/Pd/Ni}}\text{  }\!\![\!\!\text{ }{{\text{H}}_{\text{2}}}\text{C=C}{{\text{H}}_{\text{2}}}\text{ }\!\!]\!\!\text{  }\xrightarrow{{{\text{H}}_{\text{2}}}}\text{ C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{3}}}\] 


67. Convert 2-Butyne to Trans-2-butene.

Ans: The reaction is given below:


Conversion of 2-Butyne to Trans-2-butene

 

68.  How will you prepare 3-methyl but -1 – yne by starting with ethyne?

Ans: The reaction is given below:

\[\text{HC}\equiv \text{CH}\xrightarrow{\text{Na,N}{{\text{H}}_{\text{3}}}}\text{HC}\equiv {{\text{C}}^{\text{-}}}\text{N}{{\text{a}}^{\text{+}}}\] 


preparing 3-methyl but -1 – yne by starting with ethyne


69. Write the IUPAC name of the following compound

(i).


compound is 4-Phenyl but-1-ene


Ans: The IUPAC name of the compound is 4-Phenyl but-1-ene.

(ii).


Compound is 2-Methyl phenol


Ans: The IUPAC name of the compound is 2-Methyl phenol.


70.  What do you mean by delocalization?

Ans: Delocalisation – Pairs of bonding electrons that extend over three or more atoms and belong to the entire molecule are said to be delocalized. Delocalized $\text{ }\!\!\pi\!\!\text{ -orbitals}$ are much larger and therefore more stable than localised $\text{ }\!\!\pi\!\!\text{ -orbitals}$.


71.  What do you understand about Resonance energy?

Ans: Resonance energy is the difference between the energy of the most stable contributing structure and the energy of the resonance hybrid. Benzene has resonance energy of 147 KJ/mole.


72.  How is phenol reduced to benzene?

Ans: The reaction is given below:


Reaction of Phenol reduced to Benzene


73.  How is the aromaticity of a compound judged?

Ans: The following characteristics decides aromaticity of a compound:-

(i). Planarity

(ii). Complete delocalization of the $\text{ }\!\!\pi\!\!\text{ -electrons}$ in the ring.

(iii). Presence of (4n+2) $\text{ }\!\!\pi\!\!\text{ -electrons}$in the ring where n is an integer (n=0, 1, 2 ----)

This is often referred to as Huckel Rule.


74.  Give some examples of aromatic compounds.

Ans: Some examples of aromatic compounds are benzene, cyclopentadienyl anion, cycloheptatrienyl cation, etc. Their structures are given below:


Benzene


Cyclopentadienyl Anion


Cycloheptatrienyl Cation


75. How will you account for the structure of benzene?

Ans: In benzene, all six carbon atoms are $\text{s}{{\text{p}}^{\text{2}}}$ hybridized. Each carbon atom's two $\text{s}{{\text{p}}^{\text{2}}}$ hybrid orbitals overlap with the $\text{s}{{\text{p}}^{\text{2}}}$ hybrid orbitals of neighboring carbon atoms to create six C-C sigma bonds in the hexagonal plane. Each carbon atom's remaining $\text{s}{{\text{p}}^{\text{2}}}$ hybrid orbital overlaps with a hydrogen atom's s-orbital to produce six C-H sigma bonds. One hybridized p-orbital perpendicular to the ring plane remains for each carbon atom.


Structure of Benzene


The unhybridized p-orbital of the carbon atoms will form the pi-bond.


76.  How is benzene prepared from aromatic acids?

Ans: When the sodium salt of benzoic acid is heated with soda lime, benzene is produced. The reaction is given below:


Benzene prepared from aromatic acids


77. How is phenol reduced to benzene?

Ans: The reaction is given below:


Phenol reduced to benzene


78.  Why is benzene extraordinarily stable though it contains three double bonds?

Ans: Due to resonance there are three double bonds which makes them extra stable.


79.  What is Friedel craft’s reaction? Give an example.

Ans: Alkyl benzene is formed when benzene or a derivative of it interacts with an alkyl halide in the presence of $\text{AlC}{{\text{l}}_{\text{3}}}$. The reaction is given below:


Friedel craft’s reaction


80.  What happens when benzene is oxidized at 770K in presence of ${{\text{V}}_{\text{2}}}{{\text{O}}_{\text{5}}}$? Give a chemical equation.

Ans: The reaction is given below:


Reaction of Benzene is oxidized at 770K


81.  How will you convert benzene to iodobenzene? Give a chemical equation.

Ans: The reaction is given below:


Benzene to Iodobenzene


82.  What are electrophilic substitution reactions?

Ans: Electrophilic substitution reactions are those in which a weaker electrophile is replaced by a stronger electrophile.


83. How will you distinguish between Ethene and benzene

Ans: While benzene does not produce bromine water color or Baeyer's reagent colour, ethene does.


84.  How is benzene converted to benzene hexachloride?

Ans: Three chlorine molecules react with benzene to form benzene hexachloride, ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\text{C}{{\text{l}}_{\text{6}}}$, commonly known as gammaxane, when exposed to ultraviolet light. The reaction is given below:


Benzene converted to Benzene Hexachloride


85.  How will you convert benzene to hexachlorobenzene?

Ans: Hexachlorobenzene (${{\text{C}}_{\text{6}}}\text{C}{{\text{l}}_{\text{6}}}$) is produced by treating benzene with chlorine in the presence of anhydrous $\text{AlC}{{\text{l}}_{\text{3}}}$ in the dark.


3 Marks Questions

86. N – pentane has higher boiling point than neopentane but the melting point of neopentane is higher than that of n – pentane.

Ans: The surface area and van der waal forces of attraction in neopentane are much less than in n-pentane due to the existence of branches. As a result, neopentane has a lower boiling point than n-pentane.

The packing of molecules in the crystal lattice determines M.P. Because neopentane is more symmetrical than n-pentane, it packs significantly closer in the crystal lattice than n-pentane, giving it a substantially higher m.p. than n-pentane.


87. The dipole moment of trans 1,2-dichloroethane is less than the cis – isomer. Explain.

Ans: In comparison to the cis – isomer, the trans isomer has a more symmetrical structure. The dipole moments of the polar C-Cl bonds are expected to cancel them out in the trans – isomer, resulting in a molecule with a dipole moment of practically zero. These, however, do not cancel in the cis –isomer. As a result, the cis isomer has a particular moment, whereas the trans isomer has none. The structures are given below:


Cis- 1,2- Dichloroethane


Trans- 1,2- Dichloroethane


88. Explain wurtz reaction with an example.

Ans: Wurtz reaction – This reaction is used to create higher alkanes from lower alkane halides. Lower alkane halides are treated in ether with sodium metal. The general reaction and an example is given below:

\[\text{RX + 2 Na + XR }\!\!'\!\!\text{  }\xrightarrow{\text{ether}}\text{ R-R }\!\!'\!\!\text{  + 2 NaX}\] 

\[\text{C}{{\text{H}}_{\text{3}}}\text{I + 2 Na + C}{{\text{H}}_{\text{3}}}\text{I }\xrightarrow{\text{ether}}\text{ C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{\text{3}}}\text{ + 2 NaI}\] 


89. Discuss the hybridization of carbon atoms in alkene ${{\text{C}}_{\text{3}}}{{\text{H}}_{\text{4}}}$ and show the $\text{ }\!\!\pi\!\!\text{ -orbital}$ overlaps.

Ans: The structure is given below:


Hybridization of carbon atoms in alkene


The carbon atoms 1 and 3 are $\text{s}{{\text{p}}^{\text{2}}}$ hybridized since each one of them is joined by a double bond. In contrast, carbon atom 2 is sp hybridized since it has two double bonds thus the two double bonds in alkenes are perpendicular to each other.


90. Write IUPAC name of the products obtained by addition reactions of HBr to hex – 1 – ene.

(i). In the absence of peroxide

Ans: The product formed will be 2-Bromohexane. The reaction is given below:


HBr to hex – 1 – ene in the absence of peroxide


(ii). In the presence of peroxide

Ans: The product will be 1-Bromohexane. The reaction is given below:


HBr to hex – 1 – ene in the presence of peroxide


91. Explain the term polymerization with two examples.

Ans: Polymerization is the process of combining two or more molecules of unsaturated compounds to produce a larger complex given the right conditions. The resulting product is known as a polymer, and the process is known as polymerization.

(a) Additional polymerization: Nothing is wasted during the process since the larger molecule (polymer) is an exact multiple of the smaller molecule.

(b) Condensation polymerization: Molecules such as water, hydrochloric acid, and others are typically lost during this process. The polymer is not an exact multiple of the smaller molecule during polymerization.


92. Draw the orbital picture of ethyne showing

(a) Sigma overlaps

Ans: This is shown below:


Ethyne showing Sigma overlaps


(b) Pi overlaps

Ans: This is shown below:


Ethyne showing Pi overlaps


93. Give the different isomers formed by ${{\text{C}}_{\text{5}}}{{\text{H}}_{\text{8}}}$  along with their IUPAC name.

Ans: The structures are given below:

Pent-1-yne: $\text{HC}\equiv \text{C-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{3}}}$

Pent-2-yne: $\text{C}{{\text{H}}_{\text{3}}}\text{-C}\equiv \text{C-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{3}}}$ 

3-Methylbut-1-yne: 


3-Methylbut-1-yne


94. Write structures of different isomers formed by ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{10}}}$. Also write IUPAC names of all the isomers. 

Ans: All the structures and names are given below:

Hex-1-yne: $\text{HC}\equiv \text{C-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{3}}}$ 

Hex-2-yne: $\text{C}{{\text{H}}_{\text{3}}}\text{-C}\equiv \text{C-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{3}}}$ 

Hex-3-yne: $\text{C}{{\text{H}}_{\text{3}}}\text{-C}{{\text{H}}_{2}}\text{-C}\equiv \text{C-C}{{\text{H}}_{\text{2}}}\text{-C}{{\text{H}}_{\text{3}}}$ 

3-Methylpent-1-yne:


3-Methylpent-1-yne


4-Methylpent-1-yne:


4-Methylpent-1-yne


4-Methylpent-2-yne:


4-Methylpent-2-yne


3,3-Dimethylbut-1-yne:


3,3-Dimethylbut-1-yne


95.  Ethyne is acidic in nature in comparison to ethene and ethane. Why is it so?

Ans: In ethyne, hydrogen atoms are connected to sp hybridized carbon atoms, but in ethene, they are attached to $\text{s}{{\text{p}}^{\text{2}}}$  hybridized carbon atoms and in ethane, they are attached to $\text{s}{{\text{p}}^{3}}$ hybridized carbons.

The sp hybridized orbitals of carbon atoms in ethyne molecules have the highest electronegativity (50%) due to the highest percentage of s – character (50%): This attracts the shared pair of the C-H bond of ethyne to a greater extent than the $\text{s}{{\text{p}}^{\text{2}}}$ hybridized orbitals of carbon in ethene and the $\text{s}{{\text{p}}^{3}}$  hybridized orbital of carbon in ethane. In comparison to ethene and ethane, hydrogen atoms may be released as protons more easily in the ethyne molecule.


96.  Butanone is formed when an alkyne is passed through a dil. sol. of sulphuric acid at 330K in presence of mercuric sulphate. Write the possible structure of the alkyne.

Ans: Because butanone has four carbon atoms, both but – 1- yne and but – 2 – yne will form butanone when hydrated. The reactions are given below:


Butanone has four carbon atoms


Butanone has four carbon atoms


97.  How would you convert ethanoic acid into benzene?

Ans: The reaction is given below:

(Image will be uploaded soon)


98.  Name some carcinogenic hydrocarbons.

Ans: Some of the examples are 1, 2-Benzanthracene, 1, 2-Benzpyrene, 9, 10-Dimethyl-1,2-benzanthracene. There structures are given below:


1,2 Benzanthracene



1,2 Benzpyrene



9,10- Dimethyl- 1,2- Benzanthracene



4 Marks Question

99. How would you prepare benzene from lime?

Ans: There are many steps involved in the formation of benzene from lime. The reactions are given below:

\[\text{CaC}{{\text{O}}_{\text{3}}}\xrightarrow{\text{ }\!\!\Delta\!\!\text{ }}\text{CaO+C}{{\text{O}}_{\text{2}}}\] 

\[\text{CaO + C }\xrightarrow{\text{electric discharge}}\text{ Ca}{{\text{C}}_{\text{2}}}\] 

\[\text{Ca}{{\text{C}}_{\text{2}}}\text{ + 2}{{\text{H}}_{\text{2}}}\text{O }\xrightarrow{\text{Furnace}}\text{ }{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\text{ + Ca(OH}{{\text{)}}_{\text{2}}}\] 

\[\text{3}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\xrightarrow{\text{red hot Cu tube}}\text{ }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\] 


100. p-chloro nitro benzene has less dipole moment (2.4 D) than p-nitrotoluene (4.4 D). Why?

Ans: Individual moments in p-chloronitrobenzene are in opposing directions and so partially cancel. Both moments in p-nitro toluene are in the same direction and so contribute to each other. These are given below:


p-chloro nitro benzene has less dipole moment (2.4 D) than p-nitrotoluene (4.4 D)



p-chloro nitro benzene has less dipole moment (2.4 D) than p-nitrotoluene (4.4 D)


8 Marks Questions

101. How will you convert the following compounds to benzene?

(i). Acetylene 

Ans: Ethyne polymerizes into benzene when heated to a higher temperature. The reaction is given below:

\[\text{3}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\xrightarrow{\text{red hot Cu tube}}\text{ }{{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}\]

(ii). Benzoic acid

Ans: When benzoic acid is heated and treated with $\text{N}{{\text{H}}_{\text{3}}}$, it transforms to amide, which then converts to aniline, which then converts to diazonium salt, which then turns to benzene when acid is hydrolyzed. The reaction is given below:


Conversion of Benzoic acid into Benzene


(iii). Cyclohexane 

Ans: When cyclohexane is exposed to iron or quartz in a red hot tube, it oxidises to produce benzene. The reaction is given below:


Conversion of Cyclohexane into Benzene


(iv). Benzene diazonium chloride.

Ans: Benzene diazonium chloride is transformed to benzene in the presence of hypophosphorous acid. The reaction is given below:


Conversion of Benzene diazonium chloride into Benzene


102. How will you convert benzene into

(i). p – Nitro bromo benzene

Ans: Benzene will be treated with bromine to form bromobenzene. This bromobenzene will be treated with concentrated sulfuric acid and nitric acid to form p-Nitrobromobenzene. The reaction is given below:


Conversion of benzene into p – Nitro bromo benzene


(ii). m – Nitrochloro benzene

Ans: Benzene will be treated with concentrated sulfuric acid and nitric acid to form nitrobenzene. This Nitrobenzene will be treated with chlorine to form m-Nitrochloro benzene. The reaction is given below:


Conversion of benzene into m – Nitrochloro benzene


(iii). p – Nitrotoluene

Ans: Benzene will be treated with chloromethane to form Toluene. This toluene will be treated with concentrated sulfuric acid and nitric acid to form p-Nitrotoluene. The reaction is given below:  



Conversion of benzene into p – Nitrotoluene


(iv). Acetophenone

Ans: When benzene will be treated with acetyl chloride, it will form Acetophenone. The reaction is given below:


Conversion of benzene into Acetophenone



Related Study Materials for Class 11 Chemistry Chapter 9 Hydrocarbons


CBSE Class 11 Chemistry Chapter-wise Important Questions

CBSE Class 11 Chemistry Chapter-wise Important Questions and Answers cover topics from all 9 chapters, helping students prepare thoroughly by focusing on key topics for easier revision.


Additional Study Materials for Class 11 Chemistry

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FAQs on Important Questions for CBSE Class 11 Chemistry Chapter 9 - Hydrocarbons

1. What are the most important topics to focus on for Hydrocarbons Class 11 important questions as per CBSE 2025–26?

The key areas for Hydrocarbons Class 11 important questions include:

  • Classification of hydrocarbons (alkanes, alkenes, alkynes, aromatic)
  • IUPAC nomenclature and isomerism in hydrocarbons
  • Preparation and properties of alkanes, alkenes, alkynes
  • Reactions: Substitution, addition, elimination, and unique aromatic reactions
  • Concepts of resonance and aromaticity (Huckel’s Rule, benzene structure, delocalization)
  • Mechanisms of key name reactions (Markovnikov, anti-Markovnikov, Wurtz, Friedel–Crafts)
  • Practical applications and difference between aliphatic and aromatic hydrocarbons
Mastering these ensures you are exam-ready for CBSE 2025–26 patterns.

2. Which types of questions are most frequently asked from Hydrocarbons in CBSE Class 11 exams?

  • IUPAC naming and structure drawing (1–2 mark questions)
  • Mechanisms (stepwise process, especially addition and substitution reactions)
  • Conversions and interconversions (e.g., converting one hydrocarbon type to another)
  • Distinguishing tests for different hydrocarbons (like Baeyer's reagent, bromine water)
  • Explaining resonance, aromaticity, and exceptional stability
  • Reasoning-based HOTS (Higher Order Thinking Skills) regarding physical properties and reactivity trends
More weightage is given to conceptual and multi-step conversion questions in the latest CBSE exams.

3. How can understanding isomerism help in solving important questions in Hydrocarbons for Class 11?

Isomerism is crucial as many important questions require:

  • Identifying structural isomers for a given formula
  • Explaining differences in boiling/melting points due to chain branching
  • Distinguishing between positional, chain, and functional isomers
Clarity in drawing and naming isomers is often tested in both 2-mark and 3–4 mark long questions.

4. What are common exam pitfalls students face in Class 11 Hydrocarbons important questions?

  • Incorrect IUPAC nomenclature due to skipping parent chain rules
  • Confusing reaction mechanisms (for example, mixing up Markovnikov and anti-Markovnikov additions)
  • Forgetting special tests to distinguish similar compounds (methane vs ethene; alkene vs alkyne)
  • Ignoring resonance effects in aromatic stability
  • Neglecting to write balanced chemical equations for conversions
Practicing varied question types with reasoning cuts these errors.

5. How are high-order/application-based questions framed for Hydrocarbons in CBSE Class 11 exams?

These questions often ask students to:

  • Predict products or explain observations based on molecular structure or conditions
  • Distinguish mechanisms given similar reactants but different catalysts
  • Relate structure and hybridization to reactivity or acidity order (e.g., ethyne vs ethene)
  • Analyze aromaticity using Huckel's rule or predict resonance structures
Focusing on underlying logic, not just memorization, is essential for top scores.

6. What are the marking trends for Hydrocarbons Chapter 9 in CBSE Class 11 Chemistry (2025–26)?

  • Short answer (1–2 marks): Nomenclature, definitions, simple tests
  • Medium (3–4 marks): Mechanisms, multi-step conversions, isomer listing
  • Long/HOTS (5+ marks): Comparative reasoning, aromaticity, conversion cycles, resonance explanation
In recent boards, conceptual and application-based questions dominate. Always cite chemical principles and show working to capture full marks, per CBSE guidelines.

7. Why is aromaticity a frequently asked concept in Class 11 Hydrocarbons important questions?

Aromaticity is central because:

  • It underpins the stability and reactivity of compounds like benzene, toluene, and aniline
  • Exam questions often test you on Huckel’s Rule, delocalization, and resonance energy
  • Distinguishing between aromatic, non-aromatic, and anti-aromatic is common in HOTS
Understanding aromaticity is necessary for both direct and application-based questions under CBSE exam patterns.

8. What are some misconceived ideas students have about Markovnikov and anti-Markovnikov additions in Hydrocarbons Chapter 9?

Common misconceptions include:

  • Believing Markovnikov’s rule applies to all electrophilic additions — it does not when peroxides are present
  • Confusing the direction of positive/negative species in addition to alkenes and alkynes
  • Assuming alkynes always behave like alkenes in additions (reaction conditions matter)
To avoid mistakes, always check reagents (peroxides for anti-Markovnikov), and revise mechanism steps as per CBSE marking scheme.

9. How should students prioritize 5-mark and 3-mark questions while preparing for Hydrocarbons important questions?

Strategy includes:

  • For 5-mark: Focus on multistep conversions, mechanism explanation (e.g., resonance in benzene, reactions of alkenes/alkynes), and comparative reasoning (e.g., stability, acidity order)
  • For 3-mark: Practice structural isomer drawing, explaining results of distinguishing tests, and explaining boiling/melting point trends
Always support answers with diagrams and correct chemical equations, as per the latest CBSE evaluation pattern.

10. What role do named reactions (like Wurtz, Friedel–Crafts, Kolbe’s) play in Class 11 Hydrocarbons important questions?

Such reactions are commonly targeted in both direct and indirect formats:

  • Questions may require writing balanced chemical equations or outlining the mechanism
  • Contextual application: Using these in conversions between different types of hydrocarbons or for synthesis problems
  • Explaining the conditions and possible outcomes if starting or end compounds change
Knowledge of at least two–three named reactions, their steps, and exceptions is vital for Board/NEET/JEE-level questions.

11. How can I distinguish between alkanes, alkenes, and alkynes using chemical tests for CBSE Class 11 important questions?

  • Alkanes: Generally unreactive; do not decolorize bromine water or Baeyer's reagent
  • Alkenes/Alkynes: Decolorize bromine water (alkynes react more slowly)
  • Baeyer's reagent (KMnO4): Turns colorless with alkenes/alkynes, no effect with alkanes
These are classic exam questions, and writing the correct observations is necessary for full marks, as per CBSE guidelines.

12. What are the key differences between resonance and aromaticity, and why do examiners emphasize them in Hydrocarbons important questions?

Resonance describes delocalization of electrons across atoms in a molecule, stabilizing the structure. Aromaticity is a special case where a cyclic, planar molecule exhibits continuous resonance, leading to extra stability (as in benzene).

  • Exam questions test understanding by requiring explanations of why benzene is more stable than expected (due to resonance energy), or by asking which compounds fulfill aromaticity criteria (planarity, cyclic, delocalized electrons, Huckel number).

13. What strategies help solve Hydrocarbons Class 11 important questions with conversions and reaction sequences?

  • Map out reagents and interconversions stepwise
  • Memorize key reagents for preparation and transformation of alkanes, alkenes, alkynes, and aromatic compounds
  • Pay attention to reaction conditions (e.g., presence/absence of peroxides, catalysts, temperature)
  • Double-check the structures and IUPAC names at each step
These steps ensure logical, full-credit answers as required in CBSE and competitive exams.

14. Why is General Organic Chemistry (GOC) considered essential before solving Hydrocarbons Class 11 important questions?

GOC builds the foundation for understanding the behavior of hydrocarbons:

  • It covers bonding, hybridization, electron movement, and intermediate types
  • Questions often require knowledge of GOC for correctly predicting product formation, intermediate stability, and reactivity order
  • CBSE and NEET/JEE questions interlink GOC with specific hydrocarbon concepts (e.g., acidity of alkynes)
Thorough grasp of GOC is a pre-requisite for confidently tackling complex or high-weightage hydrocarbon questions.