Class 11 Chemistry Hydrocarbons Exam Questions

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Study Material for Class 11 Chemistry Chapter 13 Hydrocarbons

Class 11 Chemistry students should refer to the following Pdf for Chapter 13 Hydrocarbons in Class 11. These notes and test paper with questions and answers for Class 11 Chemistry will be very useful for exams and help you to score good marks

Class 11 Chemistry Chapter 13 Hydrocarbons

LEVEL – 1

1. Why is cyclproprane more reactive than propane?

Ans: In Cyclopropane,bond angle is 60o which is much less compared to the normal tetrahedral bond angle of 109.5o for sp3 hybridized carbon. Therefore, the molecule is very much strained and hence reactive.

 

2. Why is Wurtz reaction not preferred for preparation of alkanes containing odd no of carbon atoms?

Ans: In this type of preparation we get mixture of hydrocarbons and they cannot be separated out due to about same boiling points.

 

3. In the presence of peroxide addition of HBr to prepare propene takes place according to anti Markovnikov’s rule but peroxide effect if not seen in the case of HCl and HI. Explain.

Ans: Homolysis of HCl does not takes place and in the case of HI homolysis takes place but the iodide free radical combine with the similar free radical to form I2 molecule .Thus peroxide effect is observed only in case of HBr.

4. Arrange the different conformations of ethane in decreasing order of stability?

Ans: Staggered> Skew >eclipsed.

 

5. What is the basic differences between conformational isomers and configurational (like geometrical isomers) isomers?

Ans: One conformational isomer is changes into other without any bond rearrangement i.e. due to single bond rotation while geometrical isomers are converted in to one form to other by bond rearrangement i.e. bond breaking and bond making.

 

6.Which of the following has the highest boiling point?
(i) 2-Methylpentane (ii) 2,3-Dimethylbutane
Ans: 2- Methylpentane has highest boiling point because it has the least branched chain structure as compared to 2,3-Dimethylbutane . Therefore, it has largest surface area and hence has the highest boiling point.
 
7. Name two tests to test the presence of double bond in a compound.
Ans: . (i) Decolourises brown colour of bromine water
(ii) Decolourising pink colour of Baeyer’s reagent.
 
8. Name the process which may be used to locate the position of triple bond.
Ans: .Ozonolysis
 
9.Why benzene is extraordinarily stable though it contains three double bonds?
Ans: Due to resonance in benzene the π –electron cloud gets delocalized resulting stability of molecule.
 
 
LEVEL -2
 
1. Why alkynes do not show geometrical isomerism?
Ans: Alkynes have linear shape and therefore, do not show geometrical isomerism.
 
 
2.How many isomers are possible for monosubstituted and disubstituted benzene?
Ans: There is one mono substituted benzeneand three di substituted benzene i.e. ortho, para and metal.
 
3. Arrange benzene,n-hexane and ethyene in decreasing order of acidic behaviour.
Ans: Ethyne> benzene> n-hexane 
 

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

Ans: CH3-CH=C(C2H5)2 , 3-Ethylpent-2-ene.

 

5. Out of benzene, m-dinitro benzene and toluene which will undergonitration most easily and why?

Ans: CH3- is a ERG while –NO2 is a EWG .Therefore electron density is more in toluene than in benzene and the electron density in m-dinitrobenzene will be less than in benzene therefore toluene will undergo easily in nitration reaction.

 

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

Ans: On increasing branching surface area decreases and approaches that of a sphere .Since sphere has minimum surface area, therefore, van der Walls forces of attraction are minimum and hence boiling point decreases with branching.

 

7.How will you prepare acetaldehyde from acetylene?

Ans:

Class 11 Chemistry Hydrocarbons Exam Questions

8. Why alkanes do not dissolved in water but dissolve in benzene?

Ans: Because alkanes are non-polar and water is a polar solvent.

 

9.What is the function of CaO in soda lime?

Ans: It helps in the fusion of reaction mixture.


10.Can pyridine be regarded as an aromatic compound?

Ans: . Yes because it has planar structure and has (4n+2 )π electrons i.e 6 π electrons.

LEVEL-3


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

Ans:

(i) The molecule should contain cyclic cloud of delocalized electrons above and below the plane of the molecule.

(ii) For the delocalization of pi electrons the ring must be planer to allow cyclic overlap of p- orbitals.

(iii)It should contain (4n+2) π electrons where n=0,1,2,3........ This is known as Huckel rule.


2. Name the alkane which cannot be prepared by Wurtz Reaction?

Ans: Methane.

 

3.What is the cause of geometrical isomerism in alkanes?

Ans: Alkanes have π –bond and the restricted rotation around the π-bond gives rise the geometrical isomerism.

 

4.Why alkanes are called paraffins?

Ans: They have little affinity for chemical reactions so they are called as paraffins.

 

5.What is Markownikoff’s rule?

Ans: During the addition across unsymmetrical multiple bond ,the negative part of the addendum joins with that double bonded carbon which has less no of hydrogen.

 

6. Find out terminal and non-terminal alkynes from the following: Ethyne, Pent -2-yne ,but-1-yne , propyne.

Ans: Ethyne,but-1-yne , propyne because these molecules have triple bond at the end of the carbon chain.

 

7.What is the use of BHC (benzene hexachloride)?

Ans: It is used as an insecticide under the name of Gammexane or lindane.

 

8.What is the product of cyclic polymerization of ethyne?

Ans: . Benzene.

 

9. Name the two extreme type of conformation of ethane.

Ans: Staggered and eclipsed.

 

10. Suggest name of any other Lewis acid instead of anhydrous aluminium chloride which can be used for ethylation of benzene?

Ans: .FeCl3

Chapter 04 Chemical Bonding and Molecular Structure
Class 11 Chemistry Chemical Bonding and Molecular Structure Exam Questions

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