Check the below NCERT MCQ Questions for Class 11 Chemistry Chapter 13 Hydrocarbons with Answers Pdf free download. MCQ Questions for Class 11 Chemistry with Answers were prepared based on the latest exam pattern. We have provided Hydrocarbons Class 11 Chemistry MCQs Questions with Answers to help students understand the concept very well.
Hydrocarbons Class 11 MCQs Questions with Answers
Hydrocarbons Class 11 MCQ Question 1.
A dibromo derivative of an alkane reacts with sodium metal to form an alicyclic hydrocarbon. The derivative is ______.
(a) 2, 2-dibromobutane
(b) 1, 1-dibromopropane
(c) 1, 4-dibromobutane
(d) 1, 2-dibromoethane
Answer
Answer: (c) 1, 4-dibromobutane
Explanation:
The derivative is 1, 4 dibromobutane. This on reaction with sodium metal gives cyclobutane.
This reaction is an example of internal Wurtz reaction.
MCQ On Hydrocarbons Class 11 Question 2.
The position of double bond in alkenes can be located by :
(a) Hydrogenation of oil
(b) Ozonolysis
(c) Photolysis
(d) Hydration
Answer
Answer: (b) Ozonolysis
Explanation:
Ozonolysis is the cleavage of an alkene or alkyne with ozone to form organic compounds in which the multiple carbon-carbon bonds have been replaced by a double bond to oxygen. The outcome of the reaction depends on the type of multiple bonds being oxidized.
Bromine water can be also used to identify the position of a double bond. In this reaction, red-brown colour of bromine gets turned into coulorless indicating that there is a double bond.
Hydrocarbons Class 11 MCQ Pdf Question 3.
Nitrobenzene on reaction with conc. HNO3/H2SO4 at 80 – 100°C forms which one of the following products?
(a) 1, 2-Dinitrobenzene
(b) 1, 3-Dinitrobenzene
(c) 1, 4-Dinitrobenzene
(d) 1, 2, 4-Trinitrobenzene
Answer
Answer: (b) 1, 3-Dinitrobenzene
Explanation:
NO2 is an m-directing group and hence, 1, 3-dinitrobenzene is formed.
MCQ Of Hydrocarbons Class 11 Question 4.
Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
(a) –C ≡ N
(b) -SO3H
(c) -COOH
(d) -NO2
Answer
Answer: (d) -NO2
Explanation:
The correct order of deactivation is
−NO2 > −C ≡ N > −SO32 H > −COOH
Hydrocarbon MCQ Class 11 Question 5.
Which of the following compounds will exhibit geometrical isomerism?
(a) 1 – Phenyl – 2 – butene
(b) 3 – Phenyl – 1 – butene
(c) 2 – Phenyl – 1 butene
(d) 1, 1 – Diphenyl – propene .
Answer
Answer: (a) 1 – Phenyl – 2 – butene
Explanation:
C6H5CH2−CH = CH−CH3 (1-phenyl-2-butene) exhibits the phenomenon of geometrical isomerism due to Cab = Cad structure, so its two isomers are possible which are given are follow :
Hydrocarbons MCQs With Answers Question 6.
The order of decreasing reactivity towards an electrophilic reagent, for the following : (i) Benzene (ii) Toluene (iii) Chlorobenzene (iv) Phenol would be:
(a) (i) > (ii) > (iii) > (iv)
(b) (ii) > (iv) > (i) > (iii)
(c) (iv) > (iii) > (ii) > (i)
(d) (iv) > (ii) > (i) > (iii)
Answer
Answer: (d) (iv) > (ii) > (i) > (iii)
Explanation:
Benzene having any activating group i.e., OH, R etc., undergoes electrophilic substitution very easily as compared to benzene itself. Thus toluene (C6H5CH3), phenol (C6H5OH) undergo electrophilic substitution very readily than benzene. Chlorine with +E and +M effect deactivates the ring due to strong -I effect. So, it is difficult to carry out the substitution in chlorobenzene than in benzene, so correct order is Phenol (iv) > Toluene (ii) > Benzene (i) > Chlorobenzene (iii)
Hydrocarbons MCQs Class 11 Question 7.
Pure methane can be prepared by
(a) Soda lime decarboxylation
(b) Kolbes electrolytic method
(c) Wurtz reaction
(d) Reduction with H2
Answer
Answer: (a) Soda lime decarboxylation
Explanation:
Methane cannot be prepared by either Wurtz reaction, Kolbes electrolytic method or by reduction of alkenes with H2. While acetic acid salt on heating with soda lime gives methane.
CH3COONa– → CH4 + Na2CO3
Hydrocarbon Class 11 MCQ Question 8.
Hydrocarbon containing following bond is most reactive
(a) C ≡ C
(b) C = C
(c) C-C
(d) All of these
Answer
Answer: (a) C ≡ C
Explanation:
−C ≡ C−is most reactive because sp-hybridization.
MCQs On Hydrocarbons Class 11 Question 9.
The compound C3H4 has a triple bond, which is indicated by its reaction with
(a) Bromine water
(b) Bayers reagent
(c) Fehling solution
(d) Ammonical silver nitrate
Answer
Answer: (d) Ammonical silver nitrate
Explanation:
CH3−C ≡ C−H + AgNO3 → CH3 −C ≡ C−Ag
Propyne Ammonical Siver salt of Propyne
Hydrocarbons MCQs Question 10.
Isopropyl bromide on Wurtz reaction gives
(a) Hexane
(b) Propane
(c) 2, 3 – Dimethylbutane
(d) Neohexane
Answer
Answer: (c) 2, 3 – Dimethylbutane
Explanation:
CHMe2 − Br + 2Na + Br − CHMe2 + Br − CHMe2 \(\underrightarrow { dryEther } \) Me2CH − CHMe2 + 2NaBr
Hydrocarbons Class 11 MCQs Question 11.
Nitrobenzene on reaction with conc. HNO3/H2SO4 at 80 – 100°C forms which one of the following products?
(a) 1, 2-Dinitrobenzene
(b) 1, 3-Dinitrobenzene
(c) 1, 4-Dinitrobenzene
(d) 1, 2, 4-Trinitrobenzene
Answer
Answer: (b) 1, 3-Dinitrobenzene
Explanation:
NO2 is an m-directing group and hence, 1, 3-dinitrobenzene is formed.
Class 11 Hydrocarbons MCQs Question 12.
The angle strain in cyclobutane is
(a) 24°44
(b) 29°16
(c) 19°22
(d) 9°44
Answer
Answer: (d) 9°44
Explanation:
According to Baeyers strain theory, the amount of the strain is directly proportional to the angle through which a valency bond has deviated from its normal position . i.e., Amount of deviation) in cyclobutane (d) = (109∘28 − 90∘)/(2) = 9∘44
MCQ On Chapter Hydrocarbons Class 11 Question 13.
The first fraction obtained during the fractionation of petroleum is:
(a) Gasoline
(b) Diesel Oil
(c) Hydrocarbon Gases
(d) Kerosene Oil
Answer
Answer: (c) Hydrocarbon Gases
Explanation:
During fractionation or distillation of petroleum, gases light hydrocarbons in the form of gases are obtained from first fraction.
Hydrocarbons Class 11 Important Questions With Answers Pdf Question 14.
2 – Phenylpropene on acidic hydration gives
(a) 2 – Phenyl – 2 – propanol
(b) 2 – Phenyl – 1 – propanol
(c) 3 – Phenyl – 1 – propanol
(d) 1 – Phenyl – 2 – propanol
Answer
Answer: (a) 2 – Phenyl – 2 – propanol
Explanation:
Acidic hydration of 2-phenyl propene follows electrophilic reaction mechanism forming an intermediate 3° carbocation (more stable), there by forming 2-phenyl-2-propanol.
Class 11 Chemistry Chapter 13 MCQ Question 15.
The coal tar fraction which contains phenol is:
(a) Heavy Oil
(b) Light Oil
(c) Middle Oil
(d) Green Oil
Answer
Answer: (c) Middle Oil
Explanation:
The second fraction of coal-tar distillation which distills at 170°−230°C is called middle-oil or olic oil. It contains mainly phenol, naphthalene etc.
Class 11 Chemistry Hydrocarbons MCQs Question 16.
Which one of these is not true for benzene?
(a) There are three carbon-carbon single bonds and three carbon-carbon double bonds.
(b) Heat of hydrogenation of benzene is less than the theoretical value
(c) It forms only one type of mono substituted product
(d) The bond angle between carbon-carbon bonds is 120 Degree
Answer
Answer: (a) There are three carbon-carbon single bonds and three carbon-carbon double bonds.
Explanation:
There are three carbon-carbon single bonds and three carbon-carbon double bonds.
Hydrocarbons Class 11 MCQ Pdf Downloa Question 17.
The catalyst used in Friedel – Crafts reaction is
(a) Aluminium Chloride
(b) Anhydrous Aluminium Chloride
(c) Ferric Chloride
(d) Copper .
Answer
Answer: (b) Anhydrous Aluminium Chloride
Explanation:
It is a catalyst-based electrophilic substitution reaction. Lewis acids like anhydrous AlCl3, anhydrous FeCl3, BF3 and BiCl3 can be used as catalysts. The catalyst facilitates more effective attack by the electrophile (R+, RCO+ etc.) on the benzene ring.
Question 18.
Alkyl halides react with dialkyl copper reagents to give?
(a) Alkanes
(b) Alkenes
(c) Hydrogen
(d) Carbon
Answer
Answer: (a) Alkanes
Explanation:
Alkyl halides react with dialkyl copper reagents to give Alkanes.
Question 19.
The lowest alkene, that is capable of exhibiting geometrical isomerism is
(a) Ethene
(b) But – 1- ene
(c) But – 2 – ene
(d) Propene.
Answer
Answer: (c) But – 2 – ene
Explanation:
The lowest alkene which is capable of exhibiting geometrical isomerism is 2-Butene
Question 20.
Presence of a nitro group in a benzene ring
(a) Activates the ring towards electrophilic substitution
(b) Renders the ring basic
(c) Deactivates the ring towards nucleophilic substitution
(d) Deactivates the ring towards electrophilic substitution
Answer
Answer: (d) Deactivates the ring towards electrophilic substitution
Explanation:
−NO2 group shows −M effect, so withdraws the electron density from the ring and hence deactivate the ring towards electrophilic aromatic substitution.
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