Class 11 Chemistry MCQs Hydrocarbons for JEE/NEET

Here, you will get Class 11 Chemistry MCQ of Chapter 13 Hydrocarbons for cracking JEE and NEET/AIIMS. By solving these MCQs of Class 11 Chemistry Chapter 13 Hydrocarbons, you will get the confidence to crack JEE or NEET. Practice MCQ Questions for Class 11 Chemistry with Answers on a daily basis and score well in exams. 

Hydrocarbons Class 11 MCQs Questions with Answers

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: (b) 1, 3-Dinitrobenzene


Some meta-directing substituents in aromatic substitution are given. Which one is most deactivating?
(a) –C ≡ N
(b) -SO3H
(c) -COOH
(d) -NO2

Answer: (d) -NO2


The orth/para-drecting group among the following is
(a) COOH
(b) CN
(c) COCH3
(d) NHCOCH3

Answer: (d) NHCOCH3


Tetrabromoethane on heating with Zn gives
(a) Ethyl bromide
(b) Ethane
(c) Ethene
(d) Ethyne

Answer: (d) Ethyne


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: (d) (iv) > (ii) > (i) > (iii)


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: (d) Ammonical silver nitrate


An organic compound has a triple bond and not double bond. It can be test by
(a) Bromine water
(b) Baeyer reagent
(c) Fehling solution
(d) Ar nonical silver nitrate

Answer: (d) Ar nonical silver nitrate


The angle strain in cyclobutane is
(a) 24°44
(b) 29°16
(c) 19°22
(d) 9°44

Answer: (d) 9°44


Tetrabromoethane on heating with Zn gives
(a) Ethyl bromide
(b) Ethane
(c) Ethene
(d) Ethyne

Answer: (d) Ethyne


A dibromo derivative of an alkane reacts with sodium metal to form an alicyclic hydrocarbon. The derivative is __
(a) 1, 4-dibromobutane
(b) 1, 2-dibromoethane
(c) carbon
(d) none of the above

Answer: (a) 1, 4-dibromobutane


Hydrocarbon containing following bond is most reactive
(a) C ≡ C
(b) C = C
(c) C-C
(d) All of these

Answer: (a) C ≡ C


The first fraction obtained during the fractionation of petroleum is:
(a) Gasoline
(b) Diesel Oil
(c) Hydrocarbon Gases
(d) Kerosene Oil

Answer: (c) Hydrocarbon Gases


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: (a) There are three carbon-carbon single bonds and three carbon-carbon double bonds


Which alkene on ozonolysis gives CH3CH2CHO and CH3COCH3?
(a) CH2 = CH CH = C(CH3)2
(b) CH3 CH2CH = CH CH2CH3
(c) CH3CH2 CH = CH – CH3
(d) (CH3)2 C = CH CH3

Answer: (a) CH2 = CH CH = C(CH3)2


In the compound CH2 = CH – CH2 – CH2C = CH2 the C2 – C3 bond is of the type
(a) sp – sp²
(b) sp³ – sp³
(c) sp – sp³
(d) sp² – sp³

Answer: (d) sp² – sp³


The number and type of bonds between two carbon atoms in CaC2 are
(a) One sigma (σ) and one pi (π) bonds
(b) One sigma (σ) and two pi (π) bonds
(c) One sigma (σ) and one and a half pi (π) bonds
(d) One sigma (σ) bond.

Answer: (b) One sigma (σ) and two pi (π) bonds


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: (a) 2 – Phenyl – 2 – propanol


The coal tar fraction which contains phenol is
(a) Heavy Oil
(b) Light Oil
(c) Middle Oil
(d) Green Oil

Answer: (c) Middle Oil


Which branched chain isomer of the hydrocarbon with molecular mass 72u gives only one isomer of mono substituted alkyl halide?
(a) Neopentane
(b) Carbon
(c) Isohexane
(d) Neohexane

Answer: (a) Neopentane


Aromatic compounds burn with a sooty flame because
(a) They have a ring structure of carbon atoms
(b) They have a relatively high percentage of hydrogen
(c) They have a relatively high percentage of carbon
(d) They resist reaction with oxygen of air

Answer: (c) They have a relatively high percentage of carbon


The catalyst used in Friedel – Crafts reaction is
(a) Aluminium Chloride
(b) Anhydrous Aluminium Chloride
(c) Ferric Chloride
(d) Copper .

Answer: (b) Anhydrous Aluminium Chloride


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: (d) Deactivates the ring towards electrophilic substitution


The number of structural isomers for C6H14 is
(a) 3
(b) 4
(c) 5
(d) 6

Answer: (c) 5


Alkyl halides react with dialkyl copper reagents to give
(a) Alkanes
(b) Alkenes
(c) Hydrogen
(d) Carbon

Answer: (a) Alkanes


The lowest alkene, that is capable of exhibiting geometrical isomerism is
(a) Ethene
(b) But – 1- ene
(c) But – 2 – ene
(d) Propene

Answer: (c) But – 2 – ene


On mixing certain alkane with chlorine and irradiating it with ultraviolet light, one forms only one monochloro alkane. The alkane could be
(a) Neopentane
(b) Propane
(c) Pentane
(d) Isopentane

Answer: (a) Neopentane


The position of double bond in alkenes can be located by
(a) Hydrogenation of oil
(b) Ozonolysis
(c) Photolysis
(d) Hydration

Answer: (b) Ozonolysis


Isopropyl bromide on Wurtz reaction gives
(a) Hexane
(b) Propane
(c) 2, 3 – Dimethylbutane
(d) Neohexane

Answer: (c) 2, 3 – Dimethylbutane


Pure methane can be prepared by
(a) Soda lime decarboxylation
(b) Kolbes electrolytic method
(c) Wurtz reaction
(d) Reduction with H2

Answer: (a) Soda lime decarboxylation


Butene-1 may be converted to butane by reaction with
(a) Pd/H2
(b) Zn – HCl
(c) Sn – HCl
(d) Zn – Hg

Answer: (a) Pd/H2


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