Hydrocarbons Previous Year Questions with Solutions

NEET and JEE Main previous year solved questions on Hydrocarbons are available here. Practise Hydrocarbons questions- score higher ranks for NEET JEE

Hydrocarbons Previous Year Questions with Solutions


NEET Previous Year Papers Questions of Chemistry With Solutions

NEET 2017 Predict the correct intermediate and product in the following reaction.

Concept Questions :- Alkanes, Alkenes and Alkynes - Chemical Properties
Solution: d

NEET 2015 In the reaction with HCI, an alkene reacts in accordance with the Markownikoff's rule, to give a product 1 -chloro-l-methylcyclohexane. The possible alkane is
Concept Questions :- Alkanes, Alkenes and Alkynes - Chemical Properties
Solution : (c)

NEET 2015  2, 3-dimethyI-2-butene can be prepared by heating which of the following compounds with a strong acid?
Solution= (b)

NEET 2015  The oxidation of benzene by V205 in the presence of air produces
1. benzoic anhydride
3. benzoic acid
2. maleic anhydride
4. benzaldehyde

Ans= 2

NEET 2014  What products are formed when the following compound is treated with Br2 in the presence of FeBr3?
Concept Questions :- Aromatic Hydrocarbons, Directive Influence in Monosubstituted Benzene
Solution: (c)    NEET Question setter made him self confused !

NEET 2014 Identity Z in the sequence of reactions,
Ans= 1

NEET 2014  Which of the following organic has same hybridisation as its combustion product -(C02)?
(a) Ethane (b) Ethyne (c) Ethene (d) Ethanol
Ans= (b)

NEET 2013 Nitrobenzene on reaction with conc. HN03/H2S04 at 80-1000C forms which one of the following products?
(a) 1,2-dinitrobenzene
(b) 1,3-dinitrobenzene
(c) 1,4-dinitrobenzene
(d) 1,2,4-trinitrobenzene
Ans= (b)

NEET 2013  Which of the following compounds will not undergo Friedel-Craft's reaction easily?
1. Cumene
2. Xylene
3. Nitrobenzene
4. Toluene
Ans= (3)

NEET 2012  In the following reaction, What will be the major product?
Ans = (a)

NEET 2011 In the following reactions:-
Ans= (a)

NEET 2010 Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by:-
(a) oxidation
(b) cracking
(c) distillation under reduced pressure
(d) hydrolysis

Ans=(b)

NEET 2009  Nitrobenzene can be prepared from benzene by using a mixture of conc HN03 and conc. H2S04. In the mixture,nitric acid acts as a/an:
(a) reducing agent
(b) acid
(c) base
(d) catalyst

Ans= (c)

NEET 2009  Benzene reacts with CH3Cl in the presence of anhydrous AICl3 to form.
1. toluene
2. chlorobenzene
3. benzylchloride
4. xylene

Ans = 1

NEET 2008

Ans = 2

NEET 2007  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)

Ans - 4


JEE Main Previous Year Papers Questions of Chemistry With Solutions

JEE Main 2020 (Online) 6th September Evening Slot

Ans = b

JEE Main 2020 (Online) 5th September Evening Slot


JEE Main Online 2019 Many Set        Download JEE Main 2019 Hydrocarbons

Q. The trans-alkenes are formed by the reduction of alkynes with : JEE Main Ofline 2018
(1) NaBH4 (2) Na/liq. NH 3 (3) Sn-HCI (4) H2 - Pd/C, BaSO4 
Solution: 2

JEE Main 2016 (Online) 10th April Morning Slot
Bromination of cyclohexene under conditions given below yields :

Ans - b

JEE Main 2016 (Offline)
At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume for complete combustion. After combustion the gases occupy 345 mL. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is:
a)C3H8
b)C4H8
c)C4H10
d)C3H9
Ans- a

Q.Which compound would give 5-keto-2-methyl hexanal upon ozonlysis?  JEE Main Ofline 2015
So;ution:

Q. In the presence of peroxide, HC: and HI do not give anti-Markownikoff s addition to alkenes because   JEE Main Ofline 2014
(1) All the steps are exothermic in HCI and HI 
(2) One of the steps is endothermic in HCI and HI 
(3) HCI is oxidizing and the HI is reducing 
(4) Both HCI and HI are strong acids 
Solution: 2 
relectivity ratio for bromination is 1° : 2° : 3° : 1 : 82 : 1600 Hence 3° product will be major product. 

Bromides will take part in radical based reactions in the presence of organic peroxides. Fission of the peroxide O-O linkage causes a Br radical which behaves differently to Bromide, by adding to the less substituted side of the alkene(anti – markovnikov). HI and HCl does not do this for energetic reasons. The addition of Cl and I radicals to the alkene in an anti-markovnikov fashion is an endothermic reaction. Hence it is unfavourable. All hydrogen halides will add according to the markovnikov rule, without the presence of peroxide,

The major product obtained in the photo catalysed bromination of 2-methylbutane is :-  JEE Main Ofline 2014
(1 ) 2-bromo-2-methylbutane 
(2) 2-bromo-3-methylbutane 
(3) 1-bromo-2-methyl butane 
(4) 1-bromo-3-methyl butane 
Solution: 1

JEE Main Ofline 2014
Solution:
(1)

The reagent needed for converting     JEE Main Ofline 2014

(1) H2 / lindlar Cat.

(2) Catalytic Hydrogenation

(3) LiAlH4

(4) Li/ NH3

Solution:  Li/NH3 is the birch reagent. It reduces the alkyne to trans alkene. Hence option (4) is the answer.

The gas liberated by the electrolysis of Dipotassium succinate solution is :   JEE Main Ofline 2014
(1) Ethyne
(2) Ethene
(3) Propene
(4) Ethane
Solution:
(CH2COO)2 → CH2 = CH2 + 2CO2 (g) + 2e
2H2O + 2e → 2OH + H2 (g)
So the gas generated during electrolysis of Dipotassium succinate solution is ethene.
Hence option (2) is the answer.

Which one of the following classes of compounds is obtained by polymerization of acetylene?  JEE Main Ofline 2014
(1) Poly-ene
(2) Poly-yne
(3) Poly-amide
(4) Poly-ester
Solution:
nHC ≡ CH → ( CH =CH)n poly-yne
Hence option (2) is the answer.

In the hydroboration – oxidation reaction of propene with diborane, H2O2 and NaOH, the organic compound formed is :   JEE Main Ofline 2014

(1) CH3CH2CH2OH
(2) (CH3)3COH
(3) CH3CHOHCH3
(4) CH3CH2OH
Solution:
The hydroboration–oxidation reaction is a two-step hydration reaction that converts an alkene into alcohol. It is an anti-Markovnikov reaction. The organic compound formed in the hydroboration – oxidation reaction of propene with diborane, H2O2 and NaOH is CH3CH2CH2OH.
Hence option (1) is the answer.

The number and type of bonds in C22− ion in CaC2 are:    JEE Main Ofline 2014
(1) Two σ bonds and one π – bond
(2) Two σ bonds and two π – bonds
(3) One σ bond and two π – bonds
(4) One σ bond and one π bond
Solution:
Ca+2 [C≡C] -2
One σ bond and two π – bonds are there in C22- ion in CaC2.
Hence option (3) is the answer. ion in CaC2.
Hence option (3) is the answer.

The major organic compound formed by the reaction of 1, 1, 1-trichloroethane with silver powder is :-   JEE Main Ofline 2013
(1) 2-Butyne (2) 2-Butene (3) Acetylen (4) Ethene 
Solution: 1

In the given transformation, which of the following is the most appropriate reagent? AIEEE 2012

(1) NaBH4

(2) NH2-NH2, OH

(3) Zn – Hg / HCl

(4) Na, Liq.NH3

Solution: 2

2-Hexyne gives trans-2-Hexene on treatment with :-   AIEEE 2012
(1) LiAlH4 (2) Pt/H2
(3) Li/NH3 (4) Pd/BaSO4 
Solution: 3

Q. Ozonolysis of an organic compound 'A' produces acetone and propionaldehyde in equimolar mixture. Identify 'A' from the following compounds :-  AIEEE 2011
(1) 2-Methyl-1-pentene (2) 1-Pentene (3) 2-Pentene (4) 2-Methyl-2-pentene 
Solution: 4

Ozonolysis of an organic compound gives formaldehyde as one of the products. This confirms the presence of:- AIEEE 2011
(1) an isopropyl group
(2) an acetylenic triple bond
(3) two ethylenic double bonds
(4) a vinyl group
Solution:
In Ozonolysis, the presence of the vinyl group gives formaldehyde as one of the products.
Hence option (4) is the answer.

The main product of the following reaction is AIEEE 2010

Ans - 2

One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is:- AIEEE 2010
(1) Ethene
(2) Propene
(3) 1-Butene
(4) 2-Butene
Solution:
2-Butene on ozonolysis gives 2 moles of acetaldehyde with the molecular mass of 44gm/mol.
Hence option (4) is the answer.

JEE 2012 2014



JEE 2010 2011


JEE 2008

(JEE 2007). Presence of a nitro group in a benzene ring 

l)activates the ring towards electrophilic substitution
2)renders the ring basic .
3)deactivates the ring towards nucleophilic substitution
4)deactivates the ring towards electrophilic substitution
Ans.(4) -N02 group shows -M effect. It withdraws the electron density from the ring and hence deactivate the ring towards electrophilic aromatic substitution.

(JEE 2007) .The reaction of toluene with Cl2 in presence of FeCl3 gives predominantly

1)benzoyl chloride
2)benzyI chloride
3)o-and p-chlorotoluene
4)m-chIorotoIuene
Ans =3

(JEE 2007) .Which of the following reactions will yield 2, 2-dibromopropane ?

1)CH3 - C ≡ CH + 2HBr
2)CH3CH = CHBr + HBr
3)CH = CH + 2HBr
4)CH3 - CH = CH2 + HBr
Ans = 1

JEE 2006. HBr with CH2 = CH — OCH3 under anhydrous conditions at room temperature to give

1)CH3CHO and CH3Br
2)BrCH2CHO and CH30H
3)BrCH2 - CH2 - OCH3
4)H3C - CHBr - OCH3
Ans.(4) Electrophilic addition reaction is more favourable.

(JEE 2005) .Acid catalyzed hydration of except ethene leads to the formation of
1)Primary alcohol
2)Secondary or tertiary alcohol
3)Mixture of primary and secondary alcohols
4)Mixture of secondary and tertiary alcohols
Ans.(4) concept of hydration.

( JEE 2005) .Elimination of bromine from 2-bromobutane results in the formation of-
1)Equimolar mixture of 1 and 2-butene
2)Predominantly 2-butene
3)Predominantly I-butene
4)Predominantly 2-butyne
Ans.(2) Saytzeff's product.

(JEE 2005) .2-methylbutane on reacting with bromine in the presence of sunlight gives mainly
1)1-bromo-2-methylbutane
2)2-bromo-2-methylbutane
3)2-bromo-3-methylbutane
4)1-bromo-3-methylbutane
Ans- (2)

(JEE 2004) . Which one of the following reduced with zinc and hydrochloric acid to give the corresponding hydrocarbon ? 
1) Ethyl acetate
3) Acetamide
2) Butan-2-one
4) Acetic acid
Ans.(2) Total reduction of carbonyl.

(JEE 2004) .Which one of the following has the minimum boiling point? 
1) n-butane
3) 1-butene
2) isobutane
4) 1-butyne
Ans.(2) Lower van der Waals forces.

( JEE 2003) . Butene-I may be converted to butane by reaction with 
1) Sn-HCl
2) zn-Hg
3) Pd/H2
4) zn- HCI
Ans.(3) Hydrogenation.

JEE 2003   Bottles containing C6H5I and C6H5CH2I lost their original tables. They were labelled A and B for testing. A and B were separately taken in test tubes and boiled with NaOH solution. The end solution in each tubewas made acidic with dilute HNO3 and then some AgN03 solution was added. Substance B gave a yellow precipitate. Which one of the following statements is true for this experiment?

1)A and C6H5CH2I

2)B and C6H5I

3)Addition of HN03 was unnecessary

4) A was C6H5I

Ans.(4) C6H5I will not respond to silver nitrate test because C-I bond has a partial double bond character.

JEE 2002 Which of these will not react with acetylene
1) NaOH
2) Ammonical AgB03
3) Na
4) HCI

NEET 2002   What is the product when acetylene reacts with hypochlorous acid?

1) CH3COCl

3) Cl2CHCHO

2) CICH2CHO

4) CICHCOOH



Advantage of Jee Mains chemistry chapter wise questions with solutions

NEET & JEE Main  Sample MCQ on Chemistry Hydrocarbons

The important topics related to hydrocarbons are classification, properties, preparation and uses. As far as the NEET, JEE , WBJEE  exam is concerned, hydrocarbon is an very important topic. This article gives an idea of what type of questions can be expected from this topic. advancedchemistry.in provides solutions designed by Mudassar Husain. Students can easily join on whatsapp group to download the solutions in PDF format for free of cost .

1. Which branched chain isomer of the hydrocarbon with molecular mass 72 u gives only one isomer of mono substituted alkyl halide?
(1) Neopentane
(2) Isohexane
(3) Neohexane
(4) Tertiary butyl chloride
Solution:
Molecular mass indicates that it is pentane. Neopentane can only form one mono substituted alkyl halide as all the hydrogens are equivalent in neopentane.

Hence option (1) is the answer.

Q 2. Which one of the following has the minimum boiling point?
(1) n-Butane
(2) 1-Butyne
(3) 1-Butene
(4) Isobutene
Solution:
Among the isomeric alkanes, the normal isomer has a higher boiling point than the branched-chain isomer. The higher the branching of the chain, the lower is the boiling point. The n-alkanes have more surface area in comparison to branched-chain isomers. So, intermolecular forces are weaker in branched-chain isomers. Hence they have lower boiling points in comparison to straight-chain isomers.
Hence option (4) is the answer.

Question 3: Which of the following molecules, in pure form, is (are) unstable at room temperature? (IIT JEE 2013)
Answer: b and c
Solution: b and c are anti-aromatic molecules as these contain 4n i.e. 8 TT electrons in ring and hence, these are highly unstable.Anti aromatic
compounds are those for which number of ring TT electrons is equal to 4n (i.e. 4n = 4 or

Question 4: In allene (C3H4), the type(s) of hybridization of the carbon atoms is (are)  (IIT JEE-2011)

a. sp & sp3
b. sp and sp2
c. only sp2
d. sp2 and sp3
Answer: b
Solution: Allene are the molecules in which at least one C has double bonds with each of the adjacent carbon atoms..Structure of the allene C3H4 is
The central carbon forms two sigma and two pi bonds with the two adjacent carbon atoms, so the central carbon is sp and the two terminal carbons are sp2 hybridised.
Hence, the correct option is B.

Thank You !

Question Bank for NEET Chemistry Hydrocarbons (Practice Set)


JEE Main Previous Year Papers Questions With Solutions Chemistry Alkanes ,Alkenes ,Alkynes and Arenes
Solved Examples on Hydrocarbons
Topic wise Question on Hydrocarbons: MCQs (Multiple Choice Questions)


















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Advanced Chemistry: Hydrocarbons Previous Year Questions with Solutions
Hydrocarbons Previous Year Questions with Solutions
NEET and JEE Main previous year solved questions on Hydrocarbons are available here. Practise Hydrocarbons questions- score higher ranks for NEET JEE
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