Test: Single Correct MCQs: Hydrocarbons | JEE Advanced


31 Questions MCQ Test Class 11 Chemistry 35 Years JEE Mains &Advance Past yr Paper | Test: Single Correct MCQs: Hydrocarbons | JEE Advanced


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QUESTION: 1

Marsh gas mainly contains (1980)

Solution:
QUESTION: 2

Which of the following decolourises alkaline KMnO4 solution(1980)

Solution:

Unsaturated hydrocarbons decolourise alk. KMnO4 solution; C2H4 (H2C = CH2) is an alkene.

QUESTION: 3

The compound with the highest boiling point is (1982 - 1 Mark)

Solution:

In a homologous series, higher the number of C-atoms, higher is the b.p.

QUESTION: 4

The maximum number of isomers for an alkene with the molecular formula C4H8 is (1982 - 1 Mark)

Solution:

Four isomers

QUESTION: 5

When propyne is treated with aqueous H2SO4 in presence of HgSO4 the major product is (1983 - 1 Mark)

Solution:

QUESTION: 6

Which of the following compounds does not dissolve in conc. H2SO4 even on warming? (1983 - 1 Mark)

Solution:

CH2 = CH2 + H2SO4 → CH3CH2OSO3H
C6H6 + H2SO4 → C6H5SO3H + H2O
C6H14 + H2SO4 → No reaction

Only hexane does not dissolve in conc. H2SO4 even on warming.

QUESTION: 7

Baeyer ’s reagent is : (1984 - 1 Mark)

Solution:
QUESTION: 8

Acidic hydrogen is present in : (1985 - 1 Mark)

Solution:

Acidic hydrogen is present in alkynes, attached to the triply bonded C-atoms. They can be easily removed by means of a strong base.

QUESTION: 9

Anti-Markovnikoff addition of HBr is not observed in : (1985 - 1 Mark)

Solution:

TIPS/Formulae : Anti-Markovnikoff’s addition of HBr is observed only with unsymmetrical alkenes, a, b, and d.

QUESTION: 10

The highest boiling point is expected for : (1986 - 1 Mark)

Solution:

For isomeric alkanes, th one having longest straight chain has highest b.p. because of larger surface area.

QUESTION: 11

Which of the following will have least hindered rotation about carbon-carbon bond? (1987 - 1 Mark)

Solution:

Ethylene has restricted rotation [due to C = C], acetylene no rotation [due to C ≡ C], hexachloroethane has more rotation than ethylene but less than ethane because of greater size of the substituent (chlorine) than in ethane (substituent is hydrogen).

QUESTION: 12

When cyclohexane is poured on water, it floats, because:

Solution:
QUESTION: 13

The product(s) obtained via oxymercuration (HgSO4 + H2SO4) of 1- butyne would be (1999 - 2 Marks)

Solution:

TIPS/Formulae : Hydration of alkynes via mercuration takes place in accordance with Markovnikov's manner rule

QUESTION: 14

Propyne and propene can be distinguished by (2000S)

Solution:

In propyne (CH3C ≡CH), the terminal hydrogen is acidic and reacts with ammonical AgNO3.

QUESTION: 15

Which one of the following will react fastest with H2 under catalytic hydrogenation condition ? (2000S)

Solution:

TIPS/Formulae : The relative rates of hydrogenation decreases with increase of steric hinderance.
R2C = CH2 > RCH = CHR > R2C = CHR > R2C = CR2
Among the four olefins, (a) and (b) are less stable (Saytzeff  rule). Further in (a), the  bulky alkyl groups are on same side (cis-isomer), hence it is less stable.

QUESTION: 16

In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti-Markovnikov addition to alkenes because(2001S)

Solution:

TIPS/Formulae : Peroxide effect is effective only in case of HBr and not in case of HCl and HI.

For HCl, Step-I (b) is endothermic while step-II is exothermic but for HI, Step-I(b) is exothermic while Step-II is endothermic.

QUESTION: 17

Hydrogenation of the above compound in the presence of poisoned palladium catalyst gives (2001S)

Solution:

TIPS/Formulae : Addition on triple bond takes place by the syn-addition of hydrogen.
Since the configuration of the double bond already present is cis, the compound formed will have a plane of symmetry and hence optically inactive.

QUESTION: 18

The reaction of propene with HOCl proceeds via the addition of(2001S)

Solution:

TIPS/Formulae : Alkenes undergo electrophilic addition reactions.
HOCl undergoes self-ionization

So, it is the Cl+ that attacks in the first step.

QUESTION: 19

The nodal plane in the π-bond of ethene is located in

Solution:

TIPS/Formulae : The π bond is formed by the sideways overlapping of two p-orbitals of the two carbon atoms.
The molecular plane does not have any π electron density as the p-orbitals are perpendicular to the plane containing the ethene molecule. The nodal plane in the π-bond of ethene is located in the molecular plane.

QUESTION: 20

Consider the following reaction (2002S)

Identify the structure of the major product 'X'

Solution:

Br is less reactive and more selective and so the most stable free radical (3°) will be the major product.

QUESTION: 21

Identify the reagent from the following list which can easily distinguish between 1-butyne and 2-butyne (2002S)

Solution:

TIPS/Formulae : In 1-butyne terminal hydrogen is acidic where as in 2-butyne there is no terminal hydrogen. Thus 2-butyne will not react with ammonical Cu2Cl2. While 1-butyne, being terminal alkyne, will give red ppt. with ammonical cuprous chloride

QUESTION: 22

(2003S)

Solution:

QUESTION: 23

Which of the following is used for the conversion of 2-hexyne into trans-2-hexene ? (2004S)

Solution:

H2/Pd/BaSOreduces  an alkyne to cis-alkene, H2/Pt reduces it to alkane, NaBH4 does not reduce an alkyne.
Reduction of an alkyne by active metal in liq. NH3 gives trans-alkene.

QUESTION: 24

On monochlorination of 2-methylbutane, the total number of chiral compounds formed is (2004S)

Solution:

                                                   

(i) Chlorination at C-2 and C-4 produces no chiral compounds
(ii) Chlorination at C-3 produces a chiral carbon marked with star (d and l form).
(iii) Chlorination at C-1 also produces a chiral carbon marked with star (d and l form).

QUESTION: 25

Identify the product, P in the following reaction : CH3 - CH = CH2 + NOCl —→ P (2006 - 3M, –1)

Solution:

Nitrosyl chloride adds on olefins accor ding to Markovnikof’s rule, where NO+ constitutes the positive part of the addendum.

QUESTION: 26

Cyclohexene on ozonolysis followed by reaction with zinc dust and water gives compound E. Compound E on further treatment with aqueous KOH yields compound F.
Compound F is (2007)

Solution:

QUESTION: 27

The synthesis of 3-octyne is achieved by adding a bromoalkane into a mixture of sodium amide and an alkyne.The bromoalkane and alkyne respectively are (2010)

Solution:

Only (d) can form 3-Octyne

                                         

CH3CH2C ≡ CCH2CH2CH2CH3+ NaBr

QUESTION: 28

The bond energy (in kcal mol–1) of a C–C single bond is approximately (2010)

Solution:

C – C bond energy = 348 kJ/mol = kcal/mol
= 82.85 kcal/mol ≈ 100 kcal/mol.

QUESTION: 29

In allene (C3H4), the type(s) of hybridisation of the carbon atoms is (are) :     (2012)

Solution:

Allene (C3H4) is

QUESTION: 30

The number of optically active products obtained from the complete ozonolysis of the given compound is :      (2012)

Solution:

QUESTION: 31

Isomers of hexane, based on their branching, can be divided into three distinct classes as shown in the figure. (JEE Adv. 2014)

The correct order of their boiling point is

Solution:

Greater the extent of branching, lesser is the boiling point of the hydrocarbon, so order of b.p is III > II > I.