JEE Advanced (Single Correct MCQs): Hydrocarbons - JEE MCQ

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

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

Four isomers

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

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

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

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

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

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).

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

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

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.

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.

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.

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

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

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.

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

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

H₂/Pd/BaSO₄ reduces  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.

(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).

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

Only (d) can form 3-Octyne

CH₃CH₂C ≡ CCH₂CH₂CH₂CH₃+ NaBr

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

Allene (C₃H₄) is

Ozonolysis is an organic reaction in which the unsaturated bonds of alkenes, alkynes, or azo compounds are cleaved by an ozone molecule (O₃).
In ozonolysis of alkenes, multiple carbon-carbon double bonds are cleaved by ozone molecules and are replaced by carbon-oxygen bonds i.e., carbonyl groups. Mechanism of ozonolysis includes two steps.

In the first step, ozone being electrophilic in nature, reacts with the alkene rapidly and cleaves the carbon-carbon double bond to produce a cyclic peroxide known as ozonide.

In the second step, the decomposition of ozonide (or work-up reaction) is carried out by reductive work-up or oxidative work-up. The reductive work-up yields aldehydes or ketones whereas oxidative work-up gives carboxylic acids as the products and this is the complete ozonolysis. The products of the reaction depends on the type of multiple bonds being oxidised and the work-up conditions.

Given compound: 
Complete ozonolysis reaction of given compound is as follows:

Now, for a molecule to be optically active, there must be a chiral carbon centre in the molecule which means the carbon atom must be attached to the four different groups. But the products obtained do not contain any chiral carbon centre, hence the products obtained are optically inactive. Thus, the number of optically active products obtained from the complete ozonolysis of the given compound are zero.