JEE Chemistry Hydrocarbons MCQs Set 28

Answer: (d) 2-butyne

Explanation:
1. Silver abstracts chlorine in a Wurtz-type coupling: \(\ce{2CH3CCl3 + 6Ag -> CH3C#CCH3 + 6AgCl}\).
2. The product is 2-butyne.

Teacher's Note:
1. Silver or copper powder removes halogens from gem-trihalides to give alkynes.
2. Two molecules couple to join the carbons.

Answer: (b) a carbocation

Explanation:
1. SbCl5 is a Lewis acid that removes chloride to give a planar benzylic carbocation.
2. The nucleophile can attack from either face, giving a racemic mixture.

Teacher's Note:
1. A planar carbocation is achiral.
2. Racemisation is a feature of SN1 reactions.

Answer: (d) \(\ce{Li/NH3}\)

Explanation:
1. Dissolving metal reduction (Na or Li in liquid ammonia) gives anti addition of hydrogen, forming trans alkenes.
2. Pt/H2 gives the alkane and Pd/BaSO4 (Lindlar) gives the cis alkene.

Teacher's Note:
1. Li/NH3 gives trans, Lindlar gives cis.
2. Na/NH3 reduction proceeds through a radical anion.

Answer: (b) 2-methyl-2-pentene

Explanation:
1. Acetone comes from \(\ce{(CH3)2C=}\) and propionaldehyde from \(\ce{=CH-CH2CH3}\).
2. Joining the fragments: \(\ce{(CH3)2C=CH-CH2CH3}\), which is 2-methylpent-2-ene.

Teacher's Note:
1. Join the carbonyl carbons with a double bond to rebuild the alkene.
2. Reductive ozonolysis gives carbonyl compounds.

Answer: (a) alkanes

Explanation:
1. The Corey-House synthesis: \(\ce{R2CuLi + R'X -> R-R' + RCu + LiX}\).
2. The product is an alkane.

Teacher's Note:
1. This method makes unsymmetrical alkanes.
2. R2CuLi is a Gilman reagent.

Answer: (c) 2,3-dimethylbutane

Explanation:
1. 2,3-Dimethylbutane has two types of hydrogen: the 12 methyl hydrogens and the 2 tertiary hydrogens.
2. So it gives two monochloro products. 2,2-dimethylbutane gives three, n-hexane three and 2-methylpentane five.

Teacher's Note:
1. Count the types of equivalent hydrogens.
2. Symmetry reduces the number of products.

Answer: (b) 1-bromo-2-butene under thermodynamically controlled conditions

Explanation:
1. At low temperature (-80 degrees C) the 1,2-adduct, 3-bromobut-1-ene, is the kinetic product.
2. At higher temperature (40 degrees C) the more stable 1,4-adduct, 1-bromobut-2-ene, is the thermodynamic product and predominates.

Teacher's Note:
1. Low temperature gives the kinetic product; high temperature gives the thermodynamic product.
2. The 1,4-adduct has a more substituted double bond.

Answer: (a) Pd/H2

Explanation:
1. Catalytic hydrogenation over Pd (or Pt/Ni) adds H2 across the double bond.
2. Zn/HCl and Sn/HCl do not reduce alkenes, and Lindlar catalyst is poisoned for alkynes.

Teacher's Note:
1. Lindlar catalyst stops at the alkene stage for alkynes.
2. Alkene to alkane needs catalytic hydrogenation.

Answer: (d) III > II > IV > I

Explanation:
1. Boiling point rises with molecular mass, and among isomers it falls with branching.
2. n-Pentane (309 K) > 2-methylbutane (301 K) > neopentane (283 K) > n-butane (273 K).

Teacher's Note:
1. More surface area means stronger dispersion forces.
2. A spherical shape has the lowest boiling point.

Answer: (d) \(\ce{I2 < Br2 < Cl2 < F2}\)

Explanation:
1. Reactivity of halogens in free radical halogenation is F2 > Cl2 > Br2 > I2.
2. Fluorination is violent, iodination is endothermic and reversible.

Teacher's Note:
1. Reactivity falls as the halogen gets bigger.
2. Iodination needs an oxidising agent to remove HI.

Answer: (b) \(\ce{R-Cl < R-Br < R-I}\)

Explanation:
1. Reduction of alkyl halides by Zn/HCl depends on the ease of C-X bond cleavage.
2. Bond strength decreases in the order C-Cl > C-Br > C-I, so the reactivity increases in the order R-Cl < R-Br < R-I.

Teacher's Note:
1. Weaker bond means easier reduction.
2. Alkyl iodides are the most reactive.

Answer: (b) II > I > III

Explanation:
1. Carbanions on sp carbons (50 per cent s character) are more stable than on sp3 carbons.
2. Between the two acetylides, the methyl group in I donates electrons and destabilises the anion, so II > I > III.

Teacher's Note:
1. More s character stabilises the carbanion.
2. Electron-donating groups destabilise carbanions.

Answer: (d) \(\ce{CH4 + O2 -> C + 2H2O}\)

Explanation:
1. Incomplete combustion gives carbon (soot) or CO instead of CO2.
2. \(\ce{CH4 + O2 -> C + 2H2O}\) produces carbon black.
3. (a) and (b) are controlled oxidations and (d) is complete combustion.

Teacher's Note:
1. Complete combustion gives CO2 and H2O only.
2. Carbon black is used in making ink and tyres.

Answer: (c) Victor Meyer test

Explanation:
1. Hydration of acetylene gives acetaldehyde, P = CH3CHO.
2. It gives Tollens' test (aldehyde), Brady's test (carbonyl) and the iodoform test (CH3CO- group).
3. The Victor Meyer test is used to distinguish 1, 2 and 3 degree alcohols and is not given by it.

Teacher's Note:
1. Acetaldehyde is the only aldehyde that gives the iodoform test.
2. Victor Meyer's test is for alcohols.

Answer: (d) Maleic acid

Explanation:
1. Maleic acid (cis-butenedioic acid) is planar and symmetrical, with no chiral centre.
2. Lactic acid, tartaric acid and most alpha-amino acids have chiral carbons.

Teacher's Note:
1. Maleic and fumaric acid show only geometrical isomerism.
2. Glycine is the achiral amino acid.

Answer: (c) \(\ce{CH3COCH3 + CH3COOH}\)

Explanation:
1. The Lemieux-von Rudloff oxidation cleaves the double bond and oxidises the fragments.
2. The \(\ce{(CH3)2C=}\) end gives acetone (ketone) and the \(\ce{=CHCH3}\) end gives acetic acid (oxidised further).

Teacher's Note:
1. Aldehydes are oxidised to acids under these conditions.
2. Compare with ozonolysis (reductive), which gives acetaldehyde.

Answer: (d) branched hydrocarbon

Explanation:
1. Branched hydrocarbons (for example iso-octane) have a high octane number and burn smoothly without knocking.

Teacher's Note:
1. Octane number of iso-octane is 100; n-heptane is 0.
2. Branching reduces knocking.

Answer: (b) 6

Explanation:
1. Each sp2 carbon (2 carbons) has 1 unhybridised p orbital, which gives 2.
2. Each sp carbon (2 carbons) has 2 unhybridised p orbitals, which gives 4.
3. Total = 6.

Teacher's Note:
1. sp carbon: 2 p orbitals left; sp2: 1 left; sp3: none.
2. These form the pi bonds.

Answer: (d) 3

Explanation:
1. The internal alkynes of C6H10 are hex-2-yne, hex-3-yne and 4-methylpent-2-yne.
2. Each gives one trans alkene, so there are 3.

Teacher's Note:
1. Li/NH3 reduction is anti addition giving trans alkenes.
2. Terminal alkynes give no stereoisomers.

Answer: (a) 1

Explanation:
1. Hydration of acetylene gives acetaldehyde, \(\ce{CH3CHO}\).
2. It has one C=O double bond, which contains 1 pi bond.

Teacher's Note:
1. Kucherov reaction: HgSO4/H2SO4 hydration.
2. The enol first formed tautomerises to the aldehyde.