JEE Chemistry Hydrocarbons MCQs Set 07

Answer: (b) cis-stilbene

Explanation:
1. Chlorine adds to give 1,2-dichloro-1,2-diphenylethane (X).
2. Two moles of \(\ce{NaNH2}\) remove 2 HCl to give diphenylacetylene (Y).
3. Partial hydrogenation over the poisoned catalyst adds hydrogen from the same face, giving cis-stilbene (Z).

Teacher's Note:
1. Poisoned palladium (Lindlar-type) gives cis alkenes.
2. The catalyst stops at the alkene stage.

Answer: (c) cyclic halonium ion

Explanation:
1. The halogen molecule is polarised by the pi bond, and one halogen forms a bridged (three-membered) cyclic halonium ion.
2. The halide ion then attacks from the opposite side, giving anti addition.

Teacher's Note:
1. A bridged halonium ion explains anti addition.
2. This is why no rearrangement is seen in halogen addition.

Answer: (b) a ketone

Explanation:
1. Hydration of propyne is Markovnikov and gives the enol \(\ce{CH3-C(OH)=CH2}\).
2. The enol tautomerises to propanone, \(\ce{CH3COCH3}\), which is a ketone.

Teacher's Note:
1. Every alkyne except ethyne gives a ketone on hydration.
2. Ethyne gives ethanal.

Answer: (b) 3

Explanation:
1. The alkene must have an unbranched six carbon chain.
2. The double bond can be at position 1, 2 or 3: hex-1-ene, hex-2-ene and hex-3-ene.
3. Geometrical isomers are not counted as structural isomers, so the answer is 3.

Teacher's Note:
1. Only the position of the double bond varies in a straight chain.
2. Cis and trans forms are stereoisomers.

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

Explanation:
1. Electrophilic addition forms a benzylic carbocation, so electron-donating groups on the ring speed up the reaction.
2. The \(\ce{-OCH3}\) group is the strongest donor (+M), then \(\ce{-CH3}\) (+I, hyperconjugation).
3. Chlorine is weakly deactivating and \(\ce{-NO2}\) is strongly deactivating, giving I > II > III > IV.

Teacher's Note:
1. Rate depends on the stability of the carbocation intermediate.
2. Donors stabilise, acceptors destabilise.

Answer: (d) III > II > I

Explanation:
1. Heat of combustion rises with the number of carbon atoms, so propene (3 carbons) is lowest.
2. Between the two isomeric butenes, the less stable isomer releases more heat.
3. 2-Methylpropene is the more stable of the two (more hyperconjugation), so trans-but-2-ene has the higher heat of combustion. Order: III > II > I.

Teacher's Note:
1. Isomers differ in energy: the less stable one releases more heat on burning.
2. More carbons also means more heat.

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

Explanation:
1. Acid-catalysed hydration starts by protonation to a carbocation, so the more stable the cation the faster the reaction.
2. The methoxy group stabilises the cation strongly by delocalisation of its lone pair (+M), and styrene gives a benzylic cation that is also delocalisation stabilised.
3. Propene gives a simple secondary cation, and fluorine withdraws electrons strongly (-I), so II is least reactive. Order: I > IV > III > II.

Teacher's Note:
1. Delocalisation-stabilised carbocations form fastest.
2. Strongly electron-withdrawing groups slow electrophilic addition.

Answer: (c) Friedel-Crafts reaction

Explanation:
1. Acylation of an aromatic ring with an acyl halide and a Lewis acid catalyst is the Friedel-Crafts acylation, giving acetophenone.
2. Wurtz-Fittig couples an aryl and alkyl halide with sodium, Gattermann gives aryl halides from diazonium salts or aldehydes, and Schotten-Baumann acylates amines or alcohols.

Teacher's Note:
1. Alkyl or acyl halide with AlCl3 means Friedel-Crafts.
2. Acylation does not rearrange.

Answer: (c) 4

Explanation:
1. Reduction by \(\ce{Zn/HCl}\) replaces the halogen by hydrogen without changing the carbon skeleton.
2. Isopentane (2-methylbutane) has four types of C - H, so there are four monohalides: 1-halo-2-methylbutane, 2-halo-2-methylbutane, 2-halo-3-methylbutane and 1-halo-3-methylbutane.
3. So four halides.

Teacher's Note:
1. Count the number of different monohalides of the alkane.
2. Zn/HCl reduction keeps the skeleton intact.

Answer: (d) pentane

Explanation:
1. A mixed Wurtz reaction gives three products: \(\ce{C2H5-C2H5}\) (butane), \(\ce{(CH3)2CH-CH(CH3)2}\) (2,3-dimethylbutane) and the cross product \(\ce{C2H5-CH(CH3)2}\) (2-methylbutane).
2. Pentane would need a three-carbon straight chain joined to a two-carbon chain, which cannot form from these halides.

Teacher's Note:
1. A mixture of two halides gives three alkanes.
2. Branching in the halide is retained in the product.

Answer: (c) homologues

Explanation:
1. Soda lime decarboxylation removes the COOH carbon: \(\ce{CH3CH2CH2COOH -> CH3CH2CH3}\) (propane).
2. Red P and HI reduce the acid to the alkane with the same number of carbons: butane.
3. Propane and butane differ by a \(\ce{CH2}\) group, so they are homologues.

Teacher's Note:
1. Decarboxylation loses one carbon, reduction keeps all carbons.
2. Homologues differ by CH2.

Answer: (c) sodium acetate heated with soda lime

Explanation:
1. Heating sodium acetate with soda lime removes \(\ce{CO2}\) and gives methane, \(\ce{CH3COONa + NaOH -> CH4 + Na2CO3}\).
2. Kolbe electrolysis of acetate gives ethane, and Wurtz coupling of methyl bromide also gives ethane.
3. The Wurtz reaction of ethyl bromide gives butane.

Teacher's Note:
1. Soda lime decarboxylation shortens the chain by one carbon.
2. Coupling reactions double the alkyl group.

Answer: (a) \(\ce{HIO3}\)

Explanation:
1. Iodination is reversible, because the HI formed reduces the alkyl iodide back to the alkane.
2. An oxidising agent such as \(\ce{HIO3}\) destroys the HI as it forms (\(\ce{5HI + HIO3 -> 3I2 + 3H2O}\)).
3. This drives the reaction forward.

Teacher's Note:
1. Iodination needs an oxidising agent to remove HI.
2. HIO3 or HNO3 are used for this.

Answer: (b) isopentane

Explanation:
1. A tertiary C - H bond can be oxidised by \(\ce{KMnO4}\) to a tertiary alcohol.
2. Isopentane (2-methylbutane) has a tertiary hydrogen and gives 2-methylbutan-2-ol.
3. The other alkanes have no tertiary C - H bond.

Teacher's Note:
1. Only alkanes with tertiary hydrogen give alcohols with KMnO4.
2. Neopentane has no tertiary hydrogen.

Answer: (b) HCHO

Explanation:
1. Controlled oxidation of methane with oxygen in the presence of molybdenum oxide gives methanal (formaldehyde).
2. With a copper tube at high pressure the product is methanol instead.

Teacher's Note:
1. Catalyst decides the product of controlled oxidation.
2. Mo2O3 gives HCHO.

Answer: (c) synthesis gas

Explanation:
1. Steam reforming of methane over nickel at 1273 K gives \(\ce{CO + 3H2}\).
2. This mixture is called synthesis gas (syngas), used for making methanol.
3. Water gas (\(\ce{CO + H2}\), 1:1) is made from coke and steam, and producer gas is \(\ce{CO + N2}\).

Teacher's Note:
1. The ratio CO : H2 = 1 : 3 is synthesis gas.
2. Water gas has equal CO and H2.

Answer: (a) 2-bromo-2-methyl-1-phenylpropane and 1-bromo-2-methyl-1-phenylpropane

Explanation:
1. With peroxide, the bromine radical adds to the \(\ce{C(CH3)2}\) carbon to give the more stable benzylic radical, so A is \(\ce{C6H5CH2-C(CH3)2Br}\).
2. Without peroxide the proton adds to the \(\ce{C(CH3)2}\) carbon to give the benzylic carbocation, so B is \(\ce{C6H5-CHBr-CH(CH3)2}\).
3. So A and B are the compounds in option (a).

Teacher's Note:
1. Radical addition puts Br where the more stable radical forms.
2. Ionic addition puts H where the more stable cation forms.

Answer: (c) 2,3-dimethylbutan-2-ol

Explanation:
1. Protonation gives a secondary carbocation next to a quaternary carbon.
2. A methyl shift gives the more stable tertiary carbocation \(\ce{(CH3)2C+-CH(CH3)2}\).
3. Water adds to it and the loss of a proton gives 2,3-dimethylbutan-2-ol.

Teacher's Note:
1. A secondary carbocation next to a quaternary carbon rearranges by methyl shift.
2. This is the typical pinacolyl-type rearrangement.

Answer: (b) \(\ce{CH2=C=CH-CH3}\)

Explanation:
1. In a propadiene-type compound the central carbon is joined to its neighbours by two double bonds.
2. Ozonolysis cleaves both, so the central carbon ends up bonded only to oxygen atoms, as \(\ce{CO2}\).
3. The other three compounds give only aldehydes (HCHO, \(\ce{CH3CHO}\), \(\ce{OHC-CHO}\)).

Teacher's Note:
1. The central carbon of an propadiene-type compound becomes CO2.
2. Aldehydes come from =CHR ends.

Answer: (a) 3-ethylpent-2-ene

Explanation:
1. Join the carbonyl carbon of propanal, \(\ce{CH3CH2CH=O}\), to the carbonyl carbon of pentan-3-one, \(\ce{(CH3CH2)2C=O}\), by a double bond.
2. This gives \(\ce{CH3CH2CH=C(CH2CH3)2}\), which is 3-ethylpent-2-ene.

Teacher's Note:
1. Reverse ozonolysis: join the carbonyl carbons.
2. Count the carbons to check: 3 + 5 = 8 carbons.

Answer: (d) catalytic hydrogenation of acetylene

Explanation:
1. Sodium benzoate with soda lime, phenol with zinc dust, and benzene diazonium chloride with \(\ce{H3PO2}\) all give benzene.
2. Hydrogenation of acetylene adds hydrogen and gives ethene or ethane, not benzene.
3. Benzene from acetylene needs cyclic trimerisation through a red hot tube.

Teacher's Note:
1. Hydrogenation never forms a ring.
2. Trimerisation of ethyne gives benzene.

Answer: (c) mesitylene

Explanation:
1. Ethylbenzene, cumene and sec-butylbenzene have a single side chain with a benzylic hydrogen, so each gives benzoic acid.
2. Mesitylene (1,3,5-trimethylbenzene) has three methyl groups, and all are oxidised to give benzene-1,3,5-tricarboxylic acid (trimesic acid), not benzoic acid.

Teacher's Note:
1. One alkyl group gives benzoic acid.
2. Several groups give polycarboxylic acids.

Answer: (b) acetophenone

Explanation:
1. Acetic anhydride is an acylating agent like acetyl chloride in Friedel-Crafts acylation.
2. The acylium ion \(\ce{CH3CO+}\) substitutes a hydrogen of benzene to give methyl phenyl ketone (acetophenone).

Teacher's Note:
1. Acid anhydrides also give Friedel-Crafts acylation.
2. Product keeps the CH3-C=O group.

Answer: (b) salicylic acid

Explanation:
1. Phenylacetic acid gives toluene, o-toluic acid gives toluene, and benzoic acid gives benzene on soda lime decarboxylation.
2. Salicylic acid (2-hydroxybenzoic acid) loses \(\ce{CO2}\) to give phenol, which is not a hydrocarbon.

Teacher's Note:
1. A hydrocarbon contains only carbon and hydrogen.
2. Decarboxylation leaves other substituents in place.