Hydrocarbons MCQs

The IUPAC name of (CH3)3C-CH2-CH3 is 2,2-dimethylbutane.

Longest continuous carbon chain: 4 carbons (butane)

Numbering: Numbering from the left gives the lowest locant numbers (2,2 vs. 3,3)

Substituents: Two methyl groups attached at carbon-2 (2,2-dimethyl)

Homolytic cleavage of a diatomic molecule by heat or light generates radicals that start the chain.

In free‑radical chlorination, the first radicals must be chlorine atoms.

Thus the initiation step is the homolysis of chlorine gas into two chlorine radicals.

Identify the longest continuous carbon chain – it contains five carbons, so the parent hydrocarbon is a pentene.

Number the chain to give the double bond the lowest possible locant; numbering from the left gives the C=C between C‑2 and C‑3, i.e., the double bond at carbon 2.

Hence the IUPAC name is Pent-2-ene.

Identify the β‑hydrogen atoms on C‑1 and C‑3 of 2‑bromobutane; removal with base can give a C1=C2 or C2=C3 double bond.

The C2=C3 alkene (but‑2‑ene) is disubstituted, whereas the C1=C2 alkene (but‑1‑ene) is only monosubstituted.

Saytzeff’s rule favors the more substituted alkene, so the major product is but‑2‑ene.

In acid‑catalysed hydration the double bond is protonated to give the more stable carbocation; for propene this is the secondary carbocation at C‑2. Water attacks this carbocation and after deprotonation yields the corresponding alcohol. Hence the product formed is Propan‑2‑ol.

Calculate DU = (2C + 2 − H)/2 for a hydrocarbon.

Substitute C = 6 and H = 12: (2·6 + 2 − 12)/2 = (12 + 2 − 12)/2 = 2/2.

The degree of unsaturation equals 1, meaning one ring or one pi bond.

Answer: 1

The first elimination with a base gives a vinyl halide (haloalkene).

A second elimination requires a very strong, non‑nucleophilic base to remove the remaining β‑hydrogen from the sp² carbon.

Sodium amide (NaNH₂) in liquid ammonia provides the needed strong base, effecting the second dehydrohalogenation to form the alkyne. Sodium amide (NaNH₂)

Propyne undergoes electrophilic addition of H⁺ (from H₂SO₄) in the presence of Hg²⁺ to give the more substituted vinyl cation, which after water attack forms an enol (CH₃‑C(OH)=CH₂).

The enol rapidly tautomerises to the corresponding carbonyl compound, shifting the double bond and a proton to give the more stable ketone.

Thus the product of hydration of CH₃‑C≡CH is propanone (acetone).

UV light causes homolytic cleavage of Cl₂ giving two Cl· radicals.

The chlorine radical abstracts H from CH₄ forming CH₃· and HCl, propagating the radical chain.

Both propagation steps involve radicals, characteristic of a free‑radical substitution mechanism.

Open‑chain hydrocarbon with one C=C double bond is an alkene, whose general formula is CₙH₂ₙ.

For five carbon atoms, n = 5, so the hydrogen count is 2×5 = 10, giving C₅H₁₀.

Thus the molecular formula is C5H10.

Strong base removes the β‑hydrogen while the leaving group departs in a concerted E2 step.

A non‑aqueous medium prevents competing SN1/SN2 substitution, so KOH dissolved in ethanol supplies OH⁻ without water.

Hence dehydrohalogenation of an alkyl halide to form an alkene requires alcoholic KOH.

Proton adds to the double bond giving the more substituted carbon a positive charge.

The intermediate formed is a positively charged carbon species.

Stability increases with greater substitution, favoring the more substituted carbocation.

Thus the governing intermediate in Markovnikov addition is a carbocation.

Moles of ethene given = 0.25 mol.

Molar volume at STP = 22.4 L per mole.

Volume = moles × molar volume = 0.25 × 22.4 L = 5.6 L.

5.6 L

Vicinal dihalides undergo successive elimination of HX; the first equivalent of alcoholic KOH removes one halogen and a β‑hydrogen to give a vinyl halide (alkene).

A second elimination is required to remove the remaining HX; a strong base such as NaNH₂ abstracts the acidic alkyne hydrogen and eliminates the second halide, furnishing the alkyne.

Thus the conversion is achieved by dehydrohalogenation with alcoholic KOH followed by NaNH₂.

Dilute H₂SO₄/HgSO₄ adds water across the C≡C bond of ethyne to form the enol CH₃CH=OH.

The enol is unstable and rapidly undergoes keto‑enol tautomerism, converting the –C=OH group to a carbonyl.

Thus the product is ethanal (acetaldehyde).

Each carbon in benzene is sp² hybridised, forming three σ‑bonds (two C–C and one C–H).

The remaining unhybridised p‑orbitals overlap to give a continuous π‑electron cloud over the ring.

This delocalisation makes every C–C bond have a bond order of 1.5, giving equal lengths of about 139 pm – Delocalisation of pi electrons (resonance).

Identify the longest continuous carbon chain: six carbons → hexane backbone.

Number the chain from the end that gives the substituent the lowest possible locant; numbering from the left gives a methyl group on carbon 3.

Thus the compound is named 3‑methylhexane.

Complete oxidation of an alkane requires enough O₂ to convert every C atom to CO₂ and every H atom to H₂O.

For CₙH₂ₙ₊₂, the balanced equation is CₙH₂ₙ₊₂ + (3n+1)/2 O₂ → n CO₂ + (n+1) H₂O.

Since all carbon and hydrogen are fully oxidized, the products are carbon dioxide and water.

CO₂ and H₂O.

Identify the longest carbon chain containing the C=C bond; it has three carbons → propene backbone.

Number the chain so the double bond gets the lowest possible locant; numbering from the end nearer the C=C gives the double bond at carbon 1.

The remaining methyl group is attached to carbon 2, giving a 2‑methyl substituent.

Hence the IUPAC name is 2-Methylprop-1-ene.

Zn dust reduces the C–Br bonds, forming ZnBr₂ and generating a carbanion intermediate at each carbon.

The adjacent carbanions eliminate the two bromide ions, forming a C=C double bond between the two carbons.

Thus 1,2‑dibromoethane is converted to ethene.