JEE Chemistry Hydrocarbons MCQs Set 04

Answer: (b) \( R\text{-}I > R\text{-}Br > R\text{-}Cl \)

Explanation:
1. The C-X bond becomes weaker and the halide more reactive with sodium from Cl to I.

Teacher's Note:
1. Same trend as the leaving-group ability.

Answer: (a) butane, 2,5-dimethylhexane and 2-methylpentane

Explanation:
1. Propanoate gives an ethyl radical and 3-methylbutanoate \( (CH_3)_2CHCH_2COO^- \) gives an isobutyl radical.
2. Et-Et is butane, iBu-iBu is 2,5-dimethylhexane and Et-iBu is 2-methylpentane (isohexane).

Teacher's Note:
1. A mixture of two salts always gives three alkanes.

Answer: (d) lithium, then cuprous iodide, and then reacting the product with another alkyl halide

Explanation:
1. \( RX + 2Li \to RLi \); \( 2RLi + CuI \to R_2CuLi \); \( R_2CuLi + R'X \to R-R' \).

Teacher's Note:
1. This gives unsymmetrical alkanes.

Answer: (c) 2,2-dimethylpropanoic acid

Explanation:
1. \( (CH_3)_3C\text{-}COOH \to (CH_3)_3CH + CO_2 \). Isobutane is 2-methylpropane, which has 4 carbons, so the acid has 5.

Teacher's Note:
1. The acid has one carbon more than the alkane.

Answer: (b) isobutane

Explanation:
1. Wurtz coupling of two identical alkyl groups gives symmetrical even-carbon alkanes. Isobutane would need a (CH3)2CH group joined to CH3, giving a mixture or an alkene.

Teacher's Note:
1. Branching at the halide carbon gives elimination.

Answer: (a) isobutane

Explanation:
1. Products are Et-Et (butane), Et-Pr (pentane) and Pr-Pr (hexane). Isobutane is branched and cannot form.

Teacher's Note:
1. The carbon skeleton remains straight chain.

Answer: (d) n-octane

Explanation:
1. Boiling point increases with molar mass and decreases with branching. n-Octane is the longest unbranched alkane here.

Teacher's Note:
1. 2,2,3,3-Tetramethylbutane is very compact and has a lower boiling point.

Answer: (c) free radical elimination reaction (cracking)

Explanation:
1. At high temperature C-C and C-H bonds break homolytically to give smaller alkanes and alkenes.

Teacher's Note:
1. Cracking is used to make petrol.

Answer: (b) tetraethyl lead (TEL)

Explanation:
1. TEL, Pb(C2H5)4, raises the octane number of petrol.

Teacher's Note:
1. Unleaded petrol uses other additives because lead is toxic.

Answer: (a) \( CH_4 + O_2 \to C + 2H_2O \)

Explanation:
1. Limited oxygen gives carbon (soot); the other equations show complete combustion and controlled oxidations to methanol and methanal.

Teacher's Note:
1. Incomplete combustion gives a smoky flame.

Answer: (d) 2-methylpropanal and methanal

Explanation:
1. \( (CH_3)_2CH\text{-}CH=CH_2 \xrightarrow{O_3;\,Zn/H_2O} (CH_3)_2CHCHO + HCHO \).

Teacher's Note:
1. Each end of the double bond becomes a C=O.

Answer: (c) 2,3-dimethylbut-2-ene

Explanation:
1. \( (CH_3)_2C=C(CH_3)_2 \to 2(CH_3)_2C=O \). The molecule is symmetric about the double bond.

Teacher's Note:
1. A symmetric tetrasubstituted alkene gives one ketone.

Answer: (b) acetone and carbon dioxide

Explanation:
1. \( (CH_3)_2C=CH_2 \xrightarrow{KMnO_4/H^+} (CH_3)_2C=O + CO_2 + H_2O \): the =CH2 group is oxidised through formic acid to CO2.

Teacher's Note:
1. Terminal CH2 becomes CO2 with hot acidic permanganate.

Answer: (a) ozonolysis

Explanation:
1. Ozonolysis splits the molecule at the double bond; the carbonyl products show where it was.

Teacher's Note:
1. Identify the fragments and join their carbonyl carbons.

Answer: (d) ethanal and methanal

Explanation:
1. A is propene \( CH_3CH=CH_2 \). Ozonolysis gives \( CH_3CHO + HCHO \).

Teacher's Note:
1. Elimination first, ozonolysis second.

Answer: (c) trans-2-butene

Explanation:
1. Heat of hydrogenation measures alkene energy: the more stable the alkene the smaller it is. trans-2-Butene is a disubstituted alkene with the least strain.

Teacher's Note:
1. Stability: trans-2-butene > cis-2-butene > 1-butene.

Answer: (b) but-2-ene

Explanation:
1. Three C-C sigma bonds means four carbons; with one pi bond the formula is C4H8.
2. Two moles of aldehyde of mass 44 (ethanal) come from the symmetric but-2-ene.

Teacher's Note:
1. CH3CH=CHCH3 gives 2 CH3CHO.

Answer: (a) pent-1-ene

Explanation:
1. Only terminal alkynes (acidic C-H) give a white precipitate of the silver acetylide. Alkenes do not.

Teacher's Note:
1. Hint: Tollens' test among hydrocarbons is for terminal alkynes.

Answer: (d) \( C_2H_2 \) and \( CO_2 \)

Explanation:
1. Maleate \( \text{-OOC-CH=CH-COO}^- \) loses two \( CO_2 \) to give acetylene, \( HC\equiv CH \).

Teacher's Note:
1. Hydrogen is produced at the cathode.

Answer: (c) increases with time

Explanation:
1. At the cathode water is reduced: \( 2H_2O + 2e^- \to H_2 + 2OH^- \), so the solution becomes more basic.

Teacher's Note:
1. Anode products are hydrocarbon and CO2.

Answer: (b) ethyne

Explanation:
1. Terminal alkynes form copper(I) acetylide: \( HC\equiv CH + 2[Cu(NH_3)_2]^+ \to CuC\equiv CCu\downarrow \) (red).

Teacher's Note:
1. Ammoniacal silver nitrate gives a white precipitate.

Answer: (a) ammoniacal silver nitrate

Explanation:
1. Both are unsaturated and react with bromine, permanganate and ozone, but only propyne (terminal alkyne) gives a white silver acetylide.

Teacher's Note:
1. Terminal alkyne test.

Answer: (d) \( C_6H_5MgBr + H_3O^+ \)

Explanation:
1. Grignard reagents are protonated by water to give the hydrocarbon. The others give phenols.

Teacher's Note:
1. Benzenediazonium chloride gives benzene only with H3PO2.

Answer: (c) tert-butylbenzene with acidic \( KMnO_4 \)

Explanation:
1. A benzylic hydrogen is needed for side-chain oxidation. tert-Butylbenzene has none.

Teacher's Note:
1. Ethylbenzene and cumene give benzoic acid.

Answer: (b) Clemmensen reduction

Explanation:
1. Zn-Hg with conc. HCl is the Clemmensen reduction; hydrazine with KOH is Wolff-Kishner.

Teacher's Note:
1. Use Clemmensen for base-sensitive and Wolff-Kishner for acid-sensitive compounds.