JEE Advanced (Subjective Type Questions): Hydrocarbons- 1 | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q. 1. Give one characteristic test which would distinguish. CH₄ from C₂H₂ (1979) 

Ans. Sol.  Bromine water test : C₂H₂ decolourises bromine water while CH₄ does not decolourises bromine water.

Q. 2. Write the structural formula of the major product in each of the following cases : (i) the compound obtained by the hydration of ethyne is treated with dilute alkali (1981 - ½ Mark) (ii) ethene mixed with air is passed under pressure over a silver catalyst at 250ºC. (1981 - ½ Mark)

(2000 - 1 Mark) 

Ans. Sol. 

(i) 

NOTE : that the 1° carbocation,  formed during reaction rearranges to the more stable, 3° carbocation,  and hence the above product isformed.] (see also ix part)

Explanation :

(vii) 

Q. 3. Outline the reaction sequence for the conversion of ethene to ethyne (the number of steps should not be more than two). (1981 - 1 Mark) 

Ans. Sol. 

Q. 4. State with balanced equations, what happens when propene is bubbled through a hot aqueous solution of potassium permanganate. (1982 - 1 Mark) 

Ans. Sol. 

[NOTE : Colour of KMnO₄ is discharged]

Q. 5. Give reasons for the following : (i) Methane does not react with chlorine in the dark. (1983 - 1 Mark)
 (ii) Propene reacts with HBr to give isopropyl bromide but does not give n-propyl bromide. (1983 - 1 Mark)
 (iii) Although benzene is highly unsaturated, normally it does not undergo addition reaction. (1983 - 1 Mark)
 (iv) Toluene reacts with bromine in the presence of light to give benzyl bromide while in presence of FeBr₃ it gives p-bromotoluene. Give explanation for the above obser vations. (1996 - 2 Marks)
 (v) The central carbon-carbon bond in 1, 3 – butadiene is shorter than that in n-butane. (1998 -  2 Marks)
 (vi) tert-Butylben zen e does n ot give ben zoic acid on treatment with acidic KMnO₄. (2000 -  1 Mark)

(2005 - 1 Mark)

Ans. Sol. (i) TIPS/Formulae : Chlorination of methane is a free radical substitution reaction.
In dark, chlorine is unable to be converted into free radicals, hence the reaction does not occur.

(ii) TIPS/Formulae : Addition of unsymmetrical addendum (HBr in present case) to unsymmetrical olefin (CH₃CH = CH₂, in present case) takes place according to Markownikoff rule.

(iii) Unlike olefins, p-electrons of benzene are delocalised (resonance) and hence these are unreactive towards addition reactions. Moreover, addition reaction leads to destruction of the benzenoid ring. (iv) In presen ce of li gh t, toluene un dergoes side chain bromination through a free radical mechanism.

In  presence of FeBr₃, toluene undergoes electrophilic substitution in the benzene ring.

[NOTE :–CH₃ is o-, p-directing]

(v) TIPS/Formulae : 1, 3 - Butadiene is a conjugated diene and is a reasonance hybrid:

Thus resonance induces some double bond character in the central C-C bond leading to the shortening of this bond. Alternatively, all the four C atoms of 1, 3– butadiene are sp²  hybridised and thus their C – C bond length will be lower than that of n- butane in which all the four C atoms are sp³ hybridised.
(vi) tert-Butyl benz en e does n ot gi ve ben zoic a cid on treatment with acidic KMnO₄ because it does not contain any hydrogen atom on the key carbon atom.
(vii) Reduction of cental ring to form A involves reduction of all the three cyclobutadiene rings (which are antiaromatic as they have 4p electrons each), i.e. antiaromatic rings are converted into nonaromatic rings.
On the other hand, reduction of the terminal ring to form B involves reduction of only one antiaromatic ring.
Remember that antiaromatic rings impart unstability.

Q. 6. (i) 2-Methylpropene can be converted into isobutyl bromide by hydrogen bromide, is true under what condition s? (1984 - 1 Mark)
 (ii) 'Ethyne and its derivatives will give white precipitate with ammonical silver nitrate solution', is true under what conditions. (1984 - 1 Mark) 

Ans. Sol. (i) NOTE : Under normal conditions, ter-butyl bromide is formed, isobutyl bromide is formed in presence of peroxide.

(ii) Ethyne (HC ≡ CH) and only those derivatives which have at least one acetylenic hydrogen atom (≡ C – H) i.e. terminal alkynes will give white precipitate with ammonical silver nitrate solution.

Q. 7. Write down the reactions involved in the preparation of the following, using the reagents indicated against it in parenthesis.
 Ethylbenzene from benzene [C₂H₅OH, PCl₅, anhydrous AlCl₃]. (1984 - 2 Marks)

Ans. Sol. 

Q. 8. A certain hydrocarbon A was found to contain 85.7 percent carbon and 14.3 per cent hydrogen. This compound consumes 1 molar equivalent of hydrogen to give a saturated hydrocarbon B. 1.00 g of hydrocarbon A just decolourized 38.05 g of a 5 per cent solution (by weight) of Br₂ in CCl₄. Compound A, on oxidation with concentrated KMnO₄, gave compound C (molecular formula C₄H₈O) and acetic acid. Compound C could easily be prepared by the action of acidic aqueous mercuric sulphate on 2- butyne. Determine the molecular formula of A and deduce the structure of A, B and C. (1984 - 6 Marks)

Ans. Sol. Calculation of molecular formula of A.

∴ Empirical formula of A = CH₂ Determination of molecular weight of A  
1g of A consumes = 38.05 g of 5% Br₂  (in CCl₄)

  = 1.90 g of 100% Br₂.
Now since 1.90 g of Br₂ is consumed by 1 g of compound A
∴ 160g (1 mole) of Br₂  will be consumed by

= 84.2 g of  A = 84.0 (app.)g of A 

∴ Molecular weight of   A = 84

∴ Molecular formula of  A = (CH₂)₆ = C₆H₁₂

Since the hydrocarbon A consumes 1 molar equivlaent of hydrogen, it must contain one double bond. Oxidation of compound A with KMnO₄ to form compound C (C₄H₈O) and acetic acid indicates = CH.CH₃ fragment in A, i.e.

Now the fragment C₄H₈ of A on oxidation forms the compound 'C' (C₄H₈O) which may be easily obtained from butyne-2 and acidic aq. HgSO₄, the compound 'C' must be ethylmethyl ketone.

The formation of ketone 'C' from C₄H₈ fragment of 'A' can be explained by the following structure of A

Hence formation of 'B' can be represented as below.

Q. 9. How would you distinguish between (i) 2-butyne and 1-butyne. (1985 - 1 Mark) (ii) cyclohexane and cyclohexene. (1988 - 1 Mark)

Ans. Sol. (i) By amm. AgNO₃ or by acidic-H tests : Terminal alkynes give white precipitate with amm. AgNO₃ or red ppt. with amm. Cu₂Cl₂ (H atom attached on sp hybridized carbon is acidic). 2CH₃CH₂C ≡ CH + Ag₂O → 2CH₃CH₂C ≡ CAg + H₂O
CH₃ – C ≡ C – CH₃ + Ag₂O → No reaction
NOTE : Only terminal alkynes respond to these reactions. (ii) Cyclohexene gives positive response to bromine water test and Baeyer’s test while cyclohexane does not respond to these reagents. 

Q. 10. n-Butane is produced by the monobromination of ethane followed by the Wurtz reaction. Calculate the volume of ethane at NTP required to produce 55 g n-butane, if the bromination takes place with 90 per cent yield and the Wurtz reaction with 85 per cent yield. (1989 - 3 Marks) 

Ans. Sol.     (yield 90%) (given)

  (yield 85%) (given)Moles of n-butane to be produced

= 0.948 mol   (∵ molecular mass of C₄H₁₀ = 58)

Amount of C₂H₅Br required to obtain 0.948 mol. of C₄H₁₀ = 2 × 0.948 mol.

Hence, the amount of C₂H₅Br required

  ...(1)    [∵ yield is 85% only]

Further 1 mole of C₂H₆ gives one mole of C₂H₅Br, hence number of moles of C₂H₆ reqd. for C₂H₅Br in (1)

= 2.48 mol    [∵ yield is 90%]

∴ Required volume of ethane at NTP = 22400 × 2.48 = 55552 ml. = 55.55 litres

Q. 11. Identify, B(C₄H₈) which adds on HBr in the presence and in the absence of peroxide to give the same product, C₄H₉Br. (1993 - 1 Mark) 

Ans. Sol. TIPS/Formulae : A symmetric alkene does not follow Markovnikoff and antiMarkovnikoff 's rule (Peroxide effect).
B has to be a symmetric alkene (butene-2) CH₃CH = CHCH₃ as it will give the same product CH₃ – CH(Br) – CH₂– CH₃ in presence /absence of peroxide.

Q. 12. Identify, D(C₆H₁₂), an optically active hydrocarbon which on catalytic hydrogenation gives an optically inactive compound, C₆H₁₄. (1993 - 1 Mark) 

Ans. Sol. An optically active hydrocarbon will have an asymmetric C-atom. This means D(C₆H₁₂) should have an asymmetric C-atom & C₆H₁₄ will have no asymmetric C-atom, hence D would be 3-methylpentene-1,

Q. 13. Draw the stereochemical structures of the products in the following reactions : (1994 - 4 Marks)

Ans. Sol. (i) SN₂ reaction leads to inversion in configuration.

NOTE :
(i) Lindlar's catalyst is Pd supported over CaCO₃ which is partially poisoned by (CH₃COO)₂Pb. It can restrict the hydrogenation of alkyne to alkene stage. It yields a cis-alkene. (ii) Reduction of alkynes to alkene stage can also be carried out with sodium or lithium in liquid NH₃. Here transalkene is major product.