1. Acetylenic hydrogens are acidic because
(a) Sigma electron density of C– Hbond in acetylene is nearer to carbon, which has 50% scharacter
(b) Acetylene has only open hydrogen in each carbon
(c) Acetylene contains least number of hydrogens among the possible hydrocarbons having two carbons
(d) Acetylene belongs to the class of alkynes with molecular formula, CnH2n –
2. Which is the most suitable reagent among the following to distinguish compound (3) from rest of the compound ?
(1) CH3 – C ≡ C – CH3
(2) CH3 – CH2 – CH2 – CH3
(3) CH3 – CH2C ≡ CH
(4) CH3 – CH = CH2
(a) Bromine in carbon tetrachloride
(b) Bromine in acetic acid
(c) Alk KMnO4
(d) Ammonical silver nitrate.
3. Select the true statement about benzene amongst the following
(a) Because of unsaturation benzene easily undergoes addition
(b) There are two types of C– Cbonds in benzene molecule
(c) There is cyclic delocalisation of pi-electrons in benzene
(d) Monosubstitution of benzene gives three isomeric products.
(a) Na
(b) HCl in H2O
(c) KOH in C2H5OH
(d) Zn in alcohol.
5. Reduction of 2-butyne with sodium in liquid ammonia gives predominantly:
(a) cis – 2 – Butene
(b) No reaction
(c) trans – 2 – Butene
(d) n-Butane.
6. Acompound is treated with NaNH2 to give sodium salt. Identify the compound
(a) C2H2
(b) C6H6
(c) C2H6
(d) C2H4.
7. Reactivity of hydrogen atoms attached to different carbon atoms in alkanes has the order
(a) Tertiary > Primary > Secondary
(b) Primary > Secondary > Tertiary
(c) Both
(d) Tertiary > Secondary > Primary
8. When hydrochloric acid gas is treated with propene in presence of benzoyl peroxide, it gives
(a) 2-Chloropropane
(b) Allyl chloride
(c) No reaction
(d) n-Propyl chloride.
9. Which of the following compounds has the lowest boiling point?
(a) CH3CH2CH2CH2CH3
(b) CH3CH= CHCH2CH3
(c) CH3CH= CH– CH= CH2
(d) CH3CH2CH2CH3
10. When CH3Cl and AlCl3 are used in FriedelCrafts reaction, the electrophile is
(a) Cl+
(b) AlCl4–
(c) CH3+
(d) AlCl2+
11. In the free radical chlorination of methane, the chain initiating step involves the formation of
(a) Chlorine free radical
(b) Hydrogen chloride
(c) Methyl radical
(d) Chloromethyl radical.
12. The alkene R– CH= CH2 reacts readily with B2H6 and formed the product Bwhich on oxidation with alkaline hydrogen peroxides produces
(a) R – CH2 – CHO
(b) R – CH2 – CH2 – OH
(c)
13. When 3, 3-dimethyl 2-butanol is heated with H2 SO4, the major product obtained is
(a) 2,3-dimethyl 2-butene
(b) 3, 3-dimethyl 1- butene
(c) 2, 3-dimethyl 1- butane
(d) cis & trans isomers of 2, 3-dimethyl 2-butene
14. In the presence of platinum catalyst, hydrocarbon Aadds hydrogen to form n-hexane. When hydrogen bromide is added to Ainstead of
hydrogen, only a single bromo compound is formed. Which of the following is A?
(a) CH3 — CH2 — CH = CH— CH2 — CH3
(b) CH3 — CH2 — CH2 — CH = CH— CH3
(c) CH3 — CH = CH— CH2 — CH2 — CH3
(d) CH2 = CH— CH2 — CH2 — CH2 — CH3
15. In reaction sequence
(a) CH3CH2Cl and NaOH
(b) CH3CH2OHand H2 SO4
(c) CH2Cl – CH2OH and aqueous NaHCO3
16. Which one of the following reactions is expected to readily give a hydrocarbon product in good yields?
17. In commercial gasolines the type of hydrocarbons which are more desirable, is
(a) branched hydrocarbons
(b) straight-chain hydrocarbons
(c) aromatic hydrocarbons such as toluene
(d) linear unsaturated hydrocarbons
18. A hydrocarbon Aon chlorination gives B which on heating with alcoholic potassium hydroxide changes into another hydrocarbon C.
The latter decolourises Baeyer’s reagent and on ozonolysis forms formaldehyde only. A is
(a) Ethane
(b) Butane
(c) Methane
(d) Ethene
19. The reaction of ethyl magnesium bromide with water would give
(a) Ethane
(b) Ethyl alcohol
(c) Ethyl bromide
(d) Ethyl ether
20. Which of the following reagents convert propene to 1-propanol?
(a) H2O, H2 SO4
(b) aqueous KOH
(c) MgSO4, NaBH4 /H2O
(d) B2H6, H2O2, OH–
21. For the formation of toluene by Friedal Craft reaction, reactants used in presence of anhydrous AlCl3 are
(a) C2H2 and CCl4
(b) CH4 and CaCN2
(c) C6H6 and CH3Cl
(d) C2H5 OH and Zn
22. In preparation of alkene from alcohol using Al2O3 which is effective factor?
(a) Temperature
(b) Concentration
(c) Surface area of Al2O3
(d) Porosity of Al2O3
23. Which alkene on ozonollysis gives CH3CH2CP and
(b) CH3CH2CH = CHCH2CH3
(c) CH3CH2CH = CHCH3
24. The compound On reaction with NaIO4 in the presence of KMnO4 gives
(a) CH3CHO+ CO2
(b) CH3COCH3
(c) CH3COCH3 + CH3COOH
(d) CH3COCH3 + CH3CHO
25. Reaction of HBr with propene in the presence of peroxide gives
(a) isopropyl bromide
(b) 3-bromo propane
(c) allyl bromide
(d) n-propyl bromide
26. Using anhydrous AlCl3 as catalyst, which one of the following reactions produces ethylbenzene (PhEt) ?
(a) H3C – CH2OH + C6H6
(b) CH3 – CH CH2 + C6H6
(c) H2C = CH2 + C6H6
(d) H3C – CH3 + C6H6
27. Which one of the following alkenes will react faster with H2 under catalytic hydrogenation conditions? (R= Alkyl Substituent)
28. Products of the following reaction :
(a) CH3COOH +CO2
(b) CH3COOH +HOOC.CH2CH3
(c) CH3CHO + CH3CH2CHO
(d) CH3CHO + CH3COCH3
29. Predict the product C obtained in the following reaction of butyne – 1.
CH3CH2 – C CH +HCl ® B C
30. Which of the compounds with molecular formula C5H10 yields acetone on ozonolysis?
(a) 3-methyl-1-butene
(b) cyclopentane
(c) 2-methyl-1-butene
(d) 2-methyl-2-butene.
32. The IUPAC name of the compound having the formula CH º C – CH = CH2 is
(a) 1-butyn-3-ene
(b) but-1-yne-3-ene
(c) 1-butene-3-yne
(d) 3-butene-1-yne
33. Liquid hydrocarbons can be converted to a mixture of gaseous hydrocarbons by:
(a) Oxidation
(b) Cracking
(c) Distillation under reduced pressure
(d) Hydrolysis
(a) C6H5CH2CH2C6H5
(b) C6H5CH2OH
(c) C6H5CH3
(d) none of the above
35. The IUPAC name of the compound CH3CH= CHC= CH is
(a) Pent-l-yn-3-ene
(b) Pent-4-yn-2-ene
(c) Pent-3-en-1-yne
(d) Pent-2-en-4-yne
36. In the following reactions,
37. In the following reaction :
The major products is :
38. Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne?
(a) NaNH2
(b) HCl
(c) O2
(d) Br2
39.The radical ,
(a) 7 p-orbitals and 6 unpaired electrons
(b) 7 p-orbitals and 7 unpaired electrons
(c) 6 p-orbitals and 7 unpaired electrons
(d) 6 p-orbitals and 6 unpaired electrons
(a) Iodoformtest
(b) Tollen’s reagent test
(c) Brady’s reagent test
(d) Victor Meyer test
41. Which of the following chemical system is non aromatic?
42. 1-Butyne and cold alkaline KMnO4 react to produce
(a) CH3 CH2COOH + HCOOH
(b) CH3 CH2CH2COOH
(c) CH3 CH2COOH
(d) CH3 CH2 COOH + CO2
43. Predict the correct intermediate and product in the following reaction :
44. Which of the following can be used as the halide component for Friedel-Crafts reaction?
(a) Chlorobenzene
(b) Isopropyl chloride
(c) Bromobenzene
(d) Chloroethene
45. Propene on hydroboration and oxidation produces
(a) CH3CH2CH2OH
(b) CH3CH2CHO
(c) CH3CHOHCH3
(d) CH3CHOHCH2OH
46. In the given reaction the product P is
47. Consider the nitration of benzene using mixed conc. H2SO4 and HNO3 . If a large amount of KHSO4 is added to the mixture, the rate of nitration will be
(a) unchanged
(b) slower
(c) doubled
(d) faster
48. The boiling points of four saturated hydrocarbons are given below. Which boiling point suggests maximum number of carbon atoms in its molecule :
(a) – 162° C
(b) – 42.2° C
(c) – 88.6° C
(d) – 0.5° C
49. What products are formed when the following compound is treated with Br2 in the presence of FeBr3?
50. The most important method of preparation of hydrocarbons of lower carbon number is
(a) Sabatier and Senderen’s reaction
(b) Direct synthesis
(c) Pyrolysis of higher carbon number of hydrocarbons
(b) Electrolysis of salts of fatty acids
51. Which of the following compounds will not undergo Friedel-Crafts reaction easily?
(a) Cumene
(b) Nitrobenzene
(c) Toluene
(d) Xylene
52. In the following the most stable conformation of n-butane is
53. The reaction of toluene with Cl2 in presence of FeCl3 gives X and reaction in presence of light gives Y. Thus, X and Y are
(a) X = Benzal chloride, Y = o-chlorotoluene
(b) X = o- and p-chlorotoluene. Y = Trichloronrethyl benzene
(c) X – m-chlorotoluene, Y = p-chlorotoluene
(d) X = Benzyl chloride, Y = m-chlorotoluene
54. The state of hybridisation of C2, C3, C5, and C6, of the hydrocarbon,
is in the following sequence
(a) sp3, sp2, sp2 and sp
(b) sp, sp2, sp2 and sp3
(c) sp, sp2, sp3 and sp2
(d) sp, sp3, sp2 and sp3
55. Which of the following conformers for ethylene glycol is most stable?
(a)
56. Benzene reacts with CH3CI in the presence of anhydrous AlCl3 to form
(a) xylene
(b) toluene
(c) chlorobenzene
(d) benzyl chloride
57. In the hydrocarbon,
The state of hybridization of carbons 1, 3 and 5 are in the following sequence
(a) sp, sp2, sp3
(b) sp2, sp, sp3
(c) sp3, sp2, sp
(d) sp, sp3, sp2
58. Predict the product C obtained in the following reaction of 1-b
59. Which of the compound with molecular formula C5H10 yields acetone on ozonolysis?
(a) 2-Methyl-1-butene
(b) 2-Methyl-2-butene
(c) 3-Methyl-1-butene
(d) Cyclopentane
60. Which is maximum stable?
- 1-Butene
- trans-2-Butene
(c) cis-2-Butene
(d) All have same stability
61. Which of the following compounds will be most easily attacked by an electrophile?
62. Which one is the correct order of acidity ?
(1) CH CH > CH3 – C CH > CH2 = CH2 > CH3 – CH3
(2) CH CH > CH2 = CH2 > CH3 – C CH > CH3 – CH3
(3) CH3 – CH3 > CH2 = CH2 > CH3 – C CH > CH CH
(4) CH2 = CH2 > CH3 – CH = CH2 > CH3 – C CH > CH CH
63. Predict the correct intermediate and product in the following reaction
64. With respect to the conformers of ethane, which of the following statements is true ?
(1) Bond angle changes but bond length remains same
(2) Both bond angle and bond length change
(3) Both bond angles and bond length remains same
(4) Bond angle remains same but bond length changes
65. Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is
(1) CH CH
(2) CH2 = CH2
(3) CH4
(4) CH3 – CH3
66. An alkene “A” on reaction with O3 and Zn–H2O gives propanone and ethanal in equimolar ratio. Addition of HCl to alkene “A” gives “B” as the major product. The structure of product “B” is:
67. An alkene on ozonolysis gives methanal as one of the product. Its structure is
68. The correct structure of 2, 6-Dimethyl-dec-4-ene is
69. The major product formed in dehydrohalogenation reaction of 2-Bromo pentane is Pent-2-ene. This product formation is based on?
(1) Huckel’s Rule
(2) Saytzeff’s Rule
(3) Hund’s Rule
(4) Hofmann Rule
x
Solutions :
1.Ans: (a)The acidity of acetylene or 1 – alkynes can be explained on the basis of molecular orbital concept according toe which formation of C- H bond inacetylene involves sp – hybridised carbon atom. Now since s electrons are closer to the nucleus than p electrons, electrons present in a bond having more s character n a bond having more s character will be correspondingly more closer to the nucleus. Thus owing to high s character of C- H bond alkynes (s = 50 % ) thne electrons constitutinjg this bond are mre strongly held by the carbon nucleus. (i.e. the ace3tylenic carbon atom or the sp orbital acts as mre electronegative species than the sp2 and sp3 with the result the hydrogen present on such a carbon atom ( can be easily removed as a proton.
2.Ans: (d)Br2 in CCl4(a) , Br2 in CH3 COOH(b) and alk KMnO4(c) will react with all unsaturated compounds, i.e. 1,3, and 4 while ammonical AgNO3(d) reacts only terminal alkynes i.e. 3 and hence compound 3 can be distinguished from 1,2 and 4 by ammonical AgNO3(d)
3.Ans: (c)Benzene dont show addition reaction like other unsaturated hydrocarbons. However it shows substitution reactions. Due to resonance all the C- C bonds have the same nature, which is responsible all the C –C bonds have the same nature, which is possible because of the cyclic delocalisation of π electrons in benezene. Monhosubstitution will give only a single product.
4.Ans: (c)
This reaction is an example of dehydrohalogenation. Hence, alcoholic KOH is used as a reagent.
5.Ans: (c)Reduction of non-terminal alkynes with Na in liq. NH3 at 195 – 200 K gives trans-2-butene.
6.Ans: (a) Only C2H2(acetylene) has acidic H-atoms and hence reacts with NaNH2 to form sodium salt., i.e.
HC CH +NaNH2 -> HC can+NH3
7.Ans: (d)Tertiary alkanes are more reactive because they form tertiary free radical which is more stable whereas primary alkanes are less reactive because they form 1∘ free radicals which are less stable.
8.Ans: (a)Kharasch effect or peroxide effect is only observed in case of addition of HBr to unsymmetrical alkenes, so the addition of HC1 with propene takes place as usual by Markownikoff- rule
9.Ans: (d) Of all the options listed CH3CH2CH3 has the least number of C- atoms and hence has the lowest b.p
12.Ans: (b)Alkenes,when treated with diborane gives, alkylboranes and alkylboranes on the oxidation with alkaline Hydrogen peroxide gives alcohols.For propene, mechanism is :6CH3CH=H2+(BH3)2→2(CH3CH2CH2)3B+H2O2/OH− → 6CH3CH2CH2OH So product is R−CH2CH2OH
16.Ans: (a)The reaction which is expected to readily give a hydrocarbon product in good yield is:RCOOK RH + CO2 + KOH
17.Ans: (a)Gasoline (petrol) is a mixture of alkanes, alkenes and aromatic hydrocarbons. The quality of gasoline is determined by the amount of branched chain hydrocarbons (2,2,4 – trimethypente , commonly known as iso – octane) present in it.
18.Ans: (a)Ethane on chlorination, gives 1-chloroethane, which reacts with alcoholic KOH to give a dehydro-halogenation product (Ethene). The ethene formed from this undergoes ozonolysis to break the double bond and form two molecules of formaldehyde.
‘a’ – Ethane
‘b’ – 1-chloroethane
‘c’ – Ethene
19.Ans: (a)
25.Ans: (d)The formation of n-propyl bromide in the presence of peroxide can be explained as follows.
Step-1: Peroxide undergo fission to give free radicals R−O−O−R→2−R−O˙
Step-2: HBr combines with free radicals to form bromine free radicals R−O˙+HBr→R−OH+Br˙
Step-3: Br˙ attacks the double bond of the alkene to form more stable free radicals.
Step-4: More stable free radical attacks the HBr CH3−CH−CH2−Br+HBr→n-propyl bromideCH3CH2CH2Br+Br˙
Step-5: Br˙+Br˙→Br2
26.Ans: (c)
27.Ans: (a)During catalytic hydrogenation, the hydrogens are transferred from the catalyst to the same side of the double bond thereby cis-alkenes. Evidently smaller the number of R substituents, lesser is the steric hindrance and hence faster is the rate of hydrogenation.Stability of alkene is inversely proportional to the heat of hydrogenation of alkene.Catalytic hydrogenation takes place in 3rd compound. The energy required to break the double bond is lesser in 3rd compound than any other compound.Greater the number of alkyl groups attached to the doubly bonded carbon atoms, more stable is the alkene. (less reactive)
28.Ans: (b)The given reactant is 1-butyne. It on reaction with ozone undergoes ozonolysis to form ozonide which is followed by hydrolysis to form acetic acid and acetone as a product.
29.Ans: (c)This reaction occurs according to Markownikoff’s rule which state when an unsymmetrical alkene undergo hydroalogenation, the negative part goes to that C-atom which contains lesser no. of H –atom.
30.Ans: (d)2-methyl-2-butene (molecular formula C5H10) yields acetone on ozonolysis.
31.Ans: (b)The reaction proceeds via the formation of the more stable carbocation. Hence, the major product
32.Ans: (c) As triple bond is present at carbon no. 3 and double bond is present at carbon no. 1 so IUPAC name of compound is 1 – butene – 3- yne. In numbering double bond is given preference if from both side alkene and alkyne functional group are present at same numbers.
33.Ans: (b) Lower hydrocarbons exist in gaseous state while higher ones are in liquid state or solid state.On cracking or pyrolysis, the hydrocarbon with higher molecular mass gives a mixture of hydrocarbons having lower molecular mass. Hence, we can say that by cracking a liquid hydrocarbon can be converted into a mixture of gaseous hydrocarbons.
34.Ans: (d)
35.Ans: (c)According to the rules of IUPAC,In an organic compound having multiples bonds the order of preference is given as,Triple bond>Double bond>Single bond So, We should number the carbon from the right. As it can be seen from the figure the correct order of numbering and nomenclature is right one. The left part violated the IUPAC rule stated above and hence the correct IUPAC name of the compound will be pent-3-en-1-yne.
36.Ans: (b)
37.Ans: (a)
38.Ans: (a) 1– Butyne and 2 –butyne are distinguish by NaNH2 because 1- butyne react with NaNH2 due to presence of terminal hydrogen.
39.Ans: (d) Presence of 6p orbitals, each containing one unpaired electron, in a six memebered cyclic structure is in accordance with Huckel rule of aromaticity.
40.Ans: (d)
41.Ans: (a)Huckle rule is not obeyed. It has only four electrons. Further it does not have continuous conjuctions.
42. (d)
43(c)
44(b) Isopropyl chloride
45 (a) CH3CH2CH2OH
46 (c)
47(b) If a large amount of KHSO4 is added then conc. of HSO4- ions increases and the reaction will be shifted in backward direction hence, the rate of nitration will be slower.
48 (d) The heavier the molecule, greater is the boiling point. So molecule with boiling point – 0.5°C will have maximum number of carbon atoms.
49(c)m-xylene on bromination with bromine in presence of ferric bromide gives 1-bromo-2,4-dimethylbenzene.This is represented by option C which shows a pair of identical compounds.This is an electrophilic aromatic substitution reaction and since methyl group is ortho para directing, the substitution occurs at a position which is ortho to one methyl group and para to other methyl group.
50 (c)
51 (b) Nitrobenzene is strongly deactivated, hence will not undergo Friedel-Crafts reaction
52 (c)The conformation in which the heavier groups are present at maximum possible distances, so that the forces of repulsion get weak, is more stable.Among the given conformations of n-butane, the conformation shown in option (b), i.e. anti conformation is most stable as in it the bulkier group, i.e. – CH3 group are present at maximum possible distance and get lower energy.
53 (b)The reaction of Cl2, in presence of FeCl3, with benzene yields a ring substitution product.
In presence of sunlight, free radical reaction takes place.
54 (d)
C2 having →2−σ bond →sp
C3 having →4−σ bond →sp3
C5 having →3−σ bond →sp2
C6 having →4−σ bond →sp3
55 (d) This conformation is most stable due to intermolecular H-bonding.
56 (b)
57(d)
The state of hybridization of carbon in 1, 3 and 5 position sp , sp3, sp2.
58(c)According to Markownikoff’s rule, during hydrohalogenation to unsymmetrical alkene, the negative part of the addendum adds to less hydrogenated (i.e. more substituted) carbon atom.
59 (b) 2-methyl-2-butene (molecular formula (C5H10)) yields acetone on ozonolysis.
60 (b) This is most stable as the repulsion between two methyl groups is least.
61 (a) – OH, – Cl and – CH3 groups in benzene are ortho-para directing groups and activate the ring towards electrophilic substitution reaction. But among these -OH group is strongly activating while -CH3 is weakly activating and -Cl is deactivating. Thus, phenol will be most easily attacked by an electrophile.
62 (1) correct order
CH CH > CH3 – C CH > CH2 = CH2 > CH3 – CH3
63 (3) Hydration of alkynes gives keynotes
64 (3) No change takes place in the bond angles and the bond lengths of conformations of ethane. The only change taking place is the change in the dihedral angle.
65 (3)
66 (3)
68 (2)
69(2)
