Hydrocarbons
CHEMISTRY
- Which one of the following is an electrophile?
(A) Boron trifluoride (B) Ethylene
(C)Benzene (D) Water
- Which of the following substituted benzene derivatives would furnish three isomers when one more substituent is introduced?
(A) | (B) | ||
(C) | (D) |
- Which one of the following species is isoelectronic with ammonia?
(A) CH3 (B) CH3+
(C) CH3– (D) CH2
- Cyclooctatetraene is not aromatic. The reason for this is that
(A) It has eight pi electrons
(B) It is a planar molecule
(C) Its structure is not that of a regular octagon
(D) Its structure can not be described by more than two canonical forms
- Which of the following compounds can be optically active
(A) | (B) | ||
(C) | CH2 = CH – CH3 | (D) |
- Which one of the following ions is aromatic?
(A) | (B) | ||
(C) | (D) |
- Which one of the following molecules has all the effects, namely inductive, mesomeric and hyper conjugative?
(A) CH3Cl (B) CH3 – CH = CH2
(C) CH3 – CH = CH – – CH3 (B) CH2 = CH – CH = CH2
- Which of the following reactions would generate an electrpphile?
- C2H5Cl + anhydrous AlCl3
- C6H5Br + CuBr2
- C6H5COOH + H3O+
- Fuming HNO3 + conc. H2SO4. Select the correct answer using the codes given below:
(A) 1 and 3 (B) 1, 2 and 3
(C)1, 3 and 4 (D) 2, 3 and 4
- Consider the following reaction
HO–[CH3 – – CH2 – CH2 – CH3] X + Y. The products X and Y in the above reaction
(A) CH3 – CH2 – CH2 – CH3, CH3OH (B) CH2 = CH – CH3, N(CH3)3
(C) CH3 – CH2 – CH2 – OH, N(CH3)3 (D) CH3 – – CH3, N(CH3)3
- Among the following compounds, the pair which does not show the resonance
(A) CH2 = CH – – H; – CH = – H
(B) CH2 = CH – – Cl; – CH = CH – Cl
(C) (CH3)2 CH – – O–; (CH3)2 CH – = O
(D) CH3 – CH2 – – CH3; CH3 – CH2 – – CH3
- Cyclohexene can be converted to 3-bromocyclohexene by
(A) Bromine in acetic acid
(B) Bromine water and dichloromethane
(C) N-bromosuccinimide in methanol and aqueous H2SO4
(D) N-bromosuccinimide in methanol and benzoyl peroxide
- Sulfonation of benzene differs from most other electrophilic substitution reactions in that the reaction.
(A) is reversible
(B) requires the presence of a Lewis acid catalyst
(C) Takes place with explosive violence
(D) Requires elevated temperatures
- Consider the following statements about conformational isomers
- They are interconverted by rotation about single bond
- The energy barrier separating them is less than 15 kcal/mole
- They are best represented by means of Fischer projection formulae. Of these statements
(A) 1, 2 and 3 are correct (B) 2 and 3 are correct
(C) 1 and 3 are correct (D) 1 and 2 are correct
- Consider the following compounds
Their relative reactivity towards nitronium ion is such that
(A) II < I < III (B) III < I < II
(C) I < III < II (D) III < II < I
15. | ||||
(A) | (B) | |||
(C) | (D) |
- Which one of the following reagents can be used to convert trans-2-butene to racemic-2, 3-butane diol?
(A) Periodic acid (B) Cold aqueous potassium permanganate
(C) Chromic acid (D) Peroxy acetic acid
- The number of streoisomers possible for HOOC – (CHOH)4 – COOH is
(A) 32 (B) 16
(C) 14 (D) 12
- Which one of the following pair represents a set of electrophiles?
(A) Br+ and (B) AlCl3 and Cl–
(C) H+ and H2O (D) CN– and NH3
- The no. of isomers of 1, 3, 5 tribromo benzene is
(A) 1 (B) 2
(C) 3 (D) 4
- Which of the following statements is correct?
(A) The enol tautomer of 2, 4-pentanedione is lesser stable than the corresponding tautomer of monocarbonyl compound.
(B) The enol content of 2, 4-pentane dione in water is more than in hexane.
(C) The enol content of 2, 4-pentanedione in water is lesser than in hexane.
(D) The enol content of 2, 4-pentanedione in water and hexane are identical.
- Which of the following compound on oxidation with hot acidified KMnO4 gives succinic acid.
(A) 1-butene (B) 2-butene
(C) 2-methyl butene (D) cyclobutene
- In the following reactions
- CH3 – N = N CH3
- CH2N2
III. (CH3)3C – OH
- (CH3)3C – Cl
The reaction intermediates formed will be
I II III IV
(A) (CH3)3C– (CH3)3C–
(B) (CH3)3C– (CH3)3C–
(C) (CH3)3C+ (CH3)3C+
(D) (CH3)3C+ (CH3)3C–
- The carbon-carbon single bond distance in 1, 2-dimethylacetylene and ethane is 1.40Å and 1.54Å respectively. The shorter C – C bond distance of the former is best explained interms of
(A) Inductive effect (B) Its acidic nature
(C) Hyperconjugation (D) Coplanarity
- Consider the following pairs of compounds
(A) | |
(B) |
of these pairs
(A) Both A and B are resonance forms
(B) both A and B are tautomeric forms
(C) A is a resonance pair while B is a pair of tautomers
(D) A is a pair of tautomers while B is a resonance pair
- The reaction least likely to occur is
(A) | |
(B) | |
(C) | |
(D) |
- Which of the following statements is wrong regarding nitration of aromatic compounds?
(A) Toluene is more reactive than benzene
(B) Anisole is more reactive than benzene
(C) Chloro benzene is less reactive than benzene
(D) Deuterobenznee is more reactive than benzene
- In which of the reactions listed below the bonding changes does not take place heterolytically?
(A) | CH3Cl + NaOH ⎯→ CH3OH + NaCl | (B) | CH2 = CH2 + HBr ⎯→ CH3CH2Br |
(C) | Cl2 + CH4 ⎯→ CH3Cl + HCl | (D) |
- Resonance is due to
(A) Delocalisation of sigma electrons (B) Delocalisation of π-electrons
(C) Migration of H-atoms (D) Migration of protons
- The Markownikoff rule is used in connection with
(A) Stereochemistry of elimination reactions
(B) Stability of free radicals
(C) Activity of enzymes
(D) Addition of acids to double bonds.
- Match List I and List-II and select the correct answer using the codes given below the lists:
List – I | List – II |
A. Mesocompound | 1. An equimolar mixture of enantiomers |
B. Enantiomers | 2. Stereoisomers that are not mirror images |
C. Diastereomers | 3. Non-superimposable mirror images |
D. Racemates | 4. An optically inactive compound whose molecules are achiral eventhough they contain chiral centres. |
A B C D
(A) 3 4 1 2
(B) 3 4 2 1
(C) 4 3 1 2
(D) 4 3 2 1
- On being heated with alcoholic KOH neopentyl bromide gives mainly
(A) 2 methyl 2-butene (B) 2 methyl 1- butene
(C) 2 butene (D) 2, 2dimethyl-butene
- In CH3CH2OH, the bond that undergoes heterolytic cleavage most readily is
(A) C – C (B) C – O
(C) C – H (D) O – H
- Meso-tartaric acid is optically inactive due to presence of
(A) Molecular assymmetry (B) External compensation
(C) Internal compensation (D) Two asymmetric carbon atoms
- Which of the following compound is optically active?
(A) CH3CH2CO2H (B) CH3CHOHCO2H
(C) HO2C – CH2 – CO2H (D) CH3COCO2H
- Antimarkownikov reaction takes place in which of the following reactions
(A) CH3CH = CH2 + HBr (B) CH3CH = CH2 + HCl
(C) CH3CH = CH2 + HOCl (D) None
- The number of optically active isomers observed in 2, 3-dichlorbutane is
(A) 0 (B) 2
(C) 3 (D) 4
- Bromobenzene and methyl bromide react in ethereal solution in presence of sodium metal to give
(A) Dibromobenzene (B) Xylene
(C) Toluene (D) Methyl phenyl ether
38. |
A is
(A) CH3 – CH = CH2 (B) CH3 – C = CH2
|
CH3
(C) CH2= CH2 (D) None
- Which of the following is fast debrominated?
- The three resonating structures of vinyl chloride are given
The decreasing order of stability of resonating structures of vinyl chloride is
(A) A > B > C (B) A > C > B
(C) B > A > C (D) C > B > A
- HBO process converts the following alkene into A
(D) None of these |
- Which of the following alkane has least boiling point?
(A) CH3CH2CH2CH2CH3 (B) H3C – CH – CH2 – CH3
|
CH3
(B) CH3 (D) all having same boiling point.
|
H3C – C – CH3
|
CH3
43. | A & B are | |
44. |
(A) n-propane (B) H3C – CH – NH2
|
CH3
(C) n-butane (D) None
- Major product on monochlorination of CH3 – is
(A) (B) Cl
(C) Both are formed equally (D) None
46. | Y is | |
(D) None |
47. | A is | |
(D) None |
- H2C = CH2
Y is
(A) 2 moles of a HCO2H (B) 1 mole HCHO + 1 mole HCO2H
(C) 2 moles of HCHO (D) None
49. | Z is | |
50.
Y is
(A) Racemic mixture of 2, 3 dihydroxy butane
(B) meso 2,3dihydroxy butane
(C) 2hydroxy butane
(D) 1, hydroxy butane
ANSWERS
- A 2. A
3. C 4. A - D 6. C
- C 8. C
- B 10. D
- D 12. A
- C 14. B
- A 16. B
- A 18. A
- A 20. C
- D 22. C
- C 24. C
- B 26. D
- C 28. B
- D 30. D
- A 32. D
- C 34. B
- A 36. B
- C 38. C
- B 40. B
- A 42. C
- C 44. A
- A 46. A
- C 48. C
- D 50. A
HINTS & SOLUTIONS
- BF3 is a Lewis acid and hence acts as an electrophile.
- This is an example of Hofmann elimination reaction.
10. | Allylic bromination can be done with NBS in presence of peroxide to get a major yield. |
- –OCH3 is the most activating group and –Cl is a deactivating group.
- This process adds H2O to an alkene according to Anti-markonikow’s rule.
- In trans alkene, syn addition makes the resulting compound racemic.
- Carbenes are also electrophilic.
- Resonance does not involve the shift of atoms (or) groups whereas tautomerism involves the movement of them.
- Photochemical halogenation is not possible in an aromatic ring.
- Since removal of H+ (or) D+ does not affect the rate of the reaction, the rates of benzene and deutero benzene are same.
- In Photochemical reactions bond cleavage takes place homolytically.
- The reaction proceeds via E1 elimination and hence rearrangement of carbocation is possible.
- O – H bond cleaves readily as the electronegativity difference between them is high.
- Anti-markonikov’s addition takes place with HBr in presence of peroxides.
39. | Since the product is stabilized by resonance, it will be debrominated faster. |
40. |
- It is an example of Wölff Kishner reduction of carbonyls.
- Photochemical chlorination, as it is less selective substitutes the 1° H atom.
49. |
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