JEE NEET Chemistry Question Bank - Some Basic Principle of Organic Chemistry

20 Some Basic Principles of Organic Chemistry
• Tetravalency of carbon hybridization and shape of molecules
• In excited state electonic configuration of c-atom has four unpaired electrons, hence carbon can
form four covalent bonds.
• Formation of same number of orbitals having same properties of various type of orbitals (s and
p) of C-atom is called hybridization and orbitals produces are called hybridized orbitals.
• Hybrid orbitals of carbon form s- bond and those orbitals which do not take part in hybridisation
form pi (p) bond.
• Single bond is always s bond. In double bond one s and one p bonds are present, while in triple
bond one s and two p bonds are present.
• On the basis of number of s -bond in carbon atom its hybridization is determined.
• Remember :
Type of Type of hybridi shape of bond C–C
compound bond -zation of molecule angle bond length
carbon

Alkane –C – C – sp3 tetrahedral 109o 28’ 154 pm

Alkene >C = C< sp2 trigonal plannar 120o 134 pm

Alkyne –C º C– sp linear 180o 120 pm

1. In which of the following molecule carbon-carbon bond length is least ?
(A) ethane (B) ethene (C) ethyne (D) benzene
2. How many s and p bonds are present in dicyano ethene CN – CH = CH – CN
respectively ?
(A) 7 and 1 (B) 7 and 5 (C) 5 and 7 (D) 3 and 5
3. In but-l-ene-3-yne number of s and p bonds are ...... respectively.
(A) 7 and 5 (B) 7 and 3 (C) 6 and 4 (D) 6 and 3
2
4. In which of the following compound all C-atom have sp hybridization ?
(A) ethane (B) propene (C) ethyne (D) ethylene
5. What is the type of hybridization of each C in the following compound ?
CH3 – CH = CH – CN
(A) sp3, sp2, sp, sp (B) sp3, sp2, sp2, sp (C) sp, sp2, sp3, sp (D) sp2, sp2, sp3, sp
6. What is the expected bond angle in the molecule in which the central atom is sp3 hybridized ?
(A) 109o 28’ (B) 120o (C) 90o (D) 180o
7. Which hybrid orbitals are involved in the bond formation between C2 – C3 in given organic
compound ?
1
CH º 2C – 3CH2 – 4CH3
(A) sp – sp (B) sp2 – sp (C) sp2 – sp2 (D) sp – sp3

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P2O5
8. In the dehydration reaction CH3CONH2 CH3CN hybridization of carbonyl carbon changes
to what ?
(A) sp3 to sp2 (B) sp2 to sp3 (C) sp2 to sp (D) sp to sp3
9. What are the changes in hybridization of carbon in addition halogenation reaction of ethene ?
(A) sp3 to sp2 (B) sp2 to sp3 (C) sp2 to sp2 (D) sp to sp3
10. What is the correct order of C–C bond length ?
(A) C2H6 > C2H4 > C2H2 (B) C2H4 > C2H6 > C2H2
(C) C2H2 < C2H6 < C2H4 (D) C2H6 < C2H4 < C2H2
Answers : 1. (C), 2. (B), 3. (B), 4. (D), 5. (B), 6. (A), 7. (D), 8. (C), 9. (B), 10. (A)

• Classification of organic compounds on the basis of functional group
• The atom or group of atom which is responsible for characteristic reactions of organic compunds
is called functional group.
• While writing the structure of organic compounds it is necessary to remember the number of
bonds. For example carbon has four bonds, nitrogen has three bonds, oxygen has two bond and
for hydrogen and halogen single bond.
• While writing the name of organic compounds on the basis of number of C atom present in it
corresponding word root (Greekword) is used which as follows :
Number of C ® 1 2 3 4 5 6 7 8 9 10
word root ® meth eth prop but pent hex hept oct non dec
• Organic compounds classified into 14 parts based on functional group :

Type of Functional Prefix/suffix Example IUPAC
compound group in nomenclature Name

(1) Alkane –C – C – –| ane CH 3 –CH 2 –CH 3 propane

R – H

(2) Alkene >C=C< –| ene CH 3–CH=CH 2 propene

(3) Alkyne –CºC– –| yne CH3–CºCH propyne

(4) Alcohol –OH –| ol CH 3–CH 2–CH 2–OH propan-1-ol

(5) Ether –O– alkoxy |– CH 3–O–CH 2–CH 3 Methoxy ethane

(6) Aldehyde –CHO –| al CH 3 –CH 2 –CHO propanal

(7) Ketone –CO– –| one CH 3–CO–CH 3 propanone

(8) Carboxylic –COOH –| oic CH3– CH2–COOH propanoic
acid acid acid

(9) Ester –COOR –| oate CH3– COOCH3 Methyl
R=alkyl ethanoate

(10) Amine (a) –NH 2 –| amine CH 3 –CH 2–CH 2 –NH 2 propan-1
primary amine -amine

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Amine (b) –NH– –| amine CH 3–CH 2–NH–CH 3 N-Methyl
secondary amine ethanamine

Amine (c) –N– –| amine CH 3–CH 2–N–CH 3 N,N-dimethyl

tertiary amine CH3 ethanamine
(11) Amide –CONH 2 –| amide CH 3CH 2CONH 2 Propananide
(12) Nitro –NO 2 Nitro |– CH 3–CH 2–CH 2–NO 2 1-nitro propane
(13) Cyanide –CN –| nitrile CH 3–CH 2–CN propane nitrile
(14) Halide –X halo |– CH 3–CH 2–CH 2–Cl 1-chloro propane
X=F, Cl, Br, I

Note :
(1) In alkane, alkene and alkyne’s nomenclature proper suffix is added to wordroot on the basis of
number of carbon
eg., CH3 – CH = CH2 word root + suffix
prop + ene = propene
(2) In other compounds nomenclature appropriate prefix or suffix is added to alkane name.
eg., CH3 – CH – CH3 prefix + alkane
2 - chloro + propane
Cl
= 2 - chloro propane

(3) In case of ether, group with less number of carbon atom attached to oxygen is given
prefix alkoxy and group with greater number of carbon is taken as alkane.
(4) In ester (–COOR), after writing the name of alkyl group connected in place of R,
remaining carbon is taken as alkane and “oate” suffix is added.
(5) In 2 o and 3o amine, group with highest number of carbon is taken as alkane, suffix
“amine” is added and remaining groups taken as alkyl group.

11. From which of the following compounds suffix “oate” is added ?
(A) aldehyde (B) ketone (C) ester (D) ether
12. In IUPAC nomenclaure of cyanide compound which suffix is added ?
(A) cyano (B) cyanide (C) cyanate (D) nitrile
13. Compounds with which hybridization of carbon get suffix ‘ene’ in IUPAC nomenclature ?
(A) dsp 2 (B) sp2 (C) sp3 (D) sp
14. Mention the IUPAC name of CH3CH2COOCH3.
(A) butanoate (B) ethyl ethanoate (C) Methyl propanoate (D) Methyl propanoate
15. Identify the formula of butane nitrile :
(A) CH 3CH 2CH 2CN (B) CH 3CH 2CH 2NH 2
(C) CH 3CH 2CH 2CH 2CN (D) CH 3CH 2CH 2CH 2NO 2
16. In which compound having only one functional group, prefix is added in nomenclature ?
(A) alcohol (B) ether (C) amide (D) ketone

Answers : 11. (C), 12. (D), 13. (B), 14. (D), 15. (A), 16. (B)
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• Homologous series and isomerism
• Series of compounds having difference of – CH2 – between two successive member is
called homologous series. eg. alkane series.
CH4, C2H6, C3H8, C4H10, C5H12
• Compounds having same molecular formula but different properties are called isomers of
each other and this phenomenon is called isomerism.
• Classification of isomerism as follows :

Isomerism
¯

¯ ¯

Structural isomerism Stereo isomerism

¯

¯ ¯ ¯ ¯

(1) chain isomerism conformational geometrical optical
(2) position isomerism isomerism isomerism isomerism
(3) functional isomerism ¯ ¯ ¯

(4) metamerism staggered as enantiomers
(5) Tautomerism eclipsed trans diasteveomers
(6) ring-chain isomerism meso

Chain isomerism : arrangement of carbon atom is linear or branched.
eg., CH3 – CH2 – CH2 – CH3 and CH3 – CH – CH3
n – butane I
CH 3
iso butane

Position isomerism : Position of functional group is different.
eg., CH3 – CH2 – CH2 – OH and CH3 – CH – CH3
Propan-1-ol I
OH
propan – 2 – ol
Functional group isomerism : Functional groups are different.
eg., CH3 – CH2 – COOH and CH3 – COOCH3
Propanoic acid Methyl ethanoate
This isomerism is observed in ester-acid alcohol-ether, aldehyde-ketone and 1o, 2o, 3o amine.
metamerism : functional group is same but alkyl groups connected on both side have
different numbers of carbon atoms.
eg., CH3 – O – CH2CH2CH3 and CH3CH2 – O – CH2CH3
1–methoxy propane Ethoxy ethane
This isomerism is observed in ether, ester, 2o and 3o amine.

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Tautomerism : Isomers form due to migration of a – hydrogen atom.
eg., CH3 – C – CH3 CH2 = C – CH3
II I
O OH
keto form enol form
Ring chain isomerism : one isomer has cyclic structure and other isomer has linear chain struc-
ture.
eg., CH2 and CH3 – CH = CH2
Propene
CH2 CH2
Cyclo propane

17. How do successive members of homologous series differ from each other ?
(A) – CH3 group (B) – CH2 group (C) – C2H5 group (D) – CH group
18. Between successive members of alkane homologous series, difference between molecular
mass is ...... .
(A) 16 amu (B) 12 amu (C) 14 amu (D) 18 amu
19. Which of the following pair shows functional group isomerisms ?
(A) aldehyde and alcohol (B) alcohol and amine
(C) carboxylic acid and aldehyde (D) alcohol and ether
20. How many isomers does pentane chain contain ?
(A) 2 (B) 3 (C) 4 (D) 6
21. How many isomers of C3H9N compound having amine group ?
(A) 5 (B) 3 (C) 4 (D) 6
22. How many chain isomers are possible for C4H8 ?
(A) 3 (B) 5 (C) 4 (D) 6
23. Which of the following compound does not show position isomerism ?
(A) butene (B) butanal (C) but-1-amine (D) butyne
24. How many chain isomers are possible for pentanol ?
(A) 4 (B) 5 (C) 6 (D) 7
25. Which of the following is not isomer of diethyl ether ?
(A) Methyl propyl ether (B) butan-1-ol
(C) 2-Methyl-propan-1-ol (D) diethyl ketone
26. Number of isomers of C4H10 O ...... .
(A) 7 (B) 8 (C) 5 (D) 6
27. Isomer of ethyl alcohol is ...... .
(A) diethyl ether (B) dimethyl ether (C) ethanal (D) acetone
28. ...... type of isomerism is observed in urea.
(A) chain (B) position (C) geometrical (D) tautomerism

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29. CH3–O–C3H7 and C2H5–O–C2H5 shows which type of isomerism.
(A) position (B) chain (C) metamerism (D) tautomerism
30. Pair of Acetone and prop-1-en-2-ol are example of which type of isomerism ?
(A) Position (B) Tautomerism (C) Functonal group (D) Metamerism
31. Which type of isomerism is observed in but-2-ene ?
(A) functional group (B) metamerism (C) geometrical (D) optical
32. Diethyl ether and methyl propyl ether are ...... .
(A) position isomer (B) functional group isomer
(C) metamers (D) rotamers
33. Cyclo alkene and alkyne are which type of isomers ?
(A) Chain (B) Functional group (C) Metamers (D) Votamers
34. n–propyl alcohol and iso propyl alcohol are examples of which type of isomerism.
(A) position (B) chain (C) geometrical (D) stereo
35. Functional isomers of propionic acids are ...... .
(A) HCOOC2H5 and CH3COOCH3 (B) HCOOC2H5 and C3H7COOH
(C) C3H7OH and CH3COCH3 (D) CH3CH2COOH and C3H7OH
36. How many isomers of C6H14 are possible ?
(A) 3 (B) 4 (C) 5 (D) 6
37. Which isomerism is ehibited by C5H10O ?
(A) Position (B) Functional group (C) Metamerism (D) All the given
38. Which isomerism is not shown by alkene ?
(A) chain (B) geometrical (C) position (D) metamerism
39. Which type of compounds do not have metamers ?
(A) ketone (B) amine (C) ether (D) alcohol
40. Which isomerism is shown by R – C º N and R – N+ º C– ?
(A) Position (B) Functional group (C) metamerism (D) Tautomerism
Answers : 17. (B), 18. (C), 19. (D), 20. (B), 21. (C), 22. (A), 23. (B), 24. (A), 25. (D),
26. (A), 27. (B), 28. (D), 29. (C), 30. (B), 31. (C), 32. (C), 33. (D), 34. (A),
35. (A), 36. (C), 37. (D), 38. (D), 39. (D), 40. (B)

• Nomenclature of organic compounds (Common and IUPAC)
Organic compounds are named on the basis of rules of IUPAC which is called its IUPAC
name. Apart from this some organic compounds are known by common name or popular name.
Study IUPAC rules appropriately from text book.
• Simple method to write IUPAC name.
attach 1st determine longest attach
prefix of hydrocarbon root suffix of
substituted (alkane, alkene, alkyne) principal
group or branch functional
group
480

Priority order of principal functional group is as follows :
– COOH > – COOR > – CONH2 > CN > CHO > – CO > – OH >
– NH2 > C = C > C º C > Substituted groups (–OR, –X, –NO2, –R)
Substituded group or branch means which is not included in chain of hydro carbon root.
• Examples of IUPAC nomenclature
CH3
I
(1) 1CH3 – 2C – 3CH2 – 4CH – 5CH3 2, 2, 4 – trimethyl pentane
I I ¯ ¯

CH3 CH3 prefix hydrocarbon root
4
(2) CH2 = 3CH – 2CH2 – 1CH2 – OH But – 3 – en – 1 –ol
¯ ¯

hydro carbon root suffix
(3) CH3 – CH – CH2 – CH – CH2 – CH – 1CH3
7 6 5 4 3 2

I I I
Br CH2–CH3 OH
6–Bromo–4–ethyl heptan – 2 – ol
¯ ¯ ¯

prefix Hydro carbon root suffix
4
(4) CH3 – 3CH – 2CH2 – 1COOH 3–hydroxy butanoic acid
I ¯ ¯

OH prefix hydro carbon root + suffix

41. Correct IUPAC name of CH3CH2CH(CH3)CH (C2H5)2 is ...... .

(A) 4–ethyl 3–Methyl hexane (B) 3–ethyl 4–Methyl hexane

(C) 4–Methyl 3–ethyl hexane (D) 2, 4–diethyl pentane

42. Give IUPAC name of following compound :

CH3 – CH – C – CH2 – CH3

I II

CH3 CH2

(A) 2–ethyl 3–Methyl but-l-ene (B) 2–iso propyl but-1-ene

(C) 2–Methyl 3–ethyl but-3-ene (D) 2–(1–Methyl ethyl) but-l-ene

43. What is correct IUPAC name of CH2 = CH – CH2 – CH2 – CH2 – Cl ?

(A) 1–chloro pent-4-ene (B) Pent-4-ene chloride

(C) 5–chloro pent-1-ene (D) 1–chloro pentene

44. IUPAC name of CH2 = CH – CH2 – CH2 – OH is ...... .

(A) But-3-en-1-ol (B) but-1-en-3-ol

(C) 4-hydroxy but-1-ene (D) butenol

481

45. Give IUPAC name of the following compound : CH3 – CH – CH = C – CHO
OH CH3
(A) 4–hydroxy-1-Methyl pentenal (B) 4-hydroxy-2-Mehtyl-pent-2-enal
(C) 2-hydroxy-4-Methyl pent-3-en-5-al (D) 2-hydroxy-3-methyl pent-2-enal
46. IUPAC name of urea is ...... .
(A) diamino ketone (B) amino methanamide (C) amino ethanamide (D) amino acetamide
47. IUPAC name of CH3 – CH = CH – C º CH is ...... .
(A) Pent-2-en-4-yne (B) Pent-2-yn-3-ene
(C) Pent-3-en-1-yne (D) Pent-2-en-5-yne
48. Mention IUPAC name of NC – CH2 – CH – CH2 – CN
I
CH 2CN
(A) iso butane tri nitrile (B) 3–cyano methyl pentane dinitrile
(C) 2, 2–Bis (Cyano methyl) enthane nitrile (D) tri ethane nitrile methane
49. IUPAC name of HOOC – CH = CH – COOH is ...... .
(A) but-2-en-oic acid (B) butene di oic acid
(C) butene–1, 4–dioic acid (D) but-2-ene dioic acid

50. Mention IUPAC Name of :

(A) Nonane (B) etetra ethylmethane
(C) 3–ethyl pentane (D) 3, 3–diehtyl pentane

Answers : 41. (B), 42. (A), 43. (C), 44. (A), 45. (B), 46. (B), 47. (C), 48. (B), 49. (D),
50. (D)

• Electronic transfer in covalent bond
In organic compounds electron pair of covalent bond transfer by four different ways.
(1) Inductive Effect :
s – electron pair present in carbon chain transfer to more electronegative atom or group.
On moving away from electronegative atom to carbon atom away from it this effect reduces.
This effect is permanent
eg., C > C >> C d+ > > > Cd –
Inductive effect (I–effect) are of two types : (1) If electron attracting group is connected at
the end of chain, then it is called – I effect. Decreasing order of – I effect is as follows :
R3N+ > +NH3 > – NO2 > CN > SO3H > CHO > CO > COOH > COCl > COOR > CONH2 >
F > Cl > Br > I > OH > NH2 > C6H5 > H
If electron donor group is attached at the end of chain then it is called + I effect. Decreasing
order of + I effect :
R3+C > R2 +CH > R – +
CH2 > +
CH3
3o 2o 1o methyl

482

(2) Electromeric Effect :
• p-electron pair connected by double or triple bond transfer to one of the atom (mainly
electronegative atom) due to contact of attacking reagent.
• This effect is temporary.
• Electromeric effect is of two types :
(i) If p electron pair displaces towards the attacking reagent then it is called +E effect

+ +
eg. >C = C < + H >C C<
H
(ii) If p electron pair displaces away from the attacking reagent then it is called –E effect.

-
eg. >C = O + CN >C O
CN
(3) Mesomeric effect or Resonance effect :
• In some organic compounds, two or more than two structures are in vibration state. Such
structures are called resonating structures. Their real structure is shown by intermediate state
between two or more than two structures. Such structures are called resonance hybrid structure.
This effect is known as mesomeric or resonance effect.
• This effect is observed in the system having alternate s and p bond.
eg., benzene C6H6

• Molecules having resonance structure have greater stability.
• e– donor groups have +R effect and
e– attracting groups have –R effect.
(4) Hyper conjugation :
• When C – C single bond is directly bonded with C = C or benzene ring, then s e– pair of
C – H bond is attracted towards double bond. This effect called hyper conjugation.

H H

H – C ––– CH = CH2 H – C = CH ––– CH2
H H

• Order of + I effect of alkyl group is 3o > 2o > 1o > methyl but when alkyl group is
connected with double bond or benzene ring this order is reversed as methyl > 1o > 2o > 3o this is
hyper-conjugation.
• Importance of hyper conjugation is as follows :
(i) Greater the number of –CH3 group bonded with C = C greater is the stability of alkene.
(ii) Stability order of carbocation and free radical is 3o > 2o > 1o > methyl.
(iii) Bond length of C – C bond adjacent to C = C decreases.

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51. Which of the following group has least – I effect ?
(A) –NO 2 (B) –COOH (C) –F (D) –N +R 3
52. Which of the following group has highest + I effect ability ?
(A) (CH 3 ) 3C– (B) (CH 3) 2 –CH– (C) CH3 – CH2 – (D) – CH3
53. Which group has maximum hyper conjugative effect ?
(A) R 3 C– (B) R 2CH– (C) R–CH 2– (D) – CH3
54. In which of the following compound electromeric effect is not observed ?
(A) alkene (B) ether (C) aldehyde (D) ketone
55. In which of the following molecule resonance (delocalized electrons) is observed ?
(A) methane (B) ethane (C) benzene (D) cyclohexane
56. Which group has +R effect ?
(A) –CN (B) –CHO (C) –NH 2 (D) –NO 2
57. Which of the following alkene is the most stable ?
(A) CH 3–CH=CH–CH 3 (B) (CH 3) 2C=CH 2
(C) (CH3)2 C=CHCH3 (D) (CH 3) 2C=C(CH 3) 2
58. Which carbo cation is the most stable ?
(A) CH 3CH 2+ (B) CH2 = CH+ (C) CH º C+ (D) C 6H 5+
59. Which type of conjugation is hyper conjugation ?
(A) s – p (B) s – s (C) p – p (D) p – s
60. Which of the following free radical is the most stable ?
· · · ·
(A) C 6H 5CH 2–CH 2 (B) C 6H 5C HCH 3 (C) CH3 CH 2 (D) CH 3 CHCH 3

Answers : 51. (C), 52. (A), 53. (D), 54. (B), 55. (C), 56. (C), 57. (D), 58. (A), 59. (A),
60. (B)

• Fission of covalent bond
Fission of covalent bond is done in two different ways.
(1) Homolytic fission : Two atoms bonded by covalent bond are separated by gaining one
electron each and as a result free radical is formed.
hυ
eg., Cl − Cl 
→ Cl ⋅ + Cl ⋅ free radicals

(2) heterolytic fission : If during fission of covalent bond more electronegative atom
separated by gaining two electrons and as a result positive and negative ion are formed.
+
eg., CH3 – Cl ® CH3 + Cl– :
Positive ion negative ion
Positive ion or some neutral molecules which have capacity to accept electrons are called
electrophiles.
eg., +NO2, +Cl, +CH3, BF3, AlCl3, SnCl2

484

• Negative ion or some neutral molecules which have capacity to donate electrons are called
nucleophile.
− − •• ••
eg., X , OH , NH 2 , NH3, H2O
••

• Chemical species having positively charged carbon is called carbocation or carbonium ion.
+
eg., CH3
–
• Chemical species having negatively charged carbon is called carbanion eg. CH3
• Stability order of carbanion : methyl > 1o > 2o > 3o

61. What is called positive ion formed by heterolytic fission of covalent bond ?
(A) Electrophile (B) Nucleophile (C) Lewis acid (D) Both (A) and (C)
62. Which carbanion has the least stability ?

(A) CH3 (B) CH3 – CH2 (C) (CH3)2 CH (D) (CH3)3 C
63. Which of the following is nucleophile ?
(A) H 2 O (B) BF 3 (C) AlCl3 (D) SO 3
64. Which particles are obtained by homolytic fission of A – B bond ?
(A) One positive and one anion (B) two carbocation
(C) one carbocation and one carb anion (D) two free radicals
65. Which sentence is incorrect with reference to free radical ?
(A) They are paramagnetic species having unpaired electrons.
(B) They are electrically neutral and highly reactive.
(C) Carbon atom of free radical has 7 electron in valence orbital.
(D) They are highly stable and long lasting.
Answers : 61. (D), 62. (D), 63. (A), 64. (D), 65. (D)

• Types of organic reactions
Substitution reaction : Atom or group atom present in organic compound is substituted by another
atom or group of atom.
eg., CH4 + Cl2 
→ CH3Cl + HCl
Addition reaction : p bond of double or triple bond is broken and another molecule is added.
eg., CH2 = CH2 + Cl2 
→ Cl – CH2 – CH2 – Cl
Eliminaton reaction : From adjacent carbon atoms, a group of atoms is removed as molecule and as a
 Al O 
 2 3
result between two carbon new p–bond forms. eg., CH3 – CH2 – OH  
→
350 − 400o C CH2 = CH2 + H2O
Rearrangement reaction : Displacement of atom or group of atoms to other place within the same
molecule.
 AlCl 
 3→
eg., CH3 – CH2 – CH2 – CH3  CH3 – CH – CH3
n–butane I
CH 3

485

iso-butane
66. Mention the type of the following ®
alcohol reaction :

CH3 – CH2 – Cl + KOH CH2 = CH2 + KCl + H2O
(A) substitution (B) addition (C) elimination (D) rearrangement
67. Identify the type of reaction : CH3 CH2 I + acqueous KOH ® CH3CH2OH + KI.
(A) substitution (B) addition (C) elimination (D) halogenation
68. Which product is obtained by hydrogenation of ethene in presence of nickel catalyst ?
(A) ethanol (B) ethyne (C) ethane (D) methane
69. In which reaction new p– bond is formed ?
(A) substitution (B) addition (C) elimination (D) rearrangement
70. Which of the following is elimination reaction ?
(A) Chlorination of ethene (B) Hydration of ethene
(C) Conversion of But-1-ene into But-2-ene (D) Dehydration of ethanol
71. Number of s and p bonds in pent-l-en-4-yne is ....... .
(A) 3, 10 (B) 9, 4 (C) 4, 9 (D) 10, 3
72. In buta-1,2 – diene compound ....... .
(A) Only sp hybridized carbon atoms (B) Only sp2 hybridized carbon atoms
(C) sp and sp2 hybridized carbon atoms (D) sp, sp2 and sp3 hybridized carbon atoms
73. Which of the following IUPAC name is incorrect ?

(A) CH3 – C – CH – CH3 2–Methyl butan-3-one

O CH3

(B) CH3 – CH – CH – CH3 2, 3 – di methyl pentane
I I
CH3 CH2 – CH3
(C) CH3 – C º C×CH (CH3)2 4 – Methyl pent – 2 – yne
(D) CH3 – CH – CH – CH3 2 – bromo 3 – chlorobutane
I I
Cl Br
74. Match the column X with column Y and choose correct option :
Column -X Column-Y
(i) free radical (A) Lewis base
(ii) electrophile (B) Electrically neutral
(iii) nucleophile (C) Complete octet in valence orbit
(D) Lewis acid
(E) Electron octet is incomplete and in valence orbit single e–
(F) Electron octet incomplete
(A) (i)-(B),(E), (ii)-(D),(F), (iii)-(A),(C) (B) (i)-(A),(C), (ii)-(D),(F), (iii)-(B),(E)
(C) (i)-(D),(F), (ii)-(B),(E), (iii)-(A),(C) (D) (i)-(B),(E), (ii)-(A),(C), (iii)-(D),(F)
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75. From the given sentences select the number of correct sentences and choose appropriate option.
(1) IUPAC name of propyl cyanide is propane nitrile.
(2) IUPAC name of diethyl ether is ethoxy ethane.
(3) Ethanol and vinyl alcohols are tautomers.
(4) Methoxy propane and ethoxy ethane are metamers.
(5) Stability of 2, 3–dimethyl but-2-ene is more than 2–methyl but-2-ene.
(6) Stability order of carbo cation is 1o < 2o < 3o.
(7) In Elimination reaction hybridization of carbon atom do not change.
(8) Nucleophites are Lewis acid.
(A) (1), (3), (5), (7) (B) (2), (4), (6), (8) (C) (2), (4), (5), (6) (D) (2), (4), (6), (7)
76. In column I bondline structures and in column II IUPAC names are given. Match them and
select correct option.
Column-I Column-II
(1) OH (p) but-1-ene (A) (1)–(q), (2)–(p), (3)–(t)

(2) (q) 2–hydroxy butane (B) (1)–(s), (2)–(r), (3)–(u)

(3) (r) but-2-ene (C) (1)–(q), (2)–(r), (3)–(u),

(s) pentan-2-ol (D) (1)–(q), (2)–(p), (3)–(u),
(t) 3, 3–dimethyl butane
(u) 3, 3–dimethyl pentane

77. On matching common name in column-I with structural formula of column II which of the
following pair is correct :
Column-I Column-II
(P) formic anhydride (W) CH 3CHO (A) (P)-(X), (Q)-(W), (R)-(Y), (S)-(Z)
(Q) methyl acetate (X) CH3 – COOCH3 (B) (P)-(Z), (Q)-(Y), (R)-(W), (S)-(X)
(R) acetamide (Y) (HCO)2O (C) (P)-(Y), (Q)-(X), (R)-(Z), (S)-(W)
(S) acetaldehyde (Z) CH3CONH2 (D) (P)-(W), (Q)-(Z), (R)-(X), (S)-(Y)

78. Match the column X with column Y and mention correct option :
Column X Column Y
(Hydrocarbon group) (Common name)
(a) (CH3)2 CH– (p) tertiary butyl (A) (a)-(t), (b)-(q), (c)-(t), (d)-(u)
(b) CH3CH2 – CH – (q) iso butyl (B) (a)-(r), (b)-(q), (c)-(t), (d)-(u)
I
CH3
(c) (CH3)3 C– (r) vinyl (C) (a)-(r), (b)-(s), (c)-(t), (d)-(u)
(d) CH2 = CH– (s) secondary butyl (D) (a)-(t), (b)-(s), (c)-(p), (d)-(r)
(t) iso propyl
(u) ethyl

Answers : 66. (C), 67. (A), 68. (C), 69. (C), 70. (D), 71. (D), 72. (D), 73. (A), 74. (A),
75. (C), 76. (B), 77. (C) 78. (D)

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• Read the paragraph carefully and answer the questions given below paragraph :

• Paragraph : While giving IUPAC names of the organic compounds having more than one
functional group the functional group with highest priority is considered as principal functional
group and other functional groups are considered as substituted group. Carbon chain is given
numbers in such away that principal functional group gets least number. (Que. No. 79 80)

79. Give IUPAC name of the following compound :

CH3 – CH = C – CH2 – COOH
CONH2

(A) 2–ethyledene–3–carboxy propanamide (B) 3–ethyl-3-carbyl, propanoic acid
(C) 2-carboxymethyl but-2-en-1-amide (D) 3-carbamoyl pent-3-en-1-oic acid
80. IUPAC name of following compound is ....... .

OMe
NC
OMe O

(A) Methyl 4-cyano-3-methoxy pentanoate
(B) 4-methoxy-carbonyl-1-methoxy butanenitrile
(C) Methyl-4-cyano-3-methoxy butanoate
(D) 4-carbmethoxy-2-methoxy butanenitrile
Instruction : Question no. 81 to 90 are assertion (A) and reason (R) types. Their four
options are as follows. Select correct option :
(A) Assertion (A) and reason (R) both are correct and reason (R) is correct explanation of
assertion (A).
(B) Assertion (A) and reason (R) both are correct but reason (R) is not correct explanation
of assertion (A).
(C) Assertion (A) is correct while reason (R) is incorrect.
(D) Assertion (A) is incorrect while reason (R) is correct.

81. Assertion (A) : –CN is called cyclohexane carbonitrile.

Reason (R) : Suffix nitrile is given to alicyclic ring.
82. Assertion (A) : IUPAC name of CH3 CH = CH – C º CH is pent – 2 – en – 4 – yne.
Reason (R) : While determining the position of functional group rule of least set of
locant is applied.
83. Assertion (A) : IUPAC name of C2H5 – C – CH2 – OH is 2 – ethyl prop-2-en-1-ol
II
CH 2
Reason (R) : Instead of methylene ethyl is accepted as substituted group because
alphabatically ‘e’ of ethyl comes before ‘m’ of methylene.

488

84. Assertion (A) : Solubility of maleic acid in water is more than fumaric acid while melting
point of fumaric acid is more than maliec acid.
Reason (R) : Molecules of fumaric acid are more symmetric hence they can arrange
more closely packed in crystal structure.
85. Assertion (A) : In styrene geometrical isomerism is not observed.
Reason (R) : All the atoms in styrene molecule are in one plane
86. Assertion (A) : free radicals are paramagnetic.
Reason (R) : Free radicals have unpaired electrons.
87. Assertion (A) : Stability order of carbocation is 3o > 2o > 1o .
Reason (R) : As distance between alkyl group and reaction site increases +I effect
decreases.
88. Assertion (A) : Boiling point of cis isomer is more than trans isomer.
Reason (R) : Dipole moment of cis isomer is more than trans isomer.
89. Assertion (A) : Allyl free radicals are more stable than alkyl free radical.
Reason (R) : Due to resonance allyl free radicals are more stable.
90. Assertion (A) : CHBr = CHCl has geometrical isomers but CH2Br – CH2Cl has no
geometrical isomers.
Reason (R) : For geometrical isomers presence of C = C is essential..

Answers : 78. (D), 79. (D), 80. (C), 81. (C), 82. (D), 83. (C), 84. (A), 85. (B), 86. (A),
87. (B), 88. (A), 89. (A), 90. (C).

•

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