Solutions
Question 1. Answer the following :
(A) Write condensed formulae and bond line formulae for the following structures.
(B) Write dash formulae for the following bond line formulae.
(C) Write bond line formulae and condensed formulae for the following compounds
(a) 3−methyloctane
IUPAC Name : 3−methyloctane
Condensed Formula : CH3CH2CH(CH3)(CH2)4CH3
Bond line formula :
(b) hept−2−ene
IUPAC Name : hept−2−ene
Condensed Formula : CH3CH=CH(CH2)3CH3
Bond line formula :
(c) 2, 2, 4, 4− tetramethylpentane
IUPAC Name : 2, 2, 4, 4− tetramethylpentane
Condensed Formula : (CH3)3C-CH2-C(CH3)3
Bond line formula :
(d) octa−1,4−diene e. methoxyethane
IUPAC Name : octa−1,4−diene e. methoxyethane
Condensed Formula : CH2=CH-CH2-CH=CH(CH2)2CH3
Bond line formula :
(D) Write the structural formulae for the following names and also write correct IUPAC names for them.
(a) 5−ethyl−3−methylheptane
Wrong Name : 5−ethyl−3−methylheptane
Correct IUPAC Name : 3−ethyl−5−methylheptane
Structure :
(b) 2,4,5−trimethylthexane
Wrong Name : 2,4,5−trimethylthexane
Correct IUPAC Name : 2,3,5−trimethylthexane
Structure :
(c) 2,2,3−trimethylpentan−4−ol
Wrong Name : 2,2,3−trimethylpentan−4−ol
Correct IUPAC Name : 3,3,4−trimethylpentan−2−ol
Structure :
(E) Identify more favourable resonance structure from the following. Justify.
(a) Both the resonance structures (I) and (II) are equivalent to each other, and therefore, are equally stable.
(b) In structure (I) : One carbon (C+) has only 6 valence electrons, involves separation of opposite charges; the resonance structure I has - ve charge on more electronegative O and + ve charge on more electropositive C. It has more stability.
In structure (II) : Oxygen has only 6 valence electrons, involves separation of opposite charge, has - ve charge on the more electro-positive 'C and + ve charge on more electronegative O. All these factors are unfavourable. Therefore, structure (II) is least stable.
(F) Find out all the functional groups present in the following polyfunctional compounds.
(a) Dopamine a neurotransmitter that is deficient in Parkinson's disease.
Dopamine has two functional groups : (-OH)
(b) Thyroxine the principal thyroid hormone.
Thyroxine has five functional groups : (-OH), I, -O-, -NH2 and -COOH.
(c) Penicillin G a naturally occurring antibiotic
Penicillin G has four functional groups : C=O, -HN, -S-, -COOH.
(G) Find out the most stable species from the following. Justify.
(a) 
tert-butyl carbocation is most stable species.
(b) Br3 tribromide carbanion is most stable species.
(c) C+H3 methyl carbocation is most stable species.
(H) Identify the α−carbons in the following species and give the total number of α−hydrogen in each.
(I) Identify primary, secondary, tertiary and quaternary carbon in the following compounds.
Question 2. Match the pairs
| Column 'A' | Column 'B' |
| i. Inductive effect | a. delocalization of π electrons |
| ii. Hyperconjugation | b. displacement of π electrons |
| iii. Resonance effect | c. delocalization of σ electrons |
| d. displacement of σ electrons |
(i) Inductive effect-(d) displacement of σ electrons
(ii) Hyperconjugation-(c) delocalization σ electrons electrons
(iii) Resonance effect-(a) delocalization of π electrons
Question 3. What is meant by homologous series ? Write the first four members of homologous series that begins with
(A) CH3CHO (B) H−C≡C−H
Also write down their general molecular formula.
Homologous series : A series of compounds of the same family in which each member has the same type of carbon skeleton and functional group and differs from the next member by a constant difference of one methylene group (-CH2-) in its molecular and structural formula is called a homologous series.
(A)
| Homologue
Number |
Molecular formula
C2H2nO |
Structural formula | Name |
| 1st | C2H4O | CH3CHO | Acetaldehyde |
| 2nd | C3H6O | CH3CH2CHO | Propionaldehyde |
| 3rd | C4H8O | CH3-(CH2)2CHO | Butyraldehyde |
| 4th | C5H10O | CH3-(CH2)3CHO | Valeraldehyde |
(B)
| Homologue
Number |
Molecular formula
C2H2nO |
Structural formula | Name |
| 1st | C2H2 | H-C≡C-H | Ethyne |
| 2nd | C3H4 | CH3-C≡C-H | Propyne |
| 3rd | C4H6 | CH3-C≡C-CH3 | Butyne |
| 4th | C5H8 | CH3-CH2-C≡C-CH3 | Pentyne |
Question 4. Write IUPAC names of the following
(a) IUPAC name : 2, 7-dimethyl nonane
(b) IUPAC name : penet-3-en-1-amine
(c) IUPAC name : 4-Methyl Hex-1-en-5-yne
(d) IUPAC name : Ethyl proponoate
(e) IUPAC name : Phenyl ethanol
(f) IUPAC name : 3-methyl phenol
Question 5 Find out the type of isomerism exhibited by the following pairs.
(A) CH3−CH2−NH−CH2−CH3 and CH3−NH−CH2−CH2−CH3
The type of isomerism exhibited : Metamerism.
(B)
The type of isomerism exhibited : Functional group isomerism.
(C)
The type of isomerism exhibited : Keto-enol tautomerism.
(D)
The type of isomerism exhibited : Position isomerism and functional group isomerism.
Question 6. Draw resonance structures of the following :
(A) Phenol
(A) Resonance structures for phenol :
(B) Benzaldehyde
Resonance structures for benzaldehyde :
(C) Buta−1,3−diene
Resonance structures for buta−1,3−diene :
(D) Acetate ion
Resonance structures for acetate ion :
Question 7. Distinguish :
(A) Inductive effect and resonance effect
| Inductive effect | Resonance effect |
| The presence of a polar covalent bond is required. | The presence of a conjugated T electron system or species having an atom carrying p orbital attached to multiple bonds is required. |
| The polarity is induced in adjacent carbon-carbon single (covalent) bonds due to the presence of an influencing group (more electronegative atom than carbon). | The polarity is produced in the molecule by the interaction of conjugated T bonds (or that between n bond and p orbital on the adjacent atom). |
| Depending on the nature of the influencing group it is differentiated as +I effect and -I effect. | Depending on the nature of the influencing group it is differentiated as +R and -R effect. |
| The direction of the arrow head denotes the direction of the permanent electron displacement. | The delocalisation of n electrons is denoted by using curved arrows. |
(B) Electrophile and nucleophile
| Electrophile | Nucleophile |
| They are electron deficient species. | They are electron rich species. |
| They attack electron rich centre of the substrate. | They attack electron deficient centre of the substrate. |
| Negative charge centre in the substrate attracts electrophiles. | Positive charge centre in the substrate attracts nucleophiles. |
| They are Lewis acids. | They are Lewis bases. |
| They are cations or electron deficient atoms or molecules. | They are anions or molecules with atoms
having lone pair of electrons. |
| Examples : H+, NO2+, BF3. | Examples : OH-, Cl-, NH3. |
(C) Carbocation and carbanion
| Carbocation | Carbanion |
| It is an organic species in which carbon atom carries a positive charge. | It is an organic species in which a carbon atom carries a negative charge. |
| It is an electron deficient species. | It is an electron rich species. |
| It has a planar geometry. | It has a tetrahedral geometry. |
| The carbon atom involved is sp2- hybridised. | The carbon atom involved is sp3- hybridised. |
| It is a Lewis acid. | It is a Lewis base. |
| It is an electrophile.
Example : (CH3)3C+ |
It is a nucleophile.
Example : H3C- |
(D) Homolysis and heterolysis
| Homolysis | Heterolysis |
| The symmetrical breaking of a covalent bond in which each departing atom takes one electron from the bonding pair is called as homolytic fission. | The unsymmetrical breaking of a covalent bond in which one of the departing atoms retains the bonding pair is called heterolytic fission. |
| In this type of fission, the formation of free radicals (uncharged species) bearing unpaired electrons takes place. | In this type of fission, the formation of charged species called ions, like carbocation or carbonium ion takes
place. |
| The covalent bond between two atoms of the same element or two atoms having nearly the same electronegativity breaks in this manner. | The covalent bond between two atoms of the different elements or two atoms having different electronegativity values breaks in this manner. |
| This takes place favourably in a nonpolar solvent. | This takes place favourably in a polar solvent. |
| Generally, reaction takes place at high temperature or in presence of UV light or peroxides. | Heterolysis takes place in solutions (polar condition). |
Question 8. Write true or false. Correct the false stament
(A) Homolytic fission involves unsymmetrical breaking of a covalent bond.
False. (Homolytic fission involves symmetrical breaking of a covalent bond.)
(B) Heterolytic fission results in the formation of free radicals.
False. (Heterolytic fission results in the formation of charged species.)
(C) Free radicals are negatively charged species.
False. (free radicals are chargeless species.)
(D) Aniline is heterocyclic compound.
True.
Question 9. Phytane is naturally occurring alkane produced by the alga spirogyra and is a constituent of petroleum. The IUPAC name for phytane is 2,6,10,14−tetramethylhexadecane. Write zig−zag formula for phytane. How many primary, secondary, tertiary and quaternary carbons are present in this molecule.
IUPAC name for Phytane is 2,6,10,14-tetramethylhexadecane
Zig-zag formula :
Primary carbon atoms-2
Secondary carbon atoms-10
Tertiary carbon atoms-3
Quaternary carbon atom-1
Question 10. Observe the following structures and answer the questions given below.
(i) CH3 − CH2 − CH2 − CHO
(a) What is the relation between (i) and (ii) ?
(b) Write IUPAC name of (ii).
(c) Draw the functional group isomer of (i).
(a) The structure (i) is Butanal and the structure (ii) is 2-methyl propanal.
The structure (i) and (ii) are metamers of each other as they have the same molecular
formula C4HO and the same functional group (-aldehyde) but have different distribution atoms attached to aldehydic carbon.
(b) The IUPAC name : 2-methyl propanal
(c) The functional group isomer of (i) is -CHO (aldehyde).
Question 11. Observe the following and answer the questions given below
CH3−CH3 \(\underrightarrow{U.V.Light}\) CH3 + CH3
(a) Name the reactive intermediate produced
(b) Indicate the movement of electrons by suitable arrow to produce this intermediate
(c) Comment on stability of this intermediate produced.
(a) The reactive intermediate is (CH3) free radical.
(b) 
(c) A free radical is highly reactive and unstable and it has only transitory existence. This reactive species subsequently react with another radical or molecule to restore stable bonding pair.
Question 12. An electronic displacement in a covalent bond is represented by following noation.
(A) Identify the effect
(B) Is the displacement of electrons in a covalent bond temporary or permanent.
(A) It is inductive effect.
(B) The displacement of electrons in a covalent bond is permanent.
Question 13. Draw all the no−bond resonance structures of isopropyl carbocation.
Question 14. A covalent bond in tert−butylbromide breaks in a suitable polat solvent to give ions.
(A) Name the anion produced by this breaking of a covalent bond.
(B) Indicate the type of bond breaking in this case
(C) Comment on geometry of the cation formed by such bond cleavage.
(A) The anion produced by breaking covalent bond in tert-butyl bromide is bromide ion (Br-)
(B) The type of bond breaking in tert-butyl bromide : Heterolytic fission.
(C) Geometry of tert-butyl carbocation : Tetrahedral geometry.
Question 15. Choose correct options
(A) Which of the following statements are true with respect to electronic displacement in covalent bond ?
(a) Inductive effect operates through π bond
(b) Resonance effect operates through σ bond
(c) Inductive effect operates through σ bond
(d) Resonance effect operates through π bond
(i) a. and b
(ii) a and c
(iii) c and d
(iv) b and c
(iii) c and d
(B) Hyperconjugation involves overlap of ..... orbitals
(a) σ − σ
(b) σ −p
(c) p − p
(d) π – π
(b) σ −p
(C) Which type of isomerism is possible in CH3CHCHCH3?
(a) Position
(b) Chain
(c) Geometrical
(d) Tautomerism
(a) Position
(D) The correct IUPAC name of the compound 
is .....
(a) hept−3−ene
(b) 2−ethylpent−2−ene
(c) hex−3−ene
(d) 3−methylhex−3−ene
(d) 3−methylhex−3−ene
(E) The geometry of a carbocation is ......
(a) linear
(b) planar
(c) tetrahedral
(d) octahedral
(b) planar
(F) The homologous series of alcohols has general molecular formula .........
(a) CnH2n+1OH
(b) CnH2n+2OH
(c) CnH2n−2OH
(c) CnH2nOH
(a) CnH2n+1OH
(G) The delocalization of electrons due to overlap between p−orbital and sigma bond is called
(a) Inductive effect
(b) Electronic effect
(c) Hyperconjugation
(d) Resonance
(c) Hyperconjugation