Chapter 12 Organic Chemistry Some Basic Principles and Techniques Solutions
Question - 1 : - What are hybridisationstates of each carbon atom in the following compounds?
CH2=C=O, CH3CH=CH2,(CH3)2CO, CH2=CHCN, C6H6
Answer - 1 : -
(i) 
C–1 is sp2 hybridised.
C–2 is sp hybridised.
(ii) 
C–1 is sp3 hybridised.
C–2 is sp2 hybridised.
C–3 is sp2 hybridised.
(iii)
C–1 and C–3 are sp3 hybridised.
C–2 is sp2 hybridised.
(iv) 
C–1 is sp2 hybridised.
C–2 is sp2 hybridised.
C–3 is sp hybridised.
(v) C6H6
All the 6 carbonatoms in benzene are sp2 hybridised.
Indicate the σ and π bondsin the following molecules:
C6H6,C6H12, CH2Cl2, CH2 =C = CH2, CH3NO2, HCONHCH3
Answer - 2 : -
(i) C6H6
There are six C–C sigma (
) bonds, six C–H sigma (
) bonds, and three C=C pi (
) resonating bonds inthe given compound.
(ii) C6H12
There are six C–C sigma () bonds and twelve C–H sigma () bonds in the givencompound.
(iii) CH2Cl2There two C–H sigma () bonds and two C–Cl sigma (
) bonds in the givencompound. (iv) CH2 =C = CH2
There are two C–C sigma () bonds, four C–H sigma () bonds, and two C=C pi (
) bonds in the givencompound.
(v) CH3NO2.There are three C–H sigma () bonds, one C–N sigma (
) bond, one N–O sigma (
) bond, and one N=O pi (
) bond in the givencompound. (vi) HCONHCH3There are two C–N sigma () bonds, four C–H sigma () bonds, one N–H sigma bond, and one C=O pi () bond in the givencompound.
Question - 3 : - Write bond line formulasfor: Isopropyl alcohol, 2,3-Dimethyl butanal, Heptan-4-one.
Answer - 3 : -
The bond line formulaeof the given compounds are:
(a) Isopropyl alcohol
(b) 2, 3–dimethylbutanal
(c) Heptan–4–one
Question - 4 : - Give the IUPAC names ofthe following compounds:
(a)
(b)
(c)
(d)
(e)
(f) Cl2CHCH2OH
Answer - 4 : -
(a)
1–phenyl propane
(b)
3–methylpentanenitrile
(c)
2, 5–dimethyl heptane
(d)
3–bromo–3–chloroheptane
(e)
3–chloropropanal
(f) Cl2CHCH2OH
2,2- Dichloroethanol
Question - 5 : - Which of the followingrepresents the correct IUPAC name for the compounds concerned? (a)2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-Trimethyloctane or2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane(d) But-3-yn-1-ol or But-4-ol-1-yne
Answer - 5 : -
(a) The prefix di inthe IUPAC name indicates that two identical substituent groups are present inthe parent chain. Since two methyl groups are present in the C–2 of the parentchain of the given compound, the correct IPUAC name of the given compound is 2,2–dimethylpentane.
(b) Locant number 2,4, 7 is lower than 2, 5, 7. Hence, the IUPAC name of the given compound is 2,4, 7–trimethyloctane.
(c) If thesubstituents are present in the equivalent position of the parent chain, thenthe lower number is given to the one that comes first in the name according tothe alphabetical order. Hence, the correct IUPAC name of the given compound is2–chloro–4–methylpentane.
(d) Two functionalgroups – alcoholic and alkyne – are present in the given compound. Theprincipal functional group is the alcoholic group. Hence, the parent chain willbe suffixed with ol. The alkyne group is present in the C–3 of theparent chain. Hence, the correct IUPAC name of the given compound isBut–3–yn–1–ol.
Question - 6 : - Draw formulas for thefirst five members of each homologous series beginning with the followingcompounds. (a) H–COOH (b) CH3COCH3 (c) H–CH=CH2
Answer - 6 : -
The first five membersof each homologous series beginning with the given compounds are shown asfollows:
(a)
H–COOH : Methanoic acid
CH3–COOH :Ethanoic acid
CH3–CH2–COOH: Propanoic acid
CH3–CH2–CH2–COOH: Butanoic acid
CH3–CH2–CH2–CH2–COOH: Pentanoic acid
(b)
CH3COCH3 :Propanone
CH3COCH2CH3 :Butanone
CH3COCH2CH2CH3 :Pentan-2-one
CH3COCH2CH2CH2CH3 :Hexan-2-one
CH3COCH2CH2CH2CH2CH3 :Heptan-2-one
(c)
H–CH=CH2 :Ethene
CH3–CH=CH2 :Propene
CH3–CH2–CH=CH2 :1-Butene
CH3–CH2–CH2–CH=CH2 :1-Pentene
CH3–CH2–CH2–CH2–CH=CH2 :1-Hexene
Question - 7 : - Give condensed and bondline structural formulas and identify the functional group(s) present, if any,for :
(a) 2,2,4-Trimethylpentane
(b)2-Hydroxy-1,2,3-propanetricarboxylic acid
(c) Hexanedial
Answer - 7 : -
(a) 2, 2, 4–trimethylpentane
Condensed formula:
(CH3)2CHCH2C (CH3)3
Bond line formula:
(b) 2–hydroxy–1, 2,3–propanetricarboxylic acid
Condensed Formula:
(COOH)CH2C(OH)(COOH)CH2(COOH)
Bond line formula:
The functional groupspresent in the given compound are carboxylic acid (–COOH) and alcoholic (–OH)groups.
(c) Hexanedial
Condensed Formula:
(CHO) (CH2)4 (CHO)
Bond line Formula:
The functional grouppresent in the given compound is aldehyde
(–CHO).
Question - 8 : - Identify the functionalgroups in the following compounds
(a) 
(b) 
(c)
Answer - 8 : -
The functional groupspresent in the given compounds are:
(a) Aldehyde (–CHO),
Hydroxyl(–OH),
Methoxy (–OMe),
C=C double bond 
(b) Amino (–NH2);primary amine,
Ester (-O-CO-),
Triethylamine (N(C2H5)2);tertiary amine
(c) Nitro (–NO2),
C=C double bond 
Question - 9 : - Which of the two: O2NCH2CH2O– orCH3CH2O– is expected to be more stableand why?
Answer - 9 : -
NO2 groupis an electron-withdrawing group. Hence, it shows –I effect. By withdrawing theelectrons toward it, the NO2 group decreases thenegative charge on the compound, thereby stabilising it. On the other hand,ethyl group is an electron-releasing group. Hence, the ethyl group shows +Ieffect. This increases the negative charge on the compound, therebydestabilising it. Hence, O2NCH2CH2O– isexpected to be more stable than CH3CH2O–.
Question - 10 : - Explain why alkyl groupsact as electron donors when attached to a π system.
Answer - 10 : -
When an alkyl groupis attached to a π system, it acts as an electron-donor group by theprocess of hyperconjugation. To understand this concept better, let us take theexample of propene.
In hyperconjugation, thesigma electrons of the C–H bond of an alkyl group are delocalised. This groupis directly attached to an atom of an unsaturated system. The delocalisationoccurs because of a partial overlap of a sp3 –s sigmabond orbital with an empty p orbital of the π bondof an adjacent carbon atom.
The process ofhyperconjugation in propene is shown as follows:
This type of overlapleads to a delocalisation (also known as no-bond resonance) ofthe π electrons, making the molecule more stable.
