NCERT Class 11 Solutions Organic Chemistry

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NCERT Class 11 Solutions Organic Chemistry - NCERT Solutions prepared for CBSE students by the best teachers in Delhi.

Chapter 12 – Organic Chemistry Some Basic Principles and Techniques

Class XI Chemistry

Question 12.5: Which of the following represents the correct IUPAC name for the compounds concerned? (a) 2,2-Dimethylpentane or 2-Dimethylpentane (b) 2,4,7-Trimethyloctane or 2,5,7-Trimethyloctane (c) 2-Chloro-4-methylpentane or 4-Chloro-2-methylpentane (d) But-3-yn-1-ol or But-4-ol-1-yne

Answer:

(a) The prefix di in the IUPAC name indicates that two identical substituent groups are present in the parent chain. Since two methyl groups are present in the C–2 of the parent chain 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 the substituents are present in the equivalent position of the parent chain, then the lower number is given to the one that comes first in the name according to the alphabetical order. Hence, the correct IUPAC name of the given compound is 2–chloro– 4–methylpentane.

(d) Two functional groups – alcoholic and alkyne – are present in the given compound. The principal functional group is the alcoholic group. Hence, the parent chain will be suffixed with ol. The alkyne group is present in the C–3 of the parent chain. Hence, the correct IUPAC name of the given compound is But–3–yn–1–ol.

Question 12.6: Draw formulas for the first five members of each homologous series beginning with the following compounds. (a) H–COOH (b) CH3COCH3 (c) H–CH=CH2

Answer :The first five members of each homologous series beginning with the given compounds are shown as follows:

(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

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Question 12.7: Give condensed and bond line 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

Question 12.9: Which of the two: O2NCH2CH2O– or CH3CH2O– is expected to be more stable and why?

Answer: NO2 group is an electron-withdrawing group. Hence, it shows –I effect. By withdrawing the electrons toward it, the NO2 group decreases the negative charge on the compound, thereby stabilising it. On the other hand, ethyl group is an electron-releasing group. Hence, the ethyl group shows +I effect. This increases the negative charge on the compound, thereby destabilising it. Hence, O2NCH2CH2O– is expected to be more stable than CH3CH2O– .

Question 12.10: Explain why alkyl groups act as electron donors when attached to a π system.

Answer When an alkyl group is attached to a π system, it acts as an electron-donor group by the process of hyperconjugation. To understand this concept better, let us take the example of propene.


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