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CBSE Class 10 Science 2022 Outside Delhi Term II Set 2

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Question : 1 of 7
Marks: +1, -0
Consider the following molecular formulae of carbon compounds:
(i) CH3COOH\mathrm{CH}_3\mathrm{COOH}
(ii) CH3OH\mathrm{CH}_3\mathrm{OH}
(iii) C2H6\mathrm{C}_2\mathrm{H}_6
(iv) C3H4\mathrm{C}_3\mathrm{H}_4 (iv) C4H8\mathrm{C}_4\mathrm{H}_8
(a) Which one of these compounds belongs to homologous series of alcohols?
(b) Identify the compound having triple bond between carbon - carbon atoms.
(c) Write the molecular formula of the first member of the homologous series to which CH3COOH\mathrm{CH}_3\mathrm{COOH} belongs.
(d) Write the general formula of the series to which the compound C4H8\mathrm{C}_4\mathrm{H}_8 belongs.
Solution:  
(a) CH3OH\mathrm{CH}_3\mathrm{OH}, methanol belongs to homologous series of alcohols and it has the general formula of R−OHR-\mathrm{OH}.
(b) C3H4\mathrm{C}_3\mathrm{H}_4
Lewis Structure of C3H4\mathrm{C}_3\mathrm{H}_4
H−C∣H∣H−C≡C−HH-\overset{\mathrm{H}}{\overset{\vert}{\underset{\mathrm{H}}{\underset{\vert}{\mathrm{C}}}}}-C\equiv C-H
(c) Ethanoic acid is a member of homologous series with general formula CnH2n+1COOH\mathrm{C}_n\mathrm{H}_{2n+1}\mathrm{COOH}. The molecular formula is HCOOH\mathrm{HCOOH} (Methanoic acid).
(d) The general formula for alkenes can be written as CnH2n′\mathrm{C}_n\mathrm{H}_{2n^{'}} where n=2,3,4n=2,3,4 and C4H8\mathrm{C}_4\mathrm{H}_8 belongs to alkene series.
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