NCERT Class 11 Solutions Hydrocarbons - NCERT Solutions prepared for CBSE students by the best teachers in Delhi.Class XI Chapter 13 – Hydrocarbons Chemistry
Question 13.1: How do you account for the formation of ethane during chlorination of methane?
Answer Chlorination of methane proceeds via a free radical chain mechanism. The whole reaction takes place in the given three steps.
Step 1: Initiation: The reaction begins with the homolytic cleavage of Cl – Cl bond as:
Step 2: Propagation: In the second step, chlorine free radicals attack methane molecules and break down the C–H bond to generate methyl radicals as: These methyl radicals react with other chlorine free radicals to form methyl chloride along with the liberation of a chlorine free radical. Hence, methyl free radicals and chlorine free radicals set up a chain reaction. While HCl and CH3Cl are the major products formed, other higher halogenated compounds are also formed as:
Step 3: Termination: Formation of ethane is a result of the termination of chain reactions taking place as a result of the consumption of reactants as:
Question 13.3: For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of double or triple bond as indicated:
(a) C4H8 (one double bond)
(b) C5H8 (one triple bond)
Question 13.4: Write IUPAC names of the products obtained by the ozonolysis of the following compounds:
(i) Pent-2-ene (ii) 3,4-Dimethyl-hept-3-ene
(iii) 2-Ethylbut-1-ene (iv) 1-Phenylbut-1-ene
Question 13.5: An alkene ‘A’ on ozonolysis gives a mixture of ethanal and pentan-3-one. Write structure and IUPAC name of ‘A’.
Question 13.6: An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.
Answer As per the given information, ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. The formation of two moles of an aldehyde indicates the presence of identical structural units on both sides of the double bond containing carbon atoms. Hence, the structure of ‘A’ can be represented as:
XC = CX There are eight C–H σ bonds. Hence, there are 8 hydrogen atoms in ‘A’. Also, there are three C–C bonds. Hence, there are four carbon atoms present in the structure of ‘A’. Combining the inferences, the structure of ‘A’ can be represented as: ‘A’ has 3 C–C bonds, 8 C–H σ bonds, and one C–C π bond. Hence, the IUPAC name of ‘A’ is But-2-ene. Ozonolysis of ‘A’ takes place as
Question 13.7: Propanal and pentan-3-one are the ozonolysis products of an alkene? What is the structural formula of the alkene?
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