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NCERT Class 11 Chemistry Textbook Chapter 13 With Answer PDF Free Download
Chapter 13: Hydrocarbons
Hydrocarbons are of different types. Depending upon the types of carbon-carbon bonds present, they can be classified into three main categories – (i) saturated (ii) unsaturated, and (iii) aromatic hydrocarbons.
Saturated hydrocarbons contain carbon-carbon and carbon-hydrogen single bonds.
If different carbon atoms are joined together to form an open chain of carbon atoms with single bonds, they are termed alkanes as you have already studied in Unit 12.
On the other hand, if carbon atoms form a closed chain or a ring, they are termed as cycloalkanes. Unsaturated hydrocarbons contain carbon-carbon multiple bonds – double bonds, triple bonds, or both.
Aromatic hydrocarbons are a special type of cyclic compounds.
You can construct a large number of models of such molecules of both types (open chain and close chain) keeping in mind that carbon is tetravalent and hydrogen is monovalent.
For making models of alkanes, you can use toothpicks for bonds and plasticine balls for atoms. For alkenes, alkynes, and aromatic hydrocarbons, spring models can be constructed.
As already mentioned, alkanes are saturated open chain hydrocarbons containing carbon-carbon single bonds. Methane (CH4) is the first member of this family.
Methane is a gas found in coal mines and marshy places. If you replace one hydrogen atom of methane by carbon and join the required number of hydrogens to satisfy the tetravalence of the other carbon atom, what do you get? You get C2H6.
This hydrocarbon with molecular formula C2H6 is known as ethane. Thus you can consider C2H6 as derived from CH4 by replacing one hydrogen atom by -CH3 group. Go on constructing alkanes by doing this theoretical exercise
13.2.1 Nomenclature and Isomerism
You have already read about the nomenclature of different classes of organic compounds in Unit 12. Nomenclature and isomerism in alkanes can further be understood with the help of a few more examples.
Common names are given in parenthesis. The first three alkanes – methane, ethane, and propane have only one structure but higher alkanes can have more than one structure.
Let us write structures for C4H10. Four carbon atoms of C4H10 can be joined either in a continuous chain or with a branched-chain in the following two ways :
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NCERT Solutions Class 11 Chemistry Chapter 13 Hydrocarbons
What effect does the branching of an alkane chain has on its boiling point?
Alkanes encounter Van-der Waals forces between molecules. The higher the alkane’s power, the greater is the boiling point.
As the molecule branching increases, the surface area decreases which leads to a small contact area. As a result, the force of the Van-der Waals (or intermolecular force) decreases too.
Those forces can be overcome very easily at a relatively lower temperature. Thus, the boiling point of an alkane chain decreases as branching increases.
Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions with difficulty?
Benzene is a planar molecule with electrons delocalized under and above the ring plane. Hence, it is a material rich in electrons. As a consequence, electron-deficient species, i.e., electrophiles, are extremely attractive.
Benzene therefore very easily undergoes electrophilic substitution reactions. Nucleophiles, on the other hand, are also species that are rich in electrons. Therefore, benzene is repelled as compared to electrophiles. Thus, benzene suffers from difficulty with nucleophilic substitutions.
Arrange the following set of compounds in order of their increasing relative reactivity with an electrophile, E+
(a) p-nitrochlorobenzene, Chlorobenzene, 2,4-dinitrochlorobenzene,
(b) p – H3C – C6H4 – NO2, Toluene, p-O2N – C6H4 – NO2.
Electrophiles are reagents that participate in a reaction by accepting a pair of electrons to bind to nucleophiles.
The higher the density of electrons on a benzene ring, the more reactive the compound is to an electrophile, E+ (Electrophilic reaction).
(a) The electron density of the aromatic ring decreases due to the presence of an electron-withdrawing group (i.e., NO2 – and Cl –) which deactivates the aromatic ring.
Since, Cl – group is less electron-withdrawing (due to the inductive effect) than NO2 – group (due to resonance effect), the increasing order of reactivity is as follows:
2, 4 – dinitrochlorobenzene < p – nitrochlorobenzene < Chlorobenzene
(b) While NO2– the group is electron-withdrawing, CH3– is an electron-donating group.
Toluene, therefore, has the maximum density of electrons and is most easily attacked by E+. Since NO2– is an electron-removing group. Therefore, when the number of NO2 substitutes is higher, the order is the following.:
p-O2N – C6H4 – NO2 < p – H3C – C6H4 – NO2 < Toluene.
Hydrocarbons Textbook With Solutions PDF Free Download