Hydrocarbons Class 11 Notes CBSE Chemistry Chapter 13 [Free PDF Download]

Section – A (1 Mark Questions)

1. What do you mean by system?

Ans. A system in thermodynamics refers to that part of the universe in which observations are made.

2. What do you mean by surroundings?

Ans. The rest of the universe which might be in a position to exchange energy and matter with the system is called its surroundings.

3. Explain first law of thermodynamics.

Ans. The first law of thermodynamics states that ‘the energy of an isolated system is constant’.

4. ‘Coffee held in a cup’ is what type of system?

Ans. Coffee held in a cup is an open system because it can exchange matter (water vapors) and energy (heat) with the surroundings.

5. List one example of isolated system.

Ans. Coffee held in a thermos flask is an isolated system because it can neither exchange energy nor matter with the surroundings.

6. If work is done by the system, what will be the effect on internal enery of the system.

Ans. The internal energy of the system will decrease if work is done by the system.

7. In thermodynamics we discuss about open and close sytsem. So, animals and plants belongs to?

Ans. Animals and plants belongs to Open system.

8. Explain state of thermodynamic system?

Ans. The state of thermodynamic system may be defined by specifying values of state variables like temperature, pressure, volume.

9. What do you mean by Enthalpy?

Ans. Enthalpy is defined as total heat content of the system.

10. How do you define enthalpy mathematically?

Ans. Mathematically.

H = U + pV where U is internal energy.

Section – B (2 Marks Questions)

11. Internal energy is state function while work done is not. Why?

Ans. The change in internal energy during a process depends only upon the initial and final state of the system. Therefore, it is a state function. But the work is related the path followed. Therefore, it is not a state function.

12. Predict the sign of $\Delta\text{S}$ for the following reaction

$\text{CaCO}_3(\text{s})\overset{\Delta}{\rightarrow}\text{CaO}(\text{s})+\text{CO}_2(\text{g})$

Ans.

is positive because on the product side we have gaseous CO2. 

13. What do you mean by spontaneous process.

Ans. A spontaneous process is an irreversible process and may only be reversed by some external agency.

14. What do you mean by non-spontaneous process?

Ans. A process is said to be non-spontaneous if it does not occur of its own under given condition and occur only when an external force is continuously applied.

15. The standard heat of formation of Fe2O3(s) is 824.2kJ mol–1. Calculate heat change for the reaction.

4Fe(s) + 3O2(g) → 2Fe2O3(s)

Ans.

$\Delta\text{H}=\sum\Delta\text{H}\degree_{\text{f}}(\text{products})-\sum\Delta\text{H}\degree_{\text{f}}(\text{reactants})$

$=[2\times\Delta\text{H}\degree_{\text{f}}\text{Fe}_2\text{O}_3(\text{s})]-[4\Delta\text{H}\degree_{\text{f}}\text{Fe}(\text{s})+3\Delta\text{H}\degree_{\text{f}}\text{O}_2(\text{g})]$

= 2 (-824.2 kJ) - [4 x 0 + 3 x 0]

= - 1648.4 kJ

16. Explain Heat capacity and what are the units of heat capacity?

Ans. The heat capacity for one mole of the substance is the quantity of heat needed to raise the temperature of one mole by one degree Celsius.

Its unit is : J/K

17. Explain specific heat.

Ans. Specific heat/specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or one Kelvin).

18. For solids or liquids, there is no significant difference between $\Delta\text{H}$ and $\Delta\text{H}$ .

Ans. The difference between $\Delta\text{H}$ and $\Delta\text{U}$ is not usually significant for systems consisting of only solids and/or liquids because they do not suffer any significant volume changes upon heating.

19. Define extensive and intensive properties?

Ans. Extensive property is a property whose value depends on the quantity or size of matter present in the system.

Intensive property is a property which do not depend upon the quantity or size of matter present.

20. C(graphite) + O2(g) → CO2(g) ΔH= - 393.5 kJ mol–1 . This is combustion reaction of C and CO2 is forming. Write the $\Delta\text{H}\degree$ values of the two preocesses. 

Ans. (i) The standard enthalpy of formation of CO2 is –393.5 kJ per mole of CO2.

That is $\Delta\text{H}\degree(\text{CO}_2,\text{g})$ = –393.5 kJ mol–1

(ii) The standard enthalpy of combustion of carbon is 393.5 kJ per mole of carbon i.e

comb(c)

= –393.5 kJ mol–1

Access Class 11 Chemistry Chapter 13 – Hydrocarbons Notes 

ALKANES 

Alkanes are the simplest saturated organic compounds or hydrocarbons. They are also called paraffins. General formula of alkanes is${{C}_{n}}{{H}_{2n+2}}$ .The carbon atoms in alkanes are $s{{p}^{3}}$ hybridized and have a tetrahedral shape. The bond length between C–H is$1.12\overset{o}{\mathop{A}}\,$ and between C – C is $1.54\overset{o}{\mathop{A}}\,$ . The simplest alkane is methane $\left( C{{H}_{4}} \right)$.

Conformation

Alkanes have various conformers existing in space due to free rotation of the C – C single bond.

Different spatial arrangements of atoms give rise to isomers called conformers or rotamers. 

Conformation of ethane$\left( {{\text{C}}_{\text{2}}}{{\text{H}}_{\text{6}}} \right)$ 

Ethane has two types of conformation:

1. Newman Projection

Newman Projection of Ethane

Order of stability is, staggered > skew > eclipsed

 

1. SawHorse projection

Sawhorse Projection of Ethane

Methods of Preparation 

1. By catalytic hydrogenation of unsaturated hydrocarbons: Hydrogenation is the process of adding hydrogen to the unsaturated hydrocarbon in the presence of a catalyst. When the catalyst is Nickel and the temperature is $200-{{300}^{o}}C$ (Raney Ni) then the reaction is called Sabatier and Sanderson’s reduction. \[\underset{ethylene}{\mathop{C{{H}_{2}}=C{{H}_{2}}}}\,+{{H}_{2}}\xrightarrow[{{200}^{o}}C]{Ni}\underset{ethane}{\mathop{C{{H}_{3}}-C{{H}_{3}}}}\,\] 

\[\underset{acetylene}{\mathop{CH\equiv CH}}\,+2{{H}_{2}}\xrightarrow[{{300}^{o}}C]{Ni}\underset{ethane}{\mathop{C{{H}_{3}}-C{{H}_{3}}}}\,\] 

At room temperature, hydrogenation is carried out using platinum or palladium as a catalyst in place of Ni. Methane cannot be produced by this method as an unsaturated hydrocarbon does not have a single carbon atom. 

2. By the reduction of alkyl halides: 

This method uses alkyl halide which is reduced in presence of $Zn+C{{H}_{3}}COOH,Zn+HCl,Zn+NaOH,Zn+Cu$ along with an alcohol$({{C}_{2}}{{H}_{5}}OH)$ or aluminum amalgam in ${{C}_{2}}{{H}_{5}}OH$ or$LiAl{{H}_{4}}$.

\[R-X+2H\to RH+HX\] 

Alkyl halides can also be reduced by heating them with HI and red phosphorus in a sealed tube (under pressure).

\[R-I+HI\xrightarrow[{{150}^{o}},pressure]{P}RH+{{I}_{2}}\] 

 The function of red phosphorus is to remove iodine. 

3. By the reduction of alcohols, aldehydes, ketones and fatty acids: 

These compounds and their derivatives can be reduced in the presence of hot hydroiodic acid and red phosphorus at ${{150}^{o}}C$ in a sealed tube to produce alkanes. The reactions are:

\[ROH+2HI\xrightarrow{\text{Red }\text{P}}RH+{{H}_{2}}O+{{I}_{2}}\]

\[RCHO+4HI\xrightarrow{\text{Red }\text{P}}RC{{H}_{3}}+2~{{I}_{2}}+{{H}_{2}}O\]

\[R-COR+4HI\xrightarrow{\text{Red P}}R-C{{H}_{2}}-R+{{H}_{2}}O+2{{I}_{2}}\]

\[R-COOH+6HI\xrightarrow{\text{Red P}}RC{{H}_{3}}+2{{H}_{2}}O+3{{I}_{2}}\] 

Aldehydes and ketones are also reduced to alkanes in the presence of amalgamated zinc and conc. HCl. This reaction is known as Clemmensen reduction. 

\[R-CHO+2{{H}_{2}}\xrightarrow[conc.HCl]{Zn-Hg}R-C{{H}_{3}}+{{H}_{2}}O\] 

Aldehydes can also be reduced to alkanes using hydrazine and KOH at $150-{{200}^{o}}C$ . This reaction is known as Wolff−Kishner reduction. 

\[C{{H}_{3}}CHO+N{{H}_{2}}N{{H}_{2}}\to C{{H}_{3}}CH=NN{{H}_{2}}\xrightarrow{KOH}C{{H}_{3}}C{{H}_{3}}+{{N}_{2}}\] 

4. By condensing two molecules of alkyl halides: 

When two molecules of alkyl halides are treated with sodium metal in the presence of dry ether, then they are coupled to form an alkane. This reaction is known as Wurtz synthesis. 

\[R-Br+2Na+R-Br\xrightarrow{Dry\,ether}R-R+2NaBr\] 

5. By decarboxylation of carboxylic acid: when the sodium salt of carboxylic acid is strongly heated with soda lime, then an alkane is formed by elimination of $C{{O}_{2}}$as carbonate. 

\[R-COONa+NaOH\xrightarrow[CaO]{Heat}R-H+N{{a}_{2}}C{{O}_{3}}\] 

6. Kolbe’s electrolysis: This involves sodium or potassium salts of fatty acids to be electrolyzed that form higher alkanes at anode. 

\[2C{{H}_{3}}COONa+2{{H}_{2}}O\to C{{H}_{3}}-C{{H}_{3}}+2C{{O}_{2}}+2NaOH+{{H}_{2}}\] 

As the product is higher in alkanes, so methane can’t be prepared by this method. 

7. By action of water on aluminium carbide or beryllium carbide: \[\underset{aluminium\,carbide}{\mathop{{{A}_{4}}{{C}_{3}}}}\,+12{{H}_{2}}O\to 4Al{{(OH)}_{3}}+3C{{H}_{4}}\uparrow \] 

\[\underset{Beryllium\,carbide}{\mathop{B{{e}_{2}}C}}\,+4{{H}_{2}}O\to 2Be{{(OH)}_{2}}+C{{H}_{4}}\uparrow \] 

Physical Properties  

1. State: They consist of weak forces, therefore alkanes up to four carbon atoms are colourless and odourless gases, the next thirteen members are colourless and odourless liquids. Alkanes from carbon – 18  onwards are colourless and odourless solids.

In alkenes, except ethene, all are odourless and follow some trend as alkanes. Ethene has a pleasant odour. All are colourless. Alkynes also follow the same trend as alkanes.

2. Density: The density of alkanes increases steadily with the rise of molecular mass and becomes constant at 0.8. 

3. Solubility: They are insoluble in polar solvents such as water but soluble in non-polar solvents like ether, carbon tetrachloride$CC{{l}_{4}}$, benzene etc.

4. Boiling and melting points: As the number of carbons in straight chain alkanes increases, the boiling point also increases. But the increase in melting point is irregular with respect to increase in molecular mass. Alkenes and alkynes also show a gradual increase in boiling and melting points with the increase in molecular mass in homologous series. They are less volatile than alkanes, i.e., their boiling point and melting point are higher than corresponding alkanes. 

Chemical Properties 

Alkanes are highly stable compounds and inert substances due to the presence of non-polar C – C and C – H bonds. Alkanes are saturated compounds with strong sigma bonds which don't break under ordinary conditions. Therefore, Alkanes react at high temperature by a free radical mechanism.

1. Halogenation (free radical substitution): 

Alkanes react with halogens ($C{{l}_{2}},B{{r}_{2}}$ ) in the presence of light or in dark at high temperature to form corresponding substituted products. 

\[\underset{\text{ }methene\text{ }}{\mathop{C{{H}_{4}}}}\,\xrightarrow[-HCl]{C{{l}_{2}}}\underset{\text{ }methyl\text{ }chloride\text{ }}{\mathop{C{{H}_{3}}C1}}\,\xrightarrow[-HCl]{C{{l}_{2}}}\underset{\text{ }methylene\text{ }chloride\text{ }}{\mathop{C{{H}_{2}}C{{1}_{2}}}}\,\xrightarrow[-HCl]{C{{l}_{2}}}\underset{\text{ }chloroform\text{ }}{\mathop{CHC{{1}_{3}}}}\,\xrightarrow[-HCl]{C{{l}_{2}}}\underset{\text{ }carbon\,tetrachloride\text{ }}{\mathop{CC{{1}_{4}}}}\,\] 

The relative reactivity of halogens has the order ${{F}_{2}} > C{{l}_{2}} > B{{r}_{2}} > {{I}_{2}}$ and alkanes follow the order, ${{3}^{{}^\circ }} > {{2}^{{}^\circ }} > {{1}^{{}^\circ }} > C{{H}_{3}}$ 

2. Nitration: Nitration is the introduction of the nitro group. It is possible for alkanes having three or more carbon atoms. Nitration of propane yields a mixture of nitro products.

\[C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}\xrightarrow[{{400}^{o}}C]{HN{{O}_{3}}}C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}N{{O}_{2}}+C{{H}_{3}}\overset{\overset{N{{O}_{2}}}{\mathop{|}}\,}{\mathop{CH}}\,C{{H}_{3}}+C{{H}_{3}}C{{H}_{2}}N{{O}_{2}}+C{{H}_{3}}N{{O}_{2}}\] 

3. Sulphonation: Higher alkanes (C- 6 hexane onwards) undergo sulfonation when treated with fuming ${{H}_{2}}S{{O}_{4}}$. 

\[\underset{hexane}{\mathop{n-{{C}_{6}}{{H}_{14}}}}\,+HOS{{O}_{3}}H\to \underset{hexane\,sulphonic\,acid}{\mathop{{{C}_{6}}{{H}_{13}}S{{O}_{3}}H}}\,+{{H}_{2}}O\] 

4. Oxidation or combustion: Alkanes burn in the presence of oxygen to form carbon dioxide and water along with evolution of heat. 

\[C{{H}_{4}}+2{{O}_{2}}\to C{{O}_{2}}+2{{H}_{2}}O\] 

\[2{{C}_{2}}{{H}_{6}}+7{{O}_{2}}\to 4C{{O}_{2}}+6{{H}_{2}}O\] 

Illustration 1: The order of reactivity of halogens towards halogenation of alkane is 

1. $\text{ }{{F}_{2}} > B{{r}_{2}} > C{{l}_{2}}$ 

2. $\text{ }{{F}_{2}} > C{{l}_{2}} > B{{r}_{2}}$ 

3. $C{{l}_{2}} > {{F}_{2}} > B{{r}_{2}}$

4. $C{{l}_{2}} > B{{r}_{2}} > {{F}_{2}}$ 

Solution: (B). In a group, electronegativity of the atom decreases. So, the reactivity of halogen also decreases in a group. Thus, the order of reactivity is ${{F}_{2}} > C{{l}_{2}} > B{{r}_{2}}$.

Illustration 2 : Consider the following reaction: 

Reaction of Free Radical with Alkane

Identify the structure of the major product ‘X’. 

Structure of the Major Product ‘X’

Solution: (B). $B{{r}^{\centerdot }}$is less reactive and more selective and so the most stable free radical (${{3}^{o}}$ ) will be the major product.

Exercise 1

1. The compound with the highest boiling point is 

1. n- hexane

2. n- pentane

3. 2,2- dimethylpropane

4.  2- methyl butane

2. Relative reactivity of halogens on alkanes follow the order 

1. ${{\text{F}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}\text{B}{{\text{r}}_{\text{2}}}\text{}{{\text{I}}_{\text{2}}}$ 

2. $\text{C}{{\text{l}}_{\text{2}}}\text{B}{{\text{r}}_{\text{2}}}\text{}{{\text{I}}_{\text{2}}}\text{}{{\text{F}}_{\text{2}}}$ 

3. ${{\text{F}}_{\text{2}}}\text{}{{\text{I}}_{\text{2}}}\text{B}{{\text{r}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}$ 

4. ${{\text{I}}_{\text{2}}}\text{B}{{\text{r}}_{\text{2}}}\text{C}{{\text{l}}_{\text{2}}}\text{}{{\text{F}}_{\text{2}}}$ 

3. Propane is obtained from propene by which of the following methods?

1. Wurtz reaction 

2. Dehydration 

3. Frankland reaction 

4. Catalytic hydrogenation 

ALKENES 

Alkenes are unsaturated hydrocarbons having a double bond between two carbon atoms. Alkenes have the general formula${{C}_{n}}{{H}_{2n}}$.

Isomerism 

1. Structural isomerism 

Example, Butene has 3structural isomers,

\[\underset{But-1-ene}{\mathop{C{{H}_{3}}-C{{H}_{2}}-CH=C{{H}_{2}}}}\,\] 

\[\underset{But-2-ene}{\mathop{C{{H}_{3}}-CH=CH-C{{H}_{3}}}}\,\] 

Isobutene

2. Geometrical isomers

The hindered rotation around the C – C bond gives rise to stereoisomers having different spatial arrangements. Two isomers exist as follows:

Cis and Trans Isomers of but-2-ene

Methods of Preparation 

1. By dehydration of alcohol: Dehydration of alcohol in the presence of acids forms alkene. The reaction is an elimination reaction. 

\[R-C{{H}_{2}}-C{{H}_{2}}-OH\xrightarrow[\Delta ]{{{H}^{+}}}R-CH=C{{H}_{2}}+{{H}_{2}}O\] 

2. By the dehydrohalogenation of alkyl halides: it involves an alkyl halide in the presence of alcoholic KOH to yield alkene. \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow{alc.KOH}C{{H}_{3}}CH=C{{H}_{2}}+HBr\]

If dehydrogenation of alkyl halide gives two products, the major product will be according to Saytzeff’s rule that states that the most substituted alkene will be the major product. 

Dehydrohalogenation of Alkyl Halides

The ease of dehydrohalogenation has the order: 

Tertiary alkyl halide > secondary alkyl halide > primary alkyl halide. 

Alkyl halides follow the order: 

alkyl iodide > alkyl bromide > alkyl chloride. 

3. By the dehalogenation of vicinal dihalides: Dehydrohalogenation of vicinal dihalides in the presence of Zinc dust along with alcoholic solution yields pure alkene. 

Dehydrohalogenation of Vicinal Dihalides

4. Kolbe’s electrolysis method: The electrolysis of sodium or potassium salts of dicarboxylic acid produces alkene at anode. 

Kolbe’s Electrolysis Method

If, $Na/Liq.N{{H}_{3}}$ is used, trans alkene is formed, and in presence of Ni cis alkene is formed. 

Physical Properties

1. Melting point: the trans isomers have a high melting point than cis isomer due to symmetry and crystal lattice.

2. Boiling point: Cis isomer has a more dipole moment as it is more polar and therefore has a high boiling point than trans isomer of alkene.

Chemical Properties 

Alkenes are reactive due to the presence of double bonds. Due to THE presence of $\pi $ bonds alkenes are able to react towards electrophilic addition reaction. They also give free radical addition reactions. 

1. Addition reactions: 

(i) Addition of hydrogen (catalytic hydrogenation) \[C{{H}_{2}}=C{{H}_{2}}+{{H}_{2}}\xrightarrow[200-{{300}^{o}}C]{Ni}C{{H}_{3}}-C{{H}_{3}}\] 

(ii)Addition of halogens (Chlorine and bromine) 

\[C{{H}_{2}}=C{{H}_{2}}+B{{r}_{2}}\to \underset{ethylene\,dibromide\,(colorless)}{\mathop{BrC{{H}_{2}}-C{{H}_{2}}Br}}\,\] 

Addition of bromine is used as a test for detecting the presence of unsaturation (C – C double bond or triple bond). 

(ii) Addition of hydrogen halides 

\[C{{H}_{2}}=C{{H}_{2}}+HX\to C{{H}_{3}}C{{H}_{2}}X\] 

The order of reactivity among hydrogen halides is 

HI > HBr > HCl > HF 

In case of unsymmetrical alkenes, addition occurs according to Markonikov’s rule. This reaction takes place through an ionic mechanism. Electrophilic addition to a carbon–carbon double bond involves the formation of an intermediate that is the  more stable carbocation. 

Deviation from Markonikov’s rule: 

It has been observed that addition of HBr to unsymmetrical alkenes like propene in presence of air, peroxide or light yields n-propyl bromide by anti-Markovnikov's rule. The effect is also called the peroxide effect or Kharasch effect. 

Markonikov’s and Anti-Markovnikov's Rule for Halogenation of Propene 

1.  Addition of hypochlorous acid 

Reaction of Hypochlorous Acid of Ethylene

2.  Addition of sulphuric acid 

Reaction of Sulphuric Acid of Propene

Alkyl hydrogen sulphates are water soluble, when heated at about${{160}^{o}}C$ , they give olefins. In reaction with water they give alcohol. 

Formation of Alcohol from Alkyl Hydrogen Sulphates 

3.  Addition of water 

It is also in accordance with Markovnikov’s rule.

Addition of Water  to Alkene

4.  Addition of alkanes (alkylation) 

Alkylation of Alkanes in Alkene

5.  Addition of diborane (hydroboration)

Hydroboration of Alkanes in Alkene

In case of unsymmetrical alkenes, addition follows Anti Markovnikov's rule. 

\[6C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}+{{B}_{2}}{{H}_{6}}\to \underset{tributyl\,borane}{\mathop{2{{\left( C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}} \right)}_{3}}~B}}\,\] 

Trialkyl borane on oxidation $({{H}_{2}}{{O}_{2}}/O{{H}^{-}})$  gives alcohol and on reduction $(LiAl{{H}_{4}})$ gives alkane. 

6.  Oxymercuration – demercuration 

Oxymercuration – Demercuration of Alkene

7.  Addition of oxygen 

Addition of Oxygen in Alkene

1. Oxidation: 

1.  Oxidation by cold alkaline $KMn{{O}_{4}}$ (Bayer’s reagent) 

Oxidation of Alkene

It is a test for detecting double bonds in alkene. Hydroxylation by $KMn{{O}_{4}}$ is always a syn addition. The cis alkene on hydroxylation gives meso compound and trans alkene gives racemic mixture. Like Bayer’s reagent $Os{{O}_{4}}$ also gives glycol and the hydroxylation is a syn addition. 

Hydroxylation of Alkene Using OsO4

2.  Oxidation by per acids $(RC{{O}_{3}}H)$ 

Oxidation of Alkene by per Acids $(RC{{O}_{3}}H)$ 

Reaction of Acetic Acid with Alkene

This addition occurs in trans manners. The cis alkene gives racemic mixture and trans alkenes give meso compound. 

3.  Ozonolysis 

Ozonolysis of Alkene

4.  Oxidation by hot concentrated alkaline $KMn{{O}_{4}}$ 

\[RCH=C{{H}_{2}}+KMn{{O}_{4}}(conc.)\to RCOOH+C{{O}_{2}}+{{H}_{2}}O\]

1. Substitution reaction: 

Chlorination is done by treating the alkene with carbon tetrachloride in liquid phase or with chlorine gas:

Chlorination of Alkene

Allylic bromination (bromination at allylic carbon atom) is very easily achieved by treating the alkene having hydrogen atom at the allylic carbon atom with N-bromosuccinimde (NBS). 

Allylic Bromination of Alkene 

Illustration 3 : Which of the following alkene has the lowest heat of hydrogenation? 

Alkene with Lowest Heat of Hydrogenation

Solution: (B). Higher the stability of alkene, lower the heat of hydrogenation. 

Illustration 4 : 

 

Reaction of Alkene with Alkaline KMnO4

Which is true about this reaction? 

(A) A is meso 1, 2-butan-di-ol formed by syn addition. 

(B) A is meso 1, 2-butan-di-ol formed by anti addition. 

(C) A is a racemic mixture of d and l, 1, 2-butan-di-ol formed by anti addition. 

(D) A is a racemic mixture of d and l, 1, 2-butan-di-ol formed by syn addition. 

Solution: (A). On cis alkene there is syn addition of two –OH groups forming meso compound

Exercise 2

Halogenation of Alkene

Predominant A is:

Predominant Product A

Product X and Y from Alkene

X and Y are:

1. $\text{C}{{\text{H}}_{3}}-\text{C}{{\text{H}}_{2}}-\text{CH}=\text{C}{{\text{H}}_{2}},\left( \text{C}{{\text{H}}_{3}}\text{C}{{\text{H}}_{2}}\text{COOH}+\text{C}{{\text{O}}_{2}} \right)$ 

2. $\text{C}{{\text{H}}_{3}}-\text{CH}=\text{CH}-\text{C}{{\text{H}}_{3}},\text{C}{{\text{H}}_{3}}\text{COOH (2 moles) }$ 

3. $\text{C}{{\text{H}}_{3}}-\text{CH}=\text{CH}-\text{C}{{\text{H}}_{3}},\text{C}{{\text{H}}_{3}}\text{CHO (2 moles) }$ 

4. $\text{C}{{\text{H}}_{3}}-\text{C}{{\text{H}}_{2}}-\text{CH}=\text{C}{{\text{H}}_{2}},\left( \text{C}{{\text{H}}_{3}}\text{C}{{\text{H}}_{2}}\text{CHO}+\text{HCHO} \right)$ 

The reaction of propene with HOCl proceeds via the addition of:

5. ${{H}^{+}}$ in the first step

6. $C{{l}^{+}}$ the first step

7. $O{{H}^{-}}$  in the first step

8. $C{{l}^{+}}$ and $O{{H}^{-}}$ in a single step

ALKYNES 

Alkynes are characterized by the presence of a triple bond between two carbon atoms. The general formula of alkyne is${{C}_{n}}{{H}_{2n-2}}$ . 

Methods of Preparation 

1. By the dehydrohalogenation of vicinal dihalides: 

\[C{{H}_{2}}=C{{H}_{2}}\xrightarrow{B{{r}_{2}}}Br-C{{H}_{2}}-C{{H}_{2}}-Br+KOH(alc.)\to Br-CH=C{{H}_{2}}\xrightarrow{NaN{{H}_{2}}}CH\equiv CH\] 

\[C{{H}_{3}}-CHB{{r}_{2}}+KOH\text{ }(alc.)\text{ }\to C{{H}_{2}}=CH-Br\xrightarrow{NaN{{H}_{2}}}CH\equiv CH\] 

2. By dehalogenation of vicinal tetrahalides: 

Reaction with active metals like Zinc, Mg etc. gives acetylene. 

Dehalogenation of Vicinal Tetrahalides

3. By Kolbe electrolysis method: 

Kolbe Electrolysis Method

4. By heating iodoform or chloroform with silver powder or zinc: This method can be used for the preparation of only acetylene. 

\[CH{{I}_{3}}+6Ag+CH{{I}_{3}}\to CH\equiv CH+6AgI\] 

5. From acetylene: Higher alkynes can be prepared from acetylene when treated with sodium metal in liquid ammonia.

\[CH\equiv CH+Na\xrightarrow{Liq.N{{H}_{3}}}CH\equiv C-Na+{}^{1}/{}_{2}{{H}_{2}}\] 

\[\underset{sodium\,acetylide}{\mathop{CH\equiv CNa}}\,+C{{H}_{3}}Br\to \underset{propyne}{\mathop{CH\equiv C-C{{H}_{3}}}}\,+NaBr\] 

Similarly, $CH\equiv CH\xrightarrow[Liq.N{{H}_{3}}]{2Na}NaC=CNa\xrightarrow{2C{{H}_{3}}Br}C{{H}_{3}}C=CC{{H}_{3}}$ 

 

Chemical Properties 

Alkyne gives electrophilic addition reactions due to the presence of loosely held $\pi $ electrons, but electrophilic addition reactions in alkynes are slower than that of alkenes. 

Terminal hydrogen present in alkynes is acidic in nature. Since s electrons are closer to the nucleus than p electrons, the electrons present in the bond having more s character will be closer to the nucleus. The amount of s character in various types of C – H bond are as follows 

Hybridisation of Alkene, Percentage of S-Character

Relative acidities: 

\[\text{HOH}=\text{HOR} > \text{CH}\equiv \text{CR} > \overset{..}{\mathop{\text{N}}}\,{{\text{H}}_{3}} > \text{C}{{\text{H}}_{2}}=\text{C}{{\text{H}}_{2}} > \text{C}{{\text{H}}_{3}}-\text{C}{{\text{H}}_{3}}\]

Relative basicities: 

\[\text{O}{{\text{H}}^{-}}=\text{O}{{\text{R}}^{-}} < {{\text{C}}^{-}}\equiv \text{C}-\text{R} < \text{NH}_{2}^{-} < \text{C}{{\text{H}}^{-}}=\text{C}{{\text{H}}_{2}} < \text{CH}_{2}^{-}-\text{C}{{\text{H}}_{3}}\] 

1. Addition of hydrogen:

\[\underset{acetylene}{\mathop{CH\equiv CH}}\,+{{H}_{2}}\xrightarrow{Ni}\underset{ethylene}{\mathop{C{{H}_{2}}=C{{H}_{2}}}}\,\xrightarrow{{{H}_{2}}}\underset{ethane}{\mathop{C{{H}_{3}}-C{{H}_{3}}}}\,\] 

In case of alkynes where triple bond is not present at the end of the chain, on reduction gives cis or trans alkene, which depends upon the choice of reducing agent. With sodium in liquid ammonia the alkene is trans form and on catalytic reduction the alkene is cis form.

Addition of Hydrogen to Alkyne

2. Electrophilic addition: 

1. Addition of halogens 

Electrophilic Addition of Chlorine in Alkyne

The order of reactivity of halogens is $C{{l}_{2}} > B{{r}_{2}} > {{I}_{2}}$ 

2.  Addition of halogen acid

 

Electrophilic Addition Halogen Acid of Alkyne

The order of reactivity of halogen acids is HI > HBr > HCl.

In the presence of peroxide, anti Markonikov’s product is obtained. 

Halogenation of Alkyne 

3. Addition of hypohalous acids

Addition of Hypohalous Acids of Alkyne 

3. Nucleophilic addition reaction: In these reactions, the addition is initiated by a nucleophile and are generally catalysed by salt of heavy metals (e.g. $H{{g}^{2+}},P{{b}^{2+}},B{{a}^{2+}}$ )which are found to form $\pi $ compound with multiple bonds. 

\[\text{HC}\equiv \text{CH}+\text{H}{{\text{g}}^{2+}}\to \underset{\underset{H{{g}^{2+}}}{\mathop{\downarrow }}\,}{\mathop{\text{CH}=\text{CH}}}\,\] 

1. Addition of water

Addition of Water to Alkyne 

2. Addition of hydrogen cyanide

\[HC\equiv CH+HCN\xrightarrow{Ba{{(CN)}_{2}}}\underset{vinyl\,cyanide}{\mathop{C{{H}_{2}}=CHCN}}\,\]

3. Addition of acetic acid

Addition of Acetic Acid to Alkyne 

 

4. Addition of alcohol 

\[HC\equiv CH+{{C}_{2}}{{H}_{5}}OH\xrightarrow[{{130}^{o}}C]{{{H}_{2}}S{{O}_{4}}}C{{H}_{2}}=CHO{{C}_{2}}{{H}_{5}}\xrightarrow{{{H}_{2}}O}C{{H}_{3}}CHO+{{C}_{2}}{{H}_{3}}OH\] 

5. Addition of ozone and ozonolysis 

Ozonolysis of alkyne

4. Oxidation: 

1. Oxidation with alkaline $KMn{{O}_{4}}$ 

\[\underset{acetylene}{\mathop{HC\equiv CH}}\,+4[O]\xrightarrow{alk.KMn{{O}_{4}}}\underset{oxalic\,acid}{\mathop{HOOC-COOH}}\,\]

Oxidation of Alkyne

2. Oxidation with acidic ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ $KMn{{O}_{4}}$ 

\[\underset{acetylene}{\mathop{HC\equiv CH}}\,+\xrightarrow[{{H}^{+}}]{{{K}_{2}}C{{r}_{2}}{{O}_{7}}}HOOC-COOH\to \underset{acetic\,acid}{\mathop{C{{H}_{3}}COOH}}\,\] 

Oxidation of Acidic ${{K}_{2}}C{{r}_{2}}{{O}_{7}}$ / $KMn{{O}_{4}}$ 

5. Formation of metallic derivatives: 

The group –C≡C–H in alkynes is slightly acidic in nature and hence its hydrogen atom can be easily replaced by certain metals to give metallic derivatives called acetylides or alkynides. 

Acidic Hydrogen in Alkenes

1. Formation of sodium acetylides

\[HC\equiv CH+Na\xrightarrow{Liq.N{{H}_{3}}}\underset{monosodium\,acetylide}{\mathop{HC\equiv C-Na}}\,+Na\xrightarrow[{{120}^{o}}C]{NaN{{H}_{2}}}\underset{disodium\,acetylide}{\mathop{NaC=CNa}}\,\] 

Formation of Sodium Propynide

2. Formation of copper and silver acetylides 

\[HC\equiv CH+\underset{ammonical\,cuprous\,chloride}{\mathop{2CuC{{l}_{2}}+2N{{H}_{4}}OH}}\,\to \underset{copper\,acetylide\,(red\,ppt.)}{\mathop{Cu-C\equiv C-Cu\downarrow }}\,+2N{{H}_{4}}Cl+2{{H}_{2}}O\] 

\[HC\equiv CH+\underset{ammonical\,silver\,nitrate}{\mathop{2AgN{{O}_{3}}+2N{{H}_{4}}OH}}\,\to \underset{silver\,acetylide\,(white\,ppt)}{\mathop{AgC\equiv CAg\downarrow }}\,+2N{{H}_{4}}N{{O}_{3}}+2{{H}_{2}}O\] 

These reactions are used for detecting the presence of acetylenic hydrogen atoms. 

Illustration 4 : The products obtained via oxymercuration ($HgS{{O}_{4}}+{{H}_{2}}S{{O}_{4}}$) of 1–butyne would be 

Oxymercuration Product of 1–Butyne

Solution: (A)

Oxymercuration reaction of 1–butyne

Class 11 Hydrocarbons Revision Notes

This complete article deals with class 11 hydrocarbons revision notes. Firstly, the term ‘hydrocarbons’ refers to the compounds of hydrogen and carbon only. Hydrocarbons specifically play a crucial role in our lives. 

Moreover, humans obtain kerosene, diesel, and petrol by the fractional distillation of petroleum. In the same way, humans also obtain coal gas by the destructive distillation of coal. All such fuel types, which are the sources of energy, comprises hydrocarbons mixture.

One of the most important alkenes reactions certainly happens to be oxidization, combustion, aromatization, and free radical substitution. Moreover, alkynes and alkenes undergo different multiple reactions. Most notably, these reactions are of addition reactions, which are electrophilic additions mainly. Also, the aromatic hydrocarbons primarily undergo electrophilic substitution reactions despite having unsaturation.

Topics Covered Under Class 11 Chemistry

Some of the topics that fall under the Class 11 Chemistry include:

• Some basic concepts of Chemistry

• Classification of the Elements & Periodicity in Properties

• Structure of Atom

• States of Matter - Liquids and Gases

• Chemical Bonding & Molecular Structure

• Equilibrium

• Thermodynamics

• Redox Reactions

• Hydrocarbons

• Environmental Chemistry (Not for Examination point of View)

Sub-Topics Covered Under Hydrocarbons

• Alkanes - Alkanes are known as organic compounds that consist of only hydrogen and carbon atoms which are single-bonded

• Conformation of Alkanes - The conformation of alkanes deals with the isomers of alkanes whose formation is due to the slight structural changes

• Preparation of Alkenes and Nomenclature - This basically deals with the preparation of the Alkenes and its naming

• Preparation of Alkynes and Nomenclature - Naming of Alkynes occurs by counting the longest continuous chain’s number of carbons, and their preparation is mostly the same as those of the alkenes

• Preparation of Aromatic Hydrocarbons and Nomenclature - According to the IUPAC system, the naming of aromatic hydrocarbons happens as benzene derivatives. In contrast, their preparation happens by different methods, and the most common method used is the isolating coal tar.

• Properties of Alkanes - Alkanes are known to be the unsaturated form of hydrocarbons

• Properties of Alkynes - Alkynes are the third type of hydrocarbon which contains at least one triple bond between a pair of carbon atoms

• Properties of Aromatic Hydrocarbons - The aromatic Hydrocarbons are called cyclic hydrocarbons having the delocalized pi electrons between the carbon atoms of the ring

Importance of Class 11 Revision Notes Hydrocarbons

The Revision Notes of Class 11 Hydrocarbons Chemistry or the CBSE class 11 quick revision notes for Chemistry or other subjects help the students revise the entire syllabus during the exam days. This revision notes of Hydrocarbons covers all the important formulas and concepts given in the chapter. Even if the student wishes to have an overview of a chapter, the quick revision notes will help them. These notes will also save time during stressful exam days.

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