Hydrocarbons are the compounds of carbon and hydrogen only.

Classification

They are classified into three broad categories based on their structure and properties.

i) Saturated (alkanes and cycloalkanes)

ii) Unsaturated (alkenes and alkynes)

iii) Aromatic

Alkanes

Saturated acyclic hydrocarbons containing carbon - carbon and carbon - hydrogen single bonds. They are also known as paraffins because of low reactivity under normal conditions. General formula is ${\mathrm{C}}_n{\mathrm{H}}_{2n+2}$ whereas general formula for cycloalkanes or alicyclic compounds is ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2n}$.

Preparation

Alkanes are obtained from petroleum and natural gas. Synthetically they can be prepared by the following methods

1. Catalytic hydrogenation of alkenes / alkynes

$${\mathrm{CH}}_2={\mathrm{CH}}_2\underset{\text{ Ni or Pt or Pd }}{\overset{{\mathrm{H}}_2}{\to }}{\mathrm{CH}}_3{\mathrm{CH}}_3$$

If catalyst is $\mathrm{Ni}$, reaction is named as Sabatier-Senderens reaction.

Exothermic process

Reactivity of alkene towards hydrogenation depends upon the size of groups present on doubly bonded carbons. Larger the size of functional groups, smaller the reactivity.

Reaction proceeds through adsorption mechanism

2. Reduction of alkyl halides

$\mathrm{R}-\mathrm{X}⟶\mathrm{R}-\mathrm{H}$

This reaction can be accomplished by any of the following reagents.

1. ${\mathrm{LiAlH}}_4$

2. ${\mathrm{NaBH}}_4$

3. $\mathrm{Na}/{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{OH}$

4. $\mathrm{Zn}/{\mathrm{CH}}_3\mathrm{COOH}$ or dil $\mathrm{HCl}$

${\mathrm{NaBH}}_4$ reduces only $2^{\circ }$ and $3^{\circ }$ alkyl halides.

Reaction does not take place with alkyl fluorides

Alkyl halides (particulary iodides) on treatment with hydrogen iodide in presence of red phosphorus get reduced to alkanes.

$\mathrm{R}-\mathrm{I}+\mathrm{HI}\stackrel{423\mathrm{K}}{\to }\mathrm{R}-\mathrm{H}+{\mathrm{I}}_2$

$2\mathrm{P}+3{\mathrm{I}}_2⟶2{\mathrm{PI}}_3$

(phosphorus removes iodine from the reaction mixture so that it does not react with alkanes).

3. Hydrolysis of Grignard reagent

(Indirect method of reduction of alkyl halides). Alkyl magnesium halides (Grignard reagents) are obtained on treatment of alkyl halides with magnesium in ether. Grignard reagents belong to the class of organometallic compounds.

4. Wurtz reaction

$R-X+2Na+R-X\underset{\mathrm{\Delta }}{\overset{\text{ dryether }}{\to }}R-R$

Limitation

Not suitable for formation of unsymmetrical alkanes because of the formation of a mixture of products.

$R-X+R^{\prime }-X\underset{\mathrm{\Delta }}{\overset{\text{ Natether }}{\to }}R-R^{\prime }+R-R+R^{\prime }-R^{\prime }$

5. Decarboxylation of carboxylic acids

Carboxylic acids, on heating with soda-lime give alkanes by the loss of one molecule of carbon dioxide.

${\mathrm{CH}}_3\mathrm{COOH}+\mathrm{NaOH}\stackrel{630\mathrm{K}}{\to }\stackrel{\mathrm{CaO}}{\to }{\mathrm{CH}}_4+{\mathrm{Na}}_2{\mathrm{CO}}_3$

6. Kolbe’s electrolytic method

$2{\mathrm{CH}}_3\mathrm{COONa}+2{\mathrm{H}}_2\mathrm{O}\stackrel{\text{ electrolysis }}{\to }{\mathrm{CH}}_3-{\mathrm{CH}}_3+2\mathrm{NaOH}+{\mathrm{H}}_2+2{\mathrm{CO}}_2$

Mechanism

$2{\mathrm{CH}}_3\mathrm{COONa}\stackrel{\text{ ionization }}{\to }2{\mathrm{CH}}_3{\mathrm{COO}}^{-}+2{\mathrm{Na}}^{+}$

$2{\mathrm{H}}_2\mathrm{O}\stackrel{\text{ ionization }}{\to }2{\mathrm{OH}}^{-}+2{\mathrm{H}}^{+}$

At anode

$2{\mathrm{CH}}_3{\mathrm{COO}}^{-}-2{\mathrm{e}}^{-}\to 2{\mathrm{CH}}_3\mathrm{C}-\dot{\mathrm{O}}-2{\mathrm{CO}}_{2}2\dot{\mathrm{C}}{\mathrm{H}}_3⟶{\mathrm{CH}}_3-{\mathrm{CH}}_3$

At cathode

$2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_2$

$2{\mathrm{Na}}^{+}+2{\mathrm{OH}}^{-}\to 2\mathrm{NaOH}$

Limitation :Cannot be used to prepare methane and unsymmetrical alkanes.

7. Reduction of carboxylic acids

$\mathrm{R}-\mathrm{COOH}\underset{\mathrm{\Delta }}{\overset{\mathrm{HH},\mathrm{redP}}{\to }}\mathrm{R}-{\mathrm{CH}}_3$

Acid $\to$ Alkane with same no. of $\mathrm{C}$ atoms.

8. Corey House synthesis

$${({\mathrm{CH}}_3-{\mathrm{CH}}_2)}_2\mathrm{CuLi}+{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}\underset{\text{ ether }}{\overset{\text{ dry }}{\to }}\underset{\text{ n-pentane }}{{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_3}+{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Cu}+\mathrm{LiBr}$$

9. Reduction of aldehydes and ketones

a) Clemmensen reduction

b) Wolff Kishner reduction

Physical Properties

1. Non-Polar

Non-Polar molecules because of covalent nature of $\mathrm{C}-\mathrm{C}$ and $\mathrm{C}-\mathrm{H}$ bonds.

2. Physical state

${\mathrm{C}}_1-{\mathrm{C}}_4$ alkanes are gases, ${\mathrm{C}}_5-{\mathrm{C}}_{17}$ alkanes are liquids, ${\mathrm{C}}_{18}$ and above alkanes are solids.

3. Boiling point

Increases with increase in molecular mass. This is because vander Waals forces increase with increase in surface area.

propane $<$ butane $<$ pentane

Decreases with increase in branching. This is because with increase in branching, molecule acquires the shape of a sphere. As a result the surface area decreases and hence vander Waals forces decrease.

4. Melting Point

Increases with increase in molecular mass though the increase in not regular. It is seen that this increase in melting point is more when we go from an alkane having odd number of carbons to the next higher alkane and is less when we go from an alkane having even number of carbons to the next higher alkane. This is because melting point not only depends on the size but also on the arrangement of the molecule in the crystal lattice. And symmetrical structures fit well into the crystal lattice, are better packed and strongly held and thus possess high melting points.

5. Solubility

Alkanes being non polar are insoluble in polar solvents like water and alcohol and are soluble in non polar solvents like benzene, ${\mathrm{CCI}}_4$. They are also miscible with each other.

6. Density

Increases with increasing molecular mass and goes to a maximum of $0.8\mathrm{g}{\mathrm{cm}}^{-3}$. Alkanes being lighter float on water and thus water cannot extinguish fire caused by any alkane.

Chemical Properties

Alkanes are generaly unreactive towards many reagents. Some important reactions are,

1. Halogenation

One or more hydrogens may be substituted and thus a mixture of products may be obtaine(d)

$\mathrm{R}-\mathrm{H}+{\mathrm{X}}_2\to \underset{\text{ or heat hv}}{}\mathrm{R}-\mathrm{X}+\mathrm{HX}(\mathrm{X}=\mathrm{F},\mathrm{Cl},\mathrm{Br},\mathrm{l})$

$R-X$ may undergo the same set of reactions again so that another $H$ is replaced by $X$.

Order of reactivity: ${\mathrm{F}}_2>{\mathrm{Cl}}_2>{\mathrm{Br}}_2>{\mathrm{I}}_2$

Fluorine reacts explosively whereas iodine reacts very slowly.

Rate of substitution of hydrogen atoms in alkanes : $3^{\circ }>2^{\circ }>1^{\circ }$

2. Nitration

Cleavage of carbon chain takes place to give various possible nitro products

3. Sulphonation

$\mathrm{R}-\mathrm{H}+$ conc. ${\mathrm{H}}_2{\mathrm{SO}}_4\stackrel{\mathrm{\Delta }}{\to }\mathrm{R}-{\mathrm{SO}}_3\mathrm{H}$

If more than one type of hydrogens are available in alkane, all isomeric sulphonic acids are formed in comparable amounts.

4. Oxidation (combustion)

${\mathrm{C}}_n{\mathrm{H}}_{2n+2}+\left[\frac{3n+1}{2}\right]{\mathrm{O}}_2(\mathrm{g})\stackrel{\mathrm{\Delta }}{\to }{\mathrm{nCO}}_2(\mathrm{g})+(\mathrm{n}+1){\mathrm{H}}_2\mathrm{O}+$ energy

Alkanes are used as fuels because of the large amount of energy evolved

Incomplete combustion of alkanes takes place as shown in the following reaction in limited supply of oxygen.

$$\begin{array}{r}{\mathrm{CH}}_4+\underset{\text{ (limited) }}{3{\mathrm{O}}_2⟶2\mathrm{CO}+4{\mathrm{H}}_2\mathrm{O}}\\ {\mathrm{CH}}_4+\underset{\text{ (limited) }}{{\mathrm{O}}_2⟶\mathrm{C}}\mathrm{C}+2{\mathrm{H}}_2\mathrm{O}\end{array}$$

Carbon monoxide and carbon (soot) produced are major air pollutants.

Carbon black can be used in the manufacture of ink, printer ink, tyres, paints, polishes and filters.

Oxidation under special conditions

(i) $2{\mathrm{CH}}_4+{\mathrm{O}}_2\frac{\text{ Cu5523K }}{100\mathrm{atm}}2\mathrm{CH}{\mathrm{H}}_3\mathrm{OH}$

(ii) ${\mathrm{CH}}_4+{\mathrm{O}}_2\stackrel{{\mathrm{MO}}_2{\mathrm{O}}_3}{\to }\mathrm{HCHO}+{\mathrm{H}}_2\mathrm{O}$

(iii) $2{\mathrm{CH}}_3{\mathrm{CH}}_3+3{\mathrm{O}}_2\underset{\mathrm{\Delta }}{\overset{({\mathrm{CH}}_3{\mathrm{COO}}_2\mathrm{Mn}}{\to }}2{\mathrm{CH}}_3\mathrm{COOH}+2{\mathrm{H}}_2\mathrm{O}$

(iv) ${\left({\mathrm{CH}}_3\right)}_3\mathrm{CH}\stackrel{\mathrm{KMnO},0\mathrm{OH}}{\to }{\left({\mathrm{CH}}_3\right)}_3\mathrm{COH}$

(Alkanes with tertiary $\mathrm{H}$ atom)

5. Isomerisation

6. Aromatization (reforming)

• Alkanes containing six or more carbon atoms undergo this reaction.

7. Pyrolysis (cracking)

Decomposition of higher alkanes into simpler low boiling alkanes of lower molecular mass by application of heat.

It involves breaking of $\mathrm{C}-\mathrm{C}$ and $\mathrm{C}-\mathrm{H}$ bonds to give a mixture of products

$\underset{\text{ hexane }}{{\mathrm{C}}_6{\mathrm{H}}_{14}}\underset{6-7\mathrm{atm}}{\to }\underset{\text{ hexene }}{{\mathrm{C}}_6{\mathrm{H}}_{12}}+\underset{\text{ butene }}{{\mathrm{H}}_2}+\underset{\text{ ethane }}{{\mathrm{C}}_4{\mathrm{H}}_8}+\underset{\text{ propene }}{{\mathrm{C}}_2{\mathrm{H}}_6}+\underset{\text{ ethene }}{{\mathrm{C}}_3{\mathrm{H}}_6}+\underset{\text{ methane }}{{\mathrm{C}}_2{\mathrm{H}}_4}+{\mathrm{CH}}_4$

Practice Questions

1. 2-Methylbutane on reacting with bromine in the presence of sunlight gives mainly :

(a) 1-bromo-2-methylbutane

(b) 2-bromo-2-methylbutane

(c) 2-bromo-3-methylbutane

(d) 1-bromo-3-methylbutane

Show Answer

Answer: (b)

2. An alkyl bromide $[x]$ reacts with sodium in ether to form 4,5 -diethyloctane, the compound ’ $x$ ’ is

(a) ${\mathrm{CH}}_3{\left[{\mathrm{CH}}_2\right]}_3\mathrm{Br}$

(b) ${\mathrm{CH}}_3{\left[{\mathrm{CH}}_2\right]}_5\mathrm{Br}$

(c) ${\mathrm{CH}}_3{\left[{\mathrm{CH}}_2\right]}_3\mathrm{CH}[\mathrm{Br}]{\mathrm{CH}}_3$

(d) ${\mathrm{CH}}_3-{\left[{\mathrm{CH}}_2\right]}_2-\mathrm{CH}[\mathrm{Br}]-{\mathrm{CH}}_2{\mathrm{CH}}_3$

Show Answer

Answer: (d)

3. Alkyl halides react with sodium metal to give :

(a) alkenes

(b) alkyl sodium halide

(c) alkanes

(d) alkenyl halides

Show Answer

Answer: (c)

4. Of the five isomeric hexanes, the isomer which can give two monochlorinated compounds is

(a) n-hexane

(b) 2,3-dimethylbutane

(c) 2,2-dimethylbutane

(d) 2-methylpentane

Show Answer

Answer: (b)

5. Ethane is formed by the reaction of methyl iodide and sodium metal in dry ether solution. The reaction is known as

(a) Clemmensen reduction

(b) Kolbe’s reaction

(c) Wurtz reaction

(d) Cannizzaro reaction

Show Answer

Answer: (c)

6. Which one of the following cannot be prepared by Wurtz reaction?

(a) ${\mathrm{CH}}_4$

(b) ${\mathrm{C}}_2{\mathrm{H}}_6$

(c) ${\mathrm{C}}_3{\mathrm{H}}_8$

(d) ${\mathrm{C}}_4{\mathrm{H}}_{10}$

Show Answer

Answer: (a)

7. ${\mathrm{C}}_5{\mathrm{H}}_{11}\mathrm{Cl}(\mathrm{A})$ by wurtz reaction forms 2, 2, 5, 5-tetramethylhexane as main product. Hence, A can be

(a) 2,2-dimethyl-1-chloropropane

(b) 2-methyl-1-chlorobutane

(c) both are true

(d) none is true

Show Answer

Answer: (a)

8. ${\mathrm{C}}_8{\mathrm{H}}_{18}(A)$ on chlorination forms only one type of ${\mathrm{C}}_8{\mathrm{H}}_{17}\mathrm{Cl}(B),A$ can be

Show Answer

Answer: (d)

9. Heating a mixture of sodium benzoate and sodalime gives

(a) Methane

(b) Benzene

(c) Toluene

(d) Calcium benzoate

Show Answer

Answer: (b)

10. ${\mathrm{C}}_6{\mathrm{H}}_{12}(A)$ has chirality but on hydrogenation $A$ is converted into ${\mathrm{C}}_6{\mathrm{H}}_{14}(B)$ in which chirality disappears. $A$ is

(a) 3-methyl-1-pentene

(b) 2-methyl-2-pentene

(c) 2,3-dimethyl-2-butene

(d) 3,3-dimethyl-1-butene

Show Answer

Answer: (a)

11. What will be the product formed when 1-bromo-3-chlorocyclobutane reacts with two equivalents of metallic sodium in ether?

Show Answer

Answer: (d)

12. Consider the following reaction,

Identify the structure of the major product ’ $X$ ‘.

Show Answer

Answer: (b)

Alkenes

Unsaturated hydrocarbons with the general formula ${\mathrm{C}}_n{\mathrm{H}}_{2n}$. They are also known as olefins and the carbon-carbon double bond in alkenes contains one sigma and one pi bon(d) Each carbon of double bond is ${\mathrm{sp}}^2$ hybridized and is thus planar having bond angle of ${120}^{\circ }\mathrm{C}$. Restricted rotation about the double bond and planarity of molecule results in the alkenes showing geometrical isomerism.

Preparation

commercially the lower alkenes are obtained by the cracking of higher fractions of petroleum. In laboratory they are prepared by,

1. By partial hydrogenation of alkynes

Lindlar catalyst is palladium carbon deactivated with poisons like ${\mathrm{CaCO}}_3$, lead, quinoline, ${\mathrm{BaSO}}_4$ and is specific to give cis - alkenes in contrast to $\mathrm{Na}/$ liq ${\mathrm{NH}}_3$ which gives the trans alkene.

2. By dehydrohalogenation of alkyl halides

Elimination takes place according to Saytzeff rule, more substituted alkene is the major product.

Saytzeff rule states that in an elimination reaction, when there is the possibility of formation of more than one alkene by elimination of $\beta$-hydrogen from either side of carbon carrying halogen, more highly substituted alkene is forme(d)

Rate is : iodine > bromine > chlorine (for halogens)

Tertiary > secondary > primary (for alkyl groups).

For example, 1-bromo-1-methylcyclohexane on dehydrobromination, gives 1 -methylcyclohexene as the major product.

3. By dehalogenation of vicinal dihalides

The process of removal of a molecule of halogen from a dihaloalkane to form alkene is called dehalogenation.

4. By dehydration of alcohols

Alcohols on heating with conc. ${\mathrm{H}}_2{\mathrm{SO}}_4$ undergo loss of a water molecule to give alkenes.

If possibility exists, proton is lost from that side to yield the more highly substituted alkene.

ii) Dehydration of an alcohol can also be brought about by passing the vapours of alcohol over dehydrating agents like heated alumina $\left({\mathrm{Al}}_2{\mathrm{O}}_3\right)$ at $673\mathrm{K}$, or phosphorus pentoxide $\left({\mathrm{P}}_2{\mathrm{O}}_5\right)$ or phosphoric acid $\left({\mathrm{H}}_3{\mathrm{PO}}_4\right)$

For example

Order of reactivity of alcohols towards dehydration is, tertiary>secondary>primary.

5. Kolbe’s electrolytic method

Dicarboxylic acids can give alkenes as shown in the following example.

Mechanism

$2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 2{\mathrm{OH}}^{-}+{\mathrm{H}}^{+}$

At anode :

At cathode

${\mathrm{H}}^{+}+{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}⟶{\mathrm{H}}_2$

Physical Properties

Alkenes resemble alkanes in physical properties.

1. Physical state $-{\mathrm{C}}_2-{\mathrm{C}}_4$ alkenes are gases, ${\mathrm{C}}_5-{\mathrm{C}}_{18}$ are liquids, ${\mathrm{C}}_{19}$ and above are solids.

2. Boiling point and melting point-Melting and boiling points rise with increasing molecular mass. Straight chain alkenes have higher boiling points than branched chain compounds.

3. Solubility - Alkenes are insoluble in water but soluble in non - polar solvents like benzene, ether etc.

Chemical Properties

Alkenes undergo electrophilic addition reactions

Due to presence of $\pi$ electrons, they attract electrophiles and repel nucleophiles. Since electrophilic addition reactions are energetically more favourable than electrophilic substitution so alkenes undergo electrophilic addition.

Chemical Reactions

Addition Reactions

1. Addition of dihydrogen (catalytic hydrogenation)

${\mathrm{CH}}_2={\mathrm{CH}}_2+{\mathrm{H}}_2\stackrel{\text{ Nior Ptor Pd }}{\to }{\mathrm{CH}}_3-{\mathrm{CH}}_3$

2. Addition of hydrogen halides $(\mathrm{HCl},\mathrm{HBr},\mathrm{HI})$

${\mathrm{H}}_2\mathrm{C}={\mathrm{CH}}_2+\mathrm{HBr}⟶{\mathrm{H}}_3\mathrm{C}-{\mathrm{CH}}_2\mathrm{Br}$

Order of reactivity is $\mathrm{HI}>\mathrm{HBr}>\mathrm{HCl}$

If alkene is not symmetrical,

It follows Markovnikov’s rule. The addition of unsymmetrical reagents such as $\mathrm{HX},{\mathrm{H}}_2\mathrm{O},\mathrm{HOX}$, etc. to unsymmetrical alkenes occur in such a way that the negative part of the addendum [i.e., adding molecule] goes to that carbon atom of the double bond which carries lesser number of hydrogen atoms. For example propene undergoes addition of $\mathrm{HBr}$ to form 2-bromopropane as the major product. This can be explained mechanistically as follows.

Anti Markovnikov addition or peroxide effect or Kharash effect : In presence of peroxide, addition of $\mathrm{HBr}$ (but not of $\mathrm{HCl}$ or $\mathrm{HI}$ ) to unsymmetrical alkenes takes place anti to the Markovnikov’s rule.

${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2+\mathrm{HBr}\stackrel{\text{ peroxide }}{\to }{\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}$

Mechanism (Free radical mechanism)

1.

2. ${\dot{\mathrm{C}}}_6{\mathrm{H}}_5+\mathrm{H}-\mathrm{Br}\stackrel{\text{ Homolysis }}{\to }{\mathrm{C}}_6{\mathrm{H}}_5+\underset{\text{ Free radical }}{\dot{\mathrm{B}r}}$

3.

Because the $2^{\circ }$ free radical is more stable than $1^{\circ }$ free radical, it will be preferably forme(d)

4. $\underset{\text{ Preferred free radical }}{{\mathrm{CH}}_3-\dot{\mathrm{C}}}\mathrm{H}-{\mathrm{CH}}_2\mathrm{Br}+\mathrm{H}-\mathrm{Br}⟶\underset{\text{ 1-Bromopropane (Major product) }}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2\mathrm{Br}}+\dot{\mathrm{Br}}$

Peroxide effect is observed only in $\mathrm{HBr}$ and not in other halogen acids

For HBr both steps are exothermic

For $\mathrm{HCl}$ or $\mathrm{HF}$, second step involving the reaction of carbon radical is endothermic.

For $\mathrm{HI}$, the first step involving addition of iodine radical to alkene is endothermic.

$\begin{array}{lll}& \mathrm{\Delta }\mathrm{H}\text{ (in KJ / mole)}& \mathrm{\Delta }\mathrm{H}\text{ (in KJ / mole)}\\ \mathrm{X}\text{ in }\mathrm{HX}& \text{(i) }\mathrm{X}+{\mathrm{CH}}_2={\mathrm{CHCH}}_3\to {\mathrm{XCC}}_2-\mathrm{CH}-{\mathrm{CH}}_3& \text{(ii) }{\mathrm{XCH}}_2-{\mathrm{CHCH}}_3+\mathrm{HX}\to {\mathrm{XCH}}_2-{\mathrm{CH}}_2{\mathrm{CH}}_3+\mathrm{X}\\ \mathrm{F}& -209& +159\\ \mathrm{Cl}& -101& +27\\ \mathrm{Br}& -42& -37\\ \mathrm{I}& +12& -104\end{array}$

3. Addition of sulphuric acid

It follows Markovnikov’s rule

4. Addition of Halogens

Alkenes react with chlorine or bromine (but not iodine) to form vicinal dihalides,

Test for unsaturation

This reaction is used as test for the presence of unsaturation. On adding bromine in carbon tetrachloride to an unsaturated compound the reddish orange colour of bromine is discharge(d)

5. Addition of hypohalous acid

The overall reaction involves the addition of the elements of hypohalous acid $(\mathrm{HO}-\mathrm{X})$ in accordance with Markovnikov’s rule.

The order of reactivity of different hypohalous acids HOX is

$\mathrm{HOCl}>\mathrm{HOBr}>\mathrm{HOI}$

6. Acid Catalysed Hydration of alkenes

Addition of water

It follows Markovnikov’s rule

Rearrangement takes place due to the formation of carbocation intermediate

Mechanism

Another example where methyl shift takes place

7. Oxidation of Alkenes

i) Hydroxylation

Reaction with cold, dilute ${\mathrm{KMnO}}_4$ (Baeyer’s reagent)

Syn addition of $\mathrm{OH}$ on both doubly bonded carbons takes place.

Test for unsaturation : Delocolourisation of Baeyer’s reagent is used as a test for the presence of unsaturation in an organic compoun(d)

ii) Oxidative degradation

Reaction with hot ${\mathrm{KMnO}}_4$ or acidified ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$

Product depends on structure of alkene.

i. Mono substituted vinylic carbon is converted into carboxylic group.

$${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3\stackrel{{KMn{O}_4/H/\text{ or K }}_2{\mathrm{Cr}}_2{\mathrm{O}}_7/{\mathrm{H}}^{+}}{\to }{\mathrm{KH}}_3/\mathrm{COOH}+{\mathrm{CH}}_3\mathrm{COOH}$$

ii. Terminal alkene gives formic acid which oxidises to ${\mathrm{CO}}_2$ and ${\mathrm{H}}_2\mathrm{O}$.

$${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CH}}_2\stackrel{[0]}{\to }{\mathrm{CH}}_3\mathrm{COOH}+\mathrm{HCOOH}\stackrel{[0]}{\to }{\mathrm{CO}}_2+{\mathrm{H}}_2\mathrm{O}$$

iii. Disubstituted vinylic carbon gets converted into keto group

iii) Addition of ozone (Ozonolysis)

Useful for locating the position of double bond in a compound

Useful method for the synthesis of aldehydes and ketones

iv) Epoxidation

$\mathrm{R}-\mathrm{CH}={\mathrm{CH}}_2+\frac{1}{2}{\mathrm{O}}_2\underset{473-673\mathrm{K}}{\overset{\mathrm{Ag}}{\to }}\mathrm{R}-\mathrm{CH}-{\mathrm{CH}}_2$

Reaction given by lower alkenes

v) Combustion

$${\mathrm{CH}}_2={\mathrm{CH}}_2+3{\mathrm{O}}_2⟶2{\mathrm{CO}}_2+2{\mathrm{H}}_2\mathrm{O}$$

Alkenes burn when reacted with oxygen to give carbon dioxide and water along with heat.

8. Polymerisation

Reaction of alkenes with themselves at high temperature, pressure or in presence of catalyst to form giant molecules called polymers.

Basic alkene unit is known as a monomer.

$$\underset{\begin{array}{c}\text{ Ethene }\end{array}}{{\mathrm{nCC}}_2={\mathrm{CH}}_2}\underset{\text{ catalyst }}{\overset{\text{ high temp./press }}{\to }}\underset{\begin{array}{c}\text{ Polythene }\end{array}}{{{\mathrm{CH}}_2-{\mathrm{CH}}_2)}_n}$$

Useful for making numerous daily articles like plastic bottles, plastic bags, toys, raincoats, pipes, buckets etc.

9. Allylic halogenation

10. Allylic oxidation

Solved examples

1. Match column I with column II

Show Answer

Answer . The correct match is $a-r,b-q,c-s$ and $d-p$.

2. In the following sequence of reactions, the alkene affords the compound $\mathrm{B}$

${\mathrm{CH}}_3\mathrm{CH}={\mathrm{CHCH}}_3\stackrel{{\mathrm{O}}_3}{\to }\mathrm{A}\stackrel{{\mathrm{ZnNH}}_2\mathrm{O}}{\to }\mathrm{B}$.

The compound $B$ is

(a) ${\mathrm{CH}}_3{\mathrm{COCH}}_3$

(b) ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{COCH}}_3$

(c) ${\mathrm{CH}}_3\mathrm{CHO}$

(d) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHO}$

Show Answer

Answer . Since this is a symmetrical alkene, so ozonolysis forms only one product which is ethanal.

3. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of $44\mu$. The alkene is

(a) Ethene

(b) Propene

(c) 1-butene

(d) 2-butene

Show Answer

Answer . d

4. Which among the following alkenes will be oxidized by ${\mathrm{SeO}}_2$ ?

Show Answer

Answer . This is an example of allylic oxidation and only compound at option (b) has a hydrogen atom at allylic position. Thus answer is b.

Practice Questions

1. The reaction of $\mathrm{HBr}$ with ${\mathrm{CH}}_3-\underset{\mathrm{CH}}{\mathrm{C}}={\mathrm{CH}}_2$ in the presence of peroxide will give :

Show Answer

Answer: (c)

2. 1-phenylpropene on reaction with $\mathrm{HBr}$ gives

(a) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2-\mathrm{CH}(\mathrm{Br}){\mathrm{CH}}_3$

(b) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CH}(\mathrm{Br}){\mathrm{CH}}_2{\mathrm{CH}}_3$

(c) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}$

(d) ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{CH}(\mathrm{Br})-\mathrm{CH}={\mathrm{CH}}_2$

Show Answer

Answer: (b)

3. The reaction of

Show Answer

Answer: (c)

4. Observe the following reactions and predict the product (A)

(a) ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{Br}$

(b) ${\mathrm{C}}_6{\mathrm{H}}_5-{\mathrm{CH}}_2-\mathrm{Br}$

(c) ${\mathrm{C}}_6{\mathrm{H}}_5-\mathrm{CHBr}-{\mathrm{CH}}_3$

(d)

Show Answer

Answer: (c)

5. But-1-ene may be converted into butane by reaction with

(a) $\mathrm{Zn}/\mathrm{HCl}$

(b) $\mathrm{Sn}/\mathrm{HCl}$

(c) $\mathrm{Zn}-\mathrm{Hg}/{\mathrm{H}}_2\mathrm{O}$

(d) $\mathrm{Pd}/{\mathrm{H}}_2$

Show Answer

Answer: (d)

6. In the reaction sequence

$[\mathrm{X}]$ will be :

(a) 1-bromo-2-methylcyclopentane

(b) 1-bromo-1-methylcyclopentane

(c) 1-bromo-5-methylcyclopentane

(d) 5-bromo-1-methylcyclopentane

Show Answer

Answer: (b)

7. The reaction of propene with $\mathrm{HOCl}$ proceeds via the addition of

(a) ${\mathrm{H}}^{+}$in the first step

(b) ${\mathrm{Cl}}^{+}$in the first step

(c) ${\mathrm{OH}}^{-}$in the first step

(d) ${\mathrm{Cl}}^{+}$and ${\mathrm{OH}}^{-}$in single step

Show Answer

Answer: (b)

8. In the presence of peroxide, hydrogen chloride and hydrogen iodide do not give anti Markovnikov’s addition to alkenes because

(a) both are highly ionic

(b) one is oxidizing and the other is reducing

(c) One of the step is endothermic in both the cases

(d) All the steps are endothermic in both the cases

Show Answer

Answer: (c)

9. In the reaction below, $X$ is

Neopentyl alcohol $\underset{\text{ heat }}{\overset{{\mathrm{H}}_2{\mathrm{S}}_4}{\to }}\mathrm{X}$

(a) 2-methylpentane

(b) 2-methylpent-2-ene

(c) 2-methylbut-2-ene

(d) neopentane

Show Answer

Answer: (c)

10. The main product of the following reaction is

${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{CH}[\mathrm{OH}]\mathrm{CH}{\left[{\mathrm{CH}}_3\right]}_2\stackrel{\text{ Conc. }{\mathrm{H}}_2{\mathrm{SO}}_4}{\to }$

Show Answer

Answer: (c)

11. Reaction of 2-phenyl-1-bromocyclopentane on reaction with alcoholic $\mathrm{KOH}$ produces

(a) 4-phenylcyclopentane

(b) 2-phenylcyclopentane

(c) 1-phenylcyclopentene

(d) 3-phenylcyclopentene

Show Answer

Answer: (c)

12. Addition of $\mathrm{HCl}$ to 3, 3, 3-trichloropropene gives

(a) ${\mathrm{Cl}}_3{\mathrm{CCH}}_2{\mathrm{CH}}_2\mathrm{Cl}$

(b) ${\mathrm{Cl}}_3\mathrm{CCH}(\mathrm{Cl}){\mathrm{CH}}_3$

(c) ${\mathrm{Cl}}_2\mathrm{CHCH}(\mathrm{Cl}){\mathrm{CH}}_2\mathrm{Cl}$

(d) ${\mathrm{Cl}}_2{\mathrm{CHCH}}_2{\mathrm{CHCl}}_2$

Show Answer

Answer: (a)

13. In the following reaction

${\mathrm{C}}_2{\mathrm{H}}_2\underset{{\mathrm{H}}_2{\mathrm{SO}}_4\mathrm{O}{\mathrm{HSOO}}_4}{\overset{{\mathrm{H}}_0}{\to }}[\mathrm{X}]⟶{\mathrm{CH}}_3\mathrm{CHO}$, what is $[\mathrm{X}]$ ?

(a) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{OH}$

(b) ${\mathrm{CH}}_3-\mathrm{O}-{\mathrm{CH}}_3$

(c) ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{CHO}$

(d) ${\mathrm{CH}}_2=\mathrm{CHOH}$

Show Answer

Answer: (d)

14. For the given reaction

Which of the following is correct?

(a) I is less stable than II

(b) I will be formed at faster rate

(c) II will be formed at faster rate

(d) I and II will be formed at the same rate

Show Answer

Answer: (b)

15. In the given reaction

Show Answer

Answer: (d)

16. Which one of the following alkenes will not form tertiary carbocation with ${\mathrm{H}}^{+}$?

Show Answer

Answer: (b)

17. The predominant product formed when 3-methyl-2-pentene reacts with $\mathrm{HOCl}$ is

Show Answer

Answer: (d)

18. Consider the following reaction:

$[\mathrm{X}]$ is :

(a) 2,3-Dimethyl-2-butanol

(b) 2,3-Dimethyl-3-butanol

(c) 3,3-Dimethyl-2-butanol

(d) 3,3-Dimethyl-1-butanol

Show Answer

Answer: (a)

19. Consider the following reaction:

$[A]\stackrel{\mathrm{KMnO},\widehat{0\mathrm{O}}/\mathrm{\Delta }}{\to }{\mathrm{C}}_5{\mathrm{H}}_{10}{\mathrm{O}}_2$

In the above reaction $[\mathrm{A}]$, will be

(a) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{CH}={\mathrm{CH}}_2$

(b) ${\mathrm{CH}}_3-\mathrm{CH}-{\mathrm{CH}}_2-\mathrm{CH}={\mathrm{CH}}_2$

(c) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_3\mathrm{CH}-\mathrm{CH}={\mathrm{CH}}_2$

(d) all

Show Answer

Answer: (d)

20. Compound (A) on oxidation with ${\mathrm{KMnO}}_4/{\mathrm{OH}}^{-}$gives two compounds

Compound (A) will have the structure,

Show Answer

Answer: (b)

21. Consider the following reaction,

In the above reaction, the reaction intermediate is :

Show Answer

Answer: (c)

22. Product of the given reaction ${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3\underset{-{78}^{\circ }\mathrm{C}}{\overset{{\mathrm{O}}_3{\mathrm{CH}}_2{\mathrm{Cl}}_2}{\to }}$ will be

(a) ${\mathrm{CH}}_3-\mathrm{CHO}$

(b) ${\mathrm{CH}}_3-\mathrm{COOH}$

(c) $\begin{array}{c}{\mathrm{CH}}_3-\mathrm{CH}-\mathrm{CH}-{\mathrm{CH}}_3\\ \mathrm{OH}\mathrm{OH}\end{array}$

(d)

Show Answer

Answer: (d)

Alkynes

Unsaturated hydrocarbons containing a carbon - carbon triple bond and having the general formula ${\mathrm{C}}_{\mathrm{n}}{\mathrm{H}}_{2-2}$. Simplest member is acetylene (ethyne). Triple bond contains one $\sigma$ and two $\pi$ bonds. Each carbon of triple bond is sp hybridized and thus linear having bond angle of ${180}^{\circ }$.

Preparation

Industrially ethyne is prepared from calcium carbide as

${\mathrm{CaC}}_2+2{\mathrm{H}}_2\mathrm{O}\to \mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}+\mathrm{Ca}(\mathrm{OH}{)}_2$

Synthetic methods are,

1. Dehydrohalogenation of vicinal dihalides

2. Dehydrohalogenation of gem dihalides

$${\mathrm{RCH}}_2-{\mathrm{CHX}}_2\underset{{\mathrm{H}}^{\otimes }}{\overset{\stackrel{{\mathrm{NaNH}}_2,\mathrm{\Delta }}{⟶}}{\to }}\mathrm{R}-\mathrm{C}\equiv \mathrm{CH}$$

3. Dehalogenation of $\alpha ,\alpha ,\beta ,\beta$-tetrahaloalkanes

4. Synthesis of higher alkynes from acetylene

$$\begin{array}{rl}& \mathrm{HC}\equiv \mathrm{CH}+{\mathrm{NaNH}}_2\underset{196\mathrm{K}}{\overset{\stackrel{{\mathrm{LiqNH}}_3}{⟶}}{\to }}\mathrm{HC}\equiv {\mathrm{C}}^{-}{\mathrm{Na}}^{+}+{\mathrm{NH}}_3\\ & \text{ Ethyne }\text{ Sodium acetylide }\\ & \mathrm{HC}\equiv {\mathrm{C}}^{-}{\mathrm{Na}}^{+}+{\mathrm{CH}}_3-\mathrm{Br}⟶\mathrm{HC}\equiv \mathrm{C}-{\mathrm{CH}}_3+\mathrm{NaBr}\\ & \begin{array}{lll}\text{ So(d)acetylide }& \text{ Bromomethane }& \text{ Propyne }\end{array}\end{array}$$

Other terminal alkynes like $\mathrm{R}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}$ can also undergo the above series of reactions and elongate chain on the other end to give $R-C\equiv C-R^1$.

5. Kolbe’s hydrocarbon synthesis

Mechanism

$2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 2{\mathrm{OH}}^{-}+2{\mathrm{H}}^{+}$

At cathode: $2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}⟶{\mathrm{H}}_2$

Physical Properties

Similar to alkanes and alkenes

1. Physical state : ${\mathrm{C}}_2-{\mathrm{C}}_4$ alkynes are gases, ${\mathrm{C}}_5-{\mathrm{C}}_{12}$ are liquids, ${\mathrm{C}}_{13}$ and above are solids.

2. Melting and boiling point and density : Increase with increasing molecular mass.

3. Solubility: Alkynes are weakly polar and thus soluble in organic solvents like ether, benzene and ${\mathrm{CCl}}_4$. They are immiscible with water and are lighter than it.

Chemical Properties

Reactivity of alkynes vs alkenes

Due to greater electronegativity of $\mathrm{sp}$ hybridized carbon atoms of a triple bond than ${\mathrm{sp}}^2$ hybridized carbon atoms of double bond, $\pi$ electrons of alkynes are more tightly held by the carbon atoms than the $\pi$ electrons of alkenes, hence are less easily available for reaction with electrophiles.

(i) Addition of reagents to alkynes takes place in two steps. In the first step alkene is produced and on further addition of the reagent, a saturated compound is produce(d)

1. Addition of Hydrogen

$\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}\stackrel{\mathrm{Na}/\text{ liq. }{\mathrm{NH}}_3}{\to }{\mathrm{H}}_2\mathrm{C}={\mathrm{CH}}_2$

Product depends on the reagent use(d)

2. Addition of halogens

During this reaction, the reddish brown colour of ${\mathrm{Br}}_2$ is decolourised and hence this reaction is used as a test for unsaturation i.e. for the presence of double and triple bon(d)

3. Addition of ${\mathrm{H}}_2\mathrm{O}$ (Hydration of alkynes)

In case of unsymmetrical terminal alkynes, addition occurs in accordance with Markovnikov’s rule.

Acetylene is the only alkyne that gives an aldehyde on hydration.

All other alkynes give ketones.

4. Addition of hypohalous acid

Addition occurs in accordance with Markovnikov’s rule

5. Addition of Hydrogen Halides ( $\mathrm{HCl},\mathrm{HBr},\mathrm{HI}$ )

In absence of peroxide, addition takes place by Markovnikov’s rule

Reaction proceeds in two steps and can be stopped at the haloalkene stage. If it proceeds further, gem dihalide is obtaine(d)

6. Addition of Hydrogen Cyanide

$$\mathrm{HC}\equiv \mathrm{CH}+\mathrm{HCN}\underset{\begin{array}{c}\text{ or }\\ \mathrm{Ba}(\mathrm{CN}{)}_2\end{array}}{\overset{{\mathrm{CH}}_2{\mathrm{Cl}}_2/\mathrm{HCl}}{\to }}\underset{\begin{array}{c}\text{ vinyl cyanide }\\ \text{ or acrylonitrile }\end{array}}{{\mathrm{CH}}_2=\mathrm{CH}-\mathrm{CN}}$$

7. $\mathrm{HC}\equiv \mathrm{CH}+{\mathrm{CH}}_3\mathrm{OH}\underset{45\mathrm{OK}}{\overset{{\mathrm{CH}}_3\mathrm{K}}{\to }}{\mathrm{CH}}_2=\mathrm{CH}-\mathrm{O}-{\mathrm{CH}}_3$

$$\begin{gathered} Methoxyethene \\ (Methylvinylether) \end{gathered}$$

8.

(II) Acidity of alkynes

An $sp$ hybridized carbon is more electronegative than ${\mathrm{sp}}^2$ or $s{p}^3$ hybridized carbon atom. Due to this greater electronegativity, the electrons of $\mathrm{C}-\mathrm{H}$ bond are displaced more towards the carbon than towards the hydrogen atom. The hydrogen atom is less tightly held by the carbon and hence can be removed as a proton $\left[{\mathrm{H}}^{+}\right]$by a strong base. Consequently, alkynes behave as acids.

Acetylene reacts with ${\mathrm{Ag}}^{+}$and ${\mathrm{Cu}}^{+}$to form insoluble acetylide. This reaction is shown only by terminal alkynes and thus can be used to distinguish between terminal and non-terminal alkynes. For example,

Only terminal alkynes react

$$\begin{array}{rl}& \mathrm{R}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}+\underset{\text{ Tollen’s reagent }}{\mathrm{Ag}{\left[{\mathrm{NH}}_3\right]}_2^{+}}\to \underset{\text{ White ppt. }}{\mathrm{R}-\mathrm{C}\equiv \mathrm{C}\mathrm{Ag}\downarrow +{\mathrm{NH}}_3}\\ & \mathrm{R}-\mathrm{C}\equiv \mathrm{C}-\mathrm{R}+\mathrm{Ag}{\left[{\mathrm{NH}}_3\right]}_2^{+}\to \text{ No reaction }\end{array}$$

Terminal alkynes react with ammoniacal cuprous chloride to form red ppt. of copper acetylides.

$$\mathrm{R}-\mathrm{C}\equiv \mathrm{CH}+{\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_2\right]}^{+}{\mathrm{OH}}^{-}\to \underset{\text{ Red ppt }}{\mathrm{R}-\mathrm{C}}\equiv \mathrm{C}-\mathrm{Cu}\downarrow +{\mathrm{H}}_2\mathrm{O}+2{\mathrm{NH}}_3$$

Acidity order

i) $\mathrm{HC}\equiv \mathrm{CH}>{\mathrm{H}}_2\mathrm{C}={\mathrm{CH}}_2>{\mathrm{CH}}_3-{\mathrm{CH}}_3$

ii) $\mathrm{HC}\equiv \mathrm{CH}>{\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{CH}»{\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3$

Alkynes undergo some nucleophilic addition reactions also whereas alkenes do not.

Vinyl carbanion is more stable because the negative charge is present on a more electronegative ${\mathrm{sp}}^2$ hybridised carbon atom. Alkyl carbanion is not stable because the negative charge is present on a ${\mathrm{sp}}^3$ hybridised carbon atom.

(III) Polymerisation

1. Linear Polymerisation

$$\mathrm{HC}\equiv \mathrm{CH}⟶\underset{\text{ Polyethyne }}{\mathrm{CH}}=\mathrm{CH}{)}_n$$

Polyethyne conducts electricity and can be used as electrode in batteries.

2. Cyclic polymerization

(IV) Oxidation reactions

1. Oxidation with Cold dilute Potassium permanganate (Baeyer’s reagent)

$\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}\stackrel{{\mathrm{KMnO}}_4}{\to }\mathrm{COOH}-\mathrm{COOH}$

2. Oxidation with hot basic ${\mathrm{KMnO}}_4$

3. Oxidative Ozonolysis

4. Oxidation by ${\mathrm{SeO}}_2$

Solved Examples

1. Which of the following hydrocarbon will decolourise bromine water and react with ammoniacal cuprous chloride to form red precipitate

(a) ${\mathrm{H}}_2\mathrm{C}={\mathrm{CH}}_2$

(b) $\mathrm{H}-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}$

(c) ${\mathrm{CH}}_3-\mathrm{CH}={\mathrm{CH}}_2$

(d) ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3$

Show Answer

Answer - (b) All the four will decolourise bromine water but only terminal alkynes react with cuprous chloride to give red precipitate.

2. Which of the following isomeric alkenes of molecular formula ${\mathrm{C}}_5{\mathrm{H}}_{10}$ show geometrical isomerism?

i) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{CH}=\mathrm{CH}-{\mathrm{CH}}_3$

ii) ${\mathrm{CH}}_3-\mathrm{CH}-\mathrm{CH}={\mathrm{CH}}_2$

iii) $\begin{array}{c}{\mathrm{CH}}_3{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{C}={\mathrm{CH}}_2\end{array}$

iv) ${\mathrm{CH}}_3-\mathrm{C}=\mathrm{CH}-{\mathrm{CH}}_3$

a) only (i)

b) only (ii)

c) (i) & (ii)

d) both (iii) & (iv)

Show Answer

Answer -(a) only (i) contains different groups on each of the two double bonded carbon atoms.

3. Which of the following alkene on ozonolysis gives butan -2-one and 2 - methyl propanal

Show Answer

Answer . (d)

Practice Questions

1. Predict the product $\mathrm{C}$ obtained in the following reaction of butyne

Show Answer

Answer: (a)

2. In the reaction:

${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-\mathrm{H}\stackrel{\text{ (i) }{\mathrm{NaNH}}_2/{\mathrm{NH}}_3(\mathrm{l})}{\to }$ (A) $\stackrel{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{Br}}{\to }$

(A) $\stackrel{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{Br}}{\to }$

The product $(\mathrm{B})$ is :

(a) ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_3$

(b) ${\mathrm{CH}}_3-\mathrm{CH}={\mathrm{CH}}_2$

(c) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$

(d) ${\mathrm{CH}}_3-\mathrm{CH}=\mathrm{C}=\mathrm{CH}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$

Show Answer

Answer: (a)

3. The hydrocarbon which can react with $\mathrm{Na}/{\mathrm{NH}}_3$ is :

(a) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_2-{\mathrm{CH}}_2-{\mathrm{CH}}_3$

(b) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{C}\equiv \mathrm{CH}$

(c) ${\mathrm{CH}}_3-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_3$

(d) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{C}\equiv \mathrm{C}-{\mathrm{CH}}_2-{\mathrm{CH}}_3$

Show Answer

Answer: (b)

4. In the reaction sequence $[X]$ will be :

Show Answer

Answer: (a)

5. Identify the product in the reaction $\mathrm{PhC}\equiv \mathrm{C}-{\mathrm{CH}}_3\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{\circ },{\mathrm{Hg}}^{23}}{\to }$ ?

(a) ${\mathrm{PhCH}}_2{\mathrm{CH}}_2\mathrm{CHO}$

(b) ${\mathrm{PhCOCH}}_2{\mathrm{CH}}_3$

(c) ${\mathrm{PhCH}}_2{\mathrm{COCH}}_3$

(d) ${\mathrm{PhCOCOCH}}_3$

Show Answer

Answer: (b)

6. When ${\mathrm{C}}_2{\mathrm{H}}_2,{\mathrm{CH}}_4$ and ${\mathrm{C}}_2{\mathrm{H}}_4$ pass through a test tube which has ammoniacal $\mathrm{CuCl}$, find out which gas comes out unaffected from test tube?

(a) ${\mathrm{C}}_2{\mathrm{H}}_2$ and ${\mathrm{CH}}_4$

(b) ${\mathrm{C}}_2{\mathrm{H}}_2$ and ${\mathrm{C}}_2{\mathrm{H}}_4$

(c) ${\mathrm{C}}_2{\mathrm{H}}_4$ and ${\mathrm{CH}}_4$

(d) ${\mathrm{C}}_2{\mathrm{H}}_2$

Show Answer

Answer: (c)

7. Acetylene reacts with acetic acid in the presence of ${\mathrm{Hg}}^{2+}$ to give

(a) ${\mathrm{CH}}_3-\mathrm{CH}{\left({\mathrm{OCOCH}}_3\right)}_2$

(b) $\begin{array}{r}\mathrm{CH}-{\left({\mathrm{OCOCH}}_3\right)}_2\mathrm{CH}-{\left({\mathrm{OCOCH}}_3\right)}_2\end{array}$

(c) ${\mathrm{CH}}_2={\mathrm{CHOCOCH}}_3$

(d) all of these

Show Answer

Answer: (c)

8. In the following reaction

${\mathrm{C}}_2{\mathrm{H}}_2\underset{{\mathrm{HgSO}}_4/{\mathrm{H}}_2{\mathrm{SO}}_4}{\overset{{\mathrm{H}}_2\mathrm{O}}{\to }}[\mathrm{X}]\to {\mathrm{CH}}_3-\mathrm{CHO}$, what is $[\mathrm{X}]$ ?

(a) ${\mathrm{CH}}_3-{\mathrm{CH}}_2\mathrm{OH}$

(b) ${\mathrm{CH}}_3-\mathrm{O}-{\mathrm{CH}}_3$

(c) ${\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{CHO}$

(d) ${\mathrm{H}}_2\mathrm{C}=\mathrm{CHOH}$

Show Answer

Answer: (d)

9. Acetylene reacts with $\mathrm{HBr}$ to produce

(a) 1,2-dibromoethene

(b) 1,1-dibromoethene

(c) $\alpha$-bromoacetone

(d) 1,1-dibromoethane

Show Answer

Answer: (d)

10. 1 - Penten - 4 - yne reacts with 1 mole of bromine to produce

(a) 4, 4, 5, 5 - tetrabromopentene

(b) 1, 2, -dibromo-1, 4-pentadiene

(c) 1, 1, 2, 2, 4, 5-hexabromopentane

(d) 4,5 -dibromo-1-pentyne

Show Answer

Answer: (d)

11. The number of structural and configurational isomers of a bromo compound, ${\mathrm{C}}_5{\mathrm{H}}_9\mathrm{Br}$ formed by the addition of $\mathrm{HBr}$ to 2-pentyne respectively are

(a) 1 and 2

(b) 2 and 4

(c) 4 and 2

(d) 2 and 1

Show Answer

Answer: (b)

12. The decreasing order of acidic character among ethane (I), ethene (II), ethyne (III) and propyne (IV) is

(a) I > II > III $>$ IV

(b) II > III > I > IV

(c) III > IV > II > I

(d) IV > III > II > I

Show Answer

Answer: (c)

Aromatic Hydrocarbons

Also known as arenes.

Are cyclic, highly unsaturated compounds having different properties as compared to alkenes and alkynes.

Arenes containing benzene ring are called benzenoids and the ones not containing a benzene ring as non-benzenoids.

Benzene

Structure:

(Kekule Structure)

${\mathrm{C}}_6{\mathrm{H}}_6$, Cyclic, hexagonal, planar

Highly unsaturated but does not behave like alkenes or alkynes.

Forms only one monosubstituted compoun(d)

Due to resonance all $\mathrm{C}-\mathrm{C}$ bond lengths are equal i.e. 1.39 $\text{AA}$ and are between $\mathrm{C}=\mathrm{C}$ bond ength $(1.34\text{AA})$ and $C-C$ bond length $(1.54\text{AA})$.

All six carbon atoms in bezene are ${\mathrm{sp}}^2$ hybridized, and each $\mathrm{C}$ forms two $\mathrm{C}-\mathrm{C}$ sigma onds with adjacent carbons and one $\mathrm{C}-\mathrm{H}$ sigma bon(d) The remaining unhybridised $\mathrm{P}$ orbital overlaps aterally with adjacent carbons to form a pi clou(d)

Six $\pi$ electrons are delocalized along plane of ring and impart extra stability to the molecule.

Since isolated $\pi$ bonds are no longer available for addition reactions, benzene undergoes substitution rather than addition so that resonance is not diminishe(d)

Aromaticity (Huckel Rule)

According to Huckel, a molecule is aromatic if it is cyclic, planar and has a delocalized $\pi$ electron cloud of $(4n+2)$ e electrons in the ring where $n$ is an integer (i.e. $n=0,1,2\dots \dots$ ) e.g. benzene, naphthalene, anthracene, furan, thiophene, pyrrole, pyridine, cyclopentadienyl anion etc.

Preparation

Commercially obtained from coal tar and petroleum. In laboratory it is prepared as,

1. From ethyne

2. From benzenediazonium chloride

3. From benzenesulphonic acid

4. From sodium benzoate by decarboxylation with soda-lime (Laboratory method)

5. From phenol by reduction with zinc dust

6. From chlorobenzene by reduction with $\mathrm{Ni}-\mathrm{Al}$ alloy / $\mathrm{NaOH}$

7. From hexane

Physical Properties

Physical state

Monocyclic are usually colourless liquids, higher homologues and polynuclear are usually solids.

Solubility

Non polar molecules and thus immiscible with water and miscible in organic solvents like ether, hexane etc.

Melting and boiling points

Though boiling points increase with increasing molecular mass but melting points do not show a regular increase because of their dependence on symmetry of the molecule.

Chemical Reactions

Benzene ring serves as a source of electrons due to the presence of $\pi$ - electron cloud above and below the plane of ring and thus is attacked by electrophiles. Since it resists addition reactions due to resonance stabilization the main reactions of aromatic hydrocarbons are electrophilic substitution reactions. However benzene undergoes some addition and oxidation reactions as well.

Electrophilic substitution reaction

In this reaction hydrogen atom of the ring is replaced by electrophile (generally strong electrophile) which is obtained from the reagent.

Electrophilic substitution reaction generally takes place in the presence of Lewis acid as catalyst. The most common Lewis acids are neutral electrophiles like anhy. ${\mathrm{AlX}}_3$, anhy. ${\mathrm{ZnCl}}_2$, anhy. ${\mathrm{SnCl}}_4$, anhy. ${\mathrm{SbCl}}_5,{\mathrm{SbF}}_5$, and ${\mathrm{BF}}_3$.

Mechanism of Electrophilic substitution reactions

Takes place in three steps,

i) Generation of electrophile: takes place differently for different electrophiles ${\mathrm{E}}^{+}$as exemplified in the text below at appropriate place

ii) Formotion of arenium (carbocation) ion : Electrophile attacks the benzene ring to give a $\sigma$ complex (arenium ion) which is resonance stabilized and has one ${\mathrm{sp}}^3$ hybridized carbon.

Arenium ion is not aromatic because it contains one ${\mathrm{sp}}^3$ hybridized carbon which is not involved in delocalisation.

iii) Loss of proton $\left({\mathrm{H}}^{+}\right)$

${\mathrm{sp}}^3$ hybridized carbon looses ${\mathrm{H}}^{+}$to restore aromatic character in the substitution product.

Various electrophilic substitution reactions of benzene are

a) Nitration

Generation of electrophile, nitronium ion $\left({\mathrm{NO}}_2^{+}\right)$. In this case, nitration of benzene requires sulphuric acid as a catalyst. Sulphuric acid protonates nitric acid followed by loss of water from the protonated nitric acid to form a nitronium ion, the electrophile required for nitration.

b) Sulphonation

Generation of electrophile:

$2{\mathrm{H}}_2{\mathrm{SO}}_4\rightleftharpoons {\mathrm{SO}}_3+{\mathrm{HSO}}_4^{-}+{\mathrm{H}}_3{\mathrm{O}}^{+}$

Sulphur trioxide acts as an electrophile and hence attacks the benzene ring to form a carbocation which is stabilized by resonance.

c) Halogenation

Generation of electrophile:

In the first step of the reaction, bromine donates a lone pair of electrons to the Lewis acid to form a complex which on dissociation gives $\mathrm{Br}$ and ${\mathrm{Fe}}^{\mathrm{Br}}{}_4$.

The catalysts ${\mathrm{FeBr}}_3$ or ${\mathrm{AlCl}}_3$ act as a halogen carrier only and are regenerated in the last step.

Fluorination is not carried out directly because the reaction between benzene and fluorine is very vigorous.

Preparation of iodobenzene requires the presence of an oxidizing agent because the side product $(\mathrm{HI})$ produced is a strong reducing agent.

d) Friedel crafts alkylation

Generation of electrophile

Limitations of Friedel-Crafts Reaction

Reactivity of aromatic compounds : Aromatic compounds whose reactivity is comparable with or greater than that of benzene give this reaction. Compounds having strong deactivating groups do not give this reaction. Compounds having $-{\mathrm{NH}}_2,-\mathrm{NHR},{\mathrm{NR}}_2$ and $\mathrm{OH}$ groups also do not give this reaction.

The carbocation formed from alkyl halide, can rearrange to a more stable carbocation, and the major product is obtained from the most stable carbocation.

$$\begin{array}{rl}& {\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{CH}}_2-\mathrm{Cl}⟶{\mathrm{CH}}_3-{\mathrm{CH}}_2-\stackrel{\oplus }{\mathrm{C}}{\mathrm{H}}_2\stackrel{\text{ hydride shift }}{\to }{\mathrm{CH}}_3-\stackrel{\oplus }{\mathrm{C}}\mathrm{H}-{\mathrm{CH}}_3\\ & 1^{\circ }\text{ carbocation }{2}^{\circ }\text{ carbocation }\end{array}$$

e) Friedel Crafts acylation

Generation of an electrophile

The acid chloride or anhydride reacts with anhydrous aluminium chloride to form acylium ion. $(\mathrm{R}-{\mathrm{C}}^{+}=0)$ which acts as an electrophile.

Similarly anhydride also reacts with ${\mathrm{AlCl}}_3$ to give acylium ion.

Acid anhydride

Polyacylation in Friedel-Crafts reaction is not possible because acyl group is a deactivating group and only monosubstituted product is forme(d)

Addition Reactions

1. Addition of dihydrogen

2. Addition of halogens

Product obtained is 1, 2, 3, 4, 5, 6-hexachlorocyclohexane or BHC or gammaxene which is used as an insecticide.

Oxidation Reactions

1.

2.

${\mathrm{KMnO}}_4$ and ${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$ have no action on benzene but the alkyl group attached to benzene gets oxidized to $-\mathrm{COOH}$ group irrespective of its size.

3. Combustion

${\mathrm{C}}_6{\mathrm{H}}_6+\frac{15}{2}{\mathrm{O}}_2⟶6{\mathrm{CO}}_2+3{\mathrm{H}}_2\mathrm{O}$

Directive Influence of Functional Groups in Mono-Substituted Benzene

When a monosubstituted benzene derivative is subjected to electrophilic substitution, ortho, meta or para disubstituted derivatives are formed which are formed in unequal ratios.

This ability of a group already present in the benzene ring to direct the incoming group to a particular position is called the directive influence of groups.

1. $\mathbf{0},\mathbf{p}$-directing groups. The substituents or groups which direct the incoming group to ortho and para positions are called ortho, para-directing groups. For example $-{\mathrm{CH}}_3,-{\mathrm{CH}}_2{\mathrm{CH}}_3,-{\mathrm{C}}_6{\mathrm{H}}_5,-\mathrm{Cl}$, $-\mathrm{Br},-\mathrm{I},-\mathrm{OH},-{\mathrm{OCH}}_3,-{\mathrm{OCOCH}}_3,-{\mathrm{NH}}_2$.

The electrophile will attack the benzene ring at a position where the electron density is high. Since the electron density is high at $0-$ and $p$-position than at m-positions in phenol, so the electrophile will attack preferentially at 0 -and $p$-positions. Thus, $0\mathrm{H}$ group is $0,\mathrm{p}$-directing.

Similarly the directive influence of $-{\mathrm{CH}}_3$ group is explained by hyperconjugation.

Effect of $\mathbf{0},\mathrm{p}$-directing substituents on reactivity

Since $0-,\mathrm{p}$-directing groups increase the electron density in the benzene ring therefore, the ring gets activated and the further electrophilic substitution in the ring becomes easier.

The electron donating ability of some substituents follows the order:

$-0\ddot{0}>{\mathrm{NH}}_2>{\mathrm{NR}}_2>-\mathrm{O}\mathrm{H}>-0\ddot{0}{\mathrm{CH}}_3,-{\mathrm{NHCOCH}}_3>-{\mathrm{CH}}_3>-\mathrm{X}(-\mathrm{F}>-\mathrm{Cl}>-\mathrm{Br}>-\mathrm{I})$

Higher the electron donating ability of a substituent, faster is the reaction.

2. m-directing groups

The substituents or groups which direct the incoming group to the meta position are called meta-directing groups. For example.

${\left({\mathrm{CH}}_3\right)}_3{\mathrm{N}}^{+},-{\mathrm{NO}}_2,-\mathrm{CN},-{\mathrm{CF}}_3,-\mathrm{CHO},-\mathrm{COR},-\mathrm{COOH},-\mathrm{COOR},-{\mathrm{SO}}_3\mathrm{H}$

All electron withdrawing groups are m-directing

These meta directing groups withdraw electrons from the ring and this withdrawl is maximum from ortho and para positions as seen from the resonance structure. The +ve charge is at ortho and para positions.

Since electron density is comparatively more at $m$-position than at 0 - and $p$-position so further substitution takes place at m-position.

Directive influence of halogens

Halogens are deactivating due to -1 effect. They are $0,p$ directing due to $+R$ effect

As a result, halogens are $0,p-directing$. Due to combined result of $+R$ effect and $-I$ effect of halogens they are deactivating but $0,p-$ directing.

Solved examples

1. The value of $n$ for following compound as per Huckel’s rule is

(a) 3

(b) 5

(c) 7

(d) 14

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Answer . a

No. of electrons is 14 . So $\mathrm{n}$ is $3(4\times 3+2=14)$

2. Given

The major product formed in the above reaction is,

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Answer . b

${\mathrm{CF}}_3$ is a -l effect group and hence is meta directing

Practice Questions

1. In which of the following polysubstitution takes place?

(a)

(b)

(c)

(d)

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Answer: (b)

2. Which one of the following is most reactive towards electrophilic attack?

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Answer: (a)

3. The compound $X$ in the reaction

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Answer: (b)

4. Which one of the following is an intermediate in the reaction of benzene with ${\mathrm{CH}}_3\mathrm{Cl}$ in the presence of anhydrous ${\mathrm{AlCl}}_3$ ?

(a) ${\mathrm{Cl}}^{+}$

(b) ${\mathrm{CH}}_3^{-}$

(c) ${\mathrm{CH}}_3^{+}$

(d)

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Answer: (c)

5. The major product obtained on monobromination $\left({\mathrm{Br}}_2/{\mathrm{FeBr}}_3\right)$ of $(\mathrm{A})$ is

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Answer: (b)

6.

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Answer: (a)

7. In the reaction, ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_3\stackrel{\text{ Oxidation }}{\to }A\stackrel{\mathrm{NaOH}}{\to }B\stackrel{\text{ Sodalime }}{\to }\mathrm{C}$, the product $\mathrm{C}$ is

(a) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{OH}$

(b) ${\mathrm{C}}_6{\mathrm{H}}_6$

(c) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{COONa}$

(d) ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{ONa}$

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Answer: (b)

8. The treatment of benzene with isobutene in the presence of sulphuric acid gives

(a) Isobutylbenzene

(b) tert-Butylbenzene

(c) n-Butylbenzene

(d) No reaction

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Answer: (b)

9. Toluene when refluxed with bromine in the presence of light gives mainly :

(a) 0 -bromotoluene

(b) m-bromotoluene

(c) p-bromotoluene

(d) benzyl bromide

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Answer: (d)

10. The product $[\mathrm{S}]$ of the given reaction is/are,

$\underset{{\text{ FeBr }}_3}{\overset{{\mathrm{Br}}_2}{\to }}$ Product [S]

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Answer: (c)

11. Among the following compounds, the decreasing order of reactivity towards electrophilic substitution is

(a) III > I > II > IV

(b) IV > I > II > III

(c) I > II > III > IV

(d) II > I > III > IV

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Answer: (a)

12. Among the following, the compound that can be most readily sulphonated is

(a) benzene

(b) nitrobenzene

(c) toluene

(d) chlorobenzene

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Answer: (c)

13. The compound that is most reactive towards electrophilic substitution is

(a) toluene

(b) benzene

(c) benzoic acid

(d) nitrobenzene

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Answer: (a)

14. The reaction of toluene with chlorine in presence of ferric chloride $\left({\mathrm{FeCl}}_3\right)$ gives predominantly

(a) benzoyl chloride

(b) m-chlorotoluene

(c) benzyl chloride

(d) 0 -and $p-$ chlorotoluene

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Answer: (d)

15. Chlorination of toluene in the presence of light and heat followed by treatment with aqueous $\mathrm{NaOH}$ gives

(a) $0-$ cresol

(b) p-cresol

(c) 2,4-dihydroxy toluene

(d) benzoic acid

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Answer: (d)

16. Identify the correct order of reactivity in electrophilic substitution reactions of the following compounds, Benzene (1), Toluene (2), Chlorobenzene (3) and Nitrobenzene (4)

(a) $1>2>3>4$

(b) $4>3>2>1$

(c) $2>1>3>4$

(d) $2>3>1>4$

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Answer: (c)

17. Benzyl Chloride $\left({\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{CH}}_2\mathrm{Cl}\right)$ can be prepared from toluene by chlorination with

(a) ${\mathrm{SO}}_2{\mathrm{Cl}}_2$

(b) ${\mathrm{SOCl}}_2$

(c) ${\mathrm{Cl}}_2$ in presence of uv light

(d) $\mathrm{NaOCl}$

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Answer: (c)

18. In the following reaction

The structure of major product $\mathrm{X}$ is

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Answer: (b)

19.

Product on monobromination of this compound is

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Answer: (b)