POC-1ST

UNIT-2

1. ALKANES

• Alkanes are those organic compounds which are composed of hydrogen and carbon only with single bond.
• They are the simplest and most saturated type of hydrocarbons, as they contain $[C-C]$ and $[C-H]$ single covalent bond.

General formula $\rightarrow$

• The general formula for alkanes is $C_nH_{2n+2}$ where $n$ is the number of carbon atom.

Structure $\rightarrow$

• Alkanes have a linear or branched chain structure, with each carbon atom bonded to four other atoms (hydrogen/carbon).

HALOGENATION OF ALKANES

• It involves the substitution reactions, in which Halogen atom $[Cl, I, Br, F]$ replaces H-atom from the alkane.

• Chlorination, Bromination etc..

Chlorination

• Alkanes react with chlorine in the presence of ultraviolet light to produce alkyl chloride.

• The reaction does not stop at this stage, the remaining three hydrogen atoms of methyl chloride can be successively replaced by chlorine atom.

Mechanism

• The mechanism of chlorination/halogenation take place through the formation of free radical.
• It involves three steps:
  1. Chain initiation
  2. Chain propagation
  3. Chain termination

1. Chain initiation

• In this, chlorine molecules undergoes homolytic fission to give chlorine free radicals.

2. Chain propogation

• It further involves two steps:

  • a) in this, chlorine free radical attacks methane to produce methyl free radicals and $HCl$.

  

  • b) Now, in this, methyl free radical attacks chlorine molecules to give methyl chloride and chlorine free radicals.

$$CH_3^\bullet + Cl_2 \rightarrow CH_3-Cl + Cl^\bullet$$
(Methyl chloride)

3. Chain termination

• Now, in this step two free radicals combine to form a stable molecules, ending the chain reactions...

$$Cl^\bullet + Cl^\bullet \rightarrow Cl_2$$ $$CH_3^\bullet + CH_3^\bullet \rightarrow CH_3-CH_3$$ $$CH_3^\bullet + Cl^\bullet \rightarrow CH_3-Cl$$ (methyl chloride)

SP³ HYBRIDIZATION IN ALKANES

Hybridization $\rightarrow$ It is the concept of mixing of atomic orbitals to form a new hybrid orbitals.

• It is the process of combining two or more atomic orbitals of an atom to form equivalent hybrid orbitals.

sp³ hybridization $\rightarrow$

• It is a type of orbital hybridization in which one $s$ orbital and three $p$ orbitals of an atom to mix to form four equivalent hybrid orbitals.

• Each $sp^3$ hybrid orbital show 25% s-orbital characteristics and 75% p-orbital characteristics.

For Methane ($CH_4$)

• Hybridisation occurs only for central atom.
• Here, Carbon is our central atom.
• It involves several steps:

Steps $\rightarrow$

1. Electronic configuration of central atom.
   • Carbon $\rightarrow$ Atomic No 6 $\rightarrow$ $1s^2 2s^2 2p^2$ (Ground state)
   • $1s$ contains $\uparrow\downarrow$, $2s$ contains $\uparrow\downarrow$, $2p$ contains $\uparrow \uparrow$
2. At an excited state, always in outermost (orbitals)
   • $1s^2 2s^1 2p^3$
   • $2s$ contains $\uparrow$, $2p$ contains $\uparrow \uparrow \uparrow$
3. $sp^3$ hybridisation
   • overlapping of orbitals $s + p \rightarrow sp^3$

Geometry $\rightarrow$

• Alkanes show tetrahedral arrangement.
• Hydrogens have oval shaped orbitals.
• The angle b/w bond is 109.5°.

For Ethane

• $CH_3-CH_3$
• Both Carbon have $sp^3$ hybridisation, so, ethane $\rightarrow$ $sp^3-sp^3$ hybridisation.

PROPERTIES OF ALKANES

• Alkanes are insoluble in water and soluble in organic solvents.
• Alkanes from $C_1-C_4$ are colorless gases, $C_5$ to $C_{17}$ are colorless liquids and higher molecules are solid..
• They are non-polar compounds.
• Boiling point and melting point are increases with increases in molecular weights.

PARAFFINS

• Paraffins are just another name of alkanes.
• They are types of saturated hydrocarbon with a straight or a branched chain structure.

Uses of Paraffins

• They are used as fuels, such as methane (natural gas), propane and butane etc.
• They are used as lubricants in various industrial applications.
• Also used as solvents in the production of coating, paints & adhesives.
• Liquid paraffin, highly refined liquid mineral oil, so it is used as excipients in pharmaceutical products and used in cosmetics and personal care products, such as lotions, creams & Ointments.
• used to clean dry skin, constipation and eczema (skin disease).

2. ALKENES

• These are those unsaturated hydrocarbons, which contain one or more Carbon-Carbon double bond $[C=C]$.
• Also known as Olefins.

General formula $\rightarrow$ $C_nH_{2n}$

• where n is the no. of carbon atoms. (min. 2 carbon required).

Examples:

• $C_2H_4$ (Ethene): $CH_2=CH_2$
• $C_3H_6$ (Propene): $CH_2=CH-CH_3$
• Butene etc..

STABILITIES OF ALKENES

• Alkenes are unsaturated hydrocarbons with a Carbon-Carbon double bond $[C=C]$.
• They contains double bond, which make alkenes more reactive than alkanes but less stable.
• The stability of alkene depends on:-

1. Substituents $\rightarrow$ Alkyl group (e.g. methyl, ethyl) stablize the double bond by donating electrons. work as $e^-$ donating R group.

   

   $$CH_2=CH_2 < CH_3-CH=CH_2 < CH_3-CH_2-CH=CH_2$$ (Increasing Stability)

2. Trans form are more stable than Cis-form due to less steric hindrance.

• Trans-form: Same groups are at opposite side.
• Cis-form: Same groups are at same side.

HYBRIDIZATION IN ALKENES

• $sp^2$ hybridization is a type of orbital hybridization, where one ($s$) orbital and two ($p$) orbitals mix to form three equivalent hybrid orbitals.
• In alkenes, the carbon atom involved in the double bond undergoes $sp^2$ hybridization.

e.g. Ethene ($CH_2=CH_2$)

• for central atom:
  • Ground State: $1s^2 2s^2 2p^2$
  • Excited state: $1s^2 2s^1 2p^3$
  • Now, in hybridised state: $sp^2$ hybridization ($s$ and two $p$ orbitals mix, leaving one unhybridized $p$ orbital).
  • Bond angle is 120°.

E1 AND E2 REACTIONS

• $E_1$ and $E_2$ are part of Elimination reactions.
• These are those reactions which involve the removal of a leaving group from a substrate, resulting in the formation of a double bond (ethene / alkenes).
• These are those reactions, which are used in the preparation of alkenes by using dehydrohalogenation of alkyl halides. (de hydro + halogenation).

• E.g. $CH_3-CH_2Br \xrightarrow{Alc. KOH} CH_2=CH_2 + HBr$ (ethene)
• These are of two types:
  1. $E_1$ reactions
  2. $E_2$ reactions

E1 REACTIONS

• It stands for Unimolecular Elimination Reactions.
• It is Unimolecular, so it follow first-order kinetics i.e. it is depends on the concentration of substrate (reactant only).
• It is a type of reaction, which include two-step mechanism.
• Tertiary alkyl halides are more reactive towards this reaction. Order of reactivity $\rightarrow$ $3^\circ > 2^\circ > 1^\circ$

Mechanism $\rightarrow$

• It is two step mechanism, in which first step is rate-determining step:-

$$CH_3-C(CH_3)(Br)-CH_3 \xrightarrow{H_2O, \Delta} CH_2=C(CH_3)-CH_3 + HBr$$ (tertiary butyl bromide $\rightarrow$ 2-methyl propene + Hydrogen bromide)

• Step-1 It involves the formation of carbocation (intermediate). It is rate-determining step. (slow step)

$$CH_3-C(CH_3)(Br)-CH_3 \rightarrow CH_3-C^+(CH_3)-CH_3 + Br^-$$ (Alkyl halides $\rightarrow$ Carbocation)

• Step-2 It involves the deprotonation. $[-H^+]$ (fast step)

$$CH_3-C^+(CH_3)-CH_3 + H_2O \rightarrow CH_2=C(CH_3)-CH_3 + H_3O^+$$ (Note: Since, the first step is slow, and rate-determing, the overall rate of reaction depends only on the reactant/substrate.)

Kinetics $\rightarrow$

• It follow 1st order kinetics, because in $E_1$ reactions the rate of reactions depends on the concentration of the substrate/reactant. $$R = K [alkyl halides]$$ where $R =$ Rate of reaction, $K =$ Rate constant, $[alkyl halides] =$ concentration of alkyl halide

Order of Reactivity $\rightarrow$

• It follow $3^\circ > 2^\circ > 1^\circ$. In $E_1$ reactions, order of reactivity depends on the stability of Carbocation. (3° have more substituent more stable).

E2 REACTIONS

• $E_2$ represents (Bimolecular) Elimination Reactions.
• It is bimolecular, so it follow second order kinetics i.e. it depends on the concentration of both (reactant & reagent).
• It involves one-step mechanism.
• $1^\circ$ alkyl halides are more reactive toward this reactions. Order of Reactivity $\rightarrow$ $1^\circ > 2^\circ > 3^\circ$

Mechanism $\rightarrow$

• Reaction $\rightarrow$ $$CH_3-CH_2-Br + KOH (Alc.) \rightarrow CH_2=CH_2 + H_2O + KBr$$ (ethyl bromide + potassium bromide $\rightarrow$ ethene)

• It completed in one step, which include transition state.

$$HO^- + CH_3-CH_2-Br \rightarrow [HO \cdots H \cdots CH_2 \cdots CH_2 \cdots Br]^\ddagger \rightarrow H_2C=CH_2 + H_2O + Br^-$$

• In transition state (the rate-determining step), both ethyl bromide and hydroxide ion involved.

Kinetics $\rightarrow$

• It follow second order kinetics, rate of reaction depends on both reactant and reagent. $$R = K [alkyl halides][base]$$

Order of Reactivity $\rightarrow$

• 1° alkyl halides are more reactive toward $E_2$ reactions, due to less stearic hindrance. $OH^-$ can easily attack. $1^\circ > 2^\circ > 3^\circ$.

E1 VERSUS E2

FACTOR AFFECTING E1 & E2 REACTIONS

• Substrate $\rightarrow$ For $E_1$: $3^\circ > 2^\circ > 1^\circ$ (due to stability of carbocations). For $E_2$: $1^\circ > 2^\circ > 3^\circ$ (due to low stearic hindress).
• Leaving group $\rightarrow$ Good leaving group (e.g. Cl, Br, I) facilitate $E_1/E_2$ reacting.
• Solvent $\rightarrow$ For $E_1$: polar protic solvents (e.g. water). For $E_2$: polar aprotic solvents.
• Nature of Base $\rightarrow$ For $E_1$: weak base. For $E_2$: strong base.
• Temperature $\rightarrow$ high temp facilitate both $E_1$ & $E_2$ reactions.

RE-ARRANGEMENT OF CARBOCATIONS

• It is the shifting/migration of Hydrogen/methyl group from one Carbon to another to form a more stable carbocations.
• It is of two types:-

1. Hydride shift $\rightarrow$ A hydrogen atom migrates from one C to another.

$CH_3-CH^+-CH_2-CH_3 \rightarrow$ more stable. 2. Methyl shift $\rightarrow$ A alkyl group ($-CH_3$) migrates from one C to another.

 $CH_3-C(CH_3)(CH_3)-CH^+-CH_3 \rightarrow CH_3-C^+(CH_3)-CH(CH_3)-CH_3$ (more stable tertiary).

SAYTZEFF'S RULE (Zaitsev's Rule)

• It occurs in elimination reactions.
• When alkyl halides have two or more $\beta$-carbon present, it used in this case to identify the major and minor product.

• $CH_3-CH_2-CH(Br)-CH_3 \xrightarrow{dehydrohalogenation}$
  • $CH_3-CH=CH-CH_3$ (But-2-ene/2-Butene) $\rightarrow$ 80%
  • $CH_3-CH_2-CH=CH_2$ (But-1-ene/1-Butene) $\rightarrow$ 20%
• Acc to this rule, that product will be the major product, which contain more no. of substituents at double bond carbon.
• So, in above example, But-2-ene will be the major product.
• Evidences $\rightarrow$ It helps us understand and predict the outcomes of elimination reactions. predict the major product formed.

ELECTROPHILIC ADDITION REACTIONS OF ALKENES

• These are those reaction, in which electrophile ($E^+$) added to the double carbon by breaking down double bond.

$$C=C + E^+-Nu^- \rightarrow E-C-C-Nu$$

• It involves the formation of alkanes from alkenes.

$$CH_2=CH_2 + HBr \rightarrow CH_3-CH_2-Br$$ (ethene + Hydrogen bromide $\rightarrow$ Bromo ethane)

Mechanism $\rightarrow$

• It involves three steps:

1. Formation of electrophile ($E^+$) $$HBr \rightarrow H^+ + Br^-$$
2. Attack of electrophile on reactant (ethene) $$CH_2=CH_2 + H^+ \rightarrow CH_3-CH_2^+$$ (formation of carbocation)
3. Now, addition of Nucleophile ($Nu^-$) on carbocation $$CH_3-CH_2^+ + Br^- \rightarrow CH_3-CH_2-Br$$ (Bromoethane)

Now, Alkenes are of two types:-

• Symmetrical alkenes: when alkyl/hydrogen are same on both sides of double bond.
• Unsymmetrical alkenes: when alkyl/hydrogen are unequal on both sides of double bond.

Now, Markovnikoff and Anti-markovnikoff rules are applied on unsymmetrical alkenes.

MARKOVNIKOFF'S RULE * This rule is given by russian chemist markovnikov.

• Acc. to this rule: In an electrophilic addition reactions, Electrophile ($E^+$) goes to that double bond carbon, where more number of hydrogen are present.
• OR Nucleophile ($Nu^-$) goes to that double bond carbon, where less number of hydrogen are present.

$$CH_3-CH=CH_2 + H-Br \rightarrow CH_3-CH(Br)-CH_3$$ (2-bromopropane)

ANTI-MARKOVNIKOFF'S RULE

• Also known as Kharasch effect or Peroxide effects ($H_2O_2$).
• It occurs in assymetrical alkene.
• According to this rule: Hydrogen atom added to those double bond carbon, which contain less number of hydrogen/non-polar nature.

$$CH_3-CH=CH_2 + H-Br \xrightarrow{H_2O_2, \Delta} CH_3-CH_2-CH_2-Br$$ (1-bromopropane)

• It involves the free radical addition mechanism.

Mechanism $\rightarrow$

• It involve three steps:-

1. Chain initiation $\rightarrow$ It occurs in reagent and catalyst. $$H_2O_2 \xrightarrow{\Delta} 2 \cdot OH$$ $$H-Br + \cdot OH \rightarrow H_2O + Br^\bullet$$ (Bromine free radical)
2. Chain propogation $\rightarrow$ Now, Bromine free radical attack reactant. $$Br^\bullet + CH_3-CH=CH_2 \rightarrow CH_3-C^\bullet H-CH_2-Br$$
3. Chain termination $\rightarrow$ free radicals reacts with each other. $$CH_3-C^\bullet H-CH_2-Br + H-Br \rightarrow CH_3-CH_2-CH_2-Br + Br^\bullet$$ (1-bromopropane) $$Br^\bullet + Br^\bullet \rightarrow Br_2$$

OZONOLYSIS

• It is an organic reaction, where an alkene is cleaved by ozone ($O_3$) to form two carbonyl compounds (Aldehydes or ketones).

$$CH_2=CH_2 \xrightarrow{O_3} \text{Ozonide} \xrightarrow{Zn/H_2O} 2 CH_2=O$$ (ethene $\rightarrow$ 2 formaldehyde)

Mechanism $\rightarrow$

• It involves two steps:-
  1. Formation of ozonide $\rightarrow$ Ethene reacts with ozone to form an ozonide intermediate.

 $$CH_2=CH_2 + O_3 \rightarrow \text{Ozonide (cyclic structure with 3 oxygen atoms)}$$

2. Reduction $\rightarrow$ Now ozonide reduced to form two carbonyl compounds.

 $$\text{Ozonide} \xrightarrow{Zn/H_2O} 2(CH_2=O) + ZnO$$
 (formaldehyde)

CONJUGATED DIENES

• These are those type of organic compounds that contains two or more Carbon-carbon double bonds ($C=C$) separated by a single bond in alkenes.

e.g. $CH_3-CH=CH-CH=CH_2$

• General structure $\rightarrow$ $R_2C=CR-CR=CR_2$ where $R = \text{alkyl/hydrogen}$.
• they contains, alternate double bonds. e.g. $H_2C=CH-CH=CH_2$ (1,3-Butadiene)

CHEMICAL REACTIONS

It undergoes following chemical reactions:-

1. Electrophilic addition reactions
2. Free radical reactions
3. Diels-alder reactions

1. ELECTROPHILIC ADDITION REACTIONS

• These are those reaction in which electrophile ($E^+$) added to the carbon atom and breakage of one double bond ($\pi$-bond) take places.
• In this reaction, 1,3-Butadienes are react with halogens ($Br_2$) in the presence of an inert solvent ($CCl_4$) to give a mixture of two di-bromo compounds.

$$CH_2=CH-CH=CH_2 + Br_2 \rightarrow$$

• $CH_2(Br)-CH(Br)-CH=CH_2$ (1,2-addition, 3,4-dibromo-1-butene)
• $CH_2(Br)-CH=CH-CH_2(Br)$ (1,4-addition, 1,4-dibromo-2-butene)

Mechanism

1. Formation of carbocation $$CH_2=CH-CH=CH_2 + Br-Br \rightarrow CH_2(Br)-C^+H-CH=CH_2 \leftrightarrow CH_2(Br)-CH=CH-C^+H_2 + Br^-$$
2. Combination of bromide with carbocation
   • $\rightarrow$ 1,2-addition product
   • $\rightarrow$ 1,4-addition product

2. FREE RADICAL ADDITION REACTIONS

• Conjugated dienes can undergoes free radical reactions, which involves the addition of free-radical to the double bond carbon.
• It involves the addition of 1,3-butadiene to bromotrichloromethane ($BrCCl_3$) in the presence of an peroxide to yield 1,2- and 1,4- addition products.

$$CH_2=CH-CH=CH_2 + Br-CCl_3 \rightarrow$$

• $CH_2(CCl_3)-CH(Br)-CH=CH_2$ (1,2-addition)
• $CH_2(CCl_3)-CH=CH-CH_2(Br)$ (1,4-addition)

Mechanism $\rightarrow$

• It involves three steps:

1. Chain initiation $\rightarrow$ formation of free radicals. $$H_2O_2 \rightarrow 2 \cdot OH$$ $$Br-CCl_3 + \cdot OH \rightarrow H_2O + Br^\bullet + \cdot CCl_3$$
2. Chain propogation $\rightarrow$ addition of free radicals to diene. $$CH_2=CH-CH=CH_2 + \cdot CCl_3 \rightarrow CH_2(CCl_3)-C^\bullet H-CH=CH_2 \leftrightarrow CH_2(CCl_3)-CH=CH-C^\bullet H_2$$
3. formation of 1,2- and 1,4- addition products by reaction with $BrCCl_3$.

3. DIELS - ALDER REACTIONS * It involves the treatment of 1,3-Butadiene (conjugated dienes) with an alkene/alkyne to form cyclohexane.

• It does not required any catalyst.
• Diene + Dienophile (alkene/alkyne) $\rightarrow$ Adduct (formed product).
• Given by Paul Hermann Diels and Kurt Alder and get nobel prize.

$$1,3-\text{Butadiene (Diene)} + \text{Ethylene (Dienophile)} \xrightarrow{200^\circ C} \text{Cyclohexene (Adduct)}$$

Mechanism

• It involves single step process, in which dienophile interact with diene and formed adduct.

• In this, six-membered ring compounds are formed in large amount.

ALLYLIC REARRANGEMENT

• It is a type of organic reaction where an allylic group (a functional group attached to carbon atom adjacent to a double bond) shift to a nearby position, results in more stable compounds.

e.g. $C=C-C(OH) \rightarrow C(OH)-C=C$