# 16.6 Chemical Reactions of Alkenes Alkenes are characterized by their reactivity, which is centered around the carbon-carbon double bond. They primarily undergo **addition reactions**, where a molecule adds across the double bond. This process converts the pi ($\pi$) bond of the double bond into two new sigma ($\sigma$) bonds, resulting in a saturated molecule. ## Hydrogenation *Hydrogenation* is the addition of hydrogen gas ($H_2$) to an alkene in the presence of a catalyst to form a corresponding alkane. This reaction is also known as catalytic reduction. - **Conditions:** High temperature and a finely divided metal catalyst like Nickel (Ni), Platinum (Pt), or Palladium (Pd). - **Reaction:** $$ \underset{\text{Ethene}}{CH_2=CH_2} + H_2 \xrightarrow[\text{heat}]{Ni/Pt} \underset{\text{Ethane}}{CH_3-CH_3} $$ - **Application:** This process is used industrially to convert vegetable oils (which contain double bonds) into margarine (which is more saturated). ## Hydrohalogenation *Hydrohalogenation* is the addition of a hydrogen halide (like $HBr$ or $HCl$) to an alkene, forming a halogenoalkane (alkyl halide). - **Reaction with an unsymmetrical alkene (Propene):** When propene reacts with hydrogen bromide ($HBr$), a mixture of two products is formed. $$ \underset{\text{Propene}}{CH_3-CH=CH_2} + HBr \rightarrow \underset{\text{2-Bromopropane (Major Product)}}{CH_3-CH(Br)-CH_3} + \underset{\text{1-Bromopropane (Minor Product)}}{CH_3-CH_2-CH_2Br} $$ ### Mechanism and Markovnikov's Rule The outcome of this reaction is governed by **Markovnikov's Rule**. **Markovnikov's Rule:** *When an unsymmetrical reagent (like $H-X$) adds to an unsymmetrical alkene, the hydrogen atom adds to the carbon of the double bond that already has more hydrogen atoms.* 1. **Electrophilic Attack:** The hydrogen ion ($H^+$) from $HBr$ acts as an electrophile and attacks the electron-rich $\pi$-bond of propene. This can form two possible intermediate carbocations. - **Path A:** Forms a more stable **secondary (2°) carbocation**. - **Path B:** Forms a less stable **primary (1°) carbocation**. 2. **Carbocation Stability:** The stability of carbocations follows the order: $$ \text{Tertiary (3°)} > \text{Secondary (2°)} > \text{Primary (1°)} $$ This stability is due to the **positive inductive effect** of alkyl groups, which are electron-donating. More alkyl groups attached to the positively charged carbon help stabilize the charge. 3. **Nucleophilic Attack:** The bromide ion ($Br^-$) attacks the more stable secondary carbocation, leading to the formation of 2-bromopropane as the major product. ## Hydration of Alkenes *Hydration* is the addition of a water molecule ($H_2O$) across the double bond of an alkene to produce an alcohol. This reaction typically requires a strong acid catalyst (e.g., $H_2SO_4$). - **Application:** It is the primary method for the industrial production of ethanol. - **Mechanism (Two Steps):** 1. **Formation of Ethyl Hydrogen Sulphate:** Ethene reacts with cold, concentrated sulfuric acid. $$ \underset{\text{Ethene}}{CH_2=CH_2} + H_2SO_4 \rightarrow \underset{\text{Ethyl hydrogen sulphate}}{CH_3-CH_2-OSO_3H} $$ 2. **Hydrolysis:** The mixture is heated with water, and the ethyl hydrogen sulphate decomposes to form ethanol and regenerate the sulfuric acid catalyst. $$ \underset{\text{Ethyl hydrogen sulphate}}{CH_3-CH_2-OSO_3H} + H_2O \xrightarrow{\text{heat}} \underset{\text{Ethanol}}{CH_3-CH_2-OH} + H_2SO_4 $$ ## Halogenation *Halogenation* is the addition of a halogen molecule (e.g., $Cl_2$, $Br_2$) across the double bond to form a dihalogenoalkane. - **Reaction:** Ethene reacts with bromine ($Br_2$) in an inert solvent like tetrachloromethane ($CCl_4$) to produce 1,2-dibromoethane. $$ \underset{\text{Ethene}}{CH_2=CH_2} + Br_2 \xrightarrow{CCl_4} \underset{\text{1,2-Dibromoethane}}{Br-CH_2-CH_2-Br} $$ - **Test for Unsaturation:** This reaction is used as a chemical test for the presence of double or triple bonds. The reddish-brown color of bromine water disappears rapidly when added to an unsaturated compound. Another common test is the **Baeyer's Test** using cold alkaline $KMnO_4$. ## Halohydration *Halohydration* is the addition of a halogen and a hydroxyl group across a double bond. This is achieved by reacting an alkene with a hypohalous acid (HOX), which is formed by reacting a halogen with water. - **Reaction:** Ethene reacts with hypobromous acid ($HOBr$) to produce 2-bromoethanol (a halohydrin). $$ \underset{\text{Ethene}}{CH_2=CH_2} + \underset{\text{Hypobromous acid}}{HOBr} \rightarrow \underset{\text{2-Bromoethanol (Bromohydrin)}}{HO-CH_2-CH_2-Br} $$ ## Epoxidation *Epoxidation* is the reaction of an alkene with a peracid (like peroxyacetic acid, $CH_3COOOH$) to form an epoxide (a three-membered ring containing an oxygen atom). - **Reaction:** Ethene reacts with peroxyacetic acid to form epoxyethane (ethylene oxide). $$ \underset{\text{Ethene}}{CH_2=CH_2} + \underset{\text{Peroxyacetic acid}}{CH_3COOOH} \rightarrow \underbrace{CH_2-CH_2}_{\text{O}} + \underset{\text{Acetic acid}}{CH_3COOH} $$ ## Ozonolysis *Ozonolysis* is a reaction where an alkene's double bond is cleaved by ozone ($O_3$), forming aldehydes and/or ketones. - **Mechanism:** 1. Ozone adds to the alkene at low temperature ($-78^\circ C$) to form an unstable intermediate called a **molozonide**, which rearranges to a more stable **ozonide**. 2. The ozonide is treated with a mild reducing agent (like zinc dust and water) to yield carbonyl compounds (aldehydes or ketones). $$ H_2O_2 + Zn \rightarrow ZnO + H_2O $$ - **Application:** Ozonolysis is a powerful tool used to determine the exact position of a double bond within an alkene by analyzing the structures of the resulting products. ## Polymerization *Polymerization* is a process where small repeating units called **monomers** join together to form a very large molecule called a **polymer**. - **Reaction:** Ethene monomers polymerize under high temperature and pressure to form poly(ethene), also known as polyethylene. $$ n(CH_2=CH_2) \xrightarrow[1000 \text{ atm}, O_2 \text{ traces}]{400^\circ C} -[CH_2-CH_2]_n- $$ - **Ziegler-Natta Catalyst:** A mixture of triethylaluminium ($Al(C_2H_5)_3$) and titanium tetrachloride ($TiCl_4$) acts as a highly effective catalyst. It produces high-quality, unbranched poly(ethene) that is harder and has a higher melting point. Karl Ziegler and Giulio Natta were awarded the Nobel Prize in 1963 for its development. ## Concept Assessment Exercise 16.3 - **Q:** What are the products of hydrochlorination of **2-methylbut-2-ene**? **A:** The alkene is $CH_3-C(CH_3)=CH-CH_3$. This is an unsymmetrical alkene. According to Markovnikov's rule, the H from $HCl$ will add to the carbon with more hydrogens (C-3), and the Cl will add to the carbon with fewer hydrogens (C-2). - **Product:** 2-chloro-2-methylbutane, $CH_3-C(Cl)(CH_3)-CH_2-CH_3$. - **Q:** What are the products of hydrochlorination of **3,3-dimethylbut-1-ene**? **A:** The alkene is $CH_2=CH-C(CH_3)_3$. This is an unsymmetrical alkene. The H adds to C-1 to form a secondary carbocation on C-2. This secondary carbocation will rearrange via a 1,2-methyl shift to form a more stable tertiary carbocation. 1. **Initial Addition (Markovnikov):** $CH_3-CH(Cl)-C(CH_3)_3$ (2-chloro-3,3-dimethylbutane, minor product). 2. **Rearrangement:** The intermediate carbocation $CH_3-CH^{+}-C(CH_3)_3$ rearranges to $CH_3-CH(CH_3)-C^{+}(CH_3)_2$. The $Cl^-$ then attacks this more stable tertiary carbocation. - **Major Product:** 2-chloro-2,3-dimethylbutane, $CH_3-CH(CH_3)-C(Cl)(CH_3)_2$. ## Concept Assessment Exercise 16.4 - **Q:** What are the products of ozonolysis of (i) **propene** and (ii) **but-2-ene**? **A:** - (i) **Propene ($CH_3-CH=CH_2$)**: Cleavage at the double bond yields **ethanal ($CH_3CHO$)** and **methanal ($HCHO$)**. - (ii) **But-2-ene ($CH_3-CH=CH-CH_3$)**: Cleavage at the double bond yields two molecules of **ethanal ($CH_3CHO$)**. - **Q:** Name the alkenes which give the following compounds on their ozonolysis: (i) **ethanal only** and (ii) **a mixture of methanal and propanone**. **A:** - (i) **Ethanal only**: Since two molecules of ethanal ($CH_3CHO$) are formed, the starting alkene must be symmetrical with two $CH_3-CH=$ groups. The alkene is **but-2-ene**. - (ii) **Methanal ($HCHO$) and propanone ($(CH_3)_2CO$)**: Methanal comes from a $=CH_2$ group. Propanone comes from a $=C(CH_3)_2$ group. Combining these gives $CH_2=C(CH_3)_2$. The alkene is **2-methylpropene**.

16.6 Chemical Reactions Of Alkenes | Hydrocarbons | Chemistry 11