RBSE Solutions Class 11 Chemistry Chapter 13 Hydrocarbons

Question 1. The number of 3\(^\circ\), 2\(^\circ\) and 1\(^\circ\) Carbon in isopentane :
(a) 1, 9, 2
(b) 9, 1, 2
(c) 2, 1, 9
(d) 1, 2, 9
Answer: (d) 1, 2, 9
In simple words: Isopentane has one carbon that is bonded to three other carbons (3\(^\circ\)), two carbons bonded to two other carbons (2\(^\circ\)), and nine carbons bonded to only one other carbon (1\(^\circ\)).

🎯 Exam Tip: To find the degree of a carbon, count how many other carbon atoms it is directly bonded to. A primary (1\(^\circ\)) carbon is bonded to one other carbon, secondary (2\(^\circ\)) to two, and tertiary (3\(^\circ\)) to three.

Question 3. Which of the following groups do not show addition reactions?
(a) Alkanes
(b) Alkaloids
(c) Cycloalkenes
(d) Ketones
Answer: (a) Alkanes
In simple words: Alkanes are simple hydrocarbon compounds that have only single bonds between carbon atoms, so they usually do not take part in addition reactions.

🎯 Exam Tip: Addition reactions happen when a molecule has double or triple bonds (like alkenes or alkynes) that can break to add new atoms. Alkanes only have single bonds, making them saturated and less reactive towards addition.

Question 4. Which of the following will not react with methane under normal conditions?
(a) I\( _2 \)
(b) Cl\( _2 \)
(c) Br\( _2 \)
(d) F\( _2 \)
Answer: (a) I\( _2 \)
In simple words: Methane, which is an alkane, does not easily react with iodine (I\( _2 \)) when conditions are normal. The reaction with iodine is usually very slow or requires special conditions.

🎯 Exam Tip: Halogenation of alkanes (reaction with halogens) generally decreases in reactivity from fluorine to iodine. Fluorine reacts explosively, chlorine and bromine react steadily, but iodine requires high temperatures or oxidizing agents.

Question 5. In which of the following reaction of alkene with HX, carbanion will be formed quickly ?
(a) HI
(b) HBr
(c) HCl
(d) All of the options
Answer: (a) HI
In simple words: When an alkene reacts with a hydrogen halide (HX), the speed at which a carbanion might form depends on the specific halogen. Hydrogen iodide (HI) helps form carbanions faster than HBr or HCl because iodine is a better leaving group.

🎯 Exam Tip: The reactivity of hydrogen halides (HX) in addition reactions to alkenes follows the order: HI > HBr > HCl > HF. This is because the H-X bond strength decreases as the size of the halogen increases, making it easier to break the bond and form reactive intermediates.

Chapter 13 Very Short Answer Type Questions

Question 7. Write general formula of alkenes and explain general method for its preparation.
Answer: The general formula for alkenes is \( \text{C}_n\text{H}_{2n} \). These are hydrocarbons with at least one carbon-carbon double bond.
A common way to make alkenes is by dehydrating alcohols. This process involves heating the vapors of an alcohol over hot alumina at \( 623 \text{ K} \). This removes a water molecule from the alcohol, forming an alkene.
For example:
\[ \text{RCH}_2\text{CH}_2\text{OH} \xrightarrow{\substack{\text{Al}_2\text{O}_3 \\ 623 \text{ K}}} \text{RCH} = \text{CH}_2 + \text{H}_2\text{O} \]
In simple words: Alkenes have a basic formula of \( \text{C}_n\text{H}_{2n} \) and can be made by taking water out of alcohols, usually by heating them with alumina.

🎯 Exam Tip: Remember that dehydration of alcohols is an elimination reaction where the hydroxyl group (-OH) and a hydrogen atom from an adjacent carbon are removed to form a double bond.

Question 8. Write the chemical equation for the formation of ethene in one step only by-
(a) Ethanol
(b) Ethyl bromide
(c) Ethyne
(d) Ethylene dibromide
Answer: Ethene (\( \text{CH}_2=\text{CH}_2 \)) can be formed in one step from several precursors. Here are some examples from the given options:
(d) From Ethylene dibromide: When ethylene dibromide is treated with zinc and methanol, ethene is formed.
\[ \begin{array}{c} \text{CH}_2-\text{Br} \\ | \\ \text{CH}_2-\text{Br} \end{array} + \text{Zn} \xrightarrow{\text{Methanol}} \begin{array}{c} \text{CH}_2 \\ || \\ \text{CH}_2 \end{array} + \text{ZnBr}_2 \]
In simple words: Ethene can be made directly from ethylene dibromide by reacting it with zinc and methanol.

🎯 Exam Tip: Ethene can also be formed from ethanol by dehydration (Q17a,c) and from ethyne by hydrogenation. Understanding these different synthesis routes is key for organic reactions.

Question 9. Explain Markownikoff's rule and give examples also.
Answer: Markownikoff's Rule states that when an unsymmetrical reagent (like HBr or HCl) adds to an unsymmetrical alkene, the negative part of the reagent attaches to the carbon atom of the double bond that has the fewest hydrogen atoms already. This rule helps predict the main product of such reactions.
For example, when hydrogen bromide (HBr) reacts with propene (\( \text{CH}_3-\text{CH}=\text{CH}_2 \)), according to Markownikoff's rule, the bromine (the negative part) attaches to the second carbon, which has only one hydrogen, while the first carbon has two.
\[ \text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3-\text{CH}(\text{Br})-\text{CH}_3 \]
The main product formed is 2-Bromopropane.
In simple words: Markownikoff's rule says that when adding something like HBr to an uneven double bond, the "bad" part (like Br) goes to the carbon that has fewer hydrogen atoms.

🎯 Exam Tip: Always identify the unsymmetrical alkene and the unsymmetrical reagent first. Then, pinpoint the carbon of the double bond with fewer hydrogens for the negative part of the reagent to attach.

Question 10. Explain the addition of HBr to alkenes according to Markownikoffs rule.
Answer: According to Markownikoff's rule, when HBr adds to an alkene, the negative part, which is the bromine atom (\( \text{Br}^- \)), attaches to the carbon atom of the double bond that has the fewer number of hydrogen atoms. For instance, in propene, the middle carbon has fewer hydrogens than the end carbon. The electron-releasing group attached to the double bond also helps stabilize the carbocation intermediate, which guides this attachment. This leads to the formation of the more substituted product.
In simple words: When HBr adds to an alkene, the bromine atom goes to the carbon with fewer hydrogen atoms on the double bond. This happens because of how electrons are pushed around by other parts of the molecule.

🎯 Exam Tip: Remember that the driving force behind Markownikoff's rule is the formation of the more stable carbocation intermediate during the reaction mechanism.

Question 11. Write the name and formula of first three members of acetylene series.
Answer: The acetylene series consists of alkynes, which are hydrocarbons with a carbon-carbon triple bond. The first three members are:
1. Acetylene (\( \text{C}_2\text{H}_2 \)): Its structural formula is \( \text{CH}\equiv\text{CH} \).
2. Propyne (\( \text{C}_3\text{H}_4 \)): Its structural formula is \( \text{CH}_3-\text{C}\equiv\text{CH} \).
3. Butyne (\( \text{C}_4\text{H}_6 \)): Its structural formula is \( \text{CH}_3-\text{C}\equiv\text{C}-\text{CH}_3 \).
In simple words: The first three compounds in the acetylene group are acetylene itself, propyne, and butyne, all having a triple bond between carbons.

🎯 Exam Tip: The general formula for alkynes is \( \text{C}_n\text{H}_{2n-2} \). Make sure to show the triple bond correctly in their structural formulas.

Question 12. Write the general formula of alkene and alkyne and give a chemical test to distinguish them.
Answer: The general formula for an alkene is \( \text{C}_n\text{H}_{2n} \), which contains at least one carbon-carbon double bond. The general formula for an alkyne is \( \text{C}_n\text{H}_{2n-2} \), which contains at least one carbon-carbon triple bond.
To distinguish between an alkene and an alkyne, we can use ammoniacal cuprous chloride solution. When this solution is added to an alkene, there is no noticeable reaction or effect. However, when ammoniacal cuprous chloride solution is added to an alkyne (specifically, a terminal alkyne like acetylene, which has a hydrogen atom on the triple bond), it forms a red precipitate of cuprous acetylide.
For example, with acetylene:
\[ \text{CH}\equiv\text{CH} + 2\text{AgNO}_3 + 2\text{NH}_3 \rightarrow \text{AgC}\equiv\text{CAg} \downarrow (\text{red ppt.}) + 2\text{HNO}_3 \]
In simple words: Alkenes are \( \text{C}_n\text{H}_{2n} \) and alkynes are \( \text{C}_n\text{H}_{2n-2} \). You can tell them apart by adding ammoniacal cuprous chloride; alkynes with an end hydrogen will form a red solid, but alkenes will not.

🎯 Exam Tip: This test is specific for terminal alkynes (those with a hydrogen directly attached to a triple-bonded carbon) because they are acidic enough to react with silver ions.

Question 13. What are 3\(^\circ\), 2\(^\circ\) and 1\(^\circ\) hydrogen? Explain with examples.
Answer: The terms 3\(^\circ\), 2\(^\circ\), and 1\(^\circ\) hydrogen refer to hydrogen atoms that are attached to tertiary (3\(^\circ\)), secondary (2\(^\circ\)), and primary (1\(^\circ\)) carbon atoms, respectively, within a hydrocarbon molecule. A primary carbon is bonded to one other carbon, a secondary carbon to two, and a tertiary carbon to three.
Example: In isopentane (2-methylbutane), we can identify different types of hydrogens:
\[ \overset{1^\circ}{\text{CH}}_3-\overset{3^\circ}{\text{CH}}(\overset{1^\circ}{\text{CH}}_3)-\overset{2^\circ}{\text{CH}}_2-\overset{1^\circ}{\text{CH}}_3 \]
In this compound, there is one 3\(^\circ\) carbon (the central CH), which has one 3\(^\circ\) hydrogen. There is one 2\(^\circ\) carbon (the \( \text{CH}_2 \) group), which has two 2\(^\circ\) hydrogens. There are three 1\(^\circ\) carbons (the three \( \text{CH}_3 \) groups), which collectively contain nine 1\(^\circ\) hydrogens (three for each \( \text{CH}_3 \) group).
Therefore, isopentane contains one 3\(^\circ\) hydrogen, two 2\(^\circ\) hydrogens, and nine 1\(^\circ\) hydrogens.
In simple words: 1\(^\circ\), 2\(^\circ\), and 3\(^\circ\) hydrogens are named after the type of carbon they are connected to. For example, in a molecule like isopentane, you can count how many of each type of hydrogen exist.

🎯 Exam Tip: Always draw out the full structural formula to clearly identify each carbon atom's degree, and then count the hydrogens attached to them. This helps avoid errors in counting.

Question 15. Write the chemical equation for formation of acetylene in one step only by-
(a) Ethyl dibromide
(b) Trichloromethane
(c) Ethylidene dichloride
(d) Acetylene tetrabromide
Answer: Acetylene (\( \text{CH}\equiv\text{CH} \)) can be formed in one step from various compounds through elimination reactions.
(b) From Trichloromethane (Chloroform): Heating trichloromethane with sodium forms acetylene.
\[ 2\text{CHCl}_3 + 6\text{Na} \rightarrow \text{CH}\equiv\text{CH} + 6\text{NaCl} \]
(c) From Ethylidene dichloride (1,1-dichloroethane): Heating ethylidene dichloride with alcoholic KOH forms acetylene.
\[ \begin{array}{c} \text{CH}_3 \\ | \\ \text{CHCl}_2 \end{array} + 2\text{KOH} \xrightarrow{\text{alc., boiling}} \text{CH}\equiv\text{CH} + 2\text{KCl} + 2\text{H}_2\text{O} \]
(d) From Acetylene tetrabromide (1,1,2,2-tetrabromoethane): Heating acetylene tetrabromide with zinc forms acetylene.
\[ \begin{array}{c} \text{CHBr}_2 \\ | \\ \text{CHBr}_2 \end{array} + 2\text{Zn} \xrightarrow{\Delta} \begin{array}{c} \text{CH} \\ ||| \\ \text{CH} \end{array} + 2\text{ZnBr}_2 \]
In simple words: Acetylene can be made in one step from trichloromethane by reacting with sodium, from ethylidene dichloride by heating with alcoholic KOH, or from acetylene tetrabromide by heating with zinc.

🎯 Exam Tip: These are examples of dehalogenation or dehydrohalogenation reactions. Pay close attention to the reagents and conditions (like alcoholic KOH, zinc dust, or sodium metal) that lead to the formation of alkynes.

Question 16. Give two chemical tests to detect unsaturation in organic compounds.
Answer: Here are two chemical tests used to detect unsaturation (presence of double or triple bonds) in organic compounds:

• Bromine test: This test involves adding a 5% solution of bromine in carbon tetrachloride (\( \text{CCl}_4 \)) to the organic compound. If the compound contains unsaturation (a double or triple bond), the reddish-brown color of the bromine solution will disappear or become decolourized. This happens because bromine adds across the multiple bond.


• Bayer's test: In this test, the organic compound is treated with cold, dilute, alkaline potassium permanganate (\( \text{KMnO}_4 \)) solution. If unsaturation is present, the purple color of the \( \text{KMnO}_4 \) solution will disappear, and a brown precipitate of manganese dioxide might form. This indicates that the \( \text{KMnO}_4 \) has been reduced as it oxidizes the alkene to a diol.

🎯 Exam Tip: Remember the color changes for each test: bromine water changes from reddish-brown to colorless, and Bayer's reagent changes from purple to colorless (often with a brown precipitate).

Question 17. Give reaction: What happens when :
(a) Ethyl alcohol is heated with excess of concentrated sulphuric acid at 1600\(^\circ\)C?
(b) Passing ethene in cold aqueous solution of alkaline potassium permanganate.
(c) Vapours of ethyl alcohol are passed over hot Aluminium oxide at 3600\(^\circ\)C. ?
(d) Isopropyl bromide is heated with alcoholic KOH?
(e) High pressure is applied on ethylene in the presence of peroxide catalyst?
(f) Propylene is passed on hot solution of potassium permanganate solution?
(g) Ethylene reacts with hypochlorous acid?
(h) Ozone is treated with ethylene?
Answer: Here's what happens in each scenario:
(a) When ethyl alcohol is heated with excess concentrated sulphuric acid at \( 1600^\circ\text{C} \) (likely a typo, typically around \( 170^\circ\text{C} \)), ethene is formed through dehydration.
\[ \text{C}_2\text{H}_5\text{OH} \xrightarrow{\substack{\text{conc. H}_2\text{SO}_4 \\ 1600^\circ\text{C}}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \]
(b) On passing ethene into a cold aqueous solution of alkaline potassium permanganate (\( \text{KMnO}_4 \)), the purple color of \( \text{KMnO}_4 \) is decolourized, and ethane-1,2-diol (ethylene glycol) is formed. This is Bayer's test for unsaturation.
\[ \begin{array}{c} \text{CH}_2 \\ || \\ \text{CH}_2 \end{array} + \text{H}_2\text{O} + [\text{O}] \xrightarrow{\substack{\text{Alkaline KMnO}_4 \\ \text{Cold}}} \begin{array}{c} \text{CH}_2\text{OH} \\ | \\ \text{CH}_2\text{OH} \end{array} \]
(c) When vapors of ethyl alcohol are passed over hot aluminium oxide (\( \text{Al}_2\text{O}_3 \)) at \( 3600^\circ\text{C} \) (likely a typo, typically around \( 350^\circ\text{C} \)), ethene is formed by dehydration.
\[ \text{C}_2\text{H}_5\text{OH} \xrightarrow{\substack{\text{Al}_2\text{O}_3 \\ 3600^\circ\text{C}}} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O} \]
(d) When isopropyl bromide is heated with alcoholic KOH, propene is formed through a dehydrohalogenation (elimination) reaction.
\[ \text{CH}_3-\text{CH}(\text{Br})-\text{CH}_3 \xrightarrow{\text{alcoholic KOH}} \text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \]
(e) When high pressure is applied to ethylene (\( \text{CH}_2=\text{CH}_2 \)) in the presence of a peroxide catalyst, it undergoes addition polymerization to form polyethene.
\[ n\text{CH}_2=\text{CH}_2 \xrightarrow{\substack{\text{high pressure} \\ \text{peroxide}}} -[\text{CH}_2-\text{CH}_2]_n \]
(f) On passing propylene into a hot solution of potassium permanganate, the diol is initially formed. This diol then undergoes further oxidation, breaking the carbon chain to form acetaldehyde and formaldehyde. These products can be oxidized even further to form acetic acid and formic acid respectively.
\[ \begin{array}{c} {[\text{O}]} \\ \text{H}_2\text{O} \\ \text{CH}_3\text{CHO} + \text{HCHO} \\ {[\text{O}]} \quad\quad [\text{O}] \\ \text{CH}_3\text{COOH} \quad \text{HCOOH} \end{array} \]
(g) When ethylene reacts with hypochlorous acid (HOCl), it forms ethylene chlorohydrin.
\[ \begin{array}{c} \text{CH}_2 \\ || \\ \text{CH}_2 \end{array} + \text{HOCl} \rightarrow \begin{array}{c} \text{CH}_2-\text{Cl} \\ | \\ \text{CH}_2-\text{OH} \end{array} \]
(h) When ozone (\( \text{O}_3 \)) is treated with ethylene, ethylene ozonide is formed, which is an intermediate in ozonolysis.
\[ \begin{array}{c} \text{CH}_2 \\ || \\ \text{CH}_2 \end{array} + \text{O}_3 \rightarrow \begin{array}{c} \text{H}_2\text{C}-\text{O}-\text{CH}_2 \\ \quad \text{O} \\ \end{array} \]
In simple words: Heating ethyl alcohol with strong acid or alumina makes ethene. Ethene with cold permanganate makes a diol. Ethylene polymerizes under high pressure with a catalyst. Propylene with hot permanganate breaks down to smaller acids. Ethylene with hypochlorous acid makes chlorohydrin, and with ozone, it forms an ozonide.

🎯 Exam Tip: For "What happens when..." questions, always provide the chemical equation, list the reactants, conditions, and products clearly. Pay attention to specific temperature and reagent conditions.

Question 18. How following compounds are formed from ethylene :
(1) Acetylene;
(2) Formaldehyde ;
(3) Ethylene glycol;
(4) Ethylene chlorohydrin;
(5) Ethyl alcohol;
(6) Ethylene oxide ;
Answer: Here's how each compound can be formed from ethylene:
(1) Acetylene: When ethylene undergoes oxidation and then the resulting diol or aldehyde is further processed (e.g., thermal decomposition or dehydrohalogenation if starting from dibromoethane), acetylene can be formed.
\[ \begin{array}{c} \text{CH}_2 \\ || \\ \text{CH}_2 \end{array} \xrightarrow{\text{(O)}} \begin{array}{c} \text{CH} \\ ||| \\ \text{CH} \end{array} \]
(3) Ethylene glycol: When ethylene is passed into a cold aqueous solution of alkaline potassium permanganate (Bayer's reagent), it forms ethylene glycol.
\[ \begin{array}{c} \text{CH}_2 \\ || \\ \text{CH}_2 \end{array} + \text{H}_2\text{O} + [\text{O}] \xrightarrow{\substack{\text{alkaline KMnO}_4 \\ \text{cold}}} \begin{array}{c} \text{CH}_2\text{OH} \\ | \\ \text{CH}_2\text{OH} \end{array} \]
(4) Ethylene chlorohydrin: When ethylene reacts with hypochlorous acid (HOCl), it forms ethylene chlorohydrin.
\[ \begin{array}{c} \text{CH}_2 \\ || \\ \text{CH}_2 \end{array} + \text{HOCl} \rightarrow \begin{array}{c} \text{CH}_2-\text{Cl} \\ | \\ \text{CH}_2-\text{OH} \end{array} \]
(5) Ethyl alcohol: When ethylene reacts with water in the presence of dilute acid (like \( \text{H}_3\text{PO}_4 \)) at \( 300^\circ\text{C} \) and \( 65 \) atmospheric pressure, it forms ethyl alcohol.
\[ \begin{array}{c} \text{CH}_2 \\ || \\ \text{CH}_2 \end{array} + \text{H}_2\text{O} \xrightarrow{\substack{\text{H}_3\text{PO}_4 \\ 300^\circ\text{C}, 65 \text{ atm}}} \begin{array}{c} \text{CH}_3 \\ | \\ \text{CH}_2-\text{OH} \end{array} \]
(6) Ethylene oxide: When ethylene is mixed with air and passed under pressure over a silver catalyst at \( 200-420^\circ\text{C} \), it forms ethylene oxide (also known as epoxide).
\[ \begin{array}{c} \text{CH}_2 \\ || \\ \text{CH}_2 \end{array} + 1/2 \text{O}_2(\text{g}) \xrightarrow{\substack{\text{Ag catalyst} \\ 200-400^\circ\text{C}}} \begin{array}{c} \text{H}_2\text{C}-\text{O} \\ \setminus / \\ \text{H}_2\text{C} \end{array} \]
In simple words: Ethylene can be changed into many compounds: acetylene by oxidation, ethylene glycol by adding cold permanganate, chlorohydrin by reacting with HOCl, ethyl alcohol by adding water under specific heat and pressure, and ethylene oxide by reacting with oxygen over a silver catalyst.

🎯 Exam Tip: For each conversion, remember the key reagents, conditions (temperature, pressure, catalyst), and the type of reaction involved (e.g., oxidation, addition, hydration).

Question 19. Compare ethane, ethene and acetylene carbon-carbon bond on the basis of bond length, stability and reactivity.
Answer: Here is a comparison of ethane, ethene, and acetylene based on their carbon-carbon bonds:

🎯 Exam Tip: Remember that as the number of bonds between carbon atoms increases (single to double to triple), the bond length decreases, bond strength increases, but overall reactivity towards addition reactions also increases because the pi bonds are weaker than sigma bonds and more accessible.

Question 20. Write short notes on :
(a) Bromination of ethylene
(b) Polymerization of ethylene
(c) Markownikoff's rule
(d) Ozonolysis
Answer: Here are short notes on each topic:
(a) Bromination of ethylene: No content provided for this part in the source.

(b) Polymerization of ethylene: Ethylene undergoes addition polymerization when heated under high pressure in the presence of a peroxide catalyst. This process joins many ethylene molecules together to form a long chain polymer called polyethene.
\[ n\text{CH}_2=\text{CH}_2 \xrightarrow{\substack{\text{peroxide} \\ \text{polymerization}}} -[\text{CH}_2-\text{CH}_2]_n \]
(c) Markownikoff's rule: This rule states that when an unsymmetrical reagent (like HBr) adds to an unsymmetrical alkene, the negative part of the attacking reagent attaches to the carbon atom of the double bond that has the fewest hydrogen atoms. This leads to the formation of the more stable carbocation intermediate.
\[ \text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \rightarrow \text{CH}_3-\text{CH}(\text{Br})-\text{CH}_3 \]
(d) Ozonolysis: Ozonolysis is a reaction where an alkene reacts with ozone (\( \text{O}_3 \)) to form an ozonide. This ozonide is then usually treated with a reducing agent (like zinc and water) to break it down into aldehydes or ketones. For ethylene, it forms ethene ozonide.
\[ \begin{array}{c} \text{CH}_2 \\ || \\ \text{CH}_2 \end{array} + \text{O}_3 \rightarrow \begin{array}{c} \text{H}_2\text{C}-\text{O}-\text{CH}_2 \\ \quad \text{O} \end{array} \]
In simple words: Polymerization of ethylene turns many small ethylene molecules into a big plastic chain. Markownikoff's rule tells you where the new atoms will attach to an uneven double bond. Ozonolysis is when alkenes react with ozone to form a special ring structure called an ozonide.

🎯 Exam Tip: For short notes, define the concept and provide a clear, balanced chemical equation if applicable. For Markownikoff's rule, mention both the rule and an example. If information is missing (like for part a), indicate it discreetly by only listing the sub-part without an answer.

Question 21. Write the structural formula of the following:
(a) Ethylene glycol
(b) Ethylidene dibromide
(c) Ethylene dibromide
(d) Isopropyl bromide
(e) 2-methyl -3 hexene
(f) Propylene oxide
Answer: Here are the structural formulas for the given compounds:
(a) Ethylene glycol: (No content provided for this part in the source.)

(b) Ethylidene dibromide (1,1-dibromoethane): \( \text{CH}_3-\text{CHBr}_2 \)

(c) Ethylene dibromide (1,2-dibromoethane): \( \text{Br}-\text{CH}_2-\text{CH}_2-\text{Br} \)

(d) Isopropyl bromide (2-bromopropane): \( \text{CH}_3-\text{CH}(\text{Br})-\text{CH}_3 \)

(e) 2-methyl-3-hexene: \( \text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}=\text{CH}-\text{CH}_2-\text{CH}_3 \)

(f) Propylene oxide: \( \text{CH}_3-\text{CH}(\text{O})-\text{CH}_2 \) (Represented as a linear chain based on OCR, typically a cyclic ether.)
In simple words: Each of these compounds has a specific arrangement of atoms. For example, isopropyl bromide is a propane chain with a bromine attached to the middle carbon, and 2-methyl-3-hexene is a six-carbon chain with a double bond and a methyl branch.

🎯 Exam Tip: When writing structural formulas, ensure all valencies are correct (carbon forms four bonds, hydrogen one, halogens one, oxygen two). Pay attention to the position of double bonds and functional groups.

RBSE Class 11 Chemistry Chapter 13 Short Answer Type Questions

Question 22. What is peroxide effect? Explain giving one example.
Answer: The peroxide effect, also known as the Anti-Markownikoff's rule or Kharasch effect, describes the addition of HBr (but not HCl or HI) to unsymmetrical alkenes in the presence of peroxides. This reaction occurs contrary to Markownikoff's rule. According to this effect, when HBr adds to an unsymmetrical alkene in the presence of peroxides, the negative part (bromine atom) of the reagent attaches to the carbon atom of the double bond that bears the *higher* number of hydrogen atoms. This process proceeds via a free radical mechanism.
Example: The addition of HBr to propene (\( \text{CH}_3-\text{CH}=\text{CH}_2 \)) in the presence of peroxide:
\[ \text{CH}_3-\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3-\text{CH}_2-\text{CH}_2-\text{Br} \]
The product formed is n-propyl bromide (1-bromopropane), instead of the Markownikoff's product, 2-bromopropane.
In simple words: The peroxide effect is a special rule where HBr adds to a double bond differently than usual, putting the bromine on the carbon with more hydrogens, but only when peroxides are present.

🎯 Exam Tip: Remember that the peroxide effect applies specifically to the addition of HBr (not HCl or HI) to unsymmetrical alkenes and occurs via a free radical mechanism, yielding an anti-Markownikoff product.

Question 23. Draw the structure of stereoisomers obtained by reaction of cis-2-butene with bromine.
Answer: When cis-2-butene reacts with bromine (\( \text{Br}_2 \)), it undergoes anti-addition to form a racemic mixture of (2R,3R)-2,3-dibromobutane and (2S,3S)-2,3-dibromobutane. These are enantiomers (mirror images) and are stereoisomers.
One of the enantiomers, (2R,3R)-2,3-dibromobutane, can be represented as:
\[ \begin{array}{c} \text{CH}_3 \\ | \\ \text{H}-\text{C}-\text{Br} \\ | \\ \text{Br}-\text{C}-\text{H} \\ | \\ \text{CH}_3 \end{array} \]
The other enantiomer, (2S,3S)-2,3-dibromobutane, would be its mirror image. The reaction yields both in equal amounts, forming a racemic mixture.
In simple words: When cis-2-butene reacts with bromine, it creates two types of molecules that are mirror images of each other, called stereoisomers. These two types are formed equally.

🎯 Exam Tip: For addition reactions of alkenes, understand if it's syn-addition or anti-addition. Cis-alkenes undergoing anti-addition usually lead to a racemic mixture, while trans-alkenes lead to meso compounds or racemic mixtures depending on the reagent.

Question 24. Give the structure and name of the product obtained by the reaction of 1-butene with bromine.
Answer: When 1-butene reacts with bromine (\( \text{Br}_2 \)), an addition reaction occurs across the double bond, resulting in the formation of 1,2-dibromobutane.
The chemical equation is:
\[ \text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_3-\text{CH}_2-\text{CH}(\text{Br})-\text{CH}_2\text{Br} \]
The name of the product is 1,2-Dibromobutane.
In simple words: When 1-butene mixes with bromine, the bromine atoms add to the double bond, creating a new molecule called 1,2-dibromobutane.

🎯 Exam Tip: Remember that halogens like bromine add across the double bond of alkenes. The reaction is typically an anti-addition, forming vicinal dihalides.

Question 25. The central carbon-carbon bond in 1, 3-butadiene is shorter than n-butane. Why?
Answer: The central carbon-carbon bond in 1,3-butadiene is shorter than the carbon-carbon single bonds in n-butane because of the presence of conjugation (alternating double and single bonds) in 1,3-butadiene. The electrons in the double bonds are delocalized over the entire conjugated system, giving the central single bond some partial double-bond character. This partial double-bond character makes it shorter and stronger than a pure single bond, which is found in n-butane.
In simple words: The middle bond in 1,3-butadiene is shorter because the electrons in its double bonds are shared across the whole molecule, giving the single bond a bit of double bond strength. N-butane just has regular single bonds, so its bonds are longer.

🎯 Exam Tip: Conjugation (alternating single and double bonds) allows for electron delocalization, leading to properties like shorter single bond lengths, increased stability, and resonance stabilization. This is a key concept in organic chemistry.

Question 26. Explain with chemical equation. What happens when :
1. Passing acetylene gas in solution of ammoniacal cuprous chloride ?
2. Passing acetylene gas in red hot tube ?
3. Acetylene is passed in dilute sulphuric acid?
4. Acetylene is passed in hydrochloric acid in the presence of mercuric sulphate ?
5. Acetylene is passed in aqueous solution containing Hg++ and H+ ions?
6. Acetylene gas is passed in hypochlorous acid ?
Answer: Here's what happens in each reaction with acetylene:
1. When acetylene gas is passed into an ammoniacal cuprous chloride solution, a red precipitate of cuprous acetylide is formed. This is a test for terminal alkynes.
\[ \text{CH}\equiv\text{CH} + 2\text{CuCl} + 2\text{NH}_3 \rightarrow \text{CuC}\equiv\text{CCu} \downarrow (\text{red ppt.}) + 2\text{NH}_4\text{Cl} \]
2. When acetylene gas is passed through a red hot copper tube, it undergoes cyclic polymerization to form benzene.
\[ 3\text{CH}\equiv\text{CH} \xrightarrow{\text{Hot Cu tube}} \text{C}_6\text{H}_6 \]
3. When acetylene is passed into dilute sulphuric acid (usually with mercuric sulphate as catalyst), it undergoes hydration to form ethanal (acetaldehyde). The initial product, vinyl alcohol, quickly tautomerizes to ethanal.
\[ \text{CH}\equiv\text{CH} + \text{H}_2\text{SO}_4 \rightarrow \text{CH}_3-\text{CH}(\text{OSO}_2\text{OH})_2 \]
This intermediate then hydrolyzes.
\[ \begin{array}{c} \text{CH}_3 \\ | \\ \text{CH}(\text{OSO}_2\text{OH})_2 \end{array} \xrightarrow{\text{H}_2\text{O}} \text{CH}_3\text{CHO} \]
4. When acetylene is passed through hydrochloric acid in the presence of mercuric sulphate (\( \text{Hg}^{2+} \)) at \( 65^\circ\text{C} \), it undergoes addition to form vinyl chloride.
\[ \text{CH}\equiv\text{CH} + \text{HCl} \xrightarrow{\substack{65^\circ\text{C} \\ \text{Hg}^{2+}}} \text{CH}_2=\text{CHCl} \]
5. When acetylene is passed into an aqueous solution containing \( \text{Hg}^{2+} \) and \( \text{H}^+ \) ions (acidic mercuric sulphate solution), it undergoes hydration to form acetaldehyde.
\[ \text{CH}\equiv\text{CH} + \text{H}_2\text{O} \xrightarrow{\text{Hg}^{2+}/\text{H}^+} \text{CH}_3-\text{CHO} \]
6. When acetylene gas reacts with hypochlorous acid (HOCl), it undergoes addition to form dichloroacetaldehyde.
(No specific equation provided in the source OCR beyond the general reaction for ethylene, so providing the general product description).
*Self-correction*: The source actually shows: \( \text{H}_2\text{C=CH-C}\equiv\text{C-CH=CH}_2 \) for divinyl acetylene, then for Q26(3) it shows ethylidene hydrogen sulphate as an intermediate for acetylene in dilute sulphuric acid. Then Q26(5) shows acetaldehyde. I will use the actual equations and products provided in the source for Q26 parts. Let's re-do Q26 parts based on the exact source content for the answers for Q26(1-6) on page 15.
1. Passing acetylene gas in solution of ammoniacal cuprous chloride: Forms a red precipitate of cuprous acetylide. (The equation for this specific reaction is not explicitly shown in the source answer on page 15, but is well-known and consistent with Q12.) I will describe the product as indicated. 2. Passing acetylene gas in red hot tube: It polymerizes to benzene.
\[ 3\begin{array}{c} \text{CH} \\ ||| \\ \text{CH} \end{array} \xrightarrow{\text{Hot Cu tube}} \text{benzene} \]
3. Acetylene is passed in dilute sulphuric acid: It forms ethylidene hydrogen sulphate, which then hydrolyzes to acetaldehyde.
\[ \begin{array}{c} \text{CH} \\ ||| \\ \text{CH} \end{array} + \text{H}_2\text{SO}_4 \rightarrow \begin{array}{c} \text{CH}_3 \\ | \\ \text{CH}(\text{OSO}_2\text{OH})_2 \end{array} \]
4. Acetylene is passed in hydrochloric acid in the presence of mercuric sulphate: It gives vinyl chloride.
\[ \begin{array}{c} \text{CH} \\ ||| \\ \text{CH} \end{array} + \text{HCl} \xrightarrow{\substack{65^\circ\text{C} \\ \text{Hg}^{2+}}} \begin{array}{c} \text{CH}_2 \\ || \\ \text{CHCl} \end{array} \]
5. Acetylene is passed in aqueous solution containing Hg\(^{++}\) and H\(^+\) ions: It gives acetaldehyde.
\[ \begin{array}{c} \text{CH} \\ ||| \\ \text{CH} \end{array} + \text{H}_2\text{O} \xrightarrow{\text{Hg}^{++}/\text{H}^+} \begin{array}{c} \text{CH}_3 \\ | \\ \text{CHO} \end{array} \]
6. Acetylene gas is passed in hypochlorous acid: No specific equation is provided in the source for this reaction. (This part of the answer content is not fully visible on the last processed page, but it's listed in Q26. I must acknowledge the missing answer or simply output the label if no content is provided. Given the source provides the heading for Q26(6) but no chemical equation or explanation, I will list the label only.)
In simple words: Acetylene reacts in different ways: with cuprous chloride, it makes a red solid; in a hot tube, it forms benzene; with dilute sulfuric acid, it eventually makes acetaldehyde; with hydrochloric acid, it forms vinyl chloride; with mercury ions and water, it also forms acetaldehyde.

🎯 Exam Tip: Pay attention to the specific reagents and conditions for each reaction of acetylene, as the products can vary greatly depending on whether it's an addition, polymerization, or hydration reaction.

Question 26. Explain with chemical equation. What happens when :
3. Acetylene is passed in dilute sulphuric acid?
4. Acetylene is passed in hydrochloric acid in the presence of mercuric sulphate ?
5. Acetylene is passed in aqueous solution containing Hg++ and H+ ions?
6. Acetylene gas is passed in hypochlorous acid ?
Answer:
3. When acetylene is passed through dilute sulfuric acid, it changes into ethylidene hydrogen sulfate.
\( \ce{CH#CH} + \ce{H2SO4} \rightarrow \ce{CH3-CH(OSO2OH)} \)
4. When acetylene reacts with hydrochloric acid in the presence of mercuric sulfate, it forms vinyl chloride.
\( \ce{CH#CH} + \ce{HCl} \xrightarrow{\text{65°C, Hg^2+}} \ce{CH2=CHCl} \)
5. When acetylene is passed through an aqueous solution with Hg++ and H+ ions, it produces acetaldehyde.
\( \ce{CH#CH} + \ce{H2O} \xrightarrow{\text{Hg^2+/H^+}} \ce{CH3-CHO} \)
6. (Content not available in source)
In simple words: Acetylene is a reactive gas. It combines with different chemicals to form new products, often with acids or specific catalysts.

🎯 Exam Tip: When writing chemical equations, remember to include all catalysts and conditions (like temperature or specific ions) above or below the reaction arrow to get full marks.

Question 27. Distinguish between the following. Give chemical equation also.
1. Ethylene and Acetylene
2. Ethane and Ethyne
3. Saturated and unsaturated hydrocarbons
4. 1-butene and 1-butyne
Answer:
1. **Ethylene and Acetylene**
When ammoniacal silver nitrate solution is added to ethylene, there is no change. However, when the same solution is added to acetylene, a white solid called silver acetylide forms.
\( \ce{CH2=CH2} + \ce{AgNO3} \rightarrow \text{No effect} \)
\( \ce{CH#CH} + \ce{2AgNO3} \xrightarrow{\ce{NH3}} \ce{AgC#CAg} \downarrow + \ce{2HNO3} \)
(Silver acetylide, white precipitate)
2. **Ethane and Ethyne**
Ethane and ethyne can be told apart using the bromine test. If bromine water is added to ethane, nothing happens and the orange color remains. But when bromine water is added to ethyne, the color disappears because ethyne reacts with bromine.
\( \ce{C2H6} + \ce{Br2} \rightarrow \text{No effect} \)
\( \ce{CH#CH} + \ce{2Br2} \xrightarrow{\ce{CCl4}} \ce{CHBr2-CHBr2} \)
3. **Saturated and unsaturated hydrocarbons**
To distinguish between these, a 5% solution of bromine in carbon tetrachloride is used. When this solution is added to an unsaturated organic compound, its orange color disappears. This shows that the compound has double or triple bonds. Saturated hydrocarbons do not cause the bromine solution to decolorize.
4. **1-butene and 1-butyne**
These can be distinguished using the ammoniacal silver nitrate solution. When this solution is added to 1-butene, nothing happens. But when it's added to 1-butyne, a white precipitate forms.
\( \ce{CH3-CH2-CH=CH2} + \ce{AgNO3} \rightarrow \text{No effect} \)
\( \ce{CH3-CH2-C#CH} + \ce{AgNO3} \xrightarrow{\ce{NH3}} \ce{CH3-CH2-C#CAg} \downarrow + \ce{HNO3} \)
(White precipitate)
In simple words: Different hydrocarbons react differently with specific chemicals. These reactions, like changing color or forming a solid, help us tell them apart.

🎯 Exam Tip: Remember specific reagent tests, such as the bromine water test for unsaturation and the ammoniacal silver nitrate test for terminal alkynes, as they are crucial for distinguishing compounds.

Question 28. Explain industrial method for preparation of acetylene. Also give chemical equation also.
Answer:
Acetylene can be made on a large scale in two main ways industrially:
1. **From Calcium Carbide:**
First, calcium carbide is made by heating quicklime with coke (carbon) at a very high temperature of 2300 K.
\( \ce{CaO} + \ce{3C} \xrightarrow{\text{2300 K}} \ce{CaC2} + \ce{CO} \)
Then, calcium carbide reacts with water to produce acetylene.
\( \ce{CaC2} + \ce{2H2O} \rightarrow \ce{Ca(OH)2} + \ce{C2H2} \)
2. **By Pyrolysis of Methane:**
Acetylene can also be prepared industrially by heating methane to a very high temperature of about 1800 K. This process is called pyrolysis.
\( \ce{2CH4} \xrightarrow{\text{1800 K}} \ce{HC#CH} + \ce{2H2} \)
In simple words: Acetylene is made by heating special materials like calcium carbide with water, or by heating methane very strongly. These processes are used in factories to make a lot of acetylene.

🎯 Exam Tip: Industrial methods often involve high temperatures and specific raw materials; clearly state these conditions and the steps for each method.

Question 29. Draw geometry of acetylene molecule and explain the nature of bonds present in the molecule.
Answer:
The acetylene molecule (HC≡CH) has one carbon-carbon triple bond and two carbon-hydrogen single bonds. The length of the carbon-carbon triple bond is about 120 pm, and each carbon-hydrogen single bond is about 106 pm. The carbon-carbon triple bond is very strong, with a bond energy of 823 kJ/mol. The entire acetylene molecule has a linear shape.
The bonds consist of sigma (σ) bonds and pi (π) bonds. There are three sigma bonds (one C-C and two C-H) and two pi bonds (between the two carbon atoms). Each carbon atom uses sp-hybridized orbitals to form these bonds.
The geometry of acetylene is linear as shown below:

In simple words: Acetylene is a straight molecule with carbon atoms connected by a very strong triple bond, and each carbon also connects to a hydrogen atom. This triple bond has both strong direct links (sigma bonds) and sideways links (pi bonds).

🎯 Exam Tip: When describing molecular geometry, always mention bond lengths, bond angles, and the type of hybridization involved for complete answers.

Question 30. Explain with chemical equation what happens when:
1. Acetylene reacts with bromine water?
2. Hydrogen bromide is added to acetylene.
3. Acetylene is passed in cold dilute alkaline potassium permanganate solution?
4. Hydrolysis of the product obtained by reaction of acetylene and ozone in carbon tetrachloride solution?
Answer:
1. When acetylene reacts with bromine water, the bromine water loses its color (decolorizes) and forms 1,2-dibromoethene. If excess bromine is added, it will further react to form 1,1,2,2-tetrabromoethane.
\( \ce{CH#CH} + \ce{Br2} \rightarrow \ce{CHBr=CHBr} \)
(1,2-dibromoethene)
3. When acetylene is passed through a cold, dilute, alkaline potassium permanganate solution (Bayer's reagent), it gets oxidized and forms oxalic acid. The purple color of permanganate disappears.
\( \ce{CH#CH} + \ce{[O]} + \ce{O2} \xrightarrow{\text{alkaline KMnO4}} \ce{COOH-COOH} \)
(Oxalic acid)
4. When acetylene reacts with ozone, it first forms an ozonide. When this ozonide is then broken down by water (hydrolysis), it produces glyoxal and then formic acid.
\( \ce{CH#CH} + \ce{O3} \rightarrow \text{Ozonide} \)
\( \text{Ozonide} + \ce{H2O2} \rightarrow \ce{CHO-CHO} + \ce{2HCOOH} \)
(Glyoxal) (Formic acid)
In simple words: Acetylene is a very reactive chemical. It can combine with bromine to lose its color, or with oxygen in the presence of permanganate to become a different acid. It also reacts with ozone to form a special compound that breaks down into other acids.

🎯 Exam Tip: Always specify the reagents and reaction conditions, like "cold, dilute, alkaline" for potassium permanganate, as they dictate the products formed.

Question 31. The boiling point of n-pentane is higher than neopentane. Explain the reason.
Answer:
The boiling point of n-pentane is higher than neopentane (2,2-dimethylpropane) because of their molecular shapes. As the number of branches in an alkane molecule increases, its shape becomes more spherical. This spherical shape reduces the total surface area available for contact between molecules. With less contact area, the intermolecular forces (Van der Waals forces) between the molecules become weaker. Weaker intermolecular forces require less energy to overcome, resulting in a lower boiling point. So, n-pentane, being a straight chain, has more surface area and stronger forces than the more spherical neopentane, giving it a higher boiling point.
In simple words: Straight-chain molecules like n-pentane stick together better because they have more surface area. This needs more heat to boil them. Branched molecules like neopentane are rounder, so they don't stick as well, and boil at a lower temperature.

🎯 Exam Tip: When explaining boiling point differences, always link molecular structure (branching, surface area) to the strength of intermolecular forces, as this is the key concept.

RBSE Class 11 Chemistry Chapter 13 Long Answer Type Questions

Question 32. Give the structure and IUPAC name of the different alkanes obtained by reaction of sodium with 1-bromo propane and 2-bromo propane in the presence of ether. What is the name of the reaction?
Answer:
This reaction is called the **Wurtz reaction**.
When 1-bromopropane reacts with sodium in the presence of dry ether, n-hexane is formed.
\( \ce{2CH3-CH2-CH2-Br} + \ce{2Na} \xrightarrow{\text{dry ether}} \ce{CH3-CH2-CH2-CH2-CH2-CH3} + \ce{2NaBr} \)
(1-bromopropane) (n-hexane)
When 2-bromopropane reacts with sodium in the presence of dry ether, 2,3-dimethylbutane is obtained.
\( \ce{2CH3-CH(Br)-CH3} + \ce{2Na} \xrightarrow{\text{ether}} \ce{CH3-CH(CH3)-CH(CH3)-CH3} + \ce{2NaBr} \)
(2-bromopropane) (2,3-dimethylbutane)
The IUPAC name of the product from 2-bromopropane is 2,3-dimethylbutane.
In simple words: This is the Wurtz reaction. When two identical bromo-propane molecules join up with sodium, they form a bigger alkane. 1-bromopropane forms a straight chain (n-hexane), while 2-bromopropane forms a branched chain (2,3-dimethylbutane).

🎯 Exam Tip: Clearly identify the type of reaction (Wurtz reaction) and accurately draw the structures of both reactants and products to score full marks.

Question 33. What are alkanes? Write general formula of alkanes and four general methods for preparation of alkanes.
Answer:
**Alkanes:** Alkanes are a type of hydrocarbon that contain only single bonds between carbon atoms. They are saturated (meaning they have the maximum number of hydrogen atoms possible) and have an open-chain structure. They are also known as paraffins.
**General formula of alkanes:** The general formula for alkanes is \( \ce{C_nH_{2n+2}} \), where 'n' represents the number of carbon atoms.
**General methods for preparation of alkanes:**
1. **From unsaturated hydrocarbons (Hydrogenation):** Unsaturated hydrocarbons (like alkenes and alkynes) can be changed into alkanes by adding hydrogen in the presence of a catalyst such as Nickel (Ni), Platinum (Pt), or Palladium (Pd). This reaction usually happens at temperatures between 150°C and 200°C.
\( \ce{CH2=CH2} + \ce{H2} \xrightarrow{\text{Ni, 150°-200°C}} \ce{CH3-CH3} \)
(Ethene) (Ethane)
2. **From alkyl halides (Wurtz reaction):** When an alkyl halide (a compound with a halogen atom attached to an alkyl group) is treated with sodium metal in the presence of dry ether, an alkane is formed. This reaction is known as the Wurtz reaction.
\( \ce{2CH3Br} + \ce{2Na} \xrightarrow{\text{dry ether}} \ce{CH3-CH3} + \ce{2NaBr} \)
(Methyl bromide) (Ethane)
3. **By Clemmensen reduction:** When an aldehyde or ketone (organic compounds with a carbonyl group) is heated with 40% aqueous hydrochloric acid, amalgamated zinc (zinc treated with mercury), and an organic solvent like toluene, it is reduced to an alkane.
\( \ce{CH3-CO-CH3} \xrightarrow{\text{Zn-Hg/HCl, Toluene}} \ce{CH3-CH2-CH3} \)
(Acetone) (Propane)
In simple words: Alkanes are simple carbon-hydrogen chains with only single bonds. Their formula is \( \ce{C_nH_{2n+2}} \). They can be made by adding hydrogen to compounds with double or triple bonds, by making alkyl halides react with sodium, or by reducing aldehydes and ketones.

🎯 Exam Tip: When defining alkanes, always include "saturated" and "single bonds," and for preparation methods, list the key reagents and conditions alongside the balanced chemical equations.

Question 34. Write the common name, IUPAC name and formula of isomers of butane, pentane and hexane.
Answer:
Here are the isomers of butane, pentane, and hexane with their common and IUPAC names:

🎯 Exam Tip: When writing isomers, ensure you cover all possible structural arrangements and correctly apply IUPAC nomenclature rules for each.

Question 36. Explain with chemical equation what happens when ?
1. Dry sodium acetate is heated with soda lime ?
2. Reaction of sodium with methyl iodine in dry ether solution?
3. Electrolysis of concentrated aqueous solution of potassium acetate?
4. Aluminium carbide is treated with water?
Answer:
2. When methyl iodide reacts with sodium in dry ether, ethane is formed. This is a Wurtz reaction.
\( \ce{2CH3I} + \ce{2Na} \xrightarrow{\text{dry ether}} \ce{CH3-CH3} + \ce{2NaI} \)
3. When a concentrated aqueous solution of potassium acetate is electrolyzed (Kolbe's electrolysis), ethane is produced at the anode.
\( \ce{2CH3COOK(aq)} \rightleftharpoons \ce{2CH3COO-} + \ce{2K+} \)
At anode:
\( \ce{2CH3COO-} \rightarrow \ce{2CH3COO \cdot} + \ce{2e-} \)
\( \ce{2CH3COO \cdot} \rightarrow \ce{CH3-CH3} + \ce{2CO2} \)
At cathode:
\( \ce{H+} + \ce{e-} \rightarrow \ce{H \cdot} \)
\( \ce{H \cdot} + \ce{H \cdot} \rightarrow \ce{H2} \)
4. When aluminium carbide is treated with water, it produces aluminium hydroxide and methane gas.
\( \ce{Al4C3} + \ce{12H2O} \rightarrow \ce{4Al(OH)3} + \ce{3CH4} \)
In simple words: Different chemical reactions can create new substances. For example, some reactions create ethane gas, while others can produce methane gas or aluminium hydroxide, depending on what chemicals are mixed and how.

🎯 Exam Tip: For each reaction, ensure you write a balanced chemical equation and clearly state the products and any special conditions, like "dry ether" or "electrolysis".