Questions: Hydrocarbons

Q1. Give the general formulas of alkanes, alkenes and alkynes and one example of each.

Answer: Alkanes have the general formula CnH2n+2 and are saturated; methane is an example. Alkenes have the formula CnH2n when they contain one double bond; ethene is an example. Alkynes have the formula CnH2n-2 when they contain one triple bond; ethyne is an example.

Hinglish: Alkane CnH2n+2 (methane), alkene CnH2n (ethene) aur alkyne CnH2n-2 (ethyne) hote hain.

Q2. Why is methane tetrahedral and what is its H-C-H bond angle?

Answer: The carbon atom of methane is sp3 hybridised, so its four hybrid orbitals point towards the corners of a tetrahedron. The four hydrogen atoms therefore sit at the corners of a regular tetrahedron around the carbon, and the H-C-H bond angle is about 109.5 degrees. All four carbon-hydrogen bonds are equal in length.

Hinglish: Methane ka carbon sp3 hybridised hai, isliye chaar hydrogen atoms tetrahedron ke corners par hote hain aur H-C-H angle 109.5° hota hai.

Q3. Which conformation of ethane is more stable and why?

Answer: The staggered form lies at lower energy than the eclipsed form. In it the hydrogens on the two carbons sit as far apart as the geometry allows, so the repulsion between the bond electron clouds is smaller. The eclipsed form has the hydrogens close together and therefore has slightly higher energy, and the energy barrier between the two forms is small, so rotation is fast but not completely free.

Hinglish: Staggered conformation zyada stable hai kyunki H atoms door hote hain aur electron clouds ka repulsion kam hota hai; eclipsed mein H paas hote hain isliye energy zyada hoti hai, aur rotation puri tarah free nahi hoti.

Q4. Outline the free radical mechanism of the chlorination of methane.

Answer: In initiation, ultraviolet light breaks a chlorine molecule into two chlorine radicals. In propagation, a chlorine radical takes a hydrogen atom from methane to give hydrogen chloride and a methyl radical, and the methyl radical reacts with a chlorine molecule to give chloromethane and a new chlorine radical. In termination, the radicals combine with one another, for example two methyl radicals give ethane or a methyl and a chlorine radical give chloromethane, and the chain stops.

Hinglish: Initiation mein light se Cl2 do Cl radicals banata hai; propagation mein Cl radical methane se H le kar HCl aur methyl radical deta hai, jo Cl2 se milkar chloromethane aur naya Cl radical banata hai; termination mein radicals aapas mein combine hokar chain rok dete hain.

Q5. What is the Markovnikov rule and why does it hold?

Answer: When a hydrogen halide adds to an unsymmetrical alkene, the hydrogen adds to the carbon that already has more hydrogen atoms and the halogen adds to the carbon with fewer hydrogen atoms. This orientation is followed because the reaction goes through the more stable carbocation intermediate: a secondary or tertiary carbocation is more stable than a primary one, so the route that forms it is faster and gives the major product.

Hinglish: Markovnikov rule kehta hai ki H zyada hydrogen wale carbon par aur halogen kam hydrogen wale carbon par jata hai, kyunki reaction zyada stable secondary/tertiary carbocation ke through hoti hai.

Q6. What is the peroxide effect and which reagent shows it?

Answer: In the presence of a peroxide, the addition of hydrogen bromide to an unsymmetrical alkene gives the anti-Markovnikov product, with the bromine on the carbon that has more hydrogen atoms. The effect is also called the Kharasch effect, and it is observed only with HBr among the hydrogen halides. It works through a free radical chain that changes the orientation of the addition.

Hinglish: Peroxide ki presence mein HBr ka addition anti-Markovnikov product deta hai, ise Kharasch effect kehte hain, aur yeh sirf HBr ke saath hota hai kyunki free radical chain orientation badal deti hai.

Q7. What is ozonolysis and how is it used to locate a double bond?

Answer: Ozonolysis is the cleavage of a carbon-carbon double bond by ozone followed by decomposition of the ozonide with zinc and water. Each carbon of the double bond becomes a carbonyl carbon, so a carbon with two alkyl groups gives a ketone, a carbon with one alkyl group gives an aldehyde, and a terminal carbon gives formaldehyde. To locate the double bond in an unknown alkene, the structures of the carbonyl products are found and the two carbonyl carbons are joined by a double bond to rebuild the alkene.

Hinglish: Ozonolysis mein O3 double bond todta hai aur Zn/H2O ozonide decompose karta hai; har double-bond carbon carbonyl ban jata hai. Products se alkene wapas banane ke liye dono carbonyl carbons ko double bond se jodo - isse double bond ki position mil jati hai.

Q8. Why is the hydrogen of a terminal alkyne acidic?

Answer: The carbon of a terminal triple bond is sp hybridised, and an sp carbon is more electronegative than an sp3 or sp2 carbon. The C-H bond is therefore more polarised, and the hydrogen can be removed as a proton by a strong base or by sodium metal. The resulting acetylide ion is stable enough to form, and the reaction with ammoniacal silver nitrate or cuprous chloride gives a metal acetylide precipitate, which is the test for a terminal alkyne.

Hinglish: Terminal alkyne ka carbon sp hybridised aur zyada electronegative hota hai, isliye C-H bond polarised hota hai aur H proton ke roop mein nikal sakta hai; sodium se sodium acetylide aur ammoniacal AgNO3/Cu2Cl2 se precipitate milta hai jo terminal alkyne ka test hai.

Q9. What are the conditions for aromaticity according to the Huckel rule?

Answer: The molecule must have a planar ring, the ring must have complete conjugation so that every ring atom has a p orbital available, and the ring must contain 4n plus 2 pi electrons, where n is a whole number. Benzene satisfies all three conditions with six pi electrons and is aromatic; cyclooctatetraene has eight pi electrons and is not planar, so it is not aromatic.

Hinglish: Hückel rule ke teen conditions: planar ring, pura conjugation, aur (4n+2) π electrons. Benzene mein 6 π electrons hain isliye aromatic hai; cyclooctatetraene mein 8 hain aur planar nahi hai isliye aromatic nahi.

Q10. Why does benzene undergo substitution rather than addition?

Answer: Benzene is aromatic and gains extra stability from the delocalisation of its six pi electrons. An addition reaction would destroy the delocalised ring and lose that stabilisation, whereas a substitution reaction removes one hydrogen and puts the electrophile in its place while keeping the ring intact. The aromatic ring therefore prefers electrophilic substitution, and nitration, sulphonation, halogenation and Friedel-Crafts reactions are all of this type.

Hinglish: Benzene ka 6 π electron delocalisation use extra stability deta hai; addition se ring tootegi aur stability chali jayegi, isliye woh substitution karta hai jisme ring bani rehti hai. Nitration, sulphonation, halogenation aur Friedel-Crafts sab substitution reactions hain.

Quiz pointer. A five question chapter quiz follows this page. Attempt it without looking back and read the explanation for every question you miss, because the quiz retests the formulas, the named reactions and the aromatic substitution rules.