Hydrocarbons
ICSE · Class 11 · Chemistry
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Which of the following is the correct IUPAC name for the compound CH₃-CH(CH₃)-CH₂-CH₃?
In the Wurtz reaction, when ethyl iodide is treated with sodium in dry ether, the product obtained is:
Which type of hybridisation is present in the carbon atoms involved in the C≡C triple bond of ethyne?
According to Markownikoff's rule, when HBr is added to propene (CH₃-CH=CH₂) in the absence of peroxides, the major product is:
Sample Questions
When sodium benzoate (C₆H₅COONa) is heated with sodalime (NaOH + CaO), the product obtained is:
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Benzene
Step 1: This reaction is called decarboxylation — removal of CO₂ from a carboxyl group. Step 2: The reaction is: C₆H₅COONa + NaOH (sodalime) → C₆H₆ + Na₂CO₃. Step 3: The sodium benzoate reacts with NaOH. The carboxylate group (-COO⁻) loses CO₂, and the benzene ring acquires a hydrogen. Step 4: The product C₆H₆ is benzene. CaO in sodalime helps in fusion of the mixture but does not react. Toluene (C₇H₈) would require a longer chain. Phenol contains -OH group. Benzoic acid is the parent acid before salt formation.
Baeyer's test is used to detect unsaturation in organic compounds. The reagent used in Baeyer's test is:
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Cold dilute alkaline KMnO₄ solution
Step 1: Baeyer's reagent is a cold, dilute (1%) alkaline solution of potassium permanganate (KMnO₄). Step 2: When an alkene or alkyne is added, the purple colour of KMnO₄ gets decolourised — this indicates the presence of a C=C or C≡C bond. Step 3: The reaction involves hydroxylation: the alkene is oxidised to a 1,2-diol (glycol), and KMnO₄ is reduced to MnO₂. Step 4: Hot concentrated KMnO₄ causes oxidative cleavage, not the Baeyer's test. Bromine water (brown to colourless) is a different test for unsaturation. Ozone is used in ozonolysis to determine the position of double bonds.
Which of the following pairs of compounds are geometrical isomers of each other?
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cis-but-2-ene and trans-but-2-ene
Step 1: Geometrical isomers have the same molecular formula and connectivity but differ in the spatial arrangement of groups around a double bond. Step 2: For geometrical isomerism, a double bond must be present AND each carbon of the double bond must carry two different groups. Step 3: cis-but-2-ene (CH₃ groups on the same side) and trans-but-2-ene (CH₃ groups on opposite sides) satisfy this condition. Step 4: n-butane and isobutane are chain isomers (structural isomers). But-1-ene and but-2-ene are position isomers. Ethyne and but-2-yne are different compounds entirely (homologues). Only opt
During the free radical halogenation of methane with Cl₂, which step involves the breaking of the Cl-Cl bond to form chlorine free radicals?
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Initiation step
Step 1: The free radical chlorination of methane occurs in three steps: initiation, propagation, and termination. Step 2: In the initiation step, the Cl-Cl bond is broken homolytically by UV light or heat (520-670 K): Cl₂ → 2Cl• (two chlorine free radicals). Step 3: In the propagation step, Cl• attacks CH₄ to form CH₃• and HCl, and then CH₃• attacks Cl₂ to form CH₃Cl and another Cl•. These steps repeat in a chain. Step 4: In the termination step, two free radicals combine to stop the chain: Cl• + Cl• → Cl₂, or CH₃• + Cl• → CH₃Cl. Chain transfer is not a standard step in this mechanism.
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