Hydrocarbons practice · Easy

Option A (2,2-dichloropropane) is correct because the addition of HCl to propyne follows Markovnikov's rule, leading to the formation of 2-chloropropene, which then reacts with another molecule of HCl to give 2,2-dichloropropane.

Option D (All of these) is correct because 1-butene, 2-hexene, and 1-pentene all have different groups attached to the carbon atoms of the double bond, making them unsymmetrical.

Option B ($\mathrm{C}{\mathrm{H}_3}\mathrm{C}{\mathrm{H}_2}\mathrm{CH} = \mathrm{CHC}{\mathrm{H}_2}\mathrm{C}{\mathrm{H}_3}$) is correct because it adds one mole of hydrogen and gives two moles of the same carboxylic acid upon reaction with $\mathrm{KMn}{\mathrm{O}_4}$.

Option A (dipole – dipole interaction) is correct because alkyl halides are polar molecules due to the electronegativity difference between the halogen and carbon atoms, leading to dipole-dipole interactions.

Option D (All of these) is correct because Phosgene, Lewisite, and Mustard gas are all well-known poisonous gases.

Option C ($C{H_4}$) is correct because the Wurtz reaction involves the coupling of two alkyl halides to form a higher alkane, and methane cannot be formed by coupling two smaller alkyl groups.

Option A (${\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_4}pb$) is correct because Tetraethyllead (TEL), ${\left( {{\mathrm{C}_2}{\mathrm{H}_5}} \right)_4}pb$, was historically used as an antiknock agent in gasoline.

Option A (deactivates the ring towards electrophilic substitution) is correct because the nitro group is an electron-withdrawing group that deactivates the benzene ring.

The correct answer is A is correct but R is incorrect because acetylene reacts with sodamide to form sodium acetylide and ammonia, and acetylene is a stronger acid than ammonia.

Option A ($C_2H_4$, alc.KOH/$\Delta$) is correct because ethene reacts with HBr to form ethyl bromide, and ethyl bromide reacts with alcoholic KOH to form ethene.

Option C (Free radical substitution) is correct because halogenation of alkanes proceeds via a free radical mechanism.

Option A (2 – methyl – 2 - butene) is correct because ozonolysis of 2-methyl-2-butene yields one mole of acetaldehyde and one mole of acetone.

The correct answer is Butyne-2 because it decolourises KMnO4 (unsaturation) but does not have a terminal alkyne (no precipitate with ammoniacal silver nitrate).

Option A (${180^0}$ bond angle at the carbon – carbon triple bond) is correct because the linear geometry around the triple-bonded carbon atoms in alkynes prevents cis-trans isomerism.

Option A is correct because propyne is indeed less acidic than ethyne due to the electron-donating methyl group, which destabilizes the conjugate base. The stability of the acetylide ion is the direct reason for the difference in acidity.

Option B ($Pd/BaS{O_4}$ or $CaC{O_3}$ in quinoline or xylene) is correct because Lindlar's catalyst is palladium supported on barium sulfate or calcium carbonate, poisoned with quinoline or xylene.

Option A (infinite) is correct because alkanes can rotate around their carbon-carbon single bonds, leading to an infinite number of conformations.

Option C (Vegetable oils) is correct because vegetable oils are triglycerides, which are esters of glycerol and fatty acids, not mixtures of hydrocarbons.

Option B (${\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathrm{CHC}{\mathrm{H}_3}$) is correct because Wurtz reaction can produce symmetrical alkanes in good yield, but unsymmetrical alkanes are difficult to prepare singly due to side products.

Option B (Ethane) is correct because the decarboxylation of sodium propionate with soda-lime removes a carbon atom, resulting in ethane.

Option C (2-bromobutane) is the major product because, according to Markovnikov's rule, the bromine atom adds to the carbon with more alkyl substituents.

Option B (${C_2}{H_6} > {C_2}{H_4} > {C_6}{H_6} > {C_2}{H_2}$) is correct because bond length decreases with increasing bond order: single bond > benzene (intermediate) > double bond > triple bond.

The correct answer is 2-chloropropane because in the presence of peroxide, HCl undergoes Markovnikov addition to propene.

Option C (${C_6}{H_6}C{l_6}$) is correct because it represents the molecular formula of Gammaxane, also known as benzene hexachloride (BHC).

Option C (3, 5) is correct because introducing two substituents at the meta positions (3,5) with respect to the methyl group on toluene is difficult due to the directing effect of the methyl group.

3,5-dibromotoluene is most difficult to make from toluene because the methyl group is ortho/para directing, and it is difficult to direct two bromines to the meta positions.

Option D (${\mathrm{C}_2}{\mathrm{H}_4}$) is correct because the gas formed is ethene, which has a double bond and can decolourise alkaline $\mathrm{KMn}{\mathrm{O}_4}$.

Option A (Cracking) is correct because thermal decomposition of alkanes in the absence of air is specifically called cracking, breaking larger molecules into smaller ones.

The correct answer is Acetylene molecule has three ‘pi’ $\left( \pi \right)$ bonds and three sigma bonds because acetylene has 3 sigma and 2 pi bonds.

Option C (Bees wax) is correct because bees wax is a natural wax produced by honeybees, while the others are derived from petroleum.