Grade 11 : Chemistry - Hydrocarbons (Set 4) | Concept of Maths

Explanation:As branching increases, the molecule becomes more compact and spherical. This decreases the effective surface area of contact between molecules, weakening intermolecular van der Waals dispersion forces and lowering the boiling point.

Explanation:Wurtz reaction of pure $C_2H_5Br$ yields symmetric n-butane ($2 C_2H_5Br + 2Na \rightarrow C_4H_{10} + 2NaBr$). A mixture of $CH_3Br$ and $C_2H_5Br$ produces three different alkanes: ethane ($CH_3-CH_3$), propane ($CH_3-C_2H_5$), and butane ($C_2H_5-C_2H_5$).

Explanation:Conformer A ($\theta = 60^\circ$) is the staggered conformation, where $C-H$ bonds are furthest apart, minimizing electron pair repulsion (minimum torsional strain). Conformer B ($\theta = 0^\circ$) is the eclipsed conformation, where $C-H$ bonds align, resulting in maximum torsional strain.

Explanation:To find the alkene structure, join the carbonyl carbon atoms with a double bond after removing oxygen atoms: $(CH_3)_2C=O + O=CH_2 \rightarrow (CH_3)_2C=CH_2$. The IUPAC name for $(CH_3)_2C=CH_2$ is 2-methylprop-1-ene.

Explanation:Reaction I follows Markownikoff's rule via a stable $2^\circ$ carbocation to yield 2-bromopropane. Reaction II follows Kharasch (peroxide) effect via free radical addition to yield anti-Markownikoff product, 1-bromopropane.

Explanation:At the anode, acetate ions undergo oxidation and decarboxylation: $2CH_3COO^- \rightarrow C_2H_6 + 2CO_2 + 2e^-$. At the cathode, water reduction produces hydrogen gas: $2H_2O + 2e^- \rightarrow H_2 + 2OH^-$. Thus, Gas A is Ethane and Gas B is Hydrogen.

Explanation:2-methylpropane has 9 primary ($1^\circ$) hydrogens and 1 tertiary ($3^\circ$) hydrogen.
Relative rate for $1^\circ$ product $= 9 \times 1 = 9$.
Relative rate for $3^\circ$ product $= 1 \times 5 = 5$.
Total relative rate $= 9 + 5 = 14$.
Yield of 1-chloro derivative $= \frac{9}{14} \times 100% \approx 64.3%$.

Explanation:Combustion equation: $C_xH_y + (x + \frac{y}{4}) O_2 \rightarrow x CO_2 + \frac{y}{2} H_2O$.
Volume of $CO_2 = x \times 10 = 30 \implies x = 3$.
Volume of $O_2 = (x + \frac{y}{4}) \times 10 = 50 \implies 3 + \frac{y}{4} = 5 \implies \frac{y}{4} = 2 \implies y = 8$.
Formula: $C_3H_8$.

Explanation:Higher s-character increases electronegativity of the carbon atom, facilitating easier release of hydrogen as $H^+$. Thus, acidic strength increases with s-character: Ethane (25%) < Ethene (33.3%) < Ethyne (50%).

Explanation:In the Anti conformation ($\theta = 180^\circ$), the bulky methyl groups are at maximum separation, resulting in minimal steric hindrance and zero torsional strain, making it the lowest potential energy conformer.

Explanation:Hückel's rule requires $(4n+2)\pi$ delocalized electrons ($n=0, 1, 2...$ giving 2, 6, 10, 14 $\pi$-electrons). Cyclooctatetraene has 8 $\pi$-electrons ($4n$ electrons where $n=2$), failing Hückel's rule and making it non-aromatic (tub-shaped).

Explanation:Concentrated $H_2SO_4$ protonates $HNO_3$, leading to loss of water and generation of the nitronium electrophile: $HNO_3 + 2H_2SO_4 \rightarrow NO_2^+ + H_3O^+ + 2HSO_4^-$. Thus, $H_2SO_4$ acts as a stronger acid to form $NO_2^+$.

Explanation:Anhydrous $AlCl_3$ generates the primary carbocation $CH_3-CH_2-CH_2^+$, which rapidly rearranges via a 1,2-hydride shift to form the more stable secondary carbocation $(CH_3)_2CH^+$, reacting with benzene to give isopropylbenzene.

Explanation:Ethyne (terminal alkyne) decolorizes $Br_2$ water due to unsaturation and forms a white precipitate of silver acetylide ($Ag-C\equiv C-Ag$) with ammoniacal $AgNO_3$. Ethene decolorizes $Br_2$ but lacks terminal acidic hydrogen so it doesn't react with $AgNO_3$. Ethane is saturated and unreactive to both.

Explanation:Markovnikov addition of $H_2O$ across ethyne produces vinyl alcohol ($CH_2=CH-OH$), which rapidly undergoes keto-enol tautomerism to reorganize into stable ethanal ($CH_3CHO$).

Explanation:Trans-but-2-ene is more stable than cis-but-2-ene due to reduced steric crowding between methyl groups. More stable alkenes lie at a lower energy state and release less heat upon hydrogenation, resulting in a lower magnitude of $\Delta H_{\text{hyd}}$.

Explanation:Passing ethyne through a red-hot iron tube at 873 K causes cyclic polymerization, where 3 molecules of ethyne ($C_2H_2$) trimerize to yield 1 molecule of benzene ($C_6H_6$).

Explanation:Group 1 substituents donate electron density via $+M$ or $+I$ effects, activating the ring and increasing electron density at ortho and para positions. Group 2 substituents withdraw electron density via $-M$ or $-I$ effects, deactivating the ring and leaving the meta position relatively more electron-rich.

Explanation:Vigorous oxidation of any alkylbenzene possessing at least one benzylic hydrogen atom with alkaline $KMnO_4$ cleaves the alkyl side chain entirely, converting it into benzoic acid ($C_6H_5COOH$).

Explanation:Benzene forms a triozonide intermediate ($C_6H_6O_9$). Reductive cleavage with $Zn/H_2O$ breaks all three double-bond sites, yielding 3 moles of ethanedial (glyoxal, $CHO-CHO$) per mole of benzene.