HydrocarbonsNEET Chemistry · Class 11 · NCERT Chapter 9

Solution

The compound ${\text{C}}_7{\text{H}}_8$ is toluene. In the given reactions, toluene undergoes bromination, and the bromine atom typically substitutes at the para position due to the electron-donating effect of the methyl group, making the para position more reactive. Therefore, the product 'C' is p-bromotoluene, so option (d) is correct.

Solution

Pent-2-en-4-yne has the structure ${\text{CH}}_3-\text{CH}=\text{CH}-\text{C}\equiv \text{CH}$. This structure contains 10 sigma ($\sigma$) bonds and 3 pi ($\pi$) bonds, so option (b) is correct.

Solution

The Wurtz reaction involves the coupling of alkyl halides with sodium metal in ether to form higher alkanes. However, 2,3-dimethylbutane cannot be synthesized in good yield because the reaction does not easily form branched alkanes with complex substitution patterns. Therefore, option (b) is correct.

Solution

• Planar, cyclic, conjugated species containing (4n + 2)π electrons will be aromatic in nature (n is an integer) • , and are aromatic species • is not an aromatic compound

Solution

Both statement I and statement II are correct. Boiling point of n-pentane = 309 K isopentane = 301 K neopentane = 282.5 As branching increases molecules attain the shape of a sphere results in smaller area of contact thus weak intermolecular forces between spherical molecules, which are overcome at relatively lower temperature. Leading to decrease in boiling point. - 30 - NEET (UG)-2024 (Code-Q1)

Solution

(1) Decarboxylation of sodium benzoate by sodalime (NaOH + CaO) gives benzene: PhCOONa → PhH + Na₂CO₃.

(2) n-hexane over Mo₂O₃ catalyst at 773 K, 10–20 atm undergoes aromatisation (catalytic reforming) to benzene.

(3) Cyclic trimerisation of acetylene (ethyne) over red-hot iron tube at 873 K gives benzene.

(4) Benzenediazonium chloride hydrolysed in warm water gives PHENOL (not benzene): C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl.

Hence option (4) does NOT give benzene as the product.

Solution

The four monochlorination sites on isopentane (2-methylbutane) give four different mono-chloro products, and two of them have a chiral carbon (so they each give a pair of enantiomers).

The four substitution products with their isomer counts are shown below; total = 1 + 2 + 1 + 2 = 6.

Solution

The hybridisation of the carbon bearing the marked H is different in each compound:

• I (benzene ring C–H): the carbon is sp² hybridised
• II (terminal alkyne ≡C–H): the carbon is sp hybridised
• III (cyclopropene CH₂): the carbon is sp³ hybridised

Higher the percentage s-character of the carbon, shorter and stronger is the C–H bond. Hence the bond dissociation energies follow the order of s-character: sp (50%) > sp² (33%) > sp³ (25%).

Correct order of bond dissociation energy of C–H bond:

II > I > III