Hydrocarbons practice · Mixed difficulty

Option C (Glucose) is correct because acetylene is not directly used in the production of glucose.

Option B (2-chloropropane) is correct because in the absence of peroxide, HCl addition to propene follows Markovnikov's rule, where the Cl adds to the more substituted carbon.

Option B (Solid hydrocarbons) is correct because Ozokerite is a naturally occurring mineral wax composed of various solid hydrocarbons.

Option A (Ethane) is correct because it has a single bond between the carbon atoms, allowing for free rotation.

Option B (2 – methyl – 2 – butene) is correct because ozonolysis of 2-methyl-2-butene yields acetone and another aldehyde.

Option B (it is a reversible reaction) is correct because iodination of alkanes is a reversible reaction, and the HI formed can react with the iodinated alkane to regenerate the alkane and iodine. Oxidizing agents are used to consume the HI and drive the reaction forward.

Option D ($-\mathrm{NHR}$) is correct because alkylamino groups are stronger activating and directing groups than amino groups due to the electron-donating effect of the alkyl group.

Option C (It has unpleasant garlic odour) is correct because pure acetylene has an ethereal odor, not a garlic odor.

Option B (1.08) is correct because the C-H bond length in benzene is approximately 1.08 $\mathop \mathrm{A}\limits^\mathrm{o}$.

Option B (Wurtz reaction) is correct because the reaction of alkyl halides with sodium in dry ether to form alkanes with double the number of carbon atoms is known as the Wurtz reaction.

Option D (${NaCl}$) is correct because it is not a Lewis acid and cannot generate the electrophile required for Friedel-Crafts reactions.

Option C is correct because while the assertion is true (neopentane forms only one monosubstituted product), the reason is false (all 12 hydrogens are equivalent).

Option C (Gasoline) is correct because gasoline has the lowest boiling point among the given fractions of petroleum.

Option B (4) is correct because the formula for a saturated hydrocarbon (alkane) is ${\mathrm{C}_n}{\mathrm{H}_{2n + 2}}$, and solving $2n + 2 = 10$ gives $n = 4$.

Option B is correct because neopentane has only one type of hydrogen atom (all are primary), leading to only one possible monochloro derivative upon halogenation.

Option B is correct because the complete combustion of hydrocarbons like benzene in the presence of sufficient oxygen produces carbon dioxide and water.

Option A (1.5) is correct because the delocalization of pi electrons in benzene results in a bond order between a single and double bond.

Option C (Head on and sideways) is correct because sigma bonds are formed by head-on overlapping of atomic orbitals, while pi bonds are formed by sideways or lateral overlapping.

Option D (Base) is correct because in the nitration of benzene, nitric acid accepts a proton from sulfuric acid, acting as a base to generate the nitronium ion.

Option A (S-I & S-II are true) is correct because both benzene and ethyne give glyoxal on ozonolysis, and they have the same empirical formula (CH).

Option C is correct because the reaction with Zn/Alcohol yields an alkene, not an alkyne, while the other reactions can lead to alkyne formation.

Option D (All are correct) is correct because all the statements about alkanes are factually accurate.

Option D (Ammonical Silver nitrate) is correct because it reacts specifically with terminal alkynes due to the presence of acidic hydrogen atoms.

Option A is correct because heavy metal acetylides can be used to purify alkynes, and terminal alkynes form acetylides soluble in acids.

Option D is correct because n-butane and isobutane have different boiling points due to differences in their molecular structure and carbon chain arrangement, affecting intermolecular forces.

Option B (0) is correct because Sc3+ has no unpaired electrons, resulting in a zero magnetic moment.

Option B is correct because acetylene is the most acidic and has the shortest C-H bond distance due to the sp hybridization of the carbon atom.

Option D (pyridine) is correct because pyridine is a deactivated aromatic ring due to the nitrogen atom, and it forms a complex with the Lewis acid catalyst ($\mathrm{AlCl_3}$), preventing Friedel-Crafts reaction.

Option A (n - pentane) is correct because n-pentane yields only three mono chlorinated derivatives due to its symmetry.