Organic chemistry-some basic principles and techniques practice · Mixed difficulty

Option A ($19:2$) is correct because the ratio of the percentage of carbon and hydrogen is calculated based on the masses of CO2 and H2O produced.

Option B ($82.7$) is correct because the percentage of carbon is calculated based on the mass of water produced during combustion.

Option C ($\mathop C\limits^\Theta {H_2} - N{O_2}$) is correct because the nitro group ($\mathrm{NO_2}$) is a strong electron-withdrawing group that stabilizes the carbanion through resonance.

Option A ($C{H_3}C{H_2}CH(Cl)C{O_2}H$) is the stronger acid because the chlorine atom, being electronegative, exerts an electron-withdrawing inductive effect (-I effect) that stabilizes the conjugate base, thus increasing the acidity of the carboxylic acid.

Option C (2 and 4) is correct because $\mathrm{C}{\mathrm{H}_3}{\left( {\mathrm{CHOH}} \right)_2}\mathrm{COOH}$ has two asymmetric carbon atoms, leading to 2² = 4 stereoisomers.

Option A (Isopropyl propyl ether) is correct because the ether linkage connects an isopropyl group and a propyl group.

Option B ($E < D < C < A < B$) is correct because carbocation stability increases with the number of alkyl groups attached to the positively charged carbon due to hyperconjugation and inductive effects.

Option C is correct because van't Hoff proposed the tetrahedral mirror image structures to a pair of enantiomers. This was a key development in understanding stereochemistry.

Option C ($\mathrm{FCH}_2 \mathrm{COOH}$) is correct because the presence of the highly electronegative fluorine atom increases the acidity of the carboxylic acid through the inductive effect, stabilizing the conjugate base.

Option D (${\mathrm{C}_3}{\mathrm{H}_7}\mathrm{Br}$) is correct because the mole ratio of C:H:Br is approximately 3:7:1, which corresponds to the empirical formula ${\mathrm{C}_3}{\mathrm{H}_7}\mathrm{Br}$.

Option D ($N{H_3}$) is correct because in Kjeldahl's method, the nitrogen present in the organic compound is quantitatively converted into ammonia.

The correct answer is Sublimation because anthraquinone can be purified by sublimation due to its ability to transition directly from solid to gas phase upon heating, leaving impurities behind.

Option B (Short lived) is correct because free radicals are highly reactive species due to the presence of an unpaired electron.

Option C is correct because the physical state of the pure compound does not affect the separation process in chromatography, which depends on differential adsorption or partition.

Option A ($\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_3$) is the most reactive because alkyl groups are electron-donating, increasing the electron density of the alkene, which favors electrophilic attack.

Option B ($PbS$) is correct because in Lassaigne's test, the black color indicates the presence of sulfur, which reacts with lead acetate to form lead sulfide ($PbS$).

Option B (-COCl) is correct because acyl halides (-COCl) have higher priority than nitriles (-CN), aldehydes (-CHO), and amines (-NH₂) according to IUPAC nomenclature rules.

Option D (The products they give on addition to bromine) is correct because the stereochemistry of the addition of bromine to $cis$ and $trans$ isomers differs, leading to different products.

Option C ($\mathrm{S}^{2-}$) is correct because sulfide ions react with dichromate to form a green solution and with lead acetate to form a black precipitate.

Option D (Mutarotation) is correct because mutarotation is the change in optical rotation observed when freshly prepared solutions of certain sugars are allowed to stand.

Option D (The double bond goes mainly towards the least substituted carbon) is correct because Hofmann's rule states that in E2 elimination, the major product is the least substituted alkene.

Option C (Canonical structures) is correct because canonical structures are theoretical representations and do not exist as real structures.

Option B (4) is correct because the given compound has 4 meso forms due to the presence of multiple chiral centers and a plane of symmetry.

Option B (2) is the correct answer because tartaric acid has two chiral carbons, leading to three stereoisomers, but only two are optically active enantiomers.

Option D (All of these) is correct because Kjeldahl's method is not suitable for compounds where nitrogen is directly bonded to oxygen (nitro compounds), involved in rings (pyridine), or present as azo groups, as these forms of nitrogen are not easily converted to ammonium sulfate.

Option B (Conformers) is correct because conformers are isomers that interconvert through rotation around single bonds.

Option A (Dextrorotatory) is correct because dextrorotatory compounds rotate plane-polarized light in a clockwise direction.

Option B (Phase distribution) is correct because chromatography separates components based on their differential distribution between a stationary and a mobile phase.

Option B is correct because naphthalene and benzoic acid have different solubilities in suitable solvents at different temperatures, allowing separation by crystallization.

Option D ($II > I > IV > III$) is correct because the -I effect order is $N{O_2} > CN > F > N{H_2}$ based on electronegativity and other factors.