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

Option A (Twelve sigma-bonds and three pi-bonds) is correct because benzene has 12 sigma bonds (6 C-C and 6 C-H) and 3 pi bonds within the ring.

Option A ($-\mathrm{COOH},\mathrm{\;} -\mathrm{S}{\mathrm{O}_3}\mathrm{H},\mathrm{\;} -\mathrm{CON}{\mathrm{H}_2}, -\mathrm{CHO}$) is correct because it represents the correct decreasing order of priority for functional groups in IUPAC nomenclature.

Option A ($III < I < IV < II$) is correct because the stability of alkenes increases with the number of alkyl substituents on the double-bonded carbons (more substituted alkenes are more stable).

The correct answer is Nitrogen, sulphur and halogens because Lassaigne's test is specifically designed to detect these elements in organic compounds.

Option B is correct because carbocations are electron-deficient and can undergo rearrangements to form more stable carbocations.

Option B (IV > III > I > II) is correct because the leaving group ability depends on the stability of the leaving group after departure, with stronger acids being better leaving groups.

Option A is correct because conformers are isomers that are interconverted through rotation around a single bond. This rotation allows for different spatial arrangements of the atoms.

Option D (Acetic acid, methane) is correct because Kolbe synthesized acetic acid and Berthelot synthesized methane from inorganic sources.

Option C (Both of these takes place) is correct because homolytic bond fission of CH3Cl results in a methyl radical (CH3•) which is sp2 hybridized and planar.

Option D is correct because a racemic mixture is an equimolar mixture of two enantiomers (optical isomers).

Option C (9) is correct because there are 9 carbon atoms in Aspartame that are $sp^2$ hybridized, meaning they are involved in a double bond or are part of an aromatic ring.

Option B (Lewis acids) is correct because electrophiles are electron-deficient species that accept electron pairs, which is the definition of a Lewis acid.

Option D is the wrongly named compound because the correct IUPAC name should be 4-hexen-2-one.

The correct answer is Distillation under reduced pressure because glycerol has a high boiling point and can decompose at high temperatures, so reducing the pressure lowers the boiling point and prevents decomposition.

Option C (2-methyl butanoic acid) is correct because it has a chiral carbon atom, which is a carbon atom bonded to four different groups, allowing for stereoisomerism.

Option C (Positional isomers) is correct because ${\mathrm{C}_2}{\mathrm{H}_4}\mathrm{B}{\mathrm{r}_2}$ can exist as 1,1-dibromoethane or 1,2-dibromoethane, which are positional isomers.

Option C (Acetonitrile) is correct because Acetonitrile does not have an $sp^2$ hybridized carbon atom.

Option C (Dibromido ethylene diamine bis (triphenyl phosphine) nickel (II)) is correct because it follows the IUPAC naming conventions for coordination complexes, listing ligands alphabetically and indicating the oxidation state of the metal.

Option C is correct because optical activity is the property of a chiral substance to rotate the plane of polarized light.

Option A (15) is correct because the sum of sigma and pi bonds in Hex-1,3-dien-5-yne is 15.

Option C (CH3CHClCH2CH3) is correct because it has a chiral carbon atom (a carbon atom bonded to four different groups).

Option B (n-butane < ethoxyethane < 1-propanol < 1-butanol) is correct because boiling point increases with molecular weight and hydrogen bonding.

Option C (Both (a) and (b)) is correct because both lead acetate and sodium nitroprusside can be used to detect sulfur in the sodium extract.

The correct answer is Vaccum distillation because it allows for the purification of glycerine at lower temperatures, preventing decomposition.

Option D is correct because a tert-butyl group is a butyl group with three methyl groups attached to the same carbon atom, which is connected to the main chain.

Option D ($\overline O - C{H_2}C{H_2}N{H_2}$) is the most basic because the negatively charged oxygen atom donates electron density through the inductive effect, increasing the electron density on the nitrogen atom and making it more available for protonation.

Option C ($\mathrm{F}_2 \mathrm{CHCOO}^{-}$) is the most stable because the two fluorine atoms, being highly electronegative, exert a strong electron-withdrawing inductive effect (-I effect) that stabilizes the negative charge on the carboxylate ion.

Option D is correct because some organic compounds decompose or degrade before reaching their boiling points at normal pressure, necessitating distillation at lower pressures.

Option B is correct because a racemic mixture is a mixture of d- and its l-isomers in equal proportions. This results in no net rotation of plane-polarized light.

Option D (In all states) is correct because optical activity is a property of the molecule itself, and it can be observed in solution, liquid, and crystalline states as long as the chiral center is maintained.