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

Option A (2, 6 – di methyl oct – 3 – yne) is correct because it follows IUPAC naming rules, correctly identifying the longest chain and substituents.

Option C ($\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{N}$) is correct because the carbon in the cyanide group is sp hybridized, not $sp^2$ hybridized.

Option A is correct because both the assertion and reason are true, and the reason correctly explains why trans-1-chloropropene has a higher dipole moment than cis-1-chloropropene. The vector sum of bond dipoles is greater in the trans isomer.

Option C (Both A and R are true and R is the correct explanation of A) is correct because iodide is a better nucleophile than water and also a good leaving group, facilitating the reaction via SN2 mechanism.

Option B (Silver spiral) is correct because a silver spiral is used to remove halogens by forming silver halides, preventing them from interfering with the estimation of carbon and hydrogen.

Option B (A free radical) is correct because free radicals have an unpaired electron, making them paramagnetic.

Option B (Steric strain) is correct because the fully eclipsed conformation of butane has significant steric strain due to the close proximity of the bulky methyl groups.

Option C (4) is correct because the compound has one chiral center and exhibits cis/trans isomerism due to the double bond, resulting in 2 * 2 = 4 stereoisomers.

Option A (low ${R_f}$, stronger adsorption) is correct because compounds that elute first in column chromatography have a lower affinity for the stationary phase, resulting in a lower ${R_f}$ value and stronger adsorption to the mobile phase.

Option A ($\overline C {H_2} - C{H_2} - N{O_2}$) is the most stable carbanion because the nitro group (NO2) is a strong electron-withdrawing group that stabilizes the negative charge on the carbon atom through both inductive and resonance effects.

Option C (${3^0} < {2^0} < {1^0}$) is correct because carbanion stability decreases with increasing alkyl substitution due to the +I effect of alkyl groups destabilizing the negative charge.

Option A (${6 4 0 0 0} \mathrm{~a m u}$) is correct because the molecular weight is calculated by dividing the total mass of iron by its percentage and multiplying by the number of iron atoms.

Option B (2) is correct because ${\mathrm{C}_3}{\mathrm{H}_6}$ can be either propene or cyclopropane, representing two structural isomers.

Option C ($NH_4^+$) is correct because it already has a complete octet and a positive charge, making it unlikely to accept electrons and act as an electrophile.

Option C (A>C>B) is correct because electron-donating groups increase reactivity towards electrophiles, while electron-withdrawing groups decrease reactivity. Chlorobenzene is activated compared to nitrochlorobenzenes.

Option D (diastereoisomers) is correct because reacting a racemic mixture with a chiral resolving agent forms diastereomers, which can then be separated due to their different physical properties.

Option D (Half chair) is correct because the half-chair conformation of cyclohexane has the highest energy due to maximum torsional and steric strain.

Option C ($\mathop \mathrm{C}\limits^\_ \mathrm{C}{\mathrm{l}_3} > {\mathrm{C}_6}{\mathrm{H}_5}\mathop \mathrm{C}\limits^\_ {\mathrm{H}_2} > {\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathop \mathrm{C}\limits^\_ \mathrm{H} > {\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathop \mathrm{C}\limits^\_$) is correct because the stability of carbanions increases with electron-withdrawing groups and decreases with electron-donating groups.

Option B is correct because the yellow precipitate formed in the phosphorus detection test is ammonium phosphomolybdate, which has the formula (NH₄)₃PO₄ · 12MoO₃.

Option D (Cyclopropane) is correct because it has the highest ring strain due to its bond angles deviating significantly from the ideal tetrahedral angle.

Option A (Pentyne-3) is correct because the numbering should start from the end closest to the triple bond, making it Pentyne-2.

Option A (80) is correct because the molecular mass of the base is calculated from the mass of chloroplatinate and the mass of platinum obtained after ignition.

Option A is correct because benzaldehyde acts as an adsorbate, alumina as an adsorbent, and acetonitrile as the dynamic (mobile) phase in chromatography.

Option D is correct because aniline is steam volatile and can be separated from non-volatile impurities by steam distillation.

Option C ($3^\circ < 2^\circ < 1^\circ$) is correct because primary alcohols are the most reactive with sodium metal due to less steric hindrance around the hydroxyl group, making it easier for sodium to abstract the proton.

Option A (a $\mathrm{C}{\mathrm{H}_2}$ group) is correct because homologous series members differ by a methylene group.

Option A (6) is correct because a branched alkane with 8 carbons can have a maximum of 6 methyl groups attached to the main chain.

Option C (conformation) is correct because conformational isomerism arises from the rotation around C-C single bonds.

Option A (Stereochemistry) is correct because stereochemistry is the study of the three-dimensional arrangement of atoms in molecules and their effect on chemical and physical properties.

Option D (1,3,5-cyclohexatriene) is correct because it has the most conjugated double bonds, leading to the greatest resonance stabilization.