Organic chemistry-some basic principles and techniques practice · Medium

Option B (Carbonium ion) is correct because removing a hydride ion (H-) from methane (CH4) leaves a positively charged carbon species, which is a carbonium ion (CH3+).

Option A (Statement I is true but statement II is false) is correct because ethanol and AgCN can generate nucleophiles, but KCN and AgCN do not always generate nitrile nucleophiles under all conditions.

Option D ($NO_2^ - \,\,\,\,\mathrm{and}\,\,\,\,NH_2^ -$) is correct because both ions have a bent geometry and are $sp^2$ hybridized.

Option D (Either of the above) is correct because during elimination reactions, a carbon atom can change from sp3 to sp2 (forming a double bond) or from sp2 to sp (forming a triple bond).

Option C (They yield the same products on complete combustion) is correct because isomers have the same molecular formula, thus producing the same products upon complete combustion.

Option D ($\mathrm{N}{\mathrm{a}_2}\mathrm{S}$) is correct because in the Lassaigne's test, sulfur present in the organic compound is converted to sodium sulfide ($\mathrm{N}{\mathrm{a}_2}\mathrm{S}$).

Option A is correct because the basicity order follows the stability of the conjugate acid, with $CH_3CH_2^-$ being the strongest base and $OH^-$ the weakest.

Option A (2) is correct because pent-3-en-2-ol has one chiral center, resulting in two optical isomers (enantiomers).

Option C is correct because Kolbe's electrolysis involves the formation of free radicals at the anode, which then dimerize to form the alkane product.

The correct answer is A (${C_6}{H_5} - \mathop C\limits_{\mathop {||}\limits_O } - H$) because keto-enol tautomerism requires the presence of an alpha-hydrogen atom, which is not present in aldehydes directly attached to aromatic rings.

Option A (3 – chloropropanoic acid) is correct because it correctly identifies the parent chain as propanoic acid and the chlorine substituent at the 3rd position.

Option C ($\mathrm{C}{\mathrm{H}_2}$) is correct because the empirical formula is determined by calculating the mole ratio of carbon and hydrogen.

Option B (Ethyl-2-butenoate) is correct because it correctly identifies the ester functional group and the position of the double bond.

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 ($NH_4^+$) 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 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 D is correct because aniline is steam volatile and can be separated from non-volatile impurities by steam distillation.

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

Option A (Four) is correct because the compound has two double bonds, each capable of exhibiting cis/trans isomerism, leading to a total of four geometrical isomers.