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

Option A is correct because the acidity increases with the presence of electron-withdrawing groups closer to the carboxylic acid group, with alkynes being more electron-withdrawing than alkenes or alkyl groups, and alcohols being the least acidic.

Option D (Phenanthrene) is correct because it has the most stable resonance structures among the given options due to its angular structure, which reduces steric hindrance and allows for better electron delocalization.

Option C (Stereoisomers) is correct because stereoisomers are isomers that have the same connectivity but differ in their spatial arrangement, and mirror images are a type of stereoisomer.

Option C is correct because $\mathrm{CH}_3^ +$ has a central carbon atom with three sigma bonds and no lone pairs, resulting in sp2 hybridization.

Option A (Trihydric) is correct because glycerol has three hydroxyl (-OH) groups attached to the carbon chain.

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 B (${\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$) is correct because the carbocation formed after heating will be benzylic, which is stabilized by resonance with the benzene ring.

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 D (Ammonia) is correct because in Kjeldahl's method, the nitrogen in the organic compound is converted into ammonia.

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 C (electromeric effect) is a temporary effect that occurs only in the presence of an attacking reagent.

Option A ($C{H_2} = \mathrm{ }CH\mathrm{ }\mathop C\limits^ \bullet {H_2}$) is correct because the allylic free radical is stabilized by resonance.

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

Option B (${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{C}{\mathrm{O}_2}\mathrm{H},\mathrm{\;C}{\mathrm{H}_3}\mathrm{C}{\mathrm{O}_2}\mathrm{C}{\mathrm{H}_3}$) is correct because both compounds have the same molecular formula but different functional groups (carboxylic acid and ester).

Option A (But-3-ene-1-yne) is correct because the IUPAC rules prioritize numbering the carbon chain such that the alkene gets preference over alkyne when both are present and equidistant from the ends.

Option C (Cylinder) is correct because the two pi bonds in acetylene are perpendicular to each other, resulting in a cylindrical electron cloud around the sigma bond axis.

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 A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because tetracyanomethane has 4 sigma bonds and 4 pi bonds per cyano group, totaling 8 sigma and 8 pi bonds, making the ratio 1.

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 (2, 3 - dimethylbut - 2 - ene) is correct because it contains only primary hydrogen atoms.

Option A ($C{H_3}\mathop C\limits^ \bullet {H_2}\,\,\mathrm{and}\,\,\mathop C\limits^ \bullet l$) is correct because homolytic fission involves the equal sharing of electrons, resulting in free radicals.

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

Option A (-NO2) is correct because the nitro group exhibits both -I (inductive) and -M (mesomeric) effects, both of which are electron-withdrawing.

Option C ((ii) $>$ (iv) $>$ (iii) $>$ (i)) is the correct order of nucleophilicity because smaller, less hindered alkoxides are generally better nucleophiles.