Organic chemistry-some basic principles and techniques practice · Medium

Option B ($Ph - C{H_2} - \mathop C\limits_{\begin{array}{cccccccccccccccccccc} | \\ {Br} \\ \end{array}}^{\begin{array}{cccccccccccccccccccc} {Br} \\ | \\ \end{array}} - C{H_2} - C{H_2} - C{H_3}$) is correct because it correctly represents a pentane chain with a phenyl group at position 1 and two bromine atoms at position 2.

Option D (All of the above) is correct because the $S_N1$ mechanism is favored by a low concentration of nucleophile, a weak nucleophile, and a polar solvent.

Option A (Optically active mixture) is correct because the reaction of an enantiomerically pure acid with a racemic alcohol produces a mixture of diastereomeric esters, which are optically active.

Option A (9 sigma and 1 pi) is correct because the product A is CH3COCH3 (acetone), which has 9 sigma bonds and 1 pi bond.

Option A (Horizontal substituents above the plane) is correct because, in Fischer projections, horizontal lines represent bonds projecting out of the plane (towards the viewer).

Option C (Both (1) and (2)) is correct because both compounds contain chlorine that can be converted to chloride ions, which then react with silver nitrate to form a white precipitate of silver chloride.

Option B (Phosphorus) is correct because the yellow precipitate with ammonium molybdate indicates the presence of phosphorus.

Option A (Allyl carbonium ion $(C{H_2} = CH - \mathop {\mathrm{ }C}\limits^ + {H_2})$ is more stable then propyl carbonium ion) is correct because the allyl carbocation is stabilized by resonance, while the propyl carbocation is not.

Option D ($RO^−$) is the most reactive towards elimination because alkoxides are strong bases and favor elimination reactions.

Option D ($\mathrm{N}{\mathrm{a}_4}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_5}\mathrm{NOS}} \right]$) is correct because the purple color in the test for both nitrogen and sulfur is due to the formation of sodium nitroprusside, $Na_4[Fe(CN)_5NOS]$.

Option B (Benzene nitrile) is correct because it accurately reflects the IUPAC nomenclature for a benzene ring directly bonded to a nitrile group.

Option A (1) is correct because the reduction of benzil with NaBH4 produces a diol, and only one meso isomer is possible due to the symmetry of the molecule.

Option A $(({\text{i}})\,SOC{l_2}\,({\text{ii}})\,KCN\,({\text{iii}}){H_2}/Ni,Na(Hg)/{C_2}{H_5}OH)$ is correct because it outlines the correct sequence of reactions: converting the alcohol to an alkyl chloride using SOCl2, followed by nucleophilic substitution with KCN to form a nitrile, and finally reduction of the nitrile to an amine using H2/Ni or Na(Hg)/C2H5OH.

Option D is correct because the planar hexagon is the least stable conformer of cyclohexane due to significant angle strain and torsional strain. The chair form is the most stable.

Option A (4) is correct because 1-bromo-2-methyl cyclobutane has two chiral centers, leading to a maximum of 2^n = 2² = 4 stereoisomers.

Option C (If Statement-I is true but Statement-II is false) is correct because Statement-I is true, but Statement-II is false.

Option C (Gaseous ${\mathrm{N}_2}$) is correct because in Duma's method, the nitrogen content is determined by collecting and measuring the volume of nitrogen gas produced.

Option D (All of the above are correct) is correct because carbanions are capable of rearrangement, combination with cations, and addition to carbonyl groups due to their negative charge and electron-rich nature.

Option D ($s{p^3}$ - hybrid orbital) is correct because the methyl radical has three sigma bonds and one unpaired electron, resulting in $sp^3$ hybridization.

Option A ($\mathrm{NaCNS}$) is correct because in the Lassaigne's test, if both nitrogen and sulfur are present, they combine to form NaCNS, which reacts with $Fe^{3+}$ to form a blood-red color.

Option B ($\mathrm{BrC}{\mathrm{H}_2}-\mathrm{CHBr}-\mathrm{C}{\mathrm{H}_2}-\mathrm{C} \equiv \mathrm{CH}$) is correct because at low temperature, bromine adds to the alkene first, as alkynes are less reactive than alkenes towards electrophilic addition.

Option B (Hyperconjugation) is correct because 2,3-dimethylbut-2-ene has more alpha hydrogens than 2-butene, leading to greater stabilization through hyperconjugation.

Option C (Electron repelling group) is correct because if the electron density on $R_3C$ increases when H is replaced by Z, then Z must be donating electrons, thus repelling them.

Option D (Preparative TLC (Thin Layer Chromatography)) is correct because it is suitable for separating small amounts of compounds with different polarities.

Option A (Displacement of $\sigma$ - Electrons) is correct because the inductive effect involves the displacement of sigma electrons along a chain.

The correct answer is Store liquid air because Dewar flasks are designed to minimize heat transfer, making them ideal for storing cryogenic liquids like liquid air.

Option B ($7\sigma \,and\,3\pi$) is correct because but-1-ene-3-yne has 7 sigma bonds and 3 pi bonds when you draw out its structure and count them.

Option A (Propanal-1) is correct because the aldehyde group is always at the end of the chain (carbon 1), and the parent chain is propane.

Option C (8) is correct because there are eight structural isomers of C4H11N, considering different arrangements of the carbon chain and the amine group.

Option A (8) is correct because the number of stereoisomers is given by $2^n$, where n is the number of chiral centers. Here, n=3, so $2^3 = 8$.