Organic chemistry-some basic principles and techniques practice · Easy

Option B (Carbanion) is correct because a carbanion has 8 electrons around the carbon atom, fulfilling the octet rule.

Option B (partial displacement of $\sigma$ electrons) is correct because the inductive effect involves the partial displacement of sigma electrons along a chain of atoms.

Option C ($C{F_3}COOH$) is the strongest acid because the three fluorine atoms, being highly electronegative, exert a strong electron-withdrawing inductive effect (-I effect) that significantly stabilizes the conjugate base (carboxylate ion), thus increasing the acidity of the carboxylic acid.

Option D (All of these) is correct because ${\mathrm{C}_4}{\mathrm{H}_{10}}\mathrm{O}$ can exhibit metamerism, functional isomerism, and positional isomerism.

Option B ($R{S^ \ominus }$ is less basic but more nucleophilic than $R{O^{\, \ominus }}$) is correct because sulfur is larger and more polarizable than oxygen, making the thiolate anion a better nucleophile but a weaker base.

Option C (Organic compounds cannot be synthesized in the laboratory) is wrong because Wöhler's synthesis of urea disproved this statement, showing that organic compounds can indeed be synthesized in the lab.

Option D is correct because ortho and para nitrophenols have different volatilities due to intramolecular hydrogen bonding in ortho-nitrophenol, allowing separation by steam distillation.

Option C ($\alpha -$ hydroxypropanoic acid) is correct because lactic acid has a hydroxyl group attached to the alpha carbon of propanoic acid.

Option A ($\mathop C\limits^ \oplus {H_5}$) is correct because $CH_5^+$ is not a typical electrophile due to its structure and bonding; it is a hypercoordinated carbocation.

Option D (A and R both are false) is correct because SO3 is an electrophile, not a nucleophile, and it does not have a lone pair of electrons.

Option B (Delocalisation of $\pi$ electrons) is correct because the mesomeric effect involves the delocalization of pi electrons or lone pairs of electrons in a conjugated system.

Option C is correct because the rotation of plane-polarized light can be measured by a polarimeter. A polarimeter is specifically designed for this purpose.

Option C (Labile form) is correct because the labile form in tautomerism refers to the less stable tautomer.

Option D (Wöhler) is correct because Friedrich Wöhler synthesized urea in the laboratory, disproving the vitalism theory.

Option A ((ii) and (iii)) is correct because in (ii) there are both $sp^3$ and $sp^2$ hybridized carbons, and in (iii) there are $sp^3$, $sp^2$ and $sp$ hybridized carbons.

Option B ($- Cl$) is correct because chlorine can act as both a +M and -I group due to its lone pairs and electronegativity, but its +M effect is weaker than its -I effect.

Option C (Positional) is correct because the compounds have the same molecular formula but differ in the position of the chlorine atoms on the carbon chain.

The correct answer is retention factor because it quantifies the relative migration of a substance in chromatography.

Option A (Isopropyl propyl ether) is correct because the ether linkage connects an isopropyl group and a propyl group.

Option C is correct because van't Hoff proposed the tetrahedral mirror image structures to a pair of enantiomers. This was a key development in understanding stereochemistry.

Option B (Short lived) is correct because free radicals are highly reactive species due to the presence of an unpaired electron.

Option C (Canonical structures) is correct because canonical structures are theoretical representations and do not exist as real structures.

Option A (Dextrorotatory) is correct because dextrorotatory compounds rotate plane-polarized light in a clockwise direction.

Option B (Phase distribution) is correct because chromatography separates components based on their differential distribution between a stationary and a mobile phase.

Option D ($II > I > IV > III$) is correct because the -I effect order is $N{O_2} > CN > F > N{H_2}$ based on electronegativity and other factors.

Option C is correct because tautomerism involves the migration of a proton, resulting in two interconvertible structural isomers.

Option B (Conformational isomerism) is correct because different structures generated due to rotation about C-C axis are called conformers, and this phenomenon is called conformational isomerism.

Option C ($C{N^ - }$) is the strongest nucleophile because carbon is less electronegative than oxygen, making the carbon atom in cyanide more willing to donate its electrons compared to the oxygen in hydroxide or ethoxide. Also, cyanide is a weaker base than ethoxide, making it a better nucleophile.

Option A (3-Chloroprop-1-ene) is correct because the chlorine is on the 3rd carbon and the double bond starts at the 1st carbon.

Option D ($NaCl, NaI, Na_2S$) is correct because the solubility products ($K_{sp}$) of the silver, copper, and nickel salts determine the order of precipitation.