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

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

Option D (All of the above are correct) is correct because carbanions are nucleophilic and can undergo rearrangements, combine with cations, and add to carbonyl groups.

Option C (N-bromoethanamide) is correct because it correctly names the compound with the bromine atom attached to the nitrogen atom of the ethanamide group.

Option D (Planar form) is correct because cyclohexatriene, also known as benzene, has sp² hybridization for all carbon atoms, resulting in a planar structure due to resonance and stability.

Option B (Solubility of the compound in water) is correct because boiling point depends on intermolecular forces, size, and polarity, but not directly on water solubility.

The correct answer is B because $\mathrm{CH}_3 \mathrm{OCH}_2 \mathrm{CH}_2 \mathrm{CH}_3$ and $\mathrm{CH}_3 \mathrm{OCH}(\mathrm{CH}_3)_2$ are position isomers, not metamers.

Option B (Chromatography) is correct because adsorption is the core principle behind chromatographic separation techniques.

Option B ($\mathrm{F}{\mathrm{e}_4}{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_3}$) is correct because the blue color in Lassaigne's test is due to the formation of ferric ferrocyanide, also known as Prussian blue.

Option C ($\mathrm{Fe}_4{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_3}$) is correct because Prussian blue is formed by the reaction of ferric ions with ferrocyanide ions, resulting in the complex $\mathrm{Fe}_4{\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} \right]_3}$.

Option D (All of these) is correct because the Wurtz reaction can produce butane, 2,3-dimethylbutane, and 2-methylbutane from the given reactants.

Option C (But-1-ene-3-yne) is correct because the IUPAC naming prioritizes the double bond over the triple bond when numbering the carbon chain.

Option B (Destructive distillation of coal) is correct because coal-tar is a byproduct obtained during the destructive distillation of coal.

Option C is correct because the compound C₂H₄Br₂ can exist as 1,1-dibromoethane and 1,2-dibromoethane, which are positional isomers.

Option A (Reddish brown precipitate of ${HgCO}_3 .2 {HgO}$ is formed) is correct because when ${HgCl}_2$ reacts with ${CO}_3^{2-}$ anion, a reddish-brown precipitate of ${HgCO}_3 .2 {HgO}$ is formed.

Option D ($+\mathrm{R}$ effect of $\mathrm{NH}_2$ in aniline) is correct because the lone pair on the nitrogen in aniline is delocalized into the benzene ring through resonance, reducing its availability for protonation.

Option B ($16.2$) is correct because the percentage of nitrogen is calculated using the stoichiometry of the reaction and the given data.

Option D (a and b are enantiomers) is correct because compounds a and b are non-superimposable mirror images of each other.

The correct answer is B (Due to formation of CH₃COOH which acts as a weak acid) because the hydrolysis of methyl acetate produces acetic acid, which neutralizes the NaOH, causing the phenolphthalein to become colorless at the endpoint.

Option C ($C{H_3}COOH$) is correct because it is the weakest acid among the given options, and $pK_a$ is inversely proportional to acid strength.

Option D (functional group isomerism) is correct because ${C_3}{H_6}O$ can exist as an aldehyde or a ketone, which are different functional groups.

Option C ($N{H_2}CON{H_2}$) is correct because the compound's composition, molar mass, and reaction products (NH3 and a biuret-positive residue) match urea.

Option A (1, 4-pentadiene) is correct because the double bonds are not conjugated, preventing delocalization.

Option A ((d) $>$ (b) $>$ (a) $>$ ( c)) is correct because the acidity increases with the number of electron-withdrawing nitro groups attached to the carbon chain.

Option C (Vacuum distillation) is correct because glycerol has a high boiling point and vacuum distillation is used to lower the boiling point for separation.

Option D (8) is correct because with 3 double bonds and different ends, the number of geometrical isomers is 2 raised to the power of 3.

Option B (Beilstein test) is correct because the Beilstein test is used to detect the presence of halogens in organic compounds.

Option B (Polarity of solvent) is correct because the difference in products arises from the different reaction mechanisms favored by the polar (water) and non-polar (chloroform) solvents.

Option C is correct because methoxyethene (CH3OCH=CH2) contains both an ether functional group (R-O-R') and a carbon-carbon double bond, making it an unsaturated ether.

Option A (Magnesium pyrophosphate $\mathrm{M}{\mathrm{g}_2}{\mathrm{P}_2}{\mathrm{O}_7}$) is correct because phosphorus in an organic compound is estimated as magnesium pyrophosphate.

Electron-withdrawing groups decrease basicity, while electron-donating groups increase basicity. NO2 is strongly electron-withdrawing, Br is electron-withdrawing, and CH3 is electron-donating. Therefore, the correct order is I < II < III, making option A the correct answer.