Organic chemistry-some basic principles and techniques practice · Medium

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 (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 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.

Option B (Sodium) is correct because sodium is commonly used to dry organic liquids due to its ability to react with water less violently than other alkali metals like potassium and rubidium.

Option A (2, 2-dimethylpentane) is correct because it follows the IUPAC naming rules for alkanes, correctly identifying the parent chain and substituent positions.

Option C (ii > iii > iv > i) is correct because the reactivity towards nucleophilic addition decreases with increased steric hindrance and electron-donating groups.

Correct Answer + Core Reason: Option B ($N_2$) is correct because in the Dumas method, nitrogen content is determined by converting nitrogen in the organic compound to $N_2$ gas.

Option C (Both statements are correct) is correct because the SN1 reaction proceeds through a carbocation intermediate which can undergo rearrangement to form a more stable carbocation, leading to the observed product.

Option B is correct because nitric acid is added to the sodium extract to decompose any cyanide or sulfide ions present, which would otherwise interfere with the silver nitrate test for halogens by forming precipitates.

Option C (Organic synthesis) is correct because prior to the 17th century, organic synthesis as a deliberate and controlled process was not yet established.

The correct answer is A (3 forms) because the compound with two similar asymmetric carbon atoms will exist in three forms: two enantiomers and one meso form.

Option A (Steam distillation) is correct because steam distillation is used to purify substances that are steam volatile and immiscible with water.

Option A (Pi-electrons) is correct because the mesomeric effect involves the delocalization of pi electrons or lone pairs of electrons.

Option B (20 mL) is correct because the balanced equation shows that 2 volumes of O2 are required for each volume of CH4, so 40 mL of O2 react, leaving 20 mL unreacted.

Option A (9) is correct because there are nine distinct substitution products possible when ethane reacts with bromine in sunlight, ranging from mono-substituted to fully substituted.

Option D (N & S) is correct because the red color with FeSO4 and H2SO4 indicates the presence of both nitrogen and sulfur, forming [Fe(SCN)3].

Option A ($CHC{l_3}$) is correct because $\beta$-elimination requires the presence of a hydrogen atom on the $\beta$-carbon, and $CHC{l_3}$ does not have a $\beta$-carbon.

Option D (NO2- and NH2-) is correct because both NO2- and NH2- have sp2 hybridization, leading to similar geometries.

The correct answer is C (Free radical mechanism) because Kolbe's electrolysis involves the formation of free radicals at the anode.

R-O^- is most reactive towards elimination reaction because it is the strongest base among the given options, favoring E2 elimination.

Option A (If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion) is correct because the presence of two asymmetric carbons and one pseudo-asymmetric carbon in trihydroxyglutaric acid leads to the existence of four stereoisomers, including two meso forms and two optically active forms.

Option A (3-bromo-4-chloro-2-iodo-5-oxohexanoic acid) is correct because it follows the IUPAC naming rules, giving the lowest possible locant numbers to the substituents and prioritizing the functional groups.

Option B (35) is correct because it is the percentage of nitrogen calculated using the Kjeldahl method formula.

Option C (Chromatography) is correct because chromatography separates substances based on their differential adsorption on a stationary phase.

Option D ($N{H_3}$) is correct because in Kjeldahl's method, the nitrogen in the organic compound is converted to ammonium sulfate, which is then treated with a base to release ammonia ($NH_3$).

Option C (17.96) is correct because it is the percentage of nitrogen calculated after correcting the volume of nitrogen to standard conditions using the ideal gas law.

The correct answer is A (I > II > III > IV) because the priority in CIP sequence is determined by the atomic number of the directly attached atom.

The correct answer is C (${CH}_2={CH}-\overline{{O}}$) because it has a negative charge on the more electronegative oxygen atom and complete octets.