Organic chemistry-some basic principles and techniques practice · Easy

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

Option A (25 % s-character) is correct because in sp3 hybridization, one s orbital mixes with three p orbitals, resulting in 25% s-character and 75% p-character.

Option A is correct because carboxylic acids are indeed more acidic than carbolic acids (phenols), and this is due to the greater stability of the carboxylate anion compared to the phenoxide ion.

Option B ($CH_2 = CH - C \equiv CH$) is correct because it correctly represents the hybridization order $SP^2, SP^2, SP, SP$ from left to right.

Option D ($\mathrm{C}{\mathrm{H}_3} - \mathrm{CH} = \mathrm{CH} - \mathrm{C}{\mathrm{H}_2} - \mathrm{CH} = \mathrm{CH} - \mathrm{C\;} \equiv \mathrm{CH}$) is correct because it matches the given hybridization pattern from one end of the chain to the other.

Option C (Chemistry of hydrocarbons and their derivatives) is correct because organic chemistry encompasses the study of hydrocarbons and all compounds derived from them through various functional group modifications.

Option D is correct because benzene and chloroform have different boiling points, making distillation an effective separation method.

Option A (3) is correct because the degree of unsaturation is calculated using the formula and given molar mass and carbon percentage.

Option D (All of these) is correct because the reaction involves a Wurtz reaction, where alkyl halides react with sodium metal in dry ether to form alkanes. 2-bromopropane forms 2,3-dimethylbutane, bromoethane forms butane, and a cross-coupling reaction forms pentane.

Option C ($\mathrm{C}{\mathrm{H}_3}\mathrm{CH} = \mathrm{C}\left( {\mathrm{C}{\mathrm{H}_3}} \right)\mathrm{CH} = \mathrm{C}{\mathrm{H}_2}$) is correct because it represents a 5-carbon chain with double bonds at positions 1 and 3, and a methyl group at position 3.

Option C (SP - SP2) is correct because the carbon atoms in the C-C single bond of H-C≡C-CH=CH2 are sp and sp2 hybridized, respectively.

Option A ($I < II < III$) is correct because the basic strength increases with decreasing s-character of the carbon bearing the negative charge.

Option D (3) is correct because n-butane has three main rotational conformers: anti, gauche, and eclipsed (or partially eclipsed).

Option C (Homocyclic and aromatic) is correct because the compound has a ring system made of carbon atoms (homocyclic) and exhibits aromatic properties due to the presence of alternating double bonds.

Option C ($CH_2=C=CH_2$) is correct because it contains carbon atoms with different hybridization states ($sp^2$ and $sp$).

Option C (Electromeric effect) is correct because the electromeric effect involves the complete transfer of a shared pair of electrons to one of the atoms joined by a multiple bond at the requirement of the attacking reagent.

Option C (${\mathrm{C}_4}{\mathrm{H}_{10}}$) is correct because alkanes do not exhibit functional group isomerism as they lack a functional group.

Option A (3, 1, 1) is correct because isopentane has three primary (1°), one secondary (2°), and one tertiary (3°) carbon atoms.

Option A (Propionaldehyde) is correct because acetone (CH3COCH3) and propionaldehyde (CH3CH2CHO) have the same molecular formula (C3H6O) but different structural arrangements, making them isomers.

Option B (${\mathrm{C}_2}{\mathrm{H}_2} < {\mathrm{C}_2}{\mathrm{H}_4} < {\mathrm{C}_6}{\mathrm{H}_6} < {\mathrm{C}_2}{\mathrm{H}_6}$) is correct because bond length decreases with increasing bond order due to stronger attraction between the carbon atoms.

Option B (Bivalent) is correct because the benzal group (C₆H₅CH=) has two valencies available for bonding.

Option A (4) is correct because monochlorination of n-pentane yields 3 distinct monochlorinated isomers (1-chloropentane, 2-chloropentane, and 3-chloropentane).

Option D is correct because methyl iodide (CH3I) is a common methylating agent due to the relatively weak C-I bond, making the methyl group readily transferable.

Option B ($-NH_2$ group activates benzene nucleus for attack of electrophile at ortho and parasites) is correct because the amino group is an activating and ortho/para-directing group.

The correct answer is Nucleus loving because nucleophiles are species that are attracted to the positively charged nucleus.

Option A (Nathan and Baker) is correct because Nathan and Baker are credited with pointing out the concept of hyperconjugation.

Option C (pentanoic acid and 2 – methyl butanoic acid) is correct because they have the same molecular formula but different carbon chain arrangements.

Option B ($- ONO$) is correct because the +M effect is shown by groups with lone pairs that can donate electrons through resonance, and -ONO has lone pairs on oxygen.

Option A (Twelve sigma-bonds and three pi-bonds) is correct because benzene has 12 sigma bonds (6 C-C and 6 C-H) and 3 pi bonds within the ring.

Option A ($-\mathrm{COOH},\mathrm{\;} -\mathrm{S}{\mathrm{O}_3}\mathrm{H},\mathrm{\;} -\mathrm{CON}{\mathrm{H}_2}, -\mathrm{CHO}$) is correct because it represents the correct decreasing order of priority for functional groups in IUPAC nomenclature.