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

Option D (All of the above) is correct because 2-butene is the more substituted alkene (Saytzeff product), 1-butene is the less substituted alkene (Hofmann product), and the reaction favors the formation of the more substituted alkene.

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 D ($s p^2$ and $s p^3$) is correct because the first carbon (C1) is part of an aldehyde group and is $sp^2$ hybridized, while the second carbon (C2) is bonded to four groups and is $sp^3$ hybridized.

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 (Magnesium pyrophosphate $\mathrm{M}{\mathrm{g}_2}{\mathrm{P}_2}{\mathrm{O}_7}$) is correct because phosphorus is estimated as magnesium pyrophosphate in quantitative analysis.

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 (ferric ferrocyanide) is correct because the blue color in the Lassaigne's test for nitrogen is due to the formation of ferric ferrocyanide (Prussian blue).

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 C (Methyl-Acrylate) is correct because the common name for prop-2-enoic acid is acrylic acid, and its methyl ester is methyl acrylate.

Option A (Fractional distillation) is correct because fuels from crude oil are separated based on their different boiling points using fractional distillation.

Option D (23.7) is correct because it's the percentage of nitrogen calculated using the Duma's method formula and given data.

Option B (8) is correct because there are 8 structural isomers possible for ${\mathrm{C}_5}{\mathrm{H}_{11}}\mathrm{Cl}$.

Option C (d < c < b < a) is correct because boiling point increases with increasing molecular weight and strength of intermolecular forces. Ethanol (d) has the lowest molecular weight and weakest intermolecular forces, while pentan-1-ol (a) has the highest molecular weight and strongest intermolecular forces.

Option B (2 – bromo – 2- methyl butane) is correct because the most stable carbocation is formed at the 2nd carbon, leading to the major product.

Option D (Enantiomers) is correct because d-tartaric acid and l-tartaric acid are non-superimposable mirror images of each other, which defines them as enantiomers.

Option D (1, 1-Dichloroethane) is correct because \"ethylidene\" implies both chlorine atoms are attached to the same carbon in ethane.

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 is correct because the central carbon in allene (H₂C=C=CH₂) is sp hybridized, not sp², and the molecule is non-planar due to the perpendicular arrangement of the CH₂ groups.

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

Option C is correct because the keto content is related to the stability of the enol form; aldehydes have higher keto content than ketones, and diketones have the lowest due to stabilization of the enol form.

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 A is correct because 2,2-dimethylpentane is the correct IUPAC name for the given structure. The longest chain is 5 carbons (pentane), and there are two methyl groups at position 2.

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

Option D ($N,\;N$- dimethyl ethanamine) is correct because it follows IUPAC nomenclature rules, correctly identifying the longest carbon chain and the substituents attached to the nitrogen atom.

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 (7) is correct because there are 7 structural isomers of ${\mathrm{C}_4}{\mathrm{H}_{10}}\mathrm{O}$.