Organic chemistry-some basic principles and techniques practice · Easy

Option D (Neo hexyl chloride) is correct because the chlorine atom is attached to a primary carbon (a carbon bonded to only one other carbon).

Option C (Leibig's method) is correct because Leibig's method is used for the estimation of carbon and hydrogen in organic compounds.

Option B ($SP,S{P^3},S{P^2}$) is correct because carbon 1 is sp hybridized, carbon 3 is sp3 hybridized, and carbon 5 is sp2 hybridized.

Option B ($C{H_2} = CH - C \equiv CH$) is correct because the hybridization of the carbon atoms from left to right is SP2, SP2, SP, SP.

Option B (1s - 2p(σ)) is correct because sigma bonds formed by the overlap of s and p orbitals are generally stronger than those formed by s-s overlap, and much stronger than pi bonds from p-p overlap.

Option C (Stereoisomers) is correct because stereoisomers are isomers that have the same connectivity but differ in their spatial arrangement, and mirror images are a type of stereoisomer.

Option C is correct because $\mathrm{CH}_3^ +$ has a central carbon atom with three sigma bonds and no lone pairs, resulting in sp2 hybridization.

Option A (Trihydric) is correct because glycerol has three hydroxyl (-OH) groups attached to the carbon chain.

Option B (${\mathrm{C}_6}{\mathrm{H}_5}\mathrm{C}{\mathrm{H}_2}\mathrm{OH}$) is correct because the carbocation formed after heating will be benzylic, which is stabilized by resonance with the benzene ring.

Option D (Ammonia) is correct because in Kjeldahl's method, the nitrogen in the organic compound is converted into ammonia.

Option C (electromeric effect) is a temporary effect that occurs only in the presence of an attacking reagent.

Option A ($C{H_2} = \mathrm{ }CH\mathrm{ }\mathop C\limits^ \bullet {H_2}$) is correct because the allylic free radical is stabilized by resonance.

Option B (${\mathrm{C}_2}{\mathrm{H}_5}\mathrm{C}{\mathrm{O}_2}\mathrm{H},\mathrm{\;C}{\mathrm{H}_3}\mathrm{C}{\mathrm{O}_2}\mathrm{C}{\mathrm{H}_3}$) is correct because both compounds have the same molecular formula but different functional groups (carboxylic acid and ester).

Option A (But-3-ene-1-yne) is correct because the IUPAC rules prioritize numbering the carbon chain such that the alkene gets preference over alkyne when both are present and equidistant from the ends.

Option C (Cylinder) is correct because the two pi bonds in acetylene are perpendicular to each other, resulting in a cylindrical electron cloud around the sigma bond axis.

Option A (Both (A) & (R) are true and the (R) is the correct explanation of the (A)) is correct because tetracyanomethane has 4 sigma bonds and 4 pi bonds per cyano group, totaling 8 sigma and 8 pi bonds, making the ratio 1.

Option B (2, 3 - dimethylbut - 2 - ene) is correct because it contains only primary hydrogen atoms.

Option A (-NO2) is correct because the nitro group exhibits both -I (inductive) and -M (mesomeric) effects, both of which are electron-withdrawing.

Option A (2, 6 – di methyl oct – 3 – yne) is correct because it follows IUPAC naming rules, correctly identifying the longest chain and substituents.

Option C ($\mathrm{CH}_3-\mathrm{C} \equiv \mathrm{N}$) is correct because the carbon in the cyanide group is sp hybridized, not $sp^2$ hybridized.

Option B (A free radical) is correct because free radicals have an unpaired electron, making them paramagnetic.

Option B (Steric strain) is correct because the fully eclipsed conformation of butane has significant steric strain due to the close proximity of the bulky methyl groups.

Option C (4) is correct because the compound has one chiral center and exhibits cis/trans isomerism due to the double bond, resulting in 2 * 2 = 4 stereoisomers.

Option A ($\overline C {H_2} - C{H_2} - N{O_2}$) is the most stable carbanion because the nitro group (NO2) is a strong electron-withdrawing group that stabilizes the negative charge on the carbon atom through both inductive and resonance effects.

Option C (${3^0} < {2^0} < {1^0}$) is correct because carbanion stability decreases with increasing alkyl substitution due to the +I effect of alkyl groups destabilizing the negative charge.

Option A (${6 4 0 0 0} \mathrm{~a m u}$) is correct because the molecular weight is calculated by dividing the total mass of iron by its percentage and multiplying by the number of iron atoms.

Option B (2) is correct because ${\mathrm{C}_3}{\mathrm{H}_6}$ can be either propene or cyclopropane, representing two structural isomers.

Option C ($\mathop \mathrm{C}\limits^\_ \mathrm{C}{\mathrm{l}_3} > {\mathrm{C}_6}{\mathrm{H}_5}\mathop \mathrm{C}\limits^\_ {\mathrm{H}_2} > {\left( {\mathrm{C}{\mathrm{H}_3}} \right)_2}\mathop \mathrm{C}\limits^\_ \mathrm{H} > {\left( {\mathrm{C}{\mathrm{H}_3}} \right)_3}\mathop \mathrm{C}\limits^\_$) is correct because the stability of carbanions increases with electron-withdrawing groups and decreases with electron-donating groups.

Option A (a $\mathrm{C}{\mathrm{H}_2}$ group) is correct because homologous series members differ by a methylene group.

Option A (6) is correct because a branched alkane with 8 carbons can have a maximum of 6 methyl groups attached to the main chain.