Answer ten questions (All questions solved for full coverage)
:Hyperconjugation is the delocalization of sigma electrons of a C-H bond of an alkyl group directly attached to an atom of an unshared p-orbital
. It is often called "no-bond resonance".Difference: Hyperconjugation involves the delocalization of sigma (σ) electrons with pi (π) or p-orbitals, whereas the mesomeric effect involves the delocalization of pi (π) electrons or lone pairs
.In aniline, the lone pair of electrons on the nitrogen atom is delocalized into the benzene ring due to resonance, making it less available for protonation
. In cyclohexylamine, there is no such resonance, and the alkyl group exerts an +I effect, increasing electron density on nitrogen, making it a stronger base.The Wurtz reaction is primarily limited to the synthesis of symmetrical alkanes; attempting to synthesize unsymmetrical alkanes results in a mixture of products that are difficult to separate
. The Corey-House reaction allows for the preparation of both symmetrical and unsymmetrical alkanes in high yield without side products.The Diels-Alder reaction is a [4+2] cycloaddition reaction between a conjugated diene and a substituted alkene (dienophile) to form a substituted cyclohexene system
. It is a pericyclic reaction that occurs in a single concerted step.Acylation is preferred because it prevents polyalkylation (the acyl group is deactivating) and avoids carbocation rearrangements
. Example: Acylation of benzene with acetyl chloride gives only acetophenone, whereas alkylation with n-propyl chloride gives isopropylbenzene due to rearrangement.Invert sugar is an equimolar mixture of glucose and fructose obtained by the hydrolysis of sucrose
. It is so named because the optical rotation "inverts" from dextrorotatory (+66.5°) to levorotatory (-19.9°) upon hydrolysis.Mutarotation is the change in the specific rotation of a freshly prepared solution of an optically active compound (like α-D-glucose) until an equilibrium value is reached
. It occurs due to the interconversion between α and β anomers through the open-chain form.Sucrose is non-reducing because the anomeric carbons of both glucose (C1) and fructose (C2) are involved in the glycosidic linkage, leaving no free aldehyde or ketone group to reduce Tollens' or Fehling's reagent
.Answer five questions (All solved for full coverage)
:In acetic acid, the methyl group (-CH3) exerts a +I (inductive) effect, which increases the electron density on the carboxylate group, making the release of H+ difficult and destabilizing the conjugate base
. Formic acid has only a hydrogen atom, which lacks this +I effect, making it a stronger acid.Resonance involves the complete delocalization of pi electrons, while the inductive effect involves the partial displacement of sigma electrons along a saturated chain
.Phenol Acidity: Phenol is acidic because the phenoxide ion formed after losing H+ is stabilized by resonance (delocalization of the negative charge into the benzene ring)
.Increasing order: ĊH3 < CH3ĊH2 < (CH3)2ĊH < (CH3)3Ċ
.Justification: Stability increases with the number of +I effect alkyl groups and the number of hyperconjugative structures
.Carbenes are neutral, divalent carbon species with six valence electrons
. They are formed by the photolysis of diazomethane or ketene. Types: Singlet (paired electrons in sp2) and Triplet (unpaired electrons in p-orbitals).Rule: In the addition of H-X to an unsymmetrical alkene, the halogen attaches to the carbon with fewer hydrogen atoms
. In the presence of peroxides, HBr adds in an anti-Markovnikov fashion via a free-radical mechanism.Alcohol A is 3,3-dimethyl-1-butanol
. Dehydration gives B (3,3-dimethyl-1-butene). Ozonolysis of B gives C (Formaldehyde/Acetaldehyde) and butanone is not possible here; reassessing the structure for butanone product suggests A is 3,4-dimethyl-2-butanol.Toluene undergoes nitration faster because the methyl group is activating due to +I and hyperconjugation effects, increasing electron density on the ring
.Involves the electrophile SO3. Steps: Attack of benzene on SO3 to form sigma complex, followed by proton loss and final protonation.
Glucose does not give Schiff's test, does not form an addition product with NaHSO3, and exists in two anomeric forms (α and β)
. Ring size is established by methylation followed by oxidation (Haworth and Hirst method).Glucose and Fructose give the same osazone because the reaction involves only C1 and C2; the rest of the carbon chain (C3-C6) remains identical in both sugars
.