In case of test for Nitrogen actually acidification is not required, initially the sodium extract is made alkaline by adding few drops of NaOH, which reacts with ferrous sulphate to produce ferrous hydroxide which participates in further reactions.

In case of sulphur also acidification is not required.

If both N & S are present then a blood red colour appears due to the formation of ferric sulphocyanide in the test for nitrogen.

Acidification is necessary for detecting halogen in presence of N & S to expel HCN & H2S respectively,.

I suppose you are reffering to test of halogen in presence of N & or S in sodium extract.

The presence of N & S causes hinderance in the detection of halogen in the organic compound. NaCN & Na2S present in the sodium extract will give a white & black ppt of AgCN & Ag2S respectively on treatment with AgNO3. Hence sodium extract is boiled with small amount of conc. HNO3, before testing for halogens. NaCN & Na2S are both decomposed by HNO3.

NaCN + HNO3 .....> NaNO3 + HCN

Na2S + 2 HNO3 ....> 2 NaNO3 + H2S

If the solvent is aprotic ( it is not a hydrogen bond donor), the direct relationship between nucleophilicity & basicity holds. For example, both the nucleophilicities & basicities of the halogens decrease with increasing size in an aprotic solvent such as dimethylformamide.

If the solvent is protic ( it is a hydrogen bond donor such as water, alcohol,etc,), the relationship between the basicity & nucleophilicity becomes inverted : as basicity decreases nucleophilicity increases. Thus, iodide ion, which the weakest base of the halogen family, is the poorest nucleophile in an aprotic solvent and the best nucleophile in a protic solvent.

When a negatively charged species is placed in a protic solvent, the solvent molecules arrange themselves so that their partially  positively charged hydrogens point toward the negatively charged species. An aprotic solvent does not have a partially positively charged hydrogen.

The interaction between the ion & the dipole of the protic solvent is called as ion-dipole interaction. The change from a direct relationship between basicity & nucleophilicity in an aprotic solvent to an inverse relationship in a protic solvent results from the ion-dipole interactions between the nucleophile and the protic solvent. This occurs because atleast one of the ion-dipole interactions must be broken before the nucleophile can participate in an SN2 reaction. Weak bases interact weakly with protic solvents; strong bases interact more strongly because they are better at sharing their electrons. It is therefore easier to break the ion-dipole interactions between an iodide ion and the solvent than between the more basic fluoride ion and the solvent, because the latter is a stronger base.

Hydroxy acetone reduces ammoniacal silver nitrate ( Tollen's reagent), thereby  being oxidised to DL- lactic acid. Althogh ketones do not normally reduce Tollen's reagent or Fehling's solution, alpha hydroxy ketones are exceptions. Similarly, ketones form phenyl hydrazone with phenyl hydrazine, but alpha hydroxy ketones form osazones. This is because alpha hydroxy ketone appears to exist as an equilibrium mixture of hydroxy ketone and cyclic hemiacetal forms ( ring-chain tautomerism ).

Primary and secondary alkyl halides are readily reduced to alkanes by Lithium aluminium hydride; tertiary halides mainly give alkenes. On the other hand sodium borohydride reduces secondary and tertiary halides, but not primary. Whereas triphenyltin hydride ( Ph3SnH ) reduces all the three types of halides.

Hence the correct answer will be D) All

Cyclic intermediates can be formed during addition across carbon carbon double bond. The most familiar example is the formation of cyclic bromonium ion during addition of bromine across C=C. The other examples are hydroxylation with cold dil KMnO4 where cyclic mangnate ester is formed,  with osmium tetra oxide formation of cyclic osmate ester, , formation of epoxide, ozonoid can also be regardes as examples of cyclic intermediates.

The pericyclic reactions are other example of formation of cyclic intermediate.

In both the above examples the common thing is C=C; hence we may say that reactions of C=C can have cyclic intermediates.

2. Migratory aptitude of various groups : In case of unsymmetrical ketones, the group that migrates is the one which can more easily supply the electrons. The tendency of the groups to migrate from carbon to electron deficient oxygen follows the order :

tert-alkyl > sec-alkyl >phenyl > prim-alkyl > methyl.

Amongst aryl groups, the order is , p-methoxyphenyl > p-tolyl > phenyl > p-chlorophenyl.

The product actually formed during Baeyer-Villiger oxidation depends upon the migratory aptitude of groups. For example, the oxidation of actophenone with perbenzoic acid gives phenyl acetate and not methyl benzoate because phenyl group has a greater tendency to migrate to electron deficient oxygen than methyl group.

Similarly, Baeyer-Villiger oxidation of pinacolone gives tert.butly acetate and not methyl pivilate because migratory aptitude of tert. group is greater than that of methyl group.

1.Pinacol reaction : In presence of active metals such as magnesium (usually employed in the form of magnesium amalgam ) aldehydes and ketones undergo bimolecular reduction to form symmetrical glycols generally called pinacols. Reduction of aldehydes and ketones by dissolving metals produces alcohols. However, this reaction may also lead to the formation of 1,2-glycol.

I have not come across any evidence which specifically refers that this reaction is given only by ketones and not by aldehydes.

please put the other questions separately.

Ethyl methacrylate is an example of crossed conjugation.

CH₂=C(-Me)-C(=O)-Oet

The ethylenic bond does not enter into resonance with COOEt group. The contribution to resonance is only the path that leads to greater resonance energy, which in this case is the carbethoxy group.

The structure of p-benzoquinone is known as crossed conjugated system; it contains three( or more) conjugated double bonds which are not arranged in a continuous chain. It behaves as an alpha-beta unsaturated ketone, most of its addition reactions being typical 1,4-additions, followed by aromatisation. The driving force of these additions is largely due to the fact that the resonance energy of quinone is 20.9kJ per mole, and when it converts into the aromatic compound, the product has gained an increased stabilization energy of atleast 125.5kJ per mole.

Are you reffering to heat of solution?

When a solute is dissolved in a solvent to form a solution, there is frequently an evolution or absorption of heat. The heat change per mole of solute dissolved is not constant, but usually varies with the concentration of the solution. If the total change of enthalpy at constant temperature observed when m moles of solute are added to a definite quantity,e.g., 1000 grams, of solvent is plotted against m, for various values of the entalpy change, a curve is obtained. The curve reaches a limit when the solution is saturated at the experimental temperature; the solution then contains ms moles of solute to 1000grams of solvent. The height of the ordinate(enthalpy change) at any point,e.g.X, divided by the corresponding number of moles mx of solute dissolved,i.e.,enthalpy change delta Hx/mx , represents the increase of enthalpy per mole of solute when it dissolves to form a solution of a particular concentration; thisn quantity is called the integral heat of solution at the given concentration. It is observed that the integral heat of solution is, in general, approximately constant in dilute solution, but becomes smaller with increasing concentration.

The fact that the heat of solution of a solute varies with its concentration implies that there must be a change of enthalpy when a solution is diluted by the addition of solvent. The internal heat of dilution is the change in enthalpy when a solution containing 1 mole of solute is diluted from one concentration to another.

Many attempts have been made to correlate the nature of solvent with reaction rate or nucleophilicity. Huges & Ingold have proposed the following qualitative theory of solvent efects;

i)Ions & polar molecules, when dissolved in polar solvents, tend to become solvated.

ii) For a given solvent, solvation tends to increase with increasing magnitude of charge on the solute molecules or ions.

iii)For a given solute, solvation tends to increase with increasing dipole moment of the solvent.

iv)For a given magnitude of charge, solvation decreases as the charge is spread over a larger volume.

v)The decrease of solvation due to the dispersal of charge will be less than that due to its destruction.

Thus if the reaction of SN2 type is to be carried out with -SH a solvent which is less polar will be useful. In a polar solvent the SH ion may get solvated and thus its nucleophilicity may decrease.

SO3 is an electrophile hence it cannot be considered as nucleophile.

Nucleophilicity decreases with increasing electronegativity of the attacking atom. Hence the order would be RNH2 >ROR

If you are reffering to ALKYNE then the method quoted by Sandeep is correct. Alkynes on hydration with dilute sulphuric acid in presence of mercuric sylphate gives ketone.

The reaction would proceed with recemisation. When one optically active molecule inverts, it becomes a molecule of the racemic pair, the other is unreacted optically active molecule. Therefore when 50% of the starting optically active molecules invert, 100% recemisation occurs. This has been supported by study by Huges using labelled iodine.

The rate determining step of SN1 reactions involves the formation of the ions, i.e., carbocation and the halide ion. Therefore, polar protic solvents like water, alcohols, acids etc. would solvate these ions. The solvation stabilizes the ions as well as the transiton state leading to their formation. Further, the solvation process liberates a considerable amount of energy which facilitates the ionisation step further. Thus, SN1 reactions are favoured by polar protic solvents. Greater the polarity of the solvent, more is its ionizing power and hence faster is the rate of SN1 reaction.

The solvent sometimes plays such a dominant role that it completely alters the mechanistic course of the reaction. For example, methyl bromide which generally hydrolyses in aqueous alcoholic solution by SN2 mechanism can be made to hydrolyse by SN1 mechanism in formic acid containg small amount of water.

The answer posted appears to be correct. As I am not good in solving such problems I had to verify the mathematical calculations from some enjineering student. As the answer matches with the given answer we may consider it to be correct.

However, I have my own doubt as the temp. of thermal decomposition is not mentioned the final product should be the oxide ( observation of our research work on thermal decomposition of some inorganic compounds). In such case the calculations will change, and you can then modify them .