Organic Chemistry Some Basic Principles And Techniques

a. The longest chain should be identified, in this case longest chain has 7 carbons hence the name of parent chain heptane.

b. Then the substituents have to be identified, here we have 3 substituents (One ethyl group and two methyl groups). When we start numbering the parent chain it is from the side of the chain where the substituted carbons get the least number. So here the substitution at 3 and 4 carbons is correct rather than at 4 and 5.

c. Then coming to the substituents, when there are more than one type of substituents the alphabetical order is followed. So here ethyl comes first and then methyl.

d. Finally when, the same group is present more than once the alkyl name is prefixed with di(for two), tri (for three) and so on. Here we have two methyl groups hence named dimethyl. When we take alphabetical order ‘di’ is not considered as it only indicates the number but ‘m’ in methyl is taken into consideration. Hence, first ethyl is named and then methyl.

a. There are two functional groups in this compound, carboxylic acid and keto group. Out the two carboxylic acid has higher preference, so the numbering starts from carboxylic carbon.

b. There are 5 carbons in the parent chain hence the parent name pentanoic acid.

c. The keto group is at 4^th carbon. When keto group does not get the first preference it is prefixed as –oxo-. Hence the name 4-oxo-pentanoic acid.

a. the numbering should be such that all the substituents get the least numbering possible.

b. out of the substituents –Cl, -NO₂ and –CH₃ , nitro group (-NO₂ ) gets the priority, meaning nitro benzene is the parent compound and rest are substituents.

c. Least possible numbering is possible only when the carbon with –Cl is designated position one, then the carbon with -NO₂ is given two and the carbon with –CH₃ is given four.

The sp hybridized carbon is highly electronegative as it has more ‘s’ character meaning electrons are closer to nucleus hence held tightly. While other marked carbons are either sp² or sp³ hybridised. They have more of ‘p’ character. The distribution of electrons in ‘p’ orbital will be above and below the plane of a particular axis so they are not as strongly held as those in ‘s’ orbital .

Functional group isomerism is a type of isomerism where molecules that have same molecular formula but can exhibit different functional groups in their structural formula. Alcohols(-OH) and ethers (-O-)can have same molecular formula but can exhibit different functional groups. Aldehydes(-CHO) can give functional group isomer as ketones(C=O). Cyanides(-CN) can form functional group isomer iso- cyanide(-NC). While Alkyl halides can exhibit chain and position isomerism they cannot exhibit functional group isomerism.

To extract the essential oils from plants steam distillation is popularly used. It is based on the principle that essential oils are insoluble in water but miscible with water vapour.

Initially they take the parts of the plant from which the oils have to be extracted, in a still. Into this steam at high temperature is passed. The essential oils from plant parts volatilises into steam, which passes through condenser pipe. The cooled vapour separates into oil and water. The floating oil is then separated.

Thin layer chromatography is a technique used to separate the various colourful components of ink by dipping the paper in a solvent such that the ink does not get wet. The paper slowly absorbs the solvent, as the solvent moves upwards it separates the components of ink depending on the affinity of component towards the stationary phase and mobile phase(solvent)

Paper chromatography has two phases, stationary phase and mobile phase. Stationary phase corresponds to the cellulose filter paper which holds water molecules in them, while mobile phase corresponds to the solvent. When the paper with the sample is dipped in the solvent such that the sample does not touch the solvent directly. Slowly the solvent moves upwards due to capillary action of pose in cellulose paper. As the solvent moves through the sample the components get separated depending on their affinity towards water and solvent. Those with more affinity towards solvent move forward faster thus getting parted. As the components get parted the principle involved is partition.

The carbocation is stabilized due to weak +I (inductive) effect of methyl groups (-CH₃)

This carbocation is directly linked with –OCH₃ group. Oxygen has 2 lone pairs of electrons so it readily donates a pair of electrons to stabilise the positive charge on carbon through +R effect(resonance by donating electrons) which is stronger than +I effect. Hence, this cation is highly stabilized.

In this case the –OCH₃ group is attached farther from carbocation so will exhibit a weak –I effect( pulling of electrons towards itself). Hence, it is the least stable cation out the three.

To name a compound longest chain must be considered. When the compound is expanded as follows

We can see that the longest chain is with 6 carbons and the substitution of methyl groups is at 3^rd and 4^th carbon hence the name 3,4- dimethylhexane.

-Cl is an electron withdrawing group. It exhibits –I effect. –Br is also electron withdrawing but lesser than –Cl . Magnesium and –CH₃ exhibit +I effect. So the methyl carbon in ethyl chloride experiences greatest positive charge out of all the marked carbons.

Ionic species are stabilised by the dispersal of charge. When the carboxylate ion is attached to an electronegative atom the negative charge is dispersed on them due to dipole moment leading to stability of carboxylate ion. More the number of electronegative species attached more is the dipole moment and hence more dispersal leading to more stability. In the above case the 4^th option has two ‘F’ attached while 3^rd one has only one ‘F’ attached, 2^nd option has ‘Cl’ attached out of F and Cl, F is more electronegative. The 1^st option is simple carboxylate ion which is only resonance stabilized, there is no dipole moment there.

The electron density from the pi bond attracts proton(H⁺). This H⁺ can substituted at either of the carbons with the pi bond. But we observe that the H⁺ gets substituted at the terminal carbon(less substituted carbon) exclusively to give a more stable secondary carbocation. This preference to form more stable carbocation is called Markownikoff’s rule.

Heterolytic bond fission means, when the bond breaks the bonding electrons are not shared equally, but one species takes u both the electrons gaining negative charge while leaving positive charge on the other. When the bond is polar i.e., the bond is formed between to atoms of different electronegativities the electron density will be more with the one more electronegative out of the two. In this case the bond is between C and Br. Out of which Br is more electronegative, so when heterolytic bond fission occurs both the bonding electrons will be taken by Br, leaving positive charge on carbon.

To form a bond, a pair of (2) electrons are required. An electrophile is electron deficient so it is attracted by electron rich species( nucleophiles ). There is more electron density in a pi bond than an electrophile, so the pi bond attracts the electrophile to form a bond on less substituted carbon leaving the more substituted carbon electron deficient. In this case as both the carbons are equally substituted bond can be formed on either of the carbons. The should always be from electron rich to electron deficient in a mechanism.

(i) is an alkyne compound having 2 sets of triple bonds. Here each carbon has 2 sigma bonds and 2 pi bonds, therefore each carbon is ‘sp’ hybridized.

(iv) is an alkene with two sets of double bonds. Each carbon has three sigma bonds and one pi bond, each carbon exhibiting ‘sp²’ hybridization.

The given structure is a tetrahedron. It has two groups/atoms

( -CH₃ and –H) inplane, one above the plane(-Br) and one below the plane(-Cl).

When we compare the above data we observe that the spatial arrangement of (iii) is same as that of structure A while the rest 3 differ.

In AlCl₃, ‘Al’ is deficient by 2 electrons to complete the octate or attain noble gas configuration. Electronic configuration of Al is [Ne]¹⁰3s²3p¹, so aluminium has 3 electrons in the outermost shell, when it is bonding with 3 chlorines each shares an electron making a total of 6 electrons so aluminium is still deficient of 2 electron to complete octate(8) hence it act as an electrophile.

SO₃ is bonded with 3 oxygen atoms which are more electronegative than sulphur hence tend to pull the electron density towards themselves(-I effect) leaving a positive charge on sulphur making it act as an electrophile.

Rest all i.e., NO₂⁺ , CH₃⁺ , CH₃ –C⁺ = O are positively charged, hence electron deficient and act as electrophiles.

Isomers are compounds with same molecular formula but different structural arrangement. Position isomerism is a structural isomer where the place of the functional group differs. Both II and III are ketones, but one has keto-group on 2^nd carbon while other has keto-group on 3^rd carbon.

Functional group isomers means the molecules with same molecular formula but different functional groups. In this case II and III are ketones, so they are not functional isomers. Similarly I and IV are aldehydes with same formula but different structure so they are also not functional group isomers.

Nucleophile means that which attracts nucleus. Nucleus is positively charged as it contains the neutrons( which are neutral) and protons which are positive in nature. Opposite charges attract. So to attract positively charged nucleus the nucleophile should be negatively charged. Negative charge is borne by electrons so nucleophiles should be electron rich species i.e., able to donate one or more pairs of electrons.

Hyper conjugation is the interaction of pi electrons with the adjacent sigma bonds of the substituents or interaction of empty p orbitals with the sigma bonds of the carbocation in organic compounds. More the number of hyper conjugation structures more is the stability of unsaturated or carbocation system.

V and VI (metamers of ethers)

Explanation:

Metamerism is defined as the variation in arrangement of atoms around a bivalent functional group of same chain length. It is exhibited by ethers, esters, ketone, thioethers, secondary and tertiary ethers and such functional groups. Metamerism is similar to that of positional isomerism. Terminal functional groups such as alcohols, carboxylic acids, aldehydes, primary amines and such functional groups do not exhibit metamerism. So here in compounds V and VI the chain length is same but the distribution of carbons is different around oxygen atom.

I and V,

I and VI,

I and VII,

II and V,

II and VI,

II and VII,

III and V,

III and VI,

III and VII,

IV and V,

IV and VI,

IV and VII

Explanation:

All the seven compound given have the same molecular formula, so each alcohol is a functional group isomer of each ether given and visa-versa. Functional group isomersare a type of structural isomers having the same molecular formula but different functional groups.

I and II,

III and IV,

V and VI

Explanation:

Position isomerism means, where the length of the parent chain and functional group is same but the position of the functional group is different, then it is called position isomer. In case of( I and II ) the parent chain has 4 Carbons each. But the functional group –OH is on 1^st carbon in the 1^st compound and on 2^nd carbon in the 2^nd compound.In (V and VI) alsothe parent chain has 4 Carbons each. But the functional group –O- (ether) is placed between 2^nd and 3^rd carbon in compound V while the ether group is placed between 1^st and 2^ndcarbon in compound VI. In case of (III and IV) the parent chain has only 3 carbons(as it is branched), but the –OH group in compound III is on 1^st carbon while –OH group in compound IV is on 2^nd carbon. since they have the same chain length and same functional group but different positions for functional group.

I and III,

I and IV,

II and III,

II and IV,

II and IV,

V and VII,

VI and VII

Explanation:

Compounds having same molecular formula but vary in the length of the parent chain(as they are branched) is called chain isomerism. Compound I and II are linear (no branching) while III and IV are branched, so I and II are chain isomers with III and IV. Similarly, VII is ether with a branched alkyl group while V and VI are linear. So, VII is an isomer of both V and VI.

For testing halogens in an organic compound with AgNO₃ solution, sodium extract (Lassaigne’s test) is acidified with dilute HNO₃becauseif the compound contains either nitrogen or sulphur or both along with halides, they also react with silvernitrate to form precipitates, so to decompose sodium sulphide and sodium cyanine if present so that they will not precipitate masking the precipitates of silver halides nitric acid is added. If dilute H₂ SO₄is added then sodium sulphide and sodium cyanine are not decomposed thereby we will not get the correct result if halides are also present.

H₂C = C = CH₂. The terminal carbons are sp² hybridised as they form 3 (2 with H and 1 with C) sigma bonds and 1 pi bond(between carbons) while the centre carbon is sp hybridised (as it forms 2 sigma bonds, 1 with each carbon and 2 pi bonds, one with each carbon) .

There are 3 types of hybridisation for carbon in organic compounds, sp, sp² and sp³ depending on the number of sigma and pi bonds between the carbon atoms. More the ‘s’ character more is the electronegativity of the carbon as ‘s’ orbitals are closer to the nucleus, hence electrons held more closely.

When we compare the electronegativity of carbon and Magnesium we observe C is more electronegative than Mg, because of this electronegativity difference the bond between C-Mg is highly polarised. C being more electronegative the electron density is towards C hence Mg readily loses its electron to form C^-( carbanion).

CH₃—CH₂—CH₂—^δ-CH₂—^δ+Mg⁺ X^-

The above two compounds exhibit chain isomerism. Isomers are compounds that have same molecular formula but different structures. In the above structures, both have the same functional group (thioether / sulphide), but there is difference in the arrangement of atoms in the main chain. The first one is linear( propylmethylsulphide) , while the second one(isopropylmethylsulphide) is branched, i.e., there is difference in the chains attached to –S-. hence they are exhibiting chain isomerism.

The correct structure to be named is based on IUPAC norms.

a. The main(parent) chain should be the longest possible carbon chain with all the carbons with the functional groups included in them. The above selection is the only possible one with both the functional groups in main chain.

b. Out of carboxylic acid and alcohol acid gets the priority hence the numbering starts from carboxylic carbon.

c. There is a branched alkyl group substitution at second carbon

d. There is a hydroxyl group at 4^th carbon therefore it can be named as 2-(2-methyl butyl)-4-hydroxybutanoic acid.

Kjeldahl method cannot be used for the estimation of nitrogen present in DNA and RNA . This is because the nitrogen in these compounds does not convert into ammonium sulphate under the conditions of this method. To estimate the amount of nitrogen , the ring should be broken first, so that the free nitrogen gets converted to ammonia, which is dissolved in acid and estimated by direct or indirect titration. The difficulty in decomposing ring structures limits this test for DNA and RNA.

As the liquid compound decomposes at its boiling point which means that the compound is heat sensitive, hence we use “Steam distillation” for its purification.

Steam distillation is a type of distillation, which is specially done for the purification of the temperature sensitive material like natural aromatic compounds etc.

The possible resonance structures for are shown below:

As we know that according to the octet rule; the stability of the resonating structure having complete octet of all the atoms present it that structure is more, hence the structure C is more stable than that of the structure A because, in structure C, octet of all the atoms is complete, whereas in structure A, C-atom having positive charge is not having 8 electrons in its valence shell.

As we know that the, stability of carbocation can be decided by many effects, like inductive effect, resonance, hyperconjugation etc. And a combined effect leads to the following order of stability of carbocation:

As structure A is having a carbocation, which is primary and allylic carbocation, whereas the carbocation in the structure B has secondary and allylic carbocation.

Hence, structure B is more stable.

(NOTE: As in both cases the allylic carbocation is common, it will not be the deciding factor for stability, hence we will decide the stability by primary and secondary carbocation.

The triphenylmethylcation is very stable because, it is a tertiary carbocation as well as the positive charge present on the carbon atom is being stabilised by three phenyl groups via resonance.

As we know that the phenomenon of resonance, leads to enormous stability of carbocation, as there is delocalisation of electron cloud, and the phenyl ring has great amount of delocalised electrons which are very potentially stabilising the positive charge lying on the carbon.

There are four possible carbocation structures for 2-methylbutane. And they are:

Now, the stability order is found to be:

(III) > (II) > (I) > (IV)

This is because (III) is tertiary carbocation, (II) is secondary carbocation, (I) and (IV) are primary, but as the number of alpha hydrogen in (I) is more than that of (IV) (as more number of alpha hydrogen leads to more stability of carbocation; in hyper-conjugation.)

As we know that theLassaigne’s extract is prepared to detect the presence of N, S, Halogens and P.

As when the Lassaigne’s extract is prepared, it allows all the elements present in the particular compound to come into the elemental state, so that they can easily be detected.

Now, the reaction involved is for Nitrogen estimation:

Na + C + N→ NaCN

NaCN + FeSO₄ + dilute sulphuric→ Fe₄[FeCN₆]₃ (Prussian blue colour solution)

The reaction involved is for sulphur estimation:

Na + C + N + S → NaCNS

3NaCNS + FeSO₄ + dilute sulphuric→ Fe(CNS)₃ (Blood red colour) + 3Na⁺

Hence, the Ramesh has N as well as S present in his organic compound so, he was getting the Blood red colour of Fe(CNS)₃ and Manish and Rajni were having only Nitrogen in their compound, hence they got Prussian blue colour.

(i) There is a C=O group and a double bond in the structure,

According to the IUPAC rule, numbering will start from the functional group.

The longest chain is found to be of 7 carbon atoms, and there is methyl group on fourth carbon atom and an ethyl group at the 3^rd carbon atom.

The name of the compound is 3-ethyl-4-methyl-5-hepten-2-one.

(ii) The compound is having a 6-membered ring having a double bond i.e cyclohexene and a Nitro group, as the Nitro group is functional group, numbering will start from the carbon to which Nitro group is attached.

Hence, the name of the compound is; 1-nitro-cyclohex-2-ene.

(a)The structural formulae of 1-Bromoheptane is: C₇H₁₅Br (as hept is denoting presence of 7 carbon atoms, 1-bromo is showing that at 1^st carbon, there is a bromine atom attached, and “ane” shows all are singly bonded.)

(b)The structural formulae of 5-Bromoheptanoic acid is: C₇H₁₃O₂Br ( as hept is denoting presence of 7 carbon atoms, 5-bromo is showing that at 5^th carbon, there is a bromine atom attached, and “ane” shows all are singly bonded and “oic acid” is indicating –COOH group.)

The resonating structures are as follow:-

(i)

Cl atom has loan pair and the loan pair of electrons are in conjugation with the double bond, hence the molecule shows resonance.

(ii)

The two double bonds are at alternate position, hence, the condition of conjugation is satisfied and the resonance is possible in the molecule.

(iii)

The two double bond are again in conjugation, as present at alternate position, and as the negative charge is more stable at O, as Oxygen atom is more electronegative atom, hence the structure Iis more stable.

(i) ion is more stable among all, because as we know that the Br atom is an electron withdrawing atom, although it is having loan pair of electrons, so that the positive charge present on the carbon atom is stabilised by the loan pair of the Bromine atom by resonance, but the for the halogens we know that the inductive effect dominates over the resonance effect, hence the Bromine atom will destabilise the positive charge on the carbon atom.

And as has three bromine atoms, the positive charge is more destabilised as compared to others which have two bromine atom, one bromine atom and no bromine atom.

(ii) As we know that Cl atom is more electron withdrawing atom, and also it has vacant d-orbital, so the negative charge on the C, is easily stabilised by the Cl atoms, more is the number of Cl atom attached to the Carbanion, more is the stability of the carbanion.

Hence, CCl₃^-is most stable.

The differences between inductive effect and resonance effect:

CH₃OH will not exist as resonance hybrid, because CH₃OH do not have any conjugation possible (as we know that the conjugation is possible only when any atom having loan pair or positive charge or negative charge is present at the alternate position of the double bond).

As there is an Oxygen atom having loan pairs of electron, but no double bond is present for conjugation.

Whereas in R—CONH₂, has conjugation in between the C=O bond and the loan pair of electrons present at the N atom in –NH₂ group.

And in CH₃CH = CHCH₂NH, there is also conjugation in between the C=C bond and the loan pair of electrons present at the N atom in –NH₂ group.

The S=O bond is a polar bond,this is because the O-atom is more electronegative atom than that of the S atom. Hence, O pulls the bonded electrons towards itself more as compared to the S atom pulls, as a result of which the electron density is more towards the O-atom and the O-atom gets the partial negative charge and S gets partial positive charge.

As the electron density at S in SO₃is decreased, it is in need of electrons to supress the partial positive charge and hence, SO₃ act as an electrophile.

As according the octet rule, the resonating structure which is having more atoms with complete octet, is stable, now looking at the structure I, we get to know that all the atoms are having complete octet, whereas in the structure II, the C-atom having positive charge is not having the complete octet, hence the structure I is more stable than structure II.

As an alcohol and hydrocarbon mixed together do not have much difference in the boiling point, and also the alcohol easily catches the fire, hence the suitable method for the separation would be the “Steam Distillation”.

As the Steam distillation is a type of distillation, which is specially done for the purification of the temperature sensitive material, hence, it will be the best fit method.

If we draw the resonating structure of the two compounds we have:

A:

B:

We know, that the compounds having more equivalent structure are more stable.

Now, the resonance in compound A, leads to generation of the positive charge on the Oxygen atom, whereas no such generation of the positive charge is found in the acetate ion i.e compound B.

Hence, the two resonating structure in compound A, are not equivalent, whereas the compound B, having negative charge on the oxygen atom is having the two equivalent structures, hence the compound B is more stable.

(i) – (e), (ii) – (d), (iii) – (a), (iv) – (b), (v) – (c)

Explanation:

(i) Two solids which have different solubility in a solvent and which do not undergo reaction when dissolved in it, is the condition for crystallisation. It is same as when the impurity and the compounds are dissolved they do not interact with each other.

Hence, option (e) is correct match.

(ii)Liquid that decomposes at its boiling point are affected by the temperature and hence can only be separated by the distillation method, as during distillation the compound is not directly heated but is heated by putting the flask in either oil bath or water bath.

Hence, option (d) is correct match.

(iii) Steam volatile liquid are separated by stem distillation as they are temperature sensitive materials and they are separated by steam distillation.

Hence, option (a) is correct match.

(iv)When the two liquids which have boiling points close to each other than the separation technique used is fractional distillation. As fractional distillation setup is made in such a way that the two liquid having close boiling point can also be separated very easily.

Hence, option (b) is correct match.

(v) Two liquids with large difference in boiling points, than simple distillation is best which can separate the two liquids. The setup is easy and affordable.

Hence, option (c) is correct match.

(i) – (c), (ii) – (f), (iii) – (b), (iv) – (a), (v) – (d), (vi) – (e)

Explanation:

(i)Carbocation is a species which the carbon atom has formed three sigma bonds and it has an empty p-orbital, hence option (c) is correct match.

(ii) We know nucleophile is a species which have electrons for the supply to the species which is able to accept, hence option (f) is correct match.

(iii) Hyperconjugation is defined as the phenomenon of Conjugation of electrons of C–H σ bond with empty p-orbital present at adjacent positively charged carbon, hence option (b) is correct match.

(iv) Isomers are those two compounds which are having same molecular formula but has different arrangement of atoms in the molecule.

The cyclohexane and 1-hexene are “ring-chain isomers” hence, option (a) is correct match.

(v) The molecule ethyne is having two C-atoms which are forming three bonds in between them.

And as out of the three bonds two are pi-bonds and only one is sigma bond, and also the sigma bond is formed with H. Hence, we find each C-atom forms only two sigma bond, we know for two sigma bonds the hybridisation is sp, hence, option (d) is correct match.

(vi)We know that the electrophile is the species which can accepts the electrons, as the word electrophile itself means the electron loving species, hence, option (e) is correct match.

(i) – (c), (ii) – (e),(iii) – (a), (iv) – (b), (v) – (d)

Explanation:

(i) Dumas method is used for estimation of the Nitrogen, from organic compounds by heating the organic compounds Nitrogen gas is produced and the amount of the nitrogen gas is estimated, hence option (c), is correct.

(ii) Kjeldahl’s method is used for estimation of the Nitrogen but in this case the nitrogen containing organic compound is heated in sulphuricacid, and ammonium sulphate is formed, and from ammonium sulphate the amount of N is estimated, hence option (e), is correct.

(iii) Carius method is used to estimate the halogen quantity, some amount of an organic compound containing halogen is heated with fuming nitric acid in the presence of silver nitrate, as a result silver halide is formed which is analysed for finding the amount of halogen, hence option (a), is correct.

(iv) Silica gel isused for packing the column in the Chromatography, hence option (b), is correct.

(v) Homolysis is the process which leads to the formation of free radicals, hence option (d), is correct.

(i) – (a), (ii) – (a), (iii) – (b)

Explanation:

Free radical and the carbocation, both have the trigonal planar geometry, this is because, the hybridisation of the carbocation and the free radical is sp² as they are having only two p-orbitals involved in the hybridisation along with s-orbital, and one p-orbital remains vacant.

Whereas the carbanion is having an extra electron, due to which it is an anion, and the extra electron is making the hybridisation sp³, and geometry is tetrahedral. Out of position in tetrahedral geometry one position is occupied by the electron pair, hence the the shape of the carbanion becomes pyramidal.

Hence, the match will be (i) – (a), (ii) – (a), (iii) – (b)

(i) – (a) (ii) – (b) (iii) – (c) (iv) – (d)

Explanation:

in this case the loan pair of the oxygen atom satisfies the positive charge present on the C atom adjacent to O-atom.

And this movement of electrons from the O atom to the C⁺ is called Resonance, hence the given species is stable due to resonance. Hence option (a) is correct match.

in this case we know that the F atom is electronegative atom, hence, it pulls the electrons towards itself and hence it destabilises the C⁺. Hence option (b) is correct match.

in this case, there are 9 alpha hydrogen in the carbanion, hence due to more hyper-conjugative structures the given anion is more stable.

(as more the number of the alpha hydrogen, more is the number of the hyper-conjugative structure.)

in this case, we can see that the carbon having positive charge is attached to the two carbon atoms, and we know that when any carbon atom is attached to the two alkyl groups, it is known as the secondary carbon, and as the secondary carbon is having positive charge, it will be called as the secondary carbocation.

Mixture of propan-1-ol and propanone can be separated by steam distillation,as Simple distillation can be used to separate a mixture of two liquid which do not react and have boiling point difference of more than 20^oC.

Hence, both assertion and reason are correct and also reason is the correct explanation
Hence, option (i) is correct.

A is not correct but R is correct.

We know that the resonance hybrids are always more stable than any of the canonical structures because of the resonance energy.

The energy of orbitals is lowered by the delocalisation of the electrons in resonance, and as the energy is decreased the stability increased, and the amount of energy which is reduced, due to resonance, this energy is called resonance energy.

A canonical structure which is having lower energy,have a greater contribution to resonance hybrid.

Hence, Assertion is not correct but reason is correct.

Option (iv) is correct.

The two compounds i.e Pent- 1- ene and pent- 2- ene, are position isomers as they differ in the position of the position of the double bond only, as we know that the position isomers have same molecular formula but they differ in the position of functional group or a substituent or the position of the double bond.

Hence, option (i) is correct option.

In H₂C = C = CH₂ , both the terminal carbon atoms are sp2 hybridised, whereas the carbon atom present in the centre of the molecule is forming only two sigma bonds, as the sigma bond is used for calculating the hybridisation. Hence the hybridisation of the central carbon is sp.

So, the assertion is wrong.

Now, considering the Reason, In this molecule all the carbon atoms are attached to each other by double bonds, is true. Hence option (iv) is correct.

We know that the Carius method is used to estimate the halogen quantity, hence assertion is found to be wrong.

Now reason, Sulphur can be separated easily from other atoms in the molecule, by lassiagn’s test, but the yellow precipitate cannot be obtained.

As both assertion and reason is incorrect.

Option (iii) is the answer.

The chromatography paper is used in paper chromatography, contains water in it, which is the stationary phase, and the solvent used is as the mobile phase. A strip of chromatography paper spotted at the base with ink is suspended in a suitable solvent.

The two inks are spotted at the base of the paper strip, and is dipped into the mobile phase, as the solvent rises up the paper by capillary action and flows over the spot selectively the paper retains different components according to their affinity for the paper.
Hence, the coloured components of inks migrate at different rates because paper selectively retains different components according to the difference in their partition between the two phases.

The assertion and reason both are correct and reason is correct explanation of the assertion.

Option (i) is correct answer.

Hybridization is a term coined by a scientist Pauling in 1931.

It is an imaginary concept which shows the intermixing of the orbitals to form a new one, which is of approachable in energy, to make a bond.

It will be more easy to understand with an example, so let say we have to form a bond between the C atom and a H atom, we know that the bond can be formed by the overlapping of the orbitals, so for the proper overlapping of the orbitals, the energy of the orbital must be comparable, but in this case it is not, hence, the orbitals of C, i.e s and p orbitals of C get mixed and form a new sp³ orbital, which can easily overlap with the s orbital od the H-atom. Hence, the bond can be formed easily now after hybridisation.

Now, checking the hybridisation of the carbon atoms in the molecule CH₂ = C = CH₂.

The molecule CH₂ = C = CH₂ contains both sp as well as the sp² hybridised carbon atoms.

The two terminal Carbon atoms are forming three sigma bonds so, the hybridisation will be sp² , whereas the central carbon atom is forming only two sigma bonds so, the hybridisation will be sp.

As we know that the hybridisation of any atoms is predicted by considering the number of the sigma bond formed by that particular atom and the loan pair of electrons present at the atom.

The molecule will be planar as the sp² hybridisation leads to trigonal planar shape and sp leads to formation of linear geometry.

Hence, the molecule CH₂ = C = CH₂ is overall planar molecule.

According to the principle of crystallisation, we know that the solute must be less soluble in the cold water and more soluble in hot water, or any other solvent like water.

Now, Benzoic acid is soluble in water in lesser amount as it is an organic compound and is less polar, whereas the water molecule is very polar. But on increasing the temperature of the water we observed that the benzoic acid is more soluble in hot water as compared to cold water, and the other reason is that the impurities are almost insoluble in water or completely soluble in cold water, so that they either can be filtered from the benzoic acid solution (if not soluble) or they remain dissolved even after cooling the solution(if completely soluble in cold water), and do not interfere with recrystallization of the benzoic acid, as we can discard the extra solution once the crystallisation process is complete and we obtain the purified benzoic acid.

Hence the characteristic properties of Benzoic acids are:

Benzoic acid is relatively more soluble in hot water as compared tothe cold water.
(ii) Impurities present in benzoic acid are insoluble in water or more soluble in water so that they remain in the solution.

In the method of the fractional distillation, the temperature of the mixture is gradually increased, and the compound having the lowest boiling point is separated first.

This is because, at high temperature, all the liquids whose boiling point is low will also get evaporated and hence no separation is possible.

Hence, in the given case, the boiling point of liquid (A) is less than boiling point of liquid (B), hence the liquids A is expected to come out first in the distillate.

As liquid A boils at a higher temperature than B and C and boiling point of B is lower than C.

So, the order of boiling point for all three compounds is:

A > C > B

Now, as it is mentioned that there is a large difference in the boiling points of A and rest of the two liquids i.e., B and C. So, liquid A can be separated from the mixture by simple distillation.

And as the two liquids i.e B and C, are having close boiling points, than we use fractional distillation for the separation of the two.

The diagram showing set up of the apparatus for the processes are as follows:

Simple distillation setup:

Fractional distillation Setup :

The diagram of bubble plate type fractionating column:

The bubble plate type fractionating column can be used for the continuous separation of bulk quantities of liquids, its industrial application are in separation of oil, diesel oil, lubricating oil, etc.

The principle use in the bubble plate type fractionating column is Raoult's law. The fractionating column separates the mixture by cooling the mixed vapours, allowing the vapour to condense, and vaporize again.

A liquid with high boiling point decomposes on simple distillation but it can be steam distilled for its purification, because the steam distillation is used for separating substances which are volatile, insoluble in water and also has a high vapour pressure at the 100 ^oC, i.e the boiling point of water.

It is also used for purifying liquids which are temperature sensitive and it decomposes at their normal boiling point. Steam distillation is used for separating organic compounds like aromatic compounds which are mainly extracted from plant parts.