Answer

(i) $\stackrel{1}{C} H_2=\stackrel{2}{C}=0$

C- 1 is $s p^{2}$ hybridised.

C- 2 is sp hybridised.

(ii)

$\stackrel{1}{C} H_3-\stackrel{2}{C} H=\stackrel{3}{C} H_2$

C- 1 is $s p^{3}$ hybridised.

C- 2 is $s p^{2}$ hybridised.

C- 3 is $s p^{2}$ hybridised.

(iii)

C-1 and C-3 are sp³ hybridised.

C- 2 is $s p^{2}$ hybridised.

(iv)

C- 1 is $s p^{2}$ hybridised.

C- 2 is $s p^{2}$ hybridised.

C- 3 is $s p$ hybridised.

(v) $C_6 H_6$

All the 6 carbon atoms in benzene are $s p^{2}$ hybridised.

Answer

(i) $C_6 H_6$

There are six C-C sigma ( ${\sigma _{C-C}}$ ) bonds, six C-H sigma ( ${\sigma _{C-H}}$ ) bonds, and three $C=C$ pi $({ }^{C-C})$ resonating bonds in the given compound.

(ii) $C_6 H _{12}$

There are six C-C sigma ( $\sigma _{C-C}$ ) bonds and twelve C-H sigma ( $\sigma _{C-H}$ ) bonds in the given compound.

(iii) $CH_2 Cl_2$

There two C-H sigma ( $\sigma _{C-H}$ ) bonds and two C-Cl sigma ( ${\sigma _{C-Cl}}$ ) bonds in the given compound.

(iv) $CH_2=C=CH_2$

There are two C-C sigma ( ${\sigma _{C-C}}$ ) bonds, four C-H sigma ( ${\sigma _{C-H}}$ ) bonds, and two $C=C$ pi ( ${ }^{\pi _{C-C}}$ ) bonds in the given compound.

(v) $CH_3 NO_2$

There are three C-H sigma ( $\sigma _{C-H})$ bonds, one C-N sigma ( $\sigma _{C-N})$ bond, one N-O sigma ( $\sigma _{N-0})$ bond, and one $N=O$ pi ( $.\pi _{N-O})$ bond in the given compound.

(vi) $HCONHCH_3$

There are two C-N sigma ( $\sigma _{C-N}$ ) bonds, four C-H sigma ( $\sigma _{C-H}$ ) bonds, one N-H sigma bond, and one $C=O$ pi ( $\pi _{C-C}$ ) bond in the given compound.

Answer

The bond line formulae of the given compounds are:

(a) Isopropyl alcohol

$\Rightarrow$

(b) 2, 3-dimethyl butanal

$\Rightarrow$

(c) Heptan-4-one

$ H _3 C-CH _2-CH _2 -\overbrace C^{O}-CH_2-CH_2-CH_3 $

$\Rightarrow$

Answer

(a)

1-phenyl propane

(b)

3-methylpentanenitrile

(c)

2, 5-dimethyl heptane

(d)

3-bromo-3-chloroheptane

(e)

3-chloropropanal

(f) $Cl_2 CHCH_2 OH$

1, 1-dichloro-2-ethanol

Answer

(a) The prefix diin the IUPAC name indicates that two identical substituent groups are present in the parent chain. Since two methyl groups are present in the C-2 of the parent chain of the given compound, the correct IPUAC name of the given compound is 2, 2-dimethylpentane.

(b) Locant number 2, 4, 7 is lower than 2, 5, 7. Hence, the IUPAC name of the given compound is 2, 4, 7trimethyloctane.

(c) If the substituents are present in the equivalent position of the parent chain, then the lower number is given to the one that comes first in the name according to the alphabetical order. Hence, the correct IUPAC name of the given compound is 2-chloro-4-methylpentane.

(d) Two functional groups - alcoholic and alkyne - are present in the given compound. The principal functional group is the alcoholic group. Hence, the parent chain will be suffixed with ol. The alkyne group is present in the C-3 of the parent chain. Hence, the correct IUPAC name of the given compound is But-3-yn-1-ol.

Answer

The first five members of each homologous series beginning with the given compounds are shown as follows:

(a)

$H-COOH$ : Methanoic acid

$CH_3-COOH$ : Ethanoic acid

$CH_3-CH_2-COOH$ : Propanoic acid

$CH_3-CH_2-CH_2-COOH$ : Butanoic acid

$CH_3-CH_2-CH_2-CH_2-COOH$ : Pentanoic acid

(b)

$CH_3 COCH_3$ : Propanone

$CH_3 COCH_2 CH_3$ : Butanone

$CH_3 COCH_2 CH_2 CH_3$ : Pentan-2-one

$CH_3 COCH_2 CH_2 CH_2 CH_3$ : Hexan-2-one

$CH_3 COCH_2 CH_2 CH_2 CH_2 CH_3$ : Heptan-2-one

(c)

$H-CH=CH_2$ : Ethene

$CH_3-CH=CH_2$ : Propene

$CH_3-CH_2-CH=CH_2:$ 1-Butene

$CH_3-CH_2-CH_2-CH=CH_2$ : 1-Pentene

$CH_3-CH_2-CH_2-CH_2-CH=CH_2: 1-Hexene$

Answer

(a) 2, 2, 4-trimethylpentane

Condensed formula:

$(CH_3)_2 CHCH_2 C(CH_3)_3$

Bond line formula:

(b) 2-hydroxy-1, 2, 3-propanetricarboxylic acid Condensed Formula:

$(COOH) CH_2 C(OH)(COOH) CH_2(COOH)$

Bond line formula:

The functional groups present in the given compound are carboxylic acid (-COOH) and alcoholic (-OH) groups.

(c) Hexanedial

Condensed Formula:

$(CHO)(CH_2)_4(CHO)$

Bond line Formula:

The functional group present in the given compound is aldehyde $(-CHO)$.

Answer

The functional groups present in the given compounds are:

(a) Aldehyde (-CHO),

Hydroxyl(-OH),

Methoxy (-OMe),

$C=C$ double bond $(-\stackrel{1}{C}=\stackrel{1}{C}-)$

(b) Amino (- $NH_2$ ); primary amine,

Ester (-O-CO-),

Triethylamine $(N(C_2 H_5)_2)$; tertiary amine

(c) Nitro (– $NO_2$ ),

$C=C$ double bond $(-C=C-)$

Answer

$NO_2$ group is an electron-withdrawing group. Hence, it shows -I effect. By withdrawing the electrons toward it, the $NO_2$ group decreases the negative charge on the compound, thereby stabilising it. On the other hand, ethyl group is an electron-releasing group. Hence, the ethyl group shows +1 effect. This increases the negative charge on the compound, thereby destabilising it. Hence, $O_2 NCH_2 CH_2 O$ is expected to be more stable than $CH_3 CH_2 O$.

Answer

When an alkyl group is attached to a , $\neg$ system, it acts as an electron-donor group by the process of hyperconjugation. To understand this concept better, let us take the example of propene.

In hyperconjugation, the sigma electrons of the $C-H$ bond of an alkyl group are delocalised. This group is directly attached to an atom of an unsaturated system. The delocalisation occurs because of a partial overlap of a $s p^{3}$-ssigma bond orbital with an empty $p$ orbital of the - $\neg$ bond of an adjacent carbon atom.

The process of hyperconjugation in propene is shown as follows:

This type of overlap leads to a delocalisation (also known as no-bond resonance) of the - $\neg$ electrons, making the molecule more stable.

IV

III

Answer

(a) The structure of $C_6 H_5 OH$ is:

The resonating structures of phenol are represented as:

(b) The structure of $C_6 H_5 NO_2$ is:

The resonating structures of nitro benzene are represented as:

(c) $CH_3 CH=CH-CHO$

The resonating structures of the given compound are represented as:

1

II

(d) The structure of $C_6 H_5 CHO$ is:

The resonating structures of benzaldehyde are represented as:

(e) $C_6 H_5 CH_2^{-}$

The resonating structures of the given compound are:

(f) $CH_3 CH=CHCH_2^{-}$

The resonating structures of the given compound are:

Answer

An electrophile is a reagent that takes away an electron pair. In other words, an electron-seeking reagent is called an electrophile $(E^{+})$. Electrophiles are electron-deficient and can receive an electron pair.

Carbocations $(CH_3 CH_2^{+})$and neutral molecules having functional groups such as carbonyl group $(/ C=0$ ) are examples of electrophiles.

A nulceophile is a reagent that brings an electron pair. In other words, a nucleus-seeking reagent is called a nulceophile (Nu:).

For example: $OH^{-}, NC^{-} e^{e}$, carbanions $(R_3 C^{-})$, etc.

Neutral molecules such as $H_2-$ and ammonia also act as nulceophiles because of the presence of a lone pair.

Answer

Electrophiles are electron-deficient species and can receive an electron pair. On the other hand, nucleophiles are electron-rich species and can donate their electrons.

(a) $CH_3 COOH+HO^{-} \longrightarrow CH_3 COO^{-}+H_2 O$

Here, $HO^{-} e$ acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus-seeking species.

(b) $CH_3 COCH_3+\stackrel{-}{\mathbf{C}} \mathbf{N} \longrightarrow(CH_3)_2 C(CN)+(OH)$

Here, $- CN$ acts as a nucleophile as it is an electron-rich species, i.e., it is a nucleus-seeking species. (c) $C_6 H_5+\mathbf{C H}_3 \stackrel{+}{\mathbf{C}} O \longrightarrow C_6 H_5 COCH_3$

Here, $CH_3 \stackrel{+}{C} O$ acts as an electrophile as it is an electron-deficient species.

Answer

(a) It is an example of substitution reaction as in this reaction the bromine group in bromoethane is substituted by the -SH group.

(b) It is an example of addition reaction as in this reaction two reactant molecules combine to form a single product.

(c) It is an example of elimination reaction as in this reaction hydrogen and bromine are removed from bromoethane to give ethene.

(d) In this reaction, substitution takes place, followed by a rearrangement of atoms and groups of atoms.

Answer

(a) Compounds having the same molecular formula but with different structures are called structural isomers. The given compounds have the same molecular formula but they differ in the position of the functional group (ketone group).

In structure I, ketone group is at the C-3 of the parent chain (hexane chain) and in structure II, ketone group is at the $C-2$ of the parent chain (hexane chain). Hence, the given pair represents structural isomers.

(b) Compounds having the same molecular formula, the same constitution, and the sequence of covalent bonds, but with different relative position of their atoms in space are called geometrical isomers.

In structures I and II, the relative position of Deuterium (D) and hydrogen $(H)$ in space are different. Hence, the given pairs represent geometrical isomers.

(c) The given structures are canonical structures or contributing structures. They are hypothetical and individually do not represent any real molecule. Hence, the given pair represents resonance structures, called resonance isomers.

Answer

(a) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is an example of homolytic cleavage as one of the shared pair in a covalent bond goes with the bonded atom. The reaction intermediate formed is a free radical.

(b) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the carbon of propanone. The reaction intermediate formed is carbanion.

(c) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is an example of heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with the bromine ion. The reaction intermediate formed is a carbocation.

(d) The bond cleavage using curved-arrows to show the electron flow of the given reaction can be represented as

It is a heterolytic cleavage as the bond breaks in such a manner that the shared pair of electrons remains with one of the fragments. The intermediate formed is a carbocation.

Answer

Inductive effect

The permanent displacement of sigma ( $\tilde{A} E^{\prime}$ ) electrons along a saturated chain, whenever an electron withdrawing or electron donating group is present, is called inductive effect.

Inductive effect could be $+I$ effect or $-I$ effect. When an atom or group attracts electrons towards itself more strongly than hydrogen, it is said to possess - I effect. For example,

$F \longrightarrow CH_2 \longleftarrow CH_2 \backsim CH_2 \backsim CH_3$

When an atom or group attracts electrons towards itself less strongly than hydrogen, it is said to possess +1 effect. For example,

$CH_3 \longrightarrow CH_2 \to Cl$

Electrometric effect

It involves the complete transfer of the shared pair of - $\urcorner$ electrons to either of the two atoms linked by multiple bonds in the presence of an attacking agent. For example,

Electrometric effect could be $+E$ effect or $-E$ effect.

• E effect: When the electrons are transferred towards the attacking reagent

• E effect: When the electrons are transferred away from the attacking reagent

(a) $Cl_3 CCOOH>Cl_2 CHCOOH>ClCH_2 COOH$

The order of acidity can be explained on the basis of Inductive effect (- I effect). As the number of chlorine atoms increases, the - I effect increases. With the increase in - I effect, the acid strength also increases accordingly.

(b) $CH_3 CH_2 COOH>(CH_3)_2 CHCOOH>(CH_3)_3 C . COOH$

The order of acidity can be explained on the basis of inductive effect (+ I effect). As the number of alkyl groups increases, the $+I$ effect also increases. With the increase in $+I$ effect, the acid strength also increases accordingly.

Answer

(a) Crystallisation

Crystallisation is one of the most commonly used techniques for the purification of solid organic compounds.

Principle: It is based on the difference in the solubilites of the compound and the impurities in a given solvent. The impure compound gets dissolved in the solvent in which it is sparingly soluble at room temperature, but appreciably soluble at higher temperature. The solution is concentrated to obtain a nearly saturated solution. On cooling the solution, the pure compound crystallises out and is removed by filtration.

For example, pure aspirin is obtained by recrystallising crude aspirin. Approximately $2-4 g$ of crude aspirin is dissolved in about $20 mL$ of ethyl alcohol. The solution is heated (if necessary) to ensure complete dissolution. The solution is then left undisturbed until some crystals start to separate out. The crystals are then filtered and dried.

(b) Distillation

This method is used to separate volatile liquids from non-volatile impurities or a mixture of those liquids that have a sufficient difference in their boiling points.

Principle: It is based on the fact that liquids having different boiling points vapourise at different temperatures. The vapours are then cooled and the liquids so formed are collected separately.

For example, a mixture of chloroform ( $b . p=334 K$ ) and aniline ( $b . p=457 K$ ) can be separated by the method of distillation. The mixture is taken in a round bottom flask fitted with a condenser. It is then heated. Chloroform, being more volatile, vaporizes first and passes into the condenser. In the condenser, the vapours condense and chloroform trickles down. In the round bottom flask, aniline is left behind.

(c) Chromatography

It is one of the most useful methods for the separation and purification of organic compounds.

Principle: It is based on the difference in movement of individual components of a mixture through the stationary phase under the influence of mobile phase.

For example, a mixture of red and blue ink can be separated by chromatography. A drop of the mixture is placed on the chromatogram. The component of the ink, which is less adsorbed on the chromatogram, moves with the mobile phase while the less adsorbed component remains almost stationary.

Answer

Fractional crystallisation is the method used for separating two compounds with different solubilities in a solvent $S$. The process of fractional crystallisation is carried out in four steps.

(a) Preparation of the solution: The powdered mixture is taken in a flask and the solvent is added to it slowly and stirred simultaneously. The solvent is added till the solute is just dissolved in the solvent. This saturated solution is then heated.

(b) Filtration of the solution: The hot saturated solution is then filtered through a filter paper in a China dish.

(c) Fractional crystallisation: The solution in the China dish is now allowed to cool. The less soluble compound crystallises first, while the more soluble compound remains in the solution. After separating these crystals from the mother liquor, the latter is concentrated once again. The hot solution is allowed to cool and consequently, the crystals of the more soluble compound are obtained.

(d) Isolation and drying: These crystals are separated from the mother liquor by filtration. Finally, the crystals are dried.

Answer

The differences among distillation, distillation under reduced pressure,

and steam distillation are given in the following table.

Answer

Lassaigne’s test

This test is employed to detect the presence of nitrogen, sulphur, halogens, and phosphorous in an organic compound. These elements are present in the covalent form in an organic compound. These are converted into the ionic form by fusing the compound with sodium metal.

$ \begin{aligned} & Na+C+N \xrightarrow{\Delta} NaCN \\ & 2 Na+S \xrightarrow{\Delta} Na_2 S \\ & Na+X \xrightarrow{\Delta} NaX \\ & (X=Cl, Br, I) \end{aligned} $

The cyanide, sulphide, and halide of sodium formed are extracted from the fused mass by boiling it in distilled water. The extract so obtained is called Lassaigne’s extract. This Lassaigne’s extract is then tested for the presence of nitrogen, sulphur, halogens, and phosphorous.

(a) Test for nitrogen

Prussian blue colour

(Ferriferro cyanide)

Chemistry of the test

In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as

$ \begin{aligned} & 6 CN^{-}+Fe^{2+} \longrightarrow[Fe(CN)_6]^{4-} \\ & 3[Fe(CN)_6]^{4-}+4 Fe^{3+} \xrightarrow{x H_2 O} Fe_4[Fe(CN)_6]_3 x H_2 O \end{aligned} $

Prussian blue colour

(b) Test for sulphur

(i) Lassaigne’s extract + Lead acetate $\xrightarrow{\text{ acetic acid }}$ Black precipitate

Chemistry of the test

In the Lassaigne’s test for sulphur in an organic compound, the sodium fusion extract is acidified with acetic acid and then lead acetate is added to it. The precipitation of lead sulphide, which is black in colour, indicates the presence of sulphur in the compound.

$S^{2-}+Pb^{2+} \longrightarrow PbS$

(Black)

(ii) Lassaigne’s extract + Sodium nitroprusside $\longrightarrow$ Violet colour

Chemistry of the test

The sodium fusion extract is treated with sodium nitroprusside. Appearance of violet colour also indicates the presence of sulphur in the compound.

$ \begin{aligned} & S^{2-}+[.Fe(CN)_5 NO]^{2-} \longrightarrow \\ &(\text{ Violet) } \end{aligned} $

If in an organic compound, both nitrogen and sulphur are present, then instead of $NaCN$, formation of NaSCN takes place.

$Na+C+N+S$ Ã

This NaSCN (sodium thiocyanate) gives a blood red colour. Prussian colour is not formed due to the absence of free cyanide ions.

$ Fe^{3+}+SCN^{-} \longrightarrow[Fe(SCN)]^{2+} $

(Blood red)

(c) Test for halogens

Chemistry of the test

In the Lassaigne’s test for halogens in an organic compound, the sodium fusion extract is acidified with nitric acid and then treated with silver nitrate.

$ \begin{aligned} & X^{-}+Ag^{+} \longrightarrow AgX \\ &(X=Cl, Br, I) \end{aligned} $

If nitrogen and sulphur both are present in the organic compound, then the Lassaigne’s extract is boiled to expel nitrogen and sulphur, which would otherwise interfere in the test for halogens.

Answer

In Dumas method, a known quantity of nitrogen containing organic compound is heated strongly with excess of copper oxide in an atmosphere of carbon dioxide to produce free nitrogen in addition to carbon dioxide and water. The chemical equation involved in the process can be represented as

$ C x HyNz+(2 x+y / 2) CuO \longrightarrow x CO_2+y / 2 H_2 O+z / 2 N_2+(2 x+y / 2) Cu $

The traces of nitrogen oxides can also be produced in the reaction, which can be reduced to dinitrogen by passing the gaseous mixture over a heated copper gauge. The dinitrogen produced is collected over an aqueous solution of potassium hydroxide. The volume of nitrogen produced is then measured at room temperature and atmospheric pressure.

On the other hand, in Kjeldahl’s method, a known quantity of nitrogen containing organic compound is heated with concentrated sulphuric acid. The nitrogen present in the compound is quantitatively converted into ammonium sulphate. It is then distilled with excess of sodium hydroxide. The ammonia evolved during this process is passed into a known volume of $H_2 SO_4$. The chemical equations involved in the process are

The acid that is left unused is estimated by volumetric analysis (titrating it against a standard alkali) and the amount of ammonia produced can be determined. Thus, the percentage of nitrogen in the compound can be estimated. This method cannot be applied to the compounds, in which nitrogen is present in a ring structure, and also not applicable to compounds containing nitro and azo groups.

Answer

Estimation of halogens

Halogens are estimated by the Carius method. In this method, a known quantity of organic compound is heated with fuming nitric acid in the presence of silver nitrate, contained in a hard glass tube called the Carius tube, taken in a furnace. Carbon and hydrogen that are present in the compound are oxidized to form $CO_2$ and $H_2 O$ respectively and the halogen present in the compound is converted to the form of $AgX$. This $AgX$ is then filtered, washed, dried, and weighed.

Let the mass of organic compound be $m g$.

Mass of $AgX$ formed $=m_1 g$

1 mol of Agx contains 1 mol of $X$.

Therefore,

Mass of halogen inm $I_1 g$ of $AgX$

$ =\frac{\text{ Atomic mass of } X \times m_1 g}{\text{ Molecular mass of } AgX} $

Thus, $%$ of halogen will be $=\frac{\text{ Atomic mass of } X \times m_1 \times 100}{\text{ Molecular mass of } AgX \times m}$

Estimation of Sulphur

In this method, a known quantity of organic compound is heated with either fuming nitric acid or sodium peroxide in a hard glass tube called the Carius tube. Sulphur, present in the compound, is oxidized to form sulphuric acid. On addition of excess of barium chloride to it, the precipitation of barium sulphate takes place. This precipitate is then filtered, washed, dried, and weighed.

Let the mass of organic compound be $m g$.

Mass of $BaSO_4$ formed $=m_1 g$

$1 mol^{2}$ of $BaSO_4=233 g BaSO_4=32 g$ of Sulphur

Therefore, $m_1 g$ of $BaSO_4$ contains $\frac{32 \times m_1}{233} g$ of sulphur.

Thus, percentage of sulphur $=\frac{32 \times m_1 \times 100}{233 \times m}$

Estimation of phosphorus

In this method, a known quantity of organic compound is heated with fuming nitric acid. Phosphorus, present in the compound, is oxidized to form phosphoric acid. By adding ammonia and ammonium molybdate to the solution, phosphorus can be precipitated as ammonium phosphomolybdate.

Phosphorus can also be estimated by precipitating it as $MgNH_4 PO_4$ by adding magnesia mixture, which on ignition yields $Mg_2 P_2 O_7$.

Let the mass of organic compound be $m g$.

Mass of ammonium phosphomolybdate formed $=m_1 g$

Molar mass of ammonium phosphomolybdate $=1877 g$

Thus, percentage of phosphorus $=\frac{31 \times m_1 \times 100}{1877 \times m} %$

If $P$ is estimated as $Mg_2 P_2 O_7$,

Then, percentage of phosphorus $=\frac{62 \times m_1 \times 100}{222 \times m} %$

Answer

In paper chromatography, chromatography paper is used. This paper contains water trapped in it, which acts as the stationary phase. On the base of this chromatography paper, the solution of the mixture is spotted. The paper strip is then suspended in a suitable solvent, which acts as the mobile phase. This solvent rises up the chromatography paper by capillary action and in the procedure, it flows over the spot. The components are selectively retained on the paper (according to their differing partition in these two phases). The spots of different components travel with the mobile phase to different heights. The paper so obtained (shown in the given figure) is known as a chromatogram.

Answer

While testing the Lassaigne’s extract for the presence of halogens, it is first boiled with dilute nitric acid. This is done to decompose $NaCN$ to $HCN$ and $Na_2 S$ to $H_2 S$ and to expel these gases. That is, if any nitrogen and sulphur are present in the form of $NaCN$ and $Na_2 S$, then they are removed. The chemical equations involved in the reaction are represented as

$ \begin{aligned} & NaCN+HNO_3 \longrightarrow NaNO_3+HCN \\ & Na_2 S+2 HNO_3 \longrightarrow 2 NaNO_3+H_2 S \end{aligned} $

Answer

Nitrogen, sulphur, and halogens are covalently bonded in organic compounds. For their detection, they have to be first converted to ionic form. This is done by fusing the organic compound with sodium metal. This is called “Lassaigne’s test”. The chemical equations involved in the test are

$ \begin{aligned} & Na+C+N \longrightarrow NaCN \\ & Na+S+C+N \longrightarrow NaSCN \\ & 2 Na+S \longrightarrow Na_2 S \\ & Na+X \longrightarrow NaX \\ &(X=Cl, Br, I) \end{aligned} $

Carbon, nitrogen, sulphur, and halogen come from organic compounds.

Answer

The process of sublimation is used to separate a mixture of camphor and calcium sulphate. In this process, the sublimable compound changes from solid to vapour state without passing through the liquid state. Camphor is a sublimable compound and calcium sulphate is a non-sublimable solid. Hence, on heating, camphor will sublime while calcium sulphate will be left behind.

Answer

In steam distillation, the organic liquid starts to boil when the sum of vapour pressure due to the organic liquid $(p_1)$ and the vapour pressure due to water $(p_2)$ becomes equal to atmospheric pressure $(p)$, that is, $p=p_1+p_2$

Since $p_1<p_2$, organic liquid will vapourise at a lower temperature than its boiling point.

Answer

$CCl_4$ will not give the white precipitate of $AgCl$ on heating it with silver nitrate. This is because the chlorine atoms are covalently bonded to carbon in $CCl_4$. To obtain the precipitate, it should be present in ionic form and for this, it is necessary to prepare the Lassaigne’s extract of $CCl_4$.

Answer

Carbon dioxide is acidic in nature and potassium hydroxide is a strong base. Hence, carbon dioxide reacts with potassium hydroxide to form potassium carbonate and water as $2 KOH+CO_2 \longrightarrow K_2 CO_3+H_2 O$

Thus, the mass of the U-tube containing $KOH$ increases. This increase in the mass of U-tube gives the mass of $CO_2$ produced. From its mass, the percentage of carbon in the organic compound can be estimated.

Answer

Although the addition of sulphuric acid will precipitate lead sulphate, the addition of acetic acid will ensure a complete precipitation of sulphur in the form of lead sulphate due to common ion effect. Hence, it is necessary to use acetic acid for acidification of sodium extract for testing sulphur by lead acetate test.

Answer

Percentage of carbon in organic compound $=69 %$

That is, $100 g$ of organic compound contains $69 g$ of carbon.

$\therefore 0.2 g$ of organic compound will contain $100=0.138 g$ of $C$

$ =\frac{69 \times 0.2}{100}=0.138 g \text{ of } C $

Molecular mass of carbon dioxide, $CO_2=44 g$

That is, $12 g$ of carbon is contained in $44 g$ of $CO_2$.

$ \text{ Therefore, } 0.138 g \text{ of carbon will be contained in } \frac{44 \times 0.138}{12}=0.506 g \text{ of } CO_2 $

Thus, $0.506 g$ of $CO_2$ will be produced on complete combustion of $0.2 g$ of organic compound.

Percentage of hydrogen in organic compound is 4.8.

i.e., $100 g$ of organic compound contains $4.8 g$ of hydrogen.

Therefore, $0.2 g$ of organic compound will contain

$ \frac{4.8 \times 0.2}{100}=0.0096 g \text{ of } H $

It is known that molecular mass of water $(H_2 O)$ is $18 g$.

Thus, $2 g$ of hydrogen is contained in $18 g$ of water. $\therefore 0.0096 g$ of hydrogen will be contained in $\frac{18 \times 0.0096}{2}=0.0864 g$ of water

Thus, $0.0864 g$ of water will be produced on complete combustion of $0.2 g$ of the organic compound.

Answer

Given that, total mass of organic compound $=0.50 g$

$60 mL$ of $0.5 M$ solution of $NaOH$ was required by residual acid for neutralisation.

$60 mL$ of $0.5 M NaOH$ solution $=\frac{60}{2} mL$ of $0.5 M H_2 SO_4=30 mL$ of $0.5 M H_2 SO_4$

$ =\frac{60}{2} mL \text{ of } 0.5 M $

$\therefore$ Acid consumed in absorption of evolved ammonia is (50âє"30) mL $=20 mL$

Again, $20 mL$ of $0.5 MH_2 SO_4=40 mL$ of $0.5 MNH_3$

Also, since $1000 mL$ of $1 MNH_3$ contains $14 g$ of nitrogen,

$\therefore 40 mL$ of $0.5 M NH_3$ will contain $\frac{14 \times 40}{1000} \times 0.5=0.28 g$ of N

Therefore, percentage of nitrogen in $0.50 g$ of organic compound $=\frac{0.28}{0.50} \times 100=56 %$

Answer

Given that,

Mass of organic compound is $0.3780 g$.

Mass of $AgCl$ formed $=0.5740 g$

$1 mol$ of $AgCl$ contains $1 mol$ of $Cl$.

Thus, mass of chlorine in $0.5740 g$ of $AgCl$ $=\frac{35.5 \times 0.5740}{143.32}$

$=0.1421 g$

$\therefore$ Percentage of chlorine $=\frac{0.1421}{0.3780} \times 100=37.59 %$

Hence, the percentage of chlorine present in the given organic chloro compound is $37.59 %$

Answer

Total mass of organic compound $=0.468 g$ [Given]

Mass of barium sulphate formed $=0.668 g$ [Given]

$1 mol$ of $BaSO_4=233 g$ of $BaSO_4=32 g$ of sulphur

Thus, $0.668 g$ of $BaSO_4$ contains $\frac{32 \times 0.668}{233} g$ of sulphur $=0.0917 g$ of sulphur

Therefore, percentage of sulphur $=\frac{0.0197}{0.468} \times 100=19.59 %$

Hence, the percentage of sulphur in the given compound is $19.59 %$.

Answer

$\stackrel{6}C_2=\stackrel{5}{C} H-\stackrel{4}C_2-\stackrel{3}C_2-\stackrel{2}{C} \equiv \stackrel{1}{C} H$

In the given organic compound, the carbon atoms numbered as 1, 2, 3, 4, 5, and 6 are $s p, s p, s p^{3}, s p^{3}, s p^{2}$, and $s p^{2}$ hybridized respectively. Thus, the pair of hybridized orbitals involved in the formation of $C_2-C_3$ bond is $s p \hat{a} \epsilon^{\prime \prime} s p^{3}$.

Answer

In the Lassaigne’s test for nitrogen in an organic compound, the sodium fusion extract is boiled with iron (II) sulphate and then acidified with sulphuric acid. In the process, sodium cyanide first reacts with iron (II) sulphate and forms sodium hexacyanoferrate (II). Then, on heating with sulphuric acid, some iron (II) gets oxidised to form iron (III) hexacyanoferrate (II), which is Prussian blue in colour. The chemical equations involved in the reaction can be represented as

$ \begin{matrix} 6 CN^{-}+Fe^{2+} \longrightarrow[Fe(CN)_6]^{4-} \\ 3[Fe(CN)_6]^{4-}+4 Fe^{3+} \xrightarrow{x H_2 O} Fe_4[Fe(CN)_6]_3 x H_2 O \\ \text{ Prussian blue } \end{matrix} $

Hence, the Prussian blue colour is due to the formation of $Fe_4[Fe(CN)_6]_3$.

Answer

$(CH_3)_3 \stackrel{+}{C}$

is a tertiary carbocation. A tertiary carbocation is the most stable carbocation due to the electron releasing effect of three methyl groups. An increased + I effect by three methyl groups stabilizes the positive charge on the carbocation.

Answer

Chromatography is the most useful and the latest technique of separation and purification of organic compounds. It was first used to separate a mixture of coloured substances.

Answer

$CH_3 CH_2 I+KOH _{(a q)} \longrightarrow CH_3 CH_2 OH+KI$

It is an example of nucleophilic substitution reaction. The hydroxyl group of $KOH(OH^{\text{af }})$ with a lone pair of itself acts as a nucleophile and substitutes iodide ion in $CH_3 CH_2$ to form ethanol.