Question:

Which of the following carbocation is most stable? (a)(CH3​)3​C−CH2+​ (b) (CH3​)3​C+ (c) CH3​CH2​CH2+​ (d) CH3​CH+CH2​CH3​

Answer:

1. First, we need to understand what makes a carbocation stable. A carbocation is stabilized by having an electron-donating group attached to the positively charged carbon atom, which reduces the positive charge.

2. Now, we can compare the four carbocations given in the question. (a) (CH3​)3​C−CH2+​ has three methyl groups attached to the positively charged carbon, making it the most stable carbocation. (b) (CH3​)3​C+ has three methyl groups attached, but no electron-donating group, so it is less stable than (a). (c) CH3​CH2​CH2+​ has two methyl groups and one ethyl group attached, making it more stable than (b). (d) CH3​CH+CH2​CH3​ has two methyl groups and two ethyl groups attached, making it the most stable of the four options.

3. Therefore, the most stable carbocation is (d) CH3​CH+CH2​CH3​.

Question:

The reaction: CH3​CH2​I+KOH(aq)→CH3​CH2​OH+KI is a nucleophilic substitution. Explain.

Answer:

1. Nucleophilic substitution is a type of chemical reaction in which a nucleophile (an electron-rich species) displaces a leaving group (an electron-poor species) from a substrate.

2. In the reaction CH3​CH2​I+KOH(aq)→CH3​CH2​OH+KI, the nucleophile is KOH (aq) and the leaving group is I.

3. The KOH (aq) nucleophile attacks the substrate, CH3​CH2​I, by forming a bond with the carbon atom and pushing the I leaving group off.

4. The result is the formation of CH3​CH2​OH and KI, which are the products of the reaction.

5. Therefore, the reaction CH3​CH2​I+KOH(aq)→CH3​CH2​OH+KI is a nucleophilic substitution reaction.

Question:

Explain the principle of paper chromatography.

Answer:

1. Paper chromatography is a technique used to separate and identify components of a mixture.

2. It works by using a solvent, such as water or alcohol, to carry a sample of the mixture up a piece of absorbent paper.

3. As the solvent moves up the paper, it carries the components of the mixture with it.

4. The components of the mixture travel at different rates, depending on their solubility in the solvent.

5. As they travel, they separate and form distinct bands on the paper.

6. By studying the bands, scientists can identify the components of the mixture.

Question:

Draw all the possible resonance structures for Benzaldehyde.

Answer:

1. Draw the Lewis structure of Benzaldehyde, which is C6H5CHO.

2. Determine the number of possible resonance structures for Benzaldehyde. This can be done by counting the number of electrons that can be moved around the molecule.

3. Draw the resonance structures for Benzaldehyde. Start by drawing the basic structure, then move electrons around to create the various resonance structures. Make sure to keep the overall charge of the molecule the same.

4. Label each resonance structure with its corresponding resonance number. This will help to identify each structure.

5. Review the resonance structures and make sure they are correct.

Question:

Draw the resonance structures for CH3​CH=CHC+H2​. Show the electron shift using curved-arrow notation.

Answer:

1. Draw the Lewis structure for CH3CH=CHC+H2:

CH3CH=CHC+H2

2. Draw the resonance structures for CH3CH=CHC+H2:

CH3CH-CHC+H2

CH3C=CH-CH2+H2

3. Show the electron shift using curved-arrow notation:

CH3CH-CHC+H2 ⇌ CH3C=CH-CH2+H2

← ↑ ↓ →

Question:

0.3780 g of an organic chloro compound gave 0.5740 g of silver chloride in Carius estimation. Calculate the percentage of chlorine present in the compound.

Answer:

Step 1: Calculate the mass of the chloro compound.

Mass of chloro compound = 0.5740 g - 0.3780 g = 0.1960 g

Step 2: Calculate the mass of chlorine present in the compound.

Mass of chlorine = 0.1960 g x (1 mole of chlorine/71.5 g of chlorine) = 0.0027 moles of chlorine

Step 3: Calculate the percentage of chlorine present in the compound.

Percentage of chlorine = (0.0027 moles of chlorine/1 mole of chloro compound) x 100 = 3.77%

Question:

Explain why alkyl groups act as electron donors when attached to a π system.

Answer:

1. Alkyl groups are composed of carbon atoms bonded to hydrogen atoms. These carbon atoms have a high electron density due to their abundance of electrons.

2. When an alkyl group is attached to a π system, the electron-rich carbon atom donates electrons to the π system, which is electron-deficient.

3. This electron donation stabilizes the π system and increases its electron density. This makes the π system more reactive, which is beneficial for many chemical reactions.

Question:

Give a brief description of the principles of the following techniques taking an example in each case. Chromatography.

Answer:

1. Definition of Chromatography: Chromatography is a technique used to separate and identify mixtures of substances, such as proteins, pigments, and other molecules. It is based on the principle that different components of a mixture will travel at different rates through a medium when a force is applied.

2. Example: A common example of chromatography is paper chromatography. In this type of chromatography, a mixture is placed onto a piece of paper and a solvent is added. The solvent will travel up the paper and separate the components of the mixture based on their solubility.

3. Principle: The principle of chromatography is based on the fact that different components of a mixture will travel at different rates when a force is applied. This is due to the differences in the interactions between the components and the medium they are travelling through. For example, in paper chromatography, the molecules that are more soluble in the solvent will travel faster than those that are less soluble.

Question:

The reaction CH3​CH2​I+KOH(aq)→CH3​CH2​OH+KI, is classified as : A electrophilic substitution B nucleophilic substitution C elimination D addition Classify the following reaction in one of the reaction type studied in this unit.CH3​3​CCH2​OHHBrCH3​2​CBrCH2​CH2​CH3​H2​O

Answer:

A electrophilic substitution

Question:

For the following bond cleavages, use curved-arrows to show the electron flow and classify each as homolysis or heterolysis. Identify reactive intermediate produced as free radical, carbocation and carbanion: H3​CO−OCH3​⟶CH3​O˙+O˙CH3​

Answer:

1. Electron flow: H3CO-OCH3→ H3CO• + OCH3•

2. Classification: Homolysis

3. Reactive intermediate produced: Free radical (H3CO• and OCH3•)

Question:

Discuss the principle of estimation of halogens, sulphur and phosphorus present in an organic compound.

Answer:

1. Explain what is meant by the principle of estimation: The principle of estimation is a method of determining the approximate amount of a particular element or compound present in a sample. It involves the use of various analytical techniques such as gravimetric analysis, titration, and spectrophotometry.

2. Describe the elements that can be estimated using this principle: The principle of estimation can be used to estimate the amount of halogens, sulphur, and phosphorus present in an organic compound. These elements can be determined by various analytical techniques such as gravimetric analysis, titration, and spectrophotometry.

3. Explain the analytical techniques used to estimate these elements: Gravimetric analysis involves the use of a balance to measure the mass of a sample. Titration is a process that involves the addition of a measured amount of a reagent to a sample until a desired reaction occurs. Spectrophotometry is an analytical technique that measures the amount of light absorbed by a sample at different wavelengths.

4. Summarize the discussion: The principle of estimation is a method of determining the approximate amount of a particular element or compound present in a sample. Halogens, sulphur, and phosphorus can be estimated using this principle by using various analytical techniques such as gravimetric analysis, titration, and spectrophotometry.

Question:

In the organic compound CH2​=CH−CH2​−CH2​−C≡CH, the pair of hybridized orbitals involved in the formation of: C2​−C3​ bond is:

Answer:

Answer:

The pair of hybridized orbitals involved in the formation of C2​−C3​ bond is sp3 hybridized orbitals.

Question:

In the Lassaignes test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:

Answer:

Step 1: Understand the Lassaignes test for nitrogen in an organic compound.

Step 2: Understand what is Prussian blue colour.

Step 3: Research the reaction that occurs when the Lassaignes test is performed.

Step 4: Analyze the reaction and determine what is formed that causes the Prussian blue colour.

Question:

Give condensed and bond line structural formula and identify the functional group(s) present, if any, for 2,2,4-Trimethyl pentane.

Answer:

Condensed Structural Formula: CH3(CH2)3CH3

Bond Line Structural Formula:

H3C-CH2-CH2-CH2-CH3

Functional Group(s): None

Question:

Identify the reagents shown in the following equations as nucleophiles or electrophiles:C6​H6​+CH3​C+O→C6​H5​COCH3​

Answer:

1. C6H6: Electrophile

2. CH3C+: Nucleophile

3. O: Nucleophile

Question:

Classify the following reaction in one of the reaction type. CH3​CH2​Br+HO−→CH2​=CH2​+H2​O+Br−

Answer:

Answer: This is an example of an nucleophilic substitution reaction.

Question:

A solid has a structure in which, atoms of W, O and Na are located respectively at the corners, centre of the edge and at the centre of the cubic lattice. Name the compound.

Answer:

1. The compound is composed of atoms of W, O and Na.

2. The atoms are located at the corners, centre of the edge and at the centre of the cubic lattice.

3. Therefore, the compound is Sodium Tungstate (Na2WO4).

Question:

Describe the method which can be used to separate two compounds with different solubilities in a solvent S.

Answer:

1. Identify the two compounds to be separated and the solvent S.

2. Determine the solubility of each compound in the solvent S.

3. Choose a separation technique based on the solubility of the two compounds in the solvent S. Possible techniques include distillation, filtration, crystallization, extraction, and chromatography.

4. Prepare the mixture of the two compounds in the solvent S.

5. Perform the separation technique chosen in step 3.

6. Collect the separated compounds.

7. Analyze the separated compounds to ensure they are pure.

Question:

What are hybridisation states of each carbon atom in the following compounds? CH2​=C=O,CH3​CH=CH2​,(CH3​)2​CO,CH2​=CHCN,C6​H6​

Answer:

CH2​=C=O: sp hybridization CH3​CH=CH2​: sp2 hybridization (CH3​)2​CO: sp3 hybridization CH2​=CHCN: sp2 hybridization C6​H6​: sp2 hybridization

Question:

Indicate the σ and π bonds in the following molecules. C6​H6​,C6​H12​,CH2​Cl2​,CH2​=C=CH2​,CH3​NO2​,HCONHCH3​

Answer:

C6H6: σ bonds: 12, π bonds: 6 C6H12: σ bonds: 24, π bonds: 0 CH2Cl2: σ bonds: 4, π bonds: 0 CH2=C=CH2: σ bonds: 4, π bonds: 2 CH3NO2: σ bonds: 7, π bonds: 0 HCONHCH3: σ bonds: 8, π bonds: 2

Question:

Draw the formulae for the first five members of each homologous series beginning with the following compounds: a. H−COOH b. CH3​COCH3​ c. H−CH=CH2​.

Answer:

a. H−COOH: H-COOH, CH3-COOH, CH3CH2-COOH, CH3CH2CH2-COOH, CH3CH2CH2CH2-COOH

b. CH3COCH3: CH3COCH3, CH3CH2COCH3, CH3CH2CH2COCH3, CH3CH2CH2CH2COCH3, CH3CH2CH2CH2CH2COCH3

c. H−CH=CH2: H-CH=CH2, CH3-CH=CH2, CH3CH2-CH=CH2, CH3CH2CH2-CH=CH2, CH3CH2CH2CH2-CH=CH2

Question:

The best and latest technique for isolation, purification and separation of organic compounds is :

Answer:

1. The best and latest technique for isolation, purification and separation of organic compounds is chromatography.

2. Chromatography is a method of separating and analyzing mixtures by passing them through a medium, such as a column, that is selectively permeable to the components of the mixture.

3. Chromatography works by exploiting differences in the physical properties of the components of the mixture, such as their size, charge, or solubility.

4. Different types of chromatography, such as gas chromatography, liquid chromatography, and high-performance liquid chromatography, are used to separate and purify organic compounds.

5. Chromatography is an effective and efficient technique for isolating, purifying, and separating organic compounds.

Question:

Discuss the chemistry of Lassaignes test.

Answer:

1. Start by explaining what Lassaigne’s test is.
2. Describe the basic chemistry behind the test.
3. Explain the purpose of the test and what it is used to detect.
4. Discuss the specific chemical reactions that take place during the test.
5. Explain the results of the test and what they indicate.
6. Summarize your discussion of the chemistry of Lassaigne’s test.

Question:

What is the difference between distillation, distillation under reduced pressure and steam distillation ?

Answer:

Step 1: Distillation is a process in which a liquid mixture is heated until it boils and then the vapor is collected and condensed back into a liquid.

Step 2: Distillation under reduced pressure is a process in which the pressure of the liquid mixture is reduced before it is heated. This reduces the boiling point of the mixture and allows for a more efficient separation of the components.

Step 3: Steam distillation is a process in which steam is passed through the liquid mixture. The steam helps to vaporize the components and separate them from each other. The vapor is then condensed back into a liquid.

Question:

Will CCl4​ give white precipitate of AgCl on heating it with silver nitrate? Give reason for your answer.

Answer:

Answer: No, CCl4​ will not give white precipitate of AgCl on heating it with silver nitrate. This is because CCl4​ is a non-polar molecule and does not form complexes with silver nitrate. Silver nitrate only forms complexes with polar molecules like ammonia and carbonates.

Question:

In the organic compound CH2​=CH−CH2​−CH2​−C≡CH, the pair of hybridized orbitals involved in the formation of: C2​−C3​ bond is :

Answer:

Answer:

1. The C2-C3 bond is a sigma bond, which is formed by the head-on overlapping of two hybridized orbitals.

2. The hybridized orbitals involved in the formation of this bond are sp2 orbitals of carbon atoms at C2 and C3.

Question:

Give a brief description of the principles of the Crystallisation and give examples.

Answer:

1. Crystallisation is a process used to purify and separate a solid material from a liquid solution. It involves cooling the solution to a temperature where the solid material will form crystals, while the impurities remain in the liquid solution.

2. The principle of crystallisation is based on the fact that different substances have different solubilities at different temperatures. For example, sugar is more soluble in hot water than cold water, so if you cool a hot sugar solution, the sugar will crystallise out and the impurities will remain in the liquid.

3. Another principle of crystallisation is that of ‘seeding’. Seeding involves adding a small amount of the solid material to the solution to act as a nucleus for the crystallisation process. This helps to speed up the process and produce larger, more uniform crystals.

4. Examples of crystallisation processes include the manufacture of sugar, salt, and some pharmaceuticals. Crystallisation is also used to purify metals such as aluminium and copper.

Question:

Draw the resonance structures for C6​H5​−CH2+​. Also, show the electron shift using curved - arrow notation.

Answer:

1. Draw the Lewis structure of C6H5−CH2+:

C6H5−CH2+ | | H-C-C-C-C-C-H | | | H H H | | | H H H | | | H H H | | | H H H | | | H H H | | H H | H

2. Draw the two resonance structures of C6H5−CH2+:

Resonance Structure 1: C6H5−CH2+ | | H-C-C-C-C-C-H | | | H H H | | | H H H | | | H H H | | | H H H | | | H H H | | H + | H

Resonance Structure 2: C6H5−CH2+ | | H-C-C-C-C-C-H | | | H H H | | | H H H | | | H H H | | | H H H | | | H + H | | H H | H

3. Show the electron shift using curved arrow notation:

From Resonance Structure 1 to Resonance Structure 2:

↑ ← ← ← ← ↑

Question:

Identify the reagents shown in bold in the following equations as nucleophiles or electrophiles: H3​COOH3​+CN−​→(CH3​)2​C(CN)(OH)

Answer:

Reagents: H3​COOH3​ (nucleophile) and CN−​ (electrophile)

Question:

Which of the following represents the correct IUPAC name for the compounds concerned? a. 2,2−Dimethylpentane or 2−Dimethylpentane b. 2,4,7−Trimethyloctane or 2,5,7−Trimethyloctane c. 2−Chloro−4−methylpentane or 4−Chloro−2−methylpentane d. But−3−yn−1−ol or But−4−ol−1−yne. Which of the two: O2​NCH2​CH2​O−orCH3​CH2​O− is expected to be more stable and why ?

Answer:

Answer: c. 2−Chloro−4−methylpentane or 4−Chloro−2−methylpentane

The molecule expected to be more stable is CH3​CH2​O− because it has a more stable carbon-oxygen bond than the O2​NCH2​CH2​O− molecule. This is because the carbon-oxygen bond in the CH3​CH2​O− molecule is more strongly polarized due to the electronegativity of the oxygen atom, which makes it more stable.

Question:

Draw resonance structures for the following compound. Show the electron shift using curved arrow notation. C6​H5​OH

Answer:

1. Draw the Lewis structure of the compound C6H5OH.

2. Identify the atoms that can be involved in the resonance. In this case, the oxygen atom can form a double bond with either the carbon or the hydrogen atom.

3. Draw the two resonance structures of the compound. In the first structure, the oxygen atom forms a double bond with the carbon atom and in the second structure, the oxygen atom forms a double bond with the hydrogen atom.

4. Show the electron shift using curved arrow notation. In the first structure, the electron pair from the oxygen atom moves to the carbon atom and in the second structure, the electron pair from the oxygen atom moves to the hydrogen atom.

Question:

Classify the following reactions in one of the reaction types studied. a. CH3​CH2​Br+HS⊖→CH3​CH2​SH+Br⊖ b. (CH3​)2​C=CH2​+HCl→(CH3​)2​(Cl)C−CH3​ c. (CH3​)3​C−CH2​OH+HBr→(CH3​)2​C(Br)CH2​CH3​ d. CH3​CH2​Br+HO⊖→CH2​=CH2​ A a.Nucleophilic substitution and 1, 2-Methyl shift b. Nucleophilic substitution reaction c. Elimination reaction d. Addition reaction B a.Addition reaction b. Nucleophilic substitution reaction c. Elimination reaction d. Nucleophilic substitution and 1, 2-Methyl shift C a. Nucleophilic substitution reaction b. Addition reaction c. Nucleophilic substitution and 1, 2-Methyl shift d. Elimination reaction D a.Addition reaction b. Nucleophilic substitution reaction c. Nucleophilic substitution and 1, 2-Methyl shift d. Elimination reaction

Answer:

Answer: D a. Addition reaction b. Nucleophilic substitution reaction c. Nucleophilic substitution and 1, 2-Methyl shift d. Elimination reaction

Question:

The reaction: CH3​CH2​I+KOH(aq)→CH3​CH2​OH+KI is classified as :

Answer:

1. Identify the reactants: Reactants: CH3​CH2​I and KOH(aq)

2. Identify the products: Products: CH3​CH2​OH and KI

3. Classify the reaction: Reaction type: Nucleophilic Substitution

Question:

Classify the following reaction in one of the reaction type. (CH3​)2​C=CH2​+HCl→(CH3​)2​ClC−CH3​

Answer:

1. This is an example of an addition reaction.

Question:

Classify the following reaction in one of the reaction type. CH3​CH2​Br+HS−→CH3​CH2​SH+Br−

Answer:

Answer: This is a nucleophilic substitution reaction.

Question:

In the Lassaignes test for nitrogen in an organic compound, the Prussian blue colour is obtained due to the formation of:

Answer:

1. Firstly, the compound is heated in a test tube with concentrated sulphuric acid and copper turnings.

2. The mixture is then allowed to cool and filtered to remove any insoluble impurities.

3. The filtrate is then treated with ammonium hydroxide to form a precipitate of copper ammonium complex.

4. The precipitate is then heated with ammonium chloride solution to form copper chloride and ammonia.

5. The ammonia then reacts with ferrous sulphate to form Prussian blue, a ferric ferrocyanide complex.

Question:

Identify the reagents underlined in the following equations as nucleophiles or electrophiles: CH3​COOH+OH−​→CH3​COO−+H2​O

Answer:

Reagent: CH3COOH Nucleophile: OH-

Reagent: CH3COO- Electrophile: H2O

Question:

Draw the resonance structures for CH3​CH=CHCHO. Also, show the electron shift using curved-arrow notation.

Answer:

1. The Lewis structure of CH3CH=CHCHO is:

H3C-CH=CH-CHO

2. The resonance structures for CH3CH=CHCHO are:

H3C-CH=CH-CHO

H3C-C≡CH-CHO

H3C-CH-CH2-CHO

3. The electron shift using curved-arrow notation is:

H3C-CH=CH-CHO –> H3C-C≡CH-CHO

← H3C-CH-CH2-CHO →

Question:

Draw all the possible resonance structures for Phenol.

Answer:

Step 1: Draw the Lewis structure of Phenol.

Step 2: Identify all the atoms with an unshared pair of electrons.

Step 3: For each atom with an unshared pair of electrons, draw a separate resonance structure by moving the unshared pair of electrons to a different atom.

Step 4: Draw the resulting structures and compare them to the original structure to ensure that they are valid resonance structures.

Question:

The best and latest technique for isolation, purification and separation of organic compounds is ________.

Answer:

A. High-Performance Liquid Chromatography (HPLC)

Question:

The best and latest technique for isolation, purification and separation of organic compounds is:

Answer:

Answer:

1. Chromatography: Chromatography is a technique used to separate and isolate organic compounds by exploiting differences in their interaction with a stationary phase and a mobile phase.

2. Distillation: Distillation is a technique used to separate and purify organic compounds based on differences in their boiling points.

3. Extraction: Extraction is a technique used to separate and purify organic compounds based on differences in their solubility in different solvents.

4. Sublimation: Sublimation is a technique used to separate and purify organic compounds based on differences in their vapor pressures.

Question:

Explain the reason for the fusion of an organic compound with metallic sodium for testing nitrogen, sulphur and halogens.

Answer:

1. The fusion of an organic compound with metallic sodium is a method used to test for the presence of nitrogen, sulphur and halogens in an organic compound.

2. This is done by heating the organic compound with metallic sodium, which causes the compound to decompose and release nitrogen, sulphur and halogens in the form of gases.

3. The gases released are then tested to determine the presence or absence of nitrogen, sulphur and halogens in the organic compound.

4. This method is preferred over other methods because it is relatively simple and quick, and it can provide accurate results.

Question:

Give a brief description of the principles of distillation.

Answer:

1. Distillation is a process used to separate mixtures of liquids and solids by their boiling points.

2. The mixture is heated and the vapors are collected and condensed back into a liquid.

3. The liquid with the lower boiling point will boil off first, allowing it to be collected separately from the other components of the mixture.

4. This process can be repeated multiple times to further purify the liquid.

5. Distillation can also be used to separate mixtures of liquids that have different boiling points, such as water and alcohol.

Question:

Write bond line formula for 2,3-dimethyl butanal.

Answer:

1. Identify the molecular structure of 2,3-dimethyl butanal:
2. Draw the structural formula of 2,3-dimethyl butanal:

H3C–CH2–CH(CH3)–CHO

3. Write the bond line formula: H3C–CH2–CH(CH3)–COOH

Question:

Why is it necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test?

Answer:

1. Acetic acid is a weak acid and sulphuric acid is a strong acid.

2. When acidifying a sample, a weak acid is preferred because it will not cause a drastic change in the pH of the sample, which could affect the accuracy of the test results.

3. Lead acetate test is a colorimetric test, which means that the intensity of the color produced is related to the amount of sulphur present in the sample.

4. A drastic change in the pH of the sample due to the use of a strong acid could interfere with the accuracy of the colorimetric test.

5. Therefore, it is necessary to use acetic acid and not sulphuric acid for acidification of sodium extract for testing sulphur by lead acetate test.

Question:

Name a suitable technique of separation of the components from a mixture of calcium sulphate and camphor.

Answer:

1. Filtration: Filtration is a suitable technique for separating the components from a mixture of calcium sulphate and camphor. The solid particles of calcium sulphate can be filtered out using a filter paper.

2. Sublimation: Sublimation is another suitable technique for separating the components from a mixture of calcium sulphate and camphor. The camphor can be sublimed, leaving the calcium sulphate behind.

3. Distillation: Distillation is a suitable technique for separating the components from a mixture of calcium sulphate and camphor. The camphor can be distilled off from the mixture, leaving the calcium sulphate behind.

Question:

Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? CH3​CH2​COOH>(CH3​)2​CHCOOH>(CH3​)3​CCOOH

Answer:

Inductive effect: An effect that occurs when a molecule or ion has a permanent dipole moment due to the presence of an electronegative atom. This atom attracts electrons from adjacent atoms, resulting in a partial negative charge on the atom and a partial positive charge on the adjacent atom.

Electromeric effect: An effect that occurs when a molecule or ion has a temporary dipole moment due to the presence of an electron withdrawing group. This group temporarily withdraws electrons from adjacent atoms, resulting in a partial negative charge on the atom and a partial positive charge on the adjacent atom.

The order of acidity of the carboxylic acids is explained by the inductive effect. The more electronegative the atom is, the more it will attract electrons from adjacent atoms, resulting in a more negative charge on the carboxylic acid. Therefore, the acid with the most electronegative atom will be the most acidic. In this case, CH3​CH2​COOH has the most electronegative atom, followed by (CH3​)2​CHCOOH and then (CH3​)3​CCOOH.

Question:

Write the bond line formulae for the following compounds: I. Isopropyl alcohol II. 2,3−Dimethylbutanal III. Heptan−4−one.

Answer:

I. C3H7OH II. C6H12O III. C7H14O

Question:

Give the condensed and bond line structural formulae and identify the functional group(s) present, if any, for: a. 2,2,4−Trimethylpentane b. 2−Hydroxy−1,2,3−propanetricarboxylic acid c. Hexanedial.

Answer:

a. 2,2,4−Trimethylpentane Condensed Structural Formula: (CH3)3CCH2CH3 Bond Line Structural Formula:

H3C-CH2-CH2-CH2-CH3

No functional groups present.

b. 2−Hydroxy−1,2,3−propanetricarboxylic acid Condensed Structural Formula: HOOC-CH2-CH(OH)-COOH Bond Line Structural Formula:

H-O-O-C-C-H-OH-C-O-O-H

Functional group present: carboxylic acid.

c. Hexanedial Condensed Structural Formula: HOOC-CH2-CH2-CH2-CH2-CH2-COOH Bond Line Structural Formula:

H-O-O-C-C-H-H-H-H-H-H-C-O-O-H

Functional group present: carboxylic acid.

Question:

What are electrophiles and nucleophiles ? Explain with examples.

Answer:

Answer: Electrophiles are atoms, molecules, or ions that have a tendency to accept electrons and form a covalent bond. Examples of electrophiles include hydrogen ions (H+), metal cations (M+), and alkyl halides (R-X).

Nucleophiles are atoms, molecules, or ions that have a tendency to donate electrons and form a covalent bond. Examples of nucleophiles include hydroxide ions (OH-), amines (R-NH2), and carbanions (R-C:).

Question:

In the organic compound CH2​=CH−CH2​−CH2​−C≡CH, the pair of hydridized orbitals involved in the formation of: C2​−C3​ bond is

Answer:

1. Identify the two atoms in the bond: C2 and C3

2. Determine the hybridization of the two atoms: sp2 hybridization

3. Identify the hybridized orbitals involved in the formation of the bond: One sp2 orbital from C2 and one sp2 orbital from C3

4. Determine the pair of hydridized orbitals involved in the formation of the C2−C3 bond: One sp2 orbital from C2 and one sp2 orbital from C3

Question:

Differentiate between the principle of estimation of nitrogen in an organic compound by (i) Dumas method and (ii) Kjeldahls method

Answer:

(i) Dumas Method: This method involves the combustion of a sample in oxygen and the collection of the nitrogen-containing gases (nitrogen, ammonia, nitrous oxide) produced. The amount of nitrogen in the sample can then be calculated by measuring the volume of the gases collected.

(ii) Kjeldahls Method: This method involves the catalytic oxidation of the sample in a strong acid solution and the collection of the ammonia produced. The amount of nitrogen in the sample can then be calculated by measuring the amount of ammonia produced.

Question:

Why is nitric acid added to sodium extract before adding silver nitrate for testing halogens?

Answer:

1. Nitric acid is added to sodium extract before adding silver nitrate for testing halogens because nitric acid helps to dissolve any halide present in the sample.

2. Silver nitrate is then added to the mixture, which reacts with the halide to form a silver halide precipitate.

3. The presence of the silver halide precipitate confirms the presence of a halogen in the sample.

Question:

An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this substance is subjected to complete combustion.

Answer:

Step 1: Calculate the mass of carbon in 0.20 g of the organic compound.

Mass of carbon = 0.20 g × (69/100) = 0.138 g

Step 2: Calculate the mass of oxygen in 0.20 g of the organic compound.

Mass of oxygen = 0.20 g × (1 - (69/100 + 4.8/100)) = 0.072 g

Step 3: Calculate the mass of carbon dioxide produced.

Mass of carbon dioxide = 0.138 g × (2/12) = 0.023 g

Step 4: Calculate the mass of water produced.

Mass of water = 0.072 g × (2/18) = 0.012 g

Question:

In the organic compound CH2​=CH−CH2​−CH2​−C≡CH, the pair of hybridized orbitals involved in the formation of C2​−C3​ bond is:

Answer:

Answer: The pair of hybridized orbitals involved in the formation of C2​−C3​ bond in the organic compound CH2​=CH−CH2​−CH2​−C≡CH is sp2 hybridized orbitals.

Question:

Explain, why an organic liquid vaporises at a temperature below its boiling point in its steam distillation ?

Answer:

1. Vaporization is the process of a liquid changing into a gaseous state.

2. Steam distillation is a process in which steam is used to separate two immiscible liquids, such as an organic liquid and water.

3. In steam distillation, the organic liquid is heated to a temperature below its boiling point, which causes it to vaporize.

4. Vaporizing the organic liquid at a temperature below its boiling point allows for the separation of the liquid from the water, as the vaporized organic liquid is less dense than the water and will rise to the top of the distillation container.

5. This separation of the organic liquid from the water is the purpose of steam distillation.

Question:

A sample of 0.50 g of an organic compound was treated according to Kjeldahls method. The ammonia evolved was absorbed in 50 ml of 0.5M H2​SO4​. The residual acid required 60 mL of 0.5 M solution of NaOH for neutralisation. Find the percentage composition of nitrogen in the compound.

Answer:

1. Calculate the amount of NaOH used in the reaction: 60 mL x 0.5 M = 30 mmol of NaOH

2. Calculate the amount of ammonia evolved: 30 mmol of NaOH x 1 mole of NH3/1 mole of NaOH = 30 mmol of NH3

3. Calculate the amount of nitrogen in the sample: 30 mmol of NH3 x 1 mole of N/1 mole of NH3 = 30 mmol of N

4. Calculate the mass of nitrogen in the sample: 30 mmol of N x 14.01 g/1 mole of N = 420.3 g of N

5. Calculate the percentage composition of nitrogen in the sample: 420.3 g of N/0.50 g of sample x 100% = 84.06% N

Question:

Why is a solution of potassium hydroxide used to absorb carbon dioxide evolved during the estimation of carbon present in an organic compound?

Answer:

Answer:

1. Potassium hydroxide (KOH) is a strong base that reacts with carbon dioxide (CO2) to form potassium carbonate (K2CO3).

2. Carbon dioxide is evolved during the estimation of carbon present in an organic compound, usually through combustion or acid-base reactions.

3. The solution of potassium hydroxide is used to absorb the carbon dioxide produced during the estimation, so that it does not interfere with the measurement of carbon present in the sample.

Question:

In the estimation of sulphur by Carius method, 0.468 g of an organic sulphur compound afforded 0.668 g of barium sulphate. Find out the percentage of sulphur in the given compound.

Answer:

Step 1: Calculate the amount of sulphur in the barium sulphate. This can be done by using the formula: Amount of sulphur= (Mass of barium sulphate x % of sulphur in barium sulphate)/100

Amount of sulphur = (0.668 g x 48%)/100 Amount of sulphur = 0.32256 g

Step 2: Calculate the percentage of sulphur in the given compound. This can be done by using the formula: % of sulphur in the given compound = (Amount of sulphur in barium sulphate x 100)/Mass of organic sulphur compound

% of sulphur in the given compound = (0.32256 g x 100)/0.468 g % of sulphur in the given compound = 68.9%

Question:

Give condensed and bond line structural formula and identify the functional group(s) present, if any, for 2-hydroxy-1,2,3-propane tricarboxylic acid.

Answer:

Condensed structural formula: HOOC-C(OH)-COOH

Bond line structural formula:

H - O - O - C - C - O - H | | H O

Functional group(s): Carboxylic acid (COOH)

Question:

Give condensed and bond line structural formula and identify the functional group(s) present, if any, for Hexanedial.

Answer:

Condensed Structural Formula: C6H10O2

Bond Line Structural Formula:

H-C-C-C-C-C-C-O-O-H

Functional Group(s): Aldehyde (O=C-H)

Question:

Explain the terms Inductive and Electromeric effects. Which electron displacement effect explains the following correct orders of acidity of the carboxylic acids? Cl3​CCOOH>Cl2​CHCOOH>ClCH2​COOH

Answer:

Inductive effect: This is an effect in which the electron density of a molecule is affected by the presence of neighboring atoms or groups of atoms. For example, the presence of an electronegative atom or group of atoms can decrease the electron density of a molecule and make it more acidic.

Electronic effect: This is an effect in which the electron density of a molecule is affected by the presence of an electronegative atom or group of atoms. For example, the presence of an electronegative atom or group of atoms can increase the electron density of a molecule and make it more basic.

The correct order of acidity of the carboxylic acids is explained by the electronic effect. The more electronegative atom or group of atoms present in the carboxylic acid, the more acidic it will be. In the given example, Cl3​CCOOH has the most electronegative atoms (three chlorine atoms) and so it is the most acidic, followed by Cl2​CHCOOH, which has two chlorine atoms, and then ClCH2​COOH, which has only one chlorine atom.