Question:

Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+. Toluene, p−H3​CC6​H4​NO2​,p−O2​NC6​H4​NO2​.

Answer:

1. p-O2NC6H4NO2
2. p-H3CC6H4NO2
3. Toluene

Question:

Draw the cis and trans structures of hex-2-ene. Which isomer will have higher b.p. and why?

Answer:

Step 1: Draw the cis and trans structures of hex-2-ene.

Cis Structure:

H H H \ / \ / C=C C=C / \ /
H H H

Trans Structure:

H H H \ / \ / C=C C=C \ / \ / H H H

Step 2: Determine which isomer will have higher b.p.

The cis isomer will have a higher boiling point because it has a more compact structure, which makes it more difficult for the molecules to escape from each other.

Question:

Why does benzene undergo electrophilic substitution reactions easily and nucleophilic substitutions difficulty?

Answer:

1. Benzene is an aromatic hydrocarbon, meaning it has a cyclic structure with alternating double and single bonds. This structure is particularly stable and resistant to chemical reaction.

2. Electrophilic substitution reactions occur when an electron-deficient species (electrophile) reacts with the electron-rich aromatic ring of benzene. This type of reaction is favored due to the electron-withdrawing nature of the electrophile and the electron-donating properties of the aromatic ring.

3. In contrast, nucleophilic substitution reactions require the electron-rich species (nucleophile) to attack the electron-deficient carbon atom of the benzene ring. This type of reaction is much less favorable due to the stability of the benzene ring.

4. Therefore, benzene undergoes electrophilic substitution reactions easily and nucleophilic substitutions difficulty due to the stability of the aromatic ring.

Question:

Write structures of all the alkenes which on hydrogenation give 2 - methylbutane.

Answer:

1. Alkene structure 1: CH3CH=CH2
2. Alkene structure 2: CH3CH2CH=CH2
3. Alkene structure 3: CH3CH2C=CH3

Question:

How would you convert the following compound into benzene? (i) Hexane

Answer:

Step 1: Remove the hydrogen atoms from each of the six carbon atoms in hexane, leaving only the carbon atoms.

Step 2: Place a double bond between each of the six carbon atoms to form a hexagonal ring.

Step 3: Add six hydrogen atoms to each of the six carbon atoms to form benzene.

Question:

Lewis acid other than anhydrous aluminium chloride which can be used during ethylation of benzene is: A FeCl3​ B KOH C LiAlH4​ D none of these

Answer:

A. FeCl3

Question:

How do you account for the formation of ethane during chlorination of methane?

Answer:

1. Chlorination of methane is a chemical reaction in which chlorine atoms are added to methane molecules.

2. The reaction produces ethane, which is a hydrocarbon with two carbon atoms and six hydrogen atoms.

3. The reaction involves the substitution of a chlorine atom for a hydrogen atom on the methane molecule, forming a single bond between the two carbon atoms and a double bond between the chlorine and the carbon atoms.

4. This double bond creates a new molecule of ethane with two carbon atoms and six hydrogen atoms.

5. The reaction is endothermic, meaning it requires energy to proceed, and is catalyzed by light or heat.

Question:

How will you convert benzene into m−nitrochlorobenzene?

Answer:

1. Begin by adding nitric acid to benzene in the presence of sulfuric acid.

2. Heat the mixture of benzene, nitric acid, and sulfuric acid to about 80-90°C.

3. Allow the reaction to proceed for several hours until the desired product is formed.

4. Separate the m−nitrochlorobenzene from the reaction mixture using a suitable method, such as distillation or extraction.

5. Purify the m−nitrochlorobenzene using a suitable method, such as recrystallization.

Question:

What are the necessary conditions for any system to be aromatic?

Answer:

1. The system must contain a cyclic structure of conjugated double bonds.
2. The system must have a planar structure.
3. The system must contain a total of 4n+2 pi electrons, where n is an integer.
4. The system must have an odd number of pi electrons per atom.
5. The system must have a continuous cyclic delocalization of pi electrons.

Question:

What effect does branching of an alkane chain has on its boiling point?

Answer:

Step 1: Understand what an alkane chain is.

Step 2: Understand what boiling point is.

Step 3: Research the effect of branching on boiling point of an alkane chain.

Step 4: Analyze the research findings and draw a conclusion about the effect of branching on boiling point of an alkane chain.

Question:

Write down the products of ozonolysis of 1, 2 - dimethylbenzene (o -xylene). How does the result support Kekules structure for benzene?

Answer:

1. The products of ozonolysis of 1, 2 - dimethylbenzene (o -xylene) are formaldehyde, formic acid, and methyl formate.

2. Kekule’s structure of benzene proposes that benzene is composed of a ring of six carbon atoms, each of which is bonded to two hydrogen atoms and one double bond.

3. The products of ozonolysis support Kekule’s structure because the double bond is broken, resulting in two carbonyl compounds (formaldehyde and formic acid) and a carboxylic acid (methyl formate). This shows that the double bond is broken and the carbons are now bonded to two hydrogen atoms and one double bond, which is consistent with Kekule’s structure.

Question:

Arrange the following set of compounds in order of their decreasing relative reactivity with an electrophile, E+. (i) Chlorobenzene, 2,4−dinitrochlorobenzene, p−nitrochlorobenzene (ii) Toluene, p−H3​C−C6​H4​−CH3​,p−H3​C−C6​H4​−NO2​,p−O2​N−C6​H4​−NO2​

Answer:

(i) p-nitrochlorobenzene > 2,4-dinitrochlorobenzene > Chlorobenzene

(ii) p-O2N-C6H4-NO2 > p-H3C-C6H4-NO2 > p-H3C-C6H4-CH3 > Toluene

Question:

How will you convert benzene into p - nitrotoluene?

Answer:

1. Start by adding nitric acid and sulfuric acid to benzene in a reaction flask.

2. Heat the flask to a temperature of around 80°C and allow the reaction to take place.

3. Separate the p-nitrotoluene from the reaction mixture using fractional distillation.

4. Purify the p-nitrotoluene using recrystallization and collect the crystals.

5. Filter the crystals and dry them to obtain pure p-nitrotoluene.

Question:

What effect does branching of an alkane chain has on its boiling point?

Answer:

Step 1: Understand the question.

The question is asking about the effect of increasing the number of branches in an alkane chain on its boiling point.

Step 2: Research the answer.

Increasing the number of branches in an alkane chain increases its boiling point. This is because the increased branching creates more surface area, which increases the number of intermolecular forces between molecules and thus increases the boiling point.

Question:

How will you convert benzene into p - nitrobromobenzene?

Answer:

1. Start by reacting benzene with nitric acid (HNO3) in the presence of an oxidizing agent, such as sulfuric acid (H2SO4).

2. The nitric acid will convert the benzene into nitrobenzene (C6H5NO2).

3. Then, add bromine (Br2) to the nitrobenzene in the presence of an aqueous solution of sodium hydroxide (NaOH).

4. This will convert the nitrobenzene into p-nitrobromobenzene (C6H4BrNO2).

Question:

In the alkane H3​CCH2​C(CH3​)2​CH2​CH(CH3​)2​, identify 1°,2°,3° carbonatoms and give the number of H atoms bonded to each one of these.

Answer:

1° Carbon Atom: H3C-CH2-C(CH3)2 Number of H atoms bonded to 1° Carbon Atom: 3

2° Carbon Atom: H3C-CH2-C(CH3)2-CH2-CH(CH3)2 Number of H atoms bonded to 2° Carbon Atom: 2

3° Carbon Atom: H3C-CH2-C(CH3)2-CH2-CH(CH3)2 Number of H atoms bonded to 3° Carbon Atom: 4

Question:

Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1 - bromopropane. Explain and give mechanism.

Answer:

1. Propene is an alkene, meaning it has a double bond between two carbon atoms. HBr is an electrophile, meaning it is attracted to the electrons of the double bond.

2. When HBr is added to propene, a nucleophilic substitution reaction takes place, and 2-bromopropane is formed. This reaction mechanism is shown in the diagram below:

3. When benzoyl peroxide is present, the double bond of propene is broken by an addition-elimination mechanism, forming a carbocation intermediate. This intermediate is then attacked by the HBr, forming 1-bromopropane. This reaction mechanism is shown in the diagram below:

Question:

Why is Wurtz reaction not preferred for the preparation of alkanes containing odd number of carbon atoms? Illustrate your answer by taking one example.

Answer:

Answer:

1. The Wurtz reaction is a coupling reaction between two alkyl halides to form an alkane containing an even number of carbon atoms.

2. It is not preferred for the preparation of alkanes containing odd number of carbon atoms because the reaction does not work with odd-numbered alkyl halides.

3. For example, if we consider the preparation of pentane (C5H12), the Wurtz reaction will not work as there is no alkyl halide with 5 carbon atoms.

Question:

Write IUPAC names of the products obtained by the ozonolysis of the following compounds? 1.Pent-2-ene 2.3,4-Dimethylhept-3-ene 3.2-Ethylbut-1-ene 4.1-Phenylbut-1-ene

Answer:

1. Pent-2-ene: 2-Methylprop-1-ene

2. 3,4-Dimethylhept-3-ene: 3,4-Dimethylhept-2-ene

3. 2-Ethylbut-1-ene: 2-Ethylprop-1-ene

4. 1-Phenylbut-1-ene: 1-Phenylprop-1-ene

Question:

An alkene A on ozonolysis gives a mixture of ethanal and pentan - 3 - one. Write structure and IUPAC name of A.

Answer:

Structure of A:

A = CH2=CH-CH2-CH2-CH3

IUPAC Name of A: 2-Methylbut-1-ene

Question:

An alkene A contains three C−C, eight C−Hσ bonds and one C−Cπ bond. A on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of A.

Answer:

Step 1: Calculate the molecular formula of A.

Molecular formula of A = C3H8

Step 2: Determine the structure of A.

A is an alkene with three C−C bonds and eight C−Hσ bonds. It also contains one C−Cπ bond.

Step 3: Determine the IUPAC name of A.

The IUPAC name of A is propene.

Question:

Write IUPAC names of the products obtained by the ozonolysis of 3, 4 -Dimethylhept-3-ene.

Answer:

1. The product obtained from the ozonolysis of 3, 4-Dimethylhept-3-ene is an aldehyde and an alkene.

2. The aldehyde product is 3, 4-Dimethylheptanal.

3. The alkene product is 3-Methylhept-2-ene.

Question:

How will you convert benzene into acetophenone?

Answer:

1. Begin by reacting benzene with a strong oxidizing agent, such as chromic acid or potassium permanganate, to create an intermediate compound known as benzoic acid.

2. Reduce the benzoic acid using a reducing agent, such as zinc and hydrochloric acid, to form benzaldehyde.

3. React the benzaldehyde with a strong base, such as sodium hydroxide, to produce acetophenone.

Question:

Write IUPAC names of the following compound: CH3​CH=C(CH3​)2​

Answer:

The IUPAC name of the compound is 1,1-Dimethylprop-1-ene.

Question:

For the following compounds, write structural formulas and IUPAC names for all possible isomers having the number of the double or triple bond as indicated : (a) C4​H8​(one double bond) (b)C5​H8​ (one triple bond)

Answer:

(a)

Structural Formula 1:

IUPAC Name: But-1-ene

Structural Formula 2:

IUPAC Name: But-2-ene

(b)

Structural Formula 1:

IUPAC Name: Pent-1-yne

Structural Formula 2:

IUPAC Name: Pent-2-yne

Question:

Propanal and pentan - 3 - one are the ozonolysis products of an alkene. What is the structural formula of the alkene?

Answer:

1. The structural formula of an alkene includes two carbon atoms connected by a double bond.

2. The ozonolysis products of an alkene are propanal and pentan-3-one, which are organic compounds containing three and five carbon atoms respectively.

3. Therefore, the alkene must have four carbon atoms, with two of them connected by a double bond.

4. The structural formula of the alkene would be: C_2H_4

Question:

Why is benzene extra ordinarily stable though it contains three double bonds?

Answer:

1. Benzene is an aromatic hydrocarbon that contains a ring of six carbon atoms, each of which is bonded to one hydrogen atom and one of its neighbors by a single covalent bond.

2. In addition to these single bonds, benzene also contains three double bonds between its carbon atoms, which gives the molecule a unique stability.

3. The double bonds in benzene are held together by a phenomenon known as resonance. This means that the electrons in the double bonds can move freely around the ring, allowing the molecule to maintain its stability despite the presence of the double bonds.

4. This stability is further enhanced by the delocalization of electrons, which allows the electrons to spread out throughout the entire molecule, resulting in a more stable overall structure.

5. This extra stability is what makes benzene so extraordinarily stable, despite the presence of three double bonds.

Question:

How would you convert ethyne into benzene?

Answer:

Step 1: Ethyne (C2H2) can be converted to benzene (C6H6) by a process called catalytic reforming.

Step 2: This process involves passing ethyne and steam over a catalyst such as platinum or nickel at high temperatures and pressures.

Step 3: During this reaction, the C2H2 molecules are “cracked” into smaller molecules, and the hydrogen atoms are rearranged to form benzene.

Step 4: The reaction product is then separated from the catalyst and cooled, producing a mixture of benzene and other hydrocarbons.

Question:

Out of benzene, m dinitrobenzene and toluene which will undergo nitration most easily and why ?

Answer:

1. Nitration is a process of introducing nitro groups (NO2) into organic compounds.

2. Benzene is the most reactive compound amongst the three, so it will undergo nitration most easily.

3. This is because benzene is a highly aromatic compound, which is more susceptible to electrophilic substitution reactions like nitration.

4. Toluene is less reactive than benzene, so it will undergo nitration less easily.

5. M-dinitrobenzene is the least reactive compound amongst the three, so it will undergo nitration least easily.