NCERT Solutions for Class 12 Chemistry Chapter 6 Haloalkanes and Haloarenes

Question 6.1 Write structures of the following compounds:

(ii) 1-Chloro-4-ethylcyclohexane

Answer :

A cyclic structure is present in which at 1st carbon there is chlorine present and at the 4th position ethyl group is present.

The structure of 1-Chloro-4-ethylcyclohexane is given below :-

Question 6.1 Write structures of the following compounds:

(iii) 4-tert. Butyl-3-iodoheptane

Answer :

Main chain will be of seven carbon atoms. At 4th position tertiary butyl will be present and at the 3rd position iodine will be present.

The structure of 4-tert. Butyl-3-iodoheptane is given below :-

Question 6.1 Write structures of the following compounds:

(iv) 1,4-Dibromobut-2-ene

Answer :

Main chain will be of 4 carbon atoms and it is an alkene. There are two bromine atoms on the terminal carbon atoms.

The structure of 1,4-Dibromobut-2-ene is given below :-

Question 6.1 Write structures of the following compounds:

(v) 1-Bromo-4-sec. butyl-2-methylbenzene

Answer :

Main chain will be the benzene. Three substituents are present on benzene

The structure of 1-Bromo-4-sec. butyl-2-methylbenzene is shown below :-

Question 6.2 Why is sulphuric acid not used during the reaction of alcohols with KI?

Answer :

We don't use sulphuric acid because it acts as an oxidising agent and the required alkyl iodide is not produced. KI is expected to give HI on reacting with H₂SO₄ which will convert alcohols (R – OH) to alkyl iodides (R – I). However, H₂SO₄ is a strong oxidising agent and it oxidises HI formed during the reaction to I₂ which does not react with alcohol.

The reactions are given below :-

2KI + H₂ SO₄ → 2KHSO₄ + 2HI

2HI + H₂SO₄ → I₂ + SO₂ + H₂O

Question 6.3 Write structures of different dihalogen derivatives of propane.

Answer :

We obtain four dihalogen derivatives of propane :-

(i) 1,1 Dibromopropane

(ii) 2, 2 Dibromopropane

(iii) 1, 2 Dibromopropane

(iv) 1, 3 Dibromopropane

Question 6.4 Among the isomeric alkanes of molecular formula C₅H₁₂ , identify the one that on photochemical chlorination yields

(i) A single monochloride

Answer :

In this we have to find an isomer in which replacement of any hydrogen atom gives the same compound for all replacements.

So the isomer is Neopentane.

Question 6.4 Among the isomeric alkanes of molecular formula C₅H₁₂ , identify the one that on photochemical chlorination yields

(ii) Three isomeric monochlorides.

Answer :

For the given condition we must have three different hydrogens so that we can get three different monochlorides on the replacement.

Thus the isomer is n-pentane.

Question 6.4 Among the isomeric alkanes of molecular formula C₅H₁₂ , identify the one that, on photochemical chlorination, yields

(iii) Four isomeric monochlorides .

Answer :

For four monochlorides, we need four different hydrogens which can be replaced by chlorine.

Hence the required isomer is iso pentane:-

Question 6.5 Draw the structures of major monohalo products in each of the following reactions:

(i)

Answer: In this reaction the hydroxyl group will be replaced with the chlorine atom.

The final products are:-

Question 6.5 Draw the structures of major monohalo products in each of the following reactions:

Answer :

In this reaction the amine group will be replaced with methyl bromide.

The reaction is given below:

Question 6.5 Draw the structures of major monohalo products in each of the following reactions:

(iii)

Answer :

In this reaction the hydroxyl group with the methyl group will be replaced with the chlorine atom.

The reaction is given below:

Question 6.5 Draw the structures of major monohalo products in each of the following reactions:

(iv)

Answer :

In this reaction the iodine from the hydrogen iodide will attach the carbon atom having the double bond as well as methyl group.

The reaction is given below:

Question 6.5 (V) Draw the structures of major monohalo products in each of the following reactions:

$CH_3CH_2Br+\ NaI\rightarrow$

Answer :

In this reaction the iodine atom will replace the bromine in the haloalkane.

The reaction is given below:

CH₃CH₂I + NaBr

Question 6.5 Draw the structures of major monohalo products in each of the following reactions:

(vi)

Answer :

In this reaction, the bromine atom will attack the alpha-carbon atom of the double bond.

The reaction is given below:

Question 6.6 Arrange each set of compounds in order of increasing boiling points.

(i) Bromomethane, Bromoform, Chloromethane, Dibromomethane.

Answer :

It is known that boiling point increases with increase in molecular mass when the alkyl group is the same.

So the order of increasing boiling point is Chloromethane < Bromomethane < Dibromomethane < Bromoform

Question 6.6 Arrange each set of compounds in order of increasing boiling points.

(ii) 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane.

Answer :

In the given compounds the halide groups are same. In these cases, the boiling point depends on the bulkiness of the alkyl group. The boiling point increases with an increase in the chain length. Also, the boiling point decreases with an increase in branching.

So the order is :- 1- Chlorobutane > 1- Chloropropane > Isopropyl Chloride

Question 6.7 Which alkyl halide from the following pairs would you expect to react more rapidly by an $S_{N}2$ mechanism? Explain your answer.

\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{Br} or

Answer :

In this case, the rate of _{$S_{N}2$} reaction will depend on the hindrance of the substrate.

Since 1-Bromobutane is a $1^{\circ}$ alkyl halide and 2-Bromobutane is a $2^{\circ}$ alkyl halide, 2-Bromobutane gives more hindrance to the nucleophile.

Hence, 1-bromobutane reacts faster.

Question 6.7 Which alkyl halide from the following pairs would you expect to react more rapidly by an $S_{N}2$ mechanism? Explain your answer.

Answer :

The rate of _{$S_{N}2$} reaction decreases with an increase in hindrance to the attack of the nucleophile.

In the first compound the alkyl halide is secondary halide while the second compound is a tertiary halide. So 2-bromobutane will react faster than 2-bromo-2-methylpropane in the nucleophilic attack.

Question 6.7 Which alkyl halide from the following pairs would you expect to react more rapidly by an $S_{N}2$ mechanism? Explain your answer.

Answer :

Both compounds are secondary halides .In these kinds of cases, we see where is the substituent is attached i.e., how far from the halide group. It can be clearly seen that the methyl group attached in 1-bromo-2-methylbutane is near than that attached in 1-bromo-3-methyl butane.

Hence, the rate of _{$S_{N}2$} reaction will be faster in the case of 1-bromo-3-methylbutane.

Question 6.8 In the following pairs of halogen compounds, which compound undergoes faster $S_{N}1$ reaction?

(i)

Answer :

In $S_{N}1$ reactions, we see the formation of carbocation and this is the rate determining step for this kind of reactions. So the compound having more stable carbocation will react faster. For $S_N 1$ the order of ractivity is $3^{\circ}>2^{\circ}>1^{\circ}$. In the given case 2- Chloro, 2- Methylpropane we have $3^{\circ}$ carbon whereas in 3- Chloropentane we have $2^{\circ}$ carbon.

Question 6.8 In the following pairs of halogen compounds, which compound undergoes faster $S_{N}1$ reaction?

Answer :

In $S_{N}1$ reactions, we see the formation of carbocation and this is the rate determining step for these kinds of reactions.So the compound having more stable carbocation will react faster. The first compound is a secondary compound and the second compound is primary compound. Hence 2-Chloroheptane will react faster than 1- Chlorohexane.

Question 6.9 Identify A, B, C, D, E, R and R1 in the following:

Answer :

1st reaction:-

The compound A in the reaction is cyclohexylmagnesium bromide, compound B is Cyclohexane.

2nd reaction:-

3rd reaction:-

The overall reaction is given below:

Question 6.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(ii) $CH_{3}CH_{2}CH(CH_{3})CH(C_{2}H_{5})Cl$

Answer :

3-Chloro-4-methylhexane. And it is a secondary alkyl halide.

Question 6.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(iii) $CH_{3}CH_{2}C(CH_{3})_{2}CH_{2}I$

Answer :

(iii) 1-Iodo-2, 2-dimethylbutane. And it is a primary alkyl halide.

Question 6.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(iv) $(CH_{3})_{3}CCH_{2}CH(Br)C_{6}H_{5}$

Answer :

(iv) 1-Bromo-3, 3-dimethyl-1-phenylbutane. And it is secondary benzyl halide.

Question 6.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(v) $CH_{3}CH(CH_{3})CH(Br)CH_{3}$

Answer :

(v) 2-Bromo-3-methylbutane. And it is a secondary alkyl halide.

Question 6.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(vi) $CH_{3}C(C_{2}H_{5})_{2}CH_{2}Br$

Answer :

(vi) 1-Bromo-2-ethyl-2-methylbutane. And it is a primary alkyl halide.

Question 6.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(vii) $CH_{3}C(Cl)(C_{2}H_{5})CH_{2}CH_{3}$

Answer :

(vii) 3-Chloro-3-methylpentane. And it is a tertiary alkyl halide.

Question 6.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(viii) $CH_{3}CH = C(Cl)CH_{2}CH(CH_{3})_{2}$

Answer :

(viii) 3-Chloro-5-methylhex-2-ene. And it is vinyl halide

Question 6.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(ix) $CH_{3}CH = CH C (Br)( CH_{3})_{2}$

Answer :

(ix) 4-Bromo-4-methylpent-2-ene. And it is allyl halide.

Question 6.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(x) $p- Cl C_{6}H_{4}CH_{2}CH(CH_{3})_{2}$

Answer :

(x) 1-Chloro-4-(2-methylpropyl) benzene. And it is an aryl halide.

Question 6.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(xi) $m- Cl C_{6}H_{4}CH_{2}CH (CH_{3})_{2}$

Answer :

(xi) 1-Chloromethyl-3-(2, 2-dimethylpropyl) benzene. And it is a primary benzyl halide.

Question 6.1 Name the following halides according to IUPAC system and classify them as alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:

(xii) $o- Br C_{6}H_{4}CH (CH_{3})CH_{2}CH_{3}$

Answer :

(xii) 1-Bromo-2-(1-methylpropyl) benzene. And it is an aryl halide.

Question 6.2 Give the IUPAC names of the following compounds:

(i) $CH_{3}CH(Cl)CH (Br)CH_{3}$

Answer :

(i) 2-Bromo-3-chlorobutane

Question 6.2 Give the IUPAC names of the following compounds:

(ii) $CHF_{2}CBrCIF$

Answer :

1-Bromo-1-chloro-1, 2, 2-trifluroethane

Question 6.2 Give the IUPAC names of the following compounds:

(iii) $Cl CH_{2}C \equiv C CH_{2}Br$

Answer :

1-Bromo-4-chlorobut-2-yne

Question 6.2 Give the IUPAC names of the following compounds:

(iv) $(CCl_{3})_{3}CCl$

Answer :

2-(Trichloromethyl)-1, 1, 1, 2, 3, 3, 3-heptachloropropane

Question 6.2 Give the IUPAC names of the following compounds:

(v) $CH_{3}(p- Cl C_{6}H_{4})_{2}CH(Br)CH_{3}$

Answer :

2-Bromo-3, 3-bis(4-Chlorophenyl) butane

Question 6.2 Give the IUPAC names of the following compounds:

(vi) $(CH_{3})_{3}CCH = CCl C_{6}H_{4}I - p$

Answer :

1-Chloro-1-(4-iodophenyl)-3, 3-dimethylbut-1-ene

Question 6.3 Write the structures of the following organic halogen compounds.

(i) 2-Chloro-3-methylpentane

Answer :

(i)

Question 6.3 Write the structures of the following organic halogen compounds.

(ii) p-Bromochlorobenzene

Answer :

(ii)

Question 6.3 Write the structures of the following organic halogen compounds.

(iii) 1-Chloro-4-ethylcyclohexane

Answer :

(iii)

Question 6.3 Write the structures of the following organic halogen compounds.

(iv) 2-(2-Chlorophenyl)-1-iodooctane

Answer :

(iv)

Question 6.3 Write the structures of the following organic halogen compounds.

(v) 2-Bromobutane

Answer :

(v)

Question 6.3 Write the structures of the following organic halogen compounds.

(vi) 4-tert-Butyl-3-iodoheptane

Answer :

(vi)

Question 6.3 Write the structures of the following organic halogen compounds.

(vii) 1-Bromo-4-sec-butyl-2-methylbenzene

Answer :

(vii)

Question 6.3 Write the structures of the following organic halogen compounds.

(viii) 1,4-Dibromobut-2-ene

Answer :

(viii)

Question 6.4 Which one of the following has the highest dipole moment?

(i) CH₂Cl ₂ (ii) CHCl₃ (iii) CCl₄

Answer :

Below are the three-dimensional structures of the three compounds, as well as the direction of each bond's dipole moment:

The order of dipole moment will be :- CH₂Cl₂ > CHCl₃ > CCl_4.

The reason for the above order is given as CCl₄ is a symmetrical compound so its dipole moment will be zero. In case of CHCl₃ , one of the Cl atoms cancels dipole moment of the opposite Cl atom, so the net dipole moment is just due to one Cl. In the case of CH₂Cl₂ , both Cl groups contribute to the dipole moment so it has the highest dipole moment among all.

Question 6.5 A hydrocarbon C₅H₁₀ does not react with chlorine in dark but gives a single monochloro compound C₅H₉ Cl in bright sunlight. Identify the hydrocarbon.

Answer :

We are given the formula C₅H₁₀ which can be either of an alkene or of cycloalkane. Since the hydrocarbon doesn't react with chlorine in dark thus it cannot be alkene. So the only option left out is cyclopentane.

Since the cycloalkane interacts with Cl₂ in the presence of strong sunshine to form a single monochloro compound, C₅H₉Cl, all ten hydrogen atoms in the cycloalkanes must be equivalent. Thus, cyclopentane is the cycloalkane.

Question 6.6 Write the isomers of the compound having formula _{$C_{4}H_{9}Br$} .

Answer :

The isomers of the compound _{$C_{4}H_{9}Br$} are :-

(i) 1-Bromobutane

(ii) 2-Bromobutane

(iii) 1-Bromo-2-methylpropane

(iv) 2-Bromo-2-methylpropane

Question 6.7 Write the equations for the preparation of 1-iodobutane from:

(i) 1-butanol

Answer :

(i) The procedure given below can be used :-

Question 6.7 Write the equations for the preparation of 1-iodobutane from

(ii) 1-chlorobutane

Answer :

(ii) The required product can be obtained as shown below :-

Question 6.7 Write the equations for the preparation of 1-iodobutane from

(iii) but-1-ene

Answer :

(iii) The required product is obtained by following procedure :-

Question 6.8 What are ambident nucleophiles? Explain with an example.

Answer :

Ambident nucleophiles are nucleophiles that can attack at two distinct locations. It is a resonance hybrid of the following two structures (for example, the cyanide ion).Cyanide ion is a resonance hybrid of the following two structures:

$: \bar{C} \equiv N: \longleftrightarrow: C=\bar{N}:$

It may attack carbon to produce cyanide, and nitrogen to form isocyanide, depending on the chemical.

Question 6.9 Which compound in each of the following pairs will react faster in S_N2 reaction with –OH?

(i) CH₃Br or CH₃I

Answer :

In this case, we have the same alkyl group but different halide ions. For this rate of S_N2 reaction increases with an increase in atomic mass. So, CH₃I will react faster than CH₃Br.

Question 6.9 Which compound in each of the following pairs will react faster in S_N2 reaction with –OH?

(ii) _{$(CH_{3})_{3}CCl$} or _{$CH_{3}Cl$}

Answer :

In this case, the hindrance will be a deciding factor for the rate of S_N2 reaction because hindrance will directly affect the attack of the nucleophile. (CH)₃ CCl has very high steric hindrance and CH₃Cl has less steric hindrance. So _{$CH_{3}Cl$} will react faster as compared to _{$(CH_{3})_{3}CCl$ .}

Question 6.10 Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

(i) 1-Bromo-1-methylcyclohexane

Answer :

In this compound, it is clear that we have identical _$\beta$ hydrogen; therefore, dehalogenation of the given compound gives the same alkene.

Question 6.10 Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

(ii) 2-Chloro-2-methylbutane

Answer :

(ii)

In this compound we have two kinds of _$\beta$ hydrogen. So dehalogenation will give two kinds of alkenes, namely 2-Methylbut-2-ene and 2-Methylbut-1-ene.

The major product of this reaction will be 2-Methylbut-2-ene as the number of $\ alpha$-hydrogens attached to double-bonded carbon is greater in case of this compound.

Question 6.10 Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene:

(iii) 2,2,3-Trimethyl-3-bromopentane

Answer :

(iii)

In this compound, we have two types of _$\beta$ - hydrogen thus dehalogenation we get two types of products namely 3, 4, 4-Trimethylpent-2-ene and 2-Ethyl-3,3-dimethylbut-2-ene.

and

Here, 3, 4, 4-Trimethylpent-2-ene will be major product, since the $\ alpha$-hydrogen attached to the double bond are greater.

Question 6.11 How will you bring about the following conversions?

(i) Ethanol to but-1-yne

Answer :

Ethanol will react with SOCl₂ and pyridine to Chloroethane. Acetylene will react with NaNH₂ to form sodium acetylide. Now Chloroethane and Sodium acetylide will react to form But-1-yne. The reactions are given below:

The conversion will take place by following procedure:-

Now,

Question 6.11 How will you bring about the following conversions?

(ii) Ethane to bromoethene

Answer :

Ethane will react with bromine to form bromoethane. Now, this bromoethane will react with KOH to form ethene and then there will be the formation of 1, 2- Dibromoethane. Now, 1, 2-Dibromoethane will react with alcoholic KOH to form Bromoethene. The reactions are given below:

Question 6.11 How will you bring about the following conversions?

(iii) Propene to 1-nitropropane

Answer :

Propene will react with Hydrogen bromide in peroxide effect to form 1- Bromopropane. Now, 1-Bromopropane will react with silver nitrite to given 1- Nitropropane. The reaction is given below:

Question 6.11 How will you bring about the following conversions?

(iv) Toluene to benzyl alcohol

Answer :

Toluene will react with chlorine in the presence of light to give benzyl chloride and this benzyl chloride will react with aqueous NaOH to give benzyl alcohol. The reaction is given below:

Question 6.11 How will you bring about the following conversions?

(v) Propene to propyne

Answer :

Propene will react with bromine in the presence of carbon tetrachloride to give 1, 2-Dibromopropane. Now, 1, 2-Dibromopropane will react with NaNH₂ and liquid ammonia to give propyne. The reaction is given below:

Question 6.11 How will you bring about the following conversions?

(vi) Ethanol to ethyl fluoride

Answer :

Ethanol will react with PCl₅ to form ethyl chloride. Now, ethyl chloride will react with AgF to form Ethyl fluoride. The reaction is given below:

Question 6.11 How will you bring about the following conversions?

(vii) Bromomethane to propanone

Answer :

Bromomethane will react with KCN in the presence of ethanol to form acetonitrile. Acetonitrile will react with CH₃ MgBr in the presence of ether and then with water to give Propanone. The reaction is given below:

Question 6.11 How will you bring about the following conversions?

(viii) But-1-ene to but-2-ene

Answer :

But-1-ene will react with Hydrogen bromide to form 2-Bromobutane and this will react with alcoholic KOH to form But-2-ene. The reaction is given below:

Question 6.11 How will you bring about the following conversions?

(ix) 1-Chlorobutane to n-octane

Answer :

1-Chlorobutane will react with sodium and dry ether to form n-octane. The reaction is given below:

Question 6.11 How will you bring about the following conversions?

(x) Benzene to biphenyl.

Answer :

Benzene will react with Bromine in the presence of Fe/Br₂ to form Bromobenzene. Bromobenzene will react with sodium to form Biphenyl. The reaction is given below:

Question 6.12 Explain why

(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride?

Answer :

We know that the Cl-atom in chlorobenzene is attached to a sp² hybridized carbon atom whereas in cyclohexyl chloride, the Cl-atom is attached to a sp³ hybridized carbon atom. It is known that sp² hybridized carbon has more s-character than sp³ hybridized carbon atom. Thus, chlorobenzene is more electronegative than cyclohexyl chloride.

Apart from this, the - R effect of the benzene ring of chlorobenzene results in decreasing the electron density of the C-Cl bond near the Cl-atom. And the C-Cl bond in chlorobenzene becomes less polar. Due to these reasons, the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.

Question 6.12 Explain why

(ii) alkyl halides, though polar, are immiscible with water?

Answer :

For being soluble in water, we have a condition that the solute-water force of attraction must be stronger than the solute-solute and water-water forces of attraction. Alkyl halides are held together by dipole-dipole interactions and there are polar molecules. Similarly, the intermolecular force of attraction present between the water molecules is hydrogen bonding. The new force of attraction after we dissolve solute in water, i.e., between the alkyl halides and water molecules, is weaker than the alkyl halide-alkyl halide and water-water forces of attraction. That is why alkyl halides (though polar) are immiscible with water.

Question 6.12 Explain why

(iii) Grignard reagents should be prepared under anhydrous conditions?

Answer :

The reactivity of the Grignard reagent is very high. When they come in contact with water they readily react and form alkanes. The general reaction is given below:

Question 6.13 Give the uses of freon 12, DDT, carbon tetrachloride and iodoform.

Answer :

(i) Freon-12 or dichlorodifluoromethane is generally known as CFC. It is used in refrigerators and air conditioners as a refrigerant. It is also used in body sprays, hair sprays, etc. But it has environmental impacts as it damages the ozone layer.

(ii) DDT or p, p-dichlorodiphenyltrichloroethane is one of the best-known insecticides which was used very widely all over the world. It is very effective against mosquitoes, insects and lice. But it has harmful effects.

(iii) CCl₄ :- It is mostly used for manufacturing refrigerants for refrigerators and air conditioners. It is also used as a solvent in the manufacture of pharmaceutical products. In the early years, carbon tetrachloride was widely used as a cleaning fluid and a fire extinguisher.

(iv) Iodoform was used earlier as an antiseptic. And this antiseptic property of iodoform is due to the liberation of free iodine when it comes in contact with the skin.

Question 6.14 Write the structure of the major organic product in each of the following reactions:

(i) $CH_{3}CH_{2}CH_{2}Cl+NaI\xrightarrow[heat]{acetone}$

Answer :

The Chlorine atom from the alkyl halide by iodine. The major product of the reaction is 1-Iodopropane.

The obtained product is:-

$CH_{3}CH_{2}CH_{2}Cl+NaI\xrightarrow[heat]{acetone}$$\underset{\text { 1-Iodopropane }}{\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{CH}_2 \mathrm{I}}+\mathrm{NaCl}$

Question 6.14 Write the structure of the major organic product in each of the following reactions:

(ii) $(CH_{3})_{3}CBr+KOH \xrightarrow[heat]{ethanol}$

Answer :

There will be Dehydrohalogenation and the major product of the reaction is 2- methylpropene. The reaction is given below:

Question 6.14 Write the structure of the major organic product in each of the following reactions:

(iii) $CH_{3}CH(Br)CH_{2}CH_{3}+ NaOH \overset{water}{\rightarrow}$

Answer :

The bromine atom will be replaced with the hydroxyl ion. The major product will be Butan-2-ol. The reaction is given below:

Question 6.14 Write the structure of the major organic product in each of the following reactions:

(iv) $CH_{3}CH_{2}Br+KCN \overset{aq. ethanol}{\rightarrow}$

Answer :

The bromide ion will be replaced with the cyanide ion. The major product will be propanenitrile. The reaction is given below

Question 6.14 Write the structure of the major organic product in each of the following reactions:

(v) $C_{6}H_{5}ONa + C_{2}H_{5}Cl \rightarrow$

Answer :

The major product in the above reaction will be Phenetole. The reaction is given below:

Question 6.14 Write the structure of the major organic product in each of the following reactions:

(vi) $CH_{3}CH_{2}CH_{2}OH+SOCl_{2}\rightarrow$

Answer :

The hydroxyl ion in the alkyl halide will be replaced with chloride ion. The major product of the reaction will be 1-Chloropropane. The reaction is given below:

Question 6.14 Write the structure of the major organic product in each of the following reactions:

(vii) $CH_{3}CH_{2}CH = CH _{2}+HBr \overset{peroxide}{\rightarrow}$

Answer :

Since peroxide is used in the reaction, there will be Anti-Markovnikov's addition. The major product in the reaction will be 1-Bromobutane. The reaction is given below:

Question 6.14 Write the structure of the major organic product in each of the following reactions:

(viii) $CH_{3}CH = C(CH_{3})_{2}+ HBr \rightarrow$

Answer :

In this reaction there will be Markovnikoff’s addition. The major product of the reaction will be 2-Bromo-2-methylbutane. The reaction is given below:

Question 6.15 Write the mechanism of the following reaction:

$nBuBr + KCN \overset{EtOH -H_{2}O}{\rightarrow} nB_{4}CN$

Answer :

The reaction will proceed through S_N2 mechanism. The mechanism for the given reaction is shown below:-

Question 6.16 Arrange the compounds of each set in order of reactivity towards S _N 2 displacement:

(i) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane

Answer :

(i) Here, the deciding factor the rate of reaction will be a steric hindrance.

It is clear from the above that the order of hindrance is:-

1-Bromopentane < 2-bromopentane < 2-Bromo-2-methylbutane

So the order of rate of reaction will be:-

2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane

Question 6.16 Arrange the compounds of each set in order of reactivity towards S _N 2 displacement:

(ii) 1-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 2-Bromo-3-methylbutane

Answer :

(ii) In this case, also, the order of the rate of reaction will be decided from the steric hindrance factor.

It is clear from the above that the steric hindrance in 2-Bromo-2-methylbutane is highest (note that hindrance of the carbon attached to halide ion is seen). So the order of the rate of reaction is:-

2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane

Question 6.16 Arrange the compounds of each set in order of reactivity towards S_N2 displacement:

(iii) 1-Bromobutane, 1-Bromo-2,2-dimethylpropane, 1-Bromo-2-methylbutane, 1-Bromo-3-methylbutane.

Answer :

(iii) The steric hindrance is the deciding factor here.

The order of steric hindrance is :-

1-Bromobutane < 1-Bromo-3-methylbutane < 1-Bromo-2-methylbutane< 1-Bromo-2, 2-dimethylpropane

Thus, the order of the rate of reaction will be : -

1-Bromo-2, 2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3- methylbutane < 1-Bromobutane

Question 6.17 Out of $C_{6}H_{5}CH_{2}Cl$ and $C_{6}H_{5}CHCl C_{6}H_{5}$ , which is more easily hydrolysed by aqueous KOH.

Answer :

Hydrolysis by KOH will take place by the formation of a carbocation in its rate-determining step. So the compound having a stable carbocation will hydrolyse faster.

The carbocations of both the compounds are given below:-

It is clear that carbocation of _{$C_{6}H_{5}CHCl C_{6}H_{5}$} is more stable.

Hence _{$C_{6}H_{5}CHCl C_{6}H_{5}$} will hydrolyse faster than _{$C_{6}H_{5}CH_{2}Cl$ .}

Question 6.18 p-Dichlorobenzene has higher m.p. than those of o- and m-isomers. Discuss.

Answer :

The structures of o-Dichlorobenzene, m-Dichlorobenzene and p-Dichlorobenzene are given below.

We can see that p-Dichlorobenzene is a very symmetric structure, thus its packing will be maximum. As a result, more and more energy will be required to break bonds (during boiling). Thus boiling point is high for p-Dichlorobenzene.

Question 6.19 How can the following conversions be carried out?

(i) Propene to propan-1-ol

Answer :

Propene will react with hydrogen bromide in the presence of peroxide to give 1- Bromopropane. Now, this 1-Bromopropane will react with aqueous KOH to form Propan-1-ol.

The mechanism is given below:-

Question 6.19 How the following conversions can be carried out?

(ii) Ethanol to but-1-yne

Answer :

The reaction mechanism is given below :-

Question 6.19 How the following conversions can be carried out?

(iii) 1-Bromopropane to 2-bromopropane

Answer :

1-Bromopropane will react with alcoholic KOH to form propene. Propene will react with HBr to form 2-Bromopropane.

The mechanism is given below :-

Question 6.19 How the following conversions can be carried out?

(iv) Toluene to benzyl alcohol

Answer :

Toluene will react with chlorine in the presence of light to give benzyl chloride and this benzyl chloride will react with aqueous KOH to give benzyl alcohol.

The mechanism is given below :-

Question 6.19 How the following conversions can be carried out?

(v) Benzene to 4-bromonitrobenzene

Answer :

Benzene will react with Bromine in the presence of FeBr3to form Bromobenzene. Bromobenzen will react with concentrated nitric acid and concentrated sulfuric acid to form 4-Bromonitrobenzene. The reaction is given below:

Question 6.19 How the following conversions can be carried out?

(vi) Benzyl alcohol to 2-phenylethanoic acid

Answer :

Benzyl alcohol will react with PCl₅ to form Benzyl chloride. Benzyl chloride will react with KCN to form Benzyl cyanide. Now, benzyl cyanide on hydrolysis will give 2-Phenylethanoic acid. The reaction is given below:

Question 6.19 How can the following conversions be carried out?

(vii) Ethanol to propanenitrile

Answer :

Ethanol will react with bromine in the presence of phosphorus to form Bromoethane. Iodoethane will react with KCN in the presence of aqueous ethanol to give Propanenitrile. The reaction is given below:

Question 6.19 How can the following conversions be carried out?

(viii) Aniline to chlorobenzene

Answer :

Aniline will undergo diazotization to form Benzene diazonium chloride. The Benzene diazonium chloride will react with copper chloride in the presence of hydrochloric acid to give Chlorobenzene. The reaction is given below:

Question 6.19 How can the following conversions be carried out?

(ix) 2-Chlorobutane to 3, 4-dimethylhexane

Answer :

2-Chlorobutane will react with sodium in the presence of dry ether to form 3, 4- Dimethylhexane. The rection is given below:

Question 6.19 How can the following conversions be carried out?

(x) 2-Methyl-1-propene to 2-chloro-2-methylpropane

Answer :

2-Methyl-1-propene will react with Hydrogen chloride to give 2-Chloro-2- methylpropane. The reaction is given below:

Question 6.19 How can the following conversions be carried out?

(xi) Ethyl chloride to propanoic acid

Answer :

Ethyl chloride will react with KCN to give propanenitrile. Propanenitrile on hydrolysis will give propanoic acid. The reaction is given below:

Question 6.19 How can the following conversions be carried out?

(xii) But-1-ene to n-butyliodide

Answer :

But-1-ene will react with HBr in the presence of peroxide to form 1- Bromobutane. 1-Bromobutane will react with NaI in the presence of Acetone to give n-butyl iodide. The reaction is given below:

Question 6.19 How can the following conversions be carried out?

(xiii) 2-Chloropropane to 1-propanol

Answer :

2-Chloropropane will undergo dehydrohalogenation to give propene. Propene will react with HBr in the presence of peroxide to give 1-bromopropane. 1- Bromopropane will react with KOH to form 1-Propanol. The reaction is given below:

Question 6.19 How can the following conversions be carried out?

(xiv) Isopropyl alcohol to iodoform

Answer :

The proposed mechanism is:-

Question 6.19 How can the following conversions be carried out?

(xv) Chlorobenzene to p-nitrophenol

Answer :

Chlorobenzene will react with concentrated nitric acid and concentrated sulfuric to give p-Chloronitrobenzene. p-Chloronitrobenzene will react with 15% sodium hydroxide and hydrochloric acid to give p-nitrophenol. The reaction is given below:

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Question 6.19 How can the following conversions be carried out?

(xvi) 2-Bromopropane to 1-bromopropane

Answer :

2-Bromopropane will react with alcoholic KOH to form Propene. Now, propene will react with HBr in the presence of peroxide to form 1-Bromopropane. The reaction is given below:

Question 6.19 How can the following conversions be carried out?

(xvii) Chloroethane to butane

Answer :

Chloroethane will react with sodium in the presence of butane. The reaction is given below:

Question 6.19 How can the following conversions be carried out?

(xviii) Benzene to diphenyl

Answer :

Benzene will react with Bromine in the presence of FeBr3to form Bromobenzene. Bromobenzene will react with sodium to form Biphenyl. The reaction is given below:

Question 6.19 How can the following conversions be carried out?

(xix) tert-Butyl bromide to isobutyl bromide

Answer :

Tert-butyl bromide will react with alcoholic KOH to form 2-Methyl-1-propene. 2-Methyl-1-propene will react with HBr in the presence of peroxide to form isobutyl bromide. The reaction is given below:

Question 6.19 How can the following conversions be carried out?

(xx) Aniline to phenylisocyanide

Answer :

Aniline will react with chloroform and alcoholic KOH to form Phenyl isocyanide. The reaction is given below:

Question 6.20 The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in the presence of alcoholic KOH, alkenes are major products. Explain.

Answer :

In an aqueous solution, KOH almost completely dissociates into OH ^- ions. We know that OH ^- ions are strong nucleophile, which leads the alkyl chloride to undergo a reaction to form alcohol. But an alcoholic solution of KOH contains alkoxide (RO^-) ion, which is a strong base. Thus, it can remove hydrogen from the β-carbon of the alkyl chloride and form an alkene. The OH ^- ion is a weaker base than RO^- ion. The basic character of OH^- ion decreases in aqueous solution. Therefore, it cannot remove hydrogen from the β-carbon.

Question 6.21 Primary alkyl halide $C_{4}H_{9}Br$ (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), $C_{8}H_{18}$ which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions.

Answer :

With the given formula we have two alkyl halides They are n - butyl bromide and isobutyl bromide.

For the first set of reaction we get two possibilities:-

Therefore, compound (d) is 2, 5-dimethylhexane.

Question 6.22 What happens when

(i) n-butyl chloride is treated with alcoholic KOH,

Answer :

When n−butyl chloride is treated with alcoholic KOH, the formation of but−l−ene takes place. This reaction is a dehydrohalogenation reaction.

Question 6.22 What happens when

(ii) bromobenzene is treated with Mg in the presence of dry ether

Answer :

When bromobenzene is treated with Mg in the presence of dry ether there will be the formation of Phenylmagnesium bromide. The reaction is given below:

Question 6.22 What happens when

(iii) chlorobenzene is subjected to hydrolysis

Answer :

Chlorobenzene does not undergo hydrolysis under normal conditions. However, it undergoes hydrolysis when heated in an aqueous sodium hydroxide solution at a temperature of 623 K and a pressure of 300 atm to form phenol.

Question 6.22 What happens when

(iv) ethyl chloride is treated with aqueous KOH

Answer :

When ethyl chloride is treated with aqueous KOH, it undergoes hydrolysis to form ethanol.

Question 6.22 What happens when

(v) methyl bromide is treated with sodium in the presence of dry ether

Answer :

When methyl bromide is treated with sodium in the presence of dry ether then Wurtz reaction will take place and the product of the reaction will be Ethane. The reaction is given below:

Question 6.22 What happens when

(vi) methyl chloride is treated with KCN?

Answer :

When methyl chloride is treated with KCN then there will be a Nucleophilic substitution reaction and the product will be Methyl cyanide. The reaction is given below: