Haloalkanes and Haloarenes Test - 16

On commercial scale, phenol is obtained from chlorobenzene. The chlorobenzene needed for the purpose is prepared by following reaction:

In Raschig's process a mixture of benzene vapours, air and hydrogen chloride is passed over heated \(\mathrm{Cu}  \mathrm{Cl}_{2}\) at \(500 \mathrm{~K}\).

\(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{CHCl}_{3}+\mathrm{O}_{2}{_{(\text{air})}} \frac{\mathrm{CuCl}_{2}}{\mathrm{500K}} 2 \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{Cl}+2 \mathrm{H}_{2} \mathrm{O}\)

Phenol is obtained from chlorobenzene as shown in the above reaction.

Hence, the correct option is (D).

Physical properties of any compound depends largely on both mass and type of intermolecular and intramolecular forces of attractions.

• Extensive properties, such as mass and volume, depend on the amount of matter that is being measured. Both extensive and intensive properties are physical properties, which means they can be measured without changing the substance's chemical identity.
• The physical state and properties of a particular compound depend in large part on the type of chemical bonding it displays.

Hence, the correct option is (C).

"Alkyl halides have higher boiling points than corresponding alcohols" is the false statement.

• Alcohols have inter-molecular Hydrogen bonding between the molecules.
• Whereas, Alkyl halides are held together by London forces which are weaker forces than Hydrogen bonding.
• Due to this reason Alcohols have higher Boiling points than Alkyl halides. 

Hence, the correct option is (B).

Let's remove halogen from all the options:

(A) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}-\mathrm{Cl} \stackrel{ \mathrm{Cl}^{-}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}^{+}\)

(B) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}-\mathrm{Br} \stackrel{\mathrm{Br}^{-}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}^{+}\)

(C) \(\mathrm{CH}_{3} \mathrm{CH}_{2}(\mathrm{Cl}) \mathrm{CH}_{3} \stackrel{\mathrm{Cl}^{-}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}^{+} \mathrm{CH}_{3}\)

(D) \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3} \stackrel{\mathrm{Br}^{-}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}^{+} \mathrm{CH}_{3}\)

\(\mathrm{Br}^{-}\)is a better leaving group than \(\mathrm{Cl}^{-}\)since its conjugate acid, HBr is weaker than HCl.

Therefore, 2-bromopropane is the correct answer as it leads to the formation of a \(2^{0}\) carbocation.

Hence, the correct option is (D).

Haloalkanes of the form \(\mathrm{R}-\mathrm{Br}\) and \(\mathrm{R}-\mathrm{I}\) develop colour when exposed to light.

• As we progress down the periodic table from fluorine to iodine, molecular size increases. 
• As a result, we also see an increase in bond length. 
• As the bond length increases, upon irradiation of light, after absorption the compound emit light in visible region. 
• Therefore, \(\mathrm{R}-\mathrm{Br}\) and \(\mathrm{R}-\mathrm{I}\)  bond exhibit color when exposed to light.

Hence, the correct option is (D).

In isomeric alkyl halides the branch chain isomer have relatively low boiling point as compared to its straight chain isomer because a branch chain molecule have less surface area in comparison of straight chain isomer due to spherical shape which reduces Van der Waals force of attraction and hence decrease boiling point.

Hence, the correct option is (B).

The reaction given is a Sandmeyer's Reaction.

• The Sandmeyer reaction is used to prepare aryl halides from aryl diazonium salts. 
• It is named after the Swiss chemist Traugott Sandmeyer.
• The reaction is a method for substitution of an aromatic amino group via preparation of its diazonium salt followed by its displacement with a nucleophile, catalyzed by copper (I) salts. 

The reaction takes place as follows:

Hence, the correct option is (B).

Alkyl halides develop colour on exposure to light.

• Alkyl bromides and iodides develop colour when exposed to light. 
• Halogens have lower bond dissociation energy. 
• This low enthalpy means that even at normal temperature or on exposure to sunlight, they decompose and form free ions, which are coloured and hence impart colour on exposure to light.

Hence, the correct option is (C).

The carbon atom is attached with one hydrogen atom and 3 chlorine atom.

• It can be seen from structure that chloroform contains 4 single bonds.
• One single bond contains one sigma. 

Therefore, it contains 4 sigma bond.

Hence, the correct option is (D).

Benzene reacts with \(\mathrm{CH}_{3} \mathrm{Cl}\) in presence of anhydrous \(\mathrm{AlCl}_{3}\) to give toulene.

• This is Friedel-Craft's Alkylation and an electrophilic substitution reaction.
• Friedel-Craft's alkylation of toluene by reaction with methyl chloride in the presence of \(\mathrm{AlCl}_{3}\) and heat produces meta-isomer that is \(\mathrm{m}-\mathrm{xylene}\) as major product.

The reaction takes place as follows:

Hence, the correct option is (A).

The \(\mathrm{C}-\mathrm{X}\) bond in haloalkanes is more polar than the \(\mathrm{C}-\mathrm{X}\) in haloarenes.

In haloarenes, due to resonance, charge distribution or rotation of charge on benzene nucleus takes place therefore, the \(\mathrm{C}-\mathrm{X}\) bond in haloalkanes ( which are not aromatic) is more polar than the \(\mathrm{C}-\mathrm{X}\) bond in haloarenes (aromatic).

Hence, the correct option is (A).

The reaction of benzene with chlorine in the presence of iron gives chlorobenzene.

When benzene reacts with chlorine gas in the presence of iron catalyst such as iron (III) chloride , it displaces one hydrogen from the ring to chlorine atom and leads to the formation of chlorobenzene.

The reaction takes place as follows:

Hence, the correct option is (B).

Chlorine atom in \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) is attached to \(\mathrm{sp}^{3}\) carbon.

The structure of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) is as follows:

As from the structure, it can be seen that chlorine atom in \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) is attached to a carbon with \(4 \sigma\) bond. 

Thus, it is attached to the \(\mathrm{sp}^{3}\) carbon atom.

Hence, the correct option is (A).

Given that,

Bond length, \(\mathrm{d=2.00}\) Å \(\mathrm{=2 \times 10^{-10}~m}\)

Theoretical Dipole Moment, \(\mathrm{ =5.12 \times10^{-30}~C-m}\)

We know that,

Charge, \(\mathrm{Q=1.6 \times10^{-19}}\)

Therefore, from:

Dipole moment \(=\) Charge \(\times\) Bond Length

\(=1.6 \times10^{-19} \times 2 \times 10^{-10}~m\)

\(=32 \times 10^{-30}\mathrm{~C-m}\)

We also know that:

\(\%\)Ionic Character \(=\frac{\text{Observed Dipole Moment}}{\text{Theoritical Dipole Moment}}\times100\)

\(=\frac{5.12 \times10^{-30}}{32 \times 10^{-30}}\times 100\)

\(=0.16 \times 100\)

\(=16\%\)

Hence, the correct option is (B).

The bond between carbon and the halogen atom \((\mathrm{C}-\mathrm{X})\) is polar. 

• Since halogen atoms are more electronegative than carbon, there is partial negative charge on halogen atom and positive charge on carbon. 
• This makes the molecule dipolar and polar nature of \((\mathrm{C}-\mathrm{X})\) bond increases with an increase in the electronegativity difference between carbon and halogen atom. 

Therefore, the order of polarity of \(\mathrm{C}-\mathrm{X}\) bond will be:

\(\mathrm{C}-\mathrm{F}>\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{I}\)

Thus, the increasing order of electronegativity of halides will be:

\(\mathrm{I,  B r, Cl, F}\)

Hence, the correct option is (B).

The monohalogen derivatives of alkanes are called as alkyl halides denoted as \(\mathrm{R}-\mathrm{X}\) where \((\mathrm{R}-)\) is the alkyl group and \(\mathrm{X}-\) is the halogen atom. Therefore, statement in option (A) is correct.

The common name of an alkyl halide is always written as two separate words. The common names of alkyl halides consist of two parts: the name of the alkyl group plus the stem of the name of the halogen, with the ending "-ide". Therefore, statement in option (B) is correct.

In IUPAC system, the mono halogen derivatives of saturated hydrocarbon alkanes are called haloalkanes. For e.g., methyl chloride has IUPAC name chloromethane. Therefore, statement in option (C) is also correct.

Hence, the correct option is (D).

The IUPAC name of the given compound is 1-methyl-4-chlorobenzene.

• When one hydrogen of benzene is replaced by methyl group, then it is called toluene. 
• Chlorine will get first priority alphabetically and will be numbered as 1 and methyl will be numbered as 4.
  
• Chlorine attaches at para position is also called p-chlorotoluene or 4-chlorotoluene.

Hence, the correct option is (B).

Compound \(\mathrm{CCl}_{3} \mathrm{CH}(\mathrm{OH})_{2}\) is stable.

Here, we noticed that two OH group attached with one carbon atom at all three compounds which is called the geminal diol compound.

(C) \(\mathrm{CCl}_{3} \mathrm{CH}(\mathrm{OH})_{2}\)

(A) \(\mathrm{CH}_{3}-\mathrm{CH}(\mathrm{OH})_{2}\)

(B) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH})_{2}\)

Dehydration reaction:

(C) \( \mathrm{CCL}_{3} \mathrm{CH}(\mathrm{OH})_{2}\underset{\mathrm{H^+}}{\stackrel{\mathrm{C}}{\longrightarrow}} \mathrm{Cl}_{3} \mathrm{CH}\left(\mathrm{OH}_{2}\right)-\mathrm{O}-\mathrm{H} \underset{-\mathrm{H}_{2} \mathrm{O}}{\stackrel{\mathrm{H^+}}{\longrightarrow}} \mathrm{CH}_{3}-\mathrm{C}-\mathrm{CHO}\)

(A) \( \mathrm{CH}_{3}-\mathrm{CH}(\mathrm{OH})_{2} \stackrel{\mathrm{H^+}}{\longrightarrow} \mathrm{CH}_{3} \mathrm{CH}\left(\mathrm{OH}_{2}^{+}\right)-\mathrm{O}-\mathrm{H} \underset{-\mathrm{H}_{2} \mathrm{O}}{\stackrel{\mathrm{H^+}}{\longrightarrow}} \mathrm{CH}_{3}-\mathrm{CHO}\)

(B) \( \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}(\mathrm{OH})_{2} \stackrel{\mathrm{H}^{+}}{\longrightarrow}\left(\mathrm{CH}_{3}\right) 2 \mathrm{C}\left(\mathrm{OH}_{2}\right)-\mathrm{O}-\mathrm{H} \underset{-\mathrm{H}_{2} \mathrm{O}}{\stackrel{\mathrm{H^+}}{\longrightarrow}} \mathrm{CH}_{3}-\mathrm{CH}_{2} \mathrm{CHO}\)

\(\because+\) I group (Alkyl group) : geminal diol compound stability decreases.

\(\because-\) I group (Halogen group) : geminal diol compound stability increases.

Hence the correct option is (C).

Ortho and para hydrogen differ in nuclear spin of two atoms.

• Ortho hydrogen molecules are those in which the spins of both the nuclei are in the same direction. 
• Molecules of hydrogen in which the spins of both the nuclei are in the opposite direction are called para hydrogen.

Generally, dihydrogen is a combination of ortho and para hydrogen that stay in the equilibrium:

Ortho \(\left(\mathrm{H}_{2}\right) \underset{\mathrm{e q .}}{\leftrightarrows}\) Para \(\left(\mathrm{H}_{2}\right)\)

Hence, the correct option is (D).

The reaction of toluene with chlorine in the presence of light gives Benzyl chloride. 

• Reaction of methylbenzene i.e., toluene with chlorine in the presence of sunlight is a free radical substitution.
• In this reaction hydrogen atoms from side chain get replaced get replaced by chlorine. 
• On monochlorination, benzyl chloride is formed on dichlorination benzal chloride and on trichlorination benzochloride is formed.

The reaction proceeds in following manner:

Hence, the correct option is (C).

For \(\mathrm{{CH}_{3} Br}\):

Atomic mass: 

\(=(12\times 1)+(1\times3)+(80\times1)\)

\(=95\)

For \(\mathrm{CH}_{3} \mathrm{Cl}\):

Atomic mass: 

\(=(12\times1)+(1\times3)+(35.5\times1)\)

\(=50.5\)

For \(\mathrm{H}_{2} \mathrm{O}\): 

Atomic mass: 

\(=(16\times1)+(2\times1)\)

\(=18\)

According to above data regarding atomic masses of compounds, the decreasing order will be: 

\(\mathrm{CH}_{3} \mathrm{Br}>\mathrm{CH}_{3} \mathrm{Cl}>\) \(\mathrm{H}_{2} \mathrm{O}\)

Hence, the correct option is (C).

\(\mathrm{CH}_{3} \mathrm{Cl}\) has highest dipole moment.

• Dipole is a physical quantity. 
• Methane has zero dipole moment. 
• Replacement of one H- atoms by Cl atom increases the dipole moment. 
• The increase in dipole moment is because the bond dipole moment of \(\mathrm{C-H}\) bond and that of \(\mathrm{C-C l}\) bond reinforce each other. 
• Replacement of another \(\mathrm{H}\) atom by \(\mathrm{Cl}\) increases the bond angle due to lone pair - lone pair repulsion between \(\mathrm{Cl}\)-atoms. 
• This reduces the dipole moment and introduces the third \(\mathrm{Cl}\) - atom. 
• So, when fourth \(\mathrm{Cl}\) - atom is introduced the molecule \(\left(\mathrm{CCl}_{4}\right)\) again becomes symmetrical and dipole moment reduces to zero.
• So, \(\mathrm{CH}_{3} \mathrm{Cl}\) has highest dipole moment.

Hence, the correct option is (A).

Ethylidene chloride is a gem-dihalide. 

• Ethylidene chloride is also known as 1,1 Dichloroethane. 
• It has 2 chlorine atoms on the same carbon atom, thus, has 1,1 relationship. 
• Therefore, it is geminal-dihalide.
• Gem-dihalide is a compound that have two halogen atoms on the same carbon atom. 

Below given the structure of Ethylidene chloride:

Hence, the correct option is (B).

Among the given compounds, compound II has highest melting point.

Compound II is most symmetrical because it has both \(\mathrm{CH}_{3}\) groups and \(\mathrm{Cl}\) atoms \(\mathrm{p}\) - to each other. Therefore, it fits in the crystal lattice better than the other two isomers and thus, it has high melting point.

Hence, the correct option is (B).

The \(\mathrm{C}-\mathrm{Cl}\) bond in \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\) is shorter than in \(\mathrm{CH}_{3} \mathrm{Cl}\) and therefore, stronger.

• The carbon in \(\mathrm{C-Cl}\) bond in chlorobenzene is \(\mathrm{sp}^{2}\) hybridised, while in \(\mathrm{CH} 3-\mathrm{Cl}\) is \(\mathrm{sp}^{3}\) hybridised. 
• In \(\mathrm{sp}^{2}\) hybrid orbitals have more of \(\mathrm{s}\)-character and hence the carbon if chlorobenzene withdraws the electron pair between \(\mathrm{C-Cl}\) with greater force. 
• As a result, \(\mathrm{C-Cl}\) bond is shorter than \(\mathrm{CH}_{3}-\mathrm{Cl}\).
  
• The lone pair on chlorine are dispersed throughout the benzene ring by resonance. 
• This gives the \(\mathrm{C-Cl}\) bond a double bond character.

Hence, the correct option is (C).

The numbering of carbon chain is done from the carbon which is having two chlorine atoms. 

The parent alkane containing two carbon atoms is called as ethane and therefore, the IUPAC name of this compound is 1,1-Dichloroethane.

Hence, the correct option is (C).

\(\mathrm{R}-\mathrm{X}\) bond polarity order is: \(\mathrm{C}-\mathrm{F}>\mathrm{C}-\mathrm{Cl}>\mathrm{C}-\mathrm{Br}>\mathrm{C}-\mathrm{I}\)

• The difference in electronegativity of the carbon-halogen bonds range from \(1.5\) in \(\mathrm{C}-\mathrm{F}\) to almost 0 in \(\mathrm{C}-\mathrm{I}\). 
• This means that the \(\mathrm{C}-\mathrm{F}\) bond is extremely polar, though not ionic, and the \(\mathrm{C}-\mathrm{I}\) bond is almost nonpolar.

Hence, the correct option is (A).

The chemical formula of halothane is \(\mathrm{CF}_{3} \mathrm{CHClBr}\).

• Halothane, the brandname Fluothane, is a general anaesthetic.  
• It can be used to induce or maintain anaesthesia.

The molecular structure of halothane is as follows:

Hence, the correct option is (A).

When methyl Iodide is treated with sodium in the presence of dry ether, Ethane is formed. 

It is called Wurtz reaction. 

\(\mathrm{CH}_{3}-\mathrm{I}+2 \mathrm{Na}+\mathrm{I}-\mathrm{CH}_{3} \overset{_\text{Dry ether}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{CH}_{3}+2 \mathrm{NaI}\)

Hence, the correct option is (D).

\(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\) has the highest boiling point.

• Branching decreases the boiling point. 
• The more sphere like the molecule, the lower its surface area will be and the fewer intermolecular Van der Waals interactions will operate. 
• Therefore, \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{Cl}\) straight chain molecule with the highest number of carbons has the highest boiling point.

Hence, the correct option is (B).