Haloalkanes and Haloarenes - Class 12 Chemistry - Chapter 6 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Haloalkanes and Haloarenes | NCERT | Chemistry | Class 12
The hybridisation of carbon atoms in diamond and graphite are respectively:
Option A) $sp^3$, $sp^3$
Option B) $sp^2$, $sp^2$
Option C) $sp^2$, $sp^3$
Option D) $sp^3$, $sp^2$
In diamond, each carbon atom forms a tetrahedral structure with four sigma bonds, indicating an $sp^3$ hybridization. In contrast, graphite has a layered structure, with each carbon atom forming three sigma bonds and one pi bond, characteristic of $sp^2$ hybridization.
Thus, the correct answer is: Option D) $sp^3$, $sp^2$
Which of the following are octa-atomic and tetra-atomic interhalogens respectively?
(A) $\mathrm{ClF}_{3}$ and $IF_{7}$
(B) $IF_{7}$ and $\mathrm{ClF}_{3}$
(C) Both (A) and (B)
(D) None
Interhalogen compounds are composed of two or more different halogens. These compounds include only halogens without other types of atoms.
Here, we are interested in identifying the octa-atomic and tetra-atomic interhalogens. Octa-atomic implies the molecule contains eight halogen atoms, and tetra-atomic indicates it has four halogen atoms.
In $\mathrm{IF}_{7}$, there is one iodine atom and seven fluorine atoms, totaling eight halogen atoms. Thus, it is an octa-atomic interhalogen.
In $\mathrm{ClF}_{3}$, there is one chlorine atom and three fluorine atoms, totaling four halogen atoms. Hence, it is a tetra-atomic interhalogen.
Given the options:
(A) $\mathrm{ClF}_{3}$ and $IF_{7}$ incorrectly assigns the number of atoms.
(B) $IF_{7}$ and $\mathrm{ClF}_{3}$ correctly identifies $IF_{7}$ as octa-atomic and $\mathrm{ClF}_{3}$ as tetra-atomic.
(C) Both (A) and (B) cannot be correct because (A) is incorrect.
(D) None is incorrect as option (B) is valid.
Therefore, the correct answer is (B) $\mathrm{IF}_{7}$ and $\mathrm{ClF}_{3}$.
Electrophilic substitution reactions are faster in:
A) Chlorobenzene
B) Phenol
C) Anisole
D) Phenetole
The correct option is D) Phenetole
In the context of electrophilic substitution reactions, the rate is influenced by the ability of substituents on the benzene ring to stabilize or destabilize the intermediate carbocation formed during the reaction. Substituents that are electron-donating through resonance (mesomeric effect) typically enhance the rate of these reactions by stabilizing the carbocation.
The $-\mathrm{OC}_2\mathrm{H}_5$ group in phenetole is a stronger +M (electron-donating) group compared to the $-\mathrm{OH}$ group in phenol and the $-\mathrm{OCH}_3$ group in anisole. This makes phenetole the most reactive among the given options towards electrophilic substitution reactions.
The number of lone pairs on the compound obtained by heating a mixture of $Xe$ and $F_{2}$ in the ratio 1:5 in a sealed nickel tube is:
When heating a mixture of Xenon ($Xe$) and Fluorine ($F_2$) in a 1:5 ratio inside a sealed nickel tube, the primary reaction product formed is Xenon tetrafluoride ($XeF_4$):
$$ Xe + 2F_2 \rightarrow XeF_4 $$
To calculate the total number of lone pairs in $XeF_4$, we consider:
Xenon ($Xe$) contributes 2 lone pairs.
Each Fluorine atom contributes 3 lone pairs. There are four fluorine atoms in $XeF_4$, leading to $4 \times 3 = 12$ lone pairs from all fluorine atoms.
Adding these together gives the total number of lone pairs in the molecule:
$$ 12 \ (from \ fluorines) + 2 \ (from \ xenon) = 14 $$
Thus, the total number of lone pairs on the compound $XeF_4$ is 14.
Anti-Markovnikov's rule is only applicable for addition of:
A. $\mathrm{HCl}$
B. $\mathrm{HBr}$
C. $\mathrm{HI}$
D. All of the above.
The correct answer is B. $\mathrm{HBr}$.
$\mathrm{HBr}$ exhibits a unique behavior in the presence of peroxides and light known as the Peroxide Effect or Kharash Effect. This effect leads to the addition of $\mathrm{HBr}$ to unsymmetrical alkenes in a manner that contradicts Markovnikov's rule, which usually predicts that the hydrogen atom will add to the carbon with more hydrogen atoms.
However, $\mathrm{HCl}$ and $\mathrm{HI}$ do not exhibit the peroxide effect:
The $\mathrm{H-F}$ and $\mathrm{H-Cl}$ bonds are especially strong, making it difficult for these molecules to undergo homolytic cleavage essential for radical formation. The cleavage of an $\mathrm{H-F}$ bond can even be explosive due to its highly exothermic nature.
For $\mathrm{HCl}$, the initiation step of bond breaking is exothermic, but the propagation step (where the radical reaction would continue) is endothermic, which is not favorable thermodynamically.
With $\mathrm{HI}$, both the initiation and propagation steps are endothermic, making the radical reaction pathway energetically unfavorable. It is more likely for iodine radicals to recombine to form $\mathrm{I}_2$ than to react further with alkenes.
In contrast, for $\mathrm{HBr}$, the radical mechanism is favored because both propagation steps are exothermic, allowing for efficient radical addition across the double bond: $$ \mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}_2 + \mathrm{HBr} \rightarrow \mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}_2\mathrm{Br} $$ Thus, Anti-Markovnikov’s rule applies only to $\mathrm{HBr}$ under peroxide effect conditions.
The type of hybrid orbitals used by the chlorine atom in $\mathrm{ClO}_{2}^{-}$ is
A $\mathrm{sp}^3$
B $\mathrm{sp}^{2}$
C $\mathrm{sp}$
D None of these
The correct option is A: $$ \text{sp}^3 $$
The chlorine atom in the $\mathrm{ClO}_2^{-}$ ion utilizes $\mathrm{sp}^3$ hybrid orbitals. However, despite this $\mathrm{sp}^3$ hybridization, the molecular geometry is angular (or bent).
Which observation in silver nitrate test of $\mathrm{HCl}$ helps us to confirm the presence of chlorine in $\mathrm{HCl$?
A. Formation of hydrogen gas B. Formation of white precipitate C. Formation of nitrogen gas D. Formation of black precipitate
The correct answer is B. Formation of white precipitate. When silver nitrate ($\mathrm{AgNO}_3$) is mixed with hydrochloric acid ($\mathrm{HCl}$), a reaction occurs leading to the formation of nitric acid ($\mathrm{HNO}_3$) in aqueous solution and silver chloride ($\mathrm{AgCl}$), which is visible as a white precipitate. This specific outcome confirms the presence of chloride ions in the mixture.
The chemical equation representing this reaction is: $$ \mathrm{AgNO}_3 + \mathrm{HCl} \rightarrow \mathrm{AgCl}(\text{s}) + \mathrm{HNO}_3(\text{aq}) $$ The appearance of a white precipitate of $\mathrm{AgCl}$ distinctly indicates the presence of chlorine ions from $\mathrm{HCl}$.
Starting from acetylene and any alkyl halide, which of the following yields cyclodecyne?
A. 1 eq. $\mathrm{NaNH}_{2}$ and $\mathrm{Br}\left(\mathrm{CH}_{2}\right)_{9} \mathrm{Br}$
B. 2 eq. $\mathrm{NaNH}_{2}$ and $\mathrm{Br}\left(\mathrm{CH}_{2}\right)_{8} \mathrm{Br}$
C. 1 eq. $\mathrm{NaNH}_{2}$ and $\mathrm{Br}\left(\mathrm{CH}_{2}\right)_{8} \mathrm{Br}$
D. Both B and C
Option B is the correct answer: 2 eq. $\mathrm{NaNH}_2$ and $\mathrm{Br(CH_2)_8Br}$.
This option uses a 9-carbon dibromide with two equivalents of sodium amide ($\mathrm{NaNH}_2$), which allows for both bromides at each end of the chain to be effectively replaced by sodium amides. This further reacts with acetylene to close into a 10-membered ring, forming cyclodecyne.
And which is an example of an allylic halide?
An allylic halide is defined as an alkyl halide in which one or more halogen atoms are attached to the carbon atoms adjacent (next) to the carbon-carbon double bond. These adjacent carbons are referred to as allylic carbons.
Here is an example image of an allylic halide:
In the image, you can observe the halogen atom (e.g., Cl, Br) bonded to the carbon atom next to the double bonded carbons. This is a typical feature of allylic halides.
An example of a super octet molecule is:
(A) $\mathrm{SF}_{6}$
(B) $\mathrm{PCl}_{5}$
(C) $\mathrm{IF}_{7}$
(D) All of these
When considering a super octet molecule, we refer to instances where atoms contain more than 8 valence electrons. This typically happens in compounds involving third period or below elements because they can employ d-orbitals for bonding.
In $\mathrm{SF}_6$, sulfur (S) breaks the octet rule by holding 12 valence electrons. It achieves this by forming six S-F bonds.
For $\mathrm{PCl}_5$, phosphorus (P) also surpasses the octet rule by holding 10 valence electrons as a result of forming five P-Cl bonds.
Lastly, in $\mathrm{IF}_7$, iodine (I) stretches its valence to encompass 14 electrons, by establishing seven I-F bonds.
Given these examples, we see that each mentioned compound does indeed violate the typical octet rule by exhibiting an expanded valence shell. This leads us to conclude that all the given options (A, B, and C) correctly represent cases of super octet molecules. Therefore, the correct answer is:
(D) All of these
In the presence of $\mathrm{CCl}_{4}$, what is the end product in this reaction $\mathrm{CH}_{2}=\mathrm{CH}_{2}+\mathrm{Cl}_{2} \rightarrow$?
A. 1,2-dichloroethane
B. 1,1-dichloroethane
C. 1-chloroethane
D. pentachloroethane
The correct option is A. 1,2-dichloroethane.
This reaction is an example of an addition reaction specifically halogenation, where the alkene ($\mathrm{CH}_2=\mathrm{CH}_2$) reacts with diatomic halogen molecules ($\mathrm{Cl}_2$), resulting in dihalogenated alkanes. Among halogens, iodine typically reacts less readily in similar conditions. Thus, the end product of the reaction $\mathrm{CH}_2=\mathrm{CH}_2 + \mathrm{Cl}_2$ in the presence of $\mathrm{CCl}_4$ is $1,2$-dichloroethane.
Which one of the following compounds shows geometrical isomerism?
A 1,1-dibromoprop-1-ene
B propene
C 2-methylprop-1-ene
D 1,2-dibromobut-2-ene
The correct option is D 1,2-dibromobut-2-ene.
Conditions for geometrical isomers:
Free rotation must be restricted.
Both the terminals must be attached with different groups.
Terminals must not be in perpendicular planes.
Option A (1,1-dibromoprop-1-ene) has two identical groups (Br) attached to the terminal carbon of the double bond, thereby disqualifying it for geometrical isomerism.
Option B (propene) has two identical groups (H) attached to the terminal carbon of the double bond, so it also cannot exhibit geometrical isomerism.
Option C (2-methylprop-1-ene) has two identical groups (H) attached to the terminal carbon of the double bond, and therefore cannot display geometrical isomerism.
Option D (1,2-dibromobut-2-ene) has different groups attached to the terminal carbons of the double bond, enabling it to exhibit geometrical isomerism.
Therefore, compound D, 1,2-dibromobut-2-ene, shows geometrical isomerism due to the presence of different substituents on the carbons of the double bond, which fulfill the requirements for different spatial arrangements.
The Dow's process is used for the preparation of
A) Chlorobenzene
B) Fluorobenzene
C) Bromobenzene
D) Phenol
The correct answer is D) Phenol.
Dow's Process involves heating chlorobenzene at $623$K under a pressure of $300$ atm with a $10%$ NaOH solution. This reaction forms sodium phenate, which upon hydrolysis with HCl, yields phenol. Thus, Dow's process is primarily used for the preparation of phenol.
In allene (C3H4), the type(s) of hybridisation of the carbon atoms is (are):
A $sp$ and $sp^{3}$
B $sp$ and $sp^{2}$
C only $sp^{3}$
D $sp^{2}$ and $sp^{3}$
The correct answer is B $sp$ and $sp^2$.
Allene, also known as propadiene, is represented by the molecular formula $H_2C=C=CH_2$. This molecule is interesting due to the hybridization states of its carbon atoms.
Hybridization is determined by counting the number of sigma ($\sigma$) bonds and lone pairs around a given atom. Pi ($\pi$) bonds, which arise from the sideways overlap of p-orbitals, are not involved in hybridization.
In the allene molecule:
Terminal Carbons: Each has three $\sigma$ bonds (two to hydrogen and one to the central carbon), leading to an $sp^2$ hybridization.
Central Carbon: Has two $\sigma$ bonds (one to each terminal carbon), which corresponds to an $sp$ hybridization.
Therefore, the types of hybridization present in allene are $sp$ for the central carbon and $sp^2$ for the terminal carbons.
Out of the following, which one causes "Ozone Depletion"?
A) Land degradation
B) High levels of chlorine and bromine compounds in the stratosphere
C) Increase in greenhouse gas concentrations
D) Deforestation
The correct answer is B) High levels of chlorine and bromine compounds in the stratosphere.
Ozone depletion occurs primarily due to the presence of chlorine and bromine compounds in the stratosphere. These compounds lead to the breakdown of ozone molecules, thereby thinning the ozone layer that protects the earth from harmful ultraviolet radiation.
Polarization is an important phenomenon that is helpful in the determination of the covalent nature of a compound. With the help of Fajan's rule, one can easily predict the nature of a compound.
Which of the following halides possesses the highest covalent character?
A) $\mathrm{LiF}$
B) $\mathrm{LiCl}$
C) $\mathrm{LiBr}$
D) $\mathrm{LiI}$
The correct option is D) $\mathrm{LiI}$.
$\mathrm{Li}^+$ has high polarizing power due to its small size and high charge density. According to Fajan's rule, the greater the polarizability of the anion, the greater will be the covalent character. Among the given halides, $\mathrm{I}^-$ (iodide) is the largest and therefore has the highest polarizability.
Consequently, $\mathrm{LiI}$ possesses the highest covalent character among the listed lithium halides.
Which of the following reagent is preferred to prepare alkyl halides from alcohols?
(A) $\mathrm{PCl}_{5}$
(B) $\mathrm{HCl}$ gas in the presence of anhydrous $\mathrm{ZnCl}_{2}$ (C) $\mathrm{SOCl}_{2}$ (D) $\mathrm{PCl}_{3}$
The correct answer is (C) $\mathrm{SOCl}_2$.
Thionyl chloride ($\mathrm{SOCl}_2$) is preferred for converting alcohols to alkyl halides because the reaction produces alkyl chlorides along with sulfur dioxide ($\mathrm{SO}_2$) and hydrogen chloride (HCl) as by-products, both of which are gases that can be easily removed from the reaction mixture. Here is the chemical reaction involved:
$$ \mathrm{R-OH} + \mathrm{SOCl}_2 \rightarrow \mathrm{R-Cl} + \mathrm{SO}_2 \uparrow + \mathrm{HCl \uparrow} $$
The escaping gases (marked by the upward arrows) simplify the purification process, making $\mathrm{SOCl}_2$ a highly effective reagent for this transformation. Additionally, the use of $\mathrm{SOCl}_2$ generally leads to high yields of the desired alkyl halide product.
In compound X, all the bond angles are exactly $109^{\circ} 28^{\prime}$. X is
A) Chloromethane
B) Iodoform
C) Carbon tetrachloride
D) Chloroform
The correct answer is Option C: Carbon tetrachloride.
The compound in question, Carbon tetrachloride (denoted as $\mathrm{CCl}_4$), features sp(^3) hybridization. This hybridization leads to the formation of four equivalent orbitals that are arranged in a regular tetrahedral geometry. Each angle in a perfect tetrahedron measures exactly $109^{\circ} 28'$, matching the bond angles described in the question.
For the other compounds listed, even though they may also exhibit sp(^3) hybridization, the presence of different atoms and their respective electron cloud distortions mean that the bond angles slightly deviate from the ideal tetrahedral angle. This is why Carbon tetrachloride is the specific compound where all bond angles are consistently at $109^{\circ} 28'$.
Which of the following alkali metal chlorides is soluble in benzene?
A $\mathrm{CsCl}$
B $\mathrm{LiCl}$
C $\mathrm{NaCl}$
D $\mathrm{KCl}$
The correct answer is B $\mathrm{LiCl}$.
This is because $\mathrm{LiCl}$ exhibits a higher covalent character compared to other alkali metal chlorides, making it more soluble in non-polar solvents like benzene.
When a $\mathrm{CS}_{2}$ layer containing bromine and iodine is shaken with excess of chlorine water, the violet colour due to iodine disappears and orange colour of bromine appears. The disappearance of the violet colour is due to the formation of
(A) $\mathrm{I}_{3}^{-}$
(B) $\mathrm{HIO}_{3}$
(C) $\mathrm{ICl}_{2}$
(D) $\mathrm{I}^{-}$
The correct answer is (B) $\mathrm{HIO}_{3}$. This is derived from the reaction of $\mathrm{I}_2$ with chlorine water. In an excess of chlorine water, iodine reacts according to the following chemical equation:
$$ 5 \mathrm{Cl}_2 + \mathrm{I}_2 + 6 \mathrm{H}_2O \rightarrow 2 \mathrm{HIO}_3 + 10 \mathrm{HCl} $$
In this reaction, iodine ($\mathrm{I}_2$) is oxidized by chlorine ($\mathrm{Cl}_2$) in the presence of water ($\mathrm{H}_2O$) to form __iodic acid ($\mathrm{HIO}_3$)__ and __hydrochloric acid ($\mathrm{HCl}$)__. This results in the disappearance of the violet color of iodine, leading to the correct option being **(B) $\mathrm{HIO}_{3}$**.
$\mathrm{BF}_{3} + \mathrm{LiH} \rightarrow \mathrm{A} + \mathrm{LiBF}_{4}$. $\mathrm{A}$ is as follows:
(A) $B_{2}H_{6}$ (B) $B_{4}H_{10}$ (C) $\mathrm{BH}_{3}$ (D) $B_{5}H_{9}$
The correct answer is (A) $B_{2}H_{6}$.
Diborane, represented as $B_{2}H_{6}$, is the product formed when boron trifluoride ($\mathrm{BF}_{3}$) reacts with __lithium hydride__($\mathrm{LiH}$). This reaction can be summarized by the chemical equation: $$ \mathrm{BF}_{3} + \mathrm{LiH} \rightarrow B_{2}H_{6} + \mathrm{LiBF}_{4} $$ In this scenario, $B_{2}H_{6}$ is the correct identification of compound $\mathrm{A}$.
Sort the following compounds according to stability:
A. Cyclopropane
B. Cyclobutane
C. Cyclohexane
D. Cyclopentane
Cycloalkanes resemble alkanes in terms of reactivity, with smaller ones, particularly cyclopropane, exhibiting higher reactivity. This increased reactivity is mostly due to the bond angles within these rings. A typical carbon atom forms bonds at approximately $109.5^\circ$ in its most relaxed state, due to the tetrahedral shape.
For cyclopropane, the internal bond angles are drastically reduced to $60^\circ$. Such sharp angles lead to significant electron pair repulsion, weakening the carbon-carbon bonds and hence reducing stability.
On the contrary, larger cycloalkanes like cyclobutane and cyclopentane have bond angles of $90^\circ$ and $108^\circ$ respectively. These values approach the ideal tetrahedral angle, therefore exhibiting lesser strain and greater stability than cyclopropane.
The most stable among them, cyclohexane, benefits from a bond angle very close to the ideal $109.5^\circ$, found in its chair conformation. This structure not only minimizes angular strain but also eliminates torsional strain, presenting a fully staggered arrangement of atoms.
Thus, the order of stability based on the least to most strained structures is:
$$ \text{Cyclopropane} < \text{Cyclobutane} < \text{Cyclopentane} < \text{Cyclohexane} $$
Cyclohexane stands out as the most stable due to its near-perfect bond angles and lack of torsional strain.
What is the correct increasing order of a gas being liquifiable?
(A) $\mathrm{H}_{2} < \mathrm{N}_{2} < \mathrm{CH}_{4} < \mathrm{CO}_{2}$
B $\mathrm{H}_{2} < \mathrm{CO}_{2} < \mathrm{CH}_{4} < \mathrm{N}_{2}$ (c) $\mathrm{CO}_{2} < \mathrm{CH}_{4} < \mathrm{N}_{2} < \mathrm{H}_{2}$ (D) $\mathrm{CO}_{2} < \mathrm{CH}_{4} < \mathrm{H}_{2} < \mathrm{N}_{2}$
The correct order is (A) $$\mathrm{H}_2 < \mathrm{N}_2 < \mathrm{CH}_4 < \mathrm{CO}_2$$.
The ability of a gas to liquefy depends heavily on the value of the van der Waals' constant '$a$'. This constant '$a$' represents the measure of intermolecular forces of attraction.
Gases with weaker intermolecular forces are more challenging to liquefy because they have a lower value of '$a$'. Hydrogen ($\mathrm{H}_2$) and Nitrogen ($\mathrm{N}_2$), having lower values of 'a', are more difficult to liquefy.
On the other hand, Methane ($\mathrm{CH}_4$) and Carbon Dioxide ($\mathrm{CO}_2$), which have stronger intermolecular forces, can be liquefied more easily. The critical temperature of $\mathrm{CO}_2$ is higher than that of $\mathrm{CH}_4$, enhancing its ability to be liquefied under less severe conditions. Thus, the order from hardest to easiest to liquefy is as shown in option A.
Below $900^{\circ}\mathrm{C}$, $\mathrm{BeCl}_{2}$ in vapor phase exists as:
(A) $\mathrm{BeCl}_{2}$ (B) $\mathrm{Be}_{2}\mathrm{Cl}_{4}$
(C) $\mathrm{BeCl}_{2}$ and $\mathrm{Be}_{2}\mathrm{Cl}_{4}$
(D) $\mathrm{Be}_{2}\mathrm{Cl}_{4}$
The correct solution to the question is:
Below $900^{\circ}\mathrm{C}$, beryllium chloride ($\mathrm{BeCl}_{2}$) in the vapor phase predominantly exists as a mixture of both __monomeric__ ($\mathrm{BeCl}_{2}$) form and dimeric ($\mathrm{Be}_{2}\mathrm{Cl}_{4}$) form. Thus, the correct answer would be:
(C) $\mathrm{BeCl}_{2}$ and $\mathrm{Be}_{2}\mathrm{Cl}_{4}$
Which of the following compounds would yield trialkyl borane as shown below when treated with $\mathrm{BH}_{3}$ and THF?
A) 2-Methylbut-1-ene
B) 2-Methylbut-2-ene
C) 3-Methylbut-1-ene
D) 3-Methylbut-1-ene
The correct answer is Option D: 3-Methylbut-1-ene
Reaction Explanation:
When treating alkenes with $\mathrm{BH}_3$ and THF, the process known as hydroboration occurs. This reaction typically follows the Anti-Markovnikov rule, where the boron from $\mathrm{BH}_3$ adds to the less substituted carbon of the alkene (the carbon with more hydrogen atoms).
Selecting the Correct Alkene:
Out of the options given, we need to identify the alkene that would lead to the formation of the trialkyl borane when treated with $\mathrm{BH}_3$/THF. The reaction of 3-Methylbut-1-ene with $\mathrm{BH}_3$ and THF can produce a trialkyl borane where the boron attaches to the carbon having a higher number of hydrogen atoms initially, consistent with Anti-Markovnikov addition.
Hence, the correct option is: Option D) 3-Methylbut-1-ene
What is the name of the compound with the formula $NF_{3}$, $CF_{4}$, and $S_{2}F_{10}$?
The question requires us to name the compounds given by the formulas $NF_3$, $CF_4$, and $S_2F_{10}$. Let's analyze each formula and determine the correct chemical names:
$NF_3$ - This compound consists of nitrogen (N) and fluorine (F). The chemical name is derived by naming the element nitrogen first, followed by the prefix for the number of fluorine atoms. Since there are three fluorine atoms, the prefix "tri" is used, leading to the name nitrogen trifluoride.
$CF_4$ - This compound is made of carbon (C) and fluorine (F). There is one carbon atom and four fluorine atoms. For four fluorine atoms, the prefix "tetra" is used. Thus, the name of this compound is carbon tetrafluoride.
$S_2F_{10}$ - Composed of sulfur (S) and fluorine (F), this compound has two sulfur atoms and ten fluorine atoms. The prefix "di" denotes two sulfur atoms, and "deca" (not dec) for ten fluorine atoms. Therefore, the correct name is disulfur decafluoride.
In conclusion, the names of the compounds are as follows:
$NF_3$: Nitrogen Trifluoride
$CF_4$: Carbon Tetrafluoride
$S_2F_{10}$: Disulfur Decafluoride
The IUPAC name of is:
Option 1) 1-bromo-2-chloro-3-fluoro-6-iodobenzene Option 2) 2-bromo-1-chloro-5-fluoro-3-iodobenzene Option 3) 4-bromo-2-chloro-5-fluoro-1-iodobenzene Option 4) 2-bromo-3-chloro-1-fluoro-5-iodobenzene
The correct option is Option 2: 2-bromo-1-chloro-5-fluoro-3-iodobenzene.
For a benzene ring with multiple substituents, the numbering of the ring atoms should be done in a way that minimizes the numbers assigned to the substituent groups. Specifically, the first substituent encountered should be given the lowest possible number.
Additionally, the substituents should be listed in alphabetical order.
Hence, the IUPAC name for the given compound is:
2-bromo-1-chloro-5-fluoro-3-iodobenzene.
The electrophile involved in the above reaction is:
A. dichloromethylcation $\left(\stackrel{\oplus}{\mathrm{C}} \mathrm{HCl}_{2}\right)$
B. dichlorocarbene $\left(: \mathrm{CCl}_{2}\right)$
C. trichloromethyl anion $\left(\mathrm{CCl}_{3}\right)$
D. formyl cation $\left(\stackrel{\oplus}{\mathrm{C}} \mathrm{HO}\right)$
The correct answer is B. Dichlorocarbene ($: \mathrm{CCl}_{2}$).
Let's examine the reaction:
$$ \begin{aligned} &\mathrm{CHCl}_{3} + \mathrm{OH}^{-} \rightarrow \mathrm{CCl}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} \ &\mathrm{CCl}_{3}^{-} \rightarrow :\mathrm{CCl}_{2} + \mathrm{Cl}^{-} \end{aligned} $$
In the first step, chloroform ($\mathrm{CHCl}_{3}$) reacts with a hydroxide ion ($\mathrm{OH}^{-}$), producing $\mathrm{CCl}_{3}^{-}$ and water ($\mathrm{H}_{2} \mathrm{O}$).
In the subsequent step, $\mathrm{CCl}_{3}^{-}$ decomposes to form dichlorocarbene ($: \mathrm{CCl}_{2}$) and a chloride ion ($\mathrm{Cl}^{-}$).
Hence, the electrophile involved in this reaction is dichlorocarbene ($:\mathrm{CCl}_{2}$).
Consider the following statements:
India has successfully achieved the complete phase out of Hydrochlorofluorocarbon (HCFC)-141b, which is a chemical used by foam manufacturing enterprises and the air conditioning industry.
The complete phase out of HCFC 141b from the country is as agreed under the Montreal Protocol on Substances that Deplete the Ozone Layer and Kyoto Protocol.
Hydrochlorofluorocarbon (HCFC)-141b is not produced in the country, and all the domestic requirements are met through imports.
Which of the statements given above is/are correct?
Correct Option: B
Statement 3 only is correct.
Explanation:
India has successfully phased out the use of Hydrochlorofluorocarbon (HCFC)-141b. This chemical is utilized by foam manufacturing enterprises and is one of the most potent ozone-depleting chemicals after Chlorofluorocarbons (CFCs). However, because India has phased out HCFC-141b, statement 1 is incorrect.
HCFC-141b is primarily used as a blowing agent in the production of rigid polyurethane (PU) foams.
HCFC-141b is not produced within India, and domestic requirements are entirely met through imports. With the prohibition on importing HCFC-141b, India has successfully eliminated this key ozone-depleting substance. Thus, statement 3 is correct.
The phase-out of HCFC-141b in the foam sector is among the first at this scale for Article 5 parties (developing countries) under the Montreal Protocol. However, this action is not related to the Kyoto Protocol, making statement 2 incorrect.
Summary:
Statement 1: Incorrect (India has phased out HCFC-141b)
Statement 2: Incorrect (The phase-out is under the Montreal Protocol, not the Kyoto Protocol)
Statement 3: Correct (HCFC-141b is imported, not produced domestically)
Which gas is called super halogen?
A $F_{2}$
B $\mathrm{Cl}_{2}$
C $\mathrm{Br}_{2}$
D $I_{2}$
To determine which gas is called a super halogen, let's first understand what halogens are. Halogens are chemical elements that belong to Group 17 of the periodic table. This group includes:
Fluorine ($\mathrm{F_2}$)
Chlorine ($\mathrm{Cl_2}$)
Bromine ($\mathrm{Br_2}$)
Iodine ($\mathrm{I_2}$)
Astatine ($\mathrm{At}$)
Among these halogens, fluorine ($\mathrm{F_2}$) stands out due to its high reactivity. Fluorine is the most electronegative element and has the ability to combine directly with nearly all other elements, except some noble gases and oxygen. This makes fluorine more reactive and electronegative than the other halogens.
Because of these properties, fluorine is referred to as a super halogen.
Hence, the correct answer is:
A) $\mathrm{F_2}$
Fluorine $(\mathrm{F_2})$ is called a super halogen because it is highly reactive and electronegative, allowing it to combine directly with almost all other elements.
The unit of the Van der Waals constant 'a' is:
(A) $ \text{atm} \cdot \text{L}^2 \cdot \text{mol}^{-2} $
(B) $ \text{dyne} \cdot \text{cm}^2 \cdot \text{mol}^{-2} $
(C) $ \text{dyne} \cdot \text{m}^2 \cdot \text{mol}^{-2} $
(D) $ \text{dm}^3 \cdot \text{atm} \cdot \text{mol}^{-1} $
Correct Answer is:
(A) In $\text{Pa m}^6 \text{mol}^{-2}$
To derive the unit of the van der Waals constant 'a', let's consider the van der Waals equation:
$$ P = \frac{n^2 a}{V^2} $$
Rearranging for ( a ):
$$ a = \frac{P V^2}{n^2} $$
Here,
$P$ represents pressure,
$V$ represents volume, and
$n$ is the number of moles.
Given that:
The unit of pressure ($P$) is Pascal (Pa)
The unit of volume ($V$) is cubic meters (m³)
The unit of the number of moles ($n$) is simply moles (mol)
Substituting these units into the equation, we get:
$$ a = \frac{(\text{Pa}) (\text{m}^3)^2}{(\text{mol})^2} = \frac{\text{Pa} \cdot \text{m}^6}{\text{mol}^2} $$
Hence, the unit of ($a$) is:
$$ \text{Pa} \cdot \text{m}^6 \cdot \text{mol}^{-2} $$
Therefore, the correct unit of the van der Waals constant ($a$) is Pa m⁶ mol⁻², which matches the option (A).
Final Answer: A
When a vacuum vessel is empty, its mass is found to be 50 grams. When it is filled with a liquid of density 0.47 grams per milliliter and weighed, its mass is found to be 144 grams. When it is filled with an ideal gas at 760 milliliters pressure and 300 K temperature and weighed, its mass is found to be 50.5 grams. What is the molar mass of the ideal gas? (Given: $ R = 0.0821 \ \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} $)
A. 61.575
B. 130.98
C. 123.75
D. 47.87
The correct answer is: A
Mass of the vessel filled with liquid: $$ 144 - 50 = 94 \text{ grams} $$
Volume of the liquid (which is equal to the volume of the vessel): $$ \frac{94}{0.47} = 200 \text{ milliliters} $$
Therefore, the volume of the gas ( V ): $$ 200 \text{ milliliters} = 0.2 \text{ liters} $$
Given:
Pressure $ P = 760 \text{ mm} \rightarrow 1 \text{ atm} $
Temperature $ T = 300 \text{ K} $
Mass of the gas: $$ 50.5 - 50 = 0.5 \text{ grams} $$
Using the ideal gas equation: $$ P V = n R T = \frac{w}{M} R T $$ where:
( w ) is the mass of the gas
( M ) is the molar mass of the gas
Solving for the molar mass ( M ): $$ M = \frac{w R T}{P V} $$
Substituting the known values: $$ M = \frac{0.5 \times 0.0821 \times 300}{1 \times 0.2} = 61.575 $$
Thus, the molar mass of the ideal gas is $ \boxed{61.575} $.
Final Answer: A
In a nucleophilic substitution reaction, the order of halogens as incoming (attacking) nucleophiles is -
$F > Br^{-} > Cl^{-}$
The order of halogens as leaving nucleophiles should be -
$Br^{-} > I^{-} > Cl^{-}$
$I^{-} > Br^{-} > Cl^{-}$
$Cl^{-} > Br^{-} > I^{-}$
$Cl^{-} > I^- < Br^-$
In nucleophilic substitution reactions, the ability of nucleophiles (attackers) in terms of halogens follows the sequence:
$$ F^- > Br^- > Cl^- $$
When considering leaving nucleophiles, the order of halogens changes based on their nucleophilic abilities, which is as follows:
$$ I^- > Br^- > Cl^- $$
Therefore, the correct option is:
B. $ I^- > Br^- > Cl^- $
The thermal decomposition of alkenes is called -
Option 1) Pyrolytic decomposition
Option 2) Cracking
Option 3) Both of the above
Option 4) None
Answer: Option 3: Both of the above
The process of thermal decomposition of alkenes is referred to by two terms: pyrolytic decomposition and cracking.
Pyrolytic decomposition involves the breaking down of a compound by heat. Specifically, in the context of alkenes, this results in the formation of simpler molecules.
Cracking is a broader term commonly used in the petroleum industry, where larger hydrocarbon molecules are decomposed at high temperatures to produce smaller alkenes and other products.
Therefore, since both pyrolytic decomposition and cracking describe the thermal decomposition of alkenes, the correct answer is Option 3: Both of the above.
The oxidation state of Xe in XeO3 and bond angle in it are:
A. +6, 109°
B. +8, 103°
C. +6, 103°
D. +8, 120°
To determine the oxidation state of Xenon (Xe) in $ \text{XeO}_3 $, follow these steps:
Let the oxidation state of Xenon (Xe) be ( x ).
Oxygen (O) typically has an oxidation state of (-2).
The compound $ \text{XeO}_3 $ contains one Xenon atom and three Oxygen atoms. Therefore, we can set up the equation: $$ x + 3(-2) = 0 $$
Simplifying this: $$ x - 6 = 0 $$ $$ x = +6 $$
Therefore, the oxidation state of Xe in $ \text{XeO}_3 $ is +6.
Next, let’s determine the bond angle in $ \text{XeO}_3 $:
$ \text{XeO}_3 $ consists of three bond pairs and one lone pair on Xenon.
This configuration gives it sp^3 hybridization.
The shape of the molecule is pyramidal due to the lone pair on Xenon.
The presence of a lone pair results in bond angles being slightly less than the ideal tetrahedral angle of 109.5°.
Therefore, the bond angle in $ \text{XeO}_3 $ is approximately 103°.
Based on the oxidation state of +6 and the bond angle of 103°, the correct answer is:
C. +6, 103°
Compound (C) was prepared in a three step sequence from ethyl trifluoroacetate. The first step in a sequence involved treating ethyl trifluoroacetate with NH3 to give a compound (A), which on heating with (X) gives (B). (B) on treatment with an organometallic (Y), followed by hydrolysis produces (C). Based on the above passage attempt the following questions:
(X) should be: A. $\mathrm{BaO}_2$
B. $\mathrm{H}_2\mathrm{O}_2$
C. $\mathrm{P}_4\mathrm{O}_{10}$
D. $\mathrm{N}_2\mathrm{O}$
The appropriate reagent $ (X) $ is $\mathbf{P_4O_{10}}$.
In the given sequence, ethyl trifluoroacetate is treated with ammonia (NH$_3$) in the first step to form compound $ (A) $. When compound $ (A) $ is heated with $\mathrm{P}_4\mathrm{O}_{10}$, it yields compound $ (B) $. Hence, the correct choice for $ (X) $ is $\mathrm{P}_4\mathrm{O}_{10}$.
Boron forms BF₄⁻ ion while aluminium forms AlF₆³⁻ ion, though they are in the same group. Why?
Boron is unable to expand its octet due to the non-availability of d-orbitals. The maximum covalence of boron cannot exceed 4. This is because:
The principal quantum number ($n$) represents the number of electron shells.
The azimuthal quantum number ($l$), which refers to the sub-shell, can have values from 0 to $n-1$.
In the case of boron, with the electron configuration $1s^2\ 2s^2\ 2p^1$, the principal quantum number $n$ is 2. Since $l$ ranges from 0 to $n-1$, boron can only have $s$ and $p$ orbitals (where $l=0$ for $s$ and $l=1$ for $p$). Therefore, boron does not have $d$-orbitals (where $l=2$), and thus its covalence is limited to 4. Consequently, it forms the $BF_4^-$ ion.
On the other hand, aluminium, with the electron configuration $1s^2\ 2s^2\ 2p^6\ 3s^2\ 3p^1$, has a principal quantum number $n$ of 3. This allows for $l$ values ranging from 0 to $n-1$, thus including $d$-orbitals (where $l=2$). As a result, aluminium has empty d-orbitals that can accommodate additional electrons, allowing it to expand its covalency beyond 4. Therefore, aluminium can form the $AlF_6^{3-}$ ion.
The given chemical reaction can be described as follows:
\[ \text{CH}_3 - \text{C} \equiv \text{C} - \text{CH}_3 \xrightarrow{\text{(1) H}_2/\text{Pd/CaCO}_3 \text{ or BaSO}_4} \xrightarrow{\text{(2) Br}_2,\text{CCl}_4} \text{X} \]
A 2,3-Dibromobutane B 1,3-Dibromobutane C 2-bromobutane D All of the above.
The correct option is A: 2,3-Dibromobutane.
When alkynes undergo partial reduction in the presence of Lindlar's catalyst, they add hydrogen atoms preferentially via syn addition (addition on the same side of the bond). This process ensures that the resulting product is a cis-isomer of the alkene.
After the formation of the cis-alkene, halogens like bromine or chlorine can be added to the alkene to form vicinal dihalides.
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Ask Chatterbot AINCERT Solutions - Haloalkanes and Haloarenes | NCERT | Chemistry | Class 12
Name the following halides according to IUPAC system and classify them as
alkyl, allyl, benzyl (primary, secondary, tertiary), vinyl or aryl halides:
(i) \[(\mathrm{CH_3})_2\mathrm{CHCH(Cl)CH_3}\]
(ii) \[\mathrm{CH_3CH_2CH(CH_3)CH(C_2H_5)Cl}\]
(iii) \[\mathrm{CH_3CH_2C(CH_3)_2CH_2I}\]
(iv) \[(\mathrm{CH_3})_3\mathrm{CCH_2CH(Br)C_6H_5}\]
(v) \[\mathrm{CH_3CH(CH_3)CH(Br)CH_3}\]
(vi) \[\mathrm{CH_3C(C_2H_5)_2CH_2Br}\]
(vii) \[\mathrm{CH_3C(Cl)(C_2H_5)CH_2CH_3}\]
(viii) \[\mathrm{CH_3CH=C(Cl)CH_2CH(CH_3)_2}\]
(ix) \[\mathrm{CH_3CH=CHC(Br)(CH_3)_2}\]
(x) \[\mathrm{p{-}ClC_6H_4CH_2CH(CH_3)_2}\]
(xi) \[\mathrm{m{-}ClCH_2C_6H_4CH_2C(CH_3)_3}\]
(xii) \[\mathrm{o{-}Br{-}C_6H_4CH(CH_3)CH_2CH_3}\]
Let's go through each halide one by one for IUPAC naming and classification.
(i) [ (\mathrm{CH_3})_2\mathrm{CHCH(Cl)CH_3} ]
IUPAC Name: 4-Chloro-2-methylpentane
Classification: Secondary alkyl halide
(ii) [ \mathrm{CH_3CH_2CH(CH_3)CH(C_2H_5)Cl} ]
IUPAC Name: 3-Chloro-2-ethylpentane
Classification: Secondary alkyl halide
(iii) [ \mathrm{CH_3CH_2C(CH_3)_2CH_2I} ]
IUPAC Name: 1-Iodo-2,2-dimethylbutane
Classification: Primary alkyl halide
(iv) [ (\mathrm{CH_3})_3\mathrm{CCH_2CH(Br)C_6H_5} ]
IUPAC Name: 1-Bromo-3-(2-methyl-2-propyl)benzene
Classification: Benzyl halide (Primary)
(v) [ \mathrm{CH_3CH(CH_3)CH(Br)CH_3} ]
IUPAC Name: 2-Bromo-3-methylbutane
Classification: Secondary alkyl halide
(vi) [ \mathrm{CH_3C(C_2H_5)_2CH_2Br} ]
IUPAC Name: 1-Bromo-3,3-dimethylbutane
Classification: Primary alkyl halide
(vii) [ \mathrm{CH_3C(Cl)(C_2H_5)CH_2CH_3} ]
IUPAC Name: 2-Chloro-3-ethylpentane
Classification: Secondary alkyl halide
(viii) [ \mathrm{CH_3CH=C(Cl)CH_2CH(CH_3)_2} ]
IUPAC Name: 4-Chloro-3-isopropyl-1-butene
Classification: Vinyl halide
(ix) [ \mathrm{CH_3CH=CHC(Br)(CH_3)_2} ]
IUPAC Name: 4-Bromo-3-methyl-1-butene
Classification: Vinyl halide
(x) [ \mathrm{p{-}ClC_6H_4CH_2CH(CH_3)_2} ]
IUPAC Name: 1-(4-Chlorophenyl)-2-methylpropane
Classification: Primary benzyl halide
(xi) [ \mathrm{m{-}ClCH_2C_6H_4CH_2C(CH_3)_3} ]
IUPAC Name: 1-(3-Chlorophenyl)-1,1-dimethylpropane
Classification: Primary benzyl halide
(xii) [ \mathrm{o{-}Br{-}C_6H_4CH(CH_3)CH_2CH_3} ]
IUPAC Name: 1-(2-Bromophenyl)-2-methylpropane
Classification: Secondary benzyl halide
Give the IUPAC names of the following compounds:
(i) \[\mathrm{CH_3CH(Cl)CH(Br)CH_3}\]
(ii) \[\mathrm{CHF_2CBrClF}\]
(iii) \[\mathrm{ClCH_2C{\equiv}CCH_2Br}\]
(iv) \[\mathrm{(CCl_3)_3CCl}\]
(v) \[\mathrm{CH_3C(p{-}ClC_6H_4)_2CH(Br)CH_3}\]
(vi) \[\mathrm{(CH_3)_3CCH=CClC_6H_4 I{-}p}\]
Let's go through each compound step-by-step to determine their IUPAC names.
(i) $\mathrm{CH_3CH(Cl)CH(Br)CH_3}$
Identify the longest chain containing the highest priority functional group, which is 2-chlorobutane.
Number the carbon chain from the end nearest the first substituent.
Thus, the IUPAC name is 2-bromo-3-chlorobutane.
(ii) $\mathrm{CHF_2CBrClF}$
Identify the longest chain which contains the highest priority functional group, which here is a two-carbon chain.
Number the carbon chain from the end closest to the substituent that comes first alphabetically and according to priority rules.
Thus, the IUPAC name is 2-bromo-1-chloro-1,1,2-trifluoroethane.
(iii) $\mathrm{ClCH_2C{\equiv}CCH_2Br}$
Identify the longest chain with the functional group (alkyne), which is butyne.
Number the carbon chain from the end nearest the triple bond.
Thus, the IUPAC name is 1-chloro-4-bromo-2-butyne.
(iv) $\mathrm{(CCl_3)_3CCl})$
Identify the longest chain, which here is just a single carbon with four substituents.
Thus, the IUPAC name is 2,2,2-tris(trichloromethyl)ethane.
(v) $\mathrm{CH_3C(p{-}ClC_6H_4)_2CH(Br)CH_3}$
Identify the longest chain, with the parent chain having the highest priority groups.
Number the carbon chain from the end closest to the substituents to obtain the lowest possible set of locants.
Thus, the IUPAC name is 2-bromo-1,1-bis(4-chlorophenyl)propane.
(vi) $\mathrm{(CH_3)_3CCH=CClC_6H_4 I{-}p}$
Identify the longest chain containing the highest priority substituent, which is a six-carbon chain with a double bond (hexene).
Number the chain so that the double bond gets the lowest number and assign substituents accordingly.
Thus, the IUPAC name is 4-(4-iodophenyl)-3-chloro-2,2-dimethylhex-3-ene.
Here is the summary of IUPAC names for the compounds:
(i) 2-bromo-3-chlorobutane
(ii) 2-bromo-1-chloro-1,1,2-trifluoroethane
(iii) 1-chloro-4-bromo-2-butyne
(iv) 2,2,2-tris(trichloromethyl)ethane
(v) 2-bromo-1,1-bis(4-chlorophenyl)propane
(vi) 4-(4-iodophenyl)-3-chloro-2,2-dimethylhex-3-ene
Write the structures of the following organic halogen compounds.
(i) 2-Chloro-3-methylpentane
(ii) p-Bromochlorobenzene
(iii) 1-Chloro-4-ethylcyclohexane
(iv) 2-(2-Chlorophenyl)-1-iodooctane
(v) 2-Bromobutane
(vi) 4-tert-Butyl-3-iodoheptane
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
(viii) 1,4-Dibromobut-2-ene
Here are the structures of the requested organic halogen compounds:
(i) 2-Chloro-3-methylpentane
CH3-CHCl-CH(CH3)-CH2-CH3
(ii) p-Bromochlorobenzene
Br
|
C6H4-Cl
(iii) 1-Chloro-4-ethylcyclohexane
Cl
|
C6H11 - CH2-CH3
(iv) 2-(2-Chlorophenyl)-1-iodooctane
I
|
CH3-(CH2)5-CH-CH-Ar-Cl
(2-chlorophenyl)
(v) 2-Bromobutane
CH3-CHBr-CH2-CH3
(vi) 4-tert-Butyl-3-iodoheptane
I
|
CH3-CH2-CH(CH3)-C(CH3)3-CH2-CH2-CH3
(vii) 1-Bromo-4-sec-butyl-2-methylbenzene
Br
|
C6H4-CH-CH3-CH3
| (sec-butyl) (2-methyl)
CH3
(viii) 1,4-Dibromobut-2-ene
Br-CH2-CH=CH-CH2-Br
These structures correspond to the given names of organic halogen compounds.
Which one of the following has the highest dipole moment?
(i) CH2Cl2
(ii) CHCl3
(iii) CCl4
To determine which molecule has the highest dipole moment, we need to consider the molecular geometry and the electronegativity differences in each compound.
Here’s a comparative analysis of the given compounds:
CH$_2$Cl$_2$:
Molecular geometry: Tetrahedral.
Chlorine atoms are placed in such a way that their dipole moments do not cancel out completely.
CHCl$_3$:
Molecular geometry: Tetrahedral.
Three chlorines and one hydrogen create some dipole cancellation, but not completely.
CCl$_4$:
Molecular geometry: Tetrahedral.
Symmetrical arrangement of chlorine atoms leads to complete dipole cancellation.
Given these points, we can conclude that CCl$_4$ has no net dipole moment (because it is symmetrical). Now, between CH$_2$Cl$_2$ and CHCl$_3:
CH$_2$Cl$_2$ has two hydrogen atoms that contribute less to the cancellation of dipole moments compared to CHCl$_3, which has three chlorines contributing to cancellation.
Thus, CH$_2$Cl$_2$ has the highest dipole moment among the given compounds.
Answer: (i) CH$_2$Cl$_2$
A hydrocarbon C5H10 does not react with chlorine in dark but gives a single monochloro compound C5H9Cl in bright sunlight. Identify the hydrocarbon.
Given that the hydrocarbon has the formula $\mathrm{C}_5\mathrm{H}_{10}$ and does not react with chlorine in the dark but forms a single monochloro compound $\mathrm{C}_5\mathrm{H}_9\mathrm{Cl}$ in bright sunlight, we can infer the following:
The hydrocarbon has the formula $\mathrm{C}_5\mathrm{H}_{10}$, which corresponds to an alkene (CnH2n) or a cycloalkane (CnH2n).
The fact that it does not react with chlorine in the dark indicates it is likely not an alkene because alkenes typically react with chlorine even in the absence of light.
The formation of a single monochloro compound under bright sunlight suggests a symmetrical cycloalkane structure, where substitution can only occur at equivalent positions.
By considering the possible cycloalkane structures, cyclopentane ($\mathrm{C}_5\mathrm{H}_{10}$) fits the criteria.
Reasoning:
Cyclopentane is a symmetrical molecule with a ring structure.
It does not react with chlorine in dark conditions because there are no double bonds.
Under bright sunlight, free radical halogenation occurs, resulting in a single monochloro compound since all hydrogen atoms in cyclopentane are equivalent.
Therefore, the hydrocarbon is cyclopentane.
Write the isomers of the compound having formula C4H9Br.
The compound with the formula $$\mathrm{C}_4\mathrm{H}_9\mathrm{Br}$$ has four isomers. Below are the structures and names of these isomers, classified as primary, secondary, and tertiary bromides:
1-Bromobutane (Primary Bromide)
[ \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2\mathrm{Br} ]
2-Bromobutane (Secondary Bromide)
[ \mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{Br})\mathrm{CH}_3 ]
1-Bromo-2-methylpropane (Primary Bromide)
[ \mathrm{CH}_3\mathrm{CH}(\mathrm{CH}_3)\mathrm{CH}_2\mathrm{Br} ]
2-Bromo-2-methylpropane (Tertiary Bromide)
[ \mathrm{C}(\mathrm{Br})(\mathrm{CH}_3)_3 ]
Here are the structures:
1-Bromobutane: The bromine atom is attached to the end carbon.
2-Bromobutane: The bromine atom is attached to the second carbon in the chain.
1-Bromo-2-methylpropane: The bromine atom is attached to carbon adjacent to a methyl-substituted carbon.
2-Bromo-2-methylpropane: The bromine atom is attached to the central carbon atom, surrounded by three methyl groups.
These different isomers highlight how the position and environment of the bromine atom can significantly change the structure of the molecule while maintaining the same molecular formula.
Write the equations for the preparation of 1-iodobutane from
(i) 1-butanol
(ii) 1-chlorobutane
(iii) but-1-ene.
Here are the equations for the preparation of 1-iodobutane from different starting materials:
(i) From 1-butanol
1-Butanol reacts with sodium or potassium iodide in the presence of phosphoric acid: $$ \text{CH}_3(\text{CH}_2)_3\text{OH} + \text{HI} \rightarrow \text{CH}_3(\text{CH}_2)_3\text{I} + \text{H}_2\text{O} $$
(ii) From 1-chlorobutane
1-Chlorobutane reacts with sodium iodide in dry acetone (Finkelstein reaction): $$ \text{CH}_3(\text{CH}_2)_3\text{Cl} + \text{NaI} \rightarrow \text{CH}_3(\text{CH}_2)_3\text{I} + \text{NaCl} $$
(iii) From but-1-ene
But-1-ene reacts with hydrogen iodide (HI): $$ \text{CH}_2=\text{CH}(\text{CH}_2)_2 + \text{HI} \rightarrow \text{CH}_3(\text{CH}_2)_3\text{I} $$
In each case, the reactions yield 1-iodobutane as the final product.
What are ambident nucleophiles? Explain with an example
Ambident nucleophiles are nucleophiles that have two different atoms through which they can donate an electron pair. This means they can attach to electrophiles in two distinct ways, depending on which atom donates the electron pair.
Example: Cyanide Ion (CN⁻)
The cyanide ion is a classic example of an ambident nucleophile. It has two nucleophilic centers - the carbon atom and the nitrogen atom.
When cyanide ion (CN⁻) reacts with alkyl halides, it can form either:
Alkyl cyanides (R-CN) through nucleophilic attack via the carbon atom, or
Alkyl isocyanides (R-NC) through nucleophilic attack via the nitrogen atom.
The type of product formed depends on the nature of the cyanide source and other reaction conditions.
Reactions:
KCN (Potassium Cyanide) Reaction:Potassium cyanide is predominantly ionic. Therefore, in the reaction with alkyl halides, the cyanide ion (CN⁻) typically attacks through the carbon atom, forming an alkyl cyanide (R-CN).
$$ \text{R-Br} + \text{KCN} \rightarrow \text{R-CN} + \text{KBr} $$
AgCN (Silver Cyanide) Reaction:Silver cyanide is covalent in nature. Thus, the nucleophilic attack typically occurs via the nitrogen atom, resulting in the formation of an alkyl isocyanide (R-NC).
$$ \text{R-Br} + \text{AgCN} \rightarrow \text{R-NC} + \text{AgBr} $$
These reactions exemplify the ambident nature of the cyanide ion, demonstrating its ability to react through either the carbon or nitrogen atom.
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Ask Chatterbot AINotes - Haloalkanes and Haloarenes | Class 12 NCERT | Chemistry
Essential Guide to Haloalkanes and Haloarenes: Class 12 Chemistry Notes
Introduction to Haloalkanes and Haloarenes
Haloalkanes and haloarenes, also known as alkyl halides and aryl halides, respectively, are fundamental organic compounds where halogens (fluorine, chlorine, bromine, or iodine) replace one or more hydrogen atoms in hydrocarbons. These compounds are notable for their diverse applications in pharmaceuticals, agriculture, and industry.
Classification of Haloalkanes and Haloarenes
Based on Number of Halogen Atoms
Haloalkanes and haloarenes are classified as mono-, di-, or polyhalogen compounds depending on the number of halogen atoms present:
Structural Classification
Alkyl Halides or Haloalkanes (R-X)
In alkyl halides, the halogen atom is bonded to an alkyl group (R). These are further subdivided into primary, secondary, or tertiary haloalkanes based on the nature of the carbon atom to which the halogen is attached.
Primary Alkyl Halide: Halogen attached to a primary carbon.
Secondary Alkyl Halide: Halogen attached to a secondary carbon.
Tertiary Alkyl Halide: Halogen attached to a tertiary carbon.
Allylic Halides
The halogen is bonded to an ( sp^3 )-hybridised carbon atom adjacent to a carbon-carbon double bond ( (C=C) ).
Benzylic Halides
The halogen is bonded to an ( sp^3 )-hybridised carbon atom attached to an aromatic ring.
Compounds Containing ( sp^2 )C-X Bond
Vinylic Halides
The halogen is attached to an ( sp^2 )-hybridised carbon atom of a carbon-carbon double bond ( (C=C) ).
Aryl Halides
In these compounds, the halogen is directly bonded to the ( sp^2 )-hybridised carbon atom of an aromatic ring.
Nomenclature of Haloalkanes and Haloarenes
Common Names
Common names are derived by naming the alkyl group followed by the type of halide (e.g., methyl chloride).
IUPAC Naming System
In the IUPAC system, haloalkanes are named as halosubstituted hydrocarbons (e.g., chloroethane). For mono halogen substituted derivatives of benzene, common and IUPAC names are the same (e.g., chlorobenzene).
Methods of Preparation
From Alcohols
Alcohols react with halogen acids (like HCl, HBr), phosphorus halides, or thionyl chloride to produce haloalkanes.
From Alkanes (Free Radical Halogenation)
Chlorine or bromine reacts with alkanes under UV light or heat to form haloalkanes.
From Alkenes (Addition Reactions)
Alkenes undergo addition reactions with hydrogen halides or halogens to form haloalkanes in the presence of catalysts.
Halogen Exchange (Finkelstein and Swarts Reaction)
Alkyl chlorides/bromides react with NaI in dry acetone (Finkelstein reaction) or alkyl chlorides/bromides with metallic fluorides (Swarts reaction) to form alkyl iodides or fluorides.
Preparation of Haloarenes
Haloarenes are prepared from hydrocarbons by electrophilic substitution or from aromatic amines via the Sandmeyer reaction.
Physical Properties of Haloalkanes and Haloarenes
Melting and Boiling Points
Haloalkanes generally have higher boiling points compared to their parent hydrocarbons due to increased intermolecular forces. Boiling points increase with the size and number of halogen atoms.
Density
Haloalkanes and polyhalogen derivatives are denser than water. Density increases with the number of carbon and halogen atoms.
Solubility
Haloalkanes are slightly soluble in water due to their inability to form hydrogen bonds but readily dissolve in organic solvents.
Chemical Properties and Reactions
Nature of C-X Bond
The C-X bond in haloalkanes is polar due to the higher electronegativity of the halogen compared to carbon. Bond length and bond enthalpy vary with the type of halogen.
Reactions of Haloalkanes
Nucleophilic Substitution Reactions
Haloalkanes undergo nucleophilic substitution where a nucleophile replaces the halogen atom. The reaction mechanism may occur via ( S_N2 ) or ( S_N1 ) pathways depending on the structure of the haloalkane.
Elimination Reactions
Haloalkanes can undergo elimination reactions to form alkenes when heated with alcoholic KOH.
Reaction with Metals (Grignard Reagent)
Haloalkanes react with magnesium in dry ether to form Grignard reagents, which are useful in various organic synthesis processes.
Reactions of Haloarenes
Nucleophilic Substitution
Less reactive due to resonance stabilisation and the shorter bond length of the C-X bond.
Electrophilic Substitution
Haloarenes undergo electrophilic substitution with the halogen acting as an ortho/para-directing group.
graph LR;
D[C6H5X] -- E+ --> F[Intermediate] --> G[C6H5E]
Polyhalogen Compounds
Dichloromethane (Methylene Chloride)
Used as a solvent and paint remover but can impair hearing and vision.
Trichloromethane (Chloroform)
A solvent for fats and alkaloids, and historically used as an anaesthetic.
Triiodomethane (Iodoform)
Used as an antiseptic but replaced due to its strong odour.
Tetrachloromethane (Carbon Tetrachloride)
Used in refrigerants and as a cleaning agent but has significant health risks.
Freons
Stable, non-toxic compounds used in refrigeration and air conditioning. Known to deplete the ozone layer.
DDT
An insecticide with significant environmental impact, including bioaccumulation and toxicity to wildlife.
Uses and Applications
In Industry
Haloalkanes and haloarenes are used in solvents, refrigerants, fire extinguishers, and as intermediates in chemical synthesis.
In Medicine
Compounds like chloramphenicol and chloroquine are used for treating diseases like typhoid and malaria.
Environmental Impact
Improper use and disposal can lead to significant environmental and health hazards, including ozone depletion and bioaccumulation in the food chain.
Frequently Asked Questions (FAQs)
Differences between Haloalkanes and Haloarenes
Haloalkanes have halogens attached to sp³ hybridised carbons, while haloarenes have halogens attached to sp² hybridised carbons.
Common Reactions and Their Mechanisms
Haloalkanes typically undergo nucleophilic substitution and elimination reactions, while haloarenes are more prone to electrophilic substitution due to resonance stabilisation.
This comprehensive guide covers essential aspects of haloalkanes and haloarenes for Class 12 chemistry, aiding students in understanding their properties, preparation, and reactions effectively.
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