Properties of Haloalkanes Questions in English

(A) Ethylene difluoride $(F-CH_2-CH_2-F)$ undergoes hydrolysis to replace the fluorine atoms with hydroxyl groups,forming ethylene glycol $(HO-CH_2-CH_2-OH)$.
Reaction: $F-CH_2-CH_2-F + 2H_2O \rightarrow HO-CH_2-CH_2-OH + 2HF$.

(B) When chloroform $(CHCl_3)$ is heated with aqueous sodium hydroxide $(NaOH)$,it undergoes hydrolysis to form an unstable intermediate,methanetriol $(CH(OH)_3)$.
This intermediate loses a water molecule to form formic acid $(HCOOH)$.
Since the reaction occurs in the presence of excess $NaOH$,the formic acid is neutralized to form sodium formate $(HCOONa)$.
The reaction sequence is:
$CHCl_3 + 3NaOH \rightarrow CH(OH)_3 + 3NaCl$
$CH(OH)_3 \rightarrow HCOOH + H_2O$
$HCOOH + NaOH \rightarrow HCOONa + H_2O$
Overall reaction: $CHCl_3 + 4NaOH \rightarrow HCOONa + 3NaCl + 2H_2O$
Thus,the final product is $HCOONa$.

(A) The reaction of ethyl bromide $(C_2H_5Br)$ with lead-sodium alloy $(Pb/Na)$ is a classic method for the preparation of tetraethyl lead $(TEL)$.
The chemical equation is:
$4C_2H_5Br + 4Pb/Na \to (C_2H_5)_4Pb + 4NaBr + 3Pb$
Thus,the product formed is tetraethyl lead.

(A) The reaction is: $C_2H_5Br + AgNO_2 \rightarrow C_2H_5NO_2 + AgBr$.
$AgNO_2$ is a covalent compound where the $Ag-O$ bond is present.
However,due to the nature of the nucleophilic substitution,the attack occurs through the nitrogen atom because the $N$ atom is more nucleophilic than the $O$ atom in this specific covalent environment.
Thus,the major product formed is nitroethane $(C_2H_5NO_2)$.

(A) Chloretone is formed by the nucleophilic addition reaction of chloroform $(CHCl_3)$ with acetone $(CH_3COCH_3)$ in the presence of a base like $KOH$ or $NaOH$.
The reaction is as follows:
$CHCl_3 + CH_3COCH_3 \xrightarrow{NaOH} Cl_3C-C(OH)(CH_3)_2$
This product is known as chloretone ($1$,$1$,$1$-trichloro$-2-$methylpropan$-2-$ol).

(A) The reaction of $1$-bromopropane $(CH_3 - CH_2 - CH_2Br)$ with alcoholic $KOH$ is a dehydrohalogenation reaction.
This is an elimination reaction ($E2$ mechanism) where a hydrogen atom is removed from the $\beta$-carbon and the bromine atom is removed from the $\alpha$-carbon.
This results in the formation of a double bond between the $\alpha$ and $\beta$ carbons.
The reaction is: $CH_3 - CH_2 - CH_2Br + KOH \text{ (alc.)} \to CH_3 - CH = CH_2 + KBr + H_2O$.
The product formed is propene $(CH_3 - CH = CH_2)$.

(A) The reaction of $1-$chlorobutane with alcoholic $KOH$ is a dehydrohalogenation reaction (an elimination reaction).
In this reaction,a hydrogen atom is removed from the $\beta$-carbon and the chlorine atom is removed from the $\alpha$-carbon,resulting in the formation of an alkene.
$CH_3CH_2CH_2CH_2-Cl + KOH(\text{alc.}) \to CH_3CH_2CH=CH_2 + KCl + H_2O$.
The product formed is $1-$butene.

(A) The reaction of ethyl chloride with silver cyanide $(AgCN)$ is as follows:
$C_2H_5Cl + AgCN \to C_2H_5NC (X) + AgCl$.
Here,$X$ is ethyl isocyanide $(C_2H_5NC)$.
The functional isomer of an isocyanide $(R-NC)$ is a cyanide $(R-CN)$.
Therefore,the functional isomer of ethyl isocyanide $(C_2H_5NC)$ is ethyl cyanide $(C_2H_5CN)$.

(B) $CHCl_3 + \frac{1}{2}O_2 \xrightarrow{\text{Sunlight}} COCl_2 + HCl$
$COCl_2$ is known as Phosgene or carbonyl chloride.

(B) The formation of a white precipitate with silver nitrate $(AgNO_3)$ after treatment with $NaOH$ indicates the presence of chloride ions $(Cl^-)$ in the solution.
$C_6H_5CH_2Cl$ is a primary benzylic halide. It undergoes nucleophilic substitution with aqueous $NaOH$ to form benzyl alcohol and $NaCl$.
$C_6H_5CH_2Cl + NaOH_{(aq)} \rightarrow C_6H_5CH_2OH + NaCl$
When $dil. HNO_3$ and $AgNO_3$ are added to the resulting solution,the $Cl^-$ ions react with $Ag^+$ to form a white precipitate of $AgCl$.
$Ag^+ + Cl^- \rightarrow AgCl_{(s)} \text{ (white ppt)}$
$C_6H_5Cl$ (chlorobenzene) does not undergo nucleophilic substitution under these mild conditions,and $C_6H_4(CH_3)Br$ would yield a yellow precipitate of $AgBr$.

(B) Dehydrohalogenation is the elimination of a hydrogen atom and a halogen atom from adjacent carbon atoms to form an alkene.
This reaction requires a haloalkane (alkyl halide) with at least one $\beta$-hydrogen atom.
$a)$ $Iso-propyl$ bromide $(CH_3CH(Br)CH_3)$ has $\beta$-hydrogens and can undergo dehydrohalogenation.
$b)$ Ethanol $(CH_3CH_2OH)$ is an alcohol,not a haloalkane,and therefore cannot undergo dehydrohalogenation.
$c)$ Ethyl bromide $(CH_3CH_2Br)$ has $\beta$-hydrogens and can undergo dehydrohalogenation.
Thus,the correct answer is $B$.

(D) The reaction of $C_2H_5Cl$ with $KCN$ is a nucleophilic substitution reaction where the cyanide ion $(CN^-)$ replaces the chloride ion $(Cl^-)$ to form propanenitrile $(C_2H_5CN)$,which is $X$.
$C_2H_5Cl + KCN \to C_2H_5CN + KCl$
Subsequent hydrolysis of the nitrile $(C_2H_5CN)$ yields a carboxylic acid $(C_2H_5COOH)$,which is $Y$.
$C_2H_5CN + 2H_2O \xrightarrow{H^+} C_2H_5COOH + NH_3$
Therefore,$X$ is $C_2H_5CN$ and $Y$ is $C_2H_5COOH$.

(B) Dehydrohalogenation is the elimination of a hydrogen atom and a halogen atom from adjacent carbon atoms in a haloalkane,resulting in the formation of an alkene.
For example,the reaction of chloroethane with alcoholic $KOH$ is:
$CH_3-CH_2Cl \xrightarrow{\text{alc. } KOH} CH_2=CH_2 + HCl$
Thus,it produces a double bond.

(B) The reaction of chloroform $(CHCl_3)$ with concentrated nitric acid $(HNO_3)$ is a nitration reaction.
$CHCl_3 + HNO_3 \rightarrow CCl_3NO_2 + H_2O$
The product formed is $CCl_3NO_2$,which is commonly known as Chloropicrin (trichloronitromethane).

(C) The correct answer is $(C)$.
Chloroform $(CHCl_3)$ is tested for purity before use as an anaesthetic by adding aqueous silver nitrate $(AgNO_3)$ solution.
Pure chloroform does not react with aqueous $AgNO_3$ and thus does not form a white precipitate of silver chloride $(AgCl)$.
If the chloroform has been oxidized to phosgene $(COCl_2)$ or contains impurities like $HCl$,it will react with $AgNO_3$ to form a white precipitate of $AgCl$.

(C) Dehydrohalogenation of an alkyl halide is an elimination reaction.
In this reaction,a hydrogen atom is removed from the $\beta$-carbon and a halogen atom is removed from the $\alpha$-carbon in the presence of an alcoholic base like $KOH$,resulting in the formation of an alkene.
$CH_3CH_2Br + \text{alc. } KOH \to CH_2 = CH_2 + KBr + H_2O$
There are two main types of elimination reactions:
$(i)$ $E_1$: Unimolecular elimination
$(ii)$ $E_2$: Bimolecular elimination

(C) $CH_3-CH(Br)-CH_2-CH_2-CH_3 + C_2H_5OK \xrightarrow{C_2H_5OH} CH_3-CH=CH-CH_2-CH_3$ (Pent$-2-$ene).
When an alkyl halide reacts with an alcoholic base like potassium ethoxide,it undergoes a dehydrohalogenation reaction (elimination reaction) following Zaitsev's rule.
The reaction produces a mixture of pent$-1-$ene and pent$-2-$ene. Pent$-2-$ene is the more substituted alkene and is thus the major product.
Between the geometric isomers of pent$-2-$ene,the $trans$ isomer is more stable than the $cis$ isomer due to reduced steric hindrance between the alkyl groups. Therefore,$trans$ pent$-2-$ene is the major product.

(C) Chloroform $(CHCl_3)$ is slowly oxidized by air in the presence of light to form an extremely poisonous gas,carbonyl chloride,also known as phosgene $(COCl_2)$.
The balanced chemical equation is:
$2CHCl_3 + O_2 \xrightarrow{hv} 2COCl_2 + 2HCl$
Note: The option $C$ provided in the original set is corrected to represent the balanced reaction products $COCl_2$ and $HCl$.

(A) Haloalkanes in the presence of alcoholic $KOH$ undergo an elimination reaction,specifically known as a dehydrohalogenation reaction.
In this reaction,a hydrogen atom is removed from the $\beta$-carbon and a halogen atom is removed from the $\alpha$-carbon,resulting in the formation of an alkene.
Example: $CH_3CH_2CH_2Br + alc. KOH \rightarrow CH_3CH = CH_2 + KBr + H_2O$

(B) Alkyl halide reacts with $Mg$ in the presence of dry ether to give alkyl magnesium halide,which is also called a Grignard reagent.
This reaction is known as the formation of a Grignard reagent.
The general reaction is: $R-X + Mg \xrightarrow{\text{dry ether}} R-Mg-X$

(B) The reaction of a geminal dihalide like $CH_3CH_2CHCl_2$ with a strong base like $NaNH_2$ leads to dehydrohalogenation.
First,one molecule of $HCl$ is eliminated to form a vinyl halide $(CH_3CH=CHCl)$.
Then,a second molecule of $HCl$ is eliminated to form an alkyne.
Thus,$CH_3CH_2CHCl_2 + 2NaNH_2 \rightarrow CH_3-C \equiv CH + 2NaCl + 2NH_3$.
The final product is propyne $(CH_3-C \equiv CH)$.

(D) $C_2H_5Br$ reacts with metallic $Na$ (Wurtz reaction) to give $n$-butane,not ethane.
$C_2H_5Br + AgNO_2 (\text{alc.}) \to C_2H_5NO_2 + AgBr$ (Nitroethane).
$C_2H_5Br + CH_3COOAg \to CH_3COOC_2H_5 + AgBr$ (Ethyl acetate).
Therefore,both $(b)$ and $(c)$ are correct statements.

(C) $Methyl$ chloride $(CH_3Cl)$ reacts with silver acetate $(CH_3COOAg)$ to undergo a nucleophilic substitution reaction,producing $Methyl$ acetate $(CH_3COOCH_3)$ and silver chloride $(AgCl)$.
The chemical reaction is: $CH_3Cl + CH_3COOAg \rightarrow CH_3COOCH_3 + AgCl$

(A) The correct option is $A$.
Chloroform $(CHCl_3)$ is oxidized by air in the presence of light to form a highly poisonous gas called phosgene $(COCl_2)$.
$2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$.
To test for the presence of phosgene,the sample is treated with silver nitrate $(AgNO_3)$ solution.
If phosgene is present,it reacts with $AgNO_3$ to form a white precipitate of silver chloride $(AgCl)$.

(C) Tertiary alkyl halides undergo alkaline hydrolysis via the $S_N1$ mechanism.
The rate law for an $S_N1$ reaction is given by: $\text{Rate} = k[R-X]$.
Since the rate depends only on the concentration of the alkyl halide and is independent of the concentration of the nucleophile (alkali),doubling the concentration of the alkali will have no effect on the reaction rate.
Therefore,the rate will remain constant.

(A) $CHCl_3$ (chloroform) is a covalent compound and does not ionize to provide $Cl^-$ ions in an aqueous solution.
Since $AgNO_3$ requires free $Cl^-$ ions to form a white precipitate of $AgCl$,no reaction occurs with $CHCl_3$.

(B) Chloroform $(CHCl_3)$ is slowly oxidized by atmospheric oxygen in the presence of light to form a highly poisonous gas known as phosgene $(COCl_2)$.
The chemical reaction is:
$2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$

(C) Alcoholic potassium hydroxide $(KOH)$ is a strong base used to perform the elimination reaction known as dehydrohalogenation on haloalkanes,resulting in the formation of alkenes.

(A) The reaction of vinyl chloride $(CH_2=CH-Cl)$ with $HCl$ follows Markovnikov's rule.
According to Markovnikov's rule,the negative part of the addendum $(Cl^-)$ attaches to the carbon atom with fewer hydrogen atoms.
$CH_2=CH-Cl + HCl \to CH_3-CHCl_2$.
The product formed is $1,1-$dichloroethane.

(A) In the given reaction,the hydroxide ion $(OH^-)$ acts as a nucleophile.
It attacks the electrophilic carbon atom bonded to the halogen $(X)$ and replaces the halide ion $(X^-)$.
Since a nucleophile replaces another nucleophile (the leaving group),this reaction is classified as a nucleophilic substitution reaction.

(C) The reaction of iodoform $(CHI_3)$ with aqueous potassium hydroxide $(KOH)$ is a hydrolysis reaction.
The balanced chemical equation is:
$CHI_3 + 4KOH_{(aq)} \to HCOOK + 3KI + 2H_2O$
Thus,the product formed is potassium formate $(HCOOK)$.

(A) methylating agent is a compound that transfers a methyl group $(-CH_3)$ to another molecule.
$CH_3I$ (methyl iodide) is an excellent methylating agent because the iodide ion $(I^-)$ is a very good leaving group,making the carbon atom highly electrophilic and susceptible to nucleophilic attack.

(C) The reaction of ethyl iodide $(C_2H_5I)$ with silver nitrate $(AgNO_3)$ is a nucleophilic substitution reaction.
$C_2H_5I + AgNO_3 \to C_2H_5ONO_2 + AgI$
The product obtained is ethyl nitrate $(C_2H_5ONO_2)$.

(D) When chloroform $(CHCl_3)$ is treated with zinc dust in the presence of water,it undergoes reduction to form dichloromethane $(CH_2Cl_2)$.
The chemical reaction is as follows:
$CHCl_3 + Zn + H_2O \rightarrow CH_2Cl_2 + Zn(OH)Cl$

(D) The density of alkyl halides increases as the atomic mass and size of the halogen atom increase.
Since the atomic mass follows the order $Cl < Br < I$,the density of the corresponding alkyl halides follows the order $RCl < RBr < RI$.

(A) The correct answer is $A$.
In $CH_2=CH-Cl$ (vinyl chloride),the lone pair of electrons on the chlorine atom participates in resonance with the $C=C$ double bond.
This results in a partial double bond character between the carbon and chlorine atoms.
Due to this partial double bond character,the $C-Cl$ bond becomes stronger and shorter,making it difficult to break.
Therefore,vinyl chloride is the least reactive towards nucleophilic substitution (hydrolysis) compared to the other options.

(B) Nucleophiles are electron-rich species that donate an electron pair to an electrophilic carbon atom. According to the $Br\o nsted-Lowry$ and $Lewis$ concepts,nucleophiles act as bases because they donate a lone pair of electrons.
In the reaction $R-X + OH^{-} \to R-OH + X^{-}$,the hydroxide ion $(OH^{-})$ acts as a nucleophile and also as a base.

(B) $1$. Toluene $(C_6H_5CH_3)$ reacts with excess $Cl_2$ in the presence of sunlight (free radical substitution) to form benzotrichloride $(C_6H_5CCl_3)$.
$2$. Benzotrichloride on hydrolysis gives benzoic acid $(C_6H_5COOH)$.
$3$. Benzoic acid reacts with $NaOH$ (an acid-base reaction) to form sodium benzoate $(C_6H_5COONa)$.
Therefore,the final product is sodium benzoate.

(A) . $C_3H_7Br + KCN \to C_3H_7CN + KBr$
In the $IUPAC$ system,the carbon atom of the functional group $(-CN)$ is included in the parent chain numbering.
$C_3H_7CN$ contains $3$ carbons from the propyl group and $1$ carbon from the nitrile group,totaling $4$ carbons,which corresponds to $butanenitrile$.

(A) Chloroform $(CHCl_3)$ is slowly oxidized by air in the presence of light to form an extremely poisonous gas,carbonyl chloride $(COCl_2)$,also known as phosgene.
$2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$
To prevent this oxidation,$1\%$ of ethanol $(C_2H_5OH)$ is added to chloroform.
Ethanol reacts with any phosgene formed to convert it into non-poisonous diethyl carbonate.
$COCl_2 + 2C_2H_5OH \rightarrow (C_2H_5O)_2CO + 2HCl$
Therefore,the correct option is $A$.

(C) The reaction of benzyl bromides with ethanol proceeds via an $S_N1$ mechanism,which involves the formation of a carbocation intermediate.
The stability of the carbocation determines the reactivity. Electron-donating groups $(EDG)$ stabilize the carbocation,while electron-withdrawing groups $(EWG)$ destabilize it.
$1$. $p-$methoxybenzyl bromide: The methoxy group $(-OCH_3)$ is a strong electron-donating group by resonance ($+M$ effect),which significantly stabilizes the carbocation.
$2$. $p-$methylbenzyl bromide: The methyl group $(-CH_3)$ is a weak electron-donating group by hyperconjugation ($+H$ effect).
$3$. $p-$chlorobenzyl bromide: The chloro group $(-Cl)$ is electron-withdrawing by induction ($-I$ effect).
$4$. $p-$nitrobenzyl bromide: The nitro group $(-NO_2)$ is a strong electron-withdrawing group by both resonance $(-M)$ and induction $(-I)$ effects.
Therefore,$p-$methoxybenzyl bromide forms the most stable carbocation and reacts most readily with ethanol.

(B) When chloroform is treated with concentrated nitric acid,its hydrogen is replaced by a nitro group.
$CHCl_{3} + HNO_{3} \to CCl_{3}NO_{2} + H_{2}O$
(Chloropicrin)

(B) Chloroform is oxidized to a poisonous gas,phosgene $(COCl_2)$,by atmospheric oxygen.
$2CHCl_3 + O_2 \rightarrow 2COCl_2 + 2HCl$

(B) . When chloroform reacts with concentrated $HNO_{3}$,it forms chloropicrin,which is known as tear gas,along with water.
The chemical reaction is:
$CHCl_{3} + HNO_{3} \to CCl_{3}NO_{2} + H_{2}O$

(A) The reaction between ethyl chloride $(CH_3CH_2Cl)$ and alcoholic $KOH$ is a dehydrohalogenation reaction.
In this reaction,an elimination process occurs where a hydrogen atom and a chlorine atom are removed from adjacent carbon atoms to form an alkene.
The chemical equation for the reaction is:
$CH_3CH_2Cl + KOH (\text{alc.}) \rightarrow CH_2=CH_2 + KCl + H_2O$.
Thus,the product obtained is ethene $(C_2H_4)$.

(D) Iodoform $(CHI_3)$ exists as a yellow crystalline solid at room temperature.

(A) The correct answer is $(A)$. Benzyl chloride $(C_6H_5CH_2Cl)$ is significantly more reactive than simple alkyl halides towards nucleophilic substitution reactions ($S_N1$ mechanism). This is because the resulting benzyl carbocation $(C_6H_5CH_2^+)$ is highly stabilized by resonance with the benzene ring. Therefore,the statement that it is less reactive than alkyl halides is incorrect.

(B) The reaction of ethyl chloride $(C_2H_5Cl)$ with aqueous $KOH$ is a nucleophilic substitution reaction ($S_N2$ mechanism).
In this reaction,the hydroxide ion $(OH^-)$ acts as a nucleophile and replaces the chloride ion $(Cl^-)$ to form ethyl alcohol $(C_2H_5OH)$.
The chemical equation is: $C_2H_5Cl + KOH_{(aq)} \to C_2H_5OH + KCl$.

(C) Chloroform $(CHCl_3)$ is slowly oxidized by atmospheric oxygen in the presence of light to form an extremely poisonous gas known as phosgene $(COCl_2)$.
The chemical reaction is: $2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$.
To prevent this oxidation,it is stored in closed,dark-coloured bottles that are completely filled to exclude air and light.