Properties of Haloalkanes Questions in English

(A) $C_6H_5CH_2I$ is an alkyl halide where the iodine is attached to an $sp^3$ hybridized carbon,making it reactive towards nucleophilic substitution with $NaOH$ to produce $I^-$ ions.
$C_6H_5I$ is an aryl halide where the iodine is attached to an $sp^2$ hybridized carbon,making it inert towards nucleophilic substitution under these conditions.
When $AgNO_3$ is added to the acidified solution,$I^-$ ions react to form a yellow precipitate of $AgI$.
Since $B$ gave a yellow precipitate,$B$ must be $C_6H_5CH_2I$.
Therefore,$A$ must be $C_6H_5I$.

(D) Benzyl chloride $(C_6H_5CH_2Cl)$ is a primary benzylic halide. The carbocation formed during the reaction is resonance-stabilized,making it highly reactive towards nucleophilic substitution reactions ($S_N1$ or $S_N2$). It readily reacts with alcoholic $AgNO_3$ to form a white precipitate of $AgCl$.
Vinyl chloride $(CH_2=CHCl)$ is much less reactive towards nucleophilic substitution because the $C-Cl$ bond acquires partial double bond character due to resonance,as shown in the image. Therefore,the statement that benzyl chloride is less reactive than vinyl chloride is incorrect.

(B) Alkyl halides are converted into alkenes primarily through an elimination reaction,specifically dehydrohalogenation,in the presence of a strong base.
$R-CH_2-CH_2-X \xrightarrow{\text{base}} R-CH=CH_2 + HX$

(B) The reaction of $CH_3CH_2CH(F)CH_3$ with a strong base like $CH_3O^-$ in $CH_3OH$ proceeds via an $E2$ elimination mechanism.
Two possible alkenes can be formed: $CH_3CH=CHCH_3$ (but-$2$-ene) and $CH_3CH_2CH=CH_2$ (but-$1$-ene).
According to Saytzeff's rule,the more substituted alkene is the major product because it is more stable.
$CH_3CH=CHCH_3$ has two alkyl substituents on the double-bonded carbons,while $CH_3CH_2CH=CH_2$ has only one.
Therefore,$CH_3CH=CHCH_3$ is the major product.

(C) The reaction of alkyl halides $(R'X)$ with Gilman reagents $(R_2CuLi)$ is known as the Corey-House synthesis.
This reaction is used to form higher alkanes by coupling two alkyl groups.
The general reaction is: $R_2CuLi + R'X \rightarrow R-R' + RCu + LiX$.

(A) The boiling points of haloalkanes increase with an increase in the size and mass of the halogen atom due to the increase in magnitude of van der Waals forces.
$CH_3F$ (boiling point: $-78.4 \ ^oC$),$CH_3Cl$ (boiling point: $-24.2 \ ^oC$),$CH_3Br$ (boiling point: $3.6 \ ^oC$),and $C_2H_5Cl$ (boiling point: $12.3 \ ^oC$) are gases at room temperature.
$CH_3I$ has a boiling point of $42.4 \ ^oC$,which makes it a liquid at room temperature.
Therefore,the correct option is $(A)$.

(D) The reactivity of haloalkanes towards nucleophilic substitution depends on the nature of the halogen atom and the structure of the alkyl group.
$1$. Nature of the halogen: The bond strength decreases as the size of the halogen increases,making the $C-X$ bond weaker. Thus,the order of reactivity is $I > Br > Cl > F$.
$2$. Structure of the alkyl group: For $S_N1$ reactions,the order is tertiary $ > $ secondary $ > $ primary. For $S_N2$ reactions,the order is primary $ > $ secondary $ > $ tertiary.
Comparing the given options:
- $1-$chloropropane (primary chloride)
- $1-$bromopropane (primary bromide)
- $2-$chloropropane (secondary chloride)
- $2-$bromopropane (secondary bromide)
Considering the leaving group ability,bromides are more reactive than chlorides. Between $1-$bromopropane and $2-$bromopropane,$2-$bromopropane is more reactive in $S_N1$ conditions,and generally,the presence of a better leaving group $(Br)$ makes it more reactive than the corresponding chlorides. Therefore,$2-$bromopropane is the most reactive among the choices provided.

(A) The reaction of styrene with $HBr$ in the absence of peroxides (or light) follows Markovnikov's rule,which is an electrophilic addition reaction,yielding $1$-bromo-$1$-phenylethane $(A)$.
In the presence of light $(h\nu)$,the reaction follows the anti-Markovnikov addition (peroxide effect/free radical mechanism),yielding $1$-bromo-$2$-phenylethane $(B)$.
Thus,the formation of $A$ is an electrophilic addition reaction,and the formation of $B$ is a free radical addition reaction.

(A) The compound $CH_3-C(OH)(CH_3)-CCl_3$ is known as Chloretone.
It is synthesized by the addition reaction of acetone with chloroform in the presence of potassium hydroxide $(KOH)$.
The reaction is: $(CH_3)_2C=O + CHCl_3 \rightarrow (CH_3)_2C(OH)CCl_3$.
It is used in medicine as a sedative and hypnotic.

(D) When $CHCl_3$ (chloroform) is boiled with aqueous $NaOH$,it undergoes nucleophilic substitution to form an intermediate,$CH(OH)_3$ (methanetriol or trihydroxy methane).
This intermediate is highly unstable due to the presence of three hydroxyl groups on a single carbon atom.
It immediately loses a water molecule $(H_2O)$ to form formic acid $(HCOOH)$.
Since the medium is basic $(NaOH)$,the formic acid reacts with $NaOH$ to form sodium formate $(HCOONa)$ and water.
The overall reaction is: $CHCl_3 + 4NaOH \rightarrow HCOONa + 3NaCl + 2H_2O$.

(C) The reaction of ethyl chloride $(CH_3CH_2Cl)$ with aqueous potassium hydroxide $(KOH_{(aq)})$ is a nucleophilic substitution reaction ($S_N2$ mechanism).
$CH_3CH_2Cl + KOH_{(aq)} \to CH_3CH_2OH + KCl$
The product formed is ethanol $(CH_3CH_2OH)$.
In ethanol,both carbon atoms are bonded to four other atoms (single bonds),which corresponds to $sp^3$ hybridization.

(A) Due to resonance,the $C-Cl$ bond in vinyl chloride $(CH_2=CH-Cl)$ acquires partial double bond character.
This makes the bond stronger and shorter,making it difficult for the nucleophile to replace the chlorine atom.
Therefore,vinyl chloride does not undergo nucleophilic substitution reactions under ordinary conditions.

(A) The $SN^2$ (Substitution nucleophilic bimolecular) reaction mechanism is highly sensitive to steric hindrance.
The reactivity order for $SN^2$ reactions is: $CH_3X > 1^o > 2^o > 3^o$.
$CH_3Cl$ is a methyl halide,which has the least steric hindrance,making it the most reactive towards the $SN^2$ mechanism.
Therefore,the correct option is $A$.

(A) $2, 2-$ dichloropropane is a geminal dihalide. When it reacts with aqueous $KOH$,the two chlorine atoms are replaced by two hydroxyl groups on the same carbon atom. This intermediate is unstable and loses a water molecule to form acetone $(CH_3-CO-CH_3)$.
$CH_3-CCl_2-CH_3 + 2KOH_{(aq)}$ $\rightarrow [CH_3-C(OH)_2-CH_3]$ $\rightarrow CH_3-CO-CH_3 + H_2O$.

(A) $S_N2$ (Bimolecular Nucleophilic Substitution) reactions involve a single-step mechanism where the nucleophile attacks the carbon atom from the back side.
This mechanism is highly sensitive to steric hindrance.
Therefore,it is most favored in methyl halides $(CH_3X)$ and primary alkyl halides $(1^\circ)$.
In the given options,$CH_3-Br$ is a methyl halide and reacts with $OH^{-}$ via the $S_N2$ mechanism.

(A) The reaction of chloroform $(CHCl_3)$ with sodium ethoxide $(C_2H_5ONa)$ upon heating leads to the formation of ethyl orthoformate $(HC(OC_2H_5)_3)$ and sodium chloride $(NaCl)$.
The chemical equation is:
$CHCl_3 + 3C_2H_5ONa \xrightarrow{\Delta} HC(OC_2H_5)_3 + 3NaCl$
Therefore,the correct option is $(A)$.

(A) . The $C-F$ bond energy is the highest among the given alkyl halides.
Due to the high bond dissociation energy of the $C-F$ bond,$CH_3F$ is the least reactive and does not readily form a Grignard reagent with $Mg$ under standard conditions.

(A) The reaction of an alkyl halide with $Na$ in dry ether is the $Wurtz$ reaction,which produces a symmetric alkane.
$A$ is $t$-butyl chloride,$(CH_3)_3C-Cl$.
Reaction: $2 (CH_3)_3C-Cl + 2 Na \xrightarrow{\text{ether}} (CH_3)_3C-C(CH_3)_3 + 2 NaCl$.
The product is $2,2,3,3$-tetramethylbutane.
In $2,2,3,3$-tetramethylbutane,all $18$ hydrogen atoms are equivalent because they are all primary hydrogens attached to methyl groups bonded to quaternary carbons.
Thus,monochlorination of this hydrocarbon yields only one chloro derivative.

(D) The reaction of haloalkanes with alcoholic $KOH$ proceeds via the $E2$ elimination mechanism.
In $CH_3CH_2CH_2Br$ (a primary alkyl halide),the presence of the alkyl group provides a $+I$ effect,which facilitates the elimination of the leaving group.
$CH_2=CHBr$ is a vinyl halide where the $C-X$ bond has partial double bond character,making it very unreactive towards elimination.
$CH_3COCH_2CH_2Br$ contains an electron-withdrawing $-CO$ group ($-I$ effect),which destabilizes the transition state for elimination compared to a simple alkyl halide.
Comparing $CH_3CH_2Br$ and $CH_3CH_2CH_2Br$,the latter has a slightly higher $+I$ effect,making it more reactive in this context.
$CH_3CH_2CH_2Br + KOH \xrightarrow{\text{alc.}} CH_3CH=CH_2 + KBr + H_2O$

(D) The electronegativity of an atom in a molecule is influenced by the electron density around it.
In these haloalkanes,the alkyl group exerts an inductive effect ($+I$ effect),which increases electron density on the chlorine atom.
The order of the $+I$ effect of alkyl groups is: $(CH_3)_3C- > (CH_3)_2CH- > CH_3CH_2- > CH_3-$.
As the $+I$ effect increases,the electron density on the chlorine atom increases.
Therefore,the chlorine atom in tert-butyl chloride $(CH_3-C(CH_3)_2-Cl)$ experiences the highest electron density due to the strongest $+I$ effect from the tert-butyl group,making it the most electronegative in this series.

(D) The reaction of $1-Bromo-3-chloro$ cyclobutane with two equivalents of metallic sodium in ether is an intramolecular Wurtz reaction.
Sodium acts as a reducing agent,donating electrons to the carbon atoms bonded to the halogen atoms.
This leads to the formation of a carbanion at the carbon bearing the bromine atom,which then performs an intramolecular nucleophilic substitution ($S_N2$ type) on the carbon bearing the chlorine atom.
This results in the formation of a new carbon-carbon bond,closing the ring to form bicyclo$[1.1.0]$butane.
The correct option is $D$.

(D) The reaction proceeds in two steps:
$1$. Chlorination of toluene in the presence of light and heat (free radical substitution) yields benzyl chloride $(C_6H_5CH_2Cl)$.
$2$. Treatment of benzyl chloride with aqueous $NaOH$ (nucleophilic substitution) replaces the chlorine atom with a hydroxyl group,resulting in the formation of benzyl alcohol $(C_6H_5CH_2OH)$.

(D) When chlorine is passed into boiling toluene,free radical substitution takes place in the side chain to form benzyl chloride $(C_6H_5CH_2Cl)$.
This benzyl chloride,upon hydrolysis with aqueous alkali ($NaOH$ or $KOH$),undergoes nucleophilic substitution to yield benzyl alcohol $(C_6H_5CH_2OH)$.

(B) The leaving group ability is inversely proportional to the basicity of the leaving group. Weaker bases are better leaving groups.
The acidity of the corresponding conjugate acids follows the order: $CF_3SO_3H > CH_3SO_3H > CH_3COOH > CH_3OH$.
Therefore,the basicity order of their conjugate bases is: $CH_3O^- > CH_3COO^- > CH_3SO_3^- > CF_3SO_3^-$.
Thus,the leaving group ability order is $IV > III > I > II$.

(B) The reaction of $CH_3-CH=CH-C_6H_4-OH$ with $HBr$ follows Markovnikov's rule.
In this reaction,the proton $(H^+)$ adds to the carbon atom of the double bond that has more hydrogen atoms,and the bromide ion $(Br^-)$ adds to the more substituted carbon atom.
The carbocation formed at the benzylic position is more stable due to resonance with the benzene ring.
Therefore,the product is $CH_3-CH_2-CHBr-C_6H_4-OH$.

(A) Hydrolysis of a gem-dichloride yields a carbonyl compound.
Since the product $C_5H_{10}O$ reacts with $NH_2OH$,it is a carbonyl compound (aldehyde or ketone).
It gives the iodoform test,which indicates the presence of a methyl ketone group $(CH_3-CO-)$.
It does not give Fehling's test,which confirms it is a ketone and not an aldehyde.
Therefore,$C_5H_{10}O$ is $2$-pentanone $(CH_3-CO-CH_2-CH_2-CH_3)$.
To obtain $2$-pentanone upon hydrolysis,the starting compound $A$ must be $2,2$-dichloropentane $(CH_3-CCl_2-CH_2-CH_2-CH_3)$.

(B) $Methyl \ magnesium \ iodide$ $(CH_3MgI)$ reacts with $ethyl \ chloroformate$ $(ClCOOC_2H_5)$ to produce $ethyl \ acetate$ $(CH_3COOC_2H_5)$.
$CH_3MgI + ClCOOC_2H_5 \rightarrow CH_3COOC_2H_5 + MgICl$

(B) $KCN$ is an ionic compound that provides $CN^-$ ions. It undergoes nucleophilic substitution reactions with alkyl halides to form alkyl cyanides.
$C_2H_5Br + KCN \xrightarrow{\text{ethanol/water}} C_2H_5CN + KBr$
Ethyl bromide $(C_2H_5Br)$ is an alkyl halide,whereas ethyl alcohol is not a suitable substrate for this reaction,and aryl halides like bromobenzene and chlorobenzene do not undergo nucleophilic substitution with $KCN$ under normal conditions.

(A) The reaction proceeds via an $S_N1$ mechanism involving the formation of a carbocation intermediate.
$1$. The initial loss of the chloride ion $(Cl^-)$ generates a secondary carbocation.
$2$. This secondary carbocation can undergo a hydride shift (rearrangement) to form a more stable benzylic carbocation.
$3$. The initial secondary carbocation can be attacked by water to form product $(K)$.
$4$. The rearranged,more stable benzylic carbocation can be attacked by water to form product $(L)$.
$5$. Therefore,the hydrolysis results in a mixture of $(K)$ and $(L)$.

(A) The $\gamma-isomer$ of benzene hexachloride $(BHC)$ is known as Lindane or Gammexane.
It is a powerful insecticide containing $99\%$ of the $\gamma-isomer$.

(A) $AgNO_3$ reacts with ionic halides to form a precipitate of silver halide $(AgX)$.
$AgNO_3 + NaBr \rightarrow AgBr(s) + NaNO_3$
$AgNO_3 + CaCl_2 \rightarrow AgCl(s) + Ca(NO_3)_2$
$AgNO_3 + NaCl \rightarrow AgCl(s) + NaNO_3$
Ethyl bromide $(C_2H_5Br)$ is a covalent organic compound and does not provide free halide ions $(Br^-)$ in aqueous solution to react with $Ag^+$ ions to form a precipitate.
Therefore,ethyl bromide will not give a precipitate with $AgNO_3$.

(B) Nucleophilic substitution reaction occurs between alkyl halides and $KCN$.
Ethyl bromide $(C_2H_5Br)$ is an alkyl halide that reacts with $KCN$ in an alcoholic medium to form ethyl cyanide $(C_2H_5CN)$ and $KBr$.
The reaction is:
$C_2H_5Br + KCN \xrightarrow{\text{alcohol}} C_2H_5CN + KBr$

(C) $N$-bromosuccinimide $(NBS)$ is a reagent used for free radical bromination at the benzylic position.
In ethylbenzene $(C_6H_5CH_2CH_3)$,the benzylic carbon is the carbon atom directly attached to the benzene ring.
The reaction proceeds via a free radical mechanism to replace a hydrogen atom on the benzylic carbon with a bromine atom.
Thus,the reaction of ethylbenzene with $NBS$ yields $1-$bromo$-1-$phenylethane $(C_6H_5CH(Br)CH_3)$.

(B) The transformation involves converting $1$-bromopropane to $2$-bromopropane.
Step $1$ $(X)$: Dehydrohalogenation of $1$-bromopropane using concentrated alcoholic $NaOH$ at $80 \ ^\circ C$ yields propene $(CH_3-CH=CH_2)$.
Step $2$ $(Y)$: Electrophilic addition of $HBr$ to propene follows Markovnikov's rule,resulting in the formation of $2$-bromopropane $(CH_3-CH(Br)-CH_3)$.

(D) $1$. The compound $(a)$ has the molecular formula $C_8H_9Br$. Since it gives a white precipitate with alcoholic $AgNO_3$,it must contain a reactive halogen atom,likely a benzylic bromide.
$2$. Oxidation of $(a)$ yields an acid $(b)$ with the formula $C_8H_6O_4$. This acid is phthalic acid (benzene$-1,2-$dicarboxylic acid) because it easily forms an anhydride upon heating.
$3$. For the oxidation of $(a)$ to yield phthalic acid,the substituents on the benzene ring must be in the ortho position. Thus,$(a)$ must be $o$-methylbenzyl bromide ($1$-bromo$-2-$methylbenzene).
$4$. The structure corresponds to option $(D)$.

(B) The hydrolysis of $2$-bromo-$3$-methylbutane via the $S_N1$ mechanism proceeds through the formation of a carbocation intermediate.
$1$. The starting material is $(CH_3)_2CH-CH(Br)-CH_3$.
$2$. Upon loss of the bromide ion,a secondary carbocation $(CH_3)_2CH-CH^+-CH_3$ is formed.
$3$. This secondary carbocation undergoes a $1,2$-hydride shift to form a more stable tertiary carbocation: $(CH_3)_2C^+-CH_2-CH_3$.
$4$. The nucleophile $(OH^-)$ then attacks the tertiary carbocation to form the major product,$2$-methylbutan-$2$-ol.

(D) The alkaline hydrolysis of an alkyl halide can proceed via $S_N1$ or $S_N2$ mechanisms.
For $S_N1$ reactions,the rate depends only on the concentration of the alkyl halide $(Rate = k[R-X])$,so the rate remains constant when the alkali concentration is doubled.
For $S_N2$ reactions,the rate depends on both the alkyl halide and the nucleophile $(Rate = k[R-X][OH^-])$,so the rate doubles when the alkali concentration is doubled.
Since the mechanism is not specified,the effect on the reaction rate cannot be definitively predicted without knowing the specific alkyl halide and reaction conditions.

(A) The $C-F$ bond in $CH_3F$ has the highest bond dissociation energy among the given alkyl halides. Due to this high bond strength,$CH_3F$ is the least reactive towards magnesium $(Mg)$ to form a Grignard reagent $(CH_3MgF)$ under standard conditions.

(B) Chloroform $(CHCl_3)$ is slowly oxidized by air in the presence of light to form an extremely poisonous gas called carbonyl chloride,also known as phosgene $(COCl_2)$.
The reaction is: $2CHCl_3 + O_2 \rightarrow 2COCl_2 + 2HCl$.
To prevent this oxidation,chloroform is stored in dark-colored bottles filled up to the brim and a small amount of ethyl alcohol $(C_2H_5OH)$ is added to it.
The ethyl alcohol converts any phosgene formed into harmless diethyl carbonate $(CO(OC_2H_5)_2)$.

(C) When $Ethyl \ bromide$ $(C_2H_5Br)$ reacts with alcoholic $Silver \ nitrite$ $(AgNO_2)$,it undergoes a nucleophilic substitution reaction.
Since $AgNO_2$ is a covalent compound,the nitrogen atom acts as the nucleophile,leading to the formation of a nitro compound.
The reaction is: $C_2H_5Br + AgNO_2 \rightarrow C_2H_5NO_2 + AgBr$.
The product formed is $Nitroethane$ $(C_2H_5NO_2)$.

(B) Chloropicrin is formed by the nitration of chloroform $(CHCl_3)$ with concentrated nitric acid $(HNO_3)$.
The chemical reaction is:
$CHCl_3 + HNO_3 \rightarrow CCl_3NO_2 + H_2O$
Here,$CCl_3NO_2$ is known as chloropicrin or nitrochloroform.

(D) Grignard reagent has the general formula $R-Mg-X$,where $R$ is an alkyl or aryl group,$Mg$ is magnesium,and $X$ is a halogen $(Cl, Br, I)$.
Grignard reagents are highly reactive towards acidic protons,such as those found in carboxylic acids ($-COOH$ group).
Therefore,a $-COOH$ group cannot be present in a Grignard reagent because it would immediately react with the reagent to form an alkane and a magnesium salt.

(B) The rate of $S_N1$ reaction depends on the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.
$A$: Forms a primary $(1^{\circ})$ carbocation,which is the least stable.
$B$: Forms an allylic carbocation $(CH_2=CH-CH^+(CH_3))$,which is resonance-stabilized and thus the most stable.
$C$: Forms a secondary $(2^{\circ})$ carbocation,which is more stable than a primary carbocation but less stable than a resonance-stabilized allylic carbocation.
Therefore,the stability order of the carbocations is $B > C > A$,which is also the order of $S_N1$ reactivity.

(D) This reaction is an intramolecular Wurtz reaction. The $2$ moles of sodium react with the $1$-bromo-$3$-chlorocyclobutane to form a carbanion at the carbon atom attached to bromine,which then performs an intramolecular nucleophilic substitution on the carbon atom attached to chlorine,resulting in the formation of bicyclo[$1.1.0$]butane.

(D) $1$. Chlorination of toluene in the presence of light (free radical substitution) yields benzyl chloride $(C_6H_5CH_2Cl)$.
$2$. When benzyl chloride is heated with aqueous $NaOH$ (nucleophilic substitution),the chlorine atom is replaced by a hydroxyl group to form benzyl alcohol $(C_6H_5CH_2OH)$.
$3$. Note: The provided options do not contain benzyl alcohol. However,if the question implies side-chain chlorination followed by hydrolysis,the product is benzyl alcohol. Given the options,there might be a discrepancy in the question's intended product or options. Based on standard chemical reactions,the product is benzyl alcohol.

(C) The reaction of chloral $(CCl_3CHO)$ with chlorobenzene $(C_6H_5Cl)$ in the presence of concentrated sulfuric acid $(H_2SO_4)$ yields $1,1,1-trichloro-2,2-bis(p-chlorophenyl)ethane$,commonly known as $DDT$ (Dichlorodiphenyltrichloroethane).
This compound is a well-known organochlorine insecticide.

(C) racemic mixture is formed during nucleophilic substitution when the reaction proceeds via an $S_N1$ mechanism,which involves the formation of a planar carbocation intermediate.
This intermediate can be attacked by the nucleophile from either side with equal probability.
Compounds that can form stable carbocations (like tertiary or secondary alkyl halides that are chiral) are more likely to undergo $S_N1$ reactions.
In compound $(a)$,$CH_3-CH(Br)-C_2H_5$ is a secondary chiral alkyl halide,which can form a stable carbocation and undergo $S_N1$ substitution.
In compound $(b)$,$CH_3-C(Br)(CH_3)-C_2H_5$ is a tertiary chiral alkyl halide,which readily forms a stable tertiary carbocation and undergoes $S_N1$ substitution.
In compound $(c)$,$CH_3-CH(C_2H_5)-CH_2Br$ is a primary alkyl halide,which typically undergoes $S_N2$ reactions (inversion of configuration) rather than $S_N1$ (racemization).
Therefore,compounds $(a)$ and $(b)$ are the ones that can form a racemic mixture.

(A) The reaction involves the nucleophilic substitution of the chlorine atom in $3$-iodobenzyl chloride by the cyanide ion $(CN^-)$ from $NaCN$.
This is an $S_N2$ reaction.
The primary alkyl halide group $(-CH_2Cl)$ is much more reactive towards $S_N2$ substitution than the aryl iodide group.
Therefore,the $Cl$ atom is replaced by the $CN$ group,while the iodine atom remains unaffected.
The product is $3$-iodophenylacetonitrile.

(A) 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 chemical reaction is:
$2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$

(A) The reaction of propyl bromide $(C_3H_7Br)$ with potassium cyanide $(KCN)$ is a nucleophilic substitution reaction.
$C_3H_7Br + KCN \rightarrow C_3H_7CN + KBr$
In $C_3H_7CN$,there are $3$ carbon atoms in the propyl group plus $1$ carbon atom in the nitrile group $(-CN)$,totaling $4$ carbon atoms.
According to $IUPAC$ nomenclature,the compound with $4$ carbon atoms and a nitrile functional group is called butane nitrile.