Properties of Haloalkanes Questions in English

(D) The reaction involves the dehydrohalogenation of a vicinal dibromide $(Ph-CH(Br)-CH(Br)-CHO)$ in the presence of a base $(K_2CO_3)$ and heat $(\Delta)$.
This is an elimination reaction (specifically $E2$ mechanism).
The base abstracts a proton from the carbon adjacent to the bromine, leading to the formation of a double bond and the elimination of $HBr$.
In the given molecule, the proton on the carbon attached to the aldehyde group ($\alpha$-carbon) is more acidic due to the electron-withdrawing effect of the $-CHO$ group.
Therefore, the elimination occurs to form the conjugated product: $Ph-CH=C(Br)-CHO$.

(A) The rate of $S_{N}1$ reaction depends on the stability of the carbocation intermediate formed in the rate-determining step.
Benzyl chloride $(C_6H_5CH_2Cl)$ forms a benzyl carbocation $(C_6H_5CH_2^+)$,which is highly stabilized by resonance with the benzene ring.
Ethyl chloride $(CH_3CH_2Cl)$ forms a primary carbocation,which is less stable.
Chlorobenzene $(C_6H_5Cl)$ does not undergo $S_{N}1$ easily due to partial double bond character between $C$ and $Cl$.
Isopropyl chloride $(CH_3CH(Cl)CH_3)$ forms a secondary carbocation,which is less stable than the resonance-stabilized benzyl carbocation.
Therefore,benzyl chloride reacts most easily via the $S_{N}1$ mechanism.

(D) The reduction of chloroform $(CHCl_3)$ with zinc $(Zn)$ and alcoholic hydrochloric acid $(HCl)$ is a partial reduction process.
The reaction is as follows:
$CHCl_3 + 2[H] \xrightarrow{Zn/HCl(alc.)} CH_2Cl_2 + HCl$
In this reaction,one chlorine atom is replaced by a hydrogen atom,resulting in the formation of methylene chloride $(CH_2Cl_2)$.

(A) When chloroform $(CHCl_3)$ is treated with concentrated nitric acid $(HNO_3)$,it undergoes a nitration reaction to form chloropicrin $(CCl_3NO_2)$,which is also known as trichloronitromethane.
The chemical equation is: $CHCl_3 + HNO_3 \rightarrow CCl_3NO_2 + H_2O$.

(D) Chloroform $(CHCl_3)$ is sensitive to light and air. When exposed to light and oxygen,it undergoes slow oxidation to form a highly poisonous gas called phosgene $(COCl_2)$.
The chemical reaction is: $2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$.
To prevent this oxidation,chloroform is stored in dark-colored bottles,which are filled up to the brim to exclude air.

(C) The preparation of chloroform $(CHCl_3)$ from ethanol $(C_2H_5OH)$ and bleaching powder $(CaOCl_2)$ involves the following steps:
$1$. Oxidation: Ethanol is oxidized to acetaldehyde $(CH_3CHO)$.
$2$. Chlorination: Acetaldehyde is chlorinated to form trichloroacetaldehyde $(CCl_3CHO)$.
$3$. Hydrolysis: Trichloroacetaldehyde is hydrolyzed by calcium hydroxide to form chloroform $(CHCl_3)$ and calcium formate.
Elimination does not occur in this reaction sequence.

(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 detect the presence of phosgene,the sample is tested with $AgNO_3$ solution. If phosgene is present,it reacts with $AgNO_3$ to form a white precipitate of $AgCl$.
$COCl_2 + 2AgNO_3 + H_2O \rightarrow 2AgCl(s) + CO_2 + 2HNO_3$
Therefore,$AgNO_3$ solution is used to test the purity of chloroform.

(D) When $CCl_4$ is boiled with an aqueous solution of $KOH$,the chlorine atoms are replaced by hydroxyl groups to form an unstable intermediate,$C(OH)_4$.
This intermediate quickly loses a water molecule to form carbonic acid $(H_2CO_3)$,which further decomposes into $CO_2$ and $H_2O$.
The overall reaction is: $CCl_4 + 4KOH$ $\rightarrow C(OH)_4 + 4KCl$ $\rightarrow H_2CO_3 + 4KCl$ $\rightarrow CO_2 + 2H_2O + 4KCl$.

(A) Carbon tetrachloride $(CCl_4)$ reacts with water vapor at $500\,^oC$ to produce carbonyl chloride $(COCl_2)$,which is also known as phosgene.
The chemical equation is: $CCl_4 + H_2O \xrightarrow{500\,^oC} COCl_2 + 2HCl$.

(C) The reaction of haloalkanes with $AgNO_3$ depends on the ease of formation of carbocations or the ease of breaking the $C-X$ bond.
In iodoform $(CHI_3)$,the $C-I$ bond is much weaker than the $C-Cl$ bond in chloroform $(CHCl_3)$ due to the larger size of the iodine atom and lower bond dissociation energy.
Therefore,$CHI_3$ easily releases $I^-$ ions which react with $Ag^+$ to form a yellow precipitate of $AgI$,whereas $CHCl_3$ does not readily release $Cl^-$ ions under these conditions.

(C) Vicinal dihalides $(CH_2X-CH_2X)$ react with $Zn$ dust to form alkenes $(CH_2=CH_2)$ and $ZnX_2$.
Geminal dihalides $(CH_3-CHX_2)$ do not undergo this reaction to form alkenes under similar conditions.
Thus,$Zn$ dust is used to distinguish between them.

(D) 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 $(COCl_2)$.
$2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$
To prevent this,a small amount of ethanol $(1\%)$ is added to the bottle.
Ethanol acts as a stabilizer by reacting with any phosgene formed to convert it into non-poisonous diethyl carbonate.
$COCl_2 + 2C_2H_5OH \rightarrow (C_2H_5O)_2CO + 2HCl$
Additionally,ethanol also helps in slowing down the oxidation process of chloroform.

(A) In chloroform $(CHCl_3)$,the carbon atom is bonded to three highly electronegative chlorine atoms. These chlorine atoms exert a strong $-I$ (inductive) effect,which pulls electron density away from the carbon atom. This makes the $C-H$ bond highly polarized,allowing the hydrogen atom to be released as a proton $(H^+)$ in the presence of a strong base. Therefore,the hydrogen atom in chloroform is acidic in nature.

(B) gem-dihalide is a compound where two halogen atoms are attached to the same carbon atom. $R-CH(X)_2 + 2KOH(aq) \rightarrow R-CH(OH)_2 + 2KX$.
Since a geminal diol (having two $-OH$ groups on the same carbon) is unstable,it loses a water molecule to form a carbonyl compound.
If the gem-dihalide is a terminal one $(R-CHX_2)$,it forms an aldehyde.
If the gem-dihalide is a non-terminal one $(R-C(X)_2-R')$,it forms a ketone.
However,in the context of general chemistry questions,the hydrolysis of gem-dihalides is typically associated with the formation of carbonyl compounds (aldehydes or ketones). Given the options,if we consider the simplest gem-dihalide like $CH_3-CHCl_2$ ($1$,$1$-dichloroethane),it yields acetaldehyde. Thus,$B$ is the most specific correct answer.

(A) $CHCl_3$ (chloroform) reacts with primary amines $(R-NH_2)$ in the presence of alcoholic $KOH$ to form isocyanides (carbylamines),which have a foul smell. This is known as the carbylamine test.
$R-NH_2 + CHCl_3 + 3KOH (alc.) \rightarrow R-NC + 3KCl + 3H_2O$
$CCl_4$ does not give this reaction.
Therefore,$CCl_4$ and $CHCl_3$ can be distinguished by using $R-NH_2 + KOH$ (alc.).

(B) The reaction of $1, 1-$ dichloroethane $(CH_3CHCl_2)$ with aqueous $KOH$ undergoes nucleophilic substitution to form a geminal diol $(CH_3CH(OH)_2)$.
Since geminal diols are unstable,they lose a water molecule to form acetaldehyde $(CH_3CHO)$.
The reaction is: $CH_3CHCl_2 + 2KOH(aq)$ $\rightarrow CH_3CH(OH)_2 + 2KCl$ $\rightarrow CH_3CHO + H_2O + 2KCl$.

(A) $1$. Friedel-Crafts acylation of benzene with acetic anhydride $(CH_3CO)_2O$ in the presence of $AlCl_3$ gives acetophenone $(X = C_6H_5COCH_3)$.
$2$. Clemmensen reduction of acetophenone $(X)$ using $Zn-Hg, HCl$ reduces the carbonyl group to a methylene group,yielding ethylbenzene $(Y = C_6H_5CH_2CH_3)$.
$3$. Reaction of ethylbenzene $(Y)$ with $N$-bromosuccinimide $(NBS)$ is a radical substitution reaction that occurs at the benzylic position,yielding $1-$bromo$-1-$phenylethane $(Z = C_6H_5CH(Br)CH_3)$.

(C) molecule is chiral if it contains at least one chiral center (a carbon atom bonded to four different groups).
$A$. Butan$-2-$ol: $CH_3-CH(OH)-CH_2-CH_3$. The carbon at position $2$ is bonded to $-H$,$-OH$,$-CH_3$,and $-CH_2CH_3$. Since all four groups are different,it is a chiral molecule.
$B$. $2$-bromopentane: $CH_3-CH(Br)-CH_2-CH_2-CH_3$. The carbon at position $2$ is bonded to $-H$,$-Br$,$-CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is a chiral molecule.
$C$. $2$-bromobutane: $CH_3-CH(Br)-CH_2-CH_3$. The carbon at position $2$ is bonded to $-H$,$-Br$,$-CH_3$,and $-CH_2CH_3$. Since all four groups are different,it is a chiral molecule.
$D$. Bromoethane: $CH_3-CH_2Br$. The carbon atom bonded to $Br$ is attached to two hydrogen atoms. Since two groups are identical,it is achiral.
Note: Based on the provided images,all options $A$,$B$,and $C$ represent chiral molecules. However,in standard multiple-choice questions of this type,$2$-bromobutane $(C)$ is the most commonly cited example of a simple chiral molecule.

(A) $S_N1$ reactions proceed via the formation of a carbocation intermediate. The rate of $S_N1$ reaction depends on the stability of the carbocation formed.
$(A)$ forms a $1^o$ carbocation $(CH_3CH_2CH_2CH_2^+)$,which is the least stable.
$(B)$ forms an allylic carbocation $(CH_2=CH-CH^+(CH_3))$,which is resonance-stabilized and therefore the most stable.
$(C)$ forms a $2^o$ carbocation $(CH_3CH_2CH^+(CH_3))$,which is more stable than a $1^o$ carbocation but less stable than an allylic carbocation.
Thus,the stability order of the carbocations is $B > C > A$. Consequently,the order of $S_N1$ reactivity is $B > C > A$.

(C) Friedel-Crafts alkylation of benzene with $n$-butyl chloride $(CH_3CH_2CH_2CH_2Cl)$ in the presence of $AlCl_3$ leads to rearrangement of the carbocation. The primary carbocation rearranges to a more stable secondary carbocation,resulting in $2-$phenylbutane as the major product.
Similarly,$but-1-ene + HF$ and $butan-2-ol + H_2SO_4$ both generate the $sec-butyl$ carbocation,which yields $2-$phenylbutane.
However,the reaction of $Butanoyl chloride + AlCl_3$ is a Friedel-Crafts acylation,which produces $1-$phenylbutan$-1-$one. Subsequent Clemmensen reduction $(Zn(Hg)/HCl)$ of this ketone yields $n$-butylbenzene ($1-$phenylbutane),not $2-$phenylbutane.

(B) The Assertion is correct because Friedel-Crafts alkylation of benzene leads to polyalkylation due to the activating nature of the alkyl group introduced,making the product more reactive than the starting material.
The Reason is also correct because acyl halides $(RCOX)$ are indeed more reactive than alkyl halides $(RX)$ in terms of electrophilic substitution reactions,as the carbonyl carbon is more electron-deficient than the alkyl carbon.
However,the Reason does not explain why alkylation is not used for preparing alkylbenzene; the limitation of alkylation is polyalkylation,not the reactivity of the halide.
Therefore,both statements are true,but the Reason is not the correct explanation of the Assertion.

(D) $2-$Bromobutane on reaction with sodium ethoxide in ethanol undergoes dehydrohalogenation to give $2-$butene as the major product according to Saytzeff's rule,which states that the more substituted alkene is the preferred product.
Thus,the assertion is incorrect because $1-$butene is the minor product.
$2-$Butene is more stable than $1-$butene due to greater substitution and hyperconjugation.
The reason statement claims $1-$butene is more stable than $2-$butene,which is false.
Therefore,both the assertion and the reason are incorrect.

(A) $S_N1$ reactions proceed via the formation of a carbocation intermediate.
The reactivity toward $S_N1$ is directly proportional to the stability of the carbocation formed.
The stability order of the corresponding carbocations is: $PhCH_2^+ > CH_2=CH-CH_2^+ > (CH_3)_2CH^+ > CH_3-CH_2^+$.
Therefore,the increasing order of reactivity is: $CH_3-CH_2-X < (CH_3)_2CH-X < CH_2=CH-CH_2-X < PhCH_2-X$.

(B) The boiling point of haloalkanes depends on the molecular mass and the extent of branching.
As the molecular mass increases,the boiling point increases.
Among isomers,the boiling point decreases with an increase in branching because branching reduces the surface area,leading to weaker van der Waals forces.
Comparing the given compounds:
$CH_3CH_2CH_2CH_2Cl$ (n-butyl chloride) has the highest molecular mass and is a straight-chain isomer with no branching,resulting in the strongest intermolecular forces.
Therefore,it has the highest boiling point.

(A) $S_N2$ (Substitution Nucleophilic Bimolecular) reactions are highly favored by methyl halides and primary $(1^\circ)$ alkyl halides because they have the least steric hindrance,allowing the nucleophile to attack the carbon atom easily from the backside.
$CH_3Br$ is a methyl halide and undergoes $S_N2$ reaction most readily among the given options.
Option $(b)$ is a secondary halide,option $(c)$ is an elimination reaction,and option $(d)$ is a tertiary halide which typically follows the $S_N1$ mechanism.

(D) $AgCN$ is a covalent compound,so the nucleophilic attack occurs through the nitrogen atom,forming isocyanide $(RNC)$.
$KCN$ is an ionic compound,so the nucleophilic attack occurs through the carbon atom,forming cyanide $(RCN)$.
$KNO_2$ is an ionic compound,so the nucleophilic attack occurs through the oxygen atom,forming alkyl nitrite $(RONO)$.
$AgNO_2$ is a covalent compound,so the nucleophilic attack occurs through the nitrogen atom,forming nitroalkane $(RNO_2)$.
Therefore,all the given pairs are correctly matched.

(C) $CHCl_3$ (chloroform) is stored in dark-colored bottles to prevent its oxidation by atmospheric oxygen in the presence of sunlight.
The reaction is: $2CHCl_3 + O_2 \xrightarrow{\text{sunlight}} 2COCl_2 + 2HCl$.
The product $COCl_2$ (phosgene) is a highly poisonous gas.
Since the oxidation occurs in the presence of light,the Reason statement claiming it is oxidized in the dark is incorrect.

(C) The hydrolysis of benzyl bromide in aqueous acetone follows the $S_{N}1$ mechanism.
$1.$ The formation of the benzyl carbocation $(C_{6}H_{5}CH_{2}^{+})$ is the rate-determining step,which is stabilized by resonance.
$2.$ The reaction proceeds via a carbocation intermediate,which is characteristic of the $S_{N}1$ pathway.
Therefore,the Assertion is correct,but the Reason is incorrect.

(C) The nucleophile $Z^{-}$ can act as a base to cause elimination or as a nucleophile to cause substitution.
$CH_{3}CH_{2}O^{-}$ $(A)$ is a smaller,less sterically hindered nucleophile/base compared to $(CH_{3})_{3}CO^{-}$ $(B)$.
Because $(CH_{3})_{3}CO^{-}$ $(B)$ is bulky,it experiences significant steric hindrance when attacking the carbon atom for substitution $(S_N2)$,making $k_s(B) < k_s(A)$.
However,$(CH_{3})_{3}CO^{-}$ $(B)$ is a strong base and can easily abstract a proton for elimination,making $k_e(B) > k_e(A)$.
Since $\mu = \frac{k_s}{k_e}$,and $k_s(B) < k_s(A)$ while $k_e(B) > k_e(A)$,it follows that $\mu_{A} = \frac{k_s(A)}{k_e(A)}$ is greater than $\mu_{B} = \frac{k_s(B)}{k_e(B)}$.
Therefore,$\mu_{A} > \mu_{B}$ and $k_e(B) > k_e(A)$.

(B) The reaction of $1$-methyl ethylene oxide (propylene oxide) with an excess of $HBr$ proceeds as follows:
$1$. The oxygen atom of the epoxide ring gets protonated by $H^+$,forming a cyclic oxonium ion.
$2$. The bromide ion $(Br^-)$ attacks the more substituted carbon atom of the epoxide ring due to the development of partial positive charge character,which is more stable at the more substituted carbon.
$3$. This leads to the opening of the ring,forming a bromohydrin intermediate $(CH_3-CH(OH)-CH_2Br)$.
$4$. Since $HBr$ is in excess,the hydroxyl group $(-OH)$ is further protonated and replaced by another bromide ion $(Br^-)$ via an $S_N2$ mechanism,resulting in a vicinal dibromide.
$5$. The final product is $1,2$-dibromopropane,which is $CH_3-CH(Br)-CH_2Br$.

(A) The Saytzeff rule states that in elimination reactions,the more substituted alkene (the one with more alkyl groups attached to the double-bonded carbons) is the major product.
$(a)$ $(CH_3)_3CCH(OH)CH_3$ $\xrightarrow{H^+} (CH_3)_3CCH^+(CH_3)$ $\xrightarrow{1,2-methyl shift} (CH_3)_2C^+(CH_2CH_3)CH_3$ $\xrightarrow{-H^+} (CH_3)_2C=C(CH_3)_2$. This is a highly substituted alkene,which is the Saytzeff product.
$(b)$ $(CH_3)_2CHCH(Br)CH_3 \xrightarrow{alc. KOH} (CH_3)_2C=CHCH_3$. This is the more substituted alkene,which is the Saytzeff product.
$(c)$ $(CH_3)_2CHCH(Br)CH_3 \xrightarrow{(CH_3)_3CO^-K^+} (CH_3)_2CHCH=CH_2$. The bulky base $(CH_3)_3CO^-$ abstracts the less hindered proton,leading to the less substituted alkene,which is the Hofmann product.
$(d)$ $(CH_3)_2C(OH)CH_2CHO \xrightarrow{\Delta} (CH_3)_2C=CHCHO$. This is the more substituted alkene (conjugated with the carbonyl group),which is the Saytzeff product.
Thus,only reaction $(c)$ does not produce the Saytzeff product as the major product.

(C) The reactivity towards $E_1$ reaction depends on the stability of the carbocation intermediate formed after the loss of the leaving group $(Cl^-)$.
$(A)$ $CH_3CH_2CH_2CH_2^+$ (Primary carbocation,least stable)
$(B)$ $CH_2=CHCH_2CH_2^+$ (Primary allylic carbocation,stabilized by resonance)
$(C)$ $CH_3CH_2CH^+CH_3$ (Secondary carbocation)
$(D)$ $CH_2=CHCH^+CH_3$ (Secondary allylic carbocation,highly stabilized by resonance)
Comparing the stability: The secondary allylic carbocation $(D)$ is more stable than the primary allylic carbocation $(B)$,which is more stable than the secondary alkyl carbocation $(C)$,which is more stable than the primary alkyl carbocation $(A)$.
Therefore,the order of reactivity is $D > B > C > A$.

(D) The bond length of the $C-Cl$ bond depends on the extent of resonance.
In $Cl-CH=CH-NO_2$,the $-NO_2$ group is a strong electron-withdrawing group ($-M$ effect),while the $Cl$ atom acts as an electron donor ($+M$ effect).
This creates a strong conjugation effect where the lone pair on $Cl$ is delocalized towards the $NO_2$ group,increasing the double bond character of the $C-Cl$ bond.
Greater double bond character leads to a shorter bond length.
Therefore,$Cl-CH=CH-NO_2$ has the shortest $C-Cl$ bond.

(A) This is a substitution reaction where the $-Br$ group is replaced by the $-SH$ group.
$(b)$ This is an addition reaction where $HCl$ adds across the double bond of isobutylene.
$(c)$ This is an elimination reaction (dehydrohalogenation) where $H$ and $Br$ are removed to form an alkene.
$(d)$ This is a substitution reaction accompanied by a carbocation rearrangement.

(N/A) $(i)$ According to Markovnikov's rule,the addition of $HBr$ to styrene $(C_6H_5CH=CH_2)$ gives $C_6H_5CH(Br)CH_3$ ($1$-phenylethyl bromide).
(ii) According to Markovnikov's rule,the addition of $HCl$ to but$-1-$ene $(CH_3-CH_2-CH=CH_2)$ gives $CH_3-CH_2-CHCl-CH_3$ ($2$-chlorobutane).
(iii) In the presence of peroxide,the addition of $HBr$ to $3-$phenylprop$-1-$ene $(C_6H_5CH_2-CH=CH_2)$ follows anti-Markovnikov addition,giving $C_6H_5CH_2-CH_2-CH_2Br$ ($3$-phenylpropyl bromide).

(N/A) $KCN$ is predominantly ionic and provides cyanide ions $(CN^-)$ in solution. Since the $C-C$ bond is more stable than the $C-N$ bond,the attack occurs mainly through the carbon atom,leading to the formation of alkyl cyanides $(R-CN)$.
Conversely,$AgCN$ is mainly covalent in nature. In this case,the carbon atom is involved in the bond with silver,leaving the nitrogen atom free to donate its lone pair of electrons to the alkyl group,resulting in the formation of isocyanides $(R-NC)$ as the major product.

(N/A) The rate of $S_{N}2$ reaction depends on the steric hindrance around the electrophilic carbon and the nature of the leaving group.
$(i)$ Cyclohexylmethyl chloride is a primary $(1^{\circ})$ alkyl halide,whereas Chlorocyclohexane is a secondary $(2^{\circ})$ alkyl halide. Since $S_{N}2$ reactions are highly sensitive to steric hindrance,the less hindered primary halide reacts faster. Thus,Cyclohexylmethyl chloride reacts faster than Chlorocyclohexane.
(ii) Both $n$-Butyl iodide and $n$-Butyl chloride are primary $(1^{\circ})$ alkyl halides. The rate of $S_{N}2$ reaction depends on the ability of the leaving group to depart. Iodide $(I^-)$ is a better leaving group than chloride $(Cl^-)$ because it is a weaker base. Therefore,$n$-Butyl iodide reacts faster than $n$-Butyl chloride.

(N/A) $(i)$ $S_{N}1$ order: $CH_{3}CH_{2}CH_{2}CH_{2}Br < (CH_{3})_{2}CHCH_{2}Br < CH_{3}CH_{2}CH(Br)CH_{3} < (CH_{3})_{3}CBr$
$S_{N}2$ order: $CH_{3}CH_{2}CH_{2}CH_{2}Br > (CH_{3})_{2}CHCH_{2}Br > CH_{3}CH_{2}CH(Br)CH_{3} > (CH_{3})_{3}CBr$
In $S_{N}1$,reactivity depends on carbocation stability. Tertiary > Secondary > Primary. Among primary,$(CH_{3})_{2}CHCH_{2}Br$ is more reactive than $CH_{3}CH_{2}CH_{2}CH_{2}Br$ due to the inductive effect. In $S_{N}2$,reactivity depends on steric hindrance. Primary > Secondary > Tertiary.
$(ii)$ $S_{N}1$ order: $C_{6}H_{5}CH_{2}Br < C_{6}H_{5}CH(CH_{3})Br < C_{6}H_{5}CH(C_{6}H_{5})Br < C_{6}H_{5}C(CH_{3})(C_{6}H_{5})Br$
$S_{N}2$ order: $C_{6}H_{5}C(CH_{3})(C_{6}H_{5})Br < C_{6}H_{5}CH(C_{6}H_{5})Br < C_{6}H_{5}CH(CH_{3})Br < C_{6}H_{5}CH_{2}Br$
In $S_{N}1$,stability of the carbocation is enhanced by resonance from phenyl groups. $C_{6}H_{5}C(CH_{3})(C_{6}H_{5})Br$ forms a tertiary benzylic carbocation,which is most stable. In $S_{N}2$,reactivity decreases as steric hindrance increases.

(N/A) $(i)$ The boiling point of alkyl halides increases with an increase in the molecular mass (or number of halogen atoms). The order is: $CH_3Cl < CH_3Br < CH_2Br_2 < CHBr_3$.
$(ii)$ For isomers,the boiling point decreases with an increase in branching. For different chain lengths,the boiling point increases with the size of the alkyl group. Thus,the order is: $\text{Isopropyl chloride} < 1-\text{Chloropropane} < 1-\text{Chlorobutane}$.

(A) $(i)$ $CH_3CH_2CH_2CH_2Br$ $(1^{\circ})$ reacts faster than $CH_3CH_2CH(Br)CH_3$ $(2^{\circ})$ because $S_N2$ reactions are highly sensitive to steric hindrance. Primary alkyl halides have less steric hindrance than secondary ones,allowing easier nucleophilic attack.
$(ii)$ $CH_3CH_2CH(Br)CH_3$ $(2^{\circ})$ reacts faster than $(CH_3)_3CBr$ $(3^{\circ})$ because the $3^{\circ}$ alkyl halide is extremely sterically hindered due to three bulky methyl groups,making the $S_N2$ pathway very slow.
$(iii)$ $CH_3CH(CH_3)CH_2CH_2Br$ reacts faster than $CH_3CH_2CH(CH_3)CH_2Br$. Although both are primary,in $CH_3CH_2CH(CH_3)CH_2Br$,the methyl group is on the $\beta$-carbon (closer to the reaction site),causing greater steric hindrance compared to $CH_3CH(CH_3)CH_2CH_2Br$ where the methyl group is further away.

(N/A) $(i)$ $S_N1$ reaction proceeds via the formation of a carbocation. The alkyl halide $(I)$ is $3^{\circ}$ while $(II)$ is $2^{\circ}$. Therefore,$(I)$ forms a $3^{\circ}$ carbocation while $(II)$ forms a $2^{\circ}$ carbocation. The greater the stability of the carbocation,the faster is the rate of $S_N1$ reaction. Since a $3^{\circ}$ carbocation is more stable than a $2^{\circ}$ carbocation,$(I)$,i.e.,$2$-chloro-$2$-methylpropane,undergoes a faster $S_N1$ reaction than $(II)$,i.e.,$3$-chloropentane.
$(ii)$ The alkyl halide $(I)$ is $2^{\circ}$ while $(II)$ is $1^{\circ}$. $A$ $2^{\circ}$ carbocation is more stable than a $1^{\circ}$ carbocation. Therefore,$(I)$,$2$-chloroheptane,undergoes a faster $S_N1$ reaction than $(II)$,$1$-chlorohexane.

(B) $1.$ $Cyclohexyl-Br + Mg$ $\xrightarrow{dry\,ether} Cyclohexyl-MgBr (A)$ $\xrightarrow{H_2O} Cyclohexane (B)$
$2.$ $CH_3-CH(Br)-CH_3 (R) + Mg$ $\xrightarrow{dry\,ether} CH_3-CH(MgBr)-CH_3 (C)$ $\xrightarrow{D_2O} CH_3-CH(D)-CH_3$
$3.$ $2(CH_3)_3C-X (R^1-X) \xrightarrow{Na/ether} (CH_3)_3C-C(CH_3)_3$ (Wurtz reaction)
$4.$ $(CH_3)_3C-X$ $\xrightarrow{Mg} (CH_3)_3C-MgX (D)$ $\xrightarrow{H_2O} (CH_3)_3CH (E)$
Thus:
$A = Cyclohexylmagnesium\,bromide$
$B = Cyclohexane$
$C = CH_3-CH(MgBr)-CH_3$
$D = (CH_3)_3C-MgX$
$E = (CH_3)_3CH$
$R = CH_3-CH(Br)-CH_3$
$R^1 = (CH_3)_3C-$

(N/A) Ambident nucleophiles are nucleophiles that possess two nucleophilic sites through which they can attack.
For example,the nitrite ion $(NO_2^-)$ is an ambident nucleophile.
It can attack through the oxygen atom to form alkyl nitrites $(R-O-N=O)$,or it can attack through the nitrogen atom to form nitroalkanes $(R-NO_2)$.

(C) $(i)$ In $S_N2$ reactions,the rate depends on the leaving group ability. The order of leaving group ability is $I^{-} > Br^{-} > Cl^{-} > F^{-}$. Since $I^{-}$ is a better leaving group than $Br^{-}$,$CH_3-I$ reacts faster than $CH_3-Br$.
$(ii)$ $S_N2$ reactions are highly sensitive to steric hindrance. The reactivity order for $S_N2$ is $\text{methyl} > \text{primary} > \text{secondary} > \text{tertiary}$. $CH_3-Cl$ is a methyl halide,while $CH_3-C(CH_3)_2-Cl$ is a tertiary halide. Therefore,$CH_3-Cl$ reacts faster due to significantly less steric hindrance.

(N/A) $(i)$ In $1-$bromo$-1-$methylcyclohexane,all $\beta$-hydrogen atoms are equivalent. Thus,dehydrohalogenation gives only one alkene: $1-$methylcyclohexene.
$(ii)$ In $2-$chloro$-2-$methylbutane,there are two sets of $\beta$-hydrogens. Dehydrohalogenation yields two alkenes:
$CH_3-C(Cl)(CH_3)-CH_2-CH_3 \xrightarrow{C_2H_5ONa/C_2H_5OH} CH_3-C(CH_3)=CH-CH_3 \text{ (major, } 2-$methylbut$-2-$ene$)$ $+ CH_2=C(CH_3)-CH_2-CH_3 \text{ (minor, } 2-$methylbut$-1-$ene$)$
According to Saytzeff's rule,the more substituted alkene ($2-$methylbut$-2-$ene$)$ is the major product.
$(iii)$ In $2,2,3-$trimethyl$-3-$bromopentane,there are two sets of $\beta$-hydrogens. Dehydrohalogenation yields two alkenes:
$CH_3-C(CH_3)_2-C(Br)(CH_3)-CH_2-CH_3$ $\xrightarrow{C_2H_5ONa/C_2H_5OH} CH_3-C(CH_3)_2-C(CH_3)=CH-CH_3 \text{ (major, } 3,4,4-$trimethylpent$-2-$ene$)$ $+ CH_3-C(CH_3)_2-C(=CH_2)-CH_2-CH_3 \text{ (minor, } 2,3,3-$trimethylpent$-1-$ene$)$
According to Saytzeff's rule,the more substituted alkene ($3,4,4-$trimethylpent$-2-$ene$)$ is the major product.

(N/A) $(i)$ In chlorobenzene,the $Cl$ atom is linked to a $sp^{2}$ hybridized carbon atom. In cyclohexyl chloride,the $Cl$ atom is linked to a $sp^{3}$ hybridized carbon atom. The $sp^{2}$ hybridized carbon has more $s$-character than the $sp^{3}$ hybridized carbon atom,making it more electronegative. Consequently,the electron density of the $C-Cl$ bond near the $Cl$ atom is less in chlorobenzene than in cyclohexyl chloride. Furthermore,the $-I$ effect of the benzene ring in chlorobenzene decreases the electron density of the $C-Cl$ bond near the $Cl$ atom. As a result,the polarity of the $C-Cl$ bond in chlorobenzene decreases,leading to a lower dipole moment compared to cyclohexyl chloride.
$(ii)$ To be miscible with water,the solute-water force of attraction must be stronger than the solute-solute and water-water forces of attraction. Alkyl halides are polar molecules held together by dipole-dipole interactions,while water molecules are held together by strong hydrogen bonds. The new force of attraction between alkyl halides and water molecules is weaker than the existing alkyl halide-alkyl halide and water-water forces of attraction. Hence,alkyl halides are immiscible with water.
$(iii)$ Grignard reagents $(R-Mg-X)$ are highly reactive. In the presence of moisture,they react with water to form alkanes:
$R-Mg-X + H_{2}O \rightarrow R-H + Mg(OH)X$
Therefore,Grignard reagents must be prepared under anhydrous conditions to prevent decomposition.

(N/A) The given reaction is:
$n-BuBr + KCN \xrightarrow{EtOH/H_2O} n-BuCN$
This reaction proceeds via an $S_{N}2$ mechanism.
$1$. The nucleophile $CN^{-}$ attacks the electrophilic carbon atom of $n-butyl$ bromide $(n-BuBr)$ from the side opposite to the leaving group $(Br^{-})$.
$2$. This results in the formation of a transition state where the $C-CN$ bond is partially formed and the $C-Br$ bond is partially broken.
$3$. Finally,the $Br^{-}$ ion leaves,resulting in the formation of $n-butyl$ cyanide $(n-BuCN)$.
$4$. The $CN^{-}$ ion is an ambident nucleophile,but in the presence of $KCN$,the attack occurs primarily through the carbon atom due to the higher nucleophilicity of the carbon center.

(N/A) $(i)$ $1-Bromopentane$ $(1^{\circ})$,$2-Bromopentane$ $(2^{\circ})$,$2-Bromo-2-methylbutane$ $(3^{\circ})$. The $S_{N}2$ reactivity decreases as steric hindrance increases. The order is: $2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane$.
$(ii)$ $1-Bromo-3-methylbutane$ $(1^{\circ})$,$2-Bromo-3-methylbutane$ $(2^{\circ})$,$2-Bromo-2-methylbutane$ $(3^{\circ})$. The order of reactivity is: $2-Bromo-2-methylbutane < 2-Bromo-3-methylbutane < 1-Bromo-3-methylbutane$.
$(iii)$ All are $1^{\circ}$ alkyl halides. Reactivity decreases as branching near the reaction site increases. The order is: $1-Bromo-2,2-dimethylpropane < 1-Bromo-2-methylbutane < 1-Bromo-3-methylbutane < 1-Bromobutane$.