Properties of Haloalkanes Questions in English

(B) $(P)$ $(-)-1$-Bromo-$2$-ethylpentane undergoes $S_N2$ reaction. Since the chiral center is at $C_2$ and the reaction occurs at $C_1$,the configuration at the chiral center remains unchanged,leading to retention of configuration $(P \rightarrow 2)$.
$(Q)$ $(-)-2$-Bromopentane undergoes $S_N2$ reaction at the chiral center,resulting in Walden inversion,i.e.,inversion of configuration $(Q \rightarrow 1)$.
$(R)$ $(-)-3$-Bromo-$3$-methylhexane undergoes $S_N1$ reaction. The carbocation formed is planar,leading to a racemic mixture,i.e.,a mixture of enantiomers $(R \rightarrow 3)$.
$(S)$ The substrate has two chiral centers. $S_N1$ reaction at the chiral center creates a new chiral center,resulting in a mixture of diastereomers $(S \rightarrow 5)$.
Thus,the correct match is $P$ $\rightarrow 2, Q$ $\rightarrow 1, R$ $\rightarrow 3, S$ $\rightarrow 5$.

(B) $I$ is a primary benzylic halide,which is highly reactive towards both $S_{N}1$ (due to resonance-stabilized carbocation) and $S_{N}2$ (due to low steric hindrance).
$II$ is a primary alkyl halide,which primarily follows $S_{N}2$ mechanism.
$III$ is a tertiary alkyl halide,which follows $S_{N}1$ mechanism due to the formation of a stable tertiary carbocation.
$IV$ is a secondary benzylic halide,which follows $S_{N}1$ mechanism primarily.
Analysis of statements:
$A$. $I$ (benzylic) and $III$ (tertiary) can both undergo $S_{N}1$ reactions. This is correct.
$B$. $I$ (primary benzylic) and $II$ (primary alkyl) are both primary halides and are highly reactive towards $S_{N}2$. This is correct.
$C$. $IV$ is a chiral secondary halide. In $S_{N}2$ conditions,it would undergo inversion,but it typically reacts via $S_{N}1$ (racemization). However,if forced to undergo $S_{N}2$,it would show inversion. Given the context of standard questions,$C$ is considered correct as a property of $S_{N}2$ pathways.
$D$. Reactivity order for $S_{N}1$ is $IV > I > III$ is incorrect; the order is $IV > III > I$ or similar depending on conditions. Thus,$D$ is incorrect.
Therefore,$A, B, C$ are correct.

(B) The reaction sequence involves the conversion of an alcohol $(X)$ to an azide with inversion of configuration at the chiral center.
$1$) Treatment of the alcohol with $PBr_3$ in $Et_2O$ converts the $-OH$ group into a $-Br$ group with inversion of configuration (via $S_N2$ mechanism).
$2$) Treatment with $NaI$ in $Me_2CO$ (Finkelstein reaction) converts the $-Br$ group into a $-I$ group with further inversion of configuration.
$3$) Treatment with $NaN_3$ in $HCONMe_2$ $(DMF)$ performs an $S_N2$ substitution of the iodide with the azide ion $(-N_3)$,resulting in a third inversion of configuration.
Since there are three successive $S_N2$ inversions,the final product has the same relative configuration as the starting material $X$.
The final product shown is a cyclopentane ring with a methyl group (wedge) and an azide group (dash). Since the starting material $X$ must have the same configuration as the product,$X$ must have the methyl group (wedge) and the hydroxyl group (wedge). This corresponds to structure $(B)$.

(B) The rate of $S_N2$ reaction depends on steric hindrance and the presence of electron-withdrawing groups at the $\alpha$-carbon.
$1$. $S$ $(C_6H_5-CO-CH_2-Cl)$: The carbonyl group $(C=O)$ is strongly electron-withdrawing due to the $-I$ and $-M$ effects,which stabilizes the transition state,making it highly reactive.
$2$. $R$ $(CH_2=CH-CH_2-Cl)$: This is an allylic halide. The transition state is stabilized by conjugation with the double bond,making it more reactive than primary alkyl halides.
$3$. $P$ $(CH_3-Cl)$: This is a primary alkyl halide with minimal steric hindrance.
$4$. $Q$ $((CH_3)_2CH-Cl)$: This is a secondary alkyl halide,which is the least reactive due to higher steric hindrance compared to the others.
Therefore,the order of reactivity is $S > R > P > Q$.

(A) The compound $(CH_3)_3C-CH_2-Cl$ is neopentyl chloride.
Statement $I$ is incorrect because neopentyl chloride is a primary halide and forms a highly unstable primary carbocation upon the loss of the chloride ion,making $S_N1$ reactions extremely unfavorable.
Statement $II$ is correct because,although it is a primary halide,the bulky tert-butyl group adjacent to the reaction center causes significant steric hindrance,which makes the $S_N2$ transition state very difficult to achieve,resulting in a very slow reaction rate.

(A) The reaction sequence $RBr$ $\xrightarrow{(i) Mg, \text{dry ether}} RMgBr$ $\xrightarrow{(ii) H_2O} RH$ represents the formation of an alkane from an alkyl bromide via a Grignard reagent.
To obtain $2-$methylbutane $(C_5H_{12})$ from $RBr$,the alkyl group $R$ must be a $2-$methylbutyl group.
The possible structural isomers of $2-$methylbutyl bromide are:
$1.$ $1-$bromo$-2-$methylbutane
$2.$ $2-$bromo$-2-$methylbutane
$3.$ $2-$bromo$-3-$methylbutane
$4.$ $1-$bromo$-3-$methylbutane
All these isomers,upon reaction with $Mg$ in dry ether followed by hydrolysis with $H_2O$,will yield $2-$methylbutane.
Therefore,the total number of such structural isomers is $4$.

(A) The rate of solvolysis (which proceeds via an $S_{N}1$ mechanism) is directly proportional to the stability of the intermediate carbocation formed after the departure of the leaving group (bromide ion).
$1$. Compound $C$ forms a diphenylmethyl carbocation,which is highly stabilized by resonance with two phenyl rings.
$2$. Compound $B$ forms a benzylic carbocation (specifically,a $1-$tetralyl type),which is stabilized by resonance with one benzene ring.
$3$. Compound $A$ forms a tertiary carbocation in a cyclohexane ring,which is stabilized by inductive effects and hyperconjugation.
$4$. Compound $D$ forms a secondary carbocation,which is the least stable among the given options.
Thus,the stability order of the carbocations is $C > B > A > D$. Consequently,the ascending order of the rate of solvolysis is $D < A < B < C$.

(A) Statement $I$ represents an $S_N2$ reaction of a primary alkyl halide $(1^\circ)$. $S_N2$ reactions are generally favored in less polar or polar aprotic solvents because polar protic solvents solvate the nucleophile,reducing its reactivity.
Statement $II$ represents an $S_N1$ reaction of a tertiary alkyl halide $(3^\circ)$. $S_N1$ reactions are favored in more polar (polar protic) solvents because the polar medium stabilizes the carbocation intermediate and the transition state.
Therefore,both statements are true.

(C) The reaction involves the nucleophilic substitution of a primary alkyl chloride with $AgCN$.
$AgCN$ is a covalent compound. In such cases,the nitrogen atom of the cyanide group acts as the nucleophile because it has a lone pair of electrons available for bonding,while the carbon atom is less nucleophilic due to the covalent nature of the $Ag-C$ bond.
Therefore,the reaction follows an $S_N2$ mechanism where the $-Cl$ group is replaced by an isocyanide $(-NC)$ group.
The reaction is:
$Ar-CH_2Cl + AgCN \rightarrow Ar-CH_2NC + AgCl$
where $Ar$ is the $3-$bromo$-5-$iodophenyl group.
Thus,the major product is $1-$bromo$-3-$iodobenzyl isocyanide.

(C) The compound $(Et)_2N-CH_2CH_2Cl$ contains a nitrogen atom with a lone pair of electrons. This lone pair can attack the carbon atom attached to the chlorine atom,leading to the formation of an aziridinium ion intermediate through an intramolecular nucleophilic substitution reaction (anchimeric assistance).
This process is much faster than the intermolecular nucleophilic substitution that would occur in $(Et)_2CH-CH_2CH_2Cl$,which lacks such an internal nucleophile.
Therefore,Statement $I$ is correct because the rate of hydrolysis is faster for the nitrogen-containing compound.
Statement $II$ is also correct as it describes the mechanism of this intramolecular substitution involving the lone pair on the nitrogen atom.

(D) $AgNO_2$ is a covalent compound. The nitrogen atom has a lone pair of electrons,which attacks the alkyl halide to form a nitroalkane $(R-NO_2)$.
Therefore,$CH_3-CH_2-CH_2-Br + AgNO_2 \rightarrow CH_3-CH_2-CH_2-NO_2 (A) + AgBr$.
$AgCN$ is also a covalent compound. The nitrogen atom has a lone pair,but the carbon atom is the nucleophilic site in the cyanide ion. However,due to the covalent nature of $Ag-CN$ bond,the nitrogen atom attacks the alkyl halide to form an isocyanide $(R-NC)$.
Therefore,$CH_3-CH_2-CH_2-Br + AgCN \rightarrow CH_3-CH_2-CH_2-NC (B) + AgBr$.
Thus,the products are $CH_3-CH_2-CH_2-NO_2$ and $CH_3-CH_2-CH_2-NC$.

(D) Step $1$: The reaction of $1-(4-methylphenyl)prop-1-ene$ with $HCl$ follows Markovnikov's rule. The electrophilic addition of $H^+$ to the double bond forms a stable benzylic carbocation at the carbon adjacent to the benzene ring. Then,$Cl^-$ attacks this carbocation to form $1-(4-methylphenyl)-1-chloropropane$ as the major product $(A)$.
Step $2$: The reaction of the alkyl halide $(A)$ with $AgCN$ is a nucleophilic substitution reaction. Since $AgCN$ is a covalent compound,the nitrogen atom acts as the nucleophile,leading to the formation of an isocyanide (isonitrile) as the major product $(B)$.
Thus,the final product is $1-(4-methylphenyl)propyl$ isocyanide.

(D) The stability of a carbocation is determined by resonance,hyperconjugation,and the inductive effect.
In the given options,the carbocations formed are:
$A$: Allyl carbocation $(CH_2=CH-CH_2^+)$
$B$: Benzyl carbocation $(C_6H_5-CH_2^+)$
$C$: Cyclohexenyl carbocation
$D$: Triphenylmethyl carbocation $((C_6H_5)_3C^+)$
Among these,the triphenylmethyl carbocation is the most stable because the positive charge is delocalized over three phenyl rings through extensive resonance.
Therefore,$(C_6H_5)_3C-Br$ generates the most stable carbocation.

(B) Statement $(I) :$ When alkyl chlorides $(R-Cl)$ are treated with aqueous $KOH$,they undergo nucleophilic substitution $(S_N)$ reaction to form alcohols $(R-OH)$. This is not an elimination reaction. Thus,Statement $(I)$ is incorrect.
Statement $(II) :$ When alkyl chlorides are treated with alcoholic $KOH$,they undergo dehydrohalogenation (an elimination reaction) where the base abstracts a hydrogen atom from the $\beta-$carbon to form an alkene. Thus,Statement $(II)$ is correct.

(A) Statement $(I)$ is correct. $cis-1,2-dichloroethene$ exhibits greater polarity than $cis-1,2-dibromoethene$ due to the stronger $p\pi-d\pi$ back-bonding in the $C-Cl$ bond compared to the $C-Br$ bond,which enhances the dipole moment.
Statement $(II)$ is incorrect. While the boiling point of $trans-1,2-dibromoethene$ is indeed lower than that of $cis-1,2-dibromoethene$ (because the $cis$ isomer has a net dipole moment $\mu \neq 0$ while the $trans$ isomer has $\mu = 0$),the $trans$ isomer is $NOT$ more polar than the $cis$ isomer. The $trans$ isomer is non-polar $(\mu = 0)$,whereas the $cis$ isomer is polar $(\mu \neq 0)$.

(A) The rate of $S_N2$ reaction is directly proportional to the leaving group ability of the nucleophile.
Iodide ion $(I^-)$ is a better leaving group than chloride ion $(Cl^-)$ because of its larger size,which results in a weaker $C-I$ bond compared to the $C-Cl$ bond.
Therefore,$R-I$ reacts faster than $R-Cl$ in $S_N2$ reactions.
Both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A)$.

(A) The rate of $S_N1$ reaction is directly proportional to the stability of the carbocation intermediate formed after the departure of the leaving group $(Br^-)$.
$1$. $C_6H_5CH_2^+$: Primary benzylic carbocation (stabilized by one phenyl ring).
$2$. $C_6H_5CH(CH_3)^+$: Secondary benzylic carbocation (stabilized by one phenyl ring and one methyl group).
$3$. $(C_6H_5)_2CH^+$: Secondary benzylic carbocation (stabilized by two phenyl rings).
$4$. $(C_6H_5)_2C(CH_3)^+$: Tertiary benzylic carbocation (stabilized by two phenyl rings and one methyl group).
Comparing stability: $C_6H_5CH_2^+ < C_6H_5CH(CH_3)^+ < (C_6H_5)_2CH^+ < (C_6H_5)_2C(CH_3)^+$.
Thus,the reactivity order is $C_6H_5CH_2Br < C_6H_5CH(CH_3)Br < (C_6H_5)_2CHBr < (C_6H_5)_2C(CH_3)Br$.

(C) The reaction of alkyl chlorides or bromides with sodium iodide $(NaI)$ in the presence of acetone is known as the Finkelstein reaction.
This is a halogen exchange reaction used to prepare alkyl iodides.

(A) The reaction involves a strong base/nucleophile,sodium isopropoxide $(CH_3-CH(CH_3)-ONa)$,and a tertiary alkyl halide,tert-butyl chloride $((CH_3)_3C-Cl)$.
Since the alkyl halide is tertiary and the reagent is a strong base,the reaction will proceed via an $E2$ elimination mechanism rather than an $S_N2$ substitution mechanism due to steric hindrance.
The base will abstract a proton from the tert-butyl chloride,leading to the formation of isobutylene ($2$-methylpropene) as the major product.
The reaction is: $(CH_3)_3C-Cl + CH_3-CH(CH_3)-ONa \xrightarrow{\Delta} CH_3-C(CH_3)=CH_2 + CH_3-CH(CH_3)-OH + NaCl$.

(A) The rate of hydrolysis of a halide is directly proportional to the stability of the carbocation formed after the departure of the leaving group $(Br^-)$.
$1.$ Cycloheptatrienyl bromide forms a cycloheptatrienyl cation $(C_7H_7^+)$,which is aromatic ($6\pi$ electrons,Huckel's rule $4n+2$ where $n=1$).
$2.$ Cyclopentadienyl bromide forms a cyclopentadienyl cation,which is anti-aromatic ($4\pi$ electrons).
$3.$ Cyclopropyl bromide forms a cyclopropyl cation,which is highly unstable due to ring strain and $sp^2$ hybridization at the ring carbon.
$4.$ Bromobenzene forms a phenyl cation,which is highly unstable.
Since the cycloheptatrienyl cation is aromatic and highly stable,cycloheptatrienyl bromide undergoes hydrolysis most rapidly.

(D) The rate of $S_N2$ reactions is inversely proportional to the steric hindrance around the electrophilic carbon atom.
As the number of alkyl groups attached to the carbon bearing the chlorine atom increases,the steric hindrance increases,making the approach of the nucleophile more difficult.
The order of steric hindrance is: $CH_3Cl$ (methyl) < $CH_3CH_2Cl$ (primary) < $(CH_3)_2CHCl$ (secondary) < $(CH_3)_3CCl$ (tertiary).
Therefore,the order of reactivity in $S_N2$ reactions is: $CH_3Cl > CH_3CH_2Cl > (CH_3)_2CHCl > (CH_3)_3CCl$.

(D) Step $1$: Benzene reacts with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ (Friedel-Crafts alkylation) to form Toluene $(A)$.
Step $2$: Toluene $(A)$ reacts with $Cl_2$ in the presence of $h\nu$ (free radical substitution) to form Benzyl chloride $(B = C_6H_5CH_2Cl)$.
Step $3$: Benzyl chloride $(B)$ reacts with $AgCN$. Since $AgCN$ is a covalent compound,the nitrogen atom acts as the nucleophile,leading to the formation of Benzyl isocyanide $(C = C_6H_5CH_2NC)$.
Therefore,the correct option is $D$.

(D) The rate and yield of the $S_{N}2$ reaction are highly dependent on the steric hindrance around the electrophilic carbon atom. The reactivity order for $S_{N}2$ is: $\text{Methyl} > \text{Primary} > \text{Secondary} > \text{Tertiary}$.
- Option $A$ and $C$ involve tertiary alkyl halides,which undergo $E2$ elimination rather than $S_{N}2$ substitution due to high steric hindrance.
- Option $B$ involves a secondary alkyl halide,which is also prone to elimination.
- Option $D$ involves a primary alkyl halide $(CH_3CH_2-Br)$ in a polar aprotic solvent $(DMSO)$,which is the ideal condition for an $S_{N}2$ reaction. Therefore,the yield via $S_{N}2$ is maximum in this case.

(C) In option $A$,the reaction of methylenecyclohexane with $HCl$ follows Markovnikov's rule,forming a tertiary carbocation intermediate,which then reacts with $Cl^-$ to give $1-$chloro$-1-$methylcyclohexane. This is correct.
In option $B$,the reaction of $1-$methylcyclohexene with $HCl$ also proceeds via a tertiary carbocation intermediate at the $C-1$ position,leading to $1-$chloro$-1-$methylcyclohexane. This is correct.
In option $C$,the reaction of $3-$methylcyclohexene with $HCl$ involves the formation of a secondary carbocation. Due to the possibility of carbocation rearrangement (hydride shift),the major product is actually $1-$chloro$-1-$methylcyclohexane,not $3-$chloro$-1-$methylcyclohexane. Therefore,the product shown in option $C$ is incorrect.

(A) The rate of $S_{N}2$ reaction depends on the steric hindrance and the nature of the leaving group.
Steric hindrance follows the order: $primary > secondary > tertiary$.
In the given options:
$(I)$ is $1$-iodobutane (primary alkyl halide with a good leaving group $I^-$).
$(II)$ is $1$-chlorobutane (primary alkyl halide with a poorer leaving group $Cl^-$).
$(III)$ is chlorocyclohexane (secondary alkyl halide).
$(IV)$ is cyclohexylmethyl chloride (primary alkyl halide,but with steric hindrance from the cyclohexane ring near the reaction center).
Comparing $(I)$ and $(II)$,both are primary,but $I^-$ is a better leaving group than $Cl^-$,making $(I)$ more reactive.
Comparing $(I)$ and $(IV)$,$(I)$ is less sterically hindered than $(IV)$.
Therefore,$(I)$ undergoes $S_{N}2$ reaction at the fastest rate.

(D) Reaction $(A)$ is Hydroboration-Oxidation,which follows Anti-Markownikoff addition of water across the double bond.
Reaction $(B)$ is Oxymercuration-Demercuration,which follows Markownikoff addition of water across the double bond.
In reaction $(B)$,the starting material $CH_3-CH=CH-Ph$ is an unsymmetrical alkene. The addition of $H_2O$ follows the Markownikoff rule,resulting in the formation of a chiral center at the carbon atom attached to the phenyl group.
Since the reaction proceeds through a cyclic mercurinium ion intermediate,the attack of water can occur from either side,leading to the formation of a racemic mixture of the product $CH_3-CH(OH)-CH_2-Ph$ (or $CH_3-CH_2-CH(OH)-Ph$ depending on regioselectivity,but the formation of a chiral center leads to racemization).

(B) Chloroform $(CHCl_3)$ is slowly oxidized by air in the presence of light to form an extremely poisonous gas called phosgene $(COCl_2)$.
$2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$
To prevent this oxidation,chloroform is stored in dark-colored bottles,and a small amount of ethanol $(C_2H_5OH)$ is added to it.
Ethanol acts as a stabilizer by converting any phosgene formed into harmless diethyl carbonate:
$COCl_2 + 2C_2H_5OH \rightarrow (C_2H_5O)_2CO + 2HCl$.

(C) The reaction of an alkyl halide with aqueous $KOH$ is a nucleophilic substitution reaction.
$C_2H_5Br + \text{aq. } KOH \rightarrow C_2H_5OH + KBr$
Here,the hydroxide ion $(OH^-)$ acts as a nucleophile and replaces the bromide ion $(Br^-)$ to form ethanol $(C_2H_5OH)$.

(D) Racemization occurs during the alkaline hydrolysis of an alkyl halide if the reaction proceeds via an $S_N1$ mechanism,which involves the formation of a planar carbocation intermediate. This requires the starting alkyl halide to be chiral (optically active). Among the given options,$CH_3-CH_2-CH(Cl)-CH_3$ ($2$-chlorobutane) is a chiral molecule because the carbon atom attached to the chlorine is bonded to four different groups $(-H, -Cl, -CH_3, -CH_2CH_3)$. Therefore,it can form a planar carbocation and undergo racemization.

(C) The rate of $S_{N}2$ reactions is significantly enhanced in polar aprotic solvents.
Polar aprotic solvents like $DMSO$ (Dimethyl sulfoxide) do not form hydrogen bonds with the nucleophile,leaving it 'naked' and more reactive.
In contrast,polar protic solvents like $CH_{3}OH$ and $H_{2}O$ solvate the nucleophile through hydrogen bonding,which decreases its nucleophilicity and slows down the $S_{N}2$ reaction.

(C) In the given reaction,the bromine atom $(-Br)$ in tert-butyl bromide is replaced by a hydroxyl group $(-OH)$ from water.
Since the incoming group is a nucleophile $(-OH^-)$,this is a nucleophilic substitution reaction.
Specifically,it follows an $S_N1$ mechanism because the substrate is a tertiary alkyl halide.

(C) The reaction of toluene with $Cl_{2}$ in the presence of $hv$ (sunlight) is a free radical substitution reaction that occurs at the side chain,producing benzyl chloride $(A)$.
$C_{6}H_{5}CH_{3} + Cl_{2} \xrightarrow{hv} C_{6}H_{5}CH_{2}Cl + HCl$
Here,$A$ is benzyl chloride $(C_{6}H_{5}CH_{2}Cl)$.
When benzyl chloride $(A)$ is treated with $H_{2}O$ (hydrolysis),it undergoes nucleophilic substitution to form benzyl alcohol $(B)$.
$C_{6}H_{5}CH_{2}Cl + H_{2}O \xrightarrow{\Delta} C_{6}H_{5}CH_{2}OH + HCl$
Thus,$A$ is benzyl chloride and $B$ is benzyl alcohol.

(C) The reaction of an alkyl halide $(R-X)$ with dry silver oxide $(Ag_2O)$ upon heating is a method for the preparation of ethers (Williamson ether synthesis variant).
The chemical equation is:
$2 R-X + Ag_2O \xrightarrow{\Delta} R-O-R + 2 AgX$
Comparing this with the given reaction $2 A + \text{dry silver oxide} \xrightarrow{\Delta} \text{ether} + 2 \operatorname{Ag} X$,we can conclude that $A$ represents an alkyl halide $(R-X)$.

(B) Since the obtained compound reduces Tollen's reagent,it must be an aldehyde.
Hydrolysis of a gem-dihalide (where both halogen atoms are on the same carbon) yields a gem-diol,which is unstable and loses a water molecule to form an aldehyde or ketone.
For an aldehyde to be formed,the dihaloalkane must be a terminal gem-dihalide (i.e.,the $-Cl$ atoms are at the $C_{1}$ position).
Thus,$CH_{3}CH_{2}CHCl_{2}$ ($1,1$-dichloropropane) on hydrolysis gives propanal $(CH_{3}CH_{2}CHO)$,which reduces Tollen's reagent.
The reaction is:
$CH_{3}CH_{2}CHCl_{2}$ $\xrightarrow{H_{2}O} CH_{3}CH_{2}CH(OH)_{2}$ $\xrightarrow{-H_{2}O} CH_{3}CH_{2}CHO$ $\xrightarrow{\text{Tollen's reagent}} CH_{3}CH_{2}COOH + Ag \downarrow$

(C) $KCN$ is an ionic compound,so it provides $CN^-$ ions which act as an ambident nucleophile. The carbon atom is more nucleophilic,leading to the formation of alkyl cyanide $(C_2H_5CN)$.
Therefore,the reaction $C_2H_5Br + KCN \rightarrow C_2H_5NC + KBr$ does not occur as written; it should produce $C_2H_5CN$ (ethyl cyanide).
$AgCN$ is covalent,so the nitrogen atom is the nucleophilic site,leading to the formation of alkyl isocyanide $(C_2H_5NC)$.
$AgNO_2$ is covalent,leading to nitroalkane $(C_2H_5NO_2)$.
$KNO_2$ is ionic,leading to alkyl nitrite $(C_2H_5ONO)$.

(A) The $C-Br$ bond length in methyl bromide $(CH_3Br)$ is experimentally determined to be approximately $193 \ pm$. This value is consistent with the covalent radii of carbon $(77 \ pm)$ and bromine $(114 \ pm)$.

(C) The boiling point of haloalkanes depends on the magnitude of van der Waals forces,which increase with an increase in the molecular mass of the halogen atom.
Since the molecular mass of iodine is the highest among the given halogens,$CH_3I$ has the strongest intermolecular forces.
Therefore,the order of boiling points is $CH_3I > CH_3Br > CH_3Cl > CH_3F$.
Thus,$CH_3I$ (Iodomethane) has the highest boiling point.

(B) For a given alkyl group,the boiling point increases with the increasing atomic mass of the halogen atom.
As the size of the halogen atom increases,the magnitude of van der Waals forces increases,leading to a higher boiling point.
Therefore,the boiling point of the given alkyl halides decreases in the order: $CH_3I > CH_3Br > CH_3Cl > CH_3F$.
Thus,$CH_3F$ (Fluoromethane) has the lowest boiling point.

(A) The alkaline hydrolysis of tert-butyl bromide with aqueous alkali such as $NaOH$ or $KOH$ follows an $SN^1$ mechanism.
This is a two-step process:
Step $1$: $(CH_3)_3C-Br \rightarrow (CH_3)_3C^+ + Br^-$ (Slow,rate-determining step)
Step $2$: $(CH_3)_3C^+ + OH^- \rightarrow (CH_3)_3C-OH$ (Fast step)
In an energy profile diagram for a multi-step reaction,the highest peak corresponds to the transition state of the rate-determining step.
Since the first step is the rate-determining step,the highest energy barrier (peak) in the energy profile diagram represents the transition state of the $1^{st}$ step.

(A) The Grignard reagent is an organometallic compound with the general formula $R-Mg-X$,where $R$ is an organic group and $X$ is a halogen. It is prepared by the reaction of an alkyl or aryl halide with metallic magnesium in the presence of dry ether.

(A) The boiling point of haloalkanes increases with an increase in the molecular mass due to the greater magnitude of $Van \ der \ Waals$ forces of attraction.
Since $CH_3F$ (Fluoromethane) has the lowest molecular mass among the given compounds,it experiences the weakest $Van \ der \ Waals$ forces and thus has the lowest boiling point.

(D) molecule is chiral if it contains a chiral center,which is a carbon atom bonded to four different groups.
$1$. $2-$Chloropropane: $CH_3-CHCl-CH_3$ (The central carbon is bonded to two identical methyl groups,so it is achiral).
$2$. $2-$Chloro$-2-$methylbutane: $CH_3-CCl(CH_3)-CH_2-CH_3$ (The central carbon is bonded to two identical methyl groups,so it is achiral).
$3$. $3-$Chloro$-3-$methylbutane: $CH_3-CH_2-CCl(CH_3)-CH_3$ (The central carbon is bonded to two identical methyl groups,so it is achiral).
$4$. $2-$Chloropentane: $CH_3-CHCl-CH_2-CH_2-CH_3$ (The $C-2$ carbon is bonded to four different groups: $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_2CH_3$,so it is chiral).
Therefore,the correct option is $D$.

(C) chiral carbon atom is defined as a carbon atom that is bonded to four different groups or atoms.
Let us analyze the structures:
$A$. $2-$Iodopropane: $CH_3-CH(I)-CH_3$. The central carbon is attached to two identical $-CH_3$ groups,so it is achiral.
$B$. $2-$Iodo$-2-$methylbutane: $CH_3-C(I)(CH_3)-CH_2-CH_3$. The central carbon is attached to two identical $-CH_3$ groups,so it is achiral.
$C$. $2-$Iodo$-3-$methylbutane: $CH_3-CH(I)-CH(CH_3)_2$. The carbon at position $2$ is attached to four different groups: $-H$,$-I$,$-CH_3$,and $-CH(CH_3)_2$. Thus,it is a chiral molecule.
$D$. $3-$Iodopentane: $CH_3-CH_2-CH(I)-CH_2-CH_3$. The carbon at position $3$ is attached to two identical $-CH_2CH_3$ groups,so it is achiral.
Therefore,the correct option is $C$.

(B) chiral molecule is one that contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$n$-Butyl bromide: $CH_3-CH_2-CH_2-CH_2Br$ (No chiral carbon).
$sec$-Butyl bromide: $CH_3-CH(Br)-CH_2-CH_3$. The carbon atom marked with an asterisk $({}^*)$ is bonded to four different groups: $-H, -CH_3, -Br, -CH_2CH_3$. Thus,it is chiral.
Isobutyl bromide: $(CH_3)_2CH-CH_2Br$ (No chiral carbon).
$tert$-Butyl bromide: $(CH_3)_3CBr$ (No chiral carbon).
Therefore,$sec$-butyl bromide is the chiral molecule.

(C) molecule is chiral if it contains at least one chiral carbon atom (a carbon atom bonded to four different groups).
$A$. $2-$Bromopropane: $CH_3-CH(Br)-CH_3$. The central carbon is bonded to two identical methyl groups. It is achiral.
$B$. $2-$Bromo$-2-$methylbutane: $CH_3-C(Br)(CH_3)-CH_2-CH_3$. The central carbon is bonded to two identical methyl groups. It is achiral.
$C$. $2-$Bromo$-3-$methylbutane: $CH_3-CH(Br)-CH(CH_3)_2$. The carbon at position $2$ is bonded to four different groups: $-H$,$-Br$,$-CH_3$,and $-CH(CH_3)_2$. Thus,it is chiral.
$D$. $3-$Bromopentane: $CH_3-CH_2-CH(Br)-CH_2-CH_3$. The central carbon is bonded to two identical ethyl groups. It is achiral.
Therefore,the correct option is $C$.

(A) compound is optically inactive if it does not contain a chiral center (an asymmetric carbon atom bonded to four different groups).
$1.$ $2-$Chloro-$2-$methylbutane: The $C-2$ carbon is bonded to two identical methyl groups,so it is not chiral. Thus,it is optically inactive.
$2.$ $3-$Chlorohexane: The $C-3$ carbon is bonded to $-H$,$-Cl$,$-CH_2CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is chiral and optically active.
$3.$ $2-$Chloro-$3-$methylbutane: The $C-2$ carbon is bonded to $-H$,$-Cl$,$-CH_3$,and $-CH(CH_3)_2$. Since all four groups are different,it is chiral and optically active.
$4.$ $2-$Chloropentane: The $C-2$ carbon is bonded to $-H$,$-Cl$,$-CH_3$,and $-CH_2CH_2CH_3$. Since all four groups are different,it is chiral and optically active.
Therefore,$2-$Chloro-$2-$methylbutane is the optically inactive compound.

(B) The reaction of $tert$-butyl bromide $((CH_3)_3CBr)$ with silver fluoride $(AgF)$ is a Swarts reaction,which is used for the synthesis of alkyl fluorides.
However,$tert$-butyl bromide is a tertiary alkyl halide. When heated with $AgF$,it undergoes an $E1$ elimination reaction rather than a nucleophilic substitution ($S_N1$ or $S_N2$) because the tertiary carbocation is highly stable and the fluoride ion acts as a base.
The major product formed is $2$-methylpropene (isobutylene) due to the elimination of $HBr$.