Properties of Haloalkanes Questions in English

(C) Step $1$: Dehydrohalogenation of $2$-bromopropane with alcoholic $KOH$ leads to the formation of propene $(A)$ via an elimination reaction.
$CH_3-CH(Br)-CH_3 + \text{Alc. KOH} \xrightarrow{\Delta} CH_3-CH=CH_2 (A) + KBr + H_2O$
Step $2$: The addition of $HBr$ to propene in the absence of peroxide follows Markovnikov's rule, where the negative part of the addendum $(Br^-)$ attaches to the carbon atom with fewer hydrogen atoms.
$CH_3-CH=CH_2 + HBr \xrightarrow{\text{Peroxide absent}} CH_3-CH(Br)-CH_3 (B)$
Thus, the major product '$B$' is $2$-bromopropane.

(D) $1$. The rate-determining step of an $S_N1$ reaction is the formation of a carbocation intermediate.
$2$. Polar protic solvents (like $H_2O$ or $ROH$) stabilize the carbocation intermediate and the leaving group through solvation (hydrogen bonding), thereby lowering the activation energy.
$3$. $A$ stronger nucleophile does not affect the rate of an $S_N1$ reaction because the nucleophile is involved only in the second, fast step.
$4$. Therefore, a polar protic solvent increases the rate of an $S_N1$ reaction.

(B) $1$. Dehydrohalogenation follows $Saytzeff's$ rule, which states that the more substituted alkene is the major product.
$2$. The reactant is $2-bromobutane$ $(CH_3-CH_2-CH(Br)-CH_3)$.
$3$. Elimination of $HBr$ can occur from $C_1$ or $C_3$ to form $but-1-ene$ or $but-2-ene$ respectively.
$4$. $But-2-ene$ $(CH_3-CH=CH-CH_3)$ is a disubstituted alkene, making it more stable than the monosubstituted $but-1-ene$ $(CH_3-CH_2-CH=CH_2)$.
$5$. Therefore, $CH_3-CH=CH-CH_3$ is the major product.

(A) When chloroform $(CHCl_3)$ is exposed to air and light, it undergoes slow oxidation to form an extremely poisonous gas called phosgene ($COCl_2$, carbonyl chloride).
The chemical equation is:
$2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$

(C) The reaction between $Bromoethane$ $(C_2H_5Br)$ and silver acetate $(CH_3COOAg)$ is a nucleophilic substitution reaction.
$C_2H_5Br + CH_3COOAg \rightarrow CH_3COOC_2H_5 + AgBr$
In this reaction, the acetate ion $(CH_3COO^-)$ acts as a nucleophile and replaces the bromide ion $(Br^-)$ to form ethyl acetate $(CH_3COOC_2H_5)$.

(C) $1$. The $S_N1$ reaction mechanism proceeds via the formation of a carbocation intermediate.
$2$. The rate-determining step is the formation of this carbocation.
$3$. Bulky alkyl groups around the carbocation center stabilize the positive charge through the inductive effect and hyperconjugation, making the formation of the carbocation easier.
$4$. Therefore, tertiary alkyl halides (which have bulky groups) favor $S_N1$ reactions over primary or secondary alkyl halides.
$5$. Strong nucleophiles favor $S_N2$ reactions, and polar protic solvents are required for $S_N1$ reactions.

(A) Chloroform $(CHCl_3)$ undergoes slow oxidation in the presence of air and light to form a highly poisonous gas known as phosgene $(COCl_2)$.
The chemical equation is:
$2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$

(B) $1$. The $SN^2$ mechanism is a concerted, single-step reaction.
$2$. It involves a backside attack by the nucleophile, leading to an inversion of configuration (Walden inversion), not racemization.
$3$. During the reaction, the nucleophile and the leaving group are both partially bonded to the central carbon atom in a transition state, where the carbon atom is penta-coordinated (bonded to five groups).
$4$. $SN^2$ reactions do not involve the formation of carbocation intermediates.
$5$. Therefore, the correct feature is that it includes a transition state with a penta-coordinated carbon.

(D) $1$. The rate-determining step in an $SN^1$ reaction is the formation of a carbocation by the departure of the leaving group.
$2$. The ease of departure of the leaving group depends on the strength of the $C-X$ bond.
$3$. The bond dissociation energy decreases as the size of the halogen atom increases $(I > Br > Cl > F)$.
$4$. Therefore, the $C-I$ bond is the weakest and breaks most easily, making the iodide ion the best leaving group.
$5$. Thus, $(CH_3)_3C-I$ undergoes $SN^1$ reaction most readily.

(A) The rate of $S_N2$ reaction depends on the steric hindrance around the electrophilic carbon atom.
$1$. The order of reactivity for $S_N2$ reactions is: $Methyl \ halide > Primary \ (1^\circ) > Secondary \ (2^\circ) > Tertiary \ (3^\circ)$.
$2$. $CH_3-Br$ is a methyl halide, which has the least steric hindrance.
$3$. $(CH_3)_3C-Br$ is a tertiary halide, $(CH_3)_2CH-Br$ is a secondary halide, and $CH_3-CH_2-Br$ is a primary halide.
$4$. Therefore, $CH_3-Br$ reacts most rapidly.

(A) $1$. $AgCN$ is a covalent compound.
$2$. In $AgCN$, the nitrogen atom has a lone pair of electrons available for nucleophilic attack, while the carbon atom is not free to act as a nucleophile.
$3$. Therefore, the reaction follows an $S_N2$ mechanism where the nitrogen atom attacks the alkyl halide.
$4$. The reaction is: $CH_3CH_2Br + AgCN \rightarrow CH_3CH_2NC + AgBr$.
$5$. The product $P$ is ethyl isocyanide $(CH_3CH_2NC)$.

(D) The reaction of an alkyl halide with sodium metal in the presence of dry ether is known as the Wurtz reaction.
$2-\text{chloro}-2-\text{methylbutane}$ is a tertiary alkyl halide with the structure $CH_3-CH_2-C(CH_3)(Cl)-CH_3$.
In the Wurtz reaction, two molecules of the alkyl halide couple to form a higher alkane.
Two radicals of $2-\text{methylbutan-2-yl}$ $(CH_3-CH_2-C(CH_3)_2^{\bullet})$ combine at the tertiary carbon atoms.
The resulting structure is $CH_3-CH_2-C(CH_3)_2-C(CH_3)_2-CH_2-CH_3$.
Naming the longest chain: the longest chain has $6$ carbons (hexane), with methyl groups at positions $3$ and $4$.
The $IUPAC$ name is $3,3,4,4-\text{tetramethylhexane}$.

(D) Step $1$: The reaction of an alkyl halide with an aqueous solution of $KOH$ is a nucleophilic substitution reaction.
Step $2$: In this reaction, the hydroxide ion $(OH^-)$ acts as a nucleophile and replaces the chloride ion $(Cl^-)$ from the $2$-chlorobutane.
Step $3$: The reaction is: $CH_3CH_2CHClCH_3 + KOH_{(aq)} \rightarrow CH_3CH_2CH(OH)CH_3 + KCl$.
Step $4$: The product formed is $butan-2-ol$.

(A) $1$. The reaction of alkyl halides $(R-X)$ with silver nitrite $(AgNO_2)$ yields nitroalkanes $(R-NO_2)$ as the major product.
$2$. This occurs because the $Ag-O$ bond is covalent, making the nitrogen atom the nucleophilic center.
$3$. In contrast, reaction with potassium nitrite $(KNO_2)$ is ionic and yields alkyl nitrites $(R-ONO)$ as the major product.
$4$. Therefore, reagent $B$ is $AgNO_2$.

(A) $1$. $Alkyl \ halides$ react with $Silver \ nitrite$ $(AgNO_2)$ to form $Alkyl \ nitrites$ $(R-O-N=O)$ as the major product.
$2$. This occurs because $AgNO_2$ is a covalent compound, and the oxygen atom is more nucleophilic than the nitrogen atom.
$3$. In contrast, $Potassium \ nitrite$ $(KNO_2)$ is an ionic compound, which provides the $NO_2^-$ ion, leading to the formation of $Nitroalkanes$ $(R-NO_2)$ as the major product.

(C) $1$. The $S_N2$ mechanism is a concerted, single-step reaction.
$2$. In this mechanism, the nucleophile attacks from the backside of the carbon atom, leading to an inversion of configuration.
$3$. Bond breaking and bond formation occur simultaneously through a pentacoordinate transition state.
$4$. $S_N2$ reactions do not involve the formation of a carbocation intermediate; carbocation formation is a characteristic of $S_N1$ mechanisms.
$5$. Therefore, the formation of a planar carbocation intermediate is $NOT$ a feature of the $S_N2$ mechanism.

(B) Step $1$: The reaction of $2$-bromopentane with alcoholic $KOH$ is a dehydrohalogenation reaction (an elimination reaction).
Step $2$: According to $Saytzeff's$ rule, in elimination reactions, the more substituted alkene is the major product.
Step $3$: The reaction is: $CH_3-CH(Br)-CH_2-CH_2-CH_3 + KOH (alc.) \rightarrow CH_3-CH=CH-CH_2-CH_3 + KBr + H_2O$.
Step $4$: $Pent-2-ene$ is more substituted than $Pent-1-ene$, hence it is the major product.

(A) $1$. The reaction of alkyl halides $(RX)$ with $AgCN$ is a nucleophilic substitution reaction.
$2$. $AgCN$ is a covalent compound. The carbon atom in $AgCN$ is not free to act as a nucleophile, but the nitrogen atom has a lone pair of electrons available for bonding.
$3$. Therefore, the nitrogen atom attacks the alkyl group, leading to the formation of alkyl isocyanides $(R-NC)$.
$4$. In contrast, $KCN$ is an ionic compound that provides $CN^-$ ions, where the carbon atom acts as the nucleophile, leading to the formation of alkyl cyanides $(R-CN)$.

(B) $S_N1$ reactions proceed via a planar carbocation intermediate.
For a racemic mixture to be formed, the starting alkyl halide must be chiral.
In $2$-bromobutane, the carbon atom attached to the bromine is chiral (bonded to four different groups: $-H, -CH_3, -C_2H_5, -Br$).
Upon hydrolysis, the leaving group departs to form a planar carbocation, which can be attacked by the nucleophile from either side with equal probability, resulting in a racemic mixture.