Q.Read the given passage and answer the questions that follow: The substitution reaction of alkyl halide mainly occurs by SN1 or SN2 mechanism. Whatever mechanism alkyl halides follow for the substitution reaction to occur, the polarity of the carbon halogen bond is responsible for these substitution reactions. The rate of SN1 reactions are governed by the stability of carbocation whereas for SN2 reactions steric factor is the deciding factor. If the starting material is a chiral compound, we may end up with an inverted product or racemic mixture depending upon the type of mechanism followed by alkyl halide. Cleavage of ethers with HI is also governed by steric factor and stability of carbocation, which indicates that in organic chemistry, these two major factors help us in deciding the kind of product formed. Predict the stereochemistry of the product formed if an optically active alkyl halide undergoes substitution reaction by SN1 mechanism.

Ethyl bromide is converted to propanenitrile by cyanide substitution, then reduced by $LiAlH_4$ to propan-1-amine — a one-carbon homologation en route to an amine.

Step 1 — formation of A

$KCN$ (predominantly ionic; $CN^-$ attacks through carbon, the less electronegative and more nucleophilic end) displaces bromide from ethyl bromide by an $S_N2$ mechanism:

$CH_3CH_2-Br + KCN \rightarrow CH_3CH_2-C\equiv N\ (A,\ \text{propanenitrile}) + KBr$

This adds a carbon atom to the chain (ethyl $\rightarrow$ propanenitrile), which is why nitrile formation followed by reduction is a standard method of chain-extending to make amines with one extra carbon than the starting haloalkane.

Step 2 — formation of B

$LiAlH_4$ is a powerful reducing agent that reduces the $C\equiv N$ triple bond fully to a $-CH_2-NH_2$ group:

$CH_3CH_2-C\equiv N \xrightarrow{LiAlH_4} CH_3CH_2CH_2NH_2\ (B,\ \text{propan-1-amine})$

Alcoholic silver nitrite ($\text{AgNO}_2$) is an ambident nucleophile — it can attack via either the oxygen or the nitrogen atom. With ethyl bromide, the major product is nitroethane (C–N bond formation) because the nitrogen centre is the more nucleophilic site in the polarisable $\text{NO}_2^-$ ion, and the silver ion helps drive the reaction via an $\text{S}_\text{N}1$-like pathway.

KCN with an alkyl halide gives the alkyl nitrile as the major product, via nucleophilic substitution through carbon.

This is a nucleophilic substitution (SN2) reaction. KCN is predominantly an ionic compound, so the cyanide ion (CN⁻, an ambident nucleophile) is free in solution. Since carbon is the more electronegative and more nucleophilic end of the CN⁻ ion, it attacks the electrophilic carbon of the alkyl halide preferentially through the carbon atom (not the nitrogen).

Reaction:

CH3CH2Br + KCN --(aq. ethanol)--> CH3CH2CN + KBr

The product CH3CH2CN is ethyl cyanide, IUPAC name propanenitrile — it contains one carbon more than the starting alkyl halide, which is why this reaction is used to extend a carbon chain by one carbon.

(Contrast: AgCN is largely covalent, so with AgCN the nitrogen attacks instead, giving the isocyanide/isonitrile as the major product.)

$AgCN$ reacts with alkyl halides through nitrogen (giving isocyanides) because silver's affinity for carbon ties up the $C$ end of the cyanide ion, exposing $N$ as the attacking site — the opposite regiochemistry to the ionic reagent $KCN$.

The cyanide ion, $:C\equiv N:^-$, is ambident — it can attack an electrophile through either its carbon or its nitrogen lone pair.

• $KCN$ is predominantly ionic, releasing a free $CN^-$ ion in solution. Carbon is the softer, more polarizable and kinetically preferred nucleophilic site in a free cyanide ion, so $KCN$ reacts through carbon to give alkyl cyanides (nitriles).
• $AgCN$ is predominantly covalent (Ag has a strong affinity for carbon, $Ag{-}C\equiv N$), which ties up the carbon end. This leaves the nitrogen lone pair as the available nucleophilic site, so $AgCN$ reacts through nitrogen to give alkyl isocyanides (carbylamines) as the major product.

Applying this to the given reaction:

$CH_3CH_2Br + AgCN \xrightarrow{\text{ethanol}} CH_3CH_2{-}N\equiv C + AgBr$

Peroxide-catalysed (anti-Markovnikov) addition of HBr to prop-1-ene puts $-Br$ on the terminal carbon; treating that bromide with silver nitrite then substitutes $-Br$ with $-NO_2$ (the ambident nitrite ion attacking through its more nucleophilic nitrogen atom with a covalent, largely-ionic $Ag-O$ bond), giving 1-nitropropane.

Step 1 — Anti-Markovnikov hydrobromination (Kharasch peroxide effect): In the presence of organic peroxides, addition of $HBr$ to an alkene proceeds by a free-radical chain mechanism rather than the usual ionic (Markovnikov) mechanism. The bromine radical adds first to the terminal (less substituted) carbon of the double bond, generating the more stable secondary radical at the internal carbon, and the sequence of steps places $-Br$ on the terminal carbon as the major product:

$CH_2=CH-CH_3 + HBr \xrightarrow{\text{peroxide}} CH_3-CH_2-CH_2-Br \;\;(\text{1-bromopropane, anti-Markovnikov})$

Step 2 — Nucleophilic substitution with silver nitrite: Silver nitrite ($AgNO_2$) is used (rather than an alkali-metal nitrite like $NaNO_2/KNO_2$) because the $Ag-O$ bond in $AgNO_2$ has significant covalent character, which favours the nitrite ion attacking the substrate through its nitrogen atom (giving the nitro compound as the major product) rather than through oxygen (which would give an unstable alkyl nitrite with $NaNO_2$):

$CH_3CH_2CH_2Br + AgNO_2 \rightarrow CH_3CH_2CH_2NO_2 \;\;(\text{1-nitropropane}) + AgBr\downarrow$

An optically active alkyl halide undergoing SN1 substitution gives a racemic mixture as product because the planar carbocation intermediate allows nucleophilic attack from either face with equal probability.

Both $NO_2^{-}$ and $CN^{-}$ are ambident nucleophiles (can attack through either of two atoms); which atom attacks depends on whether the metal salt used is predominantly ionic (favouring attack via O or C, the more electronegative/terminal end that carries more negative charge) or covalent (favouring attack via the other, more nucleophilic end).

(i) $C_2H_5I + KNO_2$. $NO_2^{-}$ is an ambident nucleophile that can bond through either its N atom (giving a nitro compound, $R\text{–}NO_2$) or through an O atom (giving an alkyl nitrite, $R\text{–O–N=O}$). Potassium nitrite is largely ionic, so the free nitrite ion reacts as a hard/O-nucleophile preferentially, and the major product is the alkyl nitrite:

$C_2H_5I + KNO_2 \longrightarrow C_2H_5\text{–O–N=O}\ (\text{ethyl nitrite}) + KI$

(ii) $C_2H_5I + AgCN$. $CN^{-}$ is likewise ambident, bonding through C (giving a nitrile/cyanide, $R\text{–CN}$) or through N (giving an isocyanide, $R\text{–NC}$). $AgCN$ is largely covalent (Ag–C bond has significant covalent character due to Ag's tendency to bond through soft carbon), so it is the nitrogen lone pair of the (comparatively free) cyanide that becomes the more available nucleophilic site, and the major product with $AgCN$ is the isocyanide:

$C_2H_5I + AgCN \longrightarrow C_2H_5\text{–NC}\ (\text{ethyl isocyanide}) + AgI$

(By contrast, the more ionic $KCN$ gives mainly the nitrile, $C_2H_5CN$, on reaction with alkyl halides — the reverse selectivity from $AgCN$.)

An ambident nucleophile is a nucleophile that has two or more different atoms with lone pairs that can act as the attacking site, leading to different products depending on the reaction conditions. A classic example is the nitrite ion ($\text{NO}_2^-$), which can attack through either the nitrogen atom (forming nitro compounds) or the oxygen atom (forming nitrites).