Q.Identify the products of the following: $CH_3-CH_2-Br \xrightarrow{KCN} A \xrightarrow{LiAlH_4} B$

1. Understand the reagent: ambident nucleophile

   Silver nitrite ($\text{AgNO}_2$) dissociates in alcohol to give $\text{Ag}^+$ and $\text{NO}_2^-$ ions. The nitrite ion has two nucleophilic sites: the nitrogen atom (with a lone pair) and the oxygen atoms (with negative charge). This is the classic example of an ambident nucleophile — it can attack an alkyl halide at either site, leading to different products.

2. Which site attacks?

   In $\text{NO}_2^-$, the negative charge is delocalised over the two oxygen atoms, making them harder (less polarisable). The nitrogen atom, though neutral, has a lone pair and is softer — more polarisable. Ethyl bromide is a primary alkyl halide, but the presence of $\text{Ag}^+$ changes the game. Silver ion coordinates with the bromide, weakening the C–Br bond and favouring an $\text{S}_\text{N}1$-like mechanism (even for a primary halide) because $\text{AgBr}$ precipitates. This creates a carbocation-like transition state, which is more easily attacked by the nitrogen centre (the better nucleophile in a polarisable sense).

3. The major product: nitroethane

   Attack by the nitrogen atom gives $\text{CH}_3\text{CH}_2\text{–NO}_2$, which is nitroethane. This is the major product under these conditions. The oxygen attack would give ethyl nitrite ($\text{CH}_3\text{CH}_2\text{–O–N=O}$), but that is a minor product here because the nitrogen centre is more nucleophilic in the ambident ion.

4. Why not the other options? …

The key to predicting stereochemistry in any substitution reaction lies in understanding the mechanism's intermediate. For SN1, the rate-determining step produces a carbocation — and that carbocation is planar (sp² hybridised). This flat geometry is the entire story behind the stereochemical outcome.

Let’s walk through why.

1. The SN1 mechanism has two distinct steps. First, the leaving group departs, creating a carbocation. This step is slow and rate-determining. Second, the nucleophile attacks this carbocation in a fast step. Because the carbocation forms before the nucleophile arrives, the nucleophile has no "memory" of which side the leaving group was on.

2. The carbocation intermediate is planar. A carbocation has three bonds arranged in a trigonal planar geometry (bond angles ~120°). The empty p orbital sticks out perpendicular to this plane. This means the carbocation is achiral at that carbon — it has no "handedness" left.

3. Nucleophilic attack can occur from either face. The nucleophile can approach the planar carbocation from above the plane or below it with equal ease. There is no steric or electronic bias (assuming the nucleophile is not itself chiral or the solvent is not chiral). So roughly half the attacks happen from one side, half from the other. …

Understanding Ambident Nucleophiles

The word "ambident" comes from Latin — ambi meaning "both" and dent meaning "tooth". So an ambident nucleophile literally has "two teeth" — two different atoms that can bite into an electrophile. This is not the same as having multiple identical attacking sites (like the two oxygens in acetate ion, which are equivalent by resonance). In an ambident nucleophile, the attacking atoms are chemically different, so the product you get depends on which atom does the attacking.

The key idea is that the nucleophile has a delocalised negative charge (or lone pair) spread over two or more different atoms. Which atom actually attacks depends on factors like:

• The hardness or softness of the electrophile (HSAB principle)
• The polarity of the solvent
• The steric hindrance around the attacking sites
• Temperature and other reaction conditions

Step-by-Step Explanation

1. Identify the defining feature of an ambident nucleophile

An ambident nucleophile must have at least two non-equivalent atoms that each possess a lone pair of electrons (or a negative charge) and can form a bond with an electrophile. The atoms are different elements, so the bond formed has different character depending on which atom attacks.

2. Understand why this matters in reactions

When an ambident nucleophile reacts with an alkyl halide or other electrophile, you can get two different products. This is called ambident reactivity. The product distribution is not random — it follows predictable patterns based on the reaction conditions.

3. Take the classic example: the nitrite ion ($\text{NO}_2^-$)

The nitrite ion has the following resonance structures:

The negative charge is delocalised over both oxygen atoms and the nitrogen atom. However, the nitrogen and oxygen are different elements with different properties.

4. Show the two possible attacking sites

• Attack through nitrogen: The lone pair on nitrogen forms a bond with the electrophile. This gives a nitro compound ($\text{R} - \text{NO}_2$).
• Attack through oxygen: The lone pair on one of the oxygen atoms forms the bond. This gives an alkyl nitrite ($\text{R} - \text{O} - \text{N} = \text{O}$).

5. Explain which product forms when — the HSAB principle in action

The nitrogen atom is a softer nucleophilic centre than the oxygen atom (nitrogen is less electronegative and its lone pair is more polarisable). Oxygen is a harder nucleophile.

• With hard electrophiles (like primary alkyl halides in polar protic solvents), the harder oxygen centre tends to attack, giving alkyl nitrites. …