Q.Which alkyl halide from the following pairs would you expect to react more rapidly by an $S_N2$ mechanism? Explain your answer.

Concept and Intuition

Nitrite, $NO_2^-$, is an ambident nucleophile — it can bond to an electrophile through either its nitrogen or one of its oxygens, giving two different constitutional products from the same formal reagent. The identity of the counter-cation changes which end reacts: with the more ionic alkali-metal nitrites (Na/KNO$_2$), the reaction tends toward O-attack, giving alkyl nitrites (esters, R–O–N=O); with the more covalent silver salt $AgNO_2$, the reaction instead proceeds through nitrogen, giving the nitroalkane R–NO$_2$ as the major product — mirroring the well-known $KCN$ (C-attack, nitrile) vs $AgCN$ (N-attack, isocyanide) contrast for cyanide.

Step-by-Step Solution

1. The question states the major product's formula as $C_4H_9NO_2$, i.e. a nitroalkane (R–NO$_2$), not an alkyl nitrite ester — this identifies the reagent needed as $AgNO_2$, which favours N-attack.
2. The arrow is explicitly labelled $S_N2$: a clean, single-step backside-attack substitution with inversion at the reacting carbon and, crucially, no carbocation intermediate — so no skeletal rearrangement is possible.
3. Taking $C_4H_9Br$ in its default (straight-chain) reading as $n$-butyl bromide, $S_N2$ attack by $AgNO_2$ gives 1-nitrobutane, $CH_3CH_2CH_2CH_2NO_2$, with the carbon skeleton unchanged.
4. $Sn/HCl$ is a standard reducing system for a nitro group to a primary amine (via nitroso and hydroxylamino intermediates, net six-electron reduction): $R-NO_2 \xrightarrow{Sn/HCl} R-NH_2$ — this step does not touch the carbon skeleton at all.
5. So B = $n$-butylamine, $CH_3CH_2CH_2CH_2NH_2$.

Common Mistakes

• Picking $KNO_2$ because it "sounds textbook standard" — for this ambident nucleophile, it is specifically the silver salt that gives the nitro compound as major product, the opposite pattern to what many students first guess.
• Choosing a rearranged product like tert-butylamine — that would require a carbocation intermediate ($S_N1$-type), which the explicitly labelled $S_N2$ mechanism rules out.

ANSWER: A

Concept and Intuition

Isonitriles ($\mathrm{R-NC}$) and nitriles ($\mathrm{R-CN}$) are structural isomers formed from the ambident cyanide ion $\mathrm{CN^-}$, which can attack an electrophile through either its carbon end or its nitrogen end. With KCN (essentially ionic), the more nucleophilic and less electronegative carbon end attacks preferentially, giving the nitrile as major product. With AgCN, the compound is largely covalent — silver is bonded to the carbon of $\mathrm{CN^-}$ (soft–soft Ag–C interaction), which ties up the carbon and leaves the lone pair on nitrogen free to act as the nucleophile. So AgCN + R–X gives the isocyanide as the major product. Separately, the carbylamine reaction — a 1° amine treated with chloroform and alcoholic KOH — is a completely different, classical route to isocyanides: KOH generates the electrophilic carbene $:\mathrm{CCl_2}$ from $\mathrm{CHCl_3}$, which is attacked by the amine nitrogen's lone pair, and after loss of HCl twice, gives $\mathrm{R-NC}$ directly. This reaction is in fact used as a qualitative test for primary amines (the isocyanide has a distinctive foul smell).

Step-by-Step Solution

1. Statement I: $\mathrm{CH_3CH_2Cl}$ + $\mathrm{AgCN}$. Because AgCN is covalent (Ag–C≡N), the nitrogen lone pair performs the $S_N2$ attack on the alkyl halide's carbon, displacing $\mathrm{Cl^-}$, giving $\mathrm{CH_3CH_2-N \equiv C}$ (ethyl isocyanide) as the major product. Correctly matched.
2. Statement II: $\mathrm{CH_3CHO}$ + $\mathrm{NH_2OH}$ gives the oxime $\mathrm{CH_3-CH=N-OH}$ (a condensation, loss of $\mathrm{H_2O}$). Treating the oxime with $(\mathrm{CH_3CO})_2\mathrm{O}$ (a dehydrating agent) eliminates a second water molecule across the C=N–OH system to give the nitrile $\mathrm{CH_3-C\equiv N}$ — not the isonitrile. Incorrectly matched for the stated target.
3. Statement III: $\mathrm{CH_3CH_2NH_2}$ + $\mathrm{CHCl_3}/\mathrm{OH^-}, \Delta$ is exactly the carbylamine reaction, giving $\mathrm{CH_3CH_2-NC}$ (ethyl isocyanide) directly. Correctly matched.
4. So only I and III correctly produce ethyl isonitrile as the major product.

Common Mistakes

• Confusing KCN (gives nitrile, C-attack) with AgCN (gives isonitrile, N-attack) — students often assume any cyanide salt gives the same product.
• Thinking the oxime pathway (II) also gives an isonitrile just because nitrogen is involved — it actually dehydrates straight to the nitrile, skipping the isocyanide altogether.

ANSWER: A

Concept and Intuition

Cyanide salts react with alkyl halides at either the carbon end (giving a nitrile, R-CN) or the nitrogen end (giving an isocyanide/carbylamine, R-NC), and which end attacks depends on the counter-cation. $KCN$ is largely ionic, so the more nucleophilic carbon end of $CN^-$ attacks, giving the nitrile. $AgCN$ is more covalent (Ag–C bond character ties up the carbon), forcing the alkyl halide to be attacked through nitrogen, giving the isocyanide. Isocyanides, $R-N\equiv C$, are then reduced by $H_2$/catalyst by sequential addition across both the $\pi$ bonds of the N$\equiv$C triple bond, ending in a secondary amine $R-NH-CH_3$ (the terminal carbon becomes a methyl group bonded to N).

Step-by-Step Solution

1. $C_2H_4 + HBr \rightarrow CH_3CH_2Br$ (simple electrophilic addition of HBr to ethylene; both carbons are equivalent so there's no regiochemistry issue).
2. $CH_3CH_2Br + AgCN \rightarrow CH_3CH_2-NC$ (ethyl isocyanide) — this is X. AgCN's covalent Ag–C bond makes the nitrogen end of cyanide the nucleophile, giving the isocyanide rather than the nitrile that KCN would give.
3. $CH_3CH_2-NC \xrightarrow{H_2/\text{catalyst}} CH_3CH_2-NH-CH_3$ — catalytic hydrogenation reduces the isocyanide's $N\equiv C$ triple bond fully, converting the terminal carbon (originally the isocyanide carbon) into a $CH_3$ group attached to nitrogen, and giving a secondary amine.
4. The product Y is $CH_3CH_2-NH-CH_3$, i.e. N-ethyl-N-methylamine / ethyl methyl amine, matching option (D).

Common Mistakes

• Using KCN's reaction outcome (nitrile, then reduced to a primary amine with an extra $CH_2$, n-propylamine) instead of recognising that AgCN specifically gives the isocyanide pathway — this is the classic trap (confusing with option A, n-propyl amine).
• Forgetting that reducing an isocyanide gives a secondary amine (with an N-methyl group), not a primary amine.

ANSWER: D

Concept and Intuition

Cyanide is an ambident nucleophile: with alkyl halides in $KCN$ it attacks through carbon to give a nitrile ($R-CN$), while the carbylamine (isocyanide) test attacks through nitrogen to give an isocyanide ($R-NC$) — these are structural isomers, not the same compound. Nitriles can also be made by dehydrating a primary amide (losing water from $-CONH_2$) or by dehydrating an aldoxime (itself made from an aldehyde + hydroxylamine), both of which retain the original carbon count of the starting compound.

Step-by-Step Solution

1. $C_2H_5Cl + KCN \to C_2H_5-CN$ (X = propanenitrile) via C-attack (SN2).
2. Route I: $C_2H_5NH_2 \xrightarrow{CHCl_3/KOH,\Delta} C_2H_5NC$ — this is the carbylamine reaction, producing the isocyanide $R-NC$, NOT the nitrile $R-CN$. So I does not give X.
3. Route II: $C_2H_5CONH_2$ on dehydration loses $H_2O$ to directly give $C_2H_5CN$ — same compound as X.
4. Route III: $C_2H_5CHO \xrightarrow{NH_2OH} C_2H_5CH=NOH$ (aldoxime) $\xrightarrow{(CH_3CO)_2O,-H_2O} C_2H_5CN$ — again the same nitrile as X.
5. So only II and III genuinely reproduce X; I gives an isomeric but different compound.

Common Mistakes

• Treating the carbylamine product ($R-NC$) as 'the same as' the nitrile ($R-CN$) just because they share a molecular formula — they are different functional groups (isocyanide vs nitrile) with different chemistry and different smells/toxicity/reactivity.

ANSWER: B

Concept and Intuition

Cyanide ion, $CN^-$, is an ambident nucleophile — it can attack through either its carbon or its nitrogen atom, giving two different products from the same alkyl halide:

• $R-X + CN^- \to R-CN$ (alkyl cyanide/nitrile) — attack through carbon.
• $R-X + CN^- \to R-NC$ (alkyl isocyanide/carbylamine) — attack through nitrogen.

Which product dominates depends on the counter-ion/solvent:

• KCN (or NaCN) in a polar solvent like ethanol is largely ionic, so the more nucleophilic carbon end of $CN^-$ attacks preferentially — nitrile is the major product.
• AgCN is much more covalent (Ag–C bond character), which leaves the nitrogen lone pair more available/nucleophilic — isocyanide is the major product with AgCN.

The two products differ sharply in what their hydrolysis gives:

• Nitrile hydrolysis: $R-CN + 2H_2O \xrightarrow{H^+/OH^-} RCOOH + NH_3$ — releases ammonia gas, a colourless pungent gas that turns red litmus blue (basic gas) — matching the clue in the question.
• Isocyanide hydrolysis: $R-NC + 2H_2O \to RNH_2 + HCOOH$ — releases an amine and formic acid, not free ammonia gas in the same simple sense.

Step-by-Step Solution

1. $C_3H_7Cl$ (n-propyl chloride) + $CN^-$ replaces Cl with CN, giving a 4-carbon, 1-nitrogen product $C_4H_7N$ — consistent with either $CH_3CH_2CH_2-CN$ (nitrile) or $CH_3CH_2CH_2-NC$ (isocyanide); both share the molecular formula $C_4H_7N$.
2. The clue "Y on hydrolysis released gas which turns red litmus blue" identifies free ammonia, which is released specifically from nitrile hydrolysis.
3. So Y must be the nitrile, $CH_3CH_2CH_2-CN$ (butanenitrile/butyronitrile), meaning the reaction favoured carbon-attack.
4. Carbon-attack (giving the nitrile as major product) happens with KCN in ethanol (more ionic cyanide source), not with AgCN.
5. So X = KCN/C$_2$H$_5$OH, Y = $CH_3CH_2CH_2-CN$ — option (B).

Common Mistakes

• Mixing up which cyanide source (KCN vs AgCN) favours which product — remembering "Ag is more covalent → N attacks" is the key trigger.
• Forgetting that isocyanide hydrolysis does not directly release ammonia gas, so it wouldn't fit the litmus clue.

ANSWER: B

Concept and Intuition

Cyanide is an ambident nucleophile — it can attack an electrophile through either its carbon or its nitrogen lone pair. With ionic $KCN$/$NaCN$, the free $CN^-$ ion attacks predominantly through carbon (since a C-C bond is more stable than a C-N bond), giving alkyl cyanides (nitriles) as the major product. With $AgCN$, however, the Ag-C bond is largely covalent, tying up the carbon and leaving the nitrogen lone pair free to act as the nucleophile, giving isocyanides (isonitriles) as the major product.

Step-by-Step Solution

1. Evaluate the Assertion: 1-bromopentane reacting with $AgCN$ indeed gives pentyl isocyanide ($C_5H_{11}-NC$) as the major product — this matches known behaviour of $AgCN$ in nucleophilic substitutions, so A is TRUE.
2. Evaluate the Reason: the reason claims "AgCN is mainly ionic in nature" — this is factually incorrect. $AgCN$ is predominantly covalent (Ag-CN bond has significant covalent character), which is precisely why the reaction proceeds via nitrogen attack (isocyanide) rather than carbon attack (nitrile). So R is FALSE.
3. Since A is true and R is false, the correct combination is option (C).

Common Mistakes

• Assuming the assertion and reason must both be evaluated as "linked," without independently checking the factual correctness of R — R directly contradicts the established explanation (AgCN being covalent, not ionic).
• Confusing the roles: it's $KCN$/$NaCN$ that are ionic (giving nitriles via C-attack); $AgCN$'s covalent character causes the opposite (N-attack, isocyanide) outcome.

ANSWER: C

Concept and Intuition

$CN^-$ is an ambident nucleophile — it can attack through carbon (giving a nitrile, $R-CN$) or through nitrogen (giving an isocyanide/isonitrile, $R-NC$). With the more ionic, aqueous/ethanolic KCN, C-attack (nitrile) dominates as the major product, but some N-attack (isocyanide) also occurs as a minor product. Isocyanides are hydrolysed (acid-catalyzed) to a primary amine plus formic acid — this reaction is in fact used as a diagnostic (carbylamine test) and as a method to convert an alkyl halide into a primary amine with one carbon degradation avoided (unlike nitrile hydrolysis/reduction, which keeps the extra carbon).

Step-by-Step Solution

1. 1-Bromopropane + ethanolic KCN → major product: $CH_3CH_2CH_2-CN$ (butanenitrile precursor, C-attack); minor product: $CH_3CH_2CH_2-NC$ (propyl isocyanide, N-attack).
2. Hydrolysis of the isocyanide (minor product): $CH_3CH_2CH_2-NC + 2H_2O \xrightarrow{H^+} CH_3CH_2CH_2-NH_2 + HCOOH$.
3. So the product of hydrolysing the minor product is n-propylamine, $CH_3CH_2CH_2-NH_2$.

Common Mistakes

• Hydrolysing the major product (nitrile) instead, which would give propanoic acid/butanoic acid derivatives, not an amine directly.
• Forgetting that isocyanide hydrolysis loses the extra carbon as formic acid, giving an amine with the SAME number of carbons as the original alkyl halide (not one more, as nitrile hydrolysis would).

ANSWER: A