JEE Mains Chemistry · Organic Reaction Mechanisms

Multistep Conversions and Road Maps

A multistep scheme is solved by counting carbons first, marking the steps that add or remove carbon, and then following one functional group through each reagent in turn.

All Organic Reaction Mechanisms notes JEE Mains Chemistry strategy

Why this matters

Nine PYQs, two of them asking for a number. Five build a carbon chain one step at a time: cyanide adding a carbon at each end, a Grignard reagent attacking an ester or formaldehyde, an acetylide alkylated by a bromo-alcohol, a nitrile carbanion added to a ketone, and a carbonyl protected as an acetal. Four are aromatic routes: Friedel-Crafts acylation then Clemmensen reduction, n-heptane aromatised and then oxidised by the Etard reaction, nitro groups turned into iodine through a diazonium salt, and chlorobenzene turned into phenol at 623 K and 300 atm.

Concept 1 of 2: Carbon counting in chain-building steps

Most reagents in a scheme only change a functional group. A few make or break a C–C bond, and those decide the answer. Count the carbons of the start and of each option first; the carbon-changing steps then tell you which options can be right before you work out a single structure.

Definition

Step 1: count the carbons in the starting compound and in the product or options.
Step 2: find the steps that change the count: cyanide (+1 per halogen), a Grignard reagent or acetylide (+ its own carbons), $\mathrm{CO_2}$ on a Grignard (+1); haloform, soda lime and Hofmann bromamide (−1 each).
Step 3: follow the functional group through every other reagent: PCC stops a primary alcohol at the aldehyde, $\mathrm{KMnO_4}$ or Jones reagent goes on to the acid, $\mathrm{NaBH_4}$ and $\mathrm{LiAlH_4}$ reduce C=O to CH–OH, $\mathrm{H_2/Ni}$ reduces $\mathrm{C{\equiv}N}$ to $\mathrm{CH_2NH_2}$ .
A Grignard reagent or an acetylide is destroyed by any O–H or N–H in the molecule; that is why a carbonyl that must survive is protected as an acetal first (ethylene glycol and $\mathrm{H^+}$ ), and aqueous acid removes the acetal at the end.
An acetylide is alkylated only by a primary halide; a secondary or tertiary halide gives elimination.

Reagent	Carbon count change	Group produced	Example
KCN on an alkyl halide	+1 for each halogen replaced	Nitrile; $\mathrm{H_3O^+}$ gives COOH, $\mathrm{H_2/Ni}$ gives $\mathrm{CH_2NH_2}$	$\mathrm{CH_3CH_2Br \to CH_3CH_2CN \to CH_3CH_2COOH}$
RMgX, then HCHO and $\mathrm{H_3O^+}$	+1 on R	Primary alcohol $\mathrm{RCH_2OH}$	$\mathrm{CH_3CH_2MgBr \to CH_3CH_2CH_2OH}$
RMgX, then another aldehyde R′CHO	R joins R′CHO	Secondary alcohol	$\mathrm{C_6H_5MgBr + CH_3CHO \to C_6H_5CH(OH)CH_3}$
RMgX, then a ketone	R joins the ketone	Tertiary alcohol	$\mathrm{CH_3MgBr + CH_3COCH_3 \to (CH_3)_3COH}$
Two RMgX on an ester R′COOEt	Two R groups join; OEt leaves	Tertiary alcohol with two identical R groups	$\mathrm{2\,CH_3MgBr + CH_3COOC_2H_5 \to (CH_3)_3COH}$
RMgX, then $\mathrm{CO_2}$ and $\mathrm{H_3O^+}$	+1 on R	Carboxylic acid RCOOH	$\mathrm{CH_3MgBr \to CH_3COOH}$
$\mathrm{NaNH_2}$ on a terminal alkyne, then a primary RX	+ the carbons of R	Longer internal alkyne	$\mathrm{CH_3C{\equiv}CH \to CH_3C{\equiv}CCH_2CH_3}$ with $\mathrm{CH_3CH_2Br}$
Base on $\mathrm{ArCH_2CN}$ , then a ketone	The α-carbon joins the C=O carbon	β-Hydroxy nitrile	$\mathrm{C_6H_5CH_2CN + CH_3COCH_3 \to (CH_3)_2C(OH)CH(C_6H_5)CN}$
Ethylene glycol and $\mathrm{H^+}$ on a C=O	0 (temporary)	Cyclic acetal, stable to base, Grignard reagents and $\mathrm{NaBH_4}$	Aqueous acid gives the C=O back
$\mathrm{X_2/NaOH}$ on a methyl ketone	−1	Carboxylate and $\mathrm{CHX_3}$	$\mathrm{CH_3COCH_2CH_3 \to CH_3CH_2COO^-}$
Soda lime on RCOONa	−1	Alkane RH	$\mathrm{CH_3CH_2COONa \to CH_3CH_3}$
$\mathrm{Br_2/NaOH}$ on an amide $\mathrm{RCONH_2}$	−1	Primary amine $\mathrm{RNH_2}$	$\mathrm{CH_3CH_2CONH_2 \to CH_3CH_2NH_2}$

Mark the steps that change the carbon count before anything else; every other reagent changes only the group.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q138Moderate

Example 1 · Organic Reaction Mechanisms · Multistep Conversions and Road Maps

Acid

D

formed in above reaction is:

A Grignard reagent attacks an ester twice

After the first addition the ester loses

\mathrm{OEt^-}

and becomes a ketone, which is more reactive than the ester, so a second equivalent adds at once. The product is a tertiary alcohol with two identical groups from the Grignard reagent, never a ketone.

H₂/Ni adds hydrogen, not carbon

Reducing

\mathrm{C{\equiv}N}

\mathrm{CH_2NH_2}

keeps the carbon count; the extra carbon came in with the cyanide. Count it once, at the KCN step.

An O–H in the molecule destroys the carbanion

Grignard reagents and acetylide ions are strong bases. A free OH, COOH or NH in the same reaction protonates them before any C–C bond forms, so in practice such groups are protected first.

Drill 4 more on carbon counting in chain-building steps

Concept 2 of 2: Aromatic road maps: acylation, reduction and diazonium routes

On a benzene ring, a group is put on by electrophilic substitution and then changed step by step. Two habits solve most aromatic schemes. To attach a straight carbon chain, acylate and then reduce the C=O, because direct alkylation rearranges. To place a group that cannot go on directly (I, F, CN, OH), put on a nitro group, reduce it to

\mathrm{NH_2}

, make the diazonium salt and replace

\mathrm{N_2}

Definition

Friedel-Crafts acylation (RCOCl or an anhydride with anhydrous $\mathrm{AlCl_3}$ ): an aryl ketone, with no rearrangement and only one substitution. A cyclic anhydride gives a keto acid.
Clemmensen (Zn–Hg and conc. HCl) or Wolff-Kishner (hydrazine, then KOH in ethylene glycol): C=O to $\mathrm{CH_2}$ , leaving COOH alone.
Intramolecular acylation: an aryl-butanoic acid with acid closes a six-membered ring onto the ring's ortho position (α-tetralone).
Nitro to diazonium: $\mathrm{HNO_3/H_2SO_4}$ ; then Sn/HCl or $\mathrm{H_2/Pd}$ to $\mathrm{ArNH_2}$ ; then $\mathrm{NaNO_2/HCl}$ at 0–5 °C to $\mathrm{ArN_2^+}$ . Then KI gives ArI, CuCl or CuBr gives ArCl or ArBr, CuCN gives ArCN, warm water gives ArOH, $\mathrm{H_3PO_2}$ gives ArH.
Side chains: $\mathrm{CrO_2Cl_2}$ in $\mathrm{CS_2}$ , then $\mathrm{H_3O^+}$ (Etard) turns $\mathrm{ArCH_3}$ into ArCHO; hot alkaline $\mathrm{KMnO_4}$ turns any alkyl side chain with a benzylic H into COOH.
Aromatisation: n-hexane or n-heptane over $\mathrm{Cr_2O_3}$ , $\mathrm{V_2O_5}$ or $\mathrm{Mo_2O_3}$ at 773 K and 10–20 atm gives benzene or toluene.
Dow process: chlorobenzene with NaOH at 623 K and 300 atm gives sodium phenoxide; acid gives phenol.
Order matters: $\mathrm{CH_3}$ , OH and $\mathrm{NH_2}$ direct ortho and para; $\mathrm{NO_2}$ , COOH and C=O direct meta. Friedel-Crafts fails on a ring carrying $\mathrm{NO_2}$ .

Straight-chain alkylbenzene by acylation then reduction

\mathrm{C_6H_6 \xrightarrow{RCOCl,\ AlCl_3} C_6H_5COR \xrightarrow{Zn{-}Hg,\ HCl} C_6H_5CH_2R}

Worked example

Make n-propylbenzene from benzene. Why does 1-chloropropane with

\mathrm{AlCl_3}

not do the job?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 5 Apr 2024 · Q138Moderate

Example 2 · Organic Reaction Mechanisms · Multistep Conversions and Road Maps

Identify

A

and

B

in the given chemical reaction sequence: -

Acylate, then reduce, for a straight chain

Friedel-Crafts alkylation with a primary halide rearranges and can add more than one group. Acylation gives one straight-chain ketone, and Clemmensen or Wolff-Kishner reduction turns it into the straight-chain alkylbenzene.

Clemmensen leaves COOH alone

Zn–Hg and HCl reduce a ketone or aldehyde C=O to

\mathrm{CH_2}

but do not touch a carboxylic acid. A keto acid becomes an acid with the same number of carbons.

Diazotise cold

The diazonium salt is made with

\mathrm{NaNO_2/HCl}

at 0–5 °C. Warmed in water it turns into the phenol, so a scheme that needs ArI or ArCN keeps it cold until the replacing reagent is added.

Drill 3 more on aromatic road maps: acylation, reduction and diazonium routes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (1)

Aromatic road maps: acylation, reduction and diazonium routes
Straight-chain alkylbenzene by acylation then reduction
$\mathrm{C_6H_6 \xrightarrow{RCOCl,\ AlCl_3} C_6H_5COR \xrightarrow{Zn{-}Hg,\ HCl} C_6H_5CH_2R}$

Reference tables (1)

Carbon counting in chain-building steps12 rows

Reagent	Carbon count change	Group produced	Example
KCN on an alkyl halide	+1 for each halogen replaced	Nitrile; $\mathrm{H_3O^+}$ gives COOH, $\mathrm{H_2/Ni}$ gives $\mathrm{CH_2NH_2}$	$\mathrm{CH_3CH_2Br \to CH_3CH_2CN \to CH_3CH_2COOH}$
RMgX, then HCHO and $\mathrm{H_3O^+}$	+1 on R	Primary alcohol $\mathrm{RCH_2OH}$	$\mathrm{CH_3CH_2MgBr \to CH_3CH_2CH_2OH}$
RMgX, then another aldehyde R′CHO	R joins R′CHO	Secondary alcohol	$\mathrm{C_6H_5MgBr + CH_3CHO \to C_6H_5CH(OH)CH_3}$
RMgX, then a ketone	R joins the ketone	Tertiary alcohol	$\mathrm{CH_3MgBr + CH_3COCH_3 \to (CH_3)_3COH}$
Two RMgX on an ester R′COOEt	Two R groups join; OEt leaves	Tertiary alcohol with two identical R groups	$\mathrm{2\,CH_3MgBr + CH_3COOC_2H_5 \to (CH_3)_3COH}$
RMgX, then $\mathrm{CO_2}$ and $\mathrm{H_3O^+}$	+1 on R	Carboxylic acid RCOOH	$\mathrm{CH_3MgBr \to CH_3COOH}$
$\mathrm{NaNH_2}$ on a terminal alkyne, then a primary RX	+ the carbons of R	Longer internal alkyne	$\mathrm{CH_3C{\equiv}CH \to CH_3C{\equiv}CCH_2CH_3}$ with $\mathrm{CH_3CH_2Br}$
Base on $\mathrm{ArCH_2CN}$ , then a ketone	The α-carbon joins the C=O carbon	β-Hydroxy nitrile	$\mathrm{C_6H_5CH_2CN + CH_3COCH_3 \to (CH_3)_2C(OH)CH(C_6H_5)CN}$
Ethylene glycol and $\mathrm{H^+}$ on a C=O	0 (temporary)	Cyclic acetal, stable to base, Grignard reagents and $\mathrm{NaBH_4}$	Aqueous acid gives the C=O back
$\mathrm{X_2/NaOH}$ on a methyl ketone	−1	Carboxylate and $\mathrm{CHX_3}$	$\mathrm{CH_3COCH_2CH_3 \to CH_3CH_2COO^-}$
Soda lime on RCOONa	−1	Alkane RH	$\mathrm{CH_3CH_2COONa \to CH_3CH_3}$
$\mathrm{Br_2/NaOH}$ on an amide $\mathrm{RCONH_2}$	−1	Primary amine $\mathrm{RNH_2}$	$\mathrm{CH_3CH_2CONH_2 \to CH_3CH_2NH_2}$

Mark the steps that change the carbon count before anything else; every other reagent changes only the group.

Watch out for (6)

A Grignard reagent attacks an ester twice
→ Carbon counting in chain-building steps
H₂/Ni adds hydrogen, not carbon
→ Carbon counting in chain-building steps
An O–H in the molecule destroys the carbanion
→ Carbon counting in chain-building steps
Acylate, then reduce, for a straight chain
→ Aromatic road maps: acylation, reduction and diazonium routes
Clemmensen leaves COOH alone
→ Aromatic road maps: acylation, reduction and diazonium routes
Diazotise cold
→ Aromatic road maps: acylation, reduction and diazonium routes

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