Which of the following are used to convert RCHO into $RCH_2OH$? (Two or more than two options may be correct.) (i) $H_2/Pd$ (ii) $LiAlH_4$ (iii) $NaBH_4$ (iv) reaction with RMgX followed by hydrolysis

Which of the following are used to convert RCHO into RCH₂OH? (Two or more than two options may be correct.) (i

NCERT Exemplar · Q56

RCH_2OH

(i)

H_2/Pd

(ii)

LiAlH_4

(iii)

NaBH_4

(iv) reaction with RMgX followed by hydrolysis

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Concept understanding — Electrophilic Addition Reactions

Electrophilic Aromatic Substitution – The First Meeting

Imagine you have a benzene ring — that perfect, flat hexagon of six carbons with alternating double bonds. It's stable, almost stubbornly so. You want to attach something new to it, say a bromine atom or a nitro group. But benzene doesn't react like an alkene. It doesn't just add across a double bond. Instead, it does something more elegant: it kicks out a hydrogen and keeps its aromatic ring intact.

That's the heart of Electrophilic Aromatic Substitution (EAS).

The Intuition: Why "Substitution" and Not "Addition"?

Benzene's stability comes from its delocalised $\pi$ electrons — a cloud above and below the ring. This cloud is electron-rich, so it attracts electrophiles (electron-loving species). But if an electrophile simply added to a double bond, the ring would break its aromaticity, losing that huge stabilisation. That would be energetically costly.

So benzene does something smarter: it lets the electrophile attack, temporarily breaks aromaticity to form a high-energy intermediate (the arenium ion), and then loses a proton to restore the aromatic ring. The net result? A hydrogen is replaced by the electrophile. The ring is back to its stable, aromatic self.

Note

The key trade-off: temporary loss of aromaticity is acceptable because the final product regains it. Addition reactions would permanently destroy aromaticity — benzene avoids that.

The Precise Statement

Electrophilic Aromatic Substitution is a reaction in which an electrophile ( $E^+$ ) replaces a hydrogen atom on an aromatic ring, proceeding through a sigma complex (arenium ion) intermediate, and restoring aromaticity after deprotonation.

The general equation:

\ce{Ar-H + E+ -> Ar-E + H+}

where Ar represents an aromatic ring.

The Mechanism in Three Steps

Generation of the electrophile – Many EAS reactions need a catalyst to create a strong enough $E^+$ . For example, bromination uses $\ce{FeBr3}$ to polarise $\ce{Br2}$ into $\ce{Br^+}$ .
Attack by the aromatic ring – The $\pi$ electrons of benzene attack the electrophile, forming a sigma complex (also called the arenium ion or Wheland intermediate). This intermediate is non-aromatic — it has four $\pi$ electrons delocalised over five carbons, and one $sp^3$ carbon bearing the electrophile and a hydrogen.

\text{Benzene} + E^+ \longrightarrow \text{Sigma complex (non-aromatic)}

Deprotonation – A base (often the counterion of the catalyst, like $\ce{FeBr4-}$ ) removes the proton from the $sp^3$ carbon. The pair of electrons from the C–H bond flows back into the ring, restoring the aromatic sextet.

\text{Sigma complex} + \text{Base} \longrightarrow \text{Product} + \ce{H+}

Watch out

The sigma complex is not aromatic. It's a high-energy intermediate. Students often mistakenly think it's still aromatic — it isn't. That's why the step is fast and the complex is short-lived.

Why This Matters for Exams

EAS is the gateway to understanding how to put groups onto benzene rings. The rate-determining step is usually the formation of the sigma complex (step 2). The regiochemistry (where the electrophile goes) depends on whether the ring already has a substituent — that's the topic of activating/deactivating groups and ortho/para vs. meta directors. …

Why this formula?

Electrophilic Addition Reactions: Why the Mechanism Works

Electrophilic addition is a cornerstone of alkene and alkyne chemistry. Instead of memorising the "arrow pushing," let's understand why the reaction proceeds the way it does — driven by electron density, stability, and charge.

1. The Core Idea: Why Alkenes React This Way

Alkenes have a π-bond — a cloud of electrons above and below the plane of the σ-bond. This π-electron cloud is:

Electron-rich (nucleophilic)
Exposed (not shielded by σ-bonds like in alkanes)

An electrophile (electron-lover) is attracted to this high electron density. The reaction is electrophilic addition because the electrophile attacks first.

Key principle: The π-bond acts as a Lewis base (electron donor). The electrophile is a Lewis acid (electron acceptor).

2. The General Mechanism (Two-Step)

Step 1: Formation of a Carbocation (or Bridged Intermediate)

The electrophile (E⁺) attacks the π-bond. The π-electrons form a new σ-bond to E⁺, leaving the other carbon with a positive charge — a carbocation.

\ce{C=C} + \ce{E+} \longrightarrow \ce{C+-C-E}

Why does this happen?

The π-bond is weaker than a σ-bond (~260 kJ/mol vs ~350 kJ/mol). Breaking the π-bond to form a σ-bond is energetically favourable because the new σ-bond is stronger. The carbocation is a high-energy intermediate, but it's stabilised by:

Hyperconjugation (alkyl groups donate electron density)
Inductive effect (alkyl groups push electrons toward the positive carbon)

Step 2: Nucleophilic Attack

A nucleophile (Nu⁻) attacks the carbocation, forming a second σ-bond.

\ce{C+-C-E} + \ce{Nu-} \longrightarrow \ce{C-Nu-C-E}

Why does this happen?

The carbocation is electron-deficient (positive charge). The nucleophile is electron-rich. Opposite charges attract — this is electrostatic and orbital overlap driven.

3. The Key "Formula" — Markovnikov's Rule

Statement: In the addition of HX to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogens already, and the X attaches to the carbon with fewer hydrogens.

Why does this rule hold? (The reasoning)

Consider propene: $\ce{CH3-CH=CH2}$ + HBr.

Possible carbocations:
1. Primary carbocation: $\ce{CH3-CH+-CH2Br}$ (less stable)
2. Secondary carbocation: $\ce{CH3-CHBr-CH2+}$ (more stable)

The more substituted carbocation (secondary > primary) is more stable due to:

Hyperconjugation: More alkyl groups = more C–H σ-bonds that can donate electron density into the empty p-orbital of the carbocation.
Inductive effect: Alkyl groups are electron-donating, stabilising the positive charge.

Result: The reaction proceeds via the more stable carbocation, leading to Markovnikov addition.

Markovnikov's rule is not a law — it's a consequence of carbocation stability.

4. The "Anti-Markovnikov" Exception (Why It Happens)

With HBr in the presence of peroxides (ROOR), the addition is anti-Markovnikov — Br goes to the less substituted carbon.

Why? The mechanism changes from ionic to free-radical.

Peroxide decomposes to radicals: $\ce{ROOR -> 2 RO•}$
RO• abstracts H from HBr: $\ce{RO• + HBr -> ROH + Br•}$
Br• adds to the alkene — at the less substituted carbon (because the radical formed is more stable — tertiary > secondary > primary).
The new radical abstracts H from another HBr, regenerating Br•. …

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Aldehydes are reduced to primary alcohols by catalytic hydrogenation (H2/Ni, Pt or Pd) and by hydride donors such as LiAlH4 and NaBH4. A Grignard reagent instead adds a new alkyl group across the carbonyl, giving a secondary alcohol, not RCH2OH. The correct options are (i), (ii) and (iii).

Concept

RCHO -> RCH2OH is a straightforward reduction of the carbonyl group -- it needs either a hydride source (which adds H- to carbon, then a proton to oxygen on work-up) or gaseous H2 over a metal hydrogenation catalyst.

Checking each reagent

Option (i), H2/Pd: catalytic hydrogenation over Pd (or Pt, Ni) adds H2 across the C=O bond exactly as it does across C=C, giving RCH2OH directly. This is a standard NCERT-taught method for reducing aldehydes and ketones to alcohols. Correct.
Option (ii), LiAlH4: a strong hydride donor; delivers H- to the carbonyl carbon, and aqueous work-up protonates the alkoxide to RCH2OH. Correct. …

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Method: Reduction of Aldehydes to Primary Alcohols

This is a reduction reaction — we are adding hydrogen ( $H_2$ ) to the carbonyl group ( $C=O$ ) to convert it into an alcohol ( $-CH_2OH$ ).

Key Concept

RCHO (aldehyde) → RCH₂OH (primary alcohol)
The carbonyl carbon goes from sp² to sp³ hybridisation
Two hydrogen atoms are added across the $C=O$ double bond

Step-by-Step Analysis

Step 1: Identify the functional group change

Aldehyde ( $-CHO$ ) → Primary alcohol ( $-CH_2OH$ )
This requires reduction (gain of hydrogen / loss of oxygen)

Step 2: Check each reagent

Reagent	Reduces aldehyde?	Reason
(A) $H_2/Pd$	✓ Yes	Catalytic hydrogenation reduces $C=O$ to $CH_2OH$
(B) $LiAlH_4$	✓ Yes	Strong reducing agent; reduces aldehydes to primary alcohols
(C) $NaBH_4$	✓ Yes	Mild reducing agent; selectively reduces aldehydes and ketones

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The Correct Answer

The question asks: convert RCHO (an aldehyde) into RCH₂OH (a primary alcohol).

Option (B) LiAlH₄ — Yes. A strong reducing agent that reduces aldehydes to primary alcohols.
Option (C) NaBH₄ — Yes. A milder reducing agent that also reduces aldehydes to primary alcohols.
Option (A) H₂/Pd — Yes. Catalytic hydrogenation reduces aldehydes to primary alcohols.
Option (D) RMgX + hydrolysis — No. This is a Grignard reaction, which adds an alkyl group to the carbonyl carbon, giving a secondary alcohol (not RCH₂OH).

Correct options: (A), (B), (C)

Common Mistakes Students Make

1. Thinking NaBH₄ cannot reduce aldehydes

Mistake: Some students think NaBH₄ only works on ketones or is too weak.
Truth: NaBH₄ reduces both aldehydes and ketones to alcohols. It is selective — it does not reduce esters, acids, or nitro groups.
Avoid: Remember: NaBH₄ reduces aldehydes → primary alcohols, ketones → secondary alcohols.

2. Forgetting that H₂/Pd works here

Mistake: Students often think catalytic hydrogenation only reduces alkenes/alkynes.
Truth: H₂/Pd also reduces aldehydes to primary alcohols (though it is less common in exams).
Avoid: Know that H₂/metal catalyst reduces C=O in aldehydes and ketones, but not in carboxylic acids or esters under normal conditions.

3. Confusing Grignard with reduction

Mistake: Students see “RMgX + hydrolysis” and think it gives an alcohol — which is true — but they forget it adds a carbon.
Truth: Grignard reagent adds an alkyl group to the carbonyl carbon. For an aldehyde RCHO, the product is R–CH(OH)–R' (secondary alcohol), not RCH₂OH.
Avoid: Always check: does the reagent add a carbon or just add H₂? If it adds carbon, the product changes.

4. Not reading “Two or more than two options may be correct”

Mistake: Students pick only one option (often LiAlH₄) and miss others.
Truth: The question explicitly allows multiple correct answers. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
$\mathrm{CH_2=CH-CH_3 \xrightarrow{Br_2/CCl_4}}$
›Reveal solutionSolution
The reaction of propene with bromine in carbon tetrachloride is an electrophilic addition that yields the vicinal dibromide. The major product is 1,2-dibromopropane.

The key concept here is electrophilic addition of bromine to an alkene. Bromine (Br₂) is nonpolar, but when it approaches the electron-rich double bond, the bond becomes polarized. The π electrons act as a nucleophile, attacking one bromine atom and displacing the other as a bromide ion (Br⁻). This forms a cyclic bromonium ion intermediate. The bromide ion then attacks the more substituted carbon of the three-membered ring (since that carbon can better stabilize partial positive charge), leading to anti addition of the two bromine atoms across the double bond.

Let’s walk through the steps:

Identify the alkene: Propene (CH₂=CH–CH₃) is an unsymmetrical alkene with a terminal double bond.

Formation of the bromonium ion: The π electrons attack one bromine molecule. As the Br–Br bond breaks, the bromine that gains a positive charge bridges the two carbons of the double bond, forming a cyclic bromonium ion. The other bromine becomes a free bromide ion (Br⁻) in solution.

Nucleophilic attack by bromide: The bromide ion attacks the bromonium ion. Because the bromonium ion is a three-membered ring, the attack occurs from the opposite side (anti addition). The question is: which carbon gets attacked? The bromide prefers the more substituted carbon (the one with the methyl group) because that carbon can better accommodate partial positive charge in the transition state. So the Br⁻ attacks the middle carbon (C-2) rather than the terminal carbon (C-1). …
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AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
$C_2H_5OH + HBr \longrightarrow C_2H_5Br + H_2O$
›Reveal solutionSolution
Only ethanol's acid-catalysed dehydration to ethylene creates a C=C bond, which is what actually decolourises bromine water via addition. Answer: (D).

Concept and Intuition

Bromine water is a standard qualitative test for unsaturation (C=C or C≡C bonds): $Br_2$ adds across the double bond (electrophilic addition, forming a vicinal dibromide), consuming the orange/red bromine colour. Any reaction of ethanol that does not produce an alkene or alkyne therefore cannot decolourise bromine water, no matter what other chemistry it involves.

Step-by-Step Solution

(A) $C_2H_5OH + HBr \rightarrow C_2H_5Br + H_2O$ : a substitution product, no double bond formed — no reaction with $Br_2$ water.

(B) $2\,C_2H_5OH \xrightarrow[413K]{H^+} C_2H_5-O-C_2H_5 + H_2O$ : intermolecular dehydration to diethyl ether — no double bond, no reaction.

(C) $C_2H_5OH + Na \rightarrow C_2H_5ONa + \tfrac12H_2$ : simple deprotonation/redox with sodium — no double bond, no reaction.

(D) $C_2H_5OH \xrightarrow[443K]{H^+} C_2H_4 + H_2O$ : intramolecular (unimolecular) dehydration at the higher temperature gives ethylene, $C_2H_4$ , which has a C=C bond. …
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Q.Which of the following are used to convert RCHO into RCH2OHRCH_2OHRCH2​OH? (Two or more than two options may be correct.)

Concept

Checking each reagent

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Q.Which of the following are used to convert RCHO into $RCH_2OH$ ? (Two or more than two options may be correct.)

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