A compound (A) with molecular formula $C_4H_5N$ on reduction with DIBAL-H followed by hydrolysis, gives a compound (B). Compound (B) gives positive Tollens' test but does not give iodoform test. Compound (B) can also be obtained when ethanal is treated with dilute NaOH followed by heating. Identify (A) and (B). Write the reactions of (A) with DIBAL-H followed by hydrolysis.

A compound (A) with molecular formula C₄H₅N on reduction with DIBAL-H followed by hydrolysis, gives a compound

Question

C_4H_5N

CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★

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Concept understanding — DIBAL-H Reduction

DIBAL-H Reduction: Stopping the Reduction at Aldehyde

You already know that LiAlH₄ reduces an ester all the way to a primary alcohol. But what if you want to stop at the aldehyde — the intermediate step? That is exactly what DIBAL-H does. The trick is not a stronger reagent, but a weaker, bulkier one that runs out of steam halfway.

The Intuition: A Bulky Hammer That Hits Once

Think of an ester as a two‑storey building. LiAlH₄ is a wrecking ball that smashes both floors at once — you get the alcohol. DIBAL-H is a much larger, clumsier hammer. It can swing hard enough to break the first floor (the ester → aldehyde step), but it is so bulky that it cannot easily reach the second floor (aldehyde → alcohol). So the reaction stops at the aldehyde.

The same logic applies to nitriles. A nitrile has a triple bond that can be reduced in two stages: first to an imine (C=N), then to an amine. DIBAL-H stops at the imine, which hydrolyses to an aldehyde during work‑up.

The Precise Chemistry

DIBAL-H stands for diisobutylaluminium hydride, formula $(i\text{-Bu})_2\text{AlH}$ . It is a Lewis acid (the aluminium is electron‑deficient) and a hydride donor. The key is that it is used at low temperature (typically –78 °C, dry ice/acetone bath) and with one equivalent of reagent.

For an ester (say ethyl benzoate):

DIBAL-H coordinates to the carbonyl oxygen (Lewis acid–base interaction).
A hydride is transferred to the carbonyl carbon, forming a hemiacetal intermediate — a tetrahedral species with an –OAlR₂ group and an –OR group.
At low temperature, this hemiacetal is stable. It does not collapse to the aldehyde until you add water during work‑up.
Upon aqueous work‑up (often with a mild acid like Rochelle salt or dilute HCl), the hemiacetal hydrolyses to the aldehyde. The second hydride never gets a chance to attack because the aluminium is already tied up.

Watch out

If you warm the reaction or use excess DIBAL-H, the hemiacetal can break down to the aldehyde in situ, and then DIBAL-H will reduce that aldehyde to the alcohol. Temperature control is everything.

For a nitrile (say benzonitrile):

DIBAL-H adds a hydride to the electrophilic nitrile carbon, forming an imine intermediate (R–CH=N–AlR₂).
This imine is stable at low temperature.
Aqueous work‑up hydrolyses the imine to the aldehyde (R–CHO + NH₃).

\text{RCOOR'} \xrightarrow{\text{1. DIBAL-H, –78 °C}} \text{RCHO}

Why Not Just Use LiAlH₄ and Stop?

LiAlH₄ is too reactive. It reduces the aldehyde faster than it reduces the ester — so you never see the aldehyde. DIBAL-H is milder and bulkier. The aluminium in DIBAL-H is only bonded to two bulky isobutyl groups (plus one H), so it is less nucleophilic than the AlH₄⁻ ion. It also coordinates strongly to oxygen, stabilising the hemiacetal intermediate.

A Common Exam Point …

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Compound B is an aldehyde that gives Tollens' test but not iodoform test, and forms from ethanal via aldol condensation followed by dehydration—it is but-2-enal (crotonaldehyde). Working backward, A must be but-2-enenitrile (crotononitrile), which reduces with DIBAL-H to give B.

The puzzle here is to work backward from the properties of B and then identify what A must be. Let's decode the clues systematically.

Understanding Compound B

Compound B gives a positive Tollens' test, which tells us it contains an aldehyde group ( $-CHO$ ). The aldehyde reduces Tollens' reagent (ammoniacal silver nitrate) to metallic silver.

It does not give the iodoform test. The iodoform test is positive for methyl ketones ( $R-CO-CH_3$ ) or compounds that can be oxidized to them (like secondary alcohols with a $CH_3-CHOH-$ unit, or ethanal itself). Since B is an aldehyde (not a ketone) and doesn't give iodoform, it cannot be ethanal (which would give iodoform). This means B must be a higher aldehyde without the $CH_3-CO-$ or $CH_3-CHO$ structure that leads to iodoform.

The third clue is crucial: B forms when ethanal is treated with dilute NaOH followed by heating. This is the classic aldol condensation reaction. Two molecules of ethanal undergo aldol addition, then dehydration upon heating:

2 \, CH_3CHO \xrightarrow[\text{heat}]{\text{dil. NaOH}} CH_3CH=CHCHO + H_2O

The product is but-2-enal (crotonaldehyde), $C_4H_6O$ . This aldehyde has a conjugated $\alpha,\beta$ -unsaturated system, gives Tollens' test (aldehyde present), and does not give iodoform test (no methyl ketone or ethanal structure).

So compound B is but-2-enal, $CH_3-CH=CH-CHO$ .

Working Backward to Compound A

Compound A has molecular formula $C_4H_5N$ and reduces with DIBAL-H (diisobutylaluminum hydride) followed by hydrolysis to give B ( $C_4H_6O$ ).

DIBAL-H is a selective reducing agent. At low temperature and controlled conditions, it reduces nitriles ( $-C≡N$ ) to aldehydes:

R-C≡N \xrightarrow[\text{2. } H_3O^+]{\text{1. DIBAL-H}} R-CHO

Since B is but-2-enal ( $CH_3-CH=CH-CHO$ ), compound A must be the corresponding nitrile: but-2-enenitrile (crotononitrile), $CH_3-CH=CH-C≡N$ .

Let's verify the molecular formula: $C_4H_5N$ ✓

Tip

DIBAL-H is milder than $LiAlH_4$ . While $LiAlH_4$ would reduce a nitrile all the way to a primary amine ( $RCH_2NH_2$ ), DIBAL-H stops at the aldehyde stage—perfect for this transformation.

The Reduction Mechanism

Step 1: Reduction with DIBAL-H

The nitrile carbon is electrophilic. DIBAL-H delivers a hydride ion to the nitrile carbon:

CH_3-CH=CH-C≡N + [H^-] \xrightarrow{\text{DIBAL-H}} CH_3-CH=CH-CH=N-Al(i-Bu)_2

The intermediate is an imine-aluminum complex.

Step 2: Hydrolysis …

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CBSE 2025Set 56/6/13 marks
$C_4H_5N$
›Reveal solutionSolution

Compound B is an aldehyde that gives Tollens' test but not iodoform test, and forms from ethanal via aldol condensation followed by dehydration—it is but-2-enal (crotonaldehyde). Working backward, A must be but-2-enenitrile (crotononitrile), which reduces with DIBAL-H to give B.

The puzzle here is to work backward from the properties of B and then identify what A must be. Let's decode the clues systematically.

Understanding Compound B

Compound B gives a positive Tollens' test, which tells us it contains an aldehyde group ( $-CHO$ ). The aldehyde reduces Tollens' reagent (ammoniacal silver nitrate) to metallic silver.

It does not give the iodoform test. The iodoform test is positive for methyl ketones ( $R-CO-CH_3$ ) or compounds that can be oxidized to them (like secondary alcohols with a $CH_3-CHOH-$ unit, or ethanal itself). Since B is an aldehyde (not a ketone) and doesn't give iodoform, it cannot be ethanal (which would give iodoform). This means B must be a higher aldehyde without the $CH_3-CO-$ or $CH_3-CHO$ structure that leads to iodoform.

The third clue is crucial: B forms when ethanal is treated with dilute NaOH followed by heating. This is the classic aldol condensation reaction. Two molecules of ethanal undergo aldol addition, then dehydration upon heating:
$2 \, CH_3CHO \xrightarrow[\text{heat}]{\text{dil. NaOH}} CH_3CH=CHCHO + H_2O$
The product is but-2-enal (crotonaldehyde), $C_4H_6O$ . This aldehyde has a conjugated $\alpha,\beta$ -unsaturated system, gives Tollens' test (aldehyde present), and does not give iodoform test (no methyl ketone or ethanal structure).

So compound B is but-2-enal, $CH_3-CH=CH-CHO$ .

Working Backward to Compound A

Compound A has molecular formula $C_4H_5N$ and reduces with DIBAL-H (diisobutylaluminum hydride) followed by hydrolysis to give B ( $C_4H_6O$ ).

DIBAL-H is a selective reducing agent. At low temperature and controlled conditions, it reduces nitriles ( $-C≡N$ ) to aldehydes:
$R-C≡N \xrightarrow[\text{2. } H_3O^+]{\text{1. DIBAL-H}} R-CHO$
Since B is but-2-enal ( $CH_3-CH=CH-CHO$ ), compound A must be the corresponding nitrile: but-2-enenitrile (crotononitrile), $CH_3-CH=CH-C≡N$ .

Let's verify the molecular formula: $C_4H_5N$ ✓

Tip

DIBAL-H is milder than $LiAlH_4$ . While $LiAlH_4$ would reduce a nitrile all the way to a primary amine ( $RCH_2NH_2$ ), DIBAL-H stops at the aldehyde stage—perfect for this transformation.

The Reduction Mechanism

Step 1: Reduction with DIBAL-H

The nitrile carbon is electrophilic. DIBAL-H delivers a hydride ion to the nitrile carbon:
$CH_3-CH=CH-C≡N + [H^-] \xrightarrow{\text{DIBAL-H}} CH_3-CH=CH-CH=N-Al(i-Bu)_2$
The intermediate is an imine-aluminum complex.

Step 2: Hydrolysis …
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Understanding Compound B

Working Backward to Compound A

The Reduction Mechanism

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