(A) An organic compound 'A', with molecular formula $C_2H_6O$ reacts with active metals such as sodium to give compound 'B' and hydrogen gas. 'A' on treatment with iodine and sodium hydroxide gives 'C' and in presence of $H_2SO_4$ at 413 K gives 'D' ($C_4H_{10}O$). 'D' on reaction with excess of HI gives 'E'. Identify 'A', 'B', 'C', 'D' and 'E' and write all the reactions involved. OR (B) (a) Write the reagents which are used in the given conversions : (i) Phenol to 2, 4, 6-tribromophenol (ii) Propene to propan-1-ol (iii) Butan-2-one to butan-2-ol (b) Explain the mechanism of acid catalyzed hydration of alkene to form corresponding alcohol.

> [!TLDR] > Part (a): A = ethanol, B = sodium ethoxide, C = iodoform, D = diethyl ether, E = ethyl iodide. > Part (b): reagents = bromine water; $B_2H_6$/$H_2O_2$,OH$^-$; $NaBH_4$. Acid hydration of an alkene goes protonation → carbocation → water attack → deprotonation (Markovnikov). --- ## Part (a) $C_2H_6O$ liberating $H_2$ with Na must have an $-OH$ group, so **A = ethanol** ($CH_3CH_2OH$) (dimethyl ether would not react with Na). 1. **With sodium:** $2CH_3CH_2OH + 2Na \rightarrow 2CH_3CH_2ONa + H_2\uparrow$ → **B = sodium ethoxide** ($C_2H_5ONa$). 2. **Iodoform reaction** ($I_2$/NaOH): ethanol (a $CH_3CH(OH)-$ compound) gives a yellow precipitate of iodoform: $$CH_3CH_2OH + 4I_2 + 6NaOH \rightarrow CHI_3\downarrow + HCOONa + 5NaI + 5H_2O$$ → **C = iodoform ($CHI_3$)**. 3. **Dehydration at 413 K** (intermolecular): $2CH_3CH_2OH \xrightarrow{conc.\,H_2SO_4,\,413K} C_2H_5OC_2H_5 + H_2O$ → **D = diethyl ether** ($C_4H_{10}O$). (At 443 K it would give ethene instead.) …

(A) An organic compound 'A', with molecular formula C₂H₆O reacts with active metals such as sodium to give com

Question

C_2H_6O

(OR)

(B)

(a) Write the reagents which are used in the given conversions :

(i) Phenol to 2, 4, 6-tribromophenol

(ii) Propene to propan-1-ol

(iii) Butan-2-one to butan-2-ol

(b) Explain the mechanism of acid catalyzed hydration of alkene to form corresponding alcohol.

CBSECBSE Class XII Board 2026Subjective· 5mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Part (a)Concept understanding — Haloform (Iodoform) Reaction

Haloform (Iodoform) Reaction

The haloform reaction is a characteristic reaction of methyl ketones (compounds containing the $CH_3\text{-}CO\text{-}$ group) and of ethanal and ethanol (which carry the $CH_3\text{-}CH(OH)\text{-}$ unit). When such a compound is treated with a halogen ( $X_2$ ) in the presence of a base (NaOH), the three hydrogens of the methyl group are successively replaced by halogen, and the resulting trihalomethyl carbonyl compound is then cleaved by hydroxide.

The overall result is that the $CH_3\text{-}CO\text{-}$ fragment is lost as a haloform ( $CHX_3$ ), while the rest of the molecule is converted into a carboxylate ion (and, on acidification, a carboxylic acid). For example, with iodine and NaOH a methyl ketone $R\text{-}CO\text{-}CH_3$ gives $R\text{-}COO^-Na^+$ and a yellow precipitate of iodoform, $CHI_3$ :

$R\text{-}CO\text{-}CH_3 + 3I_2 + 4NaOH \rightarrow R\text{-}COONa + CHI_3\downarrow + 3NaI + 3H_2O$ …

Part (b)Concept understanding — Markovnikov Addition

The Intuition First

Imagine you have an alkene — a carbon-carbon double bond. That double bond is like a crowded room with two doors. When a molecule like HBr comes along, it wants to break that double bond and add across it. The question is: which carbon gets the hydrogen, and which gets the bromine?

You might think it doesn't matter — after all, the two carbons look similar. But they aren't. One carbon usually has more alkyl groups (methyl, ethyl, etc.) attached to it than the other. That carbon is more "electron-rich" — it has more friends pushing electrons toward it.

The hydrogen, being small and positively charged, is picky. It goes to the carbon that already has more hydrogens. Why? Because that carbon is less crowded and can stabilise the positive charge that forms temporarily during the reaction. The bromine, being large and negatively charged, goes to the other carbon — the one with more alkyl groups.

That's the intuition: the rich get richer. The carbon with more hydrogens gets another hydrogen. The carbon with more alkyl groups gets the halogen.

The Precise Statement

Markovnikov's Rule : When an unsymmetrical reagent (like HX, H₂O, etc.) adds to an unsymmetrical alkene, the hydrogen atom attaches to the carbon of the double bond that already has the greater number of hydrogen atoms.

In other words, for an alkene like $CH_3CH=CH_2$ (propene) reacting with HBr:

Carbon 1 (the CH₂ end) has 2 hydrogens.
Carbon 2 (the CH end) has 1 hydrogen.
The H goes to carbon 1 (more hydrogens).
The Br goes to carbon 2 (fewer hydrogens).

So the product is $CH_3CHBrCH_3$ (2-bromopropane), not $CH_3CH_2CH_2Br$ (1-bromopropane).

Why Does This Happen? The Real Chemistry

The reaction proceeds through a carbocation intermediate. When the H⁺ attacks the double bond, it can form one of two possible carbocations:

A primary carbocation (if H⁺ goes to the more substituted carbon) — unstable.
A secondary carbocation (if H⁺ goes to the less substituted carbon) — more stable.

The reaction chooses the path that gives the more stable carbocation. Alkyl groups stabilise carbocations through hyperconjugation and inductive effect — they donate electron density to the positively charged carbon.

Important

The stability order of carbocations is: tertiary > secondary > primary > methyl. Markovnikov addition always proceeds through the most stable carbocation possible.

A Common Misconception

Watch out

Many students think Markovnikov's rule means "hydrogen goes to the carbon with more hydrogens" because that carbon already has more hydrogens. That's backwards. The hydrogen goes there because that path leads to a more stable carbocation — the number of hydrogens is just a convenient way to predict the outcome, not the cause.

The One Big Exception …

Why this formula?

Markovnikov Addition: Why the Rule Holds

Markovnikov's rule is not a formula in the algebraic sense — it's a predictive principle for electrophilic addition to unsymmetrical alkenes. The "why" comes from carbocation stability and reaction mechanism.

The Rule in Words

When H–X adds to an unsymmetrical alkene, the hydrogen attaches to the carbon with more hydrogen atoms already attached, and the halogen (or X group) attaches to the carbon with fewer hydrogen atoms.

Example:

Propene ( $\ce{CH3-CH=CH2}$ ) + HBr → 2-bromopropane (major product), not 1-bromopropane.

Why This Happens: The Step-by-Step Reasoning

1. The Mechanism (Electrophilic Addition)

The reaction proceeds in two steps:

Slow step (rate-determining): The alkene's $\pi$ bond attacks the electrophilic $\ce{H+}$ from H–X, forming a carbocation intermediate.
Fast step: The carbocation is attacked by the nucleophilic $\ce{X-}$ .

2. The Key: Carbocation Stability

The more stable carbocation intermediate forms faster and determines the major product.

Carbocation Type	Stability Order	Reason
Tertiary ( $3^\circ$ )	Most stable	+3 alkyl groups donate electron density via hyperconjugation and inductive effect
Secondary ( $2^\circ$ )	Intermediate	+2 alkyl groups
Primary ( $1^\circ$ )	Least stable	+1 alkyl group
Methyl ( $\ce{CH3+}$ )	Unstable	No alkyl stabilization

3. Applying to Propene + HBr

Propene: $\ce{CH3-CH=CH2}$

Two possible protonation sites:

Path A (Markovnikov): $\ce{H+}$ adds to $\ce{CH2}$ (terminal carbon) → forms secondary carbocation:

\ce{CH3-CH+-CH3} \quad (2^\circ)

Path B (Anti-Markovnikov): $\ce{H+}$ adds to $\ce{CH}$ (middle carbon) → forms primary carbocation:

\ce{CH3-CH2-CH2+} \quad (1^\circ)

Result: The secondary carbocation is more stable (by ~25–30 kJ/mol), so Path A is faster. The $\ce{Br-}$ then attacks the positively charged carbon, giving 2-bromopropane.

The "Formula" — A Stability-Based Prediction

There is no algebraic formula, but a decision rule:

\text{Major product} = \text{Product from the more stable carbocation}

For alkenes with alkyl substituents, the stability order is:

$\text{Tertiary} > \text{Secondary} > \text{Primary} > \text{Methyl}$ …

Start your 14-day free trial to unlock the full solution →

Part (a): A = ethanol, B = sodium ethoxide, C = iodoform, D = diethyl ether, E = ethyl iodide.

Part (b): reagents = bromine water; $B_2H_6$ / $H_2O_2$ ,OH $^-$ ; $NaBH_4$ . Acid hydration of an alkene goes protonation → carbocation → water attack → deprotonation (Markovnikov).

Part (a)

$C_2H_6O$ liberating $H_2$ with Na must have an $-OH$ group, so A = ethanol ( $CH_3CH_2OH$ ) (dimethyl ether would not react with Na).

With sodium: $2CH_3CH_2OH + 2Na \rightarrow 2CH_3CH_2ONa + H_2\uparrow$ → B = sodium ethoxide ( $C_2H_5ONa$ ).
Iodoform reaction ( $I_2$ /NaOH): ethanol (a $CH_3CH(OH)-$ compound) gives a yellow precipitate of iodoform:

CH_3CH_2OH + 4I_2 + 6NaOH \rightarrow CHI_3\downarrow + HCOONa + 5NaI + 5H_2O

→ C = iodoform ( $CHI_3$ ).

3. Dehydration at 413 K (intermolecular): $2CH_3CH_2OH \xrightarrow{conc.\,H_2SO_4,\,413K} C_2H_5OC_2H_5 + H_2O$ → D = diethyl ether ( $C_4H_{10}O$ ). (At 443 K it would give ethene instead.) …

Start your 14-day free trial to unlock the full solution →

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

Showing the 12 most recent of 16 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The compound among the following which gives both iodoform and Fehling's test is$
(a) ethanol
(b) propanone
(c) butan-2-ol
(d) ethanal
›Reveal solutionSolution
Checking each compound against both tests shows only ethanal satisfies the structural requirement for the iodoform test and is itself an aldehyde, so only it is positive to both.

Ethanol, $CH_3CH_2OH$ : has the $CH_3CH(OH)-$ pattern, so $I_2/NaOH$ first oxidises it to acetaldehyde and then iodoform — iodoform positive. But ethanol itself is an alcohol, not an aldehyde, so it does not reduce Fehling's solution — Fehling negative.

Propanone (acetone), $CH_3COCH_3$ : has the $CH_3CO-$ group — iodoform positive. But it is a ketone; aliphatic ketones (lacking the aldehydic C–H) do not reduce Fehling's solution — Fehling negative.

Butan-2-ol, $CH_3CH(OH)CH_2CH_3$ : has the $CH_3CH(OH)-$ pattern — iodoform positive. It is an alcohol, not an aldehyde — Fehling negative. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. An organic compound is found to form oxime, reduces Tollen's reagent and forms iodoform with I₂/aq.KOH. Identify the compound.$
›Reveal solutionSolution
Forms oxime ⇒ carbonyl compound; reduces Tollens' reagent ⇒ it is an aldehyde; gives iodoform ⇒ has a CH₃CO– / CH₃CH(OH)– group. The compound satisfying all three is CH₃CHO (ethanal).

Let us use each clue:

Forms an oxime with hydroxylamine ( $NH_2OH$ ): only compounds with a carbonyl group ( $>C=O$ ), i.e. aldehydes and ketones, form oximes ( $>C=N-OH$ ).

Reduces Tollens' reagent (ammoniacal $AgNO_3$ ) to give a silver mirror: only aldehydes ( $-CHO$ ) are oxidised easily and reduce Tollens' reagent; ketones do not. So the compound is an aldehyde. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$CH_3$
›Reveal solutionSolution

Markovnikov addition of HI to 1-methylcyclohexene puts I on the more substituted carbon (C-1, which already bears –CH $_3$ ), via the more stable tertiary carbocation, giving 1-iodo-1-methylcyclohexane.

In 1-methylcyclohexene the double bond is between ring carbons C-1 and C-2, with a –CH $_3$ group on C-1. Electrophilic addition of HI proceeds in two steps:

Step 1 (protonation): H $^+$ adds to the alkene carbon that gives the more stable carbocation. Adding H $^+$ to C-2 places the positive charge on C-1, which is a tertiary carbocation (bonded to the ring's C-2 and C-6, plus the –CH $_3$ group). Adding H $^+$ instead to C-1 would place the charge on C-2, only a secondary carbocation. Since 3° carbocations are more stable (greater hyperconjugation/inductive stabilisation from three alkyl groups) than 2°, the reaction proceeds through the C-1 cation — this is Markovnikov's rule (the H goes to the carbon already bearing more hydrogens, i.e. C-2).

$\text{1-methylcyclohexene} + H^+ \rightarrow \text{1-methylcyclohexan-1-yl cation (3°, at C-1)}$

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. Iodoform test is not given by -$
(i) Pentan-2-one
(ii) Pentan-3-one
(iii) Ethanol
(iv) Ethanal
›Reveal solutionSolution
The iodoform test is positive only for compounds containing a CH₃-CO- (methyl ketone) or CH₃-CH(OH)- group; pentan-3-one has neither.

The iodoform test (with I₂/NaOH) is given by:

Methyl ketones, i.e. compounds with a CH₃-CO- group, and

Compounds with a CH₃-CH(OH)- group (secondary alcohols with a methyl group on the carbinol carbon), including ethanol.

Acetaldehyde (ethanal, CH₃CHO) also gives a positive test since it has a CH₃-CO- group.

Checking each option:

Pentan-2-one: CH₃-CO-CH₂-CH₂-CH₃ → has CH₃-CO- → positive. …
Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$CH_3$
›Reveal solutionSolution
Markovnikov's rule: with unsymmetrical alkenes, $H^+$ from $HX$ adds to the carbon that generates the more stable (here, tertiary) carbocation, and the halide ion then bonds to that carbon.

1-Methylcyclohexene has its ring double bond between $C_1$ (bearing the $CH_3$ substituent) and $C_2$ (bearing only $H$ ). Protonation of the alkene can occur in two ways:

$H^+$ adds to $C_1$ $\Rightarrow$ carbocation forms at $C_2$ , a secondary carbocation (flanked by $C_1$ and $C_3$ , both ring carbons).

$H^+$ adds to $C_2$ $\Rightarrow$ carbocation forms at $C_1$ , a tertiary carbocation (bonded to $CH_3$ , $C_2$ and $C_6$ — three carbon substituents). …
Start your 14-day free trial to see the full solution →
CBSE 2023Set F1 markMCQ
$Q. Which of the following will not give iodoform test?$
(a) Isopropyl alcohol
(b) Ethanol
(c) Ethanal
(d) Benzyl alcohol
›Reveal solutionSolution
Iodoform test is positive only for CH3CO- or CH3CH(OH)- containing compounds; benzyl alcohol has neither.

The iodoform (haloform) test is given by:

methyl ketones and acetaldehyde (a CH3-CO- group), and

alcohols that can be oxidised by I2/NaOH to a CH3-CO- compound, i.e. those with a CH3-CH(OH)- group.

Checking each option:

Isopropyl alcohol, CH3-CH(OH)-CH3, has CH3-CH(OH)- → gives iodoform. …
Start your 14-day free trial to see the full solution →
CBSE 2022Set E1 markMCQ
$Q. Which of the following gives iodoform test ?$
(a) CH3OH
(b) (CH3)2CHOH
(c) (CH3)3COH
(d) CH3-CH2-CH2-OH
›Reveal solutionSolution
Only alcohols with the CH3-CH(OH)- unit (or a CH3-CO- group) give the iodoform test — that is propan-2-ol, (CH3)2CHOH.

The iodoform (haloform) test is positive for compounds containing either a CH3-CO- (methyl ketone) group or a CH3-CH(OH)- group, because I2/NaOH first oxidises the CH3-CH(OH)- to CH3-CO- and then cleaves it to give yellow CHI3 (iodoform).

Checking the options:

CH3OH (methanol): no CH3-CH(OH)- unit → negative. (Ethanol would be positive, but methanol is not.) …
Start your 14-day free trial to see the full solution →
CBSE 2022Set ANNUAL1 markMCQ
$Q. Iodoform test is not given by:$
(a) Ethanol
(b) Ethanal
(c) 3-Pentanone
(d) 2-Pentanone
›Reveal solutionSolution
Iodoform test is positive for compounds containing a $CH_3CO-$ (methyl ketone) group or a $CH_3CH(OH)-$ group that can be oxidised to it. 3-Pentanone lacks this group. Option (C).

The iodoform ( $CHI_3$ ) test is given by:

Ethanol ( $CH_3CH_2OH$ ) — has $CH_3CH(OH)-$ ✓

Ethanal ( $CH_3CHO$ ) — has $CH_3CO-$ ✓

2-Pentanone ( $CH_3COCH_2CH_2CH_3$ ) — has $CH_3CO-$ ✓

But:

…
Start your 14-day free trial to see the full solution →
CBSE 2020Set NC1 mark
$Q. Write the reaction with conditions for conversion of 2-methylpropene into 1-bromo-2-methylpropane.$
›Reveal solutionSolution

Ordinary Markovnikov addition of HBr to 2-methylpropene would put Br on the more-substituted carbon; getting the anti-Markovnikov product (Br on the terminal carbon) needs the peroxide-initiated free-radical mechanism (Kharasch/peroxide effect).

Target: 2-methylpropene, $(CH_3)_2C=CH_2$ (isobutylene), converted into 1-bromo-2-methylpropane, $(CH_3)_2CH\text{–}CH_2Br$ — i.e. Br ends up on the terminal (less substituted) carbon, which is the opposite regiochemistry to normal Markovnikov addition (which would instead place Br on the more substituted carbon, giving 2-bromo-2-methylpropane).

…

Start your 14-day free trial to see the full solution →
CBSE 2019Set ANNUAL1 mark
$CH_3\text{--}CH_2\text{--}CH=CH\text{--}CH_3 + HCl \rightarrow A + B$
›Reveal solutionSolution
$CH_3\text{–}CH_2\text{–}CH=CH\text{–}CH_3$ is pent-2-ene; electrophilic addition of $HCl$ proceeds via protonation to give whichever secondary carbocation is more stabilised, so the products are a mixture of 2-chloropentane (major) and 3-chloropentane (minor).

The alkene $CH_3\text{–}CH_2\text{–}CH=CH\text{–}CH_3$ is pent-2-ene, numbered $C_1H_3\text{–}C_2H_2\text{–}C_3H=C_4H\text{–}C_5H_3$ , with the double bond between $C_3$ and $C_4$ (equivalently $C_2$ – $C_3$ counting from the other end). Since this alkene is unsymmetrically substituted but both carbons of the double bond are internal, protonation of either carbon gives a secondary carbocation, so a mixture of two constitutional isomers is obtained:

$H^+$ adds to $C_3$ : leaves the cation at $C_4$ , giving $CH_3\text{–}CH_2\text{–}CH_2\text{–}CH^+\text{–}CH_3$ — this secondary cation is flanked by a $-CH_2CH_2CH_3$ (propyl) group on one side and a $-CH_3$ on the other, i.e. more hyperconjugating $\alpha$ -hydrogens overall, making it the somewhat more stabilised carbocation. $Cl^-$ then attacks here, giving $CH_3\text{–}CH_2\text{–}CH_2\text{–}CHCl\text{–}CH_3$ , i.e. 2-chloropentane (major). …
Start your 14-day free trial to see the full solution →
CBSE 2019Set ANNUAL1 markMCQ
$Q. Which does not form iodoform on heating with I2 and base?$
(a) Acetone
(b) Ethanol
(c) Methanol
(d) Acetaldehyde
›Reveal solutionSolution
Iodoform is formed only by CH3CO- or CH3CH(OH)- compounds; methanol (CH3OH) has neither, so option (c).

The iodoform reaction (with I2 and a base, i.e. NaOI) is positive for:

Methyl ketones (containing the CH3-CO- group), e.g. acetone CH3COCH3.

Compounds oxidisable to such a group, i.e. those with a CH3-CH(OH)- group, e.g. ethanol CH3CH2OH, and acetaldehyde CH3CHO.

Checking the options:

Acetone (CH3COCH3): has CH3CO- -> gives iodoform. …
Start your 14-day free trial to see the full solution →
CBSE 2018Set ANNUAL1 markMCQ
$Q. An organic compound gives iodoform test and also gives positive test with Tollens reagent. The compound is -$
(a) CH3-CHO
(b) CH3-C(=O)-CH3 (acetone)
(c) CH3-CH2OH
(d) (CH3)2CH-OH (isopropyl alcohol)
›Reveal solutionSolution
Aldehyde (Tollens⁺) + CH₃CO group (iodoform⁺) → acetaldehyde.

Tollens' reagent is reduced only by aldehydes → the compound must be an aldehyde. This rules out acetone (ketone), ethanol and isopropanol (alcohols). …
Start your 14-day free trial to see the full solution →

Part (a)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days