Q.An organic compound is found to form oxime, reduces Tollen's reagent and forms iodoform with I₂/aq.KOH. Identify the compound.

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. …

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:

1. Forms an oxime with hydroxylamine ($NH_2OH$): only compounds with a carbonyl group ($>C=O$), i.e. aldehydes and ketones, form oximes ($>C=N-OH$).
2. 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. …

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. …

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. …

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.) …

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:

…

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. …

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). …

$C_8H_8O$ with positive DNP + iodoform but negative Tollens' and no alkene is acetophenone, $C_6H_5COCH_3$.

Concept. This is a functional-group identification of the kind CBSE Class-12 aldehydes-ketones-and-carboxylic-acids questions ask.

Why (clue by clue).

• Positive 2,4-DNP (Brady's) test $\Rightarrow$ a carbonyl ($>C=O$) is present.
• Positive iodoform test $\Rightarrow$ a $CH_3\text{-C(=O)-}$ (methyl ketone) group is present.
• Does not reduce Tollens' reagent $\Rightarrow$ it is a ketone, not an aldehyde. …

The haloform reaction converts a methyl ketone (or a compound oxidisable to one) into a haloform and a carboxylate salt, using $X_2/NaOH$.

Any compound containing a $CH_3CO-$ group (or a $CH_3CH(OH)-$ group that can be oxidised to it in situ, e.g. ethanol) reacts with a halogen ($Cl_2$, $Br_2$ or $I_2$) in the presence of $NaOH$ to give a haloform ($CHCl_3$, $CHBr_3$ or $CHI_3$) and the sodium salt of the corresponding carboxylic acid, via successive halogenation of the methyl group followed by cleavage: …

The methyl ketone group of acetophenone is cleaved by the haloform reaction, converting it to benzoic acid.

Acetophenone ($C_6H_5COCH_3$) contains a $CH_3CO-$ group attached to the ring, so it undergoes the haloform reaction with $I_2$ (or $NaOI$) in the presence of $NaOH$: the methyl group is progressively iodinated and then cleaved off as iodoform, leaving the carboxylate:

$C_6H_5COCH_3 + 3I_2 + 4NaOH \rightarrow C_6H_5COONa + CHI_3\downarrow + 3NaI + 3H_2O$ …

A carbonyl compound that fails Fehling's test (a ketone) but forms a bisulphite adduct and passes the iodoform test must be a methyl ketone; both 2-pentanone and 3-methyl-2-butanone fit at MW 86.

Not reducing Fehling's solution rules out an aldehyde — the compound must be a ketone. Molecular weight 86 corresponds to $C_5H_{10}O$ ketones. A positive iodoform test requires a $CH_3-CO-$ (methyl ketone) group:

• 2-pentanone, $CH_3COCH_2CH_2CH_3$ — has a $CH_3CO-$ group → gives iodoform. ✓
• 3-pentanone, $CH_3CH_2COCH_2CH_3$ — no $CH_3CO-$ group → does NOT give iodoform. ✗ …