Q.How is acetophenone converted to benzoic acid ?

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. ✗ …