Two compounds A and B with same molecular formula $$(C_3H_6O)$$ undergo Grignard reaction with methylmagnesium bromide to give products C and D. Products C and D show following chemical tests:

Test	C	D
Ceric ammonium nitrate Test	Positive	Positive
Lucas Test	Turbidity obtained after five minutes	Turbidity obtained Immediately
Iodoform Test	Positive	Negative

C and D respectively are:

We begin by noting that both compounds A and B have the molecular formula $$C_3H_6O$$. Such a formula fits carbonyl compounds having three carbon atoms, namely

$$\text{propanal : }CH_3CH_2CHO\quad\text{(an aldehyde)}$$

$$\text{acetone : }CH_3COCH_3\quad\text{(a ketone)}$$

When a carbonyl compound is treated with a Grignard reagent, an alkyl group is added to the carbonyl carbon and the oxygen is converted into an alcohol after acidic work-up. Stating the general result first,

$$\text{R-C(=O)-R'} + CH_3MgBr \overset{H_3O^+}{\rightarrow} \text{R-C(OH)(CH}_3\text{)-R'}$$

So each carbonyl compound of three carbons will give an alcohol of four carbons, because one extra $$CH_3$$ group is introduced.

1. Reaction of propanal with $$CH_3MgBr$$

Starting carbonyl:

$$CH_3CH_2CHO$$

Addition of $$CH_3^-$$ from the Grignard reagent at the carbonyl carbon followed by protonation gives

$$CH_3CH_2C(OH)(CH_3)H \;=\;CH_3CH_2CH(OH)CH_3$$

This product is butan-2-ol, a secondary alcohol. Let us call it product C.

2. Reaction of acetone with $$CH_3MgBr$$

Starting carbonyl:

$$CH_3COCH_3$$

Addition of $$CH_3^-$$ and protonation gives

$$CH_3C(OH)(CH_3)CH_3 \;=\;(CH_3)_3C\,OH$$

This product is 2-methyl-2-propanol (tert-butyl alcohol), a tertiary alcohol. Let us call it product D.

Now we match C and D with the chemical tests given.

(i) Ceric ammonium nitrate test - any alcohol answers it, so both C and D should be positive, exactly as observed.

(ii) Lucas test - the rate of turbidity distinguishes the class of alcohol.

• Secondary alcohols turn turbid in about five minutes.
• Tertiary alcohols turn turbid immediately.

Product C is a secondary alcohol (butan-2-ol) → turbidity after five minutes.
Product D is a tertiary alcohol (tert-butyl alcohol) → instantaneous turbidity.

(iii) Iodoform test - positive for ethanol or any secondary alcohol that contains the fragment $$CH_3-CH(OH)-$$.

• Butan-2-ol possesses the $$CH_3-CH(OH)-$$ group, so C gives a positive iodoform test.
• tert-Butyl alcohol lacks that fragment, so D gives a negative iodoform test.

The behaviour predicted for butan-2-ol and tert-butyl alcohol coincides perfectly with the observations. Therefore

$$C = CH_3CH_2CH(OH)CH_3\quad(\text{butan-2-ol})$$

$$D = (CH_3)_3C\,OH\quad(\text{tert-butyl alcohol})$$

Looking at the options, these structures are listed together only in Option A:

$$C = H_3C-CH_2-CH(OH)-CH_3;\; D = H_3C-C(CH_3)_2-OH$$

Hence, the correct answer is Option A.