Carbonyl Compounds: Question 5

Worked solution

Step 1: What the tri-iodomethane test detects

Alkaline aqueous iodine (effectively $\text{NaOI}$, formed from $\text{I}_2$ and $\text{NaOH}$) gives a pale yellow precipitate of tri-iodomethane, $\text{CHI}_3$, with any compound containing:

• a methyl ketone group, $\text{CH}_3\text{CO}-$, or
• the arrangement $\text{CH}_3\text{CH(OH)}-$: a carbon bearing both an -OH group and a $\text{CH}_3$ group attached directly to it.

Compounds that contain neither arrangement give no precipitate, even though the reagent is still added and the mixture warmed.

Step 2: Checking each compound

P, butan-1-ol, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$: the carbon bearing the -OH group is a $\text{-CH}_2\text{OH}$ group, attached to a $\text{CH}_2$ group, not a $\text{CH}_3$ group. Negative.

Q, butan-2-ol, $\text{CH}_3\text{CH(OH)}\text{CH}_2\text{CH}_3$: the carbon bearing the -OH group is directly attached to a $\text{CH}_3$ group (as well as a $\text{CH}_2\text{CH}_3$ group and an H atom), this matches the $\text{CH}_3\text{CH(OH)}-$ pattern exactly. Positive.

R, butan-2-one, $\text{CH}_3\text{COCH}_2\text{CH}_3$: this is a methyl ketone, containing a $\text{CH}_3\text{CO}-$ group. Positive.

S, 2-methylpropan-1-ol, $(\text{CH}_3)_2\text{CHCH}_2\text{OH}$: the carbon bearing the -OH group is a $\text{-CH}_2\text{OH}$ group, attached to a $\text{CH(CH}_3)_2$ group. There is no $\text{CH}_3$ group directly attached to the carbon that carries the -OH. Negative.

Step 3: Counting the positives

Only Q and R give a positive result. That is 2 out of the four liquids.

Why the other options are wrong

• A (1) undercounts: both Q and R satisfy the required structural pattern, not just one.
• C (3) and D (4) overcount: neither P nor S has a $\text{CH}_3$ group directly attached to the carbon bearing the -OH group, so neither can give a positive result.

Final answer

B, 2 of the four liquids (Q and R) give a positive tri-iodomethane test.