Chemistry mcq organic question

Original post by anonymous56754

In my times, this quiz was very popular in the organic chemistry lab exam.

•

Alkaline aqueous iodine (Iodoform test) + metyl ketone group or secondary alcolhols = yellow precipitate

•

2,4-Dinitrophenylhydrazine (DNPH) + carbonyl compounds as aldehydes and ketones = orange precipitate

Propanal (A) = orange precipitate.

Propan-1-ol (B) = no reaction with both reagents because it is a primary alchol,

Propan-2-ol (C) = yellow precipitate.

Propanone (D) = reaction with DNPH with an orange precipitate and alkaline aqueous iodine with a yellow precipitate.

The correct answer is B. Propan-1-ol.
Krgds,
Sandro

(edited 1 month ago)

Reply 2

Original post by Nitrotoluene

In my times, this quiz was very popular in the organic chemistry lab exam.
Alkaline aqueous iodine (Iodoform test) + metyl ketone group or secondary alcolhols = yellow precipitate
2,4-Dinitrophenylhydrazine (DNPH) reagent + carbonyl compounds as aldehydes and ketones = orange precipitate
Propanal (A) = orange precipitate.
Propan-1-ol (B) =In my times, this quiz was very popular in the organic chemistry lab exam.

•

Alkaline aqueous iodine (Iodoform test) + metyl ketone group or secondary alcolhols = yellow precipitate

•

2,4-Dinitrophenylhydrazine (DNPH) reagent + carbonyl compounds as aldehydes and ketones = orange precipitate

Propanal (A) = orange precipitate.

Propan-1-ol (B) = no reaction with both reagents because it is a primary alchol,

Propan-2-ol (C) = yellow precipitate.

Propanone (D) = reaction with DNPH with an orange precipitate and alkaline aqueous iodine with a yellow precipitate.

The correct answer is B. Propan-1-ol.
Krgds,
Sandro

Ah thank you, so iodine in alkaline solution gives a positive result in secondary alcohols?

Reply 3

Original post by anonymous56754

Ah thank you, so iodine in alkaline solution gives a positive result in secondary alcohols?

Quote
Ah thank you, so iodine in alkaline solution gives a positive result in secondary alcohols?
Unquote
Indeed, a positive Iodoform test is produced when secondary alcohols react with alkaline aqueous iodine.

Bye,
Sandro

Reply 4

Original post by Nitrotoluene

Thanks!

Reply 5

Original post by Nitrotoluene

In my times, this quiz was very popular in the organic chemistry lab exam.

•

Alkaline aqueous iodine (Iodoform test) + metyl ketone group or secondary alcolhols = yellow precipitate

•

2,4-Dinitrophenylhydrazine (DNPH) + carbonyl compounds as aldehydes and ketones = orange precipitate

Propanal (A) = orange precipitate.

Propan-1-ol (B) = no reaction with both reagents because it is a primary alcohol,

Propan-2-ol (C) = yellow precipitate.

Propanone (D) = reaction with DNPH with an orange precipitate and alkaline aqueous iodine with a yellow precipitate.

The correct answer is B. Propan-1-ol.
Krgds,
Sandro

Your top point isn't quite correct - an iodoform test only responds to alcohols and any ketones or aldehydes that have a -COCH3 or -CH(OH)CH3 forming part of their structure. So yes, all methyl ketones will react with alkaline iodine solution, but not all secondary alcohols will (visibly) react (try it with pentan-3-ol, for example and you won't get the yellow precipitate).

The remainder of your response, however, is fine. I would probably make it ever so slightly clearer that 2,4-DNPH only reacts with aldehydes and ketones, however, as "carbonyl compounds" are quite a broad category and some will definitely not react (2,4-DNPH doesn't react with carboxylic acids, for example).

Reply 6

Original post by anonymous56754

Ah thank you, so iodine in alkaline solution gives a positive result in secondary alcohols?

Not always - one of the groups bonded to the primary/secondary carbon with the hydroxyl group must be a methyl group (CH3) for a secondary alcohol to undergo the iodoform reaction.

The iodoform reaction just tells you that you have an alcohol that definitely isn't tertiary where the C(OH) is bonded to a CH3 or that you have an aldehyde or ketone carbonyl that is bonded to a CH3.

Reply 7

Original post by TypicalNerd

Not always - one of the groups bonded to the primary/secondary carbon with the hydroxyl group must be a methyl group (CH3) for a secondary alcohol to undergo the iodoform reaction.
The iodoform reaction just tells you that you have an alcohol that definitely isn't tertiary where the C(OH) is bonded to a CH3 or that you have an aldehyde or ketone carbonyl that is bonded to a CH3.

Hi, thank you, so why won’t propan-1-ol react because you said it doesn’t have to be tertiary or is it positive if there is an oh group with a ch3 attached to it?

Reply 8

Original post by anonymous56754

Hi, thank you, so why won’t propan-1-ol react because you said it doesn’t have to be tertiary or is it positive if there is an oh group with a ch3 attached to it?

Propan-1-ol is CH3CH2CH2OH

Notice how the carbon with an -OH group is not next to a CH3? This is why it doesn’t give a positive iodoform test.

Now compare it to propan-2-ol (CH3CH(OH)CH3), which does have a CH3 next to the carbon with the -OH group. This will undergo the iodoform reaction, because:

(a) the alcohol is oxidisable to a ketone and so it will first oxidise to CH3COCH3

(b) the resulting ketone is a methyl ketone and so can react, giving CH3COONa and CHI3

Reply 9

Original post by TypicalNerd

Propan-1-ol is CH3CH2CH2OH
Notice how the carbon with an -OH group is not next to a CH3? This is why it doesn’t give a positive iodoform test.
Now compare it to propan-2-ol (CH3CH(OH)CH3), which does have a CH3 next to the carbon with the -OH group. This will undergo the iodoform reaction, because:
(a) the alcohol is oxidisable to a ketone and so it will first oxidise to CH3COCH3
(b) the resulting ketone is a methyl ketone and so can react, giving CH3COONa and CHI3

I see, but the propan-2-ol in the question isn’t being oxidised into a ketone, or is it just the possibility of it being oxidised that we are considering?

Reply 10

Original post by anonymous56754

I see, but the propan-2-ol in the question isn’t being oxidised into a ketone, or is it just the possibility of it being oxidised that we are considering?

In the iodoform reaction, if the substrate is an alcohol, it is always first oxidised to an aldehyde/ketone - that’s why 3° alcohols cannot undergo the iodoform reaction, but 1° and 2° alcohols with C(OH)CH3 forming part of their structures can. Recall that when halogens dissolve in NaOH they disproportionate - in this case into IO^- (an oxidising agent) and I^-.

The possibility of the alcohol being oxidised is one way of remembering whether an alcohol with a C(OH)CH3 can undergo the iodoform reaction.

Reply 11

Original post by TypicalNerd

In the iodoform reaction, if the substrate is an alcohol, it is always first oxidised to an aldehyde/ketone - that’s why 3° alcohols cannot undergo the iodoform reaction, but 1° and 2° alcohols with C(OH)CH3 forming part of their structures can. Recall that when halogens dissolve in NaOH they disproportionate - in this case into IO^- (an oxidising agent) and I^-.
The possibility of the alcohol being oxidised is one way of remembering whether an alcohol with a C(OH)CH3 can undergo the iodoform reaction.

Ahh ok thank you!

Reply 12

Original post by TypicalNerd

In the iodoform reaction, if the substrate is an alcohol, it is always first oxidised to an aldehyde/ketone - that’s why 3° alcohols cannot undergo the iodoform reaction, but 1° and 2° alcohols with C(OH)CH3 forming part of their structures can. Recall that when halogens dissolve in NaOH they disproportionate - in this case into IO^- (an oxidising agent) and I^-.
The possibility of the alcohol being oxidised is one way of remembering whether an alcohol with a C(OH)CH3 can undergo the iodoform reaction.

👍️from Sandro!

(edited 1 month ago)

Reply 13