18) Carbonyl compounds (2016)

No content yet, but please feel free to ask any Q’s / make any comments.

Advertisements

2 thoughts on “18) Carbonyl compounds (2016)

  1. Hello, may you explain 9701/11/O/N/2009 Question 28?

    Well the question asked for an aldehyde reacting with CH3CO2CH3 with strong Base BUT the structure given as example is like an ketone.

    Thank you!

    Like

  2. Hi there.
    I like this question (as a learning tool as opposed to having to do it in an exam) because it exposes my own personal bad habit (one of a number) of failing to understand how to use the question info properly.

    I think the trick here is to stand back and take a general view of it. The info tells us how an anion reacts with a carbonyl. That really is the key. The actual question bit requires you to react an anion with an aldehyde instead of reacting an anion with a ketone.

    We change the anion from -CN to -CH3COCH3 instead. (The anion is of course nucleophilic).

    The -CN added into the carbonyl group and the carbonyls oxygen turned into an OH. So -CH3COCH3 anion will do the same to the C=O of an aldehyde.

    The old carbonyl group now a C(CH2COCH3)(OH) will have a H on the old carbonyl C because the carbonyl was an aldehyde, and not a ketone. The old carbonyl may actually have two H’s on it, if it was methanal)

    To get the answer, subtract the -CH3COCH3 anion section from the products offered in the answers and see if the reamining atoms when you “go backwards” from the OH back into the C=O, see if that fragment was an aldehyde.

    A is wrong because you can’t subtract a CH2COCH3 fragment.
    B is correct as you can rip off a -CH3COCH3 from the left hand side and you’re left with CH(OH)CH3 which originally would be H(C=O)CH3, i.e ethanal.
    C is also right because you can pull of -CH3COCH3 from the right, and you’re left with CH3CH2CH(OH) which would have come from CH3CH2(C=O)H propanal.
    D is incorrect because ripping away -CH3COCH3 from the right hand side leaves(CH3)COH which would mean (CH3)2C=0, i.e. propanone was used as the original carbonyl, but the question wants you to start with an aldehyde.

    Why there are two correct answers? I guess CIE didn’t spot it. They are humans too.

    Like

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s