Chemistry P2 Help!

[allysaleemally]

Hii, could anyone explain me question 4 a, 4b from chemistry paper 2, may june 2009 v1.
here are the links: http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s09_qp_2.pdf
marksheet: http://www.xtremepapers.com/papers/...d AS Level/Chemistry (9701)/9701_s09_qp_2.pdf


Thank you very much

 

[TCHS]

Okay, so in 4a) you need to write the structural formula of the product formed when an aldehyde reacts with HCN. For these type of questions, you cannot solve them unless you have their reactions etched onto your brain. From your course, you know that when an aldehyde reacts with HCN, the CN- acts as a nucleophile and attacks the carbon-oxygen bond. Therefore, it forms a cynaohidrin. So basically, nucleophilic addition takes place and CN and H simply add across the Carbon atom of CH3CHO. The structural formula of the product is CH3CH(OH)CN. The ions in red are the ones that have added across. This is an ADDITION reaction therefore the reagent HCN simply ADDS on.

Now you have product E on the right. The aldehyde, ethanal, reacts with Tollen's reagent & from your course, you should recall how Tollen's reagent is an OXIDISING AGENT for Aldehydes. When aldehydes are further oxidised, they from carboxylic acids, therefore you need to remove the aldehyde suffix and add CO2H (acid group) in its place. Therefore, now CH3CHO is oxidised to CH3CO2H.


Next, let's look below product B where it is heated with dilute H2S04. Product B, as we found out, was CH3CH(OH)CN. When an amide group reacts with H2SO4 under heat, it gives an acid so now the CN group is removed and CO2H added, making the formula CH3CH(OH)CO2H.

Let's look at product D which is simple reduction of the aldehyde CH3CHO & we know that reduction of aldehydes gives PRIMARY alcohol. Therefore, CH3CH2OH!

Product C comes next (I know I am not going alphabetically so pardon me). They give us a compound with an alkene functional group and an acid functional group. This compound is treated with COLD, DILUTE MnO4. We know (or should know!) that oxidation of Alkenes wild COLD DILUTE MnO4 results in the formation of a diol ie an alcohol is formed. The double bond breaks and an OH group is added across each Carbon atom where the double bond was. So, is formed! Next this product is treated with Cr2O7. Basically the alcohol product that was formed is now being oxidised ( we know that Cr2O7 is an oxidising agent for alcohols). Now look at the carbon atoms in the product. The two carbon atoms we should be looking at are the ones where the functional group OH is attached. Those are the first 2 carbon atoms. Each carbon atom is attached to only one other carbon atom. (OH)CH2CH(OH)CO2H, the C atoms coloured red are the ones you should look at. If the C atom with the OH group is attached to only ONE other C atom, it means it is a primary alcohol & oxidation of primary alcohol forms an aldehyde. However, in the question the words heat under reflux can be seen & we know that further oxidation under reflux of aldehydes forms carboxylic acids. Therefore the end product is an acid. HO2CCOCO2H.

I know my answer is LONG but that's my best attempt at explaining & I am so tired, I cannot go on to 4b

Cheers!