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AS/A-Level Chemistry - Carbonyl compounds & cyanide, iodine

Carbonyl compounds & cyanide, iodine

· A-level Chemistry,Organic Chemistry,functional group,ketones,Aldehydes

After learning about aldehydes and ketones in A-Level Chemistry, let's see how they react with cyanide!

The electron cloud in the C=O group is distorted with the carbon atom being δ+, which can be attacked by nucleophiles. The δ- oxygen atom is not attacked by electrophiles, because the oxygen atom is itself strongly electronegative. Compounds with fewer R groups on the >C=O are more susceptible to nucleophilic attack because Cδ+ is stronger. This can be demonstrated in the reaction with hydrogen cyanide: 

· Reagents: Hydrogen cyanide HCN ( in the presence of KCN) 

· Conditions: Reflux in alkaline solution (pH 8). If the pH is too high, there are insufficient H+ ions, however if it is too low there are insufficient CN- ions.  

 

The mechanism of this reaction is: 

· The lone pair of electrons on the carbon atom of the CN-  ion forms a bond with the δ+ carbon atom. At the same time, the π-electrons in the C=O group move to the oxygen atom 

· The anion formed in the first step removes a proton from a HCN molecule to form the organic product and another CN- ion.  The –O- group in the intermediate donates a lone pair of electrons to the hydrogen in a HCN molecule as the σ-bond between the H and the CN breaks. This regenerates CN- ions, which catalyse the reaction. 

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When hydrogen cyanide adds on to an aldehyde or to an asymmetric ketone, the product is a racemic mixture. This is because the carbonyl compound is planar around the >C=O group, and therefore, the cyanide ion can attack from above or below the plane.  

Ethanal and methylketones undergo the iodoform reaction with iodine in alkali, a complicated process in which the hydrogen atoms of the CH3C=O group are replaced by iodine atoms. The alkali present in the reaction mixture then causes the C-C bond to break and a pale yellow precipitate of iodoform, CHI3, is formed. This means the substance contains either a CH3CH(OH) or CH3CO group. 

· NaOH solution is added to iodine solution to form iodate ions (IO-) 

I2+ OH- ⇨  IO- + I- 

· These substitute into the –CH3  group next to the C=O group, forming a CI3C=O group. The electron withdrawing effect of the three halogen atoms and the oxygen atom weaken the σ-bond between the two carbon atoms and this breaks forming iodoform: 

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Drafted by Eunice (Chemistry)

References

https://slideplayer.com/slide/12735192/

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