Topic 15 · page 9 of 28

Addition of hydrogen cyanide

Cyanide ion attacks the δ+ carbon of a C=O group in nucleophilic addition, making a hydroxynitrile. Because the C=O group is flat, an aldehyde gives a racemic mixture.

16 min read15.8(iii)15.5

Read first: Reducing carbonyl compounds with LiAlH₄

The reaction 15.8(iii)

Aldehydes and ketones react with hydrogen cyanide, HCN, to give a hydroxynitrile: a compound with an –OH group and a –CN group on the same carbon.

CHX3CHO+HCN→CHX3CH(OH)CN \ce{CH3CHO + HCN -> CH3CH(OH)CN}

The product from ethanal is 2-hydroxypropanenitrile. The carbon of the –CN group is carbon 1 of the chain, so the OH is on carbon 2.

MethodConditions

Use HCN in the presence of a little KCN at room temperature. HCN is a weak acid and gives very few cyanide ions, which are the attacking species. Potassium cyanide supplies the cyanide ion, CN⁻, and makes the reaction fast enough to be useful.

Hydrogen cyanide and cyanides are extremely poisonous, so this reaction is only done under strict controls and is not a student experiment.

This is nucleophilic addition. The C=O group is polar, with a δ+ carbon, so it is attacked by a nucleophile, which is an electron-pair donor. The product is formed by adding HCN across the C=O, and nothing is lost.

The mechanism 15.8(iii)

The mechanism has two steps. Curly arrows show the movement of a pair of electrons, from the tail of the arrow to its head.

Step 1: the nucleophile attacks the δ+ carbon.

Step 1 of nucleophilic addition to ethanal. The cyanide ion has a lone pair on carbon. A curly arrow starts at that lone pair and ends at the delta plus carbon of the C=O group. A second curly arrow starts at the C=O double bond and ends at the oxygen, which becomes O minus. The product is an intermediate with O minus and CN on the same carbon.
Step 1. The lone pair on the carbon of CN⁻ forms a new bond to the δ+ carbon, and the C=O pi bond moves onto oxygen.

Step 2: the oxygen picks up a proton.

Step 2 of nucleophilic addition. A curly arrow starts at a lone pair on the negatively charged oxygen of the intermediate and ends at the hydrogen of a hydrogen cyanide molecule. A second curly arrow starts at the H–C bond of HCN and ends at the carbon, forming a cyanide ion. The products are 2-hydroxypropanenitrile and a cyanide ion.
Step 2. O⁻ takes the hydrogen from HCN, so the product forms and a cyanide ion is regenerated.

The cyanide ion used in step 1 is regenerated in step 2, which is why only a little KCN is needed. Draw the lone pair on the cyanide carbon, the δ+ and δ− on the C=O, and all curly arrows from a lone pair or a bond, never from an atom.

Optical activity shows the C=O is flat 15.5

The carbon of the C=O group is trigonal planar. Cyanide ion can attack from above or below with equal chance.

Ethanal drawn flat at the C=O group. Cyanide can attack from the front or from behind with equal chance. The two products, 2-hydroxypropanenitrile with the CN on a wedge and with the CN on a dash, are mirror images and form in equal amounts, so the product is a racemic mixture.
Attack from the front and from behind are equally likely, so two enantiomers form in equal amounts.

In 2-hydroxypropanenitrile the carbon holding the OH has four different groups: H, CH₃, OH and CN. It is a chiral centre. Attack from the front gives one enantiomer and attack from behind gives the other.

The two enantiomers form in equal amounts, so the product is a racemic mixture and is optically inactive. This observation is evidence that the C=O group is planar and attacked equally from both sides.

Check yourself

  1. State the reagents and conditions for converting ethanal to 2-hydroxypropanenitrile, and write the equation.
  2. Explain why the reaction is called nucleophilic addition.
  3. Describe the two steps of the mechanism, naming where each curly arrow starts and ends.
  4. Explain why the product from ethanal is optically inactive although the molecule has a chiral centre.
  5. Write the equation for the reaction of propanone with HCN, and explain why this product is not optically active.
  6. Name the product of propanal with HCN.

Answers cover these until you have tried the questions

  1. HCN in the presence of a little KCN, at room temperature. CHX3CHO+HCN→CHX3CH(OH)CN\ce{CH3CHO + HCN -> CH3CH(OH)CN}.
  2. The attacking species, CN⁻, is a nucleophile (an electron-pair donor), and the product is formed by adding HCN across the C=O bond so that no atoms are lost.
  3. Step 1: a curly arrow from the lone pair on the carbon of CN⁻ to the δ+ carbon of C=O, and a second from the C=O double bond to the oxygen, giving an intermediate with O⁻. Step 2: a curly arrow from a lone pair on O⁻ to the hydrogen of HCN, and a second from the H–C bond of HCN to carbon, forming the hydroxynitrile and regenerating CN⁻.
  4. The C=O group is planar, so CN⁻ attacks from either face with equal probability. Equal amounts of the two enantiomers form, a racemic mixture. Their rotations cancel, so the mixture shows no optical activity.
  5. CHX3COCHX3+HCN→(CHX3)X2C(OH)CN\ce{CH3COCH3 + HCN -> (CH3)2C(OH)CN}. The carbon bearing the OH holds two identical CH₃ groups, so it does not have four different groups. The product is not chiral, and so it has no optical activity at all.
  6. CHX3CHX2CHO+HCN→CHX3CHX2CH(OH)CN\ce{CH3CH2CHO + HCN -> CH3CH2CH(OH)CN} is 2-hydroxybutanenitrile.