After completing this section, you should be able to
- write an equation to describe the reaction of an acetylide ion with an alkyl halide.
- discuss the importance of the reaction between acetylide ions and alkyl halides as a method of extending a carbon chain.
- identify the alkyne (and hence the acetylide ion) and the alkyl halide needed to synthesize a given alkyne.
- determine whether or not the reaction of an acetylide ion with a given alkyl halide will result in substitution or elimination, and draw the structure of the product formed in either case.
Make certain that you can define, and use in context, the key term below.
The alkylation of acetylide ions is important in organic synthesis because it is a reaction in which a new carbon-carbon bond is formed; hence, it can be used when an organic chemist is trying to build a complicated molecule from much simpler starting materials. The alkyl halide used in this reaction must be primary. Thus, if you were asked for a suitable synthesis of 2,2-dimethyl-3-hexyne, you would choose to attack iodoethane with the anion of 3,3- dimethyl-1-butyne
rather than to attack 2-iodo-2-methylpropane with the anion of 1-butyne.
Nucleophilic substitution reactions of acetylides
The hydrogen on a terminal alkyne is somewhat acidic, with a pKa of approximately 26. This means that, given a strong enough base, a terminal alkyne can be deprotonated, yielding a powerful carbanion nucleophile called an acetylide or alkynide. Sodium amide (NaNH2) (in liquid ammonia) or sodium hydride (NaH) are often used as the base.
The alkynyl carbanion can then be combined with a suitable electrophile, such as a primary alkyl bromide, in a carbon-carbon bond-forming SN2 displacement reaction. Secondary and tertiary alkyl bromides will usually not work in these types of reactions due to competing elimination.
Acetylide (alkynide) anions are strong bases and strong nucleophiles. Therefore, they are able to displace halides and other leaving groups in substitution reactions. The product is a substituted alkyne, as in the examples shown.
Secondary, tertiary or even bulky primary substrates will give elimination by the E2 mechanism.
1. The pKa of ammonia is 35. Estimate the equilibrium constant for the deprotonation of pent-1-yne by amide ion, as shown above.
1. Assuming the pKa of pent-1-yne is about 25, then the difference in pKas is 10. Since pentyne is more acidic, the formation of the acetylide will be favored at equilibrium, so the equilibrium constant for the reaction is about 1010
Khan Academy video: