Which of these alkyl halides cannot be used effectively in a Gabriel amine synthesis?
A)1-bromopentane
B)1-bromo-3-methylbutane
C)2-bromo-3-methylpentane
D)1-bromo-2,3-dimethylbutane
E)2-bromo-2,3-dimethylbutane

Question

Quizplus · Accepted Answer

In a Gabriel amine synthesis, the alkyl halide is reacted with potassium phthalimide to form a phthalimide anion, which then undergoes nucleophilic substitution with a strong base to form the primary amine. The essential requirement for the alkyl halide in this reaction is that it should react readily with the phthalimide anion to form a good leaving group. The leaving group ability of an alkyl halide is determined by the stability of the halide anion that is formed when it leaves. In general, more substituted alkyl halides have less stable halide anions and thus are better leaving groups. However, when the substitution becomes too crowded, steric hindrance can prevent the halide from leaving effectively. Among the choices given, E (2-bromo-2,3-dimethylbutane) has the most steric hindrance and is therefore the least effective for a Gabriel amine synthesis. The other choices contain less steric hindrance and can be used effectively in the reaction.

Which of These Alkyl Halides Cannot Be Used Effectively in a Gabriel