Answer to Question #18947 in Organic Chemistry for Butter
A student is asked to react 3-methyl-2-butanol with HBr. predict the products and explain why the experiment will not be as successful for production of an alkyl bromide as the reaction of 1-butanol with HBr.
Microscale and Miniscale Organic Chemistry Lab Experiments 2e by Schoffstall 2004 Ch 8 Critical thinking #10.
2-methyl-2-butanol reacts with HBr to form 2-bromo-2-methyl butane and water while butanol reacts with HBr to from 1-bromobutane and water. Obviously, 2-methyl-2-butanol is a tertiary alcohol while 1-butanol is primary. Upon protonation of the oxygen, the electron donating carbon atoms will release the electrons of the C-O bond making it easier to break. This donation may also interfere with the electrons of a nucleophile as well. (While note 2-methyl-2-butanol, note the reaction of the tertiary alcohol below. The mechanism of a tertiary alcohol is SN1 and involves a tertiary carbocation.) 1-butanol, the opposite will exist. Only one electron donating carbon is present on the C-OH carbon. Hydrogen atoms will retard the cleavage of the C-O bond until formation of a Br-C bond. This is the transition state of an SN2 reaction (not shown in the mechanisms below). With the formation of a Br-C bond, now the cleave of the C-O bond is facilitated. Since there is only a single electron donating carbon atom, the electrons of reactive carbon will not interfere with the attack of bromide to the same degree as a secondary or tertiary carbon.