Draw the major product expected when 1,3-butadiene is treated with one equivalent of HBr at 40ºC and show the mechanism of its formation. For the mechanism, include lone pairs and charges in your answer. Do not draw out any hydrogen explicitly. Do not use abbreviations such as Me or Ph. A) Draw step 1 of the mechanism.

The major product and the mechanism is shown below.

Explanation:

In the reaction of 1,3-butadiene, with one equivalent of HBr, the addition product 1,2 is generated primarily at 40 ° C. When 1,3-butadiene reacts with HBr, a mixture of two products is obtained. One of them, 3-bromo-1-butene, is formed by the addition of HBr to one of the double bonds of the classic structure of this compound.

In contrast, the second and major product, 1-bromo-2-butene, does not have any of the terminal double bonds but contains a double bond between the C2-C3 carbons.

the answer is either b or c

the number of protons in the nucleus of a given element. has slight luster and conducts electricity at high temperatures like metal.but like a nonmetal,boron is brittle and does not conduct electricity well at low temperatures. in graphite, carbon is a soft,brittle solid.it is reflective,ductile,or malleable.