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 molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

explanation:

in the given question, the partial pressure is calculated by using the ideal gas equation, that is, pv = nrt

here, p is pressure, v is volume = 25 l, n is number of moles = 0.0104 moles, r is gas constant = -0.0821 l.atm / mol.k

t is temperature = 273 k

p = nrt / v

p = (0.0104 mol × 0.0821 l.atm / mol.k × 273 k) / 25 l

= 0.0093 atm