Give the structure of the product you would expect from the reaction (if any) of 2-butyne with one molar equivalent of hbr. if there is no product, simply draw the reactant, 2-butyne. note: all structures should be drawn with no bonds between carbon and hydrogen.

(Z)-2-bromo-2-butene is produced here.

Explanation:

Reaction of 2-butyne with one equivalent of HBr yields (Z)-2-bromo-2-butene.

In the first step, protonation of triple bond produces a carbocation with a double bond. Two methyl groups present in carbocation remain in opposite side to avoid steric interaction.

In the second step, a bromide ion gives nucleophilic addition reaction with carbocation to give (Z)-2-bromo-2-butene.

Full reaction mechanism has been shown below.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.