# Which will react with gaseous bromine faster and why: C3H6 or C2H2?

I know that for both $\ce{C3H6}$ and $\ce{C2H2}$, it will be an addition reaction where bromine will be added through a free-radical mechanism. But I'm not sure which one will be faster.

At first, I thought it should be the $\ce{C2H2}$ because it has a triple bond and so it has a higher electron density ( i.e. it is more nucleophilic) and thus the reaction would proceed faster. But then I realized that the reaction for $\ce{C2H2}$ will proceed in steps so I believe it being more nucleophilic won't make much of a difference.

So that's why I'm stuck with which one will be faster?

P. S. : the C3H6 refers to Ethyne and NOT cyclopropane

• While I think from context it's clear what you meant, cyclopropane also has the formula $\ce{C3H6}$. – Tyberius Mar 29 '17 at 23:47

Regardless if the addition of $\ce{Br2}$ on propene or ethyne proceeds along a electrophilic, or a radical pathway, the reaction with propene will be faster, then on ethyne.
In ethyne, the carbon atoms in question for the reaction are formally $C(sp)$ hybrids; the ones in propene $C(sp^2)$. The higher the $s$-character, the higher the electronegativity of the carbon atoms in question. Hence, the $\pi$-electrons in ethyne are tighter bond to the carbon atoms, than in the case of propene. In addition, the methyl group adjacent to the initial C=C double bond offers stabilization of the intermediate carbocation (for the electrophilic addition, favouring the Markovnikov product) or radical (for the radical addition), that is absent in the instance of ethyne.
(Since your special example is about the addition of $\ce{Br2}$, Markovnikov selectivity is not an issue here. For halogen addition, choice of solvents, presence / absence of light or initiators may favour one or the other mechanism, too.)
• @Pink_NinjaTurtle The "$s$-character" refers to the concept of how atomic orbitals are (formally) mixed to obtain hybrid orbitals. Hence, a $sp$ is the result of mixing one $s$ and one $p$ atomic orbital; where each contributes 50%. In instance of $sp^2$, there are two $p$ and one $s$ AO mixed (formally), hence the contribution by $s$ is only about 33%. And so forth. – Buttonwood Mar 29 '17 at 17:49