MediumChemistryHigh School (11–12)Organic Chemistry: Some Basic Principles and Techniques › Electronic Effects and Reaction IntermediatesNumerical
Propene, CH₃CH=CH₂, reacts with bromine to give 1,2-dibromopropane, CH₃CHBrCH₂Br…
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Propene, , reacts with bromine to give 1,2-dibromopropane, . Count the sigma bonds and the pi bonds in propene and in the product.
Need a reminder? The key ideaShow
A nucleophile is electron-rich and gives a pair. An electrophile is electron-poor and takes one. There are four types of reaction. Substitution replaces a group, with two molecules in and two out. Addition joins two molecules into one and uses up a pi bond. Elimination removes a small molecule and makes a pi bond. Rearrangement moves atoms inside one molecule. When counting bonds, a double bond holds one sigma bond and one pi bond.
The worked solution
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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.
Step 1. Propene. It has two C-C bonds, one of them double, so sigma bonds and pi bond between carbons. It has C-H bonds. Sigma total: .
Step 2. The product. It has C-C sigma bonds, C-H bonds and C-Br bonds. Sigma total: . It has no pi bonds.
Answer: propene has sigma bonds and pi bond. The product has sigma bonds and no pi bond. One pi bond turned into two new sigma bonds.
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