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Flashcards in Addition reaction mechanisms Deck (12):

What happens when a bromine molecule approaches the double bond in ethene?

It becomes polarised. The electron rich double bond in ethene pushes the electrons in the bromine molecule towards the furthest away bromine atom. This atom gains a slight negative charge whilst the other gains a slight negative charge.


What does the slightly positive bromine atom in the molecule then do?

It attacks ethene and a cyclic ion intermediate is formed along with a bromide ion.


What is an electrophile?

Positively charged electron pair acceptors


What type of fission does the bromine molecule undertake?



What does the second step involve?

The Br- ion attacking the cyclic ion intermediate.


What is a nucleophile?

negatively charged ions which have atlas one pair of lone electrons that they can donate.


What happens during step 1 of hydrohalogention?

The slightly positive H atom attacks the double bond and bonds to one of the carbons of the double bond. Forming an intermediate known as a carbocation. The Halogen halide bond breaks heterolytically and a halogen ion is formed as well.


What happens during the second step of hydrohalogenation?

The Br-ion attacks the carbocation intermediate.


Why does the stability of a carbocation depend on the number of alkyl groups attached to the positively charged carbon atom?

The alkyl groups are electron donating and can push electrons onto the positively charged carbon atom, thus increasing the stability of the carbocation.


What happens during step 1 of acid-catalysed hydration of an alkene?

The hydrogen ion of the acid catalyst is an electrophile and attacks the electron-rich double bond in the propene molecule to form a carbocation.


What happens during step 2 of acid-catalysed hydration of an alkene?

The carbocation then goes under rapid nucleophilic attack by a water molecule to give a protonated alcohol.


What happens during step 3 of acid-catalysed hydration of an alkene?

The protonated alcohol is a strong acid and readily loses a proton giving the final product.