Arrange the following alkyl bromides in order from most reactive to least reactive in an SN2 reaction: 1-bromo-2-methylbutane, 1-bromo-3-methylbutane, 2-bromo-2-methylbutane, and 1-bromopentane.

Rank the alkyl bromides from most reactive to least reactive. To rank items as equivalent, overlap them.

And what makes an alkyl bromide more reactive in this case?

Primary alkyl halides tend to undergo the SN2 reaction mechanism in nucleophilic substitution since there is less steric hindrance for nucleophilic attack and the carbocations that they form are not as stable as those formed from tertiary alkyl halides.

1-bromopentane > 1-bromo 2-methylbutane > 1-bromo-3-methylbutane> 2-bromo 2-methylbutane

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joetheelite joetheelite · Answer 2 · 2019-12-30T09:01:09-05:00

Answer:

See explanation

Explanation:

First, in order to know this it's neccesary to remember how a SN2 reaction takes place. A Sn2 reaction is a bimolecular concerted reaction where all bonds are broken and making in only one step.

For this to occur, we need a strong nucleophyle (such a strong base) and a substract with a great outgoing group (The halides are great leaving groups).

The nucleophyle attacks on the back side of the molecule with the bromine, and the result is a molecule with inverted configuration and the bromine is replaced by the nucleophyle.

However, this step is fast and concerted, and in order to do this faster, the reactant must be (Ideally) with no substituent, because if the molecule is bulky, the nucleophyle's attack to the back side is hard. This doesn't mean that it will not undergo, but it will be harder and slower.

Because of this reason, we can see that from all the alkyl bromides there, the least bulky is the 1-bromopentane, so this will be the more reactive in Sn2, followed by 1-bromo-methylbutane, then the 1-bromo-2-methylbutane and finally the 2 - bromo - 2 - methylpentane.

In the picture, you have the structures of these molecules, so you can see how the steric hindrance affects this.