Explain the fact that whereas 2-bromopentane undergoes...

Explain the fact that whereas 2-bromopentane undergoes dehydrohalogenation with $\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-} \mathrm{K}^{+}$to give mainly 2-pentene (the Sayzteff product), with $\mathrm{Me}_3 \mathrm{CO}^{-} \mathrm{K}^{+}$it gives mainly 1-pentene (the anti-Sayzteff, Hofmann, product). Since $\mathrm{Me}_3 \mathrm{CO}^{-}$is a bulky base, its attack is more sterically hindered at the $2^{\circ} \mathrm{H}$ than at the $1^{\circ} \mathrm{H}$. With $\mathrm{Me}_3 \mathrm{CO}^{-}$,

Explain the fact that whereas 2-bromopentane undergoes dehydrohalogenation with $\mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-} \mathrm{K}^{+}$to give mainly 2-pentene (the Sayzteff product), with $\mathrm{Me}_3 \mathrm{CO}^{-} \mathrm{K}^{+}$it gives mainly 1-pentene (the anti-Sayzteff, Hofmann, product).
Since $\mathrm{Me}_3 \mathrm{CO}^{-}$is a bulky base, its attack is more sterically hindered at the $2^{\circ} \mathrm{H}$ than at the $1^{\circ} \mathrm{H}$. With $\mathrm{Me}_3 \mathrm{CO}^{-}$,

Key Concepts

Saytzeff's Rule

Saytzeff's rule posits that the most substituted alkene, which is generally more stable, will be the major product in an elimination reaction. This rule applies when the base is small and unhindered, allowing it to effectively abstract the proton leading to the more substituted, Saytzeff product.

Regioselectivity

Regioselectivity in elimination reactions describes the preference for forming one constitutional isomer over another, based on factors such as stability of the alkene and steric effects. The choice between forming a more substituted versus a less substituted alkene is a classic example of regioselectivity influenced by the size and reactivity of the base.

Steric Hindrance

Steric hindrance refers to the impediment of a chemical reaction due to the physical size of groups within a molecule. In the context of elimination reactions, bulky bases experience steric hindrance when approaching more substituted (more hindered) hydrogen atoms, thereby favoring reaction at less hindered sites.

Elimination Mechanisms (E2 Reaction)

In elimination reactions, particularly the E2 mechanism, a base abstracts a proton while the leaving group departs in a single, concerted step. This simultaneous process leads to the formation of a double bond. The mechanism's outcome, including the regiochemistry of the alkene, is affected by the substrate structure and the characteristics of the base used.

Hofmann Elimination

Hofmann elimination is the process where a bulky base leads to the preferential formation of the less substituted alkene. Because a bulky base is sterically hindered, it often cannot approach the more substituted, and thus more crowded, hydrogen atoms, resulting instead in the abstraction of a less hindered (typically primary) hydrogen and formation of the anti-Saytzeff or Hofmann product.

Transcript

00:01 So in this problem, we're asked to account for this observation where for relating to the bromination of cis 2 heptine.

00:14 So cis 2 heptine would look like this.

00:19 1 .1, 2, 3, 4, 5, 6, 7.

00:23 Okay.

00:24 And so for simplicity, i'm just going to abbreviate this butyl group as a bu.

00:31 So the reason that identical amounts of these stereoisomers are obtained, we can figure it out as this drawing out of the mechanism.

00:44 And so let's picture this b -u here and our methyl here and it's sticking out like this, right? when it first attacks the bromine, it can either happen from here and form this bromonium.

01:15 Sorry, these should be hydrogens, and form this bromonium.

01:32 Excuse me.

01:36 Or it can attack from down here and form this bromonium.

01:44 So there's two different possibilities of what you can form with regards to this ion.

01:55 Sorry, this should also be a plus.

02:07 And hopefully you can see these are ananchomers, right? and so when you have this bromide come in, it's going to attack the less sterically hindered side and open over here.

02:22 It's going to do that for both of them.

02:25 However, these two products are not going to be the same, right? one of them is going to look like this.

02:40 The other one is going to look like this.

02:52 Okay.

02:52 And if you look at these, these two compounds are going to be nanchomers.

03:13 And we can verify this by assigning stereocentors to these two...

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