Question

Problem: of SN,=? Warden inverson NHR % SN₂= ? a NH2 Erors [a] 30 ml de [agas oro NH gam 91/6 oll

Answer 1

The SN₂ reaction produces, as is written, Walden inversion, so the product of pure SN₂ would be:

Which has a rotating power of +30° and will be called the (+) enantiomer.

If we had pure SN₁, we would obtain a racemic mixture of both enantiomers, since it occurs via an intermediate plane carbocation and posterior attack on any of the two faces of the intermediate. Notice that all the (-) enantiomer must come from this reaction, since the SN₂ mechanism yields only the (+) enantiomer.

We can first calculate the fraction of each form present in the mixture using:

$x_{(+)}\cdot 30^{o}-x_{(-)}\cdot 30^{o}=15^{o}$

And:

$x_{(+)}+x_{(-)}=1$

From the first expression, we can rearrange:

$x_{(+)}=0.5+x_{(-)}$

We replace this in the second expression:

$0.5+x_{(-)}+x_{(-)}=1$

So we have:

$2x_{(-)}=0.5$

So, x_(-) = 0.25 and x₍₊₎ = 0.75.

Since all x_(-) must come from an SN₁ mechanism, we know we have 0.25 of the (-) enantiomer and 0.25 of the (+) enantiomer from SN₁, and the remaining 0.5 of the (+) enantiomer from SN₂. This means that the percentages of both mechanisms is 50%, since the proportion of product from both is the same.