Why does CH3 CH2 I undergo loss of HI with strong base faster than CD3 CH2 I loses Dl ? These ar

step 1 Hello everyone. Today we're doing chapter 6 problem 52. This question asks us to explain why our

Why does CH3 CH2 I undergo loss of HI with strong base...

Why does $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}$ undergo loss of $\mathrm{HI}$ with strong base faster than $\mathrm{CD}_3 \mathrm{CH}_2 \mathrm{I}$ loses $\mathrm{Dl}$ ? These are both $\mathrm{E} 2$ reactions in which the $\mathrm{C}-\mathrm{H}$ or $\mathrm{C}-\mathrm{D}$ bonds are broken in the rate-controlling step. Therefore, the H-D isotope effect accounts for the faster rate of reaction of $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}$.

Why does $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}$ undergo loss of $\mathrm{HI}$ with strong base faster than $\mathrm{CD}_3 \mathrm{CH}_2 \mathrm{I}$ loses $\mathrm{Dl}$ ?
These are both $\mathrm{E} 2$ reactions in which the $\mathrm{C}-\mathrm{H}$ or $\mathrm{C}-\mathrm{D}$ bonds are broken in the rate-controlling step. Therefore, the H-D isotope effect accounts for the faster rate of reaction of $\mathrm{CH}_3 \mathrm{CH}_2 \mathrm{I}$.

Key Concepts

Kinetic Isotope Effect

The kinetic isotope effect refers to the change in reaction rate that occurs when a particular atom in a molecule is replaced by one of its isotopes, such as hydrogen with deuterium. In reactions where the bond to the isotopically substituted atom is broken in the rate-determining step, the heavier isotope forms a stronger bond due to lower zero-point energy, resulting in a slower reaction rate compared to the lighter isotope.

E2 Reaction Mechanism

The E2 mechanism is a bimolecular elimination process in which a base abstracts a proton at the same time that a leaving group departs from the substrate. This single-step reaction involves a concerted transition state where bond-breaking and bond-forming events occur simultaneously. Its rate depends on both the substrate and the base, and the reaction is stereospecific with the requirement of an anti-periplanar geometry between the leaving group and the proton being abstracted.

Zero-Point Energy Differences

Zero-point energy differences arise because bonds involving lighter atoms vibrate at higher frequencies and possess more zero-point energy than similar bonds involving heavier isotopes. This difference means that bonds such as C–H are effectively 'weaker' and more readily broken than C–D bonds in the transition state of a reaction, contributing to the observed differences in reaction rates when isotopic substitutions are made.

Transcript

0:00 Hello everyone.

00:01 Today we're doing chapter 6 problem 52.

00:04 This question asks us to explain why our alkyne is more acidic than our ethane.

00:11 And it's saying that even though the ch bond in this alkyne has a higher bond association energy than a c .h bond in our ethane.

00:19 So in order to find out why this alkyne is more acidic, let's draw its structure and we'll draw our ethane structure as well.

00:45 So what does acidity mean? first, the score, what acidity means? so acidity means the ease of losing a proton, so the ease of giving up a proton.

00:56 So if it's saying that this molecule is more acidic than this molecule, then this molecule can go and give up its proton a lot easier to form a carbonyon than this one can do so.

01:12 So the reason is because of the s characteristics.

01:21 So let's first identify why this al -kine -c -bond has a higher bond association energy.

01:28 So if you look at this carbon, we see that this is an sp hybridized carbon, has one substitute in here, one split in here, sp.

01:38 Versus this carbon is sp3.

01:42 So we know right away that this has 50 % s characteristics, making this c -2.

01:47 H bond shorter and because it's shorter is going to be stronger.

01:52 Therefore because it's shorter and stronger it needs higher energy to the dose dissociated versus this one.

01:59 This one only has 25 % s characteristics...

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