MediumChemistryHigh School (11–12)Organic Compounds with Functional GroupsNumerical

A tertiary bromide is dissolved in aqueous ethanol…

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A tertiary bromide is dissolved in aqueous ethanol. It reacts by the $S_{\text{N}}1$ mechanism and follows first-order kinetics. Its rate constant is $3.5 \times 10^{-3}\ \text{s}^{-1}$ at the temperature of the experiment (a value chosen for this problem). Take $\ln 2 = 0.693$ . Find the half-life, the time for $87.5\ \%$ of the bromide to react, and the percentage of the bromide left after $600\ \text{s}$ . The first-order law gives the fraction left as $e^{-kt}$ .

Need a reminder? The key ideaShow

$S_{\text{N}}1$ is first order, $\text{rate} = k[\text{RX}]$ , with half-life $t_{1/2} = \dfrac{0.693}{k}$ . A planar carbocation gives a racemic or partly racemic product. The enantiomeric excess is the difference of the two shares, and the observed rotation is the specific rotation times tube length times concentration times the excess. A carbocation can rearrange to a more stable one by a hydride shift.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

← All Problems

← Sodium ethoxide is made by dissolving 2.30 g of sodium in an excess of dry ethanol…An optically pure chiral tertiary halide reacts with water by SN1… →

More onOrganic Compounds with Functional Groups

Discussion (0)

Loading…