ACS QOTD: Reaction Diagram for Friday

Which of the following reaction diagrams correspond to the reaction in the box (assume an exothermic reaction overall)?

exothermic-reaction-diagram-300dpi

 

Explanation:

For this reaction, you must know that Azide (N3+) is a strong nucleophile and a weak base. A strong nucleophile will be expected to undergo an Sn2-style reaction. Answer A is eliminated because it is endothermic (among other reasons). Since Sn2 reactions are concerted, there should only be one step in the reaction and one intermediate, leaving answer B as the correct answer. Answers C and D both have two intermediates, indicating two different steps, which is characteristic of Sn1 or E1 reactions, but not Sn2.

(734) 442-2549

Which of the following compounds would be a major product of the following reaction?

acetylide-and-secondary-halide-300dpi

 

Explanation:

The first step of this reaction would generate an acetylide ion. The acetylide ion is a strong base and a strong nucleophile. Given that the leaving group, bromine, is on a secondary carbon, it is more likely to undergo an E2 elimination than an Sn2 attack. The elimination would generate the Zeitsev product (most substituted double bond) as shown:

acetylide-and-secondary-halide-mechanism-300dpi

While the most substituted alkene is not an answer choice, acetylene is, so answer D is the correct answer.

ACS QOTD: Lucas Reagent Synthesis

Reagent X will not react with Lucas reagent (HCl, ZnCl2) at room temperature, but it will react slowly at high temperatures. Which of the following reaction sequences could generate reagent X?

product-reacts-slowly-with-lucas-reagent-300dpi

 

Make sure you try the problem before you check the solution!

 

Explanation:

Since we need to synthesize the hypothetical reagent X, we need to check each of the possible syntheses for their product. Then, with all the products, we need to determine which could be reagent X.

Here are the products after each step of each synthesis:

product-reacts-slowly-with-lucas-reagent-key-300dpi

Since reagent X will react with Lucas reagent, it must be some type of alcohol, so answer C is eliminated. Lucas reagent will readily react with tertiary alcohols, so choice D is eliminated.

While we didn’t cover the mechanism for the Lucas reagent this semester, you need to know it goes through an Sn1 mechanism. Sn1 reactions are slow at secondary carbons and nearly impossible at primary carbons. Since reagent X requires a lot of heat to react at all, it is probably a primary alcohol undergoing an Sn1 reaction. Synthesis A is correct because it generates a primary alcohol.

ACS QOTD: Polymerization Intermediate

Which is a plausible intermediate for the polymerization reaction shown?

radical-polymerization-intermediate-centered-300dpi

 

Explanation:

First, you have to know that the peroxide, CH3OOH will result in a radical polymerization, excluding answer B as a possible intermediate.

Numbering out the polymer, we can predict what the starting alkene looked like by drawing C-1 and C-2, every substituent attached to them, and placing a double bond between C-1 and C-2:

radical-polymerization-intermediate-predict-polymer

Then, by propagating a radical with this starting alkene, we get the intermediate from answer D.

radical-polymerization-intermediate-propagation-step

Answer choice C would have required a double bond between carbon 1 and the methyl attached to carbon 1.

Answer choice A would polymerize to form an alkene polymer, while the boxed polymer is an alkane, so answer A would not work.

ACS QOTD: Stereochemistry of Some Compounds

Which of the following statements is true about the compounds in the box?

stereocenters-and-chirality-thin-large-300dpi

 

Explanation:

Compound II is a meso compound (it has an internal plane of symmetry), so it must not be optically active, eliminating A. Compound III has no chiral centers (try to do R and S on the carbons to check), so it must not be chiral. III is not a meso compound, so answer C must not be correct. This leaves answer D as the correct answer.

So what’s going on here? Compound I has undefined stereochemistry (it’s not drawn in the R/S form), but the nitrogen atom is a stereocenter. Check out the possible enantiomers:
nitrogen-inversion

On the other hand, these two enantiomers can interconvert in a process called nitrogen inversion, so a solution of this compound would be optically inactive. You may need to know about nitrogen inversion for your final, but the easiest way to get the correct answer for this question is by eliminating all the other answers.

By the way, today is our ACS review session. We’ll be holding it in the MLC. Check the Facebook group for more information.

ACS QOTD: Attack of an Alkyl Halide

The reaction between tert-butyl bromide and NaSCH3 is shown in the box. If the solvent volume was doubled and the NaSCH3 concentration was halved, what would be the ratio of the original rate of reaction to the new rate?

trisubstituted-alkylbromide-sn2-300dpi

 

 

Explanation:

Since tert-butyl bromide is a tertiary halide, it cannot react in a Sn2 fashion. So this replacement reaction must occur by Sn1 mechanism. The rate of an Sn1 reaction is only dependent on the concentration of the substrate, so the rate equation would be given as:

Rate1 = k * [ (CH3)3CBr ]

The concentration of NaSCH3 has no effect on the rate of reaction, since it is not in the rate equation. Doubling the solvent concentration effectively halves the concentration of (CH3)3CBr. This can be represented in the Rate equation as:

Rate2  = k * ½[ (CH3)3CBr ]

So the new rate will be 1/2 the rate of the old rate. This can be represented as a ratio of 1 : 1/2 or 2 : 1, so answer A is the correct answer.

ACS QOTD: Friday Final Product

What is a major product of the following reaction?

halohydrin-on-1-methylpentene

 

Explanation:

This is a halohydrin formation reaction. First, you have to understand that once the bromonium bridge forms on either the top or the bottom of the compound, water will be the nucleophile which attacks and opens the bridge, eliminating choice D.

halohydrin-on-1-methylpentene-intermediate-step

Water must attack the bromonium bridge at the most substituted carbon, eliminating answer A. Because the water must attack on the opposite side of the bromonium bridge, it will be trans to bromine in the product, so answer choice B must be the correct answer.

ACS QOTD: A Reactive Epoxide

What is the product when this epoxide reacts with H3O+?

 

Explanation:

For this question, you must understand the reaction mechanism for the acid catalyzed epoxide opening and the meaning of radiolabeled elements. Radiolabeling  an element, such as 18O, is a way of keeping track of which oxygen is which. After the epoxide is protonated, water in solution will attack at the most substituted carbon.

Because the non-radiolabeled oxygen attaches to the most substituted carbon, choices A and D are eliminated. The water cannot attack on top of the bridge due to steric hindrance, so it must sneak underneath and perform a “backside” attack (also known as an Sn2 attack). This backside attack will cause the stereochemistry of the attacked carbon to flip. Answer choice C correctly represents this inversion of stereochemistry.

929-444-2272

acs-diels-alder

 

Explanation:

Because the Diels-Alder reaction always forms rings of 6 (4 + 2), product II could not have been formed from a Diels-Alder, so answers C and D cannot be correct. Product III could have formed from the reaction between a diene and an alkyne dienophile, so answer B must be the correct answer.

On Symbols and Perception

I love this snippet on symbolic vs. perceptual analysis from Sanjoy Mahajan in his book Street-Fighting Mathematics. Students often tell me that problems are so much easier after seeing them worked out rather than reading a description of the laws and rules from their textbook. Do those descriptions help to define the reaction/relationship/method? Yes. And yet, our ability to perceive and visualize is almost always faster and more effective than parsing through the symbolic method.

A symbolic description, whether a proof or an unfamiliar temperature, is unconvincing compared to an argument that speaks to our perceptual system. The reason lies in how our brains acquired the capacity for symbolic reasoning. (See Evolving Brains [2] for an illustrated, scholarly history of the brain.) Symbolic, sequential reasoning requires language, which has evolved for only 105 yr. Although 105 yr spans many human lifetimes, it is an evolutionary eyeblink. In particular, it is short compared to the time span over which our perceptual hardware has evolved: For several hundred million years, organisms have refined their capacities for hearing, smelling, tasting, touching, and seeing.

Evolution has worked 1000 times longer on our perceptual abilities than on our symbolic-reasoning abilities. Compared to our perceptual hardware, our symbolic, sequential hardware is an ill-developed latecomer. Not surprisingly, our perceptual abilities far surpass our symbolic abilities. Even an apparently high-level symbolic activity such as playing grandmaster chess uses mostly perceptual hardware [16]. Seeing an idea conveys to us a depth of understanding that a symbolic description of it cannot easily match.

[2] John Morgan Allman. Evolving Brains. W. H. Freeman, New York, 1999.

[16] Fernand Gobet and Herbert A. Simon. The role of recognition processes and look-ahead search in time-constrained expert problem solving: Evidence from grand-master-level chess. Psychological Science, 7(1):52-55, 1996.