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Solution: Manipulating Taylor series Use the Taylor series

Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett ISBN: 9780321570567 2

Solution for problem 26E Chapter 9.3

Calculus: Early Transcendentals | 1st Edition

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Calculus: Early Transcendentals | 1st Edition | ISBN: 9780321570567 | Authors: William L. Briggs, Lyle Cochran, Bernard Gillett

Calculus: Early Transcendentals | 1st Edition

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Problem 26E

Manipulating Taylor series Use the Taylor series in Table 9.5 to find the first four nonzero terms of the Taylor series for the following functions centered at 0.

Step-by-Step Solution:
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OrganicChemistry 2Exam3Study Guide There is a lot to cover, so sorry if there is reaction center up in this study guide. Reactions with Aldehydes and Ketones – these are more reactive than ketones because the sterics make them easier to approach and less electrons are involved. Aldehyde with water – This is a nucleophilic addition mechanism Aldehydes with Alcohols- Reactions of aldehydes with alcohols product hemiacetals or acetals. HCl is used to produce the acetal. These reactions are nucleophilic substitution. This is the same with ketones but they are called ketals. Aldehydes (this also is the same for ketones) with primary amines- This makes imines. Below is a handy table that shows all the ammonium derivatives and reactions. In the image below you may recognize a reaction we see a lot, be sure to know it! Thioacetals: (He used SH-C-C-SH on the practice exam). Aldehydes and Ketones with Acid or Base: Something important to know is alpha and beta carbons. A carbon next to the carbonyl carbon is called the alpha carbon, so if a carbon has a double bonded oxygen to it, the carbon adjacent is the alpha. Let’s cover Sodium Borohydride and LAH Sodium Borohydride: LAH (also shows Sodium Borohydride again) So let’s cover Aldol, Claisen, Michael and Robinson Aldol: If a structure contains two oxygen double bonds if it can make either SIX OR FIVE membered ring it will have an intramolecular reaction instead of having to react with two molecules. Dehydration of Aldol products The dehydration of the initial Aldol products can occur, it is often favoured by the following factors:  The Aldol product must still have an α-hydrogen between the carbonyl and the -OH group.  Look for extended conjugated in the product (i.e. increased stability).  Heating the reaction often favours elimination of water (dehydration).  Non-aqueous reaction conditions favour the removal of water (equilbrium shifts to dehydration products). This was another question on the practice exam. Mixed Aldol Reactions: The success of these mixed aldol reactions is due to two factors. First, aldehydes are more reactive acceptor electrophiles than ketones, and formaldehyde is more reactive than other aldehydes. Second, aldehydes lacking alpha-hydrogens can only function as acceptor reactants, and this reduces the number of possible products by half. Mixed aldols in which both reactants can serve as donors and acceptors generally give complex mixtures of both dimeric (homo) aldols and crossed aldols. Because of this most mixed aldol reactions are usually not performed unless one reactant has no alpha hydrogens. Claisen Reactions The mechanism: So if you can see once of the O-R groups is removed. Since we are talking about claisen we need to talk about Diekman Condensation as well, which is the intramolecular reaction. As you can see there is still a loss of the O-R group. Mixed Claisen Condensation - I am not sure if he will make us do this but he liked to throw curve balls so here: Same reaction but it isn’t reacting with another molecule of itself. Michael Additions and Robinson Michael EXAMPLES: As you can see there is a loss of a double bond, so the second molecule can attach. Robinson is a Michael addition followed by an aldol reaction. Below is the mechanism for an example of Robinson. LDA- reagent Salvador likes to use Trimethylsilyl chloride / CH3-I Trimethylsilyl chloride is a stronger electrophile than methyl iodide (note the electronegativity difference between silicon and chlorine). Relative to the methylation reaction, the S 2 transition state N will resemble the reactants more than the products. Consequently, reaction at the site of greatest negative charge (oxygen) will be favored. Also, the high Si–O bond energy (over 25 kcal/mole greater than Si–C) thermodynamically favors the silyl enol ether product. CH3-I: The negative charge density is greatest at the oxygen atom (greater electronegativity), and coordination with the sodium cation is stronger there. Because methyl iodide is only a modest electrophile, the S N transition state resembles the products more than the reactants. Since the C-alkylation product is thermodynamically more stable than the O-alkylated enol ether, this is reflected in the transition state energies. Since Salvador acidity on the practice test I will put a few refreshers here: Additional Notes: Example of Reaction from Practice Test: Answer/Mechanism:

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Chapter 9.3, Problem 26E is Solved
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Textbook: Calculus: Early Transcendentals
Edition: 1
Author: William L. Briggs, Lyle Cochran, Bernard Gillett
ISBN: 9780321570567

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Solution: Manipulating Taylor series Use the Taylor series