Acid Base Extraction lab 4

20:32 hydroxide 2. Using curved arrows, show the deprotonation mechanism of phenol with (OH). Also, provide the structure of the product of this deprotonation. Start answering this question by drawing the structure of phenol. Then hydroxide ion. Find your most reactive electrons and draw the curved arrow that shows where they would go to deprotonate phenol. Then "make a bond break a bond" write they question 3. Provide valid resonance structures for the product of the previous arrow(s) charge moving onto the ortho- and para positions. Use curved to show how the charge resonates to the next structure in your scheme. all the proper notations for depicting resonance. that show the in the structures Be sure to use Send a chat C

11.27: Tell which is the stronger base and why? Aniline or p-cyanoaniline b) aniline or diphenylamine WH₂ |ال 11.30: Place the following substances, which have nearly identical formula weights, in order of increasing boiling point? 1-aminobutane, 1-butanol, methyl propyl ether, pentane

Organic chemistry : Amine Please answer questions (3) and (4) just like in the pictures given below and finish it BY TODAY ASAP

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The presence of a pi bond also makes a compound a base. With this in mind, draw the conjugate acid of ethylene, CH2=CH2.

Low-molecular-weight dicarboxylic acids normally exhibit two different pK, values. Ionization of the first carboxyl group is easier than the second. This effect diminishes with molecular size, and for adipic acid and longer chain dicarboxylic acids, the two acid ionization constants differ by about one pK unit. Dicarboxylic Acid Structural Formula pK,2 a1 Охalic НООССООН 1.23 4.19 Malonic НООСCH,COOH 2.83 5.69 Succinic HOOC(CH,),COOH 4.16 5.61 Glutaric HOOC(CH,),COOH 4.31 5.41 Adipic HOOC(CH,),COOH 4.43 5.41 Why do the two pK, values differ more for the shorter chain dicarboxylic acids than for the longer chain dicarboxylic acids?

9. Consider the neutral base pyrrolidine, C4H9N, designated B. (a) Which is the predominant species, B or BH¹, at pH 11? At pH 12? (b) At what pH is [BH] = [B]? (c) What is the quotient [B]/[BH] at pH 12.00? at pH 2.00?

Consider the two following nitrophenol structures: OH OH `NO2 NO2 p-nitrophenol m-nitrophenol Predict which will be the stronger acid. (Hint: Consider possible resonance structures analogous to those given in the text for phenol.)

6. Predict the position of the equilibrium and calculate the equilibrium constant, Keq, for the following acid-base reaction. CH3NH₂+ + CH3COO- CH3NH₂ + CH3COOH methylamine acetic acid acetate ion pK₂ = 4.74 methylammonium ion pK₂ = 10.64

First, draw your chosen compounds including any non-bonding electrons and non-zero formal charges Circle and name any pH-sensitive functional groups. Choose an organic solvent that will dissolve both or your compounds and an aqueous solution (i.e. conc. aqueous NaOH, 5% aq. NaHCO3, or 3M HCl) that would successfully separate the two compounds selected.

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2. Consider the following equilibrium: H₂O(l) + HC₂H₂O₂(aq)=H₂O*(aq) + C₂H₂O₂(aq) Why will the addition of NaOH to a solution of acetic acid cause the concentration of the acetate ion to increase? ring

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The C-H bond in acetone, (CH,),C=0, has a pk, of 19.2. Draw two resonance structures for its conjugate base. Then, explain why acetone is much more acidic than propane, CH,CH,CH, (pK,= 50). Use skeletal structures for the resonance structures. Resonance structures: draw structure draw structure (negative charge on C) (negative charge on O) Acetone is much more acidic than propane because: Acetone's conjugate base has fewer resonance structures than propane's conjugate base. The negative charge on acetone's conjugate base is stabilized by resonance. The negative charge on acetone's conjugate base is stabilized by inductance, Acetone's higher molecular mass makes it more acidic than propane.

Alcohols are acidic in nature. Therefore, a strong base can abstract the acidic hydrogen atom of the alcohol in a process known as deprotonation. The alcohol forms an alkoxide ion by losing the proton attached to the oxygen atom of the hydroxyl ( -OH) group. The alkoxide formed can act as a base or a nucleophile depending on the substrate and reaction conditions. However, not all bases can abstract the acidic proton of alcohols and not all alcohols easily lose the proton. Deprotonation depends on the strength of the base and the acidity of the alcohol. Strong bases, such as NaNH2, can easily abstract a proton from almost all alcohols. Likewise, more acidic alcohols lose a proton more easily. Determine which of the following reactions would undergo deprotonation based on the strength of the base and the acidity of the alcohol. Check all that apply. ► View Available Hint(s) CH3CH,OH + NH3 →CH,CH,O-NH CH3 CH3 H3C-C-H+NH3 → H3 C-C-H OH O-NH CH3CH2OH + NaNH, → CH3CH,O-Na* + NH3 CHC12 Cl₂…

Rank the following compounds in order of increasing acidity (1 = least acidic, 3 = most acidic) and in the space provided use resonance (of the conjugate base) to explain why the compound you have labelled “3” is the most acidic.

In this lab you are performing your first organic synthesis and safety information on all the materials you are using must be obtained. Companies are required to provide material safety data sheets (MSDS) for chemical compounds they sell. Please examine the MSDS for salicylic acid, available here. You will find GHS pictograms within the sheet that summarize the main hazards of this compound. The pictograms are essentially the same as the symbols on WHMIS labels, available here for reference. Based on the pictograms, whatare the main hazards for salicylic acid? Select as many answers as appropriate Select one or more: Acute oral or inhalation toxicity Flammable material Carcinogen Self-reactive substance Gas under pressure Corrosive Oxidizing material

If 0.100 mol of phenol, C6H5OH, were dissolved in pure water to make 1.000 L of total solution, what would the concentration of C6H5O- be at equilibrium? What would the acid’s percent dissociation be?

Propose you are given a mixture of naphthalene, propanoic acid, and diethyl amine. Explain, using a flow chart similar to that in #1, how you would separate these three compounds. Include all solvents and reagents, show the exact compound that is present at each juncture, and be clear about the experimental technique employed in each step.

t" under the one which would be soluble in water. Writeeast" under the one which would be least soluble in water. butanoic acid heptanoic acid formic acid Write "lowest" under the compound with the lowest boiling point. Write "highest" under the compound with the highest boiling point. butanoic acid cyclohexane pentanol Write "most" under the one which would be most acidic. Write "least" under the one which would be least acidic. phenol benzoic acid cyclohexanol Write "best" under the one which would be likely to smell best. Write "worst' under the one which would be likely to smell worst. butanoic acid 2-pentanone methyl butanoate Write in the products. a) (C,H),CULI CH b) HOCH,CH,OH

Complete the equation to show how pyridine, C5H5NC5⁢H5N, acts as a Brønsted-Lowry base in water. equation: C5H5N+H2O↽−−⇀

4) Write the reaction between cyclohexanecarboxylic acid and thionyl chloride showing all three products (two gaseous and one liquid). Take the liquid product and treat it with 2 moles of dimethylamine in benzene as the solvent. You get two products. What are they? Draw their formulae and write their names.

Amines are weak bases. Table 10.2 in the course textbook provides a list of pKb values for several amines. The pK values for three amines and their structures are shown below. Explain why increased substitution on nitrogen results in an increase of the pKb value. H3C-NH2 PK = 3.36 H3C N-H pKb = 3.27 H₂C' H3C N-CH3 PK = 4.19 H₂C

In the presence of excess base and excess halogen, a methyl ketone is converted to a carboxylate ion. The reaction is known as the haloform reaction because one of the products is haloform (chloroform, bromoform, or iodoform). Before spectroscopy became a routine analytical tool, the haloform reaction served as a test for methyl ketones: the formation of iodoform, a bright yellow compound, signaled that a methyl ketone was present. Why do only methyl ketones form a haloform?