Organic Chemistry
Aldehydes & Ketones
Which of the following reaction do not resemble with the major product given R-C=N © SnCl2 +HCI →R-CH=0 (ii) H30+ R-C-ci (i)H2 /Pd/BaSO. (ii) H30* Rega RCHO NaBH 10 R-C-NH2 || O > R- CH2 - NH2 R - CH = NH_NBHA → R-CH2 - NH
Content WP NWP Assessment Builder UI Ap X WP NWP Assessment Player Ul App X + C O https://education.wiley.com/was/ui/v2/assessment-player/inde 6 Chapter 12 Question 31 of 51 Current Attempt in Progress Your answer is partially correct. Give the major organic product(s) of the following reaction. 1) CH3CH2MgBr ? 0- 2) H30+ OH -CH3 CH Edit Drawing Type here to search OBI ASUS ZenBook 42 FL F2 2 1 F3 co: F4 F5 F6 '' @ # 2 % 3 4 4. 5 6 7 Q W E R. T T Y S D F G H
TPS:// eter 12 Question 22 of 51 * Your answer is incorrect. Predict the product(s) of the following reaction sequence: 1. NaBH 2.H30 ? By CH3 Show stereochemistry, if any. OH H3C CH, CH3 Edit Drawing
Deprotonation of indicated acidic H in compounds given below (1-3) generates the corresponding base. Rank the bases in order of increasing base strength (least basic to most basic). le & H H H 1 2. 3 Most Basic choose your answer... Second most basic: choose your answer... Least Basic: choose your answer... Next pais
6. dc Write the letter "A" or "B" as applicable in each blank space box below: (1 point) A) Oxidation reaction produces a carboxylic acid and a silver film inside the test tube B) No Reaction iti O // CH3-CH2-C-H [O] + Tollens reagent AgNO3 O [O] CH3-C-CH2-CH3 + Tollens reagent AgNO3
Experiment 5: Carbohydrate Qualitative tests Benedict's Test: All reducing sugars give positive Benedict's Test Principle: reducing sugars have free aldehyde or ketone group, it under go converting in to enediol forms under hot alkaline condition. The enediol are strong reducing agents converting cupric ions (Cu+2) of the Benedict's solution into Cuprous ion which altimately oxide as a red precipitate red copper(1) oxide by aldehydes. Benedict's reagent contains blue copper(II) sulfate (CuSO4). 5H20 which is reduced to red copper(I) oxide by aldehydes, thus oxidizing the aldehydes to carboxylic acids. The copper oxide is insoluble in water and so precipitates. The color of the final solution ranges from green to brick red depending on how many of the copper (II) ions are present. The test is used to distinguish between monosaccharides and reduced disaccharides This reaction will detect reducing monosaccharides in the presence of disaccharides. Reagent uses copper ions to detect reducing sugars in an acidic solution. Barfoed's reagent is copper acetate in dilute acetic acid (pH 4.6) Reducing monosaccharides are oxidized by the copper ions in a weak acidic medium to form a carboxylic acid and a reddish ppt of Cu20 (cuprous oxide). Reducing disaccharides (lactose but not sucrose) undergo the same reaction but at slower rate. Therefore, boiling time is critical and positive test within ( 2min.) indicate the monosaccharides in case of disaccharides they are first converted into monosaccharide and then colored precipitate are formed. So the porolonged boiling may hydrolyses disaccharides to give a false positive test. Method: 1 ml of the solution to be tested + 2 ml of freshly prepared Barfoed's reagent. . . Place test tubes into a boiling water bath and heat for 2 minutes. Remove the tubes from the bath and allow to cool. Formation of a green, red, or yellow precipitate is a positive test for reducing monosaccharides. Do not heat the tubes longer than 3 minutes, as a positive test can be obtained with disaccharides if they are heated long enough.
Draw the mechanism of the following reaction, using the curved-arrow notation to indicate the reorganization of electron density. Show all intermediates and denote all unshared electrons, nonzero formal charges, countercharges, and reversibility or nonreversibility. :0: CH3 CH3 CH3 Que + cat. K :OH + Hö (racemic) (racemic)
Draw the structure(s) of the major organic product(s) of the following reaction. + fuming HNO, H . 0 You do not have to consider stereochemistry. You do not have to explicitly draw H atoms. Do not include lone pairs in your answer. They will not be considered in the grading If no reaction occurs, draw the organic starting material • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner. Separate multiple products using the + sign from the drop-down menu.
Draw the structure of the major aldol product (prior to possible dehydration) of the following reaction dilute aqueous NaOH CH 0-50 . You do not have to consider stereochemistry. If no reaction occurs, draw the organic starting material
Write structures for the carbonyl electrophile and enolate nucleophile that react to give the aldol below. O HO You do not have to consider stereochemistry Draw the enolate nucleophile in its carbanion form. Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner Separate multiple reactants using the + sign from the drop-down menu.
Write structures for the carbonyl electrophile and enolate nucleophile that react to give the aldol below. ze H HO • You do not have to consider stereochemistry. • Draw the enolate nucleophile in its carbanion form • Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner Separate multiple reactants using the + sign from the drop-down menu.
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![6.
dc
Write the letter "A" or "B" as applicable in each blank space box below: (1 point)
A) Oxidation reaction produces a carboxylic acid and a silver film inside the test tube
B) No Reaction
iti
O
//
CH3-CH2-C-H
[O]
+
Tollens reagent
AgNO3
O
[O]
CH3-C-CH2](https://media.kunduz.com/media/question/raw/20211215023409152024-4104762.jpeg?w=256)
