Either the reactant (X) or the major organic product is missing from the equation below. Draw the missing compound. O CH=CHCH3 Pd/C + H₂ (excess) • You do not have to consider stereochemistry. • In the case of a missing reactant, there may be more than one answer. If so, draw all possible reactants in separate windows. • Separate structures with + signs from the drop-down menu. ChemDoodle n Submit Answer Retry Entire Group 2 more group attempts remaining
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- Draw the organic product(s) of the following reaction. You do not have to consider stereochemistry. Include cationic counter-ions, e.g., Na+ in your answer, but draw them in their own sketcher. If no reaction occurs, draw the organic starting material. Separate multiple products using the + sign from the drop-down menu.Draw the structure(s) of the major organic product(s) obtained after workup of the following reaction. You do not have to consider stereochemistry. If no reaction occurs, draw the organic starting material. Include counter-ions, e.g., Na+, I-, in your submission, but draw them in their own separate sketcher. Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner.Draw the structure(s) of the major organic product(s) obtained after workup of the following reaction. You do not have to consider stereochemistry. If no reaction occurs, draw the organic starting material. Draw structure(s) of product(s) after workup to neutralize acid. Include counter-ions, e.g., Na+, I-, in your submission, but draw them in their own separate sketcher. Draw one structure per sketcher. Add additional sketchers using the drop-down menu in the bottom right corner.
- According to this video, https://www.youtube.com/watch?v=9Ng6Zv9oLzk, draw the dienophile that was used in the example explaining s-cis and explain what is different about this dienophile from common examples?The molecule does not react with bromine in chloroform but can be brominated using bromine and FeBr3, and three different monobromo products are obtained. The number of H’s in the compound have to be a multiple of what integer? Explain your answer. The number of C’s in the compound have to be a multiple of what integer? What is the maximum number of C’s that can be in the compound? Explain how you determined the maximum number of C’s in the compound What functional groups are present in the compound (alkyl, aryl, ketone, etc.)? Explain your answer by referring to the information provided above.First image directions are "Fill in the blanks with the major organic product(s). Make sure to indicate stereochemistry in your drawings where appropriate." Second image directions are "Fill in the blanks with the major organic product or starting material(s)."
- Draw the structure of the major organic product(s) of the reaction. You do not have to consider stereochemistry. All carboxyl and amino groups should be drawn in the neutral form. 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.Which of the following statements is not correct? Oxymercuration-demercuration for alcohol synthesis is complicated by carbocation rearrangement. Oxymercuration-demercuration for alcohol synthesis follows Markonikov's Rule. Sodium borohydride is used in the last step of oxymercuration-demercuration to oxidize the hydroxyalkyl mercury compound. All of the above. Both the 1st and the 3rd statements.Is the following true or false? A highly selective reaction is one that proceeds with a high turn-over number.
- Write TRUE if the BOLD word/phrase makes the statement correct. Otherwise, write the correct WORD/PHRASE that will make the statement true. If there are two bold words/phrases in a number, write your answer for EACH of the bold words/phrases. 1. The anti-staggered conformation of butane, in which the methyl groups have a dihedral angle of less than 180°, has the highest energy. 2. The formation of 2-methylpropene as side product in the synthesis of tert-butyl chloride is a nucleophilic substitution reaction. 3. The generation of tertiary carbocation is the rate-determining step in tert-butyl chloride synthesis.what is the difference between these two the aldo condensation reaction was preformed Part A NaoH= 2.5g = 0.0625 mol Benzaldehyde= 2.65 g =0.0250 mol Acetone 0.725 g = 0.0125 mol Theoretical yield=0.0125 mol Actual yield= 2.1g Percent yield=71.67% Expected melting point= 110.5-112 degree Celsius Melting point observered= 112 degree Celsius Part B acetophenone 0.5 g, 4.2 mmol Para-methylbenzaldehyde 0.5 g, 4.2 mmol NaOH=0.18g = mol Theroretical yield=0.93g =0.0519 mol Percent yield=76.8% Actual yield= 0.715 g Melting point observered= 93 degree Celsius Expected melting point =93-94.5 degree CelsiusEnters the preferred IUPAC name of the connection in Figure 29.