Draw a structural formula for the product formed when 2,2-dimethylpropanal is reduced by hydrogen in the presence of a transition-metal catalyst. • You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. | 3 0. 8) ©////-000- [F CH₂ ChemDoodle Complete the following reaction by writing the name of the product. CH₂ 0 + H₂ Ni or Pt ?

Draw a structural formula for the product formed when 2,2-dimethylpropanal is reduced by hydrogen in the presence of a transition-metal catalyst. • You do not have to consider stereochemistry. • You do not have to explicitly draw H atoms. | 3 0. 8) ©////-000- [F CH₂ ChemDoodle Complete the following reaction by writing the name of the product. CH₂ 0 + H₂ Ni or Pt ?

Organic Chemistry

8th Edition

ISBN:9781305580350

Author:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Publisher:William H. Brown, Brent L. Iverson, Eric Anslyn, Christopher S. Foote

Chapter6: Reactions Of Alkenes

Section: Chapter Questions

Problem 6.38P: Reaction of -pinene with borane followed by treatment of the resulting trialkylborane with alkaline...

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For the base-catalysed hydrolysis of 3-bromo-3-methylhexane (i.e. reaction with the nucleophile OH-): For the reaction intermediate, draw its structure and give the VSEPR description of the geometry at the reaction centre Give the names and draw the structures of the two reaction products. Explain your conclusions.
(10pts) Compound A, C10H16, was found to be optically active. On catalytic reduction over a palladium catalyst, 2 equivalents of hydrogen were absorbed, yielding compound B, CioH2o. On ozonolysis of A, two fragments were obtained. One fragment was identified as acetic acid (CHCOOH). The other fragment, compound C, was an optically active carboxylic acid, C8H14O2. Write reactions, and draw the correct structures for A-C, explain your answer in detail.
Reaction of -pinene with borane followed by treatment of the resulting trialkylborane with alkaline hydrogen peroxide gives the following alcohol. Of the four possible cis,trans isomers, one is formed in over 85% yield. (a) Draw structural formulas for the four possible cis,trans isomers of the bicyclic alcohol. (b) Which is the structure of the isomer formed in 85% yield? How do you account for its formation? Create a model to help you make this prediction.
A chemist allows some pure (2S,3R)-3-bromo-2,3-diphenylpentane to react with a solution of sodium ethoxide(NaOCH2CH3) in ethanol. The products are two alkenes: A (cis-trans mixture) and B, a single pure isomer. Under thesame conditions, the reaction of (2S,3S)-3-bromo-2,3-diphenylpentane gives two alkenes, A (cis-trans mixture) and C.Upon catalytic hydrogenation, all three of these alkenes (A, B, and C) give 2,3-diphenylpentane. Determine the structuresof A, B, and C; give equations for their formation; and explain the stereospecificity of these reactions
Compound AA has a molecular formula of C3H6O and gives a positiveresult using Tollen’s reagent. The reaction of compound AA with hotacidified potassium permanganate, KMnO4 gives compound BB. Thecatalytic hydrogenation of compound AA with nickel, Ni producedcompound CC. The reaction of compound BB with ethanamine,CH3CH2NH2 produces compound DD I) Draw the structural formula of compounds AA, BB, CC and DD. 2)Name the type of chemical reaction for the formation of compound CC.
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Write equations for the reaction of 2-butanol with reagent. Where you predict no reaction, write NR. Q)K2Cr2O7, H2SO4, H2O, heat
A graduate student was studying enzymatic reductions of cyclohexanones when she encountered some interesting chemistry. When she used an enzyme and NADPH to reduce the following ketone, she was surprised to find that the product was optically active. She carefully repurified the product so that no enzyme, NADPH, or other contaminants were present. Still, the product was optically active. If this reaction could be accomplished using H2 and a nickel catalyst, would the product be optically active? Explain.
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#B: Methyl acetate has methoxy, -OCH3 as the remaining of the alcohol part in the ester. Isopropyl acetate has isopropoxy, -OCH(CH3)2 as the remaining of the alcohol part in the ester. -OCH(CH3)2 is more electron donating than the methoxy, -OCH3 group due to the presence of two electron-donating -CH3 group in the former. Hence saponification reaction of Isopropyl acetate is much slower than methyl acetate. Hence the rate of saponification of methyl acetate, CH3CCO2CH3 is 50 times greater than that for isopropyl acetate.