1) Analysis subquestions (7 points): (a) Draw the mechanism of the reaction - remember, there are two main parts to the aldol condensation, the addition step, followed by the elimination. (b) Explain why your reaction forms the enone product, rather than the hydroxyketone intermediate. 2) Critical analysis (7 points): a) You have been given a 1H NMR spectrum of your product. Fully assign this spectrum (i. e. determine which peaks in the 1H NMR correspond to which hydrogens in the product). The peaks have been labeled 1-8 on the spectrum, and the relevant hydrogens Ha-Hh below. b) Calculate the coupling constant between He and Hf. Explain how can this can help determine the stereochemistry (i. e. cis vs. trans) of the double bond. (7) Acetone is a symmetrical molecule, so there are two positions that can react. Draw the product you would expect to obtain if you used two molar equivalents of vanillin rather than one. c) Acetone is a symmetrical molecule, so there are two positions that can react. Draw the product you would expect to obtain if you used two molar equivalents of vanillin rather than one.

the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.