Tutor profile: Danielle V.
Subject: Organic Chemistry
What is the organic product of: (CH3)2CH2MgBr + CH3CH2O -> (with acidic conditions) (a) aldehyde (b) ketone (c) carboxylic acid (d) alcohol
This is a basic grignard reaction, grignards attacking aldehydes or ketones produce alcohols. Therefore, the answer is d. To understand the mechanism draw out the reactants 1. CH3CH2MgBr is a straight 2-carbon chain with a MgBr on the end - a grignard 2. (CH3)2CH2O is a carbonyl with two methyl groups attached - a ketone Reaction explained: In this case, the carbon directly attached to the magnesium would attack the carbon of the carbonyl group. This carbon is partially negative and is seeking an partially positive environment like that of the carbon in carbonyl group. It is partially positive because the oxygen, a more electronegative atom is pulling the electron density towards itself. Therefore, these carbons react to share the electron density This would kick off the MgBr and result in the electrons of the double bond to push up onto oxygen- producing a negative formal charge. The acidic conditions would protonate the oxygen rendering it neutral, completing the reaction. The final product would be tertiary alcohol. You can see a new bond form between the carbon of the grignard and the carbon of the carbonyl - this carbon is attached to an alcohol group and three alkyl groups.
Subject: Study Skills
I am having trouble studying for my chemistry exams: I do all of the problems assigned before the test but I'm still not doing well.
My recommendation would be to make up your own problems . Likewise, repeatedly practice the assigned problems. You'll know you know it when you can successfully create and solve similar problems. Once you've solved the assigned problems, made your own, and solved those, practice teaching someone how to solve them. I like to keep in mind the quote, "If you can't teach it to a 6 year old, you don't know it well enough"
Subject: Basic Chemistry
Please identify the limiting reagent in this reaction: CBr4 + O2 -> CO2 + 2Br2 78.4g CBr4 and 32g O2
To begin, let's identify what a limiting reagent is: -A limiting reagent is the reactant in an experiment that has the smallest amount of moles. -Moles are basically like the currency of chemistry, a way for us to compare two reactants or two compounds. -If these two compounds are reacting with each other it would makes sense that the smallest amount will determine how much product we get! -We can't simply do so by comparing the number of grams of each, because the reactants are different sizes and have different elements in them! (aka: different molecular weights)- we need to convert them to chemistry currency: moles 1. Calculate Molecular Weights of Reactants: CBr4: Carbon (C) = 12 g/mol Bromine (Br) = 80g/mol x 4 = 32 g/mol Carbon Tetrabromide (CBr4) = 332 g/mol O2: O= 16g/mol x 2 = 32g/mol Dioxide (O2): 32 g/mol 2. Convert molecular weights to moles CBr4: molecular weight: 332 g/mol amount used in reaction: 78.4g (78.4g)/(332 g/mol)= 0.236 moles O2: molecular weight: 32 g/mol amount used in reaction: 32g (32g)/(32 g/mol) = 1 mole 3. Compare moles 0.236 moles (CBr4) vs. 1 mole (O2) 0.236 moles is less than 1 mole Therefore, the limiting reagent is CBr4 or Carbon Tetrabromide!
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