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Ray K.
Published Chemist with 2 BAs. This pay for my travel. Story teller.
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Organic Chemistry
TutorMe
Question:

O Chem is a lot of fun and has a lot of material to learn and memorize, but most of it will be a lot easier to understand and explain in a medium where we can draw and talk together, rather than writing paragraphs trying to explain a diagram I could draw in 10 seconds. Not to be snarky, but the 'meaty' parts of O Chem just won't fit in this text box well. Instead I will use my answer section to discuss stereochemistry and isomerism. Key concepts in O Chem which need less diagrams than most (although they would still help)

Ray K.
Answer:

Basically, two molecules are isomers if the same set of atoms come together to form differently shaped molecules, usually with similar physical properties. More formally, they have the same chemical formula, with different structures. If the same atoms are bound in the same order (but rotated around a bond for example) they are considered stereoisomers (unless they are identical of course) There are 3 main types of isomers: constitutional isomers, configurational stereoisomers and conformational stereoisomers. Configurational stereoisomers can be further broken down into subcategories, separating the different ways in which a molecule can exhibit configurational isomerism. A good basic example of constitutional isomerism is t-propane (isopropane) and butane. These are constitutional isomers because they differ in bonding order (which atoms are attached to which.) Of all the types of isomerism, constitutional isomers are the only one that can have different functional groups (ie an ether and an alcohol, or ketone and an aldehyde.) This is why we use the term isomer, not stereoisomer. Conformational stereoisomers are the next easiest to understand and probably least important type of isomers, as they are nondifferentiable in their physical or chemical properties. They come about when two molecules have the same bond order, but different orientations rotated around a single bond. Imagine our butane molecule again. If we look down the molecule along the carbon-carbon bond axis made by the 2nd and 3rd (middle two) carbons, such that the CH3 and two H's attached to carbon 2 are facing us and the CH3 is pointing straight up while two H's are diagonally facing down-left and down-right. (this is why drawing is nice in O Chem) Now looking across to the other side of our molecule (carbons 3 and 4) we can orient the remaining CH3 and H's in a number of ways. the CH3 could point in any direction, 360 degrees, relative to the direction the CH3 facing toward us is oriented (straight up.) Between any two butane molecules, if the CH3s are oriented in different angles (say one both CH3 point up and the other one points up and one down) they are conformational stereoisomers. But since molecules can rotate freely around single bonds this does not effect the physical properties of the molecule, although some conformations are higher energy than others because of steric hindrance (atoms getting in the way of each other.) If you're still reading, I will assume you find this helpful and are ready to conquer the last form of isomers, configurational stereoisomers. Stereoisomers which can be differentiated from eachother and often exhibit different chemical or optical properties (how they respond to light.) There are 3 types; optical enantiomers/diasteriomers and geometric (cis-trans) isomers. Similar to conformational isomers, 2 geometric isomers differ by rotation around a bond. The difference is the rotation must be around a double bond so they are limited to 2 conformations instead of any angle around a bond. Let's look at 2-Butene; 4 Carbons with a double bond between the middle two. Each of these middle carbons has a CH3 and an H attached. If the H's are both on the same side (both pointed up) this is a cis-isomer, if the H's and thus the CH3s are on opposite sides (one up and one down) they are called trans-isomers. Optical Isomers are the most subtle. They require (a) central molecule(s) with 4 different attached atoms/molecules. An example would be a carbon with an F, Cl, H, and CH3 all attached to it. With the CH3 pointing up, you could attach the other 3 atoms clockwise or counterclockwise to the remaining 3 positions. The result would be 2 molecules that look very similar, but are actually mirror images of each other. These would be optical enantiomers, and they will rotate polarized light clockwise or counterclockwise, leading to different results in drugs. A great example is in aspirin. One enantiomer is the active ingredient, the other does nothing. Optical diasteriomers require more than one central atom, each of which needs 3 different things attached (although all 6 do not need to be different, you could have a H on both the top and bottom central atom as long as neither has 2 H's attached to it.) An example would be to imagine the molecule from the enantiomer example, replacing the CH3 CHOHCl. Once again if we attach the 3 substituents to each central atom either clockwise or counterclockwise, we will get 2 different, but similar molecules. IF the resulting molecules are not mirror images (and they are not the same) they are optical diasteriomers. It is important to note that optical isomers with 2 central atoms can still be enantiomers if they are mirror images. If this was confusing, don't worry. I can explain this and many other concepts much more slowly, with less confusing vocab words, and more helpful diagrams, and I won't leave you behind in my explanation so you'll be with me every step of the way.

Linear Algebra
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Question:

As with Stats, there are a wide range of problems for Linear Algebra, but they all revolve around vector manipulation. Let's take the simple example: determine if this particular set of vectors is a basis. (12,6,1) (6,5,-3) and (0,-2,5)

Ray K.
Answer:

There are 2 ways to solve this equation. The easiest way is to solve for the determinant of the matrix below, and if it is not 0, the vectors are linearly independent. This is fairly straightforward, and I won't demonstrate it today. Now before we solve, let's define a basis. In this case our basis will be a basis of R3. Which means, that all our equations are linearly indepent, and can cover all of 3d space. I can pick any xyz point and some combination of my 3 vectors can make it. The classic example is (1,0,0) (0,1,0) and (0,0,1). the Second method for solving this problem involves carrying out vector manipulation to make these 3 basic vectors. This basic basis works, because each vector only moves in one direction so for any pt x y z you can add x many of v1 y many of v2 and z many v3 to get there. If you can manipulate any set of vectors to become this basic vector set, you know it is a basis. To do this we put all 3 vectors in a table as columns: 12, 6, 0 6, 5, -2 1, -3, 5 now we add and subtract combinations of the various rows (row 1 represents all x values row 2 all y etc.) to try to make the table for the basic basis: 1 0 0 0 1 0 0 0 1 start by subtracting 6 of row 3 from row 2, and dividing row 1 by 12 our new table is 1, 1/2, 0 0, 18, -32 1, -3, 5 next let's subtract our new row 1 from row 3, and divide row 2 by 18. 1, 1/2, 0 dividing row 3 by 5 gives 0, 1, -4 1, 1/2, 0 0 5/2, 5 0, 1, -4 0, 1/2, 1 Now by adding 4 of row 3 to row 2 we get rid of the -4, left with 0 3 0, dividing 3 gives 1, 1/2, 0 0, 1, 0 0, 1/2, 1 Finally, we subtract half of row 2 from both the others to get our basic basis and prove that these vectors form a basis because they are linearly independent.

Statistics
TutorMe
Question:

Well I had a nice example of a 2 proportion t test written out by hand, and an unfortunate refresh erased it. Buuuut it was complicated anyway, instead I'll go into an in depth analysis of a simpler type of problem. Jim works in fabrication making bolts. the bolts have a diameter of 2cm with an SD of .01. His job is to sample 50 bolts and figure out if the machine is working properly at the 95% confidence level.

Ray K.
Answer:

In general, this type of problem has 4 variables and you solve for 1 of them. Zscore=(Pop mean-sample mean)/(sd/sqrt n) we can use z score because we know population SD. Z score can convert to a P value which is the probability of getting a sample this extreme, assuming the machine is actually working fine. In the example I gave above it is straighforward calculate the z score, find p and determine if p is less than 5%. But this equation can be used to solve a robust variety of problems. We could solve for percentiles by converting Z scores to probabilities, or even turn this problem with Jim on it's head and say; Kim takes a sample with mean of 2.01 cm, how large would his sample size have to be, in order to reject the null hypothesis, that the machine is working fine. Or do a similar thing with fixed sample size and solving for SD. The possiblities are nearly endless. And this is true for most of the equations in statistics. There aren't actually that many of them. Most of the time the issue is not in solving the problem, it is in interpretation. How to use which equation and why (and which calculator shortcut can do it faster.) That is what I am good at explaining

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