study set 1

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meth

1 carbon-4 H methane

Pka=10.26 Ka?

10^(-10.26)=5.495E-11 pay attention to sig figs

name this molecule: CH3CH2CH2CH3 (C4H10)

2-ethylbutane

Decane

C10H22

4-ethyl-5-methyl-2-heptene

in alphabetical order double bond on carbon 2 on the long chain 2-atoms on carbon number 4 1-atom on carbon 5

pH=

pH=-log[H+] pH=7

pOH=

pOH=-log[OH-] AND 14 pH

The lower the pKa, the ____ the acid

stronger

Ar notation

Ar=aryl/rings

non-cyclic alkane is a

a saturated hydrocarbon

van der Waals repulsion

when atoms that are not bonded get close to one another in 3-D space they cause the bond angles to be pushed out due to the electrons on the outer area of the atoms.

naming alkanes with cyclo's

you can't # thru a ring. the ring has to have its own set of #'s.

What is the value of K for this aqueous reaction at 298 K? A + B <-----> C + D Delta G = 26.81 kJ/mol

∆G = -RT lnK K = e-∆G/RT ∆G/RT = 26.81kJ/mol/(.008314 kJ/mol-K)(298K) = 1.99E-5

cyclohexane

two double bonds

1) Draw the major resonance structure for the following compound; include lone pairs of electrons, formal charges, and hydrogen atoms. 2) Add curved arrows to both structures to show the delocalization of electron pairs.

#1 IS THE MOST STABLE

resonance structures are just taking electrons from the area that has excess electrons (so negative charges) and bringing it to an area that is positively charged.

+=lacking in electrons

eclipsed conformation

0, 120, 240 degree dihedral angles energy maxima

Alkynes

1 or more triple bonds a carbon compound with a carbon-carbon triple bond.

A reaction has a standard free-energy change of -20.00 kJ mol-1 (-4.780 kcal mol-1). Calculate the equilibrium constant for the reaction at 25 °C.

1) 273 + 25 C= 298 K 2) e^-(-20.00/(.008314*298 K)= =3192 =3.192E3

A reaction has a standard free-energy change of -5.11 kJ/mol at 25 °C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively?

1) 273 + 25 C= 298 K 2) e^-(-5.11/(.008314*298)= K= 7.87 3) k = products/reactants 7.87= X/(.30-x)(.40-x) 3-1)-- (.30-x)(.40-x) FOIL =X^2-0.70X+0.12 3-2) 7.87(.12)-7.87(-.70X) + 7.87X^2= 7.87X^2 +5.509X + 0.944 Do ICE table (image)

A reaction has a standard free-energy change of -14.30 kJ mol-1 (-3.418 kcal mol-1). Calculate the equilibrium constant for the reaction at 25 °C.

1) 273 + 25 C=298 K 2) e^-(-14.30/.008314*298)= =3.211E2

What is the value of K for this aqueous reaction at 298 K? The thermodynamic equilibrium constant is related to the standard Gibbs free energy of reaction via G=16.01KJ/MOL K = e^(-DeltaG/RT) R=.008314 always 8.314

1) e^(-16.01/(.008314*298K)= 2) .001562=1.562E-3 K Value=1.562E-3

naming alkanes=any organic molecule that has single bonds

1-meth 2-eth 3-prop 4-but 5-pent 6-hex 7-hept 8-oct 9-non 10-dec 11-undec 12-dodec 13-tetradec 14-butyldec 15-pentadec 16-hexadec 17-heptadec 18-octadec 19-nonadec 20-eicos

4,5-dipropyldecane

10 carbon long chain di-indicates two different branches on carbon four there is a branch of 3 atoms (indicated by propyl) on carbon 5 "" 3 atoms.

Pka to Ka value is

10^(-Pka value) double check answer by doing -log(answer)

Calculate the number of pounds of CO2 released into the atmosphere when a 11.0-gallon tank of gasoline is burned in an automobile engine. Assume that gasoline is primarily octane, C8H18, and that the density of gasoline is 0.692 g·mL-1 (this assumption ignores additives). Also assume complete combustion.

1961bs

How many rings does an alkane have if its formula is C10H18?

2(10)+2=22 22-18=4 4/2=2 degrees of unsaturation means two rings

2-ethyl-2,4,6-trimethylheptane is really

2,4,6,6-tetramethyloctane

If 2.8 g of butanoic acid, C4H8O2, is dissolved in enough water to make 1.0 L of solution, what is the resulting pH? A table with Ka values can be found here.Butanoic acid in table Pka value=1.5x10^-5

2.8G/1.0L/88G C4H8O2= .0318 (X*X)/0.0318=1.5X10-5 Ka value C4H8O2 =-6.907E-4 or radical(1.5E-5*.0831) -Log(-6.907E-4)= -3.16 pH=3.16 x's are the products

cyclobenzene

3-double bonds

If 3.9 g of butanoic acid, C4H8O2, is dissolved in enough water to make 1.0 L of solution, what is the resulting pH? A table with Ka values can be found here.Butanoic acid in table Pka value=1.5x10^-5

3.9G/1.0L/88G C4H8O2= .0443 radical(.0443*1.5E-5 ka value)= 8.153E-4 -LOG(8.153E-4)= 3.09 pH=3.09

pentagons useu have when

5 carbons with cyclo cyclopentane

3, 4-dimethylhexane

6 carbon long chain on carbon 3 there is a branch of 1 atom (indicated by methyl) on carbon 4 there is a branch of 1 atom dimethyl means 2 methals or 1 each on each branch

Name this molecule

6-benzyl-4-phenyl-tetradecane # ed from the left (shortest distance to a branch) but then put in order of alphabetical order. 0

staggered conformation

60, 180, 300 degree dihedral energy minima

What are the possible isomers of C5H12?

A good place to start is to calculate the degree of unsaturation or D.U. of the compound, you may want to look up for that if you're unsure what it means. The D.U. in this case is zero. Hence the possible isomers are: n-Pentane (Maximizing the number of carbon atoms in the main chain) Isopentane or 2-methylbutane (Shortening the main chain to four C-atoms and looking for positions to place one methyl group) Neopentane or 2,2-dimethylpropane (Main chain has three carbons in this case) So, just keep on shortening the main chain and look for the positions where you can place your methyl (sometimes even ethyl) groups.

Draw the structure of an alkane or cycloalkane that has more than three but fewer than ten carbon atoms, and only primary hydrogens. (There are several possible structures. It is enough to draw any one of them, but you may draw two or more if you want to.)

A primary hydrogen atom is a hydrogen atom that is bonded to a primary carbon atom. A primary carbon atom is a carbon atom that is bonded to one other carbon atom. A quaternary carbon atom is a carbon atom that is bonded to four other carbon atoms. The condition that all hydrogen atoms have to be primary means that the hydrogens can only belong to methyl groups, -CH3. Therefore, the structure must have only primary and quaternary carbon atoms. (Quaternary carbon atoms have all four bonds involved with other carbon atoms, so quaternary hydrogens can\'t exist.) Remember that the stucture can have 4 to 9 carbon atoms. There are three structures that satisfy the given criteria.

Alkanes

All end in "ane". Saturated hydrocarbons (only have carbon and hydrogen atoms). Saturated--meaning they don't have any double bonds.a hydrocarbon containing only single covalent bonds

eclipsed newman projection

Adjacent carbon and the bonded H's are being covered by the first carbon and its bonded atoms. there 3 different conformations -less stable -higher energy this is because the dihedral angle of the adjacent carbons bonded H's is very close. The electrons are repelling one another generating a high energy level.

Cyclo alkanes

Alkanes with carbon chains in closed loops or rings. • Name by adding the prefix cyclo

Molecular dynamics

Allows for the modeling of molecular interactions

alkyl group

An alkane with a hydrogen atom removed, e.g. CH3, C2H5; alkyl groups are often shown as 'R'.

Select the descriptor (A, B, C, D): that best describes the relationship between the \"reactants\" and \"products\" below: CH3CH2OH +H2O -->_CH3CH20^- + H30^+ Do any electrons move? Do any atoms move? Use pKa values to determine which direction equilibrium lies. pKa = -1.74 (H3O+) PKA= -2 (CH3CH2OH) note:The weaker acid (higher pKa, lower Ka) The acid with lower pka is the strong acid. The acid higher pka is the weak acid. + or neutral=acid

Arrow A indicates resonance between two structures. Arrow D indicates equal amounts of reactants and products are expected. Arrows B and C indicate an equilibrium, with the direction of the larger arrow pointing towards the major product. What are the pKa values of CH3CH2OH vs. H3O ? Which is the stronger acid? Would you expect the stronger acid to be more reactive to deprotonation or less reactive?

Pka: For acetic acid the pKa is 4.76 (stronger acid) and for ethanol (CH3CH2OH) the pKa is 15.9 (weaker acid).

Arrow A indicates resonance between two structures. Arrow D indicates equal amounts of reactants and products are expected. Arrows B and C indicate an equilibrium, with the direction of the larger arrow pointing towards the major product. What are the pKa values of acetic acid vs. ethanol? Which is the stronger acid? Would you expect the stronger acid to be more reactive to deprotonation or less reactive? answer C

undecane

C11H24 just know how many carbons use formula 2(# of carbons) + 2 CnH2(n)+2

propyne

C3H4

Butane

C4H10

butyne

C4H6

2-methylpentane

C5H

Hexane

C6H14

Heptane

C7H16

carbonyl group

C=O

Draw hydrogen atoms in a non-cyclic alkane 5-carbon chain

Colors indicate the ways you can # the carbons.

1) Draw the other significant resonance contributor for the following compound; include lone pairs of electrons, formal charges, and hydrogen atoms. 2) Add curved arrows to both structures to show the delocalization of electron pairs.

Concepts and reason The structure of a compound, that is, a neutral. To draw the resonance structure of the given compound, delocalize the electrons and the lone pair of electron either towards the electron deficient species or towards the more electronegative atom. Formal charge on the atom can be calculated by knowing the valence electron, bonding electrons, and nonbonding electrons present on the atom. The curved arrow shows the direction of movement of electrons towards the electron deficient species or towards the more electronegative atom. Fundamentals Resonance structure can be drawn by the delocalization of electron and lone pair of electrons present in a system. Resonance structures are shown by double headed arrow. The delocalization of electron and lone pair in a molecule lowers the resonance energy and thus, the molecule is stabilized. Consider the following example for drawing the resonance structure by delocalization of lone pair and double bond as shown by curved arrows: formal charge: period number/valence electrons-(# of dots individually counted + # links) the first structure is more STABLE

Draw the structure of an alkane or cycloalkane that has a molecular mass of 84.2.

Divide 84.2 by 12 (here take the approximate value): Explanation | Hint for next step Here, the maximum number of carbon atoms is calculated using the molar mass. The molar mass of the compound is divided by 12 (mass of 1 C atom), as the mass is even. The compound (alkane or cycloalkane) contains C and H. But in general, the compounds which have even mass will contain C, H and O. The approximate value of 84.2 is 84, so it is divided by 12, then 7 carbon atoms are obtained with remainder 0. So, the molecular formula would be. C7H0

Acids

Electron acceptor. The boron in BF3 is electron poor and has an empty orbital, so it can accept a pair of electrons, making it a Lewis acid. ... A Lewis acid is defined as an electron-pair acceptor. So for something to act as a Lewis acid, it needs to want electrons.

Lewis Bases

Electron pair donors. a compound or ionic species that can donate an electron pair to an acceptor compound.

Concepts and reason Electrophile is an

Electrophile is an electron loving and nucleophile is a nucleus loving. Electron pair acceptors are electrophiles, they have empty orbitals. Electron pair donors are nucleophiles, they have non-bonding electrons.

Rank these acids according to their expected pKa values. Consider the distance of the Cl atom(s) from the site of ionization as well as the number of Cl atoms. Order from Highest pKa to lowest pKa CH3CH2COOH (pKa = 4.9) ClCH2CH2COOH (pKa = 4.0) ClCH2COOH (pKa = 2.9) Cl2CHCOOH (pka = 1.3)

Explanation: Highly electronegative atoms such as chlorine stabilize the conjugate base, making ionization occur more readily (stronger acid, lower pKa). Increasing the number of Cl atoms further lowers pKa as does decreasing the distance between the Cl atoms and the site of ionization.

Complete the equations for the following equilibria and calculate Keq where the Keq expression includes [H2O]. Be sure to enter Keq in proper scientific notation.

Keq=products/reactants Print chegg

If a atom or molecule has a negative charge it is going to be a weaker acid or base and the long arrow points away from that side of the reaction in equilibrium.

HA just represents an acid

Alkenes

Have 1 double bond

Alkynes

Have 1 triple bond

Draw the organic and inorganic products for the following acid/base reaction. Include charges. CH3CH20CH2CH3 + H30 + ---> CH3CH20HCH2CH3+ + H2O

Hint: You are moving a proton (H)

Alkanes

Hydrocarbons that contain ONLY single bonds

unsaturated hydrocarbons

Hydrocarbons with double or triple bonds

Alkenes

Hydrocarbons with one or more carbon-carbon double bonds all organic molecules that have single, or double bonds.

Using the Brønsted theory, classify the following as either an acid or a base by placing the compound in the correct bin.

If a compound is positive or neutral it can accept electrons=acid If a compound is negative its a base b/c it give away electrons/can give it a proton (H) and make the charge neutral.

Identical compounds

If two structures are superimposable then they are the same compound

The Newman projections of 1,1-dichloro-2-bromoethane are shown below. Select the most stable conformation(s).

In the Newman projection, we look down the C-C bond, which is perpendicular to the image surface. The front carbon is represented by the point where the three lines meet. The back carbon is represented by the circle. Each line represents a C-X bond, where X is a substituent. Thus, both chlorines and a hydrogen are on one carbon, whereas Br and two hydrogens are on the other carbon. First consider the stability of a staggered vs. eclipsed conformation. Newman projection D is the only eclipsed conformation shown here. Eclipsed conformations are less stable than staggered conformations, due to van der Waals repulsions. Of the remaining staggered conformations, the Newman projection B is least stable since all the halogen atoms are in close proximity and repulse each other. Newman projection B has two gauche interactions between the Br and front carbons. The Newman projections A and C are equal in stability, with one gauche interaction between the Br and one Cl and one anti interaction between the other pair. Thus both A and C are the most stable conformations among these four.

conjugate base

It loses a H and decreases in charge by one

combustion of alkanes an alkane reacts with ________ producing _______ AND __________

O2 CO2 AND H2O

Combustion of alkanes reacts with

O2 to give CO2 and H20 the reaction is initiated by heat.

Calculate the number of pounds of CO2 released into the atmosphere when a 22.0-gallon tank of gasoline is burned in an automobile engine. Assume that gasoline is primarily octane, C8H18, and that the density of gasoline is 0.692 g·mL-1 (this assumption ignores additives). Also assume complete combustion. 1 Gallon=3.785 L 1 Kg= 2.204 Pounds

One liter is 1000 milliliters Total volume in liters = 22 * 3.785 = 83.27 L Total volume = 83.27 * 1000 = 83,270 ml Since we know the density, we can use the following equation to determine the mass. 22gal X [3.785 L/gal] X [1000 mL/L] X [0.692 g/mL] X [1 mole / 114g] X [8 mol CO2 / (25/2) mol C8H18] X [ 44 g/mole] X [1 kg/1000g] X [2.204 pound/kg] X [1ton/2000pounds] Mass = Volume * Density Mass = 83,270* 0.692 = 57,622.84 grams Let's determine the mass of one mole of octane. Mass = 12 * 8 + 18 = 114 grams Let's determine the number of moles of octane. n = 57,622.84 ÷ 114 This is approximately 505.46 moles. During the combustion of octane all of carbon atoms will react with oxygen to form carbon dioxide. Since octane has 8 carbon atoms and carbon dioxide has 1 carbon atom, the number of moles of carbon dioxide will be 8 times the number of moles of octane. For CO2, n = 8 * 57,622.84 ÷ 114 = 460,982.72 ÷ 114 = 4043.70807 moles Mass of one mole = 12 + 32 = 44 g Mass = 44 * 4043.70807 This is approximately 1.78 * 10^5 grams. One pound is 454 grams. Number of pounds = 44 * 4043.70807 ÷ 454 This is approximately 391.9 pounds The next step is to write the balanced equation of the combustion of octane. C8H18 + O2 → CO2 + H2O To balance the C's, make 8 CO2. C8H18 + O2 → 8 CO2 + H2O To balance the H's, make 9 H2O. C8H18 + O2 → 8 CO2 + 9 H2O To balance the O's, make 12.5 O2. C8H18 + 12.5 O2 → 8 CO2 + 9 H2O Double all of the coefficients 2 C8H18 + 25 O2 →16 CO2 + 18 H2O The coefficients in a balanced equation determine the ratio of moles of reactants to moles of products. As you look at this equation, you can see that ratio for CO2 and C8H18 is 8:1. This is why I multiplied the number of octane by 8. I hope this helps you to understand how to solve this type of problem.

Alky groups are electron donating and halogens are electron withdrawing. Consider the effect these have on the stability of the carboxylate anion.

Option 1 is correct, (CH3)3CCOOH is the least since it has electron releasing groups attached to -COOH, which will de stabilize COO- but CF3 is highly withdrawing and makes COO- stalbe increasing acidity

A benzene ring when attached to a parent chain is called a _________ (functional) group

Phenyl when a benzylic group is part of the parent chain it is called a phenyl group. when a benzylic group is NOT part of the parent chain it is called a benzyl group. Note: there is special reactivity associated with the carbon that is immediately adjacent to the benzene ring and these are called benzylic carbons

Primary carbons

Primary = a hydrogen on a carbon attached to only ONE other carbon. Secondary = a hydrogen on a carbon attached to only TWO other carbons. Tertiary = a hydrogen on a carbon attached to THREE other carbons.

Draw a plausible structure FOR C4H7, assuming that it has two tertiary carbons and all other carbons are secondary.

Primary carbons, are carbons attached to one other carbon. (Hydrogens - although usually 3 in number in this case - are ignored in this terminology, as we shall see). Secondary carbons are attached to two other carbons. Tertiary carbons are attached to three other carbons. Finally, quaternary carbons are attached to four other carbons.

A reaction A +B <=====> C has a standard free-energy change of -3.76 kJ/mol at 25 °C. What are the concentrations of A, B, and C at equilibrium if, at the beginning of the reaction, their concentrations are 0.30 M, 0.40 M, and 0 M, respectively?

R is the gas constant found in the ideal gas law (0.0821 LiterAtm MoleKelvin once you get to 1.61 do e^1.61= Kp=5.00 Δn = (Total moles of gas on the products side) - (Total moles of gas on the reactants side). Hence \( \Delta = (d + c) - (a + b)\] [The lower case numbers are the exponents] I=initial C=change E=end 7.2=x/(0.30-x)(0.40-x) foil add .30 + .40=.70 multiply .30 Mx .40M=.12

ester

R-C=O-O-R An R, O, R attached to a carbonyll group

keytone

R-C=O-R two R's attached to a carbonyl

R and Ph notation

R=alkyl Ph=phenol

Estimate the pKa values for the functional group classes represented by the given molecules. Note: If one or more values are incorrectly placed, a single red X will appear on the top left.

Recall that a lower pKa value means the compound is more acidic and the proton is easier to remove. A higher pKa means the compound is more basic and the proton is harder to remove. pKa values vary according to the atom attached to the hydrogen: H-CH3 > H-NH2 > H-OH Electron-withdrawing groups and resonance in the molecule lower the pKa. A positive charge on the atom attached to the hyDrogen lowers the pKa. print chegg

A hydrocarbon (alkane or cycloalkane) is found by combustion analysis to contain 87.17% carbon and 12.83% hydrogen by mass. Determine the empirical formula.

Take a hypothetical 100 gram sample: (87.17 g C) / (12.0108 g/mol) = 7.25763 mol C (12.83 g H) / (1.0079 g/mol) = 12.7294 mol H Divide by the smaller number of moles: 7.25763 C / 7.25763 = 1.00000 12.7294 H / 7.25763 = 1.75393 In order to make these ratios close to integers, multiply by 4, then round to the nearest whole number to find the empirical formula. So 1.75x4=7 H 1.0x4=4 carbons: empirical formula: C4H7 Alkanes have to have an EVEN number of Hydrogens. so 1.75x8=14 H 1.0x8=8 carbons C8H14

Classify each of the following reactants and products as an acid or base according to the Brønsted theory: You have to have an acid and a base on each side. pay attention to what is accepting electrons too

The acid donates a proton; the base accepts a proton. In this reaction HF donates a proton and so is an acid; H2O accepts a proton and so is a base; H3O donates a proton in the reverse reaction, so it is an acid; F- accepts a proton in the reverse reaction, so it is a base.

Pyridine and methanol can act as weak bases. Their Kb values are given below. Consider the compounds water, ethylamine, acetone, and aniline, whose Kb values can be found in the table below. Classify them by whether they can be appreciably (Keq>1) protonated by: pyridinium, CH3OH2 , both, or neither. Bases accept protons. Thus, a stronger base is more easily protonated than a weaker base. A compound can be protonated by pyridinium if it is a stronger base than pyridine. A compound can be protonated by CH3OH2 if it is a stronger base than methanol.

The bigger the # the more basic it is whereas the more negative # is more acidic ---a negative Pka will give you a positive Ka or a stronger acid.

2-pentene

The double bond is on the second carbon

Give the formula of the conjugate base of HSO4-.

The formulas of an acid-base conjugate pair differ by H . It loses a H and decreases in charge by one.

Imagine a reaction that can replace one hydrogen atom of an alkane at random with chlorine atom. If 2,2-dimethylbutane were subjected to such reaction , how many different compounds (ignoring optical isomers) would be obtained?

There are three isomers possible. (A) CH2Cl-C(CH3)2-CH2-CH3 - 1-chloro-2,2-dimethylbutane (B) CH3-C(CH3)2-CHCl-CH3 - 3-chloro-2,2-dimethylbutane (C) CH3-C(CH3)2-CH2-CH2Cl - 4-chloro-2,2-dimethylbutane Substituting a hydrogen on either of the methyl groups on C-2 produces compound (A) again.

R-3-ethylhexane S-3-ethylhexane

These are different because one is left handed and one is right handed r=right s=left

Arrange the compounds in order of decreasing pKa, highest first. a.Cl2CHCH2OH b.ClCH2CH2OH c. CH3CH2OH

To arrange in order of decreasing pKa means to arrange in order of increasing acidity. The most acidic compound on the list has the lowest pKa. CH3CH2OH is less acidic than ClCH2CH2OH, which is less acidic than Cl2CHCH2OH. Chlorine substituents increase acidity because of their electron-withdrawing polar effect; the more chlorines there are, the greater the effect.

Draw the structure of an alkane or cycloalkane that has more than three but fewer than ten carbon atoms, and only primary hydrogens.

You would have to draw a structure that has only carbon atoms and methyl groups. Explanation: A primary hydrogen is a hydrogen on a carbon attached to only ONE other carbon, i.e., a hydrogen atom of a methyl group. Two alkanes meet the criteria. One carbon atom + four methyl groups This gives 2,2-dimethylpropane (five carbon atoms)

The molecule below contains carbon atoms (shown in black), hydrogen (white), and oxygen (red). Note: only sigma-bond connectivity is shown, and there are no formal charges. Consider the structure and determine to what classes this compound belongs.

We need to figure out which functional groups (i.e., structural units) are present, and that will clue us in as to the identity of the compound class (i.e., the name of the group). Since there are no formal charges and all hydrogens are shown, let\'s consider the atoms\' arrangement to determine the location of pi bonds. Recall that an sp3-hybridized carbon atom has four atoms connected to it by single bonds. An sp2-hybridized carbon is bonded to three atoms, and one of the bonds is a double bond. A carbon atom bonded to two atoms must be sp-hybridized, with a triple bond. Note the terminal carbon atom that has only one hydrogen and one carbon attached, and the geometry is linear. These two carbons are sp-hybridized. The HC≡ C- functional group indicates that the compound is an alkyne. The in-chain oxygen atom bound to two alkyl groups constitutes an R-O-R functional group. The compound is an ether. The other oxygen is connected to only one carbon, so this bond must be a double bond. This carbonyl group is bonded to two alkyl groups, -CH2- and -CH3. The functional group R-C(O)-R indicates a ketone.

saturated hydrocarbon

When a carbon is connected to a carbon, its connected to carbon with only single bonds. With the maximum amount of hydrogens present. A hydrocarbon in which all the bonds between carbon atoms are single bonds

A certain compound was found to have the molecular formula C5H12O2. To which of the following compound classes could the compound belong? Check all the apply.

With 12 hydrogen atoms and 5 carbon atoms, this compound cannot have any double bonds or rings. Because amides, carboxylic acids, phenols, and esters each contain one or more double bonds to carbon, these groups are ruled out. An amide is ruled out also because it contains nitrogen, which is not present in the compound. The remaining possibilities are a compound with two ether groups, a compound with two alcohol groups, or a compound with both an ether and an alcohol. Examples of each are the following:

5-carbon non-cyclic alkane

You have to use the #'s (locants) for each one of the branches from a carbon, i.e. like carbon 2.

eclipsed conformation

a conformation about a carbon-carbon single bond in which the atoms or groups on one carbon are as close as possible to the atoms or groups on an adjacent carbon

staggered conformation

a conformation about a carbon-carbon single bond in which the atoms or groups on one carbon are as far apart as possible from atoms or groups on an adjacent carbon

aryl group

a group derived from an arene by removal of an H atom from an arene and given the symbol Ar-

A Brønsted base is

a species that accepts a proton (H).

a) Into a separatory funnel is poured 50 mL of CH3CH2Br (bromoethane), a water-insoluble compound with a density of 1.460 g/mL, and 50 mL of water. The funnel is stoppered and the mixture is shaken vigorously. After standing, two layers separate. Which substance is in which layer? Explain. b) Into the same funnel is poured carefully 50 mL of hexane (density = 0.660 g/mL) so that the other two layers are not disturbed. The hexane forms a third layer. The funnel is stoppered and the mixture is shaken vigorously. After standing, two layers separate. Which compound(s) are in which layer?

a)The top layer is water and the bottom layer is bromoethane... ...because bromoethane is more dense than water.

H30+ is always a H20 can be either

acid b/c its + a base or an acid

saturated hyrocarbon

alkane hydrocarbons containing only single-bonded carbon atoms

naming alkanes with cyclo's in a tie situation where locant or numbering of branches is tied or the same no matter which direction you count then

alphabetical order wins, i.e. ethyl vs. methyl

Geometric isomers have different connectivity about

an alkene.

A nucleophile is a Lewis base that donates

an electron pair to an atom other than hydrogen.

Concepts and reason Resonance structures

are the Lewis structures which have the same placement of atoms but different in the arrangement of electrons. A hybrid of the contributing structures represents the real structure of the molecule and is known as resonance hybrid. Resonance structures can be formed when the double bonds are in conjugation. It also occurs when the double bond is separated by a lone pair through a single bond. Only the electrons (pi electrons and lone pairs) will move, atoms never move. So, the number of electrons in the molecule does not change. Fundamentals A double headed arrow () is used to represent different resonating structures. More stable resonating structures have negative charge on more electronegative atom. The curved arrow shows the movement of the electron pair. Curved arrow is represented as follows:

Alkanes burn= ___CH3CH3 + ____ O2 ---->____CO2 + ____H2O

b/c they burn, lab based transformation alkanes are not very functional. They do not react very easily It is functional on a gigantic scale.

C4H8

butene

propyl groups

can only be attached to the 1 or 2 carbon

1, 2-butadiene

carbon 1 and 3 have double bonds

In the structure of 4-isopropyl-2,4,5-trimethylheptane, identify the primary, secondary, and tertiary hydrogens.

counted by the number of bonds that its carbon bond has

Constitutional isomers have

different connectivity. Recall that connectivity is the order in which the atoms are bonded.

What is the value of K for this aqueous reaction at 298 K? The thermodynamic equilibrium constant is related to the standard Gibbs free energy of reaction via G=22.48KJ/MOL K = e^(-DeltaG/RT) R=.008314 always 8.314

e^(-22.48/(.008314 x 298)= put in sci notation k value=1.147 x 10^-4 pay attention to the original # of digits

delocalized electrons

electrons that are free to move in metals

what direction is favored in a reaction note: look for the strongest base or acid

favored means that there is more of something on that side of the reaction

conjugate acid

gains a H and gains charge by one.

electrophillic

gains electrons electron deficient. (of a molecule or group) having a tendency to attract or acquire electrons.

Tertiary=

has 1 hydrogen attached to and 3 other carbons attached to it

secondary=

has 2 hydrogens attached to it and 2 other carbons attached to it

primary=

has 3 hydrogens attached to it and one carbon attached to it

Quaternary=

has not hydrogens attached to it and 4 other carbons attached to it

The equilibrium in the reaction of an acid and a base always favors

he side of the weaker acid and weaker base. Determine by comparing pKa of both acids.

boiling points The longer a carbon chain is, the __________boiling point.

higher the boiling point.

figuring out which resonance structures are most stable

hint: 1) ever single atom has a complete octet 2) minimize the number of charges ----whichever structure best fulfills these guidelines is the most stable your moving single lone pair electrons into a pie bond or moving a pie bond into a single lone pair around a atom that is either + or negatively charged *****When drawing two other resonance forms, keep in mind that the sigma connectivity of the atoms stays the same. One may delocalize only the lone pairs and the pi electrons from multiple bonds.

period=

horizontal across p.t. not down

Alkane conformations

how does the molecule occupy 3-D space?

draw out the last word (the carbons) first then do the rest

https://www.youtube.com/watch?v=v5R0k25DarQ

3,3,5- trimethyloctene

https://www.youtube.com/watch?v=v5R0k25DarQ 20:47 8 carbons on the long chain on carbon 3 it has one atom attached/bonded to either side on carbon 5 it has 1 atom bonded double bond on carbon 1

2,4,6-octatriene

https://www.youtube.com/watch?v=v5R0k25DarQ 25:40

Two compounds may be isomeric

if they have the same molecular formula. Different compounds that have the same molecular compound. If they are not superimposable but have the same formula

Pka values increasing acidity table

increases across and down the p.t.

A smaller pKa value

indicates a stronger acid. Would you expect the Ka value to be large or small for a strong acid? Consider the mathematical relationship between Ka and pKa values. Proper scientific notation involves integers, not decimals, in the exponent.

A Lewis base

is a species that donates an electron pair.

branched vs. unbranched alkane melting points branched hydrocarbons have a _________ melting point than unbranched hydrocarbons. unbranched hydrocarbons therefore freeze at a _______ temperature than branched hydrocarbons.

lower lower

reaction quotient

molar concentrations

isomers are different

molecules with the same exact chemical formula

Draw all the structural isomers for the molecular formula C4H9Br. Be careful not to draw any structures by crossing one line over another; the system needs to know where you intend connections to be between atoms.

moving double bonds

unbranched alkanes

n-alkanes or normal alkanes

2(n) + 2

n= # of carbon atoms in a structure. if you know the # of carbon atoms in a structure then, 2n + 2, is the number of hydrogen atoms in the molecule.

methykcyclopentane

no number

3-ehtylheptane

on the third carbon there will be ethyl (2 carbons) and on the longest chain there will be a total of 8 carbons.

If a chain is bigger or has more carbons than a cyclo then the bigger chain is the

parent chain or the cyclo becomes the substitutient.

Most alkanes come from

petroleum or crude oil.

carbolic acid is also known as

phenol OH and an R attached to a carbonyl group

During exercise when the body lacks an adequate supply of oxygen to support energy production, the pyruvate that is produced from the breakdown of glucose is converted into lactate. High lactate levels can lead to acidity in the muscle cells as some of the lactate hydrolyzes to lactic acid. The dissociation of lactic acid to lactate is shown in the reaction below. Lactic acid has a pKa of 3.86. A solution containing a mixture of lactic acid and lactate was found to have a pH of 3.13. Calculate the ratio of the lactate concentration to the lactic acid concentration in this solution.

ratio of the lactate concentration to the lactic acid concentration in this solution. LACTATE-LACTIC ACID= 3.13-3.86= -.73 10^(-.73)=0.19 What fraction of the lactic acid/lactate mixture is lactic acid? 1+.19= 1.19 1/1.19= .84 FRACTION OF LACTIC ACID

Single bonds have NO

restriction for rotation because the orbitals that form that bond they overlap with and end-to-end fashion. sigma bond

A ring structure is not a

saturated hydrocarbon because two of the hydrogens would have to be removed in the above structure, i.e., CH3CJ2CH2CH3

pH scale

scale with values from 0 to 14, used to measure the concentration of H+ ions in a solution; a pH of 0 to 7 is acidic, a pH of 7 is neutral, and a pH of 7 to 14 is basic. OH=14 H=7

A nucleophilic site is an

shares electrons Electron pair donors. (negative) electron-rich site that can share electrons with an electrophilic site. functional groups have a partial negative charge and seek positively charged nuclei. They donate electrons and usually "attack" groups with partial positive charges. They are also called Lewis bases.

Geometric isomers are considered

stereoisomers. same molecular formula and the same connectivity, but have different orientations of substituents about a double bond.

Petene

the carbon double bond is on the 1st carbon

Pka values

the higher the Pka value the less acidic it is or more basic. lower =more acidic 14 pH= LOW ACIDITY

staggered newman projection

the legs of the #2 and #3 carbons are not overlapping each other. this is the first and second drawings. there are 3 different conformations with the staggered. -more stable (favored) -lower energy the propeller represents a tetrahedrane. These stick out of the plane of the page=109.5 angles the bonds behind the circle are behind the plane of the page.

The conformation of a molecule is

the spatial arrangement of its atoms when all dihedral angles are specified.

In a p-orbital the pi-bonds in alkenes and alkynes

they do NOT turn. If you do it will break the bond.

identify the functional groups in these molecules NOTES: Focus on part of the molecule not the whole molecule remember a molecule can have more than one functional group

top left-alcohol middle- primary amine right-thyol lower left-aldehyde lower right-secondary amine

A p-orbital bond

twisting a pi-bond out of plane causes the orbital overlap to disappear.

Estimate the pKa values for the functional group classes represented by the given molecules. Note: If one or more values are incorrectly placed, a single red X will appear on the top left.

type in 10^-pka value then spits out answer

For the following list of acids, rank the acids in strength from weakest acid to strongest acid. HCl, H3P, H4Si, H2S

use periodic table The acidity is determined by the electronegativity of the atom that the hydrogen is bonded to. All the atoms in this series are in the same period with similar size valence orbitals. The increase in electronegativity from left to right results in greater acidity as the sigma bond becomes more and more polarized toward the more electronegative atom and allows the hydrogen to dissociate from the molecule.

The only allowable ways to add butyl (4 carbon branches to a parent chain of carbons)

way to name each from ;eft to right) 1-methylpropyl 2-methylpropyl 1,1-dimethylethyl

The position of equilibrium in a reaction will favor the side that contains the _________ acid

weak the longer arrow points to the weak acid

If a buffer solution is 0.250 M in a weak base (Kb = 4.7 × 10-5) and 0.480 M in its conjugate acid, what is the pH?

weakbase and salt of weakbase,strongacid. pOH = pKb + log [salt/base] pKb = -log kb = -log (4.7*10^(-5)) = 4.328 pOH = 4.328 + log(0.570/0.250) = 4.686 pH = 14-pOH = 14-4.686 = 9.314

The higher the pKa, the ______ the acid.

weaker


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