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Question

Diols Nomenclature

Answer 1

molecules wit two hydroxyl groups, they are also called glycols -are named with the suffix -diol

Answer 2

diols with the hydroxyl group on adjacent carbons

Answer 3

diols with hydroxyl groups on the same carbon | -are not commonly observed because they spontaneously lose water (DEHYDRATE) to produce carbonyl compounds

Answer 4

carbonyl hydrates

Answer 5

are named as derivatives of alkanes and the larger alkyl group is chosen as the backbone - has a ALKOXY- prefix - the chain is numbered to give the ether the lowest position - COMMON NAMES for ethers is when the two alkyl groups are name in alphabetical order and add the word ether

Answer 6

numbering of the ring begins at the oxygen and proceeds to provide the lowest numbers for substituents

Answer 7

are named according to the longest chain containing the functional group. - the suffix -AL replaces the -e of the corresponding alkane - the carbonyl carbon receives the lowest number, although numbers are not always necessary

Answer 8

FORMALDEHYDE ACETALDEHYDE PROPIONALDHYDE AND STRUCTURE

Answer 9

- named just like aldehydes but with -ONE as the suffix - if highest priority the carbonyl croon must be assigned the lowest possible number - in complex molecules the carbonyl group can be names as a prefix OXO- - also the individual alkyl groups may be listed in alphabetical order and followed by the word ketone

Answer 10

are names with the ending -oic acid | -are terminal functional groups and are numbered one

Answer 11

METHANOIC ACID ETHANOIC ACID PROPANOIC ACID

Answer 12

- the longest chain attached to the nitrogen is taken as the backbone. - for simple compounds name the alkane and replace with the final -e with -AMINE - more complex often use the prefix AMINO - the prefix N is used to specify the location of an additional alkyl group attached to the nitrogen

Answer 13

are chemical compounds that have the same molecular formula but differ in structure (their atom connectivity or the spatial orientation of their atoms) -Isomers may be similar or different when it comes to their physical and chemical properties

Answer 14

Also known as constitutional isomers - are compounds that share only a molecular formula - differ in where and how atoms are connected to each other and thus often have very different chemical and physical properties (MP, BP, solubility)

Answer 15

are compounds that have the same connectivity between their atoms and differ from each other only in the way their atoms are oriented in space - Different TYPES: 1) Cis-Trans Isomers 2) Enantiomers 3) Diastereromers 4) Meso Compounds 5) Conformational Isomers

Answer 16

also known as geometric isomers - are compounds that differ in the position of substituents attachéd to the two carbons that from a double bond - since the double bonds cannot rotate the substituents are fixed - If the substituents on the carbon are both on the same side of the double bond then they are called CIS - If the substituents on the carbon are on opposite sides of the double bond then they are called TRANS

Answer 17

The Highest Priority substituent has to be determined: 1) From the carbon of interest, determine the molecular weight of the first atom encountered along each bond. The group with the highest atomic weight atom has the highest priority 2) If two atoms are the same look at the next atom attached to each; the group that has the second atom with the highest molecular weight is higher priority 3) If two atoms are the same, a double bond takes priority over a single bond. This is a tie breaker ONLY - The alkene is called Z if the two highest priority subsistent on each carbon are on the SAME SIDE - the alkene is called E if the two highest priority substituents are on opposite sides

Answer 18

- a molecule that is not superimposable upon its mirror image - carbon atoms are chiral ONLY if they have four different substituents-- a carbon like this is called ASYMMETRIC since it lacks a plane or point of symmetry

Answer 19

pair of chiral molecules that are non-superimposable mirror images

Answer 20

a three dimensional molecule can be represented in 2D by fischer projections - horizontal lines indicate bonds that project out from the plane of the page - vertical lines indicate bonds behind the plane of the page

Answer 21

If only one substituent is interchanged or if the molecule is rotated by 90 degrees -- the mirror image and thus the opposite enantiomer of the original compound is obtained

Answer 22

describes the spatial arrangements of the atoms or functional groups of a sterioisomer

Answer 23

is its configuration in relation to another chiral molecule | -is compared between the R and S enantiomers

Answer 24

describes the spatial arrangements of these atoms or groups within the molecule relative to each other -is determined y using the R/S naming convention

Answer 25

-Pairs of enantiomers (opposite R and S) have identical chemical and physical properties with one exception: OPTICAL ACTIVITY - a compound is optical active if it has the ability to rotate plane-polarized light. - If a plane-polarized light is passed through a solution of an optically active compound, the molecule rotates the orientation of the polarized light by an angel alpha. The enantiomer of this compound will rotate light but he same amount but in the opposite direction

Answer 26

a compound that orates the plane of polarized light to the right or clockwise is indicated by a (+)

Answer 27

a compound that rotate light toward the left or counterclockwise is labeled (-)

Answer 28

it cannot be determined from the structure of a molecule and is not related to its R and S designation. -it must be determined experimentally!

Answer 29

the amount of rotation depends on the number of molecules that a light wave encounters. -this depends on two factors the concentration of the optically active compound and the length of the tube through which the light passes

Answer 30

or racemate, is a mixture of equal concentrations of both he (+) and (-) enantiomers -the rotations cancel each other and no optical activity is observed

Answer 31

Are stereoisomers that are not mirror images of each other | -more than one chiral center and thus multiple forms of stereoisomers exist

Answer 32

diastereomers that differ at only one carbon | -important in chemistry of carbohydrates

Answer 33

molecules with multiple chiral centers that also have an internal plane of symmetry - one half of the molecule has an S enantiomer and the other had has its matching R enantiomer - molecule is not optically active

Answer 34

are compounds that differ only by rotation about or more single bonds -these isomers represent the same compound but in a slightly different position

Answer 35

Staggered anti-when the two biggest groups are located 180 degrees from each other and there is not overlap of atoms along the line of sight -called staggered anti because the two groups are antiperiplanar to each other

Answer 36

occurs when the two groups are 60 degrees apart

Answer 37

highest energy state and is went the two groups overlap with each other -zero degrees apart

Answer 38

is when the two groups are 120 degrees apart

Answer 39

angle strain-results when bond angles deviate from their ideal values torsional strain- results when cyclic molecules must assume conformations that have eclipsed interactions nonbonded strain-(Van Der Waals Repulsion)- resulsts when atoms or groups compete for the same space

Answer 40

cycloalkanes attempt to adopt nonplanar conformation

Answer 41

puckered- which is a slight V shape

Answer 42

envelope- looks like an airplane with no wings

Answer 43

exists mainly in three conformations: charm, boat, and the twist/skewboat

Answer 44

electron is transferred from one atom to another

Answer 45

in which paris of electrons are sheared between two atoms

Answer 46

corresponds to the energy level in an atom | -essentially a measure of size

Answer 47

when two atomic orbital are combined it forms molecular orbital - are obtained mathematically by adding the wave functions of the atomic orbitals - if the sign of the wave functions are the same, a lower-energy bonding orbital is produced - if the signs are different, a high energy ANTIBODING ORBITAL is produced`

Answer 48

means they the compound has the maximum number of hydrogen atoms attached to each carbon

Answer 49

is bonded to only one other carbon atom

Answer 50

is bonded to only 2 carbon atoms

Answer 51

is bonded to only 3 carbons

Answer 52

is bonded to only 4 carbons

Answer 53

- vary in a predictable manner - in general as the molecular weight of a straight chain alkane increases, the melting point, boiling point, and density also increase - branched molecules have slightly lower boiling point and melting point than their straight-chain isomers - Greater branching reduces the surface area of the molecule, decreasing the intermolecular attractive forces (london dispersion forces)- the molecules are held togeth less tightly, thus lowering the BP

Answer 54

gases: CH4 through C4H10 Liquids: C5H12 through C16H34 Solids (waxes and harder solids): C17H36 and so on

Answer 55

are molecules that are attracted to positive charge - nucleophile means nucleus lover - are electron rich species that are often but not always negatively charged - are attracted to atoms with partial or full positive charges

Answer 56

-if a group of nucleophiles are based on the same atom (oxygen for example) the stronger the base the stronger the nucleophile. Because bases act as electron donors and stronger nucleophiles are also better electron donors. RO->HO->RCO2>ROH>H20

Answer 57

Protic solvent(a solvent that is able to form hydrogen bonds) - in a protic solvent large atoms or ions tend to be better nucleophiles - larger ions more easily shed their solvent molecules and are more polarizable CN->I->RO->HO->Br->Cl->F->H2O

Answer 58

aprotic solvent- a solvent that cannot form hydrogen bonds - the nucleophiles are "naked"- they are not solvated - nucleophile strength is directly related to basicity F->Cl->Br->I-

Answer 59

the best leaving groups are those that are weak bases, because these can accept a negative charge and dissociate to form a stable ion in solution -HALOGEN I->Br->Cl->F- other leaving groups beside halogens ex: OH group

Answer 60

-unimolecular nucleophilic substitution reaction

Answer 61

because the rate of the reaction is dependent upon only one molecule in the reaction or the rate expression is first order!

Answer 62

is the dissociation of the substrate (starting molecule) to form a stable, positively charged ion called a CARBOCATION -the formation and stabilization of the carbocation determines all other aspects of SN1

Answer 63

involves two steps: - the dissociation of a substrate molecule into a carbocation and a leaving group - followed by the carbocation with a nucleophile to form a substituted product

Answer 64

stabilized by polar solvents that have lone electrons paris available to donate (water or ethyl alcohol) -stabilized by charge delocalization throughout the molecule -the more highly substituted carbocations are more stable-due to hydrocarbon substituent groups donate electron density toward the positive charge

Answer 65

- the slowest step is the dissociation of the molecule to form a carbocation intermediate- a step that is energetically unfavorable thus the RATE LIMITING STEP - rate of the reaction depends entirely on the concentration of the substrate - THE RATE DOES NOT depend on the concentration or the nature of the nucleophile

Answer 66

highly substituted alkanes allows for distribution of the positive charge over a greater number of carbon and hydrogen atoms- thus more stable carbocations - Generally the order of reactivity is tertiary> secondary> primary> methyl - primary and methyl SUBSTRATES DO NOT REACT

Answer 67

Highly polar solvents are better at surround and isolating ions than are less polar solvents - Polar protic solvents such as WATER AND ALCOHOLS work best because they can form hydrogen bonds with the leaving groups and solvating it and preventing it from retiring to the carbocation. - also lone electron pairs on oxygen or nitrogen atoms in the solvent molecule can stabilize the carbocation intermediate

Answer 68

Weak bases dissociate more easily from the alkyl chain and thus make better leaving groups -increasing the rate of the carbocation formation

Answer 69

SN1 reactions do not require a strong nucleophile | -SN1 reactions run equally well with either strong (fully charged) or weak (electron-rich but uncharged) nucleophiles

Answer 70

SN1 reactions involve carbocation intermediates that are sp2 hybridized and have trigonal planar geometry. -the attacking nucleophile can approach the carbocation from either above or below with equal probability therefore creating either the R or S enanatiomer with equal probability -If the original compound is optically active because of the presence of a chiral center, then a racemic mixture will be produced.

Answer 71

bimolecular nucleophilic substitution - involves a nucleophile pushing its way into a compound while simultaneously displacing the leaving group - its rate determining and only step involves two molecules: the substrate and the nucleophile

Answer 72

are concerted reactions-meaning the entire mechanism occurs in single coordinated process - the nucleophile attacks the reactant from the backside of the leaving group, forming a trigonal bipyramidal TRANSITION STATE - as the reaction progresses the bond to the nucleophile strengthens and the bond to the leaving group weakens. - The leaving group is displaced as the bond to the nucleophile becomes complete

Answer 73

- the single step of SN2 reaction involves two reacting species thus the substrate and the nucleophiles concentrations play a role in determining the rate of SN2 reactions - follows SECOND ORDER KINETICS

Answer 74

rate= k{substrate}{nucleophile}

Answer 75

- the nucleophile must have unhindered access to the central carbon of the substrate - Thus SN2 reactions occur most readily with substrate with little branching in order to minimize steric hindrance The order of reactivity: methyl>primary> secondary > tertirary **tertiary substrates do not react by the SN2 mechanism

Answer 76

SN2 reactions occur most readily in polar APROTIC solvents--the solvents are unable to form hydrogen bonds - typical polar aprotic solvents include acetone and DMSO - since these solvents cannot form hydrogen bonds, they do not create a salvation shell around the nucleophile and therefore do not interfere with its attack on the substrate

Answer 77

-weak bases dissociate more easily from the alkyl chain and thus make better leaving groups, increasing the ease of displacement by the nucleophile

Answer 78

- the nucleophile must attack a neutral molecule and displace the leaving group it must then be a strong nucleophile - this means that the nucleophile is negatively charge ionic spices and can be either a strong or weak base

Answer 79

- Since the nucleophile attacks from one side of the central carbon and the leaving group departs from the opposite side, the reaction FLIPS the bonds attachéd to the carbon. - If the reactant is chiral, optical activity will be retained, but will be inverted between R and S as long as the nucleophile and leaving group have the same priority relative to the other groups

Answer 80

SN1: any nucleophile SN2: strong (charged) nucleophile

Answer 81

SN1: polar protic SN2: Polar aprotic

Answer 82

SN1: Cl-, Br-, I-, weak bases OH if protonated to form OH2^+ SN2: Cl-, Br-, I-, weak bases

Answer 83

SN1: 3>2>1>methyl SN2: methyl>1>2>3

Answer 84

alkanes can react by a free radical substation mechanism in which one or more hydrogen atoms are replaced by Cl, Br, or I atoms - Involves 3 steps 1) Initiation: diatomic halogens are either cleaved by UV light (shown as UV or hv) or peroxide (H-O-O-H or R-O-O-H), resulting in the formation of free radicals (heat can also be used, but is not specific for radical formation) - Free radicals are uncharged species with unpaired electrons. They are extremely reactive and readily attack alkanes 2) Propagation: is when radical produces another radical the can continue the reaction. A free radical reacts with an alkane, removing a hydrogen atom to form HX, and creating an alkyl radical. The alkyl radical can then react with X2 to form an alkyl valid and generate another X. thus propagation the radical 3) Termination: two free radicals combine with one another to form a stable molecule

Answer 85

reaction of an alkane with molecular oxygen (O2) to form carbon dioxide, water, and heat -combustion is often incomplete, producing significant quantities of carbon monoxide instead of carbon dioxide

Answer 86

occurs when a molecule is broken down by heat in the absence of oxygen. - also called CRACKING is most commonly used to reduce the average molecule weight of heavy oils and to increase the production of more desirable volatile compounds - in alkanes, the C-C bonds are cleaved, producing smaller chain alkyl radicals

Answer 87

a process in which a radical transfers a hydrogen atom to another radical, producing an alkane and an alkene

Answer 88

CnHm N=1/2(2n+2-m)

Answer 89

- similar to those of alkanes - melting and boiling points increase with increasing molecular weight and are similar in value to those of the corresponding alkane - Terminal alkenes usually boil at a lower temp than internal alkenes - Trans-alkenes generallly have higher melting points than cis-alkenes because their higher symmetry allows better packing in their solid state - Trans-alkenes generally have lower boiling points than cis-alkenes because they are less polar`

Answer 90

the molecule loses either HX (where X is a halide) or a molecule of water from two adjacent carbons to form a double bond.

Answer 91

unimolecular elimination - is a two step process preceding through a carbocation intermediate - rate of reaction is only dependent on the concentration of the substrate - the elimination of a leaving group and a proton results in the production of a double bond

Answer 92

- same factors favored by SN1 - protic solvents - highly branched carbon chains - good leaving groups - weak nucleophiles in low concentrations

Answer 93

high temps favor E1, but directing a reaction toward either SN1 or E1 is difficult thus the product will be a mixture of both E1 and SN1 products

Answer 94

bimolecular elimination - its rate is dependent on the concentration of tow species, the substrate and the base - there are two possible hydrogens that can be removed from carbon on either side of the leaving group, resulting in two different products. The more substituted double bond is formed preferentially`

Answer 95

1) steric hindrance does not greatly affect E2 reactions. Therefore highly substituted carbon chains, which form the most stable alkenes, undergo E2 easily and SN2 rarely 2) A strong bulky base such as t-butoxide favors E2 over SN2. While SN2 is favored over E2 by strong nucleophies that are weak bases such as CN- or I-

Answer 96

- reduction of hydrogen - addition by electrophiles and free radicals - hydroboration - a variety of oxidations - polymeriation