DR. COLLINS PCAT – ORGANIC CHEMISTRY

question

hybridization, geometry, and bonding characteristics of *carbon*
answer

-carbon always obeys the *octet* rule in NEUTRAL molecules thus carbon will have one of the following bonding patterns: -sp3: tetrahedral; 109.5; 4 single covalent bonds around C -sp2: trigonal planar; 120; 1 double and 2 single covalent bonds around C -sp: linear; 180; 1 triple and 1 single OR 2 double covalent bonds around C
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carbons may or may not obey the octet rule
answer

-sp3: tetrahedral; 109.5; obeys octet rule -sp2: trigonal planar; 120; does NOT obey octet rule -sp: linear; N/A bond angle; obeys octet rule
question

hybridization, geometry, and bonding characteristics of *nitrogen* and *oxygen*
answer

-like carbon, nitrogen and oxygen always obey the octet rule in neutral molecules -unlike carbon, nitrogen and oxygen complete their octets with lone pairs of electrons -thus their electronic geometry (geometry of electrons) is DIFFERENT from their molecular geometry (geometry of atoms only)
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nitrogen and oxygen hybridizations
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-sp3: tetrahedral electronic geometry; bent or pyramidal molecular geometry -sp2: trigonal planar electronic geometry; linear or bent molecular geometry -sp: linear electronic and molecular geometry
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molecular shape and polarity
answer

-molecules that have sp and/or sp2 hybridized atoms ONLY (other than H) will generally be flat = 2D -the shape of the molecule (in addition to the polarity of individual bonds) determines many of its physical properties -more symmetrical/less polar molecules will have lower boiling points -melting points may or may not follow this rule, as a more symmetrical molecules packs better (more tightly), requiring more energy (higher melting points) to melt the solid -when comparing trans vs cis alkenes, the more symmetrical/less polar trans-alkenes have lower boiling points and higher melting points than cis-alkenes
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bond length
answer

-the more s character an orbital has, the shorter the bond it makes is -therefore, the length of the bonds shown below increases L to R -H-H (s-s)< C=-C (sp-sp) < C=C (sp2-sp2) < C-C (sp3-sp3)
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elements of unsaturation
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-the general formula for an alkane (hydrocarbon with single covalent bonds only AKA *saturated hydrocarbon*) is CnH2n+2 -compounds with double bonds, triple bonds, or rings have less than “2n+2” hydrogens relative to the number of carbons “n” –it is said that these compounds are *unsaturated* –each element of unsaturation is 2 H’s less than the maximum (2n+2) –thus we compare the formula of a given compound to that of the fully saturated alkane with the same number of C’s, substract the H’s, and divide the difference by two (2H = 1 element of unsaturation) -that will tell us the elements of unsaturation note: -1 double bond = 1 element of unsaturation -1 triple bond = 2 elements of unsaturation -1 ring = 1 element of unsaturation EX: the formula for an alkane with 6C is C6H(2×6)+2 = C6H14 → 14 H – 10 H = 4 H/2=2 unsaturations -halogens (F, Cl, Br, I) form single bonds like H; hence in a molecule whose formula contains one or more halogens, replace those halogens by H and then follow the instructions above EX: C2H2Cl4 = C2H6 = a fully saturated alkane; thus have no unsaturations
question

what is a saturated hydrocarbon? what is an unsaturated hydrocarbon?
answer

-hydrocarbon with only single covalent bonds -hydrocarbon with double bonds, triple bonds, or a ring
question

sigma and pi bonds
answer

-*sigma bond*: head-on overlap of atomic (s, p) or hybrid (sp, sp2, sp3) orbitals -*pi bond*: side-to-side overlap of p orbitals TYPES OF COVALENT BONDS -single covalent bonds: 1 sigma, 0 pi -double covalent bonds: 1 sigma, 1 pi -triple covalent bonds: 1 sigma, 2 pi note: -any atom involved in a double bond has both sp2 orbitals (to form sigma bonds) and a p orbital (to form the pi bond in the double bond) -any atom involved in a triple bond has sp orbitals (to form sigma bonds) and two p orbitals (to form the two pi bonds in the triple bond) -H atoms bond with their unhybridized s orbitals
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hydrogen bonding
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-definition: attraction between the positive H in one molecule and the negative F, O, or N in the next molecule -molecules that have -OH or -NH groups (or HF) can hydrogen bond
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solubility
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-“like dissolves like” -polar compounds tend to dissolve polar compounds; non-polar compounds tend to dissolve non-polar compounds -since organic compounds have mostly C-C or C-H bonds (non-polar bonds), they tend to dissolve in non-polar solvents
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boiling point
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-when comparing similar molecules (e.g. all hydrocarbons or all alcohols) boiling points increase with *increasing number of carbons* (increasing molecular weight or molar mass) -a *polar group* (or a more electronegative atom) will increase the boiling point of a molecule when compared to a molecule of similar weight with a less polar group (less electronegative atom) -groups that can *hydrogen bond* (-OH, -NH, -COOH) will increase the boiling point the MOST -compounds with low boiling points (hydrocarbons and other non-polar organic compounds) have a lot of vapor associated with the liquid, or are gases at room temp → this makes them flammable/explosive/combustible
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acid/base, nucleophile/electrophile
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-*base*: bronsted-lowry base (proton acceptor) -*acid*: bronsted-lowry acid (proton donor) -*nucleophile*: lewis base (electron donor with formation of a bond) -*electrophile*: lewis acid (electron acceptor) -strong acids ionize readily; this is due to the stability of the conjugate base formed → the stronger the acid, the more stable (less reactive) the conjugate base -conjugate base stability can be due to: 1. higher EN of an atom in the conjugate base (drawing electron density away from the atom with the negative charge) 2. larger size (“polarizability”, better able to diffuse the negative charge) or 3. resonance stabilization (delocalize negative charge) **electronegativity arguments usually apply when comparing atoms in the same ROW of the periodic table, while size arguments apply when comparing atoms up and down a COLUMN in the periodic table (similar reasoning applies to electrophiles and nucleophiles)
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nomenclature of alkanes and cycloalkanes
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-*alkanes* are hydrocarbons that have single C-C covalent bonds only (saturated) -like all organic compounds, alkanes are named based on the number of carbon atoms in the “base chain” -we use specific prefixes to indicate the number of carbon atoms (meth=1, eth=2, prop=3, but=4, pent=5, hex=6, hept=7, oct=8, non=9, dec=10) -each C in an alkane must have 4 single covalent bonds -often the “base chain” will have “branches” which are called *substituents* → end in “yl” (e.g. methyl, ethyl) -there are 2 possibilities for 3C substituents: -CH2CH2CH3 = propyl; -CH(CH3)2 = isopropyl -there are 4 possibilities for 4C substituents: -CH2CH2CH2CH3 = butyl; -CH2CH(CH3)2 = isobutyl; -CH(CH3)CH2CH3 = sec-butyl; -C(CH3)3 = tert-butyl or t-butyl
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rules for naming branched alkanes
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1. locate the longest chain of consecutive C atoms – this will be the base chain 2. number the carbons in base chain, with C#1 being the one closet to the first substituent 3. name first the substituents attached to the base chain in alphabetical order, using the carbon number as the locator 4. use di-, tri-, etc. for multiples of the SAME substituent 5. halogen atoms attached to C are treated as substituents (fluoro-, chloro-, bromo-, iodo-) 6. finish the name with the base chain name
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cycloalkanes
answer

-are hydrocarbons that have a base “ring” rather than a base “chain” -the rules for naming cycloalkanes are the same as those for open-chain alkanes, except that the ring will be numbered in a way to give the LOWEST possible combination of carbon numbers to the substituents -a cycloalkane becomes a substituent when attached to a carbon chain with MORE CARBONS than the ring -as with alkyl substituents, cycloalkyl substituents have a “yl” ending (cyclopentyl, cyclohexyl, etc)
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functional groups
answer

-definition: a unique part of an organic compound (“R”) -alkane: none -alkene: C=C -alkyne: -C=-C- -nitrile: R-C=-C- -alcohol: R-OH -amines: R-NH2 -ether: R-O-R -imine: 2R-C=NR -aldehyde: RH-C=O -ketone: 2R-C=O -carboxylic acid: R-COOH -ester: R-CO2-R -amides: NH2-C=O
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nomenclature of molecules with functional groups: OPEN-CHAIN MOLECULES THAT ARE *NOT* CARBOXYLIC ACID DERIVATIVES – systematic names
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1. locate the longest chain of consecutive C atom that contains or is attached to the functional group – this is the base chain 2. number the carbons in the base chain, with C#1 being the one closest to the functional group 3. the ending of the base chain name will reflect the nature of the functional group: *-ene* for alkene, *-yne* for alkyne, *-ol* for alchol, *-al* for aldehyde, *-one* for ketone, *-oic acid* for carboxylic acid, *nitrile* for nitrile 4. the location of the functional group must be specified if unclear (aldehydes, carboxylic acids, and carboxylic acid derivatives are always on C#1) – for alkenes and alkynes, only the LOWEST carbon number in the double or triple bond is named 5. if more than one of the SAME functional group is present, use di-, tri-, etc. (diene, triene) giving the location of each functional group 6. if more than one TYPE of functional group is present, the one with the highest priority (the most oxidized) will be the base chain, and others will become substituents: -OH becomes a hydroxyl substituent, -OCH3(-OCH2CH3) are methoxy(ethoxy) substituents, -NH2 is an amino substituent, -CHO is a formyl substituent, etc.
question

nomenclature of molecules with functional groups: OPEN-CHAIN MOLECULES THAT ARE *NOT* CARBOXYLIC ACID DERIVATIVES – common names
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A. alcohols -name ends in alcohol (as a separate name) -name begins with the name of the alkyl group that is attached to the -OH B. ketones, ethers, and amines -name the alkyl substituents attached to the C=O (ketone), O (ether), or N (amine) -use di- or tri- if the alkyl substituents are the same **amines are classified as primary (1 alkyl group + 2H’s), secondary (2 alkyl groups + 1H), or tertiary amines (3 alkyl groups) → when naming amines specify the alkyl groups only and ignore the H’s (e.g. NH3 is ammonia) C. aldehydes, carboxylic acids, carboxylic acid derivatives 1 C = form- 2 C = ace- 3 C = propion- 4 C = butyr- name ends in aldehyde (aldehydes), -ic acid (carboxylic acids), -ate (esters), -amide (amides), -nitrile (nitriles)
question

what’s the systematic prefix used for -CHO? -OH -OCH3 -OCH2CH3 -NH2
answer

formyl hydroxy methoxy ethoxy amino
question

systematic and common names for carboxylic acid derivatives
answer

-the -OH in a carboxylic acid can be replaced with -OR (ester), -NH2 or NR2 (amide), etc. 1. name the carboxylic acid first (systematic or common name) 2. replace the “ic acid” in the carboxylic acid for the ending of each functional group (-ate for ester, -amide for amides, etc) 3. before the base chain name, list the alkyl substituent(s) that are attached to the O or N
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aromatic compounds
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-definition: compounds that contain the benzene ring as a “base ring” -benzene = 6C ring with three alternating double covalent bonds -as with cycloalkanes, we will number the carbons in the ring so that the substituents are in the lowest possible combination of carbon numbers; name will end in -benzene -special rule: disubstituted benzene compounds (not cycloalkanes or cycloalkenes) may be named using numbers OR using *ortho-* (o- = 1,2), *meta-* (m- = 1,3) or *para-* (p- = 1,4) **benzene ring with 1C substituent attached to it (-CHO, -COOH, -CN) starts with “benz” or “benzo” **if nitrogen replaces one of the carbons in benzene → becomes a *pyridine* (N receives #1 when numbering the ring)
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aromaticity
answer

-definition: the cyclic compound is more stable than the open-chain counterpart; the opposite is true (less stable) for anti-aromatic compounds TERMS for aromaticity 1. cyclic with conjugated pi bonds (alternating double and single bonds) 2. each atom in the ring must be sp2 hybridized, with an unhybridized p orbital 3. the p orbitals must overlap continuously around the ring (usually a planar structure) **non-cyclic, non-planar, non-conjugated compounds = anti-aromatic!
question

huckel’s rule
answer

-*aromatic*: a continuous ring of overlapping p orbitals that has *4N+2* pi electrons (N=0, 1, 2 ,3, etc) -*anti-aromatic*: a continuous ring of overlapping p orbitals hat has *4N* pi electrons **any substitued benzene compound (toluene, phenol) or any compound containing more than one benzene ring (naphthalene, anthracene) is aromatic!
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isomers
answer

-are different molecules (with different properties and different names) that share the SAME MOLECULAR FORMULA two main types 1. structural (constitutional) isomers 2. stereoisomers
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structural isomers
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-are molecules that have a *different bonding sequence*
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stereoisomers
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-are molecules that have the same bonding sequence but *different spatial orientation* TWO TYPES OF STEREOISOMERS 1. *enantiomers*: stereoisomers that are non-superimposable mirror images -a molecule with one sp3 C that has four different groups attached to it is said to have a *chiral carbon*, which will have an enantiomer (S & R) -when placed in a *polarimeter* (an instrument that irradiates the sample with polarized light), one of the enantiomers will rotate the plane of polarized light clockwise by a certain number of degrees, while the other enantiomer will rotate the plane of polarized light in the opposite direction counterclockwise by the same number of degrees -a pure sample of each of the enantiomers (100% S or 100% R) is said to be *optically active*; while a 50:50 mixture of the two enantiomers will NOT rotate the plane of polarized light (optically inactive) → this is called *racemic mixture* 2. *diastereomers*: stereoisomers that are NOT mirror images -example: cis and trans isomers (only exist if there is a double bond or a ring) → stereoisomers because same bonding sequence but different spatial orientation; diastereomers because they are not mirror images of each other
question

substitution and elimination reactions (SN1, SN2, E1, E2)
answer

-these are reactions that typically occur when using alkyl halides as a substrate -*alkyl halides*: molecules where a halogen (X = F, Cl, Br, I) is connected to an sp3 C -alkyl halides can be classified as methyl, primary, secondary, or tertiary alkyl halides -alkyl halides are good substrates for substitution or elimination reactions because halogens are good *leaving groups* (stable as a negative ion due to their large size and high EN) -the nature of the alkyl halide (methyl, primary, secondary, or tertiary) will determine the TYPE of substitution and/or elimination reaction that occurs
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nucleophilic substitution
answer

a nucleophile replaces the leaving group in the alkyl halide
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elimination
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both the halide ion and the H+ are lost from the alkyl halide; a new pi bond is formed
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both nucleophilic substitution (SN) and elimination (E) reactions can occur in one or two steps
answer

a. SN2 and E2 occur in ONE step (the “2” refers to the *2 reactants* – alkyl halide and nucleophile/base – that are involved in the rate determining step, which is the only step) b. SN1 and E1 occur in TWO steps (*1 reactant* only – the alkyl halide – is involved in the first, slow, rate determining step. since the nucleophile/base only participates in the second step, which is the fast one that is not rate-determining, it has no impact on the rate of the reaction)
question

SN2 (one step reaction)
answer

-a strong nucleophile (OH-) attacks the back side of the electrophilic C of the alkyl halide, forcing the leaving group (Br-) to leave -concerted, one step reaction (making and breaking of bonds occurs simultaneously) -back side attack results in *inversion of configuration* (R to S or S to R) if the carbon involved is chiral -the nucleophile must be able to reach the electrophilic sp3 C attached to the Br from the “back” –the smaller the groups attached to the C, the easier it will be for the nucleophile to reach it –hence the nature of the alkyl halide is one of the factors that determines the rate of SN2 reactions: (fastest) CH3X > 1 > 2 >>> 3 (~unreactive)
question

SN1 (two step reaction)
answer

1ST STEP: formation of carbocation intermediate, rate-determining or slow step 2ND STEP: fast step, with subsequent loss of H+ if necessary -the strength of the nucleophile is unimportant, since it is not involved in the rate determining step; the nucleophile will also be the solvent (a weak nucleophile) –water → hydrolysis, resulting in the formation of an alcohol –alcohols like methanol or ethanol → solvolysis, resulting in the formation of ethers -two step reaction with the formation of a reactive intermediate = the flat carbocation -*racemization occurs* -alkyl groups are electron donors; the more alkyl groups attached to the carbocation (electron deficient), the more stable that carbocation will be and the faster the reaction will proceed -the rate of SN1 reactions relative to the alkyl halide substrate is the exact OPPOSITE of that of an SN2 reaction: (fastest) 3 > 2 >>> 1 or methyl (~unreactive)
question

what is the difference in outcome between an SN1 and SN2 reaction when the reaction site is at a chiral carbon?
answer

the SN1 reaction results in racemization; the SN2 reaction results in inversion at the stereocenter
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E2 (one step reaction)
answer

-a strong base (i.e. OH-) abstracts a proton from a poor SN2 substrate (tertiary alkyl halide – sterically hindered for back-side attack by a nucleophile – is the best) -concerted, one step reaction -bromide ion leaves as the proton is abstracted -pi bond forms -the order of reactivity of alkyl halides is: 3 > 2 > 1 -if more than one alkene can form, the MORE SUBSTITUTED alkene is generally the major product
question

E1 (two step reaction)
answer

1ST STEP: formation of carbocation intermediate, rate-determining or slow step (same as in SN1) 2ND STEP: fast step -the strength of the base is unimportant, since it is not involved in the rate determining step -two step reaction with the formation of a reactive intermediate in the first, rate determining step = the flat carbocation -as with SN1, the STABILITY of the carbocation determines the rate of the reaction; thus the rate of E1 reaction relative to the alkyl halide is: (fastest) 3 > 2 >>> 1 or methyl (~unreactive) -if more than one alkene can form, the more substituted alkene is generally the major product
question

dehydration reactions
answer

-elimination reactions can occur with substrates other than alkyl halides, as long as there is a good leaving group attached to the sp3 C -in general OH- is a bad leaving group (strong base, not stable, will attack the product formed); but H2O is a fairly good leaving group -thus alcohols can undergo an elimination reaction if they are protonated in acid FIRST -the net result of the elimination reaction of an alcohol is the loss of water (dehydration) with the formation of an alkene -as with alkyl halides, the rate of reaction relative to the substrate is 3 > 2 >>> 1
question

when an alkene reacts, what type of bond is broken?
answer

pi
question

what type of reaction causes a pi bond to break into two sigma bonds?
answer

addition reaction
question

addition reactions to alkenes
answer

-pi electrons in alkenes are loosely held and nucleophilic -these pi electrons will attack an electrophile and form a carbocation (step 1) -a nucleophile (water, bromide ion, etc) will then add to the carbocation (step 2) -the NET RESULT = addition to the double bond, with the double bond becoming a single bond -many different types of addition reactions to a double bond are possible; one imrportant consideration is the *regiochemistry* of the product → to which carbon (region of the molecule) will the “E” and “Nuc” in the reaction attach? MARKOVNIKOV & ANTI-MARKOVNIKOV -the *markovnikov* product forms when the proton of an acid (H+) adds to the carbon in the double bond that already has the MOST H’s “rich gets richer” -the *anti-markovnikov* product forms when the H adds to the MORE SUBSTITUTED carbon of the double bond **under which conditions will markovnikov or anti-markovnikov products form? -markovnikov products occur under ACIDIC conditions → the first RDS involves protonation (H+ from the acid is the electrophile), with the formation of the more stable (more substituted) carbocation; addition of the nucleophile to the carbocation (with loss of H+ if necessary) gives rise to the markov product -formation of the anti-markov product is the result of a reaction mechanism that does NOT involve a carbocation (i.e. with radical reactive intermediates)
question

how is the regioselectivity in a markov type addition of HX to a double bond best explained?
answer

adding H to the C in the double bond with the most H’s directly attached to it creates the most stable carbocation intermediate
question

markov or nonmarkov summary of alkene reactions (pg 31)
answer

-HBr → markovnikov -HBr/ROOR → anti-markov -1) Hg(OAc)2, HOR 2) NaBH4 → markov -1) BH3, THF 2) H2O2, OH → anti-markov -if rather than adding HX, we add X2 (XX) to the double bond, there are NO regioselectivity issues -addition of halogens (X2, X = F, Cl, Br, I) to an alkene is possible because of the *polarizability of the halogen* –as X2 approaches the site of high electron density (the double bond), X2 is polarized (electrons shift from one end to the other) such that the nearest X atom now has some partial positive charge –the pi electrons of the double bond attack the positive X atom, breaking the X-X bond –the SMALLER the atom or molecule, the LESS polarizable it is –thus F2 is the least polarizable halogen, and the least reactive in addition reactions
question

which is the least likely halogen to add a double bond?
answer

F2, because its the smallest halogen = least polarizable, thus least reactive
question

epoxide opening
answer

-another way of obtaining 1,2 diols (besides hydroxylation) is by opening an epoxide with water (hydrolysis) -*epoxides* are 3-membered rings with two carbons and one oxygen, that are obtained by adding “O” to alkenes (the source of the “O” is peroxyacid) -the epoxide can then be opened with water under acidic conditions (or with OH- under basic conditions) to yield a *trans 1,2 diol*
question

an acid catalyzed hydrolysis of an epoxide yields what?
answer

a trans-diol
question

addition reaction to alkynes
answer

-similar addition to alkenes -pi bond becomes two sigma bonds -one molecule (triple bond → double bond) or two molecules (triple bond → single bond) may add -as with alkenes, addition of asymmetrical molecules (HX, H2O) can give markov or anti-markov products, but ONLY if the addition is to a *terminal alkyne* (internal alkyne gives mixtures of products)
question

alkyne reactions (pg 35)
answer

-HX adds TWICE to give geminal dihalide –HBr x2 → markov –HBr/ROOR x2 → anti-markov -H2O adds ONCE to give an unstable *enol* that will tautomerize (rearrange) to a *ketone* (markov product) or and *aldehyde* (anti-markov product) –HgSO4/H3O+ → markov (ketone) –1) Sia2BH 2) H2O2, NaOH → anti-markov (aldehyde)
question

the anti-markovnikov hydration of a terminal alkyne produces what functional group?
answer

aldehyde
question

oxidative cleavage
answer

besides addition reactions, double bonds (alkenes) can be involved in other types of reactions in an oxidation cleavage: -BOTH the pi and sigma bonds in a double bond break -C=C becomes C=O -mixtures of ketones and aldehydes (or carboxylic acids) are formed (the carboxylic acid is the oxidation product of the initial aldehyde under strongly oxidizing conditions)
question

oxidative cleavage reagents (pg 36)
answer

1. warm or concentrated acidic KMnO4 (strongly oxidizing) → give carboxylic acid and ketone 2. ozonolysis, followed by a reducing agent → give aldehyde and ketone
question

oxidation/reduction
answer

a molecule is oxidized or reduced when the following happens: -*oxidation*: loss of H2, gain of O, O2, or X2 -*reduction*: gain of H2 or H-, loss of O, O2, or X2 -neither: gain or loss of H+, H2O, HX (hydration/dehydration reactions, or addition of HX to double bonds, thus are NOT oxidations or reductions) AGENTS – redox rxns can also be identified by the type of reagent used -*oxidizing agents*: reagents that cause an oxidation to occur → Na2Cr2O7, CrO3/H2SO4 (H2CrO4), PCC, HNO3, HIO4, peroxides, KMnO4, O2, O3 -*reducing agents*: reagents that cause a reduction to occur → (CH3)2S, Zn, NaBH4, LiAlH4, SnCl2, H2
question

oxidation of a primary alcohol yields…? oxidation of a secondary alcohol yields…?
answer

-primary alcohol yields: aldehyde first, then carboxylic acid -secondary alcohol yields: ketone only, cannot be oxidized any further
question

combustion
answer

-the reaction of O2 with a molecule or atom -the molecule/atom that reacts with O2 is being oxidized (gaining oxygen), while O2 is being reduced
question

the combustion of gasoline is an example of what type of reaction?
answer

oxidation-reduction
question

in the breathalyzer test for intoxication, ethanol (CH3CH2OH) is converted into ethanoic acid (CH3COOH). this is an example of what reaction?
answer

oxidation
question

reactions of carboxylic acids – NEUTRALIZATION (pg 40)
answer

-like many other bases, carboxylic acids react with bases to form salts and water ex: CH3COOH + NaOH → CH3COO-Na+ + H2O acid + base → salt + water
question

reactions of carboxylic acids – ESTERIFICATION and other condensation reactions (pg 40 to 41)
answer

-a condensation reaction = joins two molecules by losing H2O A. *esterification reaction*: the condensation of a carboxylic acid and an alcohol forms an *ester* and water; reaction is catalyzed by ACID ex: R-COOH + HO-R (H+) → R-COO-R’ + H2O acid + alcohol → ester + water B. the condensation of a carboxylic acid and an amine will form an *amide* and water ex. R-COOH + NH2-R → R-CO-NH-R’ + H2O carboxylic acid + amine → amide + water C. the condensation of a carboxylic acid and a thiol will form a *thioester* and water ex. R-COOH + H-S-R → R-CO-S-R’ + H2O acid + thiol → thioester + water
question

what are carboxylic acid derivatives?
answer

-molecules that are made from carboxylic acids -examples are esters, amides, thioesters; acid or acryl chlorides (R-CO-Cl), anhydrides (R-CO-O-CO-R’), and nitriles (R-CN)
question

an ester is formed by the reaction of an alcohol and…?
answer

acid
question

the reaction between a carboxylic acid (R-COOH) and a thiol (HS-R’) will yield…?
answer

a thioester: R-CO-S-R’ + H2O
question

reactions of carboxylic acids – ACID HYDROLYSIS (pg 42)
answer

-all carboxylic acid derivatives can be hydrolyzed (under acidic, aqueous conditions) BACK to the carboxylic acid -the *reactivity* of carboxylic acid derivatives (including the RATE of hydrolysis) decreases from left to right: (most reactive) acid chloride > anhydride > ester > amide > carboxylate (least reactive) -*steric effects* (the size of the aliphatic group attached to the carbonyl carbon and/or oxygen) are also important in determining the rate of hydroysis of carboxylic acid derivatives –small substituents such as H or CH3 result in FASTER hydrolysis, while large substituents (t-butyl, phenyl) result in SLOWER hydrolysis
question

which molecule will hydrolyze into a carboxylic acid and a thiol?
answer

thioester
question

carboxylic acid derivatives interconversion (pg 44)
answer

-the more reactive carboxylic acid derivatives (acid chlorides, anhydrides) can be converted to the less reactive carboxylic acid derivatives (esters, amides) by reacting with *alcohols* (to form esters in esterification reaction) or *amines* (to form amides)
question

saponification (basic hydrolysis) (pg 45)
answer

-under ACIDIC conditions, the hydrolysis of an ester yields a carboxylic acid and an alcohol -under BASIC conditions, the hydrolysis of an ester yields the *salt* of the carboxylic acid and an alcohol
question

electrophilic aromatic substitution
answer

-electrophile substitutes for a hydrogen on the benzene ring -requires a VERY strong electrophile (a cation generated in-situ, often by addition of a lewis acid like AlCl3) -halogenation: Cl2/AlCl3 -sulfonation: SO3/H2SO4 -nitration: HNO3/H2SO4 -alkylation: CH3Cl/AlCl3 -acylation: Cl-C=O-CH3/AlCl3

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