Chem - Organic chem SL&HL Flashcards

Question

Amines

Answer 1

functional group: -NH2 -CnH2n+2NH2, RNH2 "amino+alkane"

Answer 2

functional group: carboxamide, -CONH2 combination of amino and carbonyl groups - CnH2n+1CONH2, RCNH2 - "-amide"

Answer 3

functional group: nitrile, -CN -CnH2n+1CN, RCN -"-nitrile"

Answer 4

Strength of C-H bonds - alkanes consist of C and H bonded by single bonds - without heat, it's hard to break these strong C-C and C-H covalent bonds Lack of polarity - electronegativites of C and H are almost similar - electrons shared equally - alkanes don't react with polat reagents, as they have no nucleophilic or electrophilic areas alkanes only react in combustion reactions and undergo substitutions by halogens

Answer 5

when alkanes are burnt in XS O2, complete combustion takes plane, and all C and H will oxidise to CO2 and H2O

Answer 6

when alkanes are burnt in limited supply of O2, incomplete combustion will take place and not all C-atoms fully oxidise. Alkane + Oxygen --> CO + H2O CO is a toxic gas, will bind to haemoglobin in blood which can then no longer bind to O2 CO is odorless, and found in car engines where O2

Answer 7

alkanes - halogenations u.v. light (sunlight) is needed a hydrigen atom gets substitutes by a halogen 1. inititation step - Halogen bond is broken by UV energy to form two radicals - produces two radical in a homolytic fission reaction 2. propegation step - progression of the substitution reaction in a chain type reaction - free radicals are very reactive, and will attack unreactive alkanes - a C-H bond breaks homolytically ( each atom gets an e- from covalent bond) - an alkyl free radical is produced - which can attack another halogen to form halogenoalkane and regenerate the halogen bond - this reaction is not verys uitable for preparing specific halogenoalkanes as a mixture of substitution products are formed - if there is enough halogens present, all the hydrigens in the alkane will eventually substituted 3. termination step - when the chain reaction terminates due to two free radicals reacting together and forming a single unreactive molecule - multiple products possible

Answer 8

C=C bonds gives alkanes a number of chemical properties not seen in alkanes - alkenes contain pi-bonds, it is possible to break the weaker pi-bonds and form stronger σ bonds with other species withoit forcing any atom on the molecule to break off - alkenes are able to undergo substitutions - can undergo addition - unsaturated - addition reactions are faster than substitution reaction as only weak pi-bonds are broken rather than σ bonds.

Answer 9

Br2 - yellow, orange solution, "bromine water" - unknown molecule/compound is shaken with bromine water - if compound is unsaturated - addition will take place and color is lost

Answer 10

electrophilic addition - more detail in other flashcard - C=C doubel bond is broken, and a new single bond is formed from each of the two C atoms - occurs at room temperature - halogens can be used to test if molecule is unsaturated

Answer 11

'electron seeker' has a positive charge

Answer 12

'wants to give away electrons' negative charge

Answer 13

- when alkenes are reacted in steam at 300°C, pressure at 60 atm, and sulfuric acid/phosphoric acid catalyst, water is added to the double bond - alkene ---> alcohol - reaction processes via an intermediate in which H+ and H2SO4- ions are added. across double bond - the intermediate is quickly hydrolyseed by water, reforming H2SO4

Answer 14

- most important addition reaction of alkenes - is the reaction in which many monomers containing at least one C=C double bond form long chain of polymers as the only product - the pi-bond in each C=C breaks and monomers link together to form C-C - polymer - long chain that is made up of repeating units - small, reactive molecules that react together to from the polymer are called monomers

Answer 15

- repeated unit is the smallest group of atoms that when connected one after the other make up the polymer chain - square brackets - in poly(alkenes) and substituted poly(alkenes) made up of one type of monomer the repeating unit is the same as the monomer except that the C=C double bond is changed to C-C

Answer 16

- Alcohol + Oxygen ---> CO2 and H2O - lower alcohols burn with an almost invisible flame and make good fuels - ethanol can be produced sustainably as a fuel by the fermentation of sugars - ethanol has a lower energy density is lower than gasoline, therfore blending ethanol and gasoline increases energy denisty, makes it safer in case of fires as the flames can be seen

Answer 17

oxidise to form aldehydes and then carboxylix acids oxidising agents: K2Cr2O7 or KMnO4 Potassium dichromate: K2Cr2O7 - orange - in a solution of dilute acid - orange Cr2O7 (2-) ions--> greenCr (3+) ions Potassium Manganate: KMnO4 - purple - in a solution of dilute acid - purple MnO4(-) ions --> colorless Mn(2+) In XS Oxidising agents, if the aldehyde produced isn't distilled off, it will further oxidise to carboxylic acids, however this reaction takes some time to complete and requires heating. Tests for alcohols: - oxidation using acidified dichromatic provides basis for the test for alcohols as it shows color change - orange --> green - does not work for tertiary alcohols

Answer 18

- oxidise to ketones - under sustained heating

Answer 19

- DO NOT undergo oxidation - there are only C-C bonds on functional group carbon in tertiary alcohols - no H that can break off to form water

Answer 20

ester: -COOR, sweet, fruity smell - formed via condensation reactions between Crboxylic acids and Alcohols, with concentrated H2SO4 as catalyst esterification : NUCLEOPHILLIC SUBSTITUTION naming ester: first part of alcohol, second part: carboxylic acid

Answer 21

- much more reactive than alkanes due to presence of electronegative halogens - X-C bond is polat, C - carries partial + charge, X - carries partial - charge Nucleophilic substitutions with NaOH - halogenoalkane + aq alkali --> alcohol - OH- behaves as nucleophile by donating a pair of e- to - C-atom bonded to halogen - reaction is slow at room temp. - high yield when heated under reflux

Answer 22

Arenes are very stable compounds due to delocalisation of pi e- in the ring, - e- density is spread ou over the molecules during substitution: ring is maintained during addition: aromatic stabilisation is disturbed so it's not favored

Answer 23

with Cl2 or Br2 in the presence of anahydrous AlCl3 or AlBr2 catalysts to form halogenoarene - cl/br acts as electrophile and replaces H atom in benzene - catalyst necessart due to stability of benzene structure - alkylarenes such as methylbenzene undergo halogenation on the 2 or 4 position, which is due to e- donating alkyl group which activates these positions. - ∴the halogenation of alkylarenes results in the formation of 2 products - in xs halogen, it joins at the 2 or 4 or 6 positions -

Answer 24

- substitution - a nitro (NO2) group replaces H atom on arene - benzene is reacted with mixture of conc. nitric acid and conc. sulfuric acid at 25-60℃ - occurs at 2 and 4 positions

Answer 25

- organic reaction in which nucleophile attacks a carbonyl C-atom with partial + charge. - atom attaches to carbonyl has a -charge and is replaces by nucleophile - involves halogenoalkanes, the halogen is replaced by nucleophile - the OH-, is stronger nucleophile than H2O, bc it has a full - charge - strength of nucleophile depends on it's abolity to make its e- pairs available for reactions in general: - charged ions > neutral atom conjugate base > conjugate acid OH->H2O

Answer 26

- tertiary halogenoalkanes - C attached to halogen i salso bonded to 3 alkyl groups 1 stands for the rate of reaction SN1 is 2 steps First step: - C-X bond breaks heterotically and the halogen leaves the halogenoalkane as an X- ion - slow, rate determining step - only depends on the conc. of halogenoalkane - rate= k(halogenoalkane) - forms a tertiary carbocation ( tertiary C with + charge) Second step: - the tertiary carbocation is attacked by nucleophile

Answer 27

forms anions and cations and used double headed arrows to show movement of both e- from covalent bonds

Answer 28

forms free radicals and uses single headed arrows, "fish hooks", to show the moevement of a single e- as a covalent bond breaks

Answer 29

- Primary halogenoalkanes, the C attached to the halogen is bonded to one alkyl group 2 stands for rate of reaction SN2 is one step reaction - nucleophile donates a pair of e- to the g+ C atom of the halogenoalkane to form a new bond - rate determining step depends on the conc. halogenoalkane and nucleophile - rate=k(halogenoalkane)(nucleophile) - at the same time, the C-X bond breaks and the halogen takes both e- in the bond - the halogen leaves the halogeno alkane as X- ion - nucleophile can only attack from oppisite direction of C-X bond - as C-Nucleophile bond forms, C-Br bond breaks causing X atom to leave as an ion.

Answer 30

1. nature of nucleophile 2. halogen involved 3. the structure (class) of halogenoalkane 4. protic and aprotic solvents

Answer 31

most effective nucleophiles are neutral or - charged species, that have lone pair of e- avaliable to donate to the g+ C in halogenoalkane - greater the e- density on the nucleophile ion or molecule -- the stronger the nucleophile - when nucleophiles have some charge, the electronegativty of the atom carrying the lone pair becomes the deciding factor - less electronegative the atom = stronger nucleophile - less electronegative atom has a weaker grip on lone pairs e-

Answer 32

- the halogenoalkane have different rates of substitution reactions - C-I bond is the weakest and easiest to break - C-F bond is the strongest and hardest to break - floroalkanes --> chloroalkanes --> bromoalkanes --> iodoalkanes - least reactive --> most reactive

Answer 33

- tertiary halogenoalkane --> SN1, forms stable tertiary cabocations - -secondary --> mixture of SN1 and 2, forming less stable primary carbocations - this has to do with + inductive effect of alkyl group attached to C-X - tertiary carbocation - 3 alkyl groups - primary - 1 alkyl group, less stable

Answer 34

Protic solvents - contain a H atom bonded to electronegative N or O - capable of hydrogen bonding Aprotic solvents - contain H atom but not bonded to eelctronegative atom - cannot participate in H-bonding SN1 - best conducted using protic, polar solvents SN2 - best conducted using aprotic, non-polar solvents

Answer 35

addition of an electrophile (lewis acid) to an alkene double bond, C=C - C=C , is an area of high electron density which makes it susceptible to attack by electrophiles - C=C bond breaks forming a C-C bond and 2 new bonds Electrophiles - H2(g)--> alkane - H2O(g) --> alcohol - HX, hydrogen halides --> halogenoalkane - X2, halogen --> dihalogenoalkane

Chem - Organic chem SL&HL Flashcards

(63 cards)