EK Ch3 Oxygen containing reactions (Chem)

Question 1

alcohols

Answer 1

• boiling point goes up with molecular weight and down with branching
  -melting point trend where melting point also goes up with molecular weight is not as reliable but still exists
  -branching generally lowers bp and has a less clear effect on the melting point
  -boiling/melting point of alcohols is much higher than their similar size alkanes due to their ability to HYDROGEN BOND
• hydrogen bonding inc intermolecular forces, which must be overcome to change phases*
  r-O-H 108.9 degrees bent

Question 2

electron withdrawing groups

Answer 2

• strongly electronegative and pull electron density away from the rest of hte molecule
• make alcohols more acidic by drawing electron density away from the acidic proton, inc its partial positive charge
• following deprotonation, EWGs stabilize the resulting negative charge

Question 3

Electron donating groups

Answer 3

donate electrons, stabilize positive charge

• by stabilizing conjugate acids, EDGs make a molecule more basic
• Alkyl groups are electron donating, so they inc the basicity and dec the acidity of alcohols
• Due to the greater number of EDGs the trend of alcohol acidity from strongest acid to weakest acid is : methyl > 1>2>3

Question 4

Ethers

Answer 4

-relatively non-reactive
- contain oxygen with lone pairs of electrons so they can H bond with compound s containing a H attached to N, O or F
-ether cannot H bond with itself it will have a boiling point roughly comparable to that of an alkane with a similar molecular weight
-soluble in water as alcohols of similar molecular weight
-organic compounds tend to be much more soluble in ethers than they are in alcohols because H bonds do not need to be broken for the compound to dissolve
-less polar than alcohols, many organic compounds are relatively nonpolar “likes dissolve likes” principle!
if they appear in a passage for a reaction it could be a substitution reaction or the cleavage of the ether by HI or HBr to form the corresponding alcohol and alkyl halide

Question 5

when N appears on MCAT think of these 2 rxns:

Answer 5

1. nitrogen acts as a nuc where the lone pair of electrons attacks a positive charge
2. Nitrogen can take on a fourth bond (becoming positively charged)

Question 6

Nitrogen

Answer 6

• better nuc than O and worse LG than O
• C and N are adjacent on periodic table so their electronegativities are closer in value than those of C and O so it shares it electrons more readily with carbon in a bond than O, so CN bonds once formed are stronger and less polar, and less reactive than CO bonds

Question 7

nitrogen in aa

Answer 7

• basic amino acids like glutamine and arginine have N atoms that can act as nucelophilies in the active sites of enzymes!

Question 8

electrophiles 2

Answer 8

• on mcat common electrophiles all have carbonyls!- always ketones, aldehydes, carboxylic acids because due to the partial positive charge on the carbonyl carbon!
• these are molecules with a tendency to accept electrons to form new chemical bonds. They are targets of nucleophiles
• carbs and lipids are good examples of biological molecules with electrophilic functional groups

Question 9

glucose ring closing

Answer 9

• OH, O on C 5 attacks/ acts as nucleophile and attacks carbonyl in a nucelophilic addition reaction and forming a hemiacetal
• in aqueous solutions, carbohydrates exist predominantly in ring form, though equilibruim also allows a small amount of the chain form.
• Monosaccharides can undergo intermolecular nuc substitution reactions with other monosaccarides to form acetals. This is the mechanism by which monosaccharides join to form polysaccharides.

Question 10

carb naming

Answer 10

-named for number of carbons
triose, tetrose, pentose, hexose, heptose, and so on

names are combined with aldoses when carbohydrates contain an aldehyde or ketoses when they contain ketones

glucose is= aldohexose

Question 11

D or L numbering/labeling

Answer 11

carbohydrates can be classified based on stereochemistry and are labeled D or L

1. number all carbons in a Fischer projection with number 1 being at the end of the aldehyde or the end ketone it is closest to
2. if projection the hydroxyl group on the highest numbered chiral carbon points to the right, the carbohydrate is classified as D if it points to the left it is classified as L

Question 12

D/ L carbohydrates

Answer 12

carbohydrates can be classified based on stereochemistry and are labeled D or L

-most naturally produced carbohdyrates are D epimers.
- enzymes as a result involved in carb metabolism cannot “recognize” L carbohydrates and L carbohydrates cannot be assimilated by the body.
THINK D FOR DELICIOUS!

Question 13

epimers

Answer 13

carbs that have the same structure except for the configuration around a single chiral center are epimers, or stereoisomers of each other

Question 14

anomeric carbon

Answer 14

its the carbon that was formerly the carbonyl carbon and can be identified as only carbon in sugar attached to 2 oxygens!!

• its alcohol group points upwards or downwards, in alpha glucose the hydroxyl group on anomeric carbon (carbon 1 now labeled) and the methoxy group on carbon 6 in glucose are on opposite sides of the carbon ring
• in beta glucose, the hydroxyl group and the methoxy group are on the same side of the carbon ring! think Beta- just like the two bumps on the letter B

Question 15

furanose

Answer 15

5 membered ring

Question 16

pyranose

Answer 16

6 membered ring, number of ring members contain including oxygen, so glucose ring is glucopyranose

Question 17

-oside

Answer 17

• for sugars that are formed when a sugar is attacked by an alcohol to create an acetal, the are given names that end in -oside
• if hydroxyl group on the anomeric carbon of glucose were replaced by an O-methyl group, it would become methyl glucopyranoside
• group attached to the anomeric carbon of a glycoside is called an aglycone, dissaccharides and polysac. are glycosides where the aglycone is another sugar, these linkages are called glycosidic linkages

Question 18

reducing sugar

Answer 18

• have a hemiacetal or hemiketal rather than an acetal or ketal, so are susceptible to attack, bonds are formed btw reducing sugar molecules in a dehydration reaction!
• these bonds can be broken via hydrolysis, water added to break bonds btw biological molecules
• this process occurs slowly, the hydrolysis, unless catalyzed by an enzyme

Question 19

dehydration reaction

Answer 19

• in bio its commonly called dehydration rxn becuase water is removed from the molecule , ex forming matlose
• in orgo, called acetal formation

Question 20

maltose

Answer 20

alpha-1-4’ glycosidic linkage, two glucose molecules

Question 21

sucrose

Answer 21

1-1’ glycosidic linkage, glucose and fructose

• this linkage is alpha with respect to glucose and beta with respect to fructose!
• it is more accurately called a 1,2’ linkage because the anomeric carbon on fructose is numbered 2 not 1 like glucose

Question 22

lactose

Answer 22

beta-1-4’ galactosidic linkage, galactose and glucose
-some adults lack enzyme to break this down, this causes the condition of lactose intolerance

Question 23

cellulose

Answer 23

beta-1-4’ glycosidic linkage, a chain of glucose molecules
- humans do not possess the enzyme needed to break or digest these glucosidic linkages, some animals rely on gut bacteria to break these bonds

Question 24

amylose (starch)

Answer 24

alpha-1-4’ glycosidic linkage, a chain of glucose molecules

Question 25

amylopectin

Answer 25

alpha-1-4’ glycosidic linkage, a branched chain of glucose molecules with alpha- 1,6’ glucosidic linkages forming the branches

Question 26

glycogen

Answer 26

alpha-1-4’ glycosidic linkage, a branched chain of glucose molecules with alpha-1,6’ glucosidic linkages forming the branches

Question 27

amino acid stereochemistry

Answer 27

• all amino acids are S absolute configuration around central chiral carbon
• exceptions=
• cysteine, in which priority of the sulfur R groups makes cysteine R in absolute configuration
• glycine which has two H substituents makes it achiral
• amino acids found in body are referred to as L amino acids

Question 28

gabriel synthesis

Answer 28

-amino acids can be synthesized de novo via Gabriel or stretcker syntehsis
-1. nitrogen protected in a phthalimide, analogous to a di-amide to prevent more than one alkylation from occuring, first step N acts as nuc and substitutes the bromdie on diethyl bromomalonate
2. a very unusual step, a H leaves the middle carbon of diethyl-bromomalonate
this carbon then becomes a carbanion and is very nucleophilic, stabilized by the electron withdrawing carboxylic acids on either side of it
3. nucleophilic carbon undergoes nucleophilic substitution with a new alkyl halide- this step in which the R group is added to the amino acid.
4. final step, N is hydrolyzed from the phthalimide by acid and water to form a free amino acid

Question 29

Strecker synthesis

Answer 29

-aldehyde is mixed with potassium cyanide and ammonium chloride
-first step1. cyanide anion acts as nuc toward the carbonyl. This initially results in a hydroxynitrile molecule, however, nitrogen is a better nucelophile than oxygen, and acid is present to protonate the alcohol group.
2. leads to second step, nucleophilic substitution occurs an an aminonitrile is formed
the nitrile group behaves as a carboxylic acid derivative
3. strong acid in water protonates the nitrile group turning it back into its carboxylic acid form, molecule is now an amino acid

Question 30

fatty acids

Answer 30

long even numbered carbon chains with carboxylic acid group at one end

• amphiphatic, hydrophobic and hydrophilic group on molecule
• since hydrophobic carbon chain predominates, fatty acids are nonpolar, pka of most fatty acids is around 4.5 so most fatty acids exist in their anion form in the cellular enviornment

-carbons next to carbonyl carbon is called alpha c, and on the opposite end of the chain is called omega, third c from end of fatty acid is referred to as the omega-3 carbon

Question 31

triglycerides

Answer 31

• one glycerol molecule and three fatty acids, present of carbonyl groups
• in their formation hydroxyl groups on glycerol act as nuceliophiles, attacking carbonyl carbon of a carboxylic acid group on a fatty acid creating an ester bond
• formation of lipids is called lipogenesis, reverse process is called lipolysis

Question 32

formation and hydrolysis of triglycerides

Answer 32

formation of triglycerides involves the formation of three ester bonds

Question 33

lipid breakdown

Answer 33

when lipid breakdown is base catalyzed in a process called saponification, soap is formed fatty acid salts and glycerol!

Question 34

carbon chains saturated vs unsaturated

Answer 34

-saturated= containing no double bonds 
unsaturated= containing double bonds 

all naturally occurring double bonds in fatty acids are CIS

REMEMBER - melting point increases with molecular weight and that unsaturated fatty acids have lower melting points than saturated fatty acids!
-saturated fatty acids are straight and solidify more easily, while unsaturated fatty acids have “kinks” and do not solidify as easily

Question 35

OH group creates…. in DNA

Answer 35

anhydride bond with phosphate group of another nucleotide

Question 36

in DNA long strains are created by…

Answer 36

• oxygen attached to the 3’ C of one nucleotide attacks the phosphate group attached to the 5’ carbon of another nucleotide. A hydroxy group acts as LG, making reaction a nucleophilic substitution.
• nucleotides create long strands by linking together via phosphodiester bonds

Question 37

During the Gabriel synthesis of primary amines, the secondary carbon serves as the:
A.nucleophile.
B. electrophile.
C. leaving group.
D. intermediate.

Answer 37

B. electrophile.
During Gabriel synthesis, the nitrogen of a phthalimide molecule acts like a nucleophile, attacking the electrophilic secondary carbon of an alkyl halide.