have the general formula Cn(H2O)n, where n varies from 3 to 8
Aldose: A monosaccharide containing an aldehyde group
Ketose: A monosaccharide containing a ketone group
A) aldose
B) ketose
naming monosaccharides
prefix: number of carbons in molecule
suffix: ose, indicates that it is a carbohydrate
A) triose
silver mirror test
used to detect the presence of aldehydes
based on the reduction of silver ions to silver metal, which forms the mirror
Tollens' reagent selectively oxidises aldehydes but not ketones
conclusion: if an aldehyde is present, silver forms because aldehyde is oxidised to carboxyl group ALLOWING Ag+ in Tollens' reagent to reduce to Ag solid.
on the other hand, if ketones were present, nothing would occur as it cannot undergo oxidation, preventing Ag+ to reduce.
carbohydrates
carbohydrate = hydrated carbon atom (has H & OH's)
horizontal timelapse, with O in top right (hemiacetal bond)
label the carbon numbers. 1 is the anomeric carbon, 6 is the CH2OH
order atoms/groups in axial or equatorial position (using rule)
a-glucose = -OH group in axial position on 1st carbon
b-glucose = -OH group in equatorial position on 1st carbon
A) equatorial
B) chair
C) haworth
why is b-glucose more common?
the equatorial position is always more stable
more stable by minimising steric hindrance
basically means the repulsions of electrons in b-glucose are more stable than a-glucose
this is because the -OH groups are all in equatorial positions, hence more stable and more common
mutorotation
where a solution of a cyclic compound (eg. glucose) changes its rotation due to the equilibrium between its α and β anomers (due to the ring opening and closing)
b-D-glucose rotates plane polarised light by +18.7 degrees
a-D-glucose rotates it by +112 degrees
if b-glucose is put in water, its rotation changes from 18.7 to 52.7, and a will decrease to 52.7, telling us that both forms exist but more B anomer
disaccharides - glycoside formation
-OH of the anomeric carbon is replaced by –OR (acetal formation)
the OH on glucose and OH on methanol forms an acetal group (not ether) connected by glycoside bond
cyclic acetals are NOT in equilibrium, so they do not undergo mutarotation (cannot change between a/b forms)
A) glycoside
B) anomeric
disaccharides
sugars are alcohols so they can bind to each other
a-1,4'-glycoside bond
"a-1" means the start of the bond is from the 1st carbon of alpha molecule
"4'-glycoside bond" means the bond connects from the 4th carbon on the other molecule
more examples
A) b
B) a
N-glycosides
The anomeric carbon of a cyclic hemiacetal also reacts with the N-H group of amine to form an N-glycoside
A) anomeric
B) glycoside
sucrose disaccharide
fructose + glucose
here, the glycoside bond is a & b because it's a-glucose + b-fructose
A) a
B) 1
Polysaccharides
the red squiggle means its a chain, so the molecule continues
cellulose is made of only b-glucose
A) 4
B) 1
C) starch
cellulose
provides structure in plants due to its rigid structure
has a linear chain connected by strong hydrogen bonding
the D-glucose units are joined by b-1,4 links
starch
fully digestible (unlike cellulose) and only present in plants
eg. beans, wheat, rice, potatoes
D-glucose is joined by a-1,4 links
starch - amylose
amylose makes up 20% of starch
amylose coils into helices and forms a linear chain
contains only a-1,4 links
soluble in hot water
starch - amylopectin
amylopectin makes up 80% of starch
amylopectin forms a branched structure (higher surface area)
contains a-1,4 and a-1,6 links
insoluble in water
glycogen
stores energy in animals in the liver and muscles
similar to amylopectin in being a long polymer of D-glucose with the same type of links
same as amylopectin but with MORE branches
A) plants
B) animals
mutarotation (commonly tested)
applies only when in a SOLUTION
when glucose dissolves in water, it forms an equilibrium of a and b forms because the cyclic ring opens and closes
optical rotation: the way lightbends when it passes through solution
a and b glucose bend light in different angles, helping scientists
eg. 1) b-glucose put in water, 2) -O- bond goes back to -OH (6th carbon) and -COOH (1st carbon) 3) carboxyl group flips 180 4) reform -O- bond but now it's a-glucose