Question

Arrange the following lipids:

1) a bilayer made of lipids with monosaturated 12 carbon fatty acids

2) a bilayer made of lipids with saturated 12 carbon fatty acids

3) a bilayer made of lipids with saturated 20 carbon fatty acids

4) a bilayer of lipids with polyupolyunsaturated 20 carbon fatty acids

Arrange the lipids in order of:

A) In relative order of increasing mp

B)decreasing solubility in water

C) increasing solubility in benzene

Answer 1

A) The melting point of fatty acid depends on the number of carbon atom and a double bond. Saturated fatty acids present as solid whereas unsaturated fatty acids present in liquid form. Melting point increases with an increase in the number of carbon atom whereas decreases with an increase in the number of double bonds.

Monosaturated FA = 1 double bond present

Polyunsaturated FA = more than one double bond present

The increasing order of melting point is

the bilayer of lipids with polyunsaturated 20 carbon fatty acids < A bilayer made of lipids with monosaturated 12 carbon fatty acids < bilayer made of lipids with saturated 12 carbon fatty acids < bilayer made of lipids with saturated 20 carbon fatty acids

The solubility decreases with an increase in chain length and increases with an increase in double bond as we know alkenes are more soluble in water in comparison to the alkanes.

So the decreasing order of solubility in water is as follows

the bilayer of lipids with polyunsaturated 20 carbon fatty acids > A bilayer made of lipids with monosaturated 12 carbon fatty acids > bilayer made of lipids with saturated 12 carbon fatty acids >bilayer made of lipids with saturated 20 carbon fatty acids

The solubility in benzene is opposite to the solubility in the water

A bilayer of lipids with polyunsaturated 20 carbon fatty acids < A bilayer made of lipids with monosaturated 12 carbon fatty acids < bilayer made of lipids with saturated 12 carbon fatty acids < bilayer made of lipids with saturated 20 carbon fatty acids