Blood is essentially an aqueous solution, but it must transport a variety of sub

Question

Blood is essentially an aqueous solution, but it must transport a variety of substances (hormones, fatty acids) . Certain proteins known as serum albumins facilitate this transport. Would these proteins be polar , non polar , or both ? Why? Blood is essentially an aqueous solution, but it must transport a variety of substances (hormones, fatty acids) . Certain proteins known as serum albumins facilitate this transport. Would these proteins be polar , non polar , or both ? Why? 10 35. Which of the compounds below has the strongest interaction with water? a. CH3OH CH3 CH3 d. H OCH3 e. 16 I idocaine is used as a local anesthetic and a cardiac denressant

Explanation / Answer

Ans. #1. Serum albumin have BOTH the polar and non-polar regions. Albumin binds to a non-polar molecule like steroid hormones and fatty acids with its non-polar region. The overall (albumin-steroid or albumin-FA) complex behaves as hydrophilic molecule due to bulkier polar surface of the protein. Being hydrophilic or polar, the albumin protein efficiently transports the non-polar molecules complexed with albumin through the aqueous medium of the blood.

# A non-polar molecule can’t dissolve in the aqueous medium of the blood on its own. So, such molecule require a carrier protein (here, albumin) for their transport through the aqueous medium of blood.

#34. Hydrogen bond is a weak attractive force between a highly electronegative atoms (O, N, F) and a H-atoms linked to such atoms.

The criteria for H-bonds formation is that there must be -

I. A highly electronegative atom (F, N, O-atom), and

II. A H-atoms linked to such atoms (F, N, O).

# Aldehyde (b), ketone (c) and ester do NOT have an H-atom linked to an O-atom. So, there is no more an H-atom that could form hydrogen bonds. The O-atoms may still form hydrogen bond. However, presence of a non-polar CH3 group in ketone and ester causes stearic hindrance and shields the efficiency of O-atom to form H-bonds with H-atom of water.

# Compare methanol (a) and formic acid (d): Both the molecules have a –OH group that can form H-bonds with water. Methanol has a non-polar –CH3 that may hinder the formation of H-bonds to some extent. Formic acid has a-H atom instead of –CH3 group in methanol, so steric hindrance is much reduced in formic acid. Moreover, the O-atom of formic acid can also form H-bonds with H-atom of water.

Therefore, due to owing an additional O-atom (that facilitates H-bonding) and a H-atom in place of –CH3, the interaction between formic acid and water expected to be greatest (greater than that of methanol, too).

So, correct option is- d. H-C(O)-OH

# 35: For being aromatic, the cyclic compound must have a conjugated double bonds (alternate double bonds) throughout the ring.

Note option d (4a,8a-dihydronaphthalene), the ring system is not conjugated near the junction of two rings. So, it is NOT aromatic.

# So, correct option is- d.

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