Question

Name the reagents used in the following reactions :
(i) Oxidation of a primary alcohol to an aldehyde
(ii) Oxidation of a primary alcohol to a carboxylic acid

Answer 1

(i)
A nucleotide is a three-part building block, so breaking it apart with water (hydrolysis) simply hands back its three pieces. For a DNA nucleotide that carries thymine, the pieces are the nitrogen base thymine, the sugar 2-deoxy-D-ribose, and phosphoric acid.

OR

(ii)
DNA and RNA differ in two structural details. The sugar in DNA is 2-deoxyribose, which is missing one oxygen, while in RNA the sugar is the fuller ribose. Also, where DNA uses the base thymine, RNA swaps in uracil instead.
So (i) gives thymine + 2-deoxy-D-ribose + phosphoric acid; or (ii) DNA has deoxyribose and thymine while RNA has ribose and uracil.

Answer 2

Step 1: sort vitamins by what they dissolve in.
Vitamins fall into two families depending on whether they melt into fat or into water. The fat-soluble ones gather in the body's fatty tissues, while the water-soluble ones travel in watery fluids and are not stored for long.

Step 2: give one of each.
A fat-soluble example is Vitamin A (the others in this group are D, E and K). A water-soluble example is Vitamin C (the B-group vitamins are also water-soluble).
Fat-soluble: Vitamin A; water-soluble: Vitamin C.

Answer 3

(i)
Think about what oxidation of an alcohol really needs. To turn an alcohol into a carbonyl, the oxidising agent has to pull off the hydrogen that sits directly on the carbon carrying the $\mathrm{-OH}$ group. In a tertiary alcohol that carbon is joined to three other carbon groups, so there is simply no hydrogen left on it. With no such hydrogen to remove, the gentle oxidation that works for primary and secondary alcohols cannot happen. The only way to oxidise a tertiary alcohol is to use very harsh conditions that snap carbon to carbon bonds and tear the molecule into smaller pieces, which is why we say tertiary alcohols normally resist oxidation.

(ii)
When an ether is cut by HI, the iodide ion ends up on whichever side can form the more stable positive carbon. Here the ether is $\mathrm{(CH_3)_3C\text{-}O\text{-}C_2H_5}$. The tert-butyl side can become a very stable tertiary carbocation, so the bond breaks on that side and iodide attaches there. The ethyl side leaves as an alcohol. So the products are tert-butyl iodide and ethanol.
\[ \mathrm{(CH_3)_3C\text{-}O\text{-}C_2H_5 + HI \rightarrow (CH_3)_3C\text{-}I + C_2H_5OH} \]
Final answer: (i) the carbon bearing $\mathrm{-OH}$ has no hydrogen to be removed; (ii) $\mathrm{(CH_3)_3C\text{-}I}$ and $\mathrm{C_2H_5OH}$.